STEP 2000
Busway
Table of Contents
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STEP 2000
Busway
Table of Contents
Introduction ..............................................................................2 Distribution Systems ................................................................4 Busway Purpose and Definition................................................6 Sentron Busway ..................................................................... 10 Types and Application ............................................................. 11 Design Standards and Ratings................................................ 13 Circuit Protection .................................................................... 18 Busway Construction..............................................................26 Busway System Components ................................................35 Sentron Low Amp Busway .....................................................50 Planning a Sentron Busway System .......................................53 Cable/Conduit Conversion ......................................................70 XL-U Busway...........................................................................73 XJ-L Busway ...........................................................................79 BD Busway .............................................................................82 Trol-E-Duct ..............................................................................86 Review Answers.....................................................................91 Final Exam ..............................................................................92
1
Introduction
Welcome to another course in the STEP 2000 series, Siemens Technical Education Program, designed to prepare our distributors to sell Siemens Energy & Automation products more effectively. This course covers Busway and related products. Upon completion of Busway you should be able to:
2
•
Identify the major components of several Siemens busway systems and describe their functions
•
Identify the role of busway in a distribution system
•
Explain the need for circuit protection
•
Identify feeder and plug-in busway and explain the use of each
•
Identify various organizations involved with busway design standards
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Describe selected sections of the National Electrical Code® (NEC®) as it applies to busway
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Measure and layout a basic busway system
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Identify various ratings of Siemens busway
•
Describe how a cost savings is realized when busway is selected over cable and conduit
This knowledge will help you better understand customer applications. In addition, you will be better able to describe products to customers and determine important differences between products. You should complete Basics of Electricity before attempting Busway. An understanding of many of the concepts covered in Basics of Electricity is required for Busway. If you are an employee of a Siemens Energy & Automation authorized distributor, fill out the final exam tear-out card and mail in the card. We will mail you a certificate of completion if you score a passing grade. Good luck with your efforts. I-T-E, Vacu-Break, Speedfax, and XL-U are registered trademarks of Siemens Energy & Automation, Inc. Sentron, Trol-E-Duct, BD, and XJ-L are trademarks of Siemens Energy & Automation, Inc. National Electrical Code® and NEC® are registered trademarks of the National Fire Protection Association, Quincy, MA 02269. Portions of the National Electrical Code® are reprinted with permission from NFPA 70-2002, National Electrical Code®, Copyright, 2001, National Fire Protection Association, Quincy, MA 02269. This reprinted material is not the complete and official position of the National Fire Protection Association on the referenced subject which is represented by the standard in its entirety. Underwriters Laboratories, Inc. is a registered trademark of Underwriters Laboratories, Inc., Northbrook, IL 60062. The abbreviation “UL” is understood to mean Underwriters Laboratories, Inc. National Electrical Manufacturers Association is located at 2101 L. Street, N.W., Washington, D.C. 20037. The abbreviation “NEMA” is understood to mean National Electrical Manufacturers Association. Other trademarks are the property of their respective owners.
3
Distribution Systems
A distribution system is a system that distributes electrical power throughout a building. Distribution systems are used in every residential, commercial, and industrial building. Distribution systems used in commercial and industrial locations are complex. A distribution system consists of metering devices to measure power consumption, main and branch disconnects, protective devices, switching devices to start and stop power flow, conductors, and transformers. Power may be distributed through various switchboards, transformers, and panelboards. Good distribution systems don’t just happen. Careful engineering is required so that the distribution system safely and efficiently supplies adequate electric service to both present and possible future loads.
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Distribution Example
In this example of a distribution system the incoming power is 277/480 volts, three-phase, four-wire. The utility company supplies power from a transformer. The secondary winding of the transformer produces 277/480 VAC.
Feeders
A feeder is a set of conductors that originate at a main distribution center and supplies one or more secondary, or one or more branch circuit distribution centers. Three feeders are used in this example. The first feeder is used for various types of power equipment. The second feeder supplies a group of 480 VAC motors. The third feeder is used for 120 volt lighting and receptacles.
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Busway Purpose and Definition
Commercial and industrial distribution systems use several methods to transport electrical energy. These methods may include heavy conductors run in trays or conduit. Once installed, cable and conduit assemblies are difficult to change. Power may also be distributed using bus bars in an enclosure. This is referred to as busway. Bus Bars
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A bus bar is a conductor that serves as a common connection for two or more circuits. It is represented schematically by a straight line with a number of connections made to it. Standard bus bars in Siemens busway are made of aluminum or copper.
NEMA Definition
Busway is defined by the National Electrical Manufacturers Association (NEMA) as a prefabricated electrical distribution system consisting of bus bars in a protective enclosure, including straight lengths, fittings, devices, and accessories. Busway includes bus bars, an insulating and/or support material, and a housing.
Busway Used in a Distribution System
A major advantage of busway is the ease in which busway sections are connected together. Electrical power can be supplied to any area of a building by connecting standard lengths of busway. It typically takes fewer man-hours to install or change a busway system than cable and conduit assemblies.
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The total distribution system frequently consists of a combination of busway and cable and conduit. In this example power from the utility company is metered and enters the plant through a distribution switchboard. The switchboard serves as the main disconnecting means. The feeder on the left feeds a distribution switchboard, which in turn feeds a panelboard and a 480 volt, three-phase, three-wire (3Ø3W) motor. The middle feeder feeds another switchboard, which divides the power into three, three-phase, three-wire circuits. Each circuit feeds a busway run to 480 volt motors. The feeder on the right supplies 120/208 volt power, through a step-down transformer, to lighting and receptacle panelboards. Branch circuits from the lighting and receptacle panelboards supply power for lighting and outlets throughout the plant. In many cases busway can be used in lieu of the cable/conduit feeders at a lower cost.
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Busway is used in various applications and can be found in industrial installations as well as high-rise buildings. Busway used in industrial locations can supply power to heavy equipment, lighting, and air conditioning. Busway risers (vertical busway) can be installed economically in a high-rise building where it can be used to distribute lighting and air conditioning loads.
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Sentron Busway
Throughout this course Siemens Sentron™ busway will be used to explain and illustrate principles and requirements of busway. Sentron busway will meet the needs of most busway systems with current ratings from 800 amperes to 5000 amperes. Siemens manufactures several types of busway. There are a number of reasons why different types of busway are manufactured. An existing pre-Sentron busway system, for example, may need to be expanded. Other types of Siemens busway, including significant features, and ratings will be discussed later in the course.
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Types and Application
Feeder Busway
There are two types of busway, feeder and plug-in. Feeder busway is used to distribute power to loads that are concentrated in one physical area. Industrial applications frequently involve long runs from the power source to a single load. This load may be a large machine, motor control center, panelboard, or switchboard.
Service Entrance
The service entrance is the point of entrance of supply conductors to a building or other structure. Feeder busway, which can be purchased for indoor or outdoor use, can be used as service entrance conductors to bring power from a utility transformer to a main disconnect inside the building.
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Plug-in Busway
Plug-in busway is used in applications where power requirements are distributed over a large area. Using plug-in units, load connections can be added or relocated easily. Plug-in busway is for indoor use only.
Review 1
12
1.
A ____________ ____________ distributes electrical power throughout a building.
2.
A ____________ is a set of conductors that originate at a main distribution center and supply one or more secondary, or one or more branch circuit distribution centers.
3.
____________ is a type of power distribution device that is made up of heavy bus bars in an enclosure.
4.
It typically takes fewer man-hours to install or change a ____________ system than cable and conduit assemblies.
5.
The two types of busway are ____________ and ______ ______ .
6.
____________ busway can be purchased for use indoors or outdoors, ____________ busway is for indoor use only.
Design Standards and Ratings
Several organizations maintain standards of design, construction, installation, and performance of busway. The following list includes some of these organizations: Underwriters Laboratories, Inc. (UL) The National Electrical Manufacturers Association (NEMA) International Electrotechnical Commission (IEC) The National Electrical Code® (NEC®) Organizations responsible for state and local electrical codes Sentron™ busway meets the worldwide standards of UL 857, IEC 439-1, and IEC 439-2. Underwriters Laboratories
Busway bearing the Underwriters Laboratories listing mark must pass a series of performance tests based on UL publication UL 857. These tests and standards relate to the strength and integrity of a busway system when subjected to specific operating and environmental conditions. UL 1479 provides guidelines for a fire rating. Sentron busway has been tested in accordance with UL 1479 and offers a certified two hour fire rating for gypsum wallboard and a three hour fire rating for concrete slab or cement block. These ratings are achieved by using standard Sentron busway installed with SpecSeal® sealant from Specified Technologies, Inc.
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NEMA
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NEMA standards for busway are listed in NEMA publication number BU 1.1-2000. NEMA is primarily associated with equipment used in North America. It is important to note that NEMA short-circuit ratings require a 3 cycle short-circuit rating. This means that the busway was tested and rated on the basis of successfully experiencing 3 cycles of peak current (IP). NEMA recommends the following minimum short-circuit current ratings for busway.
IEC
The International Electrotechnical Commission is associated with equipment sold in many countries, including the United States. IEC standards are found in IEC publications 439 and 529. IEC also recommends short-circuit ratings for busway. Siemens manufacturers Sentron busway with continuous current ratings from 800 amperes to 5000 amperes. The following table shows the short-circuit ratings for Sentron busway. These ratings meet IEC specifications.
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The following chart lists IEC and UL specifications for enclosure protection of busway.
Feeder or Plug-In Indoor Spray Proof
IP 40 IP 54 & IP55
Feeder Only Outdoor
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NEMA 3R
National Electrical Code®
The National Electrical Code® (sponsored by the National Fire Protection Association), once adopted by the authority having jurisdiction, stipulates installation requirements which are necessary for the safe application of electrical equipment. ® Article 368 of the NEC® specifically applies to busway, ® although other articles in the NEC® are applicable in certain ® situations. Thorough familiarization of the NEC® requirements for busway is recommended. 368.1 368.2 368.4 368.5 368.6 368.7 368.8 368.9 368.10 368.11 368.12 368.13 368.15
State and Local Codes
Scope Definition Use Support Through Walls and Floors Dead Ends Branches from Busways Overcurrent Protection Rating of Overcurrent Protection - Feeders Reduction in Ampacity Size of Busway Feeder of Branch Circuits Rating of Overcurrent Protection - Branch Circuits Marking
State and local authorities have electrical codes which are often more stringent than other organizations. You are encouraged to become familiar with this material in your local area. In addition, busway is frequently used for the main electrical service of a building, in which case the busway is connected to one or more distribution transformers owned by local electric power companies. Electrical power companies throughout the United States prefer different methods of connecting to busway. It is recommended that the local power company be contacted before applying or installing a service entrance busway run.
NEC® and National Electrical Code® are registered trademarks of the National Fire Protection Association.
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Circuit Protection
Circuit protection must be taken into consideration with any electrical circuit, including busway. Current flow in a conductor always generates a watts loss in the form of heat. As current flow increases, the conductor must be sized appropriately in order to compensate for higher watt losses. Excess heat is damaging to electrical components. For that reason, conductors have a rated continuous current carrying capacity or ampacity. Overcurrent protection devices are used to protect conductors from excessive current flow. Two devices used to protect circuits from overcurrent are fuses and circuit breakers. These protective devices are designed to limit the flow of current in a circuit to a safe level, preventing the circuit conductors from overheating.
The National Electrical Code® defines overcurrent as any current in excess of the rated current of equipment or the ampacity of a conductor. It may result from overload, short circuit, or ground fault (Article 100-definitions).
NEC® and National Electrical Code® are registered trademarks of the National Fire Protection Association. Reprinted with permission from NFPA 70-2002, the National Electrical Code®, Copyright© 2001, National Fire Protection Association, Quincy, MA 02269.
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Circuit protection would be unnecessary if overloads and short circuits could be eliminated. Unfortunately, overloads and short circuits do occur. To protect a circuit against these currents, a protective device must determine when a fault condition develops and automatically disconnect the electrical equipment from the voltage source. Inverse Time-Current Characteristic
An overcurrent protection device must be able to recognize the difference between overcurrents and short circuits and respond in the proper way. Protection devices use an inverse time-current characteristic. Slight overcurrents can be allowed to continue for some period of time, but as the current magnitude increases, the protection device must open faster. Short circuits must be interrupted instantly.
Fuse Construction
A fuse is the simplest device for interrupting a circuit experiencing an overload or a short circuit. A typical fuse, like the one shown below, consists of an element electrically connected to end blades or ferrules. The element provides a current path through the fuse. The element is enclosed in a tube and surrounded by a filler material.
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Overcurrent
Current flowing through the element generates heat, which is absorbed by the filler material. When an overcurrent occurs temperature in the element rises. In the event of a harmless transient overload condition the excess heat is absorbed by the filler material. If a sustained overload occurs the heat will eventually melt open an element segment forming a gap; thus stopping the flow of current.
Short-Circuit Current
Short-circuit current can be several thousand amperes and generates extreme heat. When a short circuit occurs several element segments can melt simultaneously, which helps remove the load from the source voltage quickly. Short-circuit current is typically cut off in less than half a cycle, before it can reach its full value.
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Nontime-Delay Fuses
Nontime-delay fuses provide excellent short circuit protection. Short-term overloads, such as motor starting current, may cause nuisance openings of nontime-delay fuses. They are best used in circuits not subject to large transient surge currents. Nontime-delay fuses usually hold 500% of their rating for approximately one-fourth second, after which the current carrying element melts. This means that these fuses should not be used in motor circuits which often have inrush (starting) currents greater than 500%.
Time-Delay Fuses
Time-delay fuses provide overload and short circuit protection. Time-delay fuses usually allow five times the rated current for up to ten seconds. This is normally sufficient time to allow a motor to start without nuisance opening of the fuse unless an overload persists.
Ampere Rating
Fuses have a specific ampere rating, which is the continuous current carrying capability of a fuse. The ampere rating of a fuse, in general, should not exceed the current carrying capacity of the circuit. For example, if a conductor is rated for 10 amperes, the largest fuse that would be selected is 10 amperes. There are some specific circumstances when the ampere rating is permitted to be greater than the current carrying capacity of the circuit. For example, motor and welder circuits can exceed conductor ampacity to allow for inrush currents and duty cycles within limits established by the NEC®.
Sentron Fusible Switches
Plug-in fusible switches are available on Siemens busway. Sentron™ fusible switches, for example, are rated from 30 - 600 amperes.
Voltage Rating
The voltage rating of a fuse must be at least equal to the circuit voltage. The voltage rating of a fuse can be higher than the circuit voltage, but never lower. A 600 volt fuse, for example, can be used in a 480 volt circuit. A 250 volt fuse could not be used in a 480 volt circuit.
Ampere Interrupting Capacity (AIC)
Fuses are also rated according to the level of fault current they can interrupt. This is referred to as ampere interrupting capacity (AIC). When applying a fuse, one must be selected which can sustain the largest potential short circuit current which can occur in the selected application. The fuse could rupture, causing extensive damage, if the fault current exceeds the fuse interrupting rating.
NEC® and National Electrical Code® are registered trademarks of the National Fire Protection Association.
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UL Fuse Classification
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Fuses are grouped into current limiting and non-current limiting classes based on their operating and construction characteristics. Fuses that incorporate features or dimensions for the rejection of another fuse of the same ampere rating but with a lower interruption rating are considered current limiting fuses. Underwriters Laboratories (UL) establishes and standardizes basic performance and physical specifications to develop its safety test procedures. These standards have resulted in distinct classes of low voltage fuses rated at 600 volts or less. The following chart lists various UL fuse classes.
Circuit Breakers
Another device used for overcurrent protection is a circuit breaker. The National Electrical Code® defines a circuit breaker as a device designed to open and close a circuit by nonautomatic means, and to open the circuit automatically on a predetermined overcurrent without damage to itself when properly applied within its rating (Article 100 - definitions). Circuit breakers provide a manual means of energizing and deenergizing a circuit. In addition, circuit breakers provide automatic overcurrent protection of a circuit. A circuit breaker allows a circuit to be reactivated quickly after a short circuit or overload is cleared. Unlike fuses which must be replaced when they open, a simple flip of the breaker’s handle restores the circuit.
Ampere Rating
Like fuses, every circuit breaker has a specific ampere, voltage, and fault current interruption rating. The ampere rating is the maximum continuous current a circuit breaker can carry without exceeding its rating. As a general rule, the circuit breaker ampere rating should match the conductor ampere rating. For example, if the conductor is rated for 20 amperes, the circuit breaker should be rated for 20 amperes. Siemens I-T-E® breakers are rated on the basis of using 60° C or 75° C conductors. This means that even if a conductor with a higher temperature rating were used, the ampacity of the conductor must be figured on its 60° C or 75° C rating. NEC® and National Electrical Code® are registered trademarks of the National Fire Protection Association. Reprinted with permission from NFPA 70-2002, the National Electrical Code®, Copyright© 2001, National Fire Protection Association, Quincy, MA 02269.
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There are some specific circumstances when the ampere rating is permitted to be greater than the current carrying capacity of the circuit. For example, motor and welder circuits can exceed conductor ampacity to allow for inrush currents and duty cycles within limits established by NEC®. Generally the ampere rating of a circuit breaker is selected at 125% of the continuous load current. This usually corresponds to the conductor ampacity which is also selected at 125% of continuous load current. For example, a 125 ampere circuit breaker would be selected for a load of 100 amperes. Sentron MCCB Plug-In Units
Plug-in devices with molded case circuit breakers (MCCB) are available on Sentron busway with circuit breaker ratings from 125 - 800 amperes.
Voltage Rating
The voltage rating of the circuit breaker must be at least equal to the circuit voltage. The voltage rating of a circuit breaker can be higher than the circuit voltage, but never lower. For example, a 480 VAC circuit breaker could be used on a 240 VAC circuit. A 240 VAC circuit breaker could not be used on a 480 VAC circuit. The voltage rating is a function of the circuit breakers ability to suppress the internal arc that occurs when the circuit breaker’s contacts open.
Fault Current Interrupting Rating
Circuit breakers are also rated according to the level of fault current they can interrupt. When applying a circuit breaker, one must be selected which can sustain the largest potential short circuit current which can occur in the selected application. Siemens circuit breakers have interrupting ratings from 10,000 to 200,000 amperes. To find the interrupting rating of a specific circuit breaker refer to the Speedfax® catalog.
Additional Information
For additional information on circuit breakers refer to the STEP 2000 course, Molded Case Circuit Breakers.
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Review 2 1.
In accordance with UL 1479, Sentron busway offers a ____________ hour fire rating for gypsum wallboard, and a ____________ hour fire rating for concrete slab when used with SpecSeal® sealant.
2.
Sentron busway, with a continuous current rating of 800 amperes and aluminum bus bars, has a 60 cycle shortcircuit rating of ____________ amperes.
3.
The highest level of enclosure protection of Sentron feeder busway is IP ____________ and the highest level of enclosure protection of Sentron plug-in busway is IP ____________ .
4.
Article ____________ in the National Electrical Code® specifically applies to busway.
5.
A Class R has an ampere interrupting capacity of ______ ______ amperes.
6.
Siemens circuit breakers have a fault current interrupting capacity of ____________ to ____________ amperes.
NEC® and National Electrical Code® are registered trademarks of the National Fire Protection Association.
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Busway Construction
Bus Bars
A better understanding of what busway is can be gained by examining its construction. A typical Siemens Sentron™ busway section has three or four formed aluminum or copper bars that function as electrical conductors. Aluminum busway can be supplied in ampacities up to 4000 amperes. Copper busway can be supplied in ampacities up to 5000 amperes.
Bus bars manufactured for use in feeder busway differ from those manufactured for use in plug-in busway. Plug-in busway will have a tab or some other form of connecting a plug-in device such as a disconnect.
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Each bus bar is referred to as a phase. Bus bars of Sentron busway are separated electrically with epoxy insulation.
Enclosure
Glass wrap tape is wrapped around the Sentron bus bars to provide additionally protection and hold the bars together. The bus bars are then installed in an enclosure. The enclosure provides protection and support.
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Bars per Pole
Sentron busway uses one bar per pole on busway rated up to 2000 amperes aluminum and 2500 amperes copper.
Sentron busway uses two bars per pole on busway rated from 2500 to 4000 amperes aluminum and 3000 to 5000 amperes copper.
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NEMA Phase Arrangement
Bus bars are required to have phases in sequence so that an installer can have the same fixed phase arrangement in each termination point. This is established by NEMA (National Electrical Manufacturers Association). The following diagram illustrates accepted NEMA phase arrangements.
The following illustration shows the proper phase arrangement of bus bars in Sentron busway.
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Number of Bus Bars
The number of bars depends on the number of phases on the power supply and whether or not a neutral or ground is used.
200% Neutral
Siemens Sentron busway is available with a 200% neutral within the bus bar housing. Certain loads on the distribution system can cause non-sinusodial current referred to as harmonics. These harmonics cause circulating currents which increase the heat in the system and shorten component life. The 200% neutral capacity minimizes overheating, thus prolonging the life of power distribution equipment.
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Ground
The National Electrical Code® requires the metal enclosure of any busway run to be grounded back at the service entrance equipment. Sentron busway has several options to meet this requirement. The busway housing is an integral ground. Under more severe industrial applications a heavier ground may be required. The following cross section drawing of Sentron busway shows a bus bar a 50% internal ground has been added. This means that the ground is rated at 50% of the ampacity of the phase bus bars.
Busway Lengths
The standard length of a plug-in busway section is 10’ (3048 mm). Sentron busway is also available in 4’ (1219 mm), 6’ (1829 mm), and 8’ (2438 mm) lengths.
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Plug-in outlets on Sentron plug-in busway are located on 2’ (610 mm) centers on both sides of the busway.
Sentron Plug-In Outlets
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The Sentron plug-in outlet features a molded guard which prevents incidental finger contact with live conductors. This meets IEC, IP 2X requirements for preventing a 0.472” (12 mm) probe from entering. This is referred to as finger safe.
Feeder Busway Lengths
In addition to the 4’ (1219 mm), 6’ (1829 mm), 8’ (2438 mm), and standard 10’ (3048 mm) lengths, Sentron feeder busway sections are available in 0.125” (3.2 mm) increments from 1’ 4.5” (419 mm) to 10’ (3048 mm). Feeder busway does not have any plug-in outlets.
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Review 3
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1.
In the Sentron busway, aluminum bus bars are available with ampacities up to ____________ amperes and copper bus bars are available with ampacities up to ____________ amperes.
2.
Identify the type of busway each of the bus bars represent in the following illustration.
3.
Plug-in busway is available in ____________ , ____________ , ____________ , and 10 foot lengths.
4.
Plug-in outlets are located on ____________ foot centers.
5.
Sentron feeder busway sections are available in ____________ inch increments from ____________ to ____________ feet.
Busway System Components
There are a number of components that make up a busway system. The various system components illustrated in this section, unless otherwise noted, will be the Siemens Sentron™ series. For more information on any component consult the Sentron Busway System Selection and Application Guide. It should also be noted that certain components available on one type busway system may not be available on another type busway system. Although components used in various busway systems perform the same or similar functions, they can’t be interchanged from one busway system to another. There are a number of reasons for this. Systems are tested and rated as a complete unit. Ratings and system integrity could not be guaranteed when components are interchanged between systems. Additionally, components from one system may not physically fit or connect to components of another system. Sections of Siemens Sentron busway, for example are clamped together with a joint stack. Siemens BD™ busway is bolted together.
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Joint Stack
The Siemens Sentron busway system, uses a single-bolt joint stack to connect busway sections. The bus bars from two busway sections are slid into a joint stack.
The assembly is clamped solidly together with the single bolt located on the joint stack. Sentron busway sections and components are supplied with required joint stacks.
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The single-joint bolt is a double-headed break-off bolt. The outer head is 5/8” and the bottom head is 3/4” . The doubleheaded bolt is tightened until the 5/8” outer head twists off (approximately 55 ft. lbs.). This eliminates the need for torque wrenches during initial installation. The bottom 3/4” head is permanent and is used for future joint maintenance. Each joint is adjustable by ± 5/8” .
Elbows
Elbows, offsets, and tees allow for turns and height changes in the busway system to made in any direction. An elbow can turn the busway system right or left, up or down. Elbows are supplied with a joint stack and covers.
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Combination Elbows
Combination elbows can turn the busway system up or down, and right or left.
Tees
Tees are used to start a new section of busway in a different direction. Tees can start a new section to the right, to the left, up, or down. Tees are supplied with two joint stacks.
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Crosses
A cross allows a busway run to expand in four directions.
Offsets
Offsets allow the busway system to continue in the same direction. Offsets can move the busway system to the right, to the left, up, or down. Offsets are supplied with a joint stack.
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Cable Tap Boxes
Tap boxes are used to connect electrical cable to the busway distribution system. End cable tap boxes can be installed at either end of the busway system. They can be used on feeder or plug-in busway.
Center, or plug-in, cable tap boxes can be installed along the length of a busway system. Plug-in cable tap boxes can only be used on plug-in busway.
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Stubs
Sentron busway standard stubs can be used to connect busway to other Siemens equipment, such as switchgear and switchboards. Sentron busway stubs can be shipped installed in Siemens switchboards and switchgear. This eliminates the field labor required to connect the busway to the switchboard, saving the installer time and money.
Flanged Ends
Flanged ends are also used to connect busway equipment such as switchgear and switchboards. These can be used with existing equipment. Siemens will furnish the outline drawings of this flanged end to the coordinating switchboard or equipment installer.
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Service Heads
Service heads are used to connect the busway to the electric service. There are two types in the Sentron series. A single service head that has all three phases, or three separate heads, one for each phase.
Riser Adaptors
In busway, a riser is a length of vertical busway. Panelboards and meter centers can be mounted directly to risers with a side-mounted adapter. When Sentron plug-in busway is used as a riser, plug-in receptacles are located only on one side.
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Phase Rotation
Some applications may require a phase rotation of the power supply to be reversed. The direction of rotation of a 3Ø AC motor, for example, is determined by the phase sequence of the power supply.
Reducers
A busway reducer is used to reduce the allowable ampere rating. Money can often be saved by using a lower rated group of sections near the end of a busway run. A branch circuit, for example, does not need as high an ampere rating as the main feeder circuit. Article 368.11 of the NEC® states that overcurrent protection shall be required where busways are reduced in ampacity. There is an exception to this article. Exception: For industrial establishments only, omission of overcurrent protection shall be permitted at points where busways are reduced in ampacity, provided that the length of the busway having the smaller ampacity does not exceed 15 m (50 ft) and has an ampacity at least equal to one-third the rating or setting of the overcurrent device next back on the line, and provided further that such busway is free from contact with combustible material. NEC® and National Electrical Code® are registered trademarks of the National Fire Protection Association. Reprinted with permission from NFPA 70-2002, the National Electrical Code®, Copyright© 2001, National Fire Protection Association, Quincy, MA 02269.
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Sentron busway offers fused reducers to meet the requirement of NEC® Article 368.11, and non-fused reducers when the exception is allowed. Illustrated below is a fused reducer.
Expansion Fittings
Expansion fittings are used when a busway system crosses an expansion joint in a building, or on long straight runs where both ends are held in a permanent fixed position. The Sentron expansion fitting contains a sliding expansion enclosure. Flexible connectors in the expansion enclosure allow a ±2” movement.
NEC® and National Electrical Code® are registered trademarks of the National Fire Protection Association.
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Bus Plugs
Sentron II busway bus plugs are available with Siemens Sentron molded case circuit breakers or Siemens fusible switch.
Sentron II bus plugs feature a visible position indicator and relocateable operating handle. Alignment and interlock tabs prevent the plug from being installed 180° out of position and ensures that the plug is in the off position during installation and removal. Clamp assemblies are used to draw the bus plug firmly against the busway housing. Sentron Bus Plugs are designed with standard wire bending space and extended wire bending space (gutter). Plug-in units can be mounted horizontally or vertically (riser).
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In-line Disconnect Cubicle
Cubicles provide a means of mounting switches or circuit breakers where power enters or leaves a busway system. Inline disconnect cubicles are used where bolted connections are preferred, or at ampere ratings exceeding the standard plug-in unit ratings.
End Closer
End closers are used to safely terminate a run of busway and protect the bus bar ends.
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Hangers
Various hangers are used to support busway. When a vertical run of busway passes through a floor, a floor support is required. Spring hangers provide secure mounting of Sentron busway in riser applications. These hangers counter the weight of the busway on each floor and compensate for minimal building movement and thermal expansion.
Several types of hangers are available to suspend the busway from the ceiling, structural steel support, or mounted to a wall.
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Flanges
Wall, ceiling, and floor flanges are designed to close off the area around the busway as it passes through a wall, ceiling, or floor. The flange does not provide an air tight seal around the busway.
Roof flanges provide a watertight seal when outdoor rated busway enters through a roof. The pitch or angle of the roof must be specified when ordering roof flanges.
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Review 4 Identify the components in the following illustration:
1.
____________
2.
____________
3.
____________
4.
____________
5.
____________
6.
____________
7.
____________
8.
____________
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Sentron Low Amp Busway
Sentron Low Amp Busway is a member of the Siemens Sentron Busway family. Sentron Low Amp Busway specifically addresses the needs of low amp power distribution applications, with ratings from 225A to 800A UL and 100A to 800A IEC. Low Amp Busway provides all the features and advantages of Sentron Busway.
Plug-In Sections
Sentron Low Amp is available as plug-in only, with aluminum or copper bus bars. Aluminum bars are available in 225 - 600 ampere sections. Copper bars are available in 225 - 800 ampere sections. Sections include an integral housing ground. An optional copper internal or isolated ground is available in most ratings. Sentron Low Amp plug-in sections are available in 4’ (1219 mm), 6’, (1829 mm) 8’ (2438 mm), and 10’ (3048 mm) sections. Plugin openings are centered on 24” (610 mm) and staggered on both sides of the busway. Plug-in outlets, like standard Sentron, are finger safe and IP 2X rated. Plug-in sections are protected to IP40 specifications. IP54 is optionally available.
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Bus Plugs
Sentron II bus plugs used with standard Sentron can also be used with Sentron Low Amp.
Joint Stack Assembly
Sentron Low Amp uses the same joint stack assembly design as standard Sentron.
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Hangers
In addition to the hanger shown below, Sentron Low Amp is available with the same type hangers and supports as available with standard Sentron.
Components
Sentron Low Amp is available with a full range of elbows, tees, offsets, tap boxes, flanged ends, expansion fittings, reducers, and phase rotation fittings.
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Planning a Sentron Busway System
There are several considerations when planning a busway run. The best route would require the fewest fittings and the maximum number of 10’ straight sections. There are a number of techniques to ensure an accurate measurement before purchasing and installing busway. The following procedures are given as an example and are useful in obtaining a correct measurement. Laser Measuring Device
Laser measuring devices, such as the one illustrated below, provide an easy and highly accurate means of measuring a busway run.
Note: Consult Busway factory for laser measuring device information.
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Laser measuring devices project a laser beam which is reflected on an object such as a wall, ceiling, floor, or piece of machinery. The measuring device is able to accurately measure the distance the beam travels.
When measuring the distance from wall-to-wall or wall-toobstruction, place the laser measuring device flat against the wall. The distance measured will be from the wall to the point the laser beam is interrupted.
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Laying Out a Run
Using the laser measuring device, determine the height and location of obstructions. Select a route requiring the fewest offsets.
The planned route can be laid out on the floor with a pencil or chalk. Transfer the position of pipes, ducts, beams, and other obstructions to the floor. It will be easier to transfer the planned busway route to paper if significant portions are laid out full scale first.
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Once the route is laid out the laser measuring device can be used along the run to measure distance.
Walls, Ceilings, Floors
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When piercing a wall, ceiling, or floor find a reference point which is common to both sides and measure from it. This may be a pipe, a wall, or a door.
Sample Layout
In the following example, a busway system, connected to a switchboard, will pass through three rooms. The floor to ceiling height is 15’ on the first floor, and 12’6” on the second floor. The overall length is 42’. Walls and floors are 6” thick. The switchboard is a standard 90” high. Various types of equipment on the second floor will be connected to the busway through plug-in outlets along the length of the room.
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It is determined that a clear space is available 13’ above the floor in the switchboard room (5’6” from the top of the switchboard). The clear space extends on the other side of the wall in the second room to the far right wall. It is also clear on the second floor along the far right wall and 10’ above the floor for the length of the second floor.
A rough sketch can now be made of the proposed busway system route.
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NEC® Requirements
An important part of applying a busway system is to be sure that the system meets the requirements of The National Electrical Code®.
Article 368.4
According to NEC® Article 368.4, busways shall be permitted to be installed where they are located as follows: (1)
Located in the open and are visible, except as permitted in 368.6 or
(2)
Installed behind access panels, provided the busways are totally enclosed, of nonventilating-type construction, and installed so that the joints between sections and at fittings are accessible for maintenance purposes. Where installed behind access panels, means of access shall be provided, and the following conditions shall be met: a. The space behind the access panels shall not be used for air-handling purposes; or b. Where the space behind the access panels is used for environmental air, other than ducts and plenums, in which case there shall be no provisions for plug-in connections, and the conductors shall be insulated.
NEC® and National Electrical Code® are registered trademarks of the National Fire Protection Association. Reprinted with permission from NFPA 70-2002, the National Electrical Code®, Copyright© 2001, National Fire Protection Association, Quincy, MA 02269.
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Article 368.4 also restricts the use of busway in conditions where it may be damaged or cause damage. Busways shall not be installed as follows: (1) Where subject to severe physical damage or corrosive vapors (2) In hoistways (3) In any hazardous (classified) location, unless specifically approved for such use (4) Outdoors or in wet or damp locations unless identified for such use. Lighting busway and trolley busway shall not be installed less than 2.5 m (8 ft) above the floor or working platform unless provided with a cover identified for the purpose. Article 364.5
Article 368.5 requires adequate support for the busway. The following drawing illustrates one type of support available for Siemens Sentron™ busway. Busways shall be securely supported at intervals not exceeding 1.5 m (5 ft) unless otherwise designed and marked. Note: Picture frame and trapeze hangers used with Sentron Busway are designed on a maximum of 3.05 m (10 ft) centers.
NEC® and National Electrical Code® are registered trademarks of the National Fire Protection Association. Reprinted with permission from NFPA 70-2002, the National Electrical Code®, Copyright© 2001, National Fire Protection Association, Quincy, MA 02269.
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Article 368.6
Article 368.6 allows busway to pass through walls and floors provided there are no section joints in the wall or floor and vertical busway extends at least 6 feet through the floor. In addition, certain applications a curb may be required around busway passing through two or more dry floors. This will help eliminate the possibility of spilled liquid from running down the busway causing damage to the electrical system. (A) Walls. Unbroken lengths of busway shall be permitted to be extended through dry walls. (B) Floors. Floor penetrations shall comply with (1) and (2): (1) Busways shall be permitted to be extended vertically through dry floors if totally enclosed (unventilated) where passing through and for a minimum distance of 1.8 m (6 ft) above the floor to provide adequate protection from physical damage. (2) In other than industrial establishments, where a vertical riser penetrates two or more dry floors, a minimum 100 mm (4 in.) high curb shall be installed around all floor openings for riser busways to prevent liquids from entering the opening. The curb shall be installed within 300 mm (12 in.) of the floor opening. Electrical Equipment shall be located so that it will not be damaged by liquids that are retained by the curb. In addition to NEC® requirements, Sentron busway requires a minimum of 7” from a wall to a joint where a new section of busway begins. Sentron busway passing through a floor requires a minimum of 16” between the floor and a joint. This space is required for the floor supports.
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Article 368.7
Article 368.7 states that a dead end of a busway shall be closed. The following drawing illustrates the end closer used on Sentron busway.
Minimum Clearance
There are certain minimum clearances required when installing busway near a wall, ceiling, or another busway run. It is beyond the scope of this course to cover in detail the minimum clearances of every component. The minimum clearances of Sentron busway components are listed in the Sentron™ Busway Systems Selection and Application Guide. Specifications for other systems are listed in their respective selection and application guides.
Dimensions
Component dimensions must also be considered when planning a busway system. The dimensions given in the following examples are for illustrative purposes. For a complete listing of Sentron busway components refer to the Sentron™ Busway Systems Selection and Application Guide. Specifications for other systems are listed in their respective selection and application guides.
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Section Dimensions
It has already been mentioned that Sentron plug-in busway is available in 4’, 6’, 8’, and 10’ sections, and feeder busway is available in lengths from 2’ to 10’ in increments of 0.125”. The height of Sentron busway is 5” . The width varies with the maximum amperage rating. The width of 800 amp, one-bar-perpole aluminum Sentron busway, for example, is 4.5” wide.
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Other Component Dimensions
Other components, such as elbows, offsets, and tees must also be considered. Right and left elbows, for example, vary from 12” x 12” to 24” x 24” . This is due to the variance of bus bar width with amperage rating. An 800 ampere aluminum system, for example, would be 12” x 12”. Up or down elbows are 10” x 10” .
In the example busway system, the busway will be connected to a switchboard. A flanged end must also be selected. The flanged end is 8” long from the flange to the joint stack.
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System Assembly
The components can now be selected for the installation. A switchboard flanged end (8”(203 mm)), a 4’ (1219 mm) length of feeder busway, and one elbow (10” (254 mm)) is selected. The total height is 5’6” (1676 mm).
The busway runs horizontally on the first floor 31’8” (9652 mm) before making its second turn. Feeder busway is selected because no equipment will be connected to it on the first floor. A second elbow and three 10’ (3048 mm) feeder sections are selected.
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It is 2’6” (762 mm) from the top of the horizontal feeder run to the second floor level. The horizontal busway run on the second floor will be installed 10’ (3048 mm) from the floor, for a total rise of 12’6” (3810 mm). One elbow is already installed on the first floor horizontal feeder busway run. A second elbow will be needed at the top of the vertical riser. Each elbow is 10” (254 mm), which is subtracted from the total rise of 12’6” (3810 mm). 10’10” (3302 mm) of vertical riser will complete the vertical rise.
In addition to the standard lengths of 4’ (1219 mm), 6’ (1829 mm), 8’ (2438 mm), and 10’ (3048 mm), Sentron feeder busway also comes in any length from 2’ (609 mm) to 10’ (3048 mm) in 0.125” (3.17 mm) increments. One solution for the vertical riser might be to select one 6’ (1829 mm) and one 4’10” (1473 mm) section.
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The busway run is completed with three 10’ (3048 mm) plug-in sections on the second floor.
An end closer, wall and floor flanges, floor support, hangers, and the desired number of plug-in units finish the system. In this example three plug-in units were used.
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Bill of Material
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We now have the dimensions needed to assemble a bill of material for our example project.
Information Needed to Order Busway
The following information is necessary when planning a busway installation or expansion:
•
Description of application
•
Type of busway
•
Voltage and number of conductors
•
Maximum current
•
Length and configuration of run
•
Location and type of power supply to busway
•
Number of hangers
•
Type and number of tap-off devices (tees, crosses)
•
Type and number of accessories
Review 5 1.
According to NEC® Article 368.7 a dead end of a busway shall be ____________ .
2.
Dimensions of Sentron busway can be found in the _____________________________________________ .
3.
A right elbow with copper bus bars, rated at 5000 amperes has an X measurement of ____________ inches and a Y measurement of ____________ inches.
4.
According to NEC® Article 368.5 busway shall be securely supported at intervals not exceeding ____________ feet unless the busway is otherwise designed and marked.
5.
According to the National Electrical Code® it ____________ to extend unbroken lengths of busway through dry walls. a. is permissible b. is not permissible
NEC® and National Electrical Code® are registered trademarks of the National Fire Protection Association.
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Cable/Conduit Conversion
Busway can be used in many applications where cable and conduit are more commonly used. The question arises, “Why use busway instead of conventional cable and conduit?”
Benefits of Busway
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There are several reasons why busway may be a better choice over cable and conduit. Busway provides greater flexibility by allowing equipment to be connected anywhere along the run on 24” centers. Equipment can be easily disconnected and moved to a new location without major rewiring.
Busway has a smaller cross section. This means less installation space is required. Sentron™ busway with aluminum bus bars rated at 1000 amperes, for example, occupies a much smaller space than a comparable cable and conduit installation. The smaller cross section also means that busway is lighter in weight, by as much as half, which means less loading on the building.
The installed cost of busway is typically less than cable and conduit. Busway is easier to install. Sections are simply hung and joined together using readily available hardware. Total installed costs associated with using Sentron busway over cable and conduit typically results in 20 - 30% lower installed cost.
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Sentron Busway Estimating Program
A software program, available from Siemens, compares the total installed price of cable and conduit to Sentron busway.
Comparison Example
The following table shows one example of the cost savings of busway over cable and conduit. The job calls for a 500 foot run of 1350 amperes. A hypothetical labor rate of $37.15 an hour is used. It will take an estimated 455 hours to install the cable and conduit. It will take an estimated 134 hours to install Sentron busway. The total savings, by using Sentron busway, is $12,693.
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XL-U Busway
XL-U® is available in both feeder and plug-in busway with ratings of 225 to 5000 amperes with aluminum bus bars or 225 to 6500 amperes with copper bus bars. Maximum voltage is 600 volts. XL-U feeder busway is available in either indoor or outdoor ventilated types. XL-U plug-in busway is indoor only and ventilated. XL-U is available in 3Ø3W and 3Ø4W.
Paired Phases
XL-U is available with a paired-phase bus bar scheme. Bus bars are grouped in pairs so that AC current in each pair is nearly equal in magnitude and opposite in direction. Two bus bars per phase are used. Phase C is paired with phase A, phase A is paired with phase B, and phase B is paired with phase C. The result is a minimized magnetic field. Current is balanced and temperature rise is kept to a minimum. Voltage drop is reduced. XL-U busway can be used on any application within its current rating but it is usually used for long runs where end-of-run voltage is critical. Due to its paired-phase design, XL-U busway is known throughout the industry as the best product available for welder loads.
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Sections and Components
XL-U feeder busway sections can be supplied in any length from 14” (356 mm) to 10’ (3048 mm). XL-U Plug-in busway is available in 4’ (1269 mm), 6’ (1828 mm), 8’ (2438 mm), and 10’ (3048 mm) sections. Elbows, tees, crosses, end closers, wall flanges, tap boxes, flanged end connections, switchboard connections, bus plugs, reducers, and hangers are available.
Joint Stack
XL-U busway uses a joint stack, similar to the Sentron™ busway, to connect sections together. The joint stack bolt is secured with a recommended 35 ft. lbs. of torque.
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One Bar Per Phase
XL-U busway is available from 225 to 6500 amperes. The number of bus bars and the dimensions depends on the maximum current rating. XL-U busway can be mounted vertically or horizontally, either edgewise or flatwise. The cross sections illustrated below are shown edgewise mounted. The “W” dimension varies with the current rating. There are two maximum current ratings for XL-U, UL and standard rating. XL-U busway is available in a one-bar-per-phase configuration for the maximum current ratings shown in the following table.
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Two Bars Per Phase
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XL-U busway is available in a two-bar-per-phase, paired- phase configuration for the maximum current ratings shown in the following table.
Four Bars Per Phase
At higher current ratings bus bars are doubled up. Four bars per phase are used in the current ratings shown in the following table. Note that paired-phasing is still used.
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Eight Bars Per Phase
To accommodate even higher current levels eight bars per phase are used.
Components
The following components are available for XL-U busway:
• • • • • • • • • •
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Hangers End closers Flanged ends Plug-in and center cable tap boxes Elbows Offsets Tees Crosses Reducers and expansion sections Bus plugs (circuit breaker, fusible)
XJ-L Busway
When an application needs a horizontal run of plug-in busway with a current rating that does not exceed 200 amperes, Siemens XJ-L™ would be a good choice. XJ-L busway is available with 100 or 200 ampere capacities, three-phase, three-wire (3Ø3W), 600 VAC or three-phase, four-wire (3Ø4W), 600 VAC. The neutral bus bar in the 3Ø4W type is rated for 100%. XJ-L busway is available in 2’ (610 mm), 5’ (1524 mm), and 10’ (3048 mm) lengths. There are up to 12 plug-ins per 10’ (3048 mm) length. Plug-ins are located in alternate positions from side-to-side.
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Installation
All XJ-L busway sections mate together end-to-end with overlapping joints which are held in place by integral spring pressure clips.
The sections are bolted together with captive screws.
Components
The following components are available for XJ-L busway:
• • • • • • •
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Hangers End closers Flanged ends Plug-in and center cable tap boxes Elbows Tees Bus plugs (circuit breaker, fusible)
XQ-R bus plug
XQ-R bus plugs serve both 120 and 240 VAC needs. This is useful for computer applications, laboratory/test facilities, schools, hospitals, and machine shops.
Review 6 1.
The installed cost of busway is typically ____________ than cable and conduit. a. b.
more less
2.
The maximum current rating available with XL-U busway with aluminum bus bars is ____________ amperes.
3.
____________ ____________ is a unique feature of XL-U busway that results in a magnetic field cancellation.
4.
To accommodate levels of current in the 3000 to 5000 ampere range, using aluminum XL-U busway, _________ ___ bars per phase are used.
5.
XJ-L is available in either ____________ ampere or ____________ ampere capacities.
6.
The ____________ is a bus plug available for XJ-L busway which provides 120 and 240 VAC.
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BD Busway
The BD™ busway is a general purpose power distribution busway of the plug-in design. BD busway is well established in the industry and has proven to be a dependable system. BD busway was first introduced in 1932, and with the exception of minor upgrades in materials, the basic design has remained unchanged. This means older systems can be expanded with today’s BD busway components.
Installation
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The bus bars of one end of a section are offset and the other end is straight. To connect two sections together match an offset end with a straight end. When ordering new BD busway to expand an existing system it is important to note if the new connection will be to an existing offset or straight end.
Bus bars are bolted together with a recommended 25 ft. lbs. of torque.
Plug-Ins
Each 10’ (3048 mm) section has ten bus plug receptacles, spaced alternately on each side (five on each side) of the busway section. Circuit breaker plugs are available in sizes from 100 to 800 amperes for voltages of 600 VAC or less. Fusible ® Vacu-Break switch plugs are available in sizes of 30 to 600 amperes, 3-pole, 600 VAC or less, or 4-pole solid neutral, 240 or 480 VAC. Capacitor and transformer bus plugs are also available. Capacitor bus plugs are used to reduce inductive heating and improving power factor. Transformer bus plugs furnish singlephase 120 or 240 VAC for lighting, small motors, or portable tools. Transformer plugs are also available with or without twopole AC magnetic contactors for plugging in on three-phase 240 or 480 VAC.
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Ratings and Dimensions
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BD busway comes in 10’ (3048 mm) lengths in current ratings from 225 to 1600 amperes. The number of bus bars per phase is determined by the current rating. A section of 225 amperes aluminum busway, for example, would have one bar per phase. A section of 1000 ampere aluminum busway would have two bus bars per phase. The following busway cross section diagram and table reflect ampere ratings and dimensions of BD busway.
Components
The following components are available for BD busway:
• • • • • • • •
Hangers End closers Flanged ends Plug-in and center cable tap boxes Elbows Tees Crosses Bus plugs (circuit breaker, fusible)
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Trol-E-Duct
Trol-E-Duct™ is part of the Siemens busway family. This busway is designed to provide power to indoor moving equipment, such as cranes, conveyors, hoists, mobile tools, and similar moving equipment.
Ratings
Trol-E-Duct is available in the following ratings: Indoor 1Ø2W, 1Ø3W, 3Ø3W, 3Ø4W 100 to 800 amperes 600 VAC
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Construction
Bus bars inside a housing provide power to a trolley. Brushes, mounted on the trolley, make contact with the bus bars. The trolley can be moved along an internal channel that is part of the housing. A section of the bus bar can be removed and replaced with a spacer that can be as small as ¼” (6.35 mm) or as large as 1’ (305 mm). It may be desirable to have a system that is powered only at specific locations. In these cases less expensive busless casings can be used throughout the unpowered sections.
Sections
Trol-E-Duct sections can be straight or curved. Curved sections are available only for 100 ampere busway. Curved sections can either be horizontal (shown) or vertical (not shown). Curved sections join with straight sections to create a continuous Trol-E-Duct system that matches any structural pattern. Curved sections are available for any radius of two feet or greater.
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Trolleys
Several types of trolleys are available for Trol-E-Duct. There are two types of tool hanger trolleys. Types TPTH and TBTH. Both types of tool hangers should not be used on a curved section with a radius of less than 5’. Type TPTH has provision for wiring a tool to the trolley. Type TBTH also has space for starters, receptacles and fuses.
Several other types of trolleys are available for a wide variety of applications. In the following example a hoist is wired to a trolley. The hoist moves along a monorail that runs parallel to a Trol-E-Duct system.
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Standard Trolleys
Standard trolleys are recommended for applications where the load current is less than 30 amperes, trolley speed is less than 5 m.p.h., and the curve radius is 5’ (1524 mm) or larger.
Heavy-Duty Trolleys
Heavy duty trolleys are used where the load current is less than 60 amperes, trolley speed is less than 5 m.p.h., and the curve radius is 5’ (1524 mm) or larger. By adding an additional pigtail between the brush and the wire grip the trolley can be rated as high as 90 amperes.
Roller-Type Trolleys
Roller-type trolleys are used on lighter duty applications. The load current is less than 20 amperes and the trolley speed does not exceed 2.5 m.p.h. Roller-type trolleys will negotiate a 5’ (1524 mm) radius curve.
Curved-Type Trolleys
Curved-type trolleys will negotiate a curve with a 3’ (914 mm) radius or greater, speeds up to 5 m.p.h., and loads as high as 60 amperes.
Button-Type Trolleys
Button-type trolleys will negotiate a curve with a 5’ (1524 mm) radius or greater, speeds up to 5 m.p.h., and loads as high as 60 amperes. Extra wheels on this trolley make it well suited for carrying a heavy weight.
Transfer Trolley
The transfer trolley is rated at 30 amperes and is suited for applications where the trolley must jump a gap in duct. Flared ends are used on the bus duct when the trolley must jump. The trolley can jump a separation as great as 3/8” (9.5 mm) with sections misaligned as much as 3/8” (9.5 mm). It will negotiate a curve with a radius of 2’ (610 mm) or greater.
Nylon Brush Trolley
Nylon brush trolleys have brushes which are used to remove abrasive material, such as metallic or cement dust, from the bus bars.
Abrasive Trolley
Abrasive trolleys are designed to clean and remove corrosive buildup from bus bars in harsh atmospheres.
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Review 7
90
1.
BD busway was first introduced in ____________ .
2.
When ordering new BD busway to expand an existing system it is important to note if the new connection will be to an existing ____________ or straight end.
3.
Fusible Vacu-Break switches are available in sizes of ____________ to ____________ amperes for BD busway.
4.
____________ is a type of busway that is designed to provide power to moving equipment such as cranes or hoists.
5.
Standard Trol-E-Duct trolleys can handle a curve radius of ____________ feet or larger.
Review Answers
Review 1
1) distribution system; 2) feeder; 3) Busway; 4) busway: 5) feeder, plug-in; 6) Feeder, plug-in.
Review 2
1) two, three; 2) 47,000; 3) 66, 54; 4) 368; 5) 200,000; 6) 10,000, 200,000.
Review 3
1) 4,000, 5,000; 2) a. feeder, b. plug-in; 3) 4’, 6’, 8’; 4) 2; 5) 0.125, 1’ 4.5” , 10’.
Review 4
1) C; 2) F; 3) A; 4) B; 5) E; 6) H; 7) G; 8) D.
Review 5
1) closed; 2) Sentron Busway Systems Selection and Application Guide; 3) 24, 24; 4) 5; 5) a.
Review 6
1) b; 2) 5000; 3) Paired phases; 4) eight; 5) 100, 200; 6) XQ-R.
Review 7
1) 1932; 2) offset; 3) 30, 600; 4) Trol-E-Duct; 5) 5.
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Final Exam
The final exam is intended to be a learning tool. The book may be used during the exam. A tear-out answer sheet is provided. A grade of 70% or better is passing. Upon successful completion of the test a certificate will be issued. Questions
1.
The two types of busway are ____________ . a. b. c. d.
2.
It typically takes ____________ man-hours to install a busway system than cable and conduit. a. b.
3.
fewer twice as many
10,000 14,000
c. d.
22,000 42,000
2X 40
c. d.
43 54
General guidelines which discuss the safe application of busway are covered in the NEC® Article ____________ . a. b.
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c. d.
According to IEC 529, a plug-in outlet that is said to be “finger safe” has an IEC code of IP ____________ . a. b.
5.
about the same more
According to NEMA publication number BU 1.1-2000, the short-circuit rating of plug-in busway with a maximum continuous current rating of 1000 amperes should be at least ____________ amperes. a. b.
4.
feeder and service entrance feeder and plug-in plug-in and service entrance indoor and outdoor
368 445
c. d.
365 450
6.
The interrupting rating of a class R fuse is ____________ amperes. a. b.
7.
50,000 200,000
2,000 3,000
c. d.
4,000 5,000
The following Sentron bus bar is used for ____________ busway.
a. b. 9.
c. d.
The maximum current rating of Sentron™ busway with aluminum bus bars is ____________ amperes. a. b.
8.
10,000 100,000
feeder plug-in
c. d.
service entrance outdoor
Plug-in outlets on Sentron busway are located on ____________ centers. a. b.
16” (406 mm) 2’ (610 mm)
c. d.
30” (762 mm) 3’ (914 mm)
10. A Sentron busway expansion fitting allows for an expansion compensation of ± ____________ . a. b.
2” (50.8 mm) 1’ (305 mm)
c. d.
18” (457 mm) 6” (152 mm)
11. ___________ are used to house Siemens molded case circuit breakers or fusible switches. a. b.
Reducers Service heads
c. d.
Bus Plugs Expansion fittings
12. ____________ are used to safely terminate a run of busway and protect the bus bar ends. a. b.
End closers Service heads
c. d.
Tees Flanged ends 93
13. A minimum distance of ____________ from the wall to a joint where the new section of busway begins is required when Sentron busway passes through a wall. a. b.
4” (102 mm) 7” (127 mm)
c. d.
6” (152 mm) 10” (254 mm)
14. The dimensions of a Sentron right elbow, with aluminum bus bars, rated for 1000 amperes is ____________ . a. b.
10” x 10” 12” x 12”
c. d.
18” x 18” 24” x 24”
15. The paired-phase bus bar scheme is used with ____________ busway. a. b.
Sentron XJ-L
c. d.
XL-U BD
16. The maximum current rating of XL-U busway with copper bus bars is ____________ amperes. a. b. 17.
3000 4000
c. d.
5000 6500
The maximum current rating of XJ-L busway is ____________ amperes. a. b.
100 5000
c. d.
200 6000
18. The type of Siemens busway that was first introduced in 1932 is ____________ busway. a. b.
Sentron XJ-L
c. d.
XL-U BD
19. ____________ is the type of Trol-E-Duct tool box hanger that provides space for starters, receptacles and fuses. a. b.
TBTH TPTH
c. d.
Transfer trolley Button-type trolley
20. Curved Trol-E-Duct sections are available for ____________ . a. b. c. d. 94
any radius of 2’ (610 mm) or greater a 2’ (610 mm) radius only any radius of 5’ (1524 mm) or greater a 5’ radius only (1524 mm)
quickSTEP Online Courses
quickSTEP online courses are available at http://www.sea.siemens.com/step. The quickSTEP training site is divided into three sections: Courses, Downloads, and a Glossary. Online courses include reviews, a final exam, the ability to print a certificate of completion, and the opportunity to register in the Sales & Distributor training database to maintain a record of your accomplishments. From this site the complete text of all STEP 2000 courses can be downloaded in PDF format. These files contain the most recent changes and updates to the STEP 2000 courses. A unique feature of the quickSTEP site is our pictorial glossary. The pictorial glossary can be accessed from anywhere within a quickSTEP course. This enables the student to look up an unfamiliar word without leaving the current work area.
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Table of Contents
Introduction ............................................................................. 2 Control Circuits........................................................................ 4 Electrical Symbols ................................................................... 6 Line Diagrams ....................................................................... 16 Overload Protection............................................................... 22 Overload Relays .................................................................... 26 Manual Control ...................................................................... 35 Magnetic Contactors and Starters ......................................... 41 Starter Ratings ....................................................................... 46 Furnas INNOVA PLUS Starters .............................................. 49 ESP100 Starters .................................................................... 50 SIRIUS Type 3R Starters ........................................................ 51 World Series Type 3TF Starters .............................................. 53 Multi-Speed and Reversing Starters ...................................... 55 Reduced-Voltage Starting ...................................................... 59 Solid-State Reduced-Voltage Controllers ............................... 64 Pilot Devices ......................................................................... 67 Control Transformers .............................................................. 78 Control Relays ....................................................................... 79 Timing Relays ........................................................................ 85 Pressure Switches ................................................................ 90 LOGO! Logic Module ............................................................ 93 Review Answers ................................................................... 96 Final Exam ............................................................................. 97
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Introduction
Welcome to another course in the STEP 2000 series, Siemens Technical Education Program, designed to prepare our distributors to sell Siemens Energy & Automation products more effectively. This course covers Basics of Control Components and related products. Upon completion of Basics of Control Components you will be able to:
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State the purpose and general principles of control components and circuits
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State the difference between manual and automatic control operation
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Identify various symbols which represent control components
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Read a basic line diagram
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Describe the construction and operating principles of manual starters
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Describe the construction and operating principles of magnetic contactors and magnetic motor starters
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Identify various Siemens and Furnas manual starters and magnetic motor starters, and describe their operation in a control circuit
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Explain the need for motor overload protection
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State the need for reduced-voltage motor starting
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Describe typical motor starting methods
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Describe the difference between normally open and normally pilot devices
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Describe the operating principles of control relays
This knowledge will help you better understand customer applications. In addition, you will be better able to describe products to customers and determine important differences between products. You should complete Basics of Electricity before attempting Basics of Control Components. An understanding of many of the concepts covered in Basics of Electricity is required for Basics of Control Components. In addition, you may want to complete the STEP 2000 course Sensors after completing Basics of Control Components. If you are an employee of a Siemens Energy & Automation authorized distributor, fill out the final exam tear-out card and mail in the card. We will mail you a certificate of completion if you score a passing grade. Good luck with your efforts. Siemens & Furnas Controls is a business unit of Siemens Energy & Automation, Inc. SIMICONT is a registered trademark of Siemens Energy & Automation, Inc. INNOVA PLUS and ESP100 are trademarks of Siemens Energy & Automation, Inc. National Electrical Code® and NEC® are registered trademarks of the National Fire Protection Association, Quincy, MA 02269. Portions of the National Electrical Code are reprinted with permission from NFPA 70-1999, National Electrical Code Copyright, 1998, National Fire Protection Association, Quincy, MA 02269. This reprinted material is not the complete and official position of the National Fire Protection Association on the referenced subject which is represented by the standard in its entirety. Underwriters Laboratories Inc. is a registered trademark of Underwriters Laboratories Inc., Northbrook, IL 60062. The abbreviation “UL” shall be understood to mean Underwriters Laboratories Inc. CAGE CLAMP is a trademark of WAGO Corporation, Brown Deer, WI 53223
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Control Circuits
The National Electrical Code® (NEC®) defines a controller as a device or group of devices that serves to govern, in some predetermined manner, the electrical power delivered to the apparatus to which it is connected (Article 100-definitions). Control
Control, as applied to control circuits, is a broad term that means anything from a simple toggle switch to a complex system of components which may include relays, contactors, timers, switches, and indicating lights. Every electrical circuit for light or power has control elements. One example of a simple control circuit is a light switch used to turn lights on and off.
Of course there are many other devices and equipment systems in industrial applications. Motor control, for example, can be used to start and stop a motor and protect the motor, associated machinery, and personnel. In addition, motor controllers might also be used for reversing, changing speed, jogging, sequencing, and pilot-light indication. Control circuits can be complex: accomplishing high degrees of automatic and precise machine operation.
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NEC® and National Electrical Code® are registered trademarks of the National Fire Protection Association. Reprinted with permission from NFPA 70-1999, the National Electrical Code®, Copyright© 1998, National Fire Protection Association, Quincy, MA 02269.
Manual Control
Control is considered to be manually operated when someone must initiate an action for the circuit to operate. For example, someone might have to flip the switch of a manual starter to start and stop a motor.
Automatic Operation
While manual operation of machines is still common practice, many machines are started and stopped automatically. Frequently there is a combination of manual and automatic control. A process may have to be started manually, but may be stopped automatically.
Control Elements
The elements of a control circuit include all of the equipment and devices concerned with the circuit function. This includes enclosures, conductors, relays, contactors, pilot devices, and overcurrent-protection devices. The selection of control equipment for a specific application requires a thorough understanding of controller operating characteristics and wiring layout. The proper control devices must be selected and integrated into the overall plan.
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Electrical Symbols
Language has been developed in order to transfer ideas and information. In order to understand the ideas and information being communicated, an understanding of the language is necessary. The language of controls consists of a commonly used set of symbols which represents control components. Contact Symbols
Contact symbols are used to indicate an open or closed path of current flow. Contacts are shown as normally open (NO) or normally closed (NC). Contacts shown by this symbol require another device to actuate them.
The standard method of showing a contact is by indicating the circuit condition it produces when the actuating device is in the deenergized or nonoperated state. For example, in the following illustration a relay is used as the actuating device. The contacts are shown as normally open, meaning the contacts are open when the relay is deenergized. A complete path of current does not exist and the light is off.
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Normally Open Contact Example
In a control diagram or schematic, symbols are usually not shown in the energized or operated state. For the purposes of explanation in this text, a contact or device shown in a state opposite of its normal state will be highlighted. For example, in the following illustration the circuit is first shown in the deenergized state. The contacts are shown in their normally open (NO) state. When the relay is energized, the contacts close completing the path of current and illuminating the light. The contacts have been highlighted to indicate they are now closed. This is not a legitimate symbol. It is used here for illustrative purposes only.
Normally Closed Contact Example
In the following illustration the contacts are shown as normally closed (NC), meaning the contacts are closed when the relay is deenergized. A complete path of current exists and the light is on. When the relay is energized, the contacts open turning the light off.
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Switch Symbols
Switch symbols are also used to indicate an open or closed path of current flow. Variations of this symbol are used to represent limit switches, foot switches, pressure switches, level switches, temperature-actuated switches, flow switches, and selector switches. Switches, like contacts, require another device or action to change their state. In the case of a manual switch someone must manually change the position of the switch.
Normally Open Switch Example
In the following illustration a battery is connected to one side of a normally open switch and a light to the other. Current is prevented from flowing to the light when the switch is open. When someone closes the switch, the path of current flow is completed and the light illuminates.
Normally Closed Switch Example
In the following illustration a battery is connected to one side of a normally closed switch and a light to the other. Current is flowing to the light when the switch is closed. When someone opens the switch, the path of current flow is interrupted and the light turns off.
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Pushbutton Symbols
There are two basic types of pushbuttons: momentary and maintained. A normally open momentary pushbutton closes as long as the button is held down. A normally closed momentary pushbutton opens as long as the button is held down. A maintained pushbutton latches in place when the button is pressed.
Normally Open Pushbutton Example
In the following illustration a battery is connected to one side of a normally open pushbutton and a light is connected to the other side. When the pushbutton is depressed, a complete path of current flow exists through the pushbutton and the light is illuminated.
Normally Closed Pushbutton Example
In the following example current will flow to the light as long as the pushbutton is not depressed. When the pushbutton is depressed, current flow is interrupted and the light turns off.
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Coil Symbols
Coils are used in electromagnetic starters, contactors, and relays. The purpose of contactors and relays is to open and close associated contacts. A letter is used to designate the coil; for example, “M” frequently indicates a motor starter and “CR” indicates a control relay. The associated contacts have the same identifying letter. Contactors and relays use an electromagnetic action which will be described later to open and close these contacts. The associated contacts can be either normally open or normally closed.
Coil Example Using Normally Open Contacts
In the following example, the “M” contacts in series with the motor are controlled by the “M” contactor coil. When someone closes the switch, a complete path of current flow exists through the switch and “M” contactor coil. The “M” contactor coil actuates the “M” contacts which provide power to the motor.
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Overload Relay Symbols
Overload relays are used to protect motors from overheating due to an overload on the driven machinery, low-line voltage, or an open phase in a three-phase system. When excessive current is drawn for a predetermined amount of time, the relay opens and the motor is disconnected from its source of power.
Pilot Light Symbols
A pilot light is a small electric light used to indicate a specific condition of a circuit. For example, a red light might be used to indicate a motor is running. The letter in the center of the pilot light symbol indicates the color of the light.
Other Symbols
In addition to the symbols discussed here, there are many other symbols used in control circuits. The following chart shows many of the commonly used symbols.
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Abbreviations
Abbreviations are frequently used in control circuits. The following list identifies a few commonly used abbreviations. AC ALM AM ARM AU BAT BR CAP CB CKT CONT CR CT D DC DISC DP DPDT DPST DT F FREQ FTS FU GEN GRD HOA IC INTLK IOL JB LS LT M
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Alternating Current Alarm Ammeter Armature Automatic Battery Brake Relay Capacitor Circuit Breaker Circuit Control Control Relay Current Transformer Down Direct Current Disconnect Switch Double-Pole Double-Pole, Double-Throw Double-Pole, Single-Throw DoubleThrow Forward Frequency Foot Switch Fuse Generator Ground Hand/Off/Auto Selector Switch Integrated Circuit Interlock Instanstaneous Overload Junction Box Limit Switch Lamp Motor Starter
MTR MN NEG NEUT NC NO OHM OL PB PH POS PRI PS R REC RES RH S SEC SOL SP SPDT SPST SS SSW T TB TD THS TR U UV VFD XFR
Motor Manual Negative Neutral Normally Closed Normally Open Ohmmeter Overload Pushbutton Phase Positive Primary Pressure Switch Reverse Rectifier Resistor Rheostat Switch Secondary Solenoid Single-Pole Single-Pole, Double Throw Single-Pole, Single Throw Selector Switch Safety Switch Transformer Terminal Board Time Delay Thermostat Switch Time Delay Relay Up Under Voltage Variable Frequency Drive Transformer
Review 1 1.
A control is ____________ operated when someone must initiate an action for the circuit to operate.
2.
Which of the following symbols represents a normally open contact?
a. 3.
b.
c.
Which of the following symbols indicates a normally open pushbutton? a.
5.
c.
Which of the following symbols represents a normally closed contact?
a. 4.
b.
b.
c.
Which of the following symbols indicates a mushroom head pushbutton? a.
b.
c.
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Line Diagrams
The method of expressing the language of control symbols is a line diagram, also referred to as a ladder diagram. Line diagrams are made up of two circuits, the control circuit and the power circuit. Electrical wires in a line diagram are represented by lines. Control-circuit wiring is represented by a lighter-weight line and power-circuit wiring is represented by a heavier-weight line. A small dot or node at the intersection of two or more wires indicates an electrical connection.
Line diagrams show the functional relationship of components and devices in an electrical circuit, not the physical relationship. For example, the following illustration shows the physical relationship of a pilot light and a pushbutton.
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The functional relationship can be shown pictorially with the following illustration.
Reading a Line Diagram
This functional relationship is shown symbolically with a line diagram. Line diagrams are read from left to right. Depressing the pushbutton would allow current to flow from L1 through the pushbutton, illuminating the pilot light, to L2. Releasing the pushbutton stops current flow turning the pilot light off.
Power Circuit and Control Circuit
The power circuit, indicated by the heavier-weight line, is what actually distributes power from the source to the connected load (motor). The control circuit, indicated by the lighter-weight line, is used to “control” the distribution of power.
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Connecting Loads and Control Devices
Control circuits are made up of control loads and control devices. The control load is an electrical device that uses electrical power. Pilot lights, relays, and contactors are examples of control loads. Control devices are used to activate the control load. Pushbuttons and switches are examples of control devices. The following illustration shows the proper connection of a pilot light (load) with a pushbutton (control device). The power lines are drawn vertically and marked L1 and L2. In this example the voltage potential between L1 and L2 is 120 VAC. The pilot light selected must be rated for 120 VAC. When the pushbutton is depressed, the full 120 volt potential is applied to the pilot light.
Connecting the Load to L2
Only one control load should be placed in any one circuit line between L1 and L2. One side of the control load is connected to L2 either directly or, in some instances, through overload relay contacts. In the following example a pilot light is directly connected to L2 on one circuit line. A contactor coil is indirectly connected through a set of overload contacts (OL) to L2 on a second circuit line. This is a parallel connection. Depressing the pushbutton would apply 120 VAC to the pilot light and the “M” contactor.
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Control loads are generally not connected in series. The following illustration shows two ways to connect a control load. In one instance the control loads are improperly connected in series. When the pushbutton is depressed, the voltage across L1 and L2 is divided across both loads, the result being that neither load will receive the full 120 volts necessary for proper operation. If one load fails in this configuration, the entire circuit is rendered useless. In the other instance the loads are properly connected in parallel. In this circuit there is only one load for each line between L1 and L2. The full 120 volts will appear across each load when the pushbutton is depressed. If one load fails in this configuration, the other load will continue to operate normally.
Connecting Control Devices
Control devices are connected between L1 and the load. The control device can be connected in series or parallel, depending on the desired results. In the following illustration, the pushbuttons are connected in parallel. Depressing either pushbutton will allow current to flow from L1, through the depressed pushbutton, through the pilot light, to L2.
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In the following illustration, two pushbuttons are connected in series. Both pushbuttons must be depressed in order to allow current to flow from L1 through the load to L2.
Line Numbering
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Numbering each line makes it easier to understand more complex line diagrams. In the following illustration, line 1 connects pushbutton 1 to pilot light 1. Line 2 connects pushbutton 2 to pilot light 1. Line 3 connects switch 1 to pilot light 2 and the “M” contactor on line 4.
Review 2 1.
Line diagrams are read from ____________ to ____________ , or L1 to L2.
2.
Match the items on the line diagram with the associated list.
A ____________ B ____________ C ____________ D ____________ E ____________ F ____________
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Overload Protection
Before discussing specific control components, it is necessary to review what an overload is and what steps can be taken to limit the damage an overload can cause. Current and Temperature
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Current flow in a conductor always generates heat due to resistance. The greater the current flow, the hotter the conductor. Excess heat is damaging to electrical components. For that reason, conductors have a rated continuous current carrying capacity or ampacity. Overcurrent protection devices are used to protect conductors from excessive current flow. Thermal overload relays are designed to protect the conductors (windings) in a motor. These protective devices are designed to keep the flow of current in a circuit at a safe level to prevent the circuit conductors from overheating.
Excessive current is referred to as overcurrent. The National Electrical Code® defines overcurrent as any current in excess of the rated current of equipment or the ampacity of a conductor. It may result from overload, short circuit, or ground fault (Article 100-definitions). Short Circuits
When two bare conductors touch, a short circuit occurs. When a short circuit occurs, resistance drops to almost zero. Shortcircuit current can be thousands of times higher than normal operating current.
Ohm’s Law demonstrates the relationship of current, voltage, and resistance. For example, a 240 volt motor with 24 ohms of resistance would normally draw 10 amps of current.
When a short circuit develops, resistance drops. If resistance drops to 24 milliohms, current will be 10,000 amps.
The heat generated by this current will cause extensive damage to connected equipment and conductors. This dangerous current must be interrupted immediately when a short circuit occurs.
Reprinted with permission from NFPA 70-1999, the National Electrical Code®, Copyright© 1998, National Fire Protection Association, Quincy, MA 02269.
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Overload Conditions
An overload occurs when too many devices are operated on a single circuit or a piece of electrical equipment is made to work harder than it is designed for. For example, a motor rated for 10 amperes may draw 20, 30, or more amperes in an overload condition. In the following illustration a package has become jammed on a conveyor causing the motor to work harder and draw more current. Because the motor is drawing more current it heats up. Damage will occur to the motor in a short time if the problem is not corrected or the circuit is not shut down by the overload relay.
Temporary Overload Due to Starting Current
Electric motors are rated according to the amount of current they will draw at full load. When most motors start, they draw current in excess of the motor’s full-load current rating. Motors are designed to tolerate this overload current for a short period of time. Many motors require 6 times (600%) the full-load current rating to start. Some newer, high-efficiency motors may require higher starting currents. As the motor accelerates to operating speed, the current drops off quickly. The time it takes for a motor to accelerate to operating speed depends on the operating characteristics of the motor and the driven load. A motor, for example, might require 600% of full-load current and take 8 seconds to reach operating speed.
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Overload Protection
Fuses and circuit breakers are protective devices used to protect circuits against short circuits, ground faults, and overloads. In the event of a short circuit, a properly sized fuse or circuit breaker will immediately open the circuit. There is, however, a dilemma that occurs when applying fuses and circuit breakers in motor control circuits. The protective device must be capable of allowing the motor to exceed its fullload rating for a short time. Otherwise, the motor will trip each time it is started. In this situation it is possible for a motor to encounter an overload condition which does not draw enough current to open the fuse or trip the circuit breaker. This overload condition could easily cause enough heat to damage the motor. In the next section we will see how overload relays are used to solve this problem.
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Overload Relays
Overload relays are designed to meet the special protective needs of motor control circuits. Overload relays:
Trip Class
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allow harmless temporary overloads, such as motor starting, without disrupting the circuit
•
will trip and open a circuit if current is high enough to cause motor damage over a period of time
•
can be reset once the overload is removed
Overload relays are rated by a trip class, which defines the length of time it will take for the relay to trip in an overload condition. The most common trip classes are Class 10, Class 20 and Class 30. Class 10, for example, has to trip the motor off line in 10 seconds or less at 600% of the full load amps. This is usually sufficient time for the motor to reach full speed. Many industrial loads, particularly high inertia loads, use Class 30. Siemens standard overload relays are Class 10 or Class 20 with Class 30 available with some starters.
Overload Relay in a Motor Circuit
The following illustration shows a motor circuit with a manual starter and overloads. Current flows through the overloads while the motor is running. Excess current will cause the overload to trip at a predetermined level, opening the circuit between the power source and the motor. After a predetermined amount of time the starter can be reset. When the cause of the overload has been identified and corrected the motor can be restarted.
Bimetal Overloads
Overload protection is accomplished with the use of a bimetal strip. This component consists of a small heater element wired in series with the motor and a bimetal strip that can be used as a trip lever. A bimetal strip is made of two dissimilar metals bonded together. The two metals have different thermal expansion characteristics, so the bimetal bends at a given rate when heated. Under normal operating conditions the heat generated by the heater element will be insufficient to cause the bimetal strip to bend enough to trip the overload relay.
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As current rises, heat also rises. The hotter the bimetal becomes, the more it bends. In an overload condition the heat generated from the heater will cause the bimetal strip to bend until the mechanism is tripped, stopping the motor.
Some overload relays that are equipped with a bimetal strip are designed to automatically reset the circuit when the bimetal strip has cooled and reshaped itself, restarting the motor. If the cause of the overload still exists, the motor will trip again and reset at given intervals. Care must be exercised in the selection of this type of overload as repeated cycling will eventually damage the motor.
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Ambient Compensated Overload Relay
In certain applications, such as a submersible pump, the motor may be installed in a location having a constant ambient temperature. The motor control, along with the overload relay, may be installed in a location with a varying ambient temperature. The trip point of the overload relay will vary with the temperature of the surrounding air as well as current flowing through the motor. This can lead to premature and nuisance tripping. Ambient compensated overload relays are designed to overcome this problem. A compensated bimetal strip is used along with a primary bimetal strip. As the ambient temperature changes, both bimetal strips will bend equally and the overload relay will not trip the motor. However, current flow through the motor and the heater element will affect the primary bimetal strip. In the event of an overload condition the primary bimetal strip will engage the trip unit.
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Class 48 Bimetal Ambient Compensated Overload Relay
The Class 48 bimetal ambient compensated overload relay is available in single-pole or three-pole designs. Unlike the melting alloy overload relay, the bimetal ambient compensated overload relay can be set for manual or self-resetting operation. An adjustment dial located on the unit allows the ampere trip setting to be adjusted by ±15%. The bimetal ambient compensated overload relay heater elements are available in Class 20 or Class 10 ratings. A normally open or normally closed auxiliary contact is available as an option.
SIRIUS 3RU11 Overload Relay
The Siemens SIRIUS 3RU11 is a bimetal type of overload relay with the heater elements as an integral part of the design. The unit comes with a Class 10 trip as standard. The SIRIUS 3RU11 features manual or automatic reset, adjustable current settings, and ambient compensation. A switch-position indicator also incorporates a test function which is used to simulate a tripped overload relay. A normally open and a normally closed auxiliary contact are included.
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Electronic Overload Relays
Electronic overload relays are another option for motor protection. The features and benefits of electronic overload relays vary but there are a few common traits. One advantage offered by electronic overload relays is a heaterless design. This reduces installation cost and the need to stock a variety of heaters to match motor ratings. Electronic relays offer phase loss protection. If a power phase is lost, motor windings can burn out very quickly. Electronic overload relays can detect a phase loss and disconnect the motor from the power source. This feature is not available on mechanical types of overload relays.
Furnas ESP 100 Electronic Overload Relay
A single ESP100 overload relay replaces at least six size ranges of heaters. Instead of installing heaters the full-load amperes (FLA) of the motor is set with a dial. The ESP100 overload relay illustrated below, for example, is adjustable from 9 to 18 amperes. NEMA Class 10, 20, and 30 trip curves are available for a variety of applications. The relay comes in either a manual or self-resetting version. Auxiliary contacts are available as an option.
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Siemens 3RB10 Electronic Overload Relay
The Siemens SIRIUS 3RB10 is an electronic overload relay with a design very similar to the ESP 100. The unit comes with a Class 10 or Class 20 trip. The 3RB10 features manual or automatic reset, adjustable current settings, and ambient compensation. A switch-position indicator also incorporates a test function which is used to simulate a tripped overload relay. A normally open and a normally closed auxiliary contact are included.
Siemens 3RB12 Electronic Overload Relay
In addition to heaterless construction and phase loss protection, the 3RB12 offers ground fault protection, phase unbalance, LED displays (ready, ground fault, and overload), automatic reset with remote capability, and selectable trip classes (5, 10, 15, 20, 25, or 30). The 3RB12 is self-monitoring and is provided with 2 normally open and 2 normally closed isolated auxiliary contacts.
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PROFIBUS DP
In any complex process the need for rapid information flow is critcal. PROFIBUS DP is an open communication system based upon international standards developed through industry associations. PROFIBUS DP allows the connection of several field devices, such as SIMOCODE-DP, on a single bus for communication to a PLC or computer. PROFIBUS DP is suitable as a replacement for costly parallel wiring.
3UF5 SIMOCODE-DP
The 3UF5 SIMOCODE-DP overload relay integrates with PROFIBUS-DP. SIMOCODE-DP protects the load against overload, phase failure, ground fault, and current imbalance. SIMOCODE-DP can be parametrized, controlled, observed, and tested from a central source such as a PC with WinSIMOCODE-DP/Professional installed, or a PLC with a PROFIBUS-DP communication processor. The 3UF50 basic unit can also be used as an autonomous solid-state overload relay for motor protection. A trip class in six steps from Class 5 to Class 30 can be selected. The basic unit (shown) is supplied with four inputs and four outputs. An available expansion unit provides eight additional inputs and four additional outputs.
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Review 3 1.
With an increase in current, heat will ____________ . a. increase b. decrease c. remain the same
2.
The National Electrical Code® defines overcurrent as any current in ____________ of the rated current of equipment or the ampacity of a conductor.
3.
An ____________ occurs when electrical equipment is required to work harder than it is rated.
4.
A Class __________ overload relay will trip an overloaded motor offline within 10 seconds at six times fullload amps. a. 10 b. 20 c. 30
5.
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A ____________ strip uses two dissimilar metals bonded together.
Manual Control
Manual control, as the name implies, are devices operated by hand. A simple knife switch, like the one shown in the following illustration, was the first manual-control device used to start and stop motors. The knife switch was eventually replaced with improved control designs, such as manual and magnetic starters.
Basic Operation
The National Electrical Code® requires that a motor control device must also protect the motor from destroying itself under overload conditions. Manual starters, therefore, consist of a manual contactor, such as a simple switch mechanism, and a device for overload protection. The following diagram illustrates a single-pole manual motor starter. Each set of contacts is called a pole. A starter with two sets of contacts would be called a two-pole starter.
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Two-Pole Manual Starter
Starters are connected between the power source and the load. For example, a two-pole or single-phase motor starter is connected to a motor. When the switch is in the “OFF” position, the contacts are open preventing current flow to the motor from the power source. When the switch is in the “ON” position, the contacts are closed and current flows from the power source (L1), through the motor, returning to the power source (L2).
This is represented with a line drawing and symbols as illustrated in the following drawing.
Low Voltage Protection
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Some manual motor starters offer low-voltage protection (LVP) as an option. LVP will automatically remove power from the motor when incoming power drops or is interrupted. The starter must be manually reset when power is restored. This protects personnel from potential injury caused by machinery that may otherwise automatically restart when power is restored.
SMF FractionalHorsepower Manual Starters
Siemens SMF fractional-horsepower starters provide overload protection and manual “ON/OFF” control for small motors. SMF starters are available in one- or two-pole versions suitable for AC motors up to 1 HP and 277 VAC. The two-pole version is suitable for DC motors up to 3/4 HP and 230 VDC. SMF manual starters are available in a variety of enclosures. A two-speed version is available.
MMS and MRS Manual Switches
Siemens MMS and MRS manual switches are similar to SMF starters but do not provide overload protection. MMS and MRS switches only provide manual “ON/OFF” control of single- or three-phase AC motors where overload protection is provided separately. These devices are suitable for use with three-phase AC motors up to 10 HP and 600 VAC and up to 1-1/2 HP and 230 VDC. The MMS and MRS manual switches are available in various enclosures. Two-speed and reversing versions are available.
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Furnas Class 11 Manual Starter and Manual Contactor
Furnas Class 11 manual starters use a melting-alloy overload relay with interchangeable heater elements and a manual reset. It has a maximum rating of 10 HP at 460 VAC (3Ø) and 5 HP at 230 VAC (1Ø). Class 11 manual starters are available in a complete line of general-purpose and industrial-duty enclosures. Class 11 manual starters may also be furnished with a low-voltage protection circuit. Class 11 manual contactors provide no overload protection.
3RV10 Motor Starter Protectors
3RV10 motor starter protectors (MSPs) are part of the Siemens SIRIUS 3R motor control product line. A thermal overload with a bimetal strip is used to provide overload protection with the 3RV10 motor starter protector. 3RV10 MSPs come in four frame sizes: 3RV101, 3RV102, 3RV103, and 3RV104. Frame 3RV101 3RV102 3RV103 3RV104
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Max Current at 460 VAC 12 Amps 25 Amps 50 Amps 100 Amps
Max HP at 460 VAC 7.5 20 40 75
The 3RV101 is available in both screw-terminal and CAGE CLAMP™ versions. The 3RV102, 3RV103, and 3RV104 are available with screw terminals.
CAGE CLAMP™
The CAGE CLAMP™ is available on many Siemens SIRIUS 3R products including the MSPs. To connect a wire, simply push an electrician blade screwdriver into the appropriate portal, insert the stripped end of the wire into the portal directly above, remove the screwdriver, and the wire is securely connected. CAGE CLAMP™ devices are especially beneficial in installations that are subject to vibration.
Enclosures and Options
Siemens 3RV10 MSPs are available in a variety of enclosures. Several options, such as indicator lights, are also available.
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Reversing Drum Controller
Manually operated drum controllers, like the Furnas Class 58 reversing drum controller, stop and change direction of reversible AC motors. Overload protection is not provided by the reversing drum controller and must be supplied by an external means. The Furnas Class 58 reversing drum controller is rated for 10 HP at 460 VAC. Another style of drum switch is used to change speed of multi-speed motors.
Master Switch
The Furnas Class 53 master switches provide single-handle control of hoists, cranes, oven pushers, and other equipment requiring speed steps of wound rotor or direct-current motors. Master switches are available with momentary or maintained contacts and up to five speed settings.
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Magnetic Contactors and Starters
Most motor applications require the use of remote control devices to start and stop the motor. Magnetic contactors, similar to the ones shown below, are commonly used to provide this function. Contactors are also used to control distribution of power in lighting and heating circuits.
Basic Contactor Operation
Magnetic contactors operate utilizing electromagnetic principles. A simple electromagnet can be fashioned by winding a wire around a soft iron core. When a DC voltage is applied to the wire, the iron becomes magnetic. When the DC voltage is removed from the wire, the iron returns to its nonmagnetic state. This principle is used to operate magnetic contactors.
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The following illustration shows the interior of a basic contactor. There are two circuits involved in the operation of a contactor: the control circuit and the power circuit. The control circuit is connected to the coil of an electromagnet, and the power circuit is connected to the stationary contacts.
The operation of this electromagnet is similar to the operation of the electromagnet we made by wrapping wire around a soft iron core. When power is supplied to the coil from the control circuit, a magnetic field is produced magnetizing the electromagnet. The magnetic field attracts the armature to the magnet, which in turn closes the contacts. With the contacts closed, current flows through the power circuit from the line to the load. When the electromagnet’s coil is deenergized, the magnetic field collapses and the movable contacts open under spring pressure. Current no longer flows through the power circuit.
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The following schematic shows the electromagnetic coil of a contactor connected to the control circuit through a switch (SW1). The contacts of the contactor are connected in the power circuit to the AC line and a three-phase motor. When SW1 is closed, the electromagnetic coil is energized, closing the “M” contacts and applying power to the motor. Opening SW1 deenergizes the coil and the “M” contacts open, removing power from the motor.
Overload Relay
Contactors are used to control power in a variety of applications. When applied in motor-control applications, contactors can only start and stop motors. Contactors cannot sense when the motor is being loaded beyond its rated conditions. They provide no overload protection. Most motor applications require overload protection. However, some smaller-rated motors have overload protection built into the motor (such as a household garbage disposal). Overload relays, similar to the one shown below, provide this protection. The operating principle, using heaters and bimetal strips, is similar to the overloads discussed previously.
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Motor Starter
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Contactors and overload relays are separate control devices. When a contactor is combined with an overload relay, it is called a motor starter.
Motor Starter in a Control Circuit
The following diagram shows the electrical relationship of the contactor and overload relay. The contactor, highlighted with the darker grey, includes the electromagnetic coil, the main motor contacts, and the auxiliary contacts. The overload relay, highlighted by the lighter grey, includes the “OL” heaters and overload contacts. The contactor and the overload relay have additional contacts, referred to as auxiliary contacts, for use in the control circuit. In this circuit a normally closed “OL” contact has been placed in series with the “M” contactor coil and L2. A normally open “M” auxiliary contact (“Ma”) has been placed in parallel with the “Start” pushbutton.
Review 4 1.
A starter with two sets of contacts would be called a ____________ -pole starter.
2.
____________ will automatically disconnect power from the motor when incoming power drops or is interrupted.
3.
The Furnas Class 11 motor starter protects motors up to ____________ HP at 460 VAC and ____________ HP at 230 VAC.
4.
The 3RV102 motor starter protector protects motors up to ____________ HP at 460 VAC.
5.
When a contactor is combined with an overload relay, it is called a ____________ ____________ .
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Starter Ratings
Starter contactors are rated according to size and type of load they handle. The National Electrical Manufacturers Association (NEMA) and the International Electrotechnical Commission (IEC) are two organizations that rate contactors and motor starters. NEMA is primarily associated with equipment used in North America. IEC, on the other hand, is associated with equipment sold in many countries including the United States. International trade agreements, market globalization, and domestic and foreign competition have made it important for controls manufacturers to be increasingly aware of international standards. NEMA
NEMA ratings are maximum horsepower ratings, according to the National Electrical Manufacturers Association ICS2 standards. NEMA starters and contactors are selected according to their NEMA size. These sizes range from size 00 to size 9. NEMA Size 00 0 1 2 3 4 5 6 7 8 9
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Continuous HP 230 VAC HP 460 VAC Amp Rating 9 1 2 18 3 5 27 7 10 45 15 25 90 30 50 135 50 100 270 100 200 540 200 400 810 300 600 1215 450 900 2250 800 1600
NEMA motor-control devices have generally become known for their very rugged, heavy-duty construction. Because of their rugged design NEMA devices are physically larger than IEC devices. NEMA motor starters and contactors can be used in virtually any application at their stated rating, from simple “ON” and “OFF” applications to more-demanding applications that include plugging and jogging. To select a NEMA motor starter for a particular motor one need only know the horsepower and voltage of the motor. If there is considerable plugging and jogging duty involved, however, even a NEMA-rated device will require some derating. Motor Matched Sizes
Siemens also has what are called Motor Matched sizes available on some Siemens motor starters. The ratings for these devices fall in between the ratings of normal NEMA sizes. This allows the user to more closely match the motor control to the actual application. The following table shows Motor Matched sizes available. MM Size Continuous HP 230 VAC HP 460 VAC AMP Rating 1¾ 2½ 3½ 4½
IEC
40 60 115 210
10 20 40 75
15 31 75 150
Not all applications require a heavy-duty industrial starter. In applications where space is more limited and the duty cycle is not severe, IEC devices represent a cost-effective solution. IEC devices are rated for maximum operational current as specified by the International Electrotechnical Commission in publication IEC 158-1. IEC does not specify sizes. Utilization categories are used with IEC devices to define the typical duty cycle of an IEC device. AC-3 and AC-4 are the categories of most interest for general motor-starting applications. Utilization Category
IEC Category Description
AC1
Non-inductive or slightly inductive loads.
AC2
Starting of slip-ring motors
AC3
Starting of squirrel-cage motors and switching off only after the motor is up to speed. (Make LRA, Break FLA)
AC4
Starting of squirrel-cage motors with inching and plugging duty. Rapid Start/Stop. (Make and break LRA)
AC11
Auxiliary (control) circuits.
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Definite Purpose
Definite Purpose (DP) contactors have certain characteristics which must be taken into consideration. DP contactors were designed for specific applications where the operating conditions are clearly defined. These operating conditions include full load amps, locked rotor amps, noninductive amps (resisitive load), number of power poles, duty cycle, and the total number of expected operations. DP contactors are sized by the motor full-load amps (FLA) and locked rotor amps (LRA). FLA is the amount of current the motor draws at full speed, under full mechanical load, at rated voltage. LRA is the maximum current the motor will draw at the instant full-line voltage is applied to the motor. DP contactors are well suited for loads found in the following areas: • Heating, Ventilating, and Air Conditioning (HVAC) • Farm Equipment and Irrigation • Environmental Control Systems • Office Equipment • Pool and Spa Controls • Welding Equipment • Medical Equipment • Food-Service Equipment
Other Organizations
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There are several other organizations that have developed standards and tests for electrical equipment. Underwriters Laboratory (UL), for example, specifies a maximum horsepower rating for which a contactor can be used. The contactor is tested by Underwriters Laboratory using test procedure U.L. 508. All Siemens contactors are rated in accordance with at least one of the previous organizations’ test procedures. Some carry multiple ratings. For example, Furnas INNOVA starters meet or exceed NEMA and UL standards. Siemens SIRIUS starters meet or exceed NEMA, IEC, and UL standards.
Furnas INNOVA PLUS Starters
Furnas INNOVA PLUS™ starters are available in NEMA sizes 0 through 4. They are available up to 100 HP. Furnas INNOVA PLUS starters are available with Class 10 or 20 ambientcompensated bimetal overload relays.
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ESP100 Starters
The Furnas ESP100™ starters use the same contactor as the INNOVA PLUS™ starters. The ESP100 starters are supplied with a Class 10, 20, or 30 ESP100 solid-state overload relay. The ESP100 overload relay protects 3Ø motors with FLA of ¼ ampere through 135 amperes. From ¼ ampere to 10 amperes the overload has a 4:1 FLA range, i.e.; 2½ - 10 amperes. Above 10 amperes the range is 2:1. The ESP100 overload relay illustrated below, for example, is adjustable from 9 to 18 amperes.The ESP100 also protects the motor against phase loss. The ESP100 trips within three seconds of loss of one of the power-supply phases.
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SIRIUS Type 3R Starters
SIRIUS 3R is a complete modular, building-block system. The system includes a structured range of contactors and overload relays covering loads up to 95 amps in four frame sizes. These four frame sizes are referred to as S00 (12A), S0 (25A), S2 (50A), and S3 (95A). A feature of the SIRIUS product line is a narrow mounting width. An S3 contactor rated at 75 HP, for example, is only 70mm (2.75”). SIRIUS 3R contactors and overload relays can operate in ambient temperatures up to 140°F (60°C). This, along with the smaller size, allows more units to be packed into a panel without overheating the components.
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CAGE CLAMP™
Size S00 contactors and overload relays are available with CAGE CLAMP™ connections on power and control-circuit terminals. Size S0, S2, and S3 contactors and overload relays have CAGE CLAMP™ connections on the control-circuit terminals only.
Overload Relays
SIRIUS 3R overload relays provide Class 10 overcurrent protection for both AC and DC motors. Ambient-compensated bimetal strips prevent the overload relay from nuisance tripping when the panel temperature is higher than the ambient temperature of the motor. The design of the overload relay also includes a differential trip bar that causes the unit to trip faster in the event of a phase-loss condition. An optional remote-reset module (not shown) is available.
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World Series Type 3TF Starters
The World Series starters are supplied with a type 3TF contactor and overload relay. World Series starters are available in horsepower ratings from 100 to 500 HP at 460 VAC. Auxiliary contacts are provided for use in the control circuit. World Series type 3TF contactors are available with various enclosures. Additional auxiliary contacts can be added. Coil voltages for the electromagnetic coil range from 24 to 600 VAC. The overload relay is a Class 10 relay that uses a bimetal strip unit and heater element to detect overloads. Each phase monitors current. The unit has a full-load amps adjustment, test button, and reset button. The full-load amps adjustment corresponds to the range of the motor full-load ampere rating. The test button is to ensure the overload relay is functioning properly. The reset button is used to reset a trip. It can be either automatic or manual reset. There is also a trip indicator.
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Overload Relay Selection
The following chart is useful in selecting the correct contactor and overload-relay combination. The chart reflects the maximum horsepower rating using Underwriters Laboratory test procedure U.L. 508 and the appropriate overload relay. Contactor Max HP (at Overload 460 VAC) Relay 3TF50 100 3UA60 3TF51 100 3UA61 3TF52 125 3UA62 3TF53 150 3UA62 3TF54 200 3UA66 3TF55 250 3UA66 3TF56 300 3UA66 3TF57 400 3UA68 3TF68 500 3UA68
Review 5
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1.
____________ is an organization primarily associated with rating equipment used in North America and ____________ is associated with rating equipment used in many countries including the U.S.
2.
A NEMA Size ____________ starter is rated for 200 HP at 460 volts .
3.
IEC utilization category ____________ applications are described as the starting of squirrel-cage motors and switching off only after a motor is up to speed.
4.
Furnas INNOVA PLUS™ starters are available in NEMA sizes 0 through ____________ .
5.
The ESP100 trips within ____________ seconds of loss of one of the power-supply phases.
6.
The maximum load current of a size S2 SIRIUS 3R starter is ____________ amps.
7.
The correct overload relay for a 3TF54 contactor is ____________ .
Multi-Speed and Reversing Starters
Full-voltage AC magnetic multi-speed controllers are designed to control squirrel-cage induction motors for operation at two, three, or four different constant speeds, depending on motor construction. The speed of a constant-speed motor is a function of the supply frequency and the number of poles and is given in the following formula:
The speed in RPM is the synchronous speed or the speed of the rotating magnetic field in the motor stator. Actual rotor speed is always less due to slip. The design of the motor and the amount of load applied determine the percentage of slip. This value is not the same for all motors. A motor with four poles on a 60 hertz AC line has a synchronous speed of 1800 RPM. This means that, after allowing for slip, the motor is likely to run at 1650 to 1750 RPM when loaded.
An induction motor with two poles on a 60 hertz AC line, however, would run at twice that speed. When motors are required to run at different speeds, the motor’s torque or horsepower characteristics will change with a change in speed. The proper motor must be selected and correctly connected for the application. In these applications, there are three categories. Constant Torque (CT) Variable Torque (VT) Constant Horsepower (CHP)
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Separate-Winding
There are two basic methods of providing multi-speed control using magnetic starters: separate-winding motors and consequent-pole motors. Separate-winding motors have a separate winding for each speed. The speed of each winding depends on the number of poles. The low-speed winding is wound for more poles than the high-speed winding.The motor cost is higher than consequent pole, but the control is simpler. There are many ways multi-speed motors can be connected depending on speed, torque, and horsepower requirements. The following schematic shows one possible connection of a twospeed, two-winding, wye-connected motor.
Consequent-Pole Motors
Consequent-pole motors have a single winding for two speeds. Taps can be brought from the winding for reconnection for a different number of poles. Two-speed, consequent-pole motors have one reconnectable winding. Low speed of a twospeed, consequent-pole motor is one half the speed of high speed. Three-speed motors have one reconnectable winding and one fixed winding. Four-speed motors have two reconnectable windings.
Speed Selection
There are three control schemes of speed selection for multispeed motors: selective control, compelling control, and progressive control. Selective control permits motor starting at any speed and to move to a higher speed the operator depresses the desired speed pushbutton. Compelling control requires the motor to be started at the lowest speed, then the operator must manually increment through each speed step to the desired speed. With progressive control the motor is started at the lowest speed and automatically increments to the selected speed.
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Reversing
Many applications require a motor to run in both directions. In order to change the direction of motor rotation, the direction of current flow through the windings must be changed. This is done on a three-phase motor by reversing any two of the three motor leads. Traditionally T1 and T3 are reversed. The following illustration shows a three-phase reversing motor circuit. It has one set of forward (F) contacts controlled by the “F” contactor, and one set of reverse (R) contacts controlled by the “R” contactor.
When the “F” contacts are closed, current flows through the motor causing it to turn in a clockwise direction.
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When the “R” contacts are closed, current flows through the motor in the opposite direction causing it to rotate in a counterclockwise direction. Mechanical interlocks prevent both forward and reverse circuits from being energized at the same time.
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Reduced-Voltage Starting
Full-Voltage Starting
The most common type of motor starting is full-voltage starting. The motor is placed directly across the line with this method.
With this type of starter the motor receives the full-line voltage. When a motor is started with full voltage, starting current can be as high as 600% of full-load current on standard squirrel cage motors. It can be even higher on high efficiency motors. There are situations where this method of starting is not acceptable. On large motors the high starting current is reflected back into the power lines of the electric utility, causing lights to flicker and in more serious situations can cause computers to malfunction. Many power companies in the U.S. require reduced-voltage starting on large-horsepower motors.
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Another potential problem with applying full-voltage starts is the high torque developed when power is first applied to the motor. This can be as high as 175% to 200% of full-load torque on a standard NEMA B type motor. Many applications require the starting torque to be applied gradually. A conveyor belt, for example, requires the starting torque to be applied gradually to prevent belt slipping or bunching.
Reduced-Voltage Starting
In general, starting methods which deviate from full-voltage starting by providing a lower starting voltage are referred to as reduced-voltage starting. Reduced-voltage starting should be used when it is necessary to limit the initial inrush of current or it is desired to reduce the starting torque of a motor. Reduced-voltage starting reduces the starting voltage of an induction motor with the purpose of confining the rate of change of the starting current to predetermined limits. It is important to remember that when the voltage is reduced to start a motor, current is also reduced, which also reduces the amount of starting torque a motor can deliver. Several methods are available for reduced-voltage starting. The application or the type of motor generally dictates the method to use. A few of the methods offered by Siemens are described in the following paragraphs.
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Autotransformer ReducedVoltage Starters
Autotransformer reduced-voltage starters provide the highest starting torque per ampere of line current and is one of the most effective means of starting a motor for an application in which starting current must be reduced with a minimum sacrifice of starting torque. Autotransformers have adjustable taps to reduce starting voltage to 50%, 65%, or 80% of full-line voltage. Applications: Blowers, Pumps, Compressors
Part-Winding Starters
Part-winding, reduced-voltage starters are used on motors with two separate parallel windings on the stator. The windings used during start draw about 65 - 80% of rated locked rotor current. During run each winding carries approximately 50% of the load current. Part-winding, reduced-voltage starters are the leastexpensive type of reduced-voltage starters and use a very simplified control circuit. However, they require special motor design and are not suitable for high inertia loads. There is no adjustment of current or torque. Applications: Pumps, Fans, Refrigeration, Compressors
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Wye-Delta Starters
Wye-delta, reduced-voltage starters are applicable only with motors having stator windings not connected internally and all six motor leads available. Connected in a wye configuration, the motor starts with reduced starting line current. The motor is reconfigured to a delta connection for run. This type of starter is a good method for applications requiring frequent starts. The starting torque is lower compared to other methods of reduced voltage starters. Applications: Central Air Conditioning Equipment, Compressors
Primary Resistance Starter
This is a simple and effective starting method. The motor is initially energized through a resistor in each of the three incoming lines. Part of the voltage is dropped through the resistors. The motor receives 70% to 80% of the full-line voltage. As the motor picks up speed, the motor sees more of the line voltage. At a preset time a time-delay relay closes a separate set of contacts, shorting out the resistors and applying full voltage to the motor. This type of reduced voltage starting is limited by the amount of heat the resistors can dissipate. Applications: Conveyors, Belt-Driven and Gear Drive Equipment
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Review 6 1.
____________ - ____________ ____________ is a method of providing multi-speed control that utilize taps brought out from a reconnectable winding.
2.
With ____________ ____________ the motor is started at the lowest speed and automatically increments to the selected speed.
3.
In general, starting methods which deviate from fullvoltage starting by providing a lower starting voltage are referred to as ___________ ___________ ___________ .
4.
____________ reduced-voltage starters have adjustable taps to reduce starting voltage to 50%, 65%, or 80% of full line voltage.
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Solid-State Reduced-Voltage Controllers
Solid-state or soft-start controllers also use reduced voltage starting. These controllers are more advanced and allow greater control of the starting, running, and stopping of an AC motor than the electromechanical starters discussed in the previous section. Reduced-voltage electromechanical starters start a motor in steps by first applying a reduced voltage followed by full voltage. Solid-state reduced-voltage controllers, however, can apply voltage gradually from some low intial volatge to 100% voltage. The following graph compares a solid-state reduced-voltage controller to a full- voltage (across-the-line) starter. By applying voltage gradually, the motor experiences reduced inrush current and speed is accelerated smoothly. In addition, just enough torque can be applied to start and accelerate the motor. This is beneficial for loads that have problems with the initial jerk and rapid acceleration of across-the-line starting.
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SIRIUS 3R Soft-Start Controls
SIRIUS 3R controllers provide gradual voltage starting and stopping.SIRIUS 3R soft-start controls are compact and compliment the rest of the SIRIUS line. The compact design allows the controls to be DIN rail mounted and integrated with any combination of other controls, such as, overloads, contactors, and motor starter protectors. SIRIUS 3R soft-start controls are available for motors up 75HP at 575 volts. A cost saving advantage of the SIRIUS 3R controller is the ability of one model to handle voltages from 200 to 460 volts.
SIKOSTART
SIKOSTART controllers are used in applications of up to 1000 horsepower. Like the SIRIUS 3R, SIKOSTART provides gradual voltage starting and stopping.
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SIKOSTART Wired Inline
SIKOSTART can easily be wired conventionally in line with the motor windings. In this configuration the controller sees full motor current.
SIKOSTART Wired Inside the Delta
On motors that have all leads available, the SIKOSTART controller can also be wired inside the delta connection of the motor. This offers a significant cost advantage. Current inside the delta of an AC motor is approximately 57% of nominal motor current. With this configuration a smaller SIKOSTART controller can be selected.
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Pilot Devices
A pilot device directs the operation of another device (pushbuttons and selector switches) or indicates the status of the operating system (pilot lights). This section discusses Siemens 3SB and Furnas Class 51/52 pushbuttons, selector switches, and pilot lights. 3SB devices are available in 22 mm diameters. Class 51/52 pilot devices are available in 30 mm diameter. The diameter refers to the size of the knockout hole required to mount the devices. Class 51 devices are rated for hazardous locations environments such as Class I, Groups C and D and Class II, Groups E, F, and G. Class 52 devices are heavy duty for harsh, industrial environments. Bifurcated Contacts
Whether one chooses the 3SB or the Class 51/52 pilot devices, the fine silver contacts have a 10 A/600 V continuous-current rating and can be used on solid state equipment. The 3SB and the Class 51/52 devices use bifurcated movable contacts. The design of the bifurcated contacts provides four different pathways for current to flow, thus improving contact reliability.
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Pushbuttons
A pushbutton is a control device used to manually open and close a set of contacts. Pushbuttons are available in a flush mount, extended mount, with a mushroom head, illuminated or nonilluminated. Pushbuttons come with either normally open, normally closed, or combination contact blocks. The Siemens 22 mm pushbuttons can handle up to a maximum of 6 circuits. The Furnas 30 mm pushbutton can handle up to a maximum fo 16 circuits.
Normally Open Pushbuttons
Pushbuttons are used in control circuits to perform various functions. For example, pushbuttons can be used when starting and stopping a motor. A typical pushbutton uses an operating plunger, a return spring, and one set of contacts. The following drawing illustrates a normally open (NO) pushbutton. Normally the contacts are open and no current flows through them. Depressing the button causes the contacts to close. When the button is released, the spring returns the plunger to the open position.
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Normally Closed Pushbuttons
Normally closed (NC) pushbuttons, such as the one shown below, are also used to open and close a circuit. In the pushbutton’s normal position the contacts are closed to allow current flow through the control circuit. Depressing the button opens the contacts preventing current flow through the circuit. These types of pushbuttons are momentary contact pushbuttons because the contacts remain in their activated position only as long as the plunger is held depressed.
Pushbuttons are available with variations of the contact configuration. For example, a pushbutton may have one set of normally open and one set of normally closed contacts so that when the button is depressed, one set of contacts is open and the other set is closed. By connecting to the proper set of contacts, either a normally open or normally closed situation exists. Using Pushbuttons in a Control Circuit
The following line diagram shows an example of how a normally open and a normally closed pushbutton might be used in a control circuit.
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Momentarily depressing the “Start” pushbutton completes the path of current flow and energizes the “M” contactor’s electromagnetic coil.
Holding Circuit Three-Wire Control
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This closes the associated normally open “M” and “Ma” contacts. When the “Start” pushbutton is released a holding circuit exists to the “M” electromagnetic coil through the auxiliary contacts “Ma” . The motor will run until the normally closed “Stop” pushbutton is depressed, breaking the path of current flow to the “M” electromagnetic coil and opening the associated “M” and “Ma” contacts. This is referred to as threewire control because there are three wires or three connection points required to connect the “Start” and “Stop” pushbuttons and the holding circuit (“Ma”). An advantage to three-wire control is low-voltage protection. If an overload causes the “OL” contacts in the control circuit to open, the “M” coil is deenergized and the motor shut down. When the overload is cleared, the motor will not suddenly restart on its own. An operator must depress the “Start” button to restart the motor.
Two-Wire Control
In comparison, a two-wire control has only two connection points for the “Start/Stop” circuit. When the contacts of the control device close, they complete the coil circuit of the contactor, causing it to be energized and connect the load to the line through the power contacts. When the contacts of the control device open, the power is removed from the motor and it stops. A two-wire control circuit provides low-voltage release but not low-voltage protection. This means that in the event of a power loss the contactor will deenergize, stopping the motor. This is low-voltage release. However, when power is restored, the motor will restart without warning if the control device is still closed. This type of control scheme is used for remote or inaccessible installations such as water-treatment plants or pumping stations. In these applications it is desireable to have an immediate return to service when power is restored.
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Selector Switches
Selector switches are also used to manually open and close contacts. Selector switches can be maintained, spring return or key operated. Selector switches are available in two-, three-, and four-position types. The basic difference between a push button and a selector switch is the operator mechanism. With a selector switch the operator is rotated to open and close contacts. Contact blocks used on pushbuttons are interchangeable with those on used on selector switches. Selector switches are used to select one of several circuit possibilities such as manual or automatic operation, low or high speed, up or down, right or left, and stop or run. The Siemens 22 mm selector switches can handle up to a maximum of 6 circuits. The Furnas 30 mm selector switch can handle up to a maximum of 16 circuits.
Two Position Selector Switch
In the following example PL1 is connected to the power source when the switch is in position 1. PL2 is connected to the power source when the switch is in position 2. In this circuit either PL1 or PL2 would be on at all times. If there were only one load, then the selector switch could be used as an On/Off switch.
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Contact Truth Tables
There are two accepted methods of indicating contact position of a selector switch in a circuit. The first method uses solid and dashed lines to denote contact position as shown in the previous example. In the second method truth tables, also known as target tables, are used. Each contact is marked with a letter. An “X” in the truth table indicates which contacts are closed for a given switch position. In this example contact A is closed, connecting PL1 to the power source, when the switch is in position 1. Contact B is closed, connecting PL2 to the power source, when the switch is in position 2.
Three-Position
A three-position selector switch can be used to select either of two sets of contacts or to disconnect both sets of contacts. Hand/Off/Auto is a typical application for a three-position selector switch used for controlling a pump. In the Hand (manual) position the pump will start when the Start pushbutton is pressed. The pump can be stopped by switching the switch to the Off position.The liquid level switch has no effect in either the Hand or Off position. When the selector switch is set to Auto, the pump will be controlled by the liquid-level switch. At a predetermined level the liquid level switch closes, starting the pump. At a predetermined level the liquid level switch opens, stopping the pump.
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Pilot Lights
Pilot lights provide visual information at a glance of the circuit’s operating condition. Pilot lights are normally used for “ON/OFF” indication, caution, changing conditions, and alarm signaling.
Pilot lights come with a color lens, such as red, green, amber, blue, white, or clear. A red pilot light normally indicates that a system is running. A green pilot light normally indicates that the system is off or deenergized. For example, a red pilot light located on a control panel would give visual indication that a motor was running. A green pilot light would give visual indication that a motor was stopped.
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Using a Pilot Light in a Control Circuit
In the following line diagram, a red pilot light is connected in parallel with the “M” electromagnetic coil.
When the coil is energized, the light will illuminate to indicate the motor is running. In the event the pilot light burns out the motor will continue to run.
In the following line diagram a green pilot light is connected through a normally closed “M” auxiliary contact (Mb). When the coil is deenergized, the pilot light is on to indicate the motor is not running.
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Depressing the “Start” pushbutton and energizing the “M” contactor opens the normally closed “Mb” contacts, turning the light off.
Signalling Columns
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Signalling columns can be mounted locally on individual machines, making it possible for the operating personnel to monitor production stations from a distance. Individual modules, or elements, are connected together. Various visual elements are available, including strobe lights, steady or flashing lights, and incandescent or LED lights. Lenses for the light elements are available in red, yellow, green, blue, and clear. Audible elements for the 8WD43 include a siren and a buzzer. Audible elements for the 8WD42 include a buzzer. In addition, a communication element is available allowing the signalling column to communicate with PLCs or computers through the Actuator Sensor Interface (ASI) network. Up to 10 elements can be used on a signalling column.
Review 7 1.
A ____________ ____________ directs the operation of another device.
2.
Which of the following circuits represents a two-wire control and which represents a three-wire control?
3.
Pilot lights provide ____________ information of the circuit’s operating condition.
4.
A ____________ pilot light normally indicates a motor is running and a ____________ pilot light normally indicates a motor is stopped.
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Control Transformers
It is often desirable to operate the control circuit at a lower voltage than the power circuit. Control transformers are used to step a voltage down to a lower level. Siemens control transformers are available in various primary and secondary voltages from 50 to 5000 VA.
In the following example, the power circuit is 460 VAC. A control transformer is used to step the voltage down to 24 VAC for use in the control circuit. The electromagnetic coil voltage must be rated for 24 VAC. Fuses on the primary and secondary windings of the transformer provide overcurrent protection.
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Control Relays
Relays are widely used in control circuits. They are used for switching multiple control circuits and for controlling light loads such as starting coils, pilot lights, and audible alarms.
Relay Operation
The operation of a control relay is similar to a contactor. In the following example a relay with a set of normally open (NO) contacts is used. When power is applied from the control circuit, an electromagnetic coil is energized. The resultant electromagnetic field pulls the armature and movable contacts toward the electromagnet closing the contacts. When power is removed, spring tension pushes the armature and movable contacts away from the electromagnet opening the contacts.
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Contact Arrangement
A relay can contain normally open, normally closed, or both types of contacts. The main difference between a control relay and a contactor is the size and number of contacts. The contacts in a control relay are relatively small because they need to handle only the small currents used in control circuits. There are no power contacts. Also, unlike a contactor, each contact in a control relay controls a different circuit. In a contactor, they all control the starting and stopping of the motor. Some relays have a greater number of contacts than are found in the typical contactor. The use of contacts in relays can be complex. There are three words which must be understood when dealing with relays.
Pole
Pole describes the number of isolated circuits that can pass through the relay at one time. A single-pole circuit can carry current through one circuit. A double-pole circuit can carry current through two circuits simultaneously. The two circuits are mechanically connected so that they open or close at the same time.
Throw
Throw is the number of different closed-contact positions per pole. This is the total number of different circuits each pole controls.
The following abbreviations are frequently used to indicate contact configurations: SPST SPDT DPST DPDT
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Single-Pole, Single-Throw Single-Pole, Double-Throw Double-Pole, Single-Throw Double-Pole, Double-Throw
Break
Break is the number of separate contacts the switch contacts use to open or close individual circuits. If the switch breaks the circuit in one place, it is a single-break. If the relay breaks the circuit in two places, it is a double-break.
The following illustrates various contact arrangements.
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Interposing a Relay
The following line diagram illustrates one way a control relay might be used in a circuit. A 24 VAC coil may not be strong enough to operate a large starter. In this example the electromagnetic coil of the “M” contactor selected is rated for 460 VAC. The electromagnetic coil of the control relay (CR) selected is 24 VAC. This is known as interposing a relay.
When the “Start” pushbutton in line 2 is momentarily depressed, power is supplied to the control relay (CR).
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The “CR” contacts in lines 1and 2 close. The “CR” contacts in line 2 maintain the start circuit. The “CR” contacts in line 1 complete the path of current to the “M” motor starter. The “M” motor starter energizes and closes the “M” contacts in the power circuit, starting the motor. Depressing the “Stop” pushbutton deenergizes the “CR” relay and “M” motor starter.
SIRIUS 3RH11 Control Relays
Siemens has a complete line of industrial-control relays. SIRIUS 3RH11 relays are available with screw terminal or Cage Clamp. The screw terminal version is shown in the following illustration. Four contacts are available in the basic device. Four additional contacts in the form of a snap-on device can be added to the front of the relay. Some SIRIUS 3RH11 relays are specifically designed to interface directly with PLCs and other solid-state control devices. SIRIUS 3RH11 relays are rated for switching both AC and DC circuits. Coil voltages range from 12 VDC to 230 VDC and 24 VAC to 600 VAC.
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General-Purpose Relays (Plug-In Relays)
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Siemens also manufactures a variety of general-purpose relays for socket and flange mounting. Coil voltages are available in 24 VAC, 120 VAC or 24 VDC. The biggest benefit of this type of relay is all the wiring stays in place with the socket if the relay needs to replaced with a new one.
Timing Relays
Timing relays, such as the SIRIUS 3RP timing relays, are used in control switching operations involving time delay. SIRIUS 3RP1 timing relays have timing ranges available from .05 seconds to 10 hours. SIRIUS 3RP15 have timing ranges from .05 seconds to 100 hours.
Time Delay
A timing relay has two major functions: On-delay and Off-delay timing. An arrow is used to denote the function of the timer. An arrow pointing up indicates an On-delay timing action. An arrow pointing down indicates an Off-delay timing action.
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On-delay and Off-delay timers can turn their connected loads on or off, depending on how the timer’s output is wired into the circuit. On-delay indicates that once a timer has received a signal to turn on, a predetermined time (set by the timer) must pass before the timer’s contacts change state. Off-delay indicates that once a timer has received a signal to turn off, a predetermined time (set by the timer) must pass before the timer’s contacts change state. On-Delay, Time Closed
The following is an example of On-delay, timed closed. For this example a set of normally open (NO) contacts is used. This is also referred to as normally open timed closed (NOTC). The timing relay (TR1) has been set for an On-delay of 5 seconds.
When S1 is closed, TR1 begins timing. When 5 seconds has elapsed, TR1 will close its associated normally open (NO) TR1 contacts, illuminating pilot light PL1. When S1 is open, deenergizing TR1, the TR1 contacts open immediately, extinguishing PL1.
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On-Delay, Timed Open
The following is an example of On-delay, timed open. For this example a set of normally closed (NC) contacts is used. This is also referred to as normally closed, timed open (NCTO). PL1 is illuminated as long as S1 remains open. The timing relay (TR1) has been set for an ON delay of 5 seconds.
When S1 is closed, timing relay TR1 is energized. After a timed delay of 5 seconds, the associated normally closed TR1 contacts open, extinguishing PL1. When S1 is open, deenergizing TR1, the TR1 contacts close immediately, illuminating PL1.
Off-Delay, Timed Open
The following is an example of Off-delay, timed open. For this example a set of normally open contacts (NO) is used. This is also referred to as normally open, timed open (NOTO). The timing relay (TR1) has been set for an off delay of 5 seconds. Closing S1 energizes TR1 causing its associated normally open TR1 contacts to close immediately, illuminating PL1.
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When S1 is opened, TR1 begins timing. When 5 seconds has elapsed, TR1 will open its associated normally open contacts, extinguishing pilot light PL1.
Off-Delay, Timed Closed
The following is an example of Off-delay, timed closed. For this example a set of normally closed (NC) contacts is used. This is also referred to as normally closed, timed closed (NCTC). The timing relay (TR1) has been set for 5 seconds. PL1 is on. Closing S1 energizes TR1 causing its associated contacts to open immediately, extinguishing PL1.
When S1 is opened, timing relay TR1 is deenergized. After 5 seconds, the associated normally closed contacts close, illuminating PL1.
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Instantaneous Contacts
Timing relays can also have normally open or normally closed instantaneous contacts. In the following example, when switch S1 is closed, the TR1 instantaneous contacts will close immediately, illuminating PL1. After a preset time delay the TR1 timing contacts will close, illuminating PL2.
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Pressure Switches
Pressure switches are control devices that respond to changes in pressure of liquid or air. The liquid or air is referred to as fluid pressure. They open or close electrical contacts in response to pressure changes by either turning on or off a motor, opening or closing louvers, or signaling a warning light or horn. For loads up to 5 HP the pressure switches may handle the current directly. For larger loads the pressure switch is used to energize relays, contactors, or motor starters, which then energize the load.
Pressure Switch Components
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The basic components of a pressure switch are illustrated below. Electrical contacts are operated by the movement of a diaphragm against the force of a spring. The contacts may be normally open (NO) or normally closed (NC). The spring setting determines how much fluid pressure is required to operate the contacts.
Application
Pressure switches are frequently used to maintain a specified pressure range in a storage tank. Storage tanks can be used to hold a liquid, such as water, or a gas, such as air.
Operation
In this example a normally closed pressure switch is used. The pump starts as soon as power is applied to the circuit. When the pressure in the storage tank has reached a predetermined level, the contacts in the pressure switch open, removing power from the pump motor. As the contents of the storage tank are used, the pressure in the tank decreases. At a predetermined level the pressure switch will close its contacts, applying power to the pump motor.
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Pressure Range
Pressure switches are designed to operate within a specified pressure range, usually given in pounds per square inch (PSI). In the following example, a Furnas Class 69ES water pressure switch operates within a range of 10 to 80 PSI. The minimum close, or cut-in pressure, is 10 PSI. This is the point at which fluid pressure on the diaphragm causes the contacts to close. The maximum open, or cut-out pressure, is 80 PSI. This is the point at which fluid pressure on the diaphragm causes the contacts to open. Pressure differential is the difference between these two settings. The Furnas Class 69ES pressure switch can have a differential range of 15-25 PSI. In this example the cut-in pressure has been set to 30 PSI. The cut-out pressure has been set to 50 PSI. The pressure differential is 20 PSI. The pressure switch will regulate the pressure between 30 and 50 PSI.
Reverse Action
Reverse action pressure switches cut-in on a rising pressure. They are designed to ground the ignition on gas engine driven pumps and compressors when the maximum pressure has been reached. In the following example a Furnas Class 69WR5 reverse action pressure switch has been selected. The 69WR5 has a minimum open (cut-out) of 10 PSI and a maximum close (cut-in) of 80 PSI. The differential is set so that the switch opens at 30 PSI and closes at 50 PSI.
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LOGO! Logic Module
LOGO! is a logic module used to perform control tasks. The module is compact and user friendly, providing a cost-effective solution for the end user.
Hard-Wired Control
In the past, many of these control tasks were solved with contactor or relay controls. This is often referred to as hard-wired control. Circuit diagrams had to be designed and electrical components specified and installed. A change in control function or system expansion could require extensive component changes and rewiring.
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Many of the same tasks can be performed with LOGO!. Initial hard-wiring, though still required, is greatly simplified. Modifying the application is as easy as changing the program via the keypad located on the front of the LOGO!. Likewise, control programs can be created and tested before implementation via a PC software program. Once the program is performing per specification, the transfer to LOGO! is as simple as plugging in a cable. Basic LOGO! Operation
LOGO! accepts a variety of digital inputs, such as pushbuttons, switches, and contacts. LOGO! makes decisions and executes control instructions based on the user-defined program. The instructions control various outputs. The outputs can be connected to virtually any type of load such as relays, contactors, lights, and small motors.
Design Features
LOGO! is available in many different versions for different supply voltages (12 VDC, 24 VDC, 24 VAC or 115/230 VAC). All models have: • • • • • • • • • •
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Relay outputs with maximum 10 amp output current (not LOGO! 24/24L models) Integrated clock (not LOGO! 24/24L models) Integrated display Integrated keypad Integrated basic and special functions Integrated EEPROM for storing programs and setpoints Optional program module Yellow Module for simple program dupication Red Module for program backup and protection Basic AND, OR, NOT, NAND, NOR, and XOR functions AND and NAND functions with positive and negative edge detection Special ON delay, latching ON delay, OFF delay, pulse relay, latching relay, clock pulse generator, and counter (up/ down) functions (total of 21 special functions)
Basic version features: • •
Six digital inputs, four digital outputs for AC models Eight digital inputs, four digital outputs for DC models with two inputs capable of accepting analog inputs
Pure version features: •
LOGO! Basic without display
L model features: • •
Twelve digital inputs, eight digital outputs Four additional inputs and outputs on the AS-i modules
The maximum possible options for every model version are as follows: • • • • • • • •
16 Timers 24 Counters Eight Time Switches Three Operating Hour Counters 42 Current Impulse Relays 42 Latching Relays Four Markers for Program Continuation 56 Total Function Blocks
Review 8 1.
___________ is the total number of different circuits each pole controls.
2.
___________ describes the number of isolated circuits that can pass through a relay at one time.
3.
An SPDT relay has ___________ pole(s) and ____________ closed contact position(s).
4.
A timing relay that has received a signal to turn on, and then delays a predetermined amount of time before an action takes place is referred to as ____________ delay.
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Review Answers
Review 1
1) manually; 2) a; 3) b; 4) b; 5) c.
Review 2
1) left to right; 2) A - Node, B - Power Circuit, C - Power Load, D- Control Circuit; E - Control Device; F - Control Load.
Review 3
1) a; 2) excess; 3) overload; 4) a; 5) bimetal.
Review 4
1) 2; 2) LVP; 3) 10, 5; 4) 20; 5) motor starter.
Review 5
1) NEMA, IEC; 2) 5; 3) AC3; 4) 4; 5) 3; 6) 50; 7) 3UA66.
Review 6
1) Consequent-pole motor; 2) progressive control; 3) reducedvoltage starting; 4) Autotransformer.
Review 7
1) pilot device; 2) Three-Wire Control, Two-Wire Control; 3) visual; 4) red, green.
Review 8
1) Throw; 2) Pole; 3) one, two; 4) ON.
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Final Exam
The final exam is intended to be a learning tool. The book may be used during the exam. A tear-out answer sheet is provided. After completing the test, mail in the answer sheet for grading. A grade of 70% or better is passing. Upon successful completion of the test a certificate will be issued. Questions
1.
The standard method of showing a contact is by indicating the circuit condition it produces when the actuating device is in the ____________ state. a. c.
2.
4.
normally open deenergized
green amber
b. d.
red white
Which of the following symbols represents a normally closed, timed open (NCTO) contact? a.
b.
c.
d.
With an increase of current, temperature will ____________ . a. c.
5.
b. d.
A motor that is running would usually be indicated by a ____________ pilot light. a. c.
3.
normally closed energized
decrease remain the same
b. d.
increase increase and decrease
The two circuits involved in the operation of a contactor are the ____________ circuits. a. b. c. d.
power and control power and armature control and electromagnetic control and starter 97
6.
A motor starter is a combination of a/an ____________ . a. b. c. d.
7.
Which of the following is not part of a contactor? a. b. c. d.
8.
b. d.
compelling consequent pole
NEMA ICS
b. d.
UL IEC
The proper overload relay for a World Series 3TF50 contactor is ____________ . a. c.
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selective progressive
The organization primarily concerned with the rating of contactors and starters used in many countries, including the U.S. is ____________ . a. c.
11.
apply torque gradually increase starting torque get motor to full speed faster run the motor at a lower speed
A type of speed selection control that requires the operator to manually increment through each speed step to get to the desired speed is ____________ control. a. c.
10.
armature electromagnetic coil overcurrent sensing device stationary contacts
One reason reduced-voltage starting may be used to start a motor is to ____________ . a. b. c. d.
9.
electromagnet and armature contactor and electromagnet contactor and overload relay overload relay and instantaneous contacts
3UA50 3UA58
b. d.
3UA54 3UA60
12.
A device used to provide visual information of the circuit’s operating condition is a ____________ . a. c.
13.
DPDT SPDT
cut-out pressure pressure range
b. d.
cut-in pressure pressure differential
200 810
b. d.
540 1600
siren buzzer
b. d.
buzzer and siren horn
Furnas INNOVA PLUS™ starters are available up to ____________ HP. a. c.
18.
b. d.
Audible elements for the 8WD42 siganlling column include a ____________ . a. c.
17.
DPST SPST
A NEMA Size 6 starter has a continuous amp rating of ____________ amps. a. c.
16.
selector switch pilot light
The point at which fluid pressure on the diaphragm of a pressure switch causes the contacts to open is referred to as ____________ . a. c.
15.
b. d.
A relay that has two isolated circuits and one closed contact position per pole is a ____________ . a. c.
14.
pushbutton proximity switch
25 100
b. d.
50 250
SIRIUS Type 3R motor starters are available for loads up to ___________ amps. a. c.
95 200
b. d.
135 270
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19.
In the following diagram, the motor will stop when ____________ . a. b. c. d.
20.
The __________ overload relay integrates with PROFIBUS-DP. a. c.
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the “Stop” button is depressed limit switch “LS1” opens the motor overload contact opens all of the above
3RB10 3UF5
b. d.
ESP 100 3RU11
Table of Contents
Introduction ............................................................................. 2 Totally Integrated Automation and DC Drives ......................... 4 Mechanical Basics ................................................................... 6 DC Motors ............................................................................. 13 Basic DC Motor Operation .................................................... 17 Types of DC Motors .............................................................. 22 DC Motor Ratings .................................................................. 26 Speed/Torque Relationships of Shunt Connected Motors .... 30 Basic DC Drives ..................................................................... 35 Converting AC to DC ............................................................. 39 Basic Drive Operation ............................................................ 44 SIMOREG 6RA70 DC MASTER Electronics .......................... 54 Parameters and Function Blocks ........................................... 69 Engineering Tools................................................................... 75 Applications ........................................................................... 77 Application Examples ............................................................ 78 Selecting a Siemens DC Drive .............................................. 81 Review Answers .................................................................... 85 Final Exam ............................................................................. 86
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Introduction
Welcome to another course in the STEP 2000 series, Siemens Technical Education Program, designed to prepare our distributors to sell Siemens Energy & Automation products more effectively. This course covers Basics of DC Drives and related products. Upon completion of Basics of DC Drives you will be able to: Explain the concepts of force, inertia, speed, and torque
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•
Explain the difference between work and power
•
Describe the operation of a DC motor
•
Identify types of DC motors by their windings
•
Identify nameplate information on a DC motor necessary for application to a DC drive
•
Identify the differences between a power module and a base drive
•
Explain the process of converting AC to DC using thyristors
•
Describe the basic construction of a DC drive
•
Explain the significant differences between 1- and 4quadrant operation in a DC drive
•
Describe features and operation of the Siemens 6RA70 DC MASTER
•
Describe the characteristics of constant torque, constant horsepower, and variable torque applications
This knowledge will help you better understand customer applications. In addition, you will be better able to describe products to customers and determine important differences between products. If you are an employee of a Siemens Energy & Automation authorized distributor, fill out the final exam tear-out card and mail in the card. We will mail you a certificate of completion if you score a passing grade. Good luck with your efforts. SIMOREG, SIMOREG DC-MASTER, SIMOVIS, and SIMOLINK are registered trademarks of Siemens Energy & Automation, Inc. Other trademarks are the property of their respective owners.
3
Totally Integrated Automation and DC Drives
Totally Integrated Automation
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Totally Integrated Automation (TIA) is a strategy developed by Siemens that emphasizes the seamless integration of automation products. The TIA strategy incorporates a wide variety of automation products such as programmable controllers, computer numerical controls, Human Machine Interfaces (HMI), and DC drives which are easily connected via open protocol networks. An important aspect of TIA is the ability of devices to communicate with each other over various network protocols such as PROFIBUS-DP.
Siemens DC Drives
SIMOREG® is the trade name for Siemens adjustable speed DC Drives. SIMOREG stands for SIemens MOtor REGulator. Siemens DC drives are an important element of the TIA strategy. DC motors were the first practical device to convert electrical energy into mechanical energy. DC motors, coupled with DC drives such as the Siemens SIMOREG 6RA70, have been widely used in industrial drive applications for years, offering very precise control.
Although AC motors and vector-control drives now offer alternatives to DC, there are many applications where DC drives offer advantages in operator friendliness, reliability, cost effectiveness, and performance. We will discuss applications later in the course.
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Mechanical Basics
Before discussing Siemens DC drives it is necessary to understand some of the basic terminology associated with the mechanics of DC drive operation. Many of these terms are familiar to us in some other context. Later in the course we will see how these terms apply to DC drives. Force
In simple terms, a force is a push or a pull. Force may be caused by electromagnetism, gravity, or a combination of physical means. The English unit of measurement for force is pounds (lb).
Net Force
Net force is the vector sum of all forces that act on an object, including friction and gravity. When forces are applied in the same direction they are added. For example, if two 10 lb forces were applied in the same direction the net force would be 20 lb.
If 10 lb of force were applied in one direction and 5 lb of force applied in the opposite direction, the net force would be 5 lb and the object would move in the direction of the greater force.
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If 10 lb of force were applied equally in both directions, the net force would be zero and the object would not move.
Torque
Torque is a twisting or turning force that tends to cause an object to rotate. A force applied to the end of a lever, for example, causes a turning effect or torque at the pivot point. Torque (τ) is the product of force and radius (lever distance). Torque (τ) = Force x Radius In the English system torque is measured in pound-feet (lb-ft) or pound-inches (lb-in). If 10 lbs of force were applied to a lever 1 foot long, for example, there would be 10 lb-ft of torque.
An increase in force or radius would result in a corresponding increase in torque. Increasing the radius to 2 feet, for example, results in 20 lb-ft of torque.
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Speed
An object in motion travels a given distance in a given time. Speed is the ratio of the distance traveled to the time it takes to travel the distance.
Linear Speed
The linear speed of an object is a measure of how long it takes the object to get from point A to point B. Linear speed is usually given in a form such as feet per second (f/s). For example, if the distance between point A and point B were 10 feet, and it took 2 seconds to travel the distance, the speed would be 5 f/s.
Angular (Rotational) Speed
The angular speed of a rotating object is a measurement of how long it takes a given point on the object to make one complete revolution from its starting point. Angular speed is generally given in revolutions per minute (RPM). An object that makes ten complete revolutions in one minute, for example, has a speed of 10 RPM.
Acceleration
An object can change speed. An increase in speed is called acceleration. Acceleration occurs when there is a change in the force acting upon the object. An object can also change from a higher to a lower speed. This is known as deceleration (negative acceleration). A rotating object, for example, can accelerate from 10 RPM to 20 RPM, or decelerate from 20 RPM to 10 RPM.
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Law of Inertia
Mechanical systems are subject to the law of inertia. The law of inertia states that an object will tend to remain in its current state of rest or motion unless acted upon by an external force. This property of resistance to acceleration/deceleration is referred to as the moment of inertia. The English system of 2 measurement is pound-feet squared (lb-ft ). If we look at a continuous roll of paper, as it unwinds, we know that when the roll is stopped, it would take a certain amount of force to overcome the inertia of the roll to get it rolling. The force required to overcome this inertia can come from a source of energy such as a motor. Once rolling, the paper will continue unwinding until another force acts on it to bring it to a stop.
Friction
A large amount of force is applied to overcome the inertia of the system at rest to start it moving. Because friction removes energy from a mechanical system, a continual force must be applied to keep an object in motion. The law of inertia is still valid, however, since the force applied is needed only to compensate for the energy lost. Once the system is in motion, only the energy required to compensate for various losses need be applied to keep it in motion. In the previous illustration, for example: these losses include: • • •
Friction within motor and driven equipment bearings Windage losses in the motor and driven equipment Friction between material on winder and rollers
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Work
Whenever a force of any kind causes motion, work is accomplished. For example, work is accomplished when an object on a conveyor is moved from one point to another.
Work is defined by the product of the net force (F) applied and the distance (d) moved. If twice the force is applied, twice the work is done. If an object moves twice the distance, twice the work is done. W=Fxd
Power
Power is the rate of doing work, or work divided by time.
In other words, power is the amount of work it takes to move the package from one point to another point, divided by the time.
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Horsepower
Power can be expressed in foot-pounds per second, but is often expressed in horsepower (HP). This unit was defined in the 18th century by James Watt. Watt sold steam engines and was asked how many horses one steam engine would replace. He had horses walk around a wheel that would lift a weight. He found that each horse would average about 550 foot-pounds of work per second. One horsepower is equivalent to 500 foot-pounds per second or 33,000 foot-pounds per minute.
The following formula can be used to calculate horsepower when torque (lb-ft) and speed (RPM) are known. It can be seen from the formula that an increase of torque, speed, or both will cause a corresponding increase in horsepower.
Power in an electrical circuit is measured in watts (W) or kilowatts (kW). Variable speed drives and motors manufactured in the United States are generally rated in horsepower (HP); however, it is becoming common practice to rate equipment using the International System of Units (SI units) of watts and kilowatts.
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Review 1 1.
____________ is the trade name for Siemens motor generators (DC drives).
2.
If 20 lb of force where applied in one direction and 5 lb of force applied in the opposite direction, the net force would be ____________ lb.
3.
If 5 lb of force were applied to a radius of 3 feet, the torque would be ____________ lb-ft.
4.
Speed is determined by ___________ . a. dividing Time by Distance b. dividing Distance by Time c. multiplying Distance x Time d. subtracting Distance from Time
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5.
Work is accomplished whenever ____________ causes motion.
6.
The law of inertia states that an object will tend to remain in its current state of rest or motion unless acted upon by an ____________ ____________ .
DC Motors
DC motors have been used in industrial applications for years. Coupled with a DC drive, DC motors provide very precise control. DC motors can be used with conveyors, elevators, extruders, marine applications, material handling, paper, plastics, rubber, steel, and textile applications to name a few.
13
Construction
DC motors are made up of several major components which include the following: • Frame • Shaft • Bearings • Main Field Windings (Stator) • Armature (Rotor) • Commutator • Brush Assembly Of these components, it is important to understand the electrical characteristics of the main field windings, known as the stator, and the rotating windings, known as the armature. An understanding of these two components will help with the understanding of various functions of a DC Drive.
14
Basic Construction
The relationship of the electrical components of a DC motor is shown in the following illustration. Field windings are mounted on pole pieces to form electromagnets. In smaller DC motors the field may be a permanent magnet. However, in larger DC fields the field is typically an electromagnet. Field windings and pole pieces are bolted to the frame. The armature is inserted between the field windings. The armature is supported by bearings and end brackets (not shown). Carbon brushes are held against the commutator.
Armature
The armature rotates between the poles of the field windings. The armature is made up of a shaft, core, armature windings, and a commutator. The armature windings are usually form wound and then placed in slots in the core.
15
Brushes
16
Brushes ride on the side of the commutator to provide supply voltage to the motor. The DC motor is mechanically complex which can cause problems for them in certain adverse environments. Dirt on the commutator, for example, can inhibit supply voltage from reaching the armature. A certain amount of care is required when using DC motors in certain industrial applications. Corrosives can damage the commutator. In addition, the action of the carbon brush against the commutator causes sparks which may be problematic in hazardous environments.
Basic DC Motor Operation
Magnetic Fields
You will recall from the previous section that there are two electrical elements of a DC motor, the field windings and the armature. The armature windings are made up of current carrying conductors that terminate at a commutator. DC voltage is applied to the armature windings through carbon brushes which ride on the commutator. In small DC motors, permanent magnets can be used for the stator. However, in large motors used in industrial applications the stator is an electromagnet. When voltage is applied to stator windings an electromagnet with north and south poles is established. The resultant magnetic field is static (nonrotational). For simplicity of explanation, the stator will be represented by permanent magnets in the following illustrations.
17
Magnetic Fields
A DC motor rotates as a result of two magnetic fields interacting with each other. The first field is the main field that exists in the stator windings. The second field exists in the armature. Whenever current flows through a conductor a magnetic field is generated around the conductor.
Right-Hand Rule for Motors
A relationship, known as the right-hand rule for motors, exists between the main field, the field around a conductor, and the direction the conductor tends to move. If the thumb, index finger, and third finger are held at right angles to each other and placed as shown in the following illustration so that the index finger points in the direction of the main field flux and the third finger points in the direction of electron flow in the conductor, the thumb will indicate direction of conductor motion. As can be seen from the following illustration, conductors on the left side tend to be pushed up. Conductors on the right side tend to be pushed down. This results in a motor that is rotating in a clockwise direction. You will see later that the amount of force acting on the conductor to produce rotation is directly proportional to the field strength and the amount of current flowing in the conductor.
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CEMF
Whenever a conductor cuts through lines of flux a voltage is induced in the conductor. In a DC motor the armature conductors cut through the lines of flux of the main field. The voltage induced into the armature conductors is always in opposition to the applied DC voltage. Since the voltage induced into the conductor is in opposition to the applied voltage it is known as CEMF (counter electromotive force). CEMF reduces the applied armature voltage.
The amount of induced CEMF depends on many factors such as the number of turns in the coils, flux density, and the speed which the flux lines are cut. Armature Field
An armature, as we have learned, is made up of many coils and conductors. The magnetic fields of these conductors combine to form a resultant armature field with a north and south pole. The north pole of the armature is attracted to the south pole of the main field. The south pole of the armature is attracted to the north pole of the main field. This attraction exerts a continuous torque on the armature. Even though the armature is continuously moving, the resultant field appears to be fixed. This is due to commutation, which will be discussed next.
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Commutation
In the following illustration of a DC motor only one armature conductor is shown. Half of the conductor has been shaded black, the other half white. The conductor is connected to two segments of the commutator. In position 1 the black half of the conductor is in contact with the negative side of the DC applied voltage. Current flows away from the commutator on the black half of the conductor and returns to the positive side, flowing towards the commutator on the white half.
In position 2 the conductor has rotated 90°. At this position the conductor is lined up with the main field. This conductor is no longer cutting main field magnetic lines of flux; therefore, no voltage is being induced into the conductor. Only applied voltage is present. The conductor coil is short-circuited by the brush spanning the two adjacent commutator segments. This allows current to reverse as the black commutator segment makes contact with the positive side of the applied DC voltage and the white commutator segment makes contact with the negative side of the applied DC voltage.
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As the conductor continues to rotate from position 2 to position 3 current flows away from the commutator in the white half and toward the commutator in the black half. Current has reversed direction in the conductor. This is known as commutation.
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Types of DC Motors
The field of DC motors can be a permanent magnet, or electromagnets connected in series, shunt, or compound. Permanent Magnet Motors
The permanent magnet motor uses a magnet to supply field flux. Permanent magnet DC motors have excellent starting torque capability with good speed regulation. A disadvantage of permanent magnet DC motors is they are limited to the amount of load they can drive. These motors can be found on low horsepower applications. Another disadvantage is that torque is usually limited to 150% of rated torque to prevent demagnetization of the permanent magnets.
Series Motors
In a series DC motor the field is connected in series with the armature. The field is wound with a few turns of large wire because it must carry the full armature current. A characteristic of series motors is the motor develops a large amount of starting torque. However, speed varies widely between no load and full load. Series motors cannot be used where a constant speed is required under varying loads. Additionally, the speed of a series motor with no load increases to the point where the motor can become damaged. Some load must always be connected to a series-connected motor. Series-connected motors generally are not suitable for use on most variable speed drive applications.
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Shunt Motors
In a shunt motor the field is connected in parallel (shunt) with the armature windings. The shunt-connected motor offers good speed regulation. The field winding can be separately excited or connected to the same source as the armature. An advantage to a separately excited shunt field is the ability of a variable speed drive to provide independent control of the armature and field. The shunt-connected motor offers simplified control for reversing. This is especially beneficial in regenerative drives.
Compound Motors
Compound motors have a field connected in series with the armature and a separately excited shunt field. The series field provides better starting torque and the shunt field provides better speed regulation. However, the series field can cause control problems in variable speed drive applications and is generally not used in four quadrant drives.
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Speed/Torque Curves
The following chart compares speed/torque characteristics of DC motors. At the point of equilibrium, the torque produced by the motor is equal to the amount of torque required to turn the load at a constant speed. At lower speeds, such as might happen when load is added, motor torque is higher than load torque and the motor will accelerate back to the point of equilibrium. At speeds above the point of equilibrium, such as might happen when load is removed, the motor’s driving torque is less than required load torque and the motor will decelerate back to the point of equilibrium.
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Review 2 1.
The field in larger DC motors is typically an ____________ .
2.
Whenever ____________ flows through a conductor a magnetic field is generated around the conductor.
3.
Voltage induced into the conductors of an armature that is in opposition to the applied voltage is known as ____________ .
4.
Identify the following motor types.
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DC Motor Ratings
The nameplate of a DC motor provides important information necessary for correctly applying a DC motor with a DC drive. The following specifications are generally indicated on the nameplate: • • • • • • • • • •
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Manufacturer’s Type and Frame Designation Horsepower at Base Speed Maximum Ambient Temperature Insulation Class Base Speed at Rated Load Rated Armature Voltage Rated Field Voltage Armature Rated Load Current Winding Type (Shunt, Series, Compound, Permanent Magnet) Enclosure
HP
Horsepower is a unit of power, which is an indication of the rate at which work is done. The horsepower rating of a motor refers to the horsepower at base speed. It can be seen from the following formula that a decrease in speed (RPM) results in a proportional decrease in horsepower (HP).
Armature Speed, Volts, and Amps
Typically armature voltage in the U.S. is either 250 VDC or 500 VDC. The speed of an unloaded motor can generally be predicted for any armature voltage. For example, an unloaded motor might run at 1200 RPM at 500 volts. The same motor would run at approximately 600 RPM at 250 volts.
The base speed listed on a motor’s nameplate, however, is an indication of how fast the motor will turn with rated armature voltage and rated load (amps) at rated flux (Φ). The maximum speed of a motor may also be listed on the nameplate. This is an indication of the maximum mechanical speed a motor should be run in field weakening. If a maximum speed is not listed the vendor should be contacted prior to running a motor over the base speed.
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Winding
The type of field winding is also listed on the nameplate. Shunt winding is typically used on DC Drives.
Field Volts and Amps
Shunt fields are typically wound for 150 VDC or 300 VDC. Our sample motor has a winding that can be connected to either 150 VDC or 300 VDC.
Field Economizing
In many applications it may be necessary to apply voltage to the shunt field during periods when the motor is stationary and the armature circuit is not energized. Full shunt voltage applied to a stationary motor will generate excessive heat which will eventually burn up the shunt windings. Field economizing is a technique used by DC drives, such as the SIMOREG® 6RA70, to reduce the amount of applied field voltage to a lower level when the armature is de-energized (standby). Field voltage is reduced to approximately 10% of rated value. A benefit of field economizing over shuting the field off is the prevention of condensation.
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Insulation Class
The National Electrical Manufacturers Association (NEMA) has established insulation classes to meet motor temperature requirements found in different operating environments. The insulation classes are A, B, F, and H. Before a motor is started the windings are at the temperature of the surrounding air. This is known as ambient temperature. NEMA has standardized on an ambient temperature of 40°C (104°F) for all classes.
Temperature will rise in the motor as soon as it is started. The combination of ambient temperature and allowed temperature rise equals the maximum winding temperature in a motor. A motor with Class F (commonly used) insulation, for example, has a maximum temperature rise of 105°C. The maximum winding temperature is 145°C (40°C ambient + 105°C rise). A margin is allowed to provide for a point at the center of the motor’s windings where the temperature is higher. This is referred to as the motor’s hot spot.
The operating temperature of a motor is important to efficient operation and long life. Operating a motor above the limits of the insulation class reduces the motor’s life expectancy. A 10°C increase in the operating temperature can decrease the life expectancy of a motor by as much as 50%. In addition, excess heat increases brush wear. 29
Speed/Torque Relationships of Shunt Connected Motors
An understanding of certain relationships within a DC motor will help us understand the purposes of various functions in a DC drive later in the course. The formulas given in the following discussion apply to all three types of DC motors (series, shunt, and compound). However, because shunt connected motors are more commonly used with DC drives focus will be on shunt connected DC motors. DC Motor Equations
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In a DC drive voltage applied (Va) to the armature circuit is received from a variable DC source. Voltage applied to the field circuit (Vf) is from a separate source. The armature of all DC motors contains some amount of resistance (Ra). When voltage is applied (Va), current (Ia) flows through the armature. You will recall from earlier discussion that current flowing through the armature conductors generates a magnetic field. This field interacts with the shunt field (Φ) and rotation results.
Armature Voltage
The following armature voltage equation will be used to demonstrate various operating principles of a DC motor. Variations of this equation can be used to demonstrate how armature voltage, CEMF, torque, and motor speed interact with each other. Va = (KtΦn) + (IaRa) Where: Va = Applied Armature Voltage Kt = Motor Design Constants Φ = Shunt Field Flux n = Armature Speed Ia = Armature Current Ra = Armature Resistance
CEMF
We have already learned that rotation of the armature through the shunt field induces a voltage in the armature (Ea) that is in opposition to the armature voltage (Va). This is counter electromotive force (CEMF). CEMF is dependent on armature speed (n) and shunt field (Φ) strength. An increase in armature speed (n) or an increase of shunt field (Φ) strength will cause a corresponding increase in CEMF (Ea). Ea = KtΦn or Ea = Va - (IaRa)
Motor Speed
The relationship between VA and speed is linear as long as flux (Φ) remains constant. For example, speed will be 50% of base speed with 50% of VA applied.
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Motor Torque
The interaction of the shunt and armature field flux produces torque (M). An increase in armature current (Ia) increases armature flux, thereby increasing torque. An increase in field current (If) increases shunt field flux (Φ), thereby increasing torque. M ≈ IaΦ
Constant Torque
Base speed corresponds to full armature voltage (Va) and full flux (Φ). A DC motor can operate at rated torque (M) at any speed up to base speed, by selecting the appropriate value of armature voltage. This is often referred to as the constant torque region. Actual torque (M) produced, however, is determined by the demand of the load (Ia).
Constant Horsepower
Some applications require the motor to be operated above base speed. Armature voltage (Va), however, cannot be higher than rated nameplate voltage. Another method of increasing speed is to weaken the field (Φ). Weakening the field reduces the amount of torque (M) a motor can produce. Applications that operate with field weakening must require less torque at higher speeds.
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Horsepower is said to be constant because speed (N) increases and torque (M) decreases in proportion.
Field Saturation
It can be seen from the speed (n) and torque (M) formulas that field flux (Φ) density has a direct effect on motor speed and available torque. An increase in field flux (Φ), for example, will cause a decrease in speed (n) and an increase in available motor torque (M).
The relationship between field current (If) and flux (Φ) is not as directly proportional as it may appear. As flux density increases the field’s ability to hold additional flux decreases. It becomes increasingly difficult to increase flux density. This is known as saturation. A saturation curve, such as the example shown below, can be plotted for a DC motor. Flux (Φ) will rise somewhat proportionally with an increase of field current (If) until the knee of the curve. Further increases of field current (If) will result in a less proportional flux (Φ) increase. Once the field is saturated no additional flux (Φ) will be developed.
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Review 3 1.
One way to increase motor speed is to ____________ armature voltage. a. increase b. decrease
2.
CEMF is zero when the armature is ____________ . a. turning at low speed b. turning at max speed c. not turning d. accelerating
3.
A ____________ - connected motor is typically used with DC drives.
4.
A DC motor, operating from zero to base speed, can be said to be operating in the constant ____________ range. a. horsepower b. torque
5.
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No additional ____________ can be developed once the field becomes saturated.
Basic DC Drives
The remainder of this course will focus on applying the SIMOREG DC MASTER® 6RA70, to DC motors and associated applications. The SIMOREG DC MASTER 6RA70 drives are designed to provide precise DC motor speed control over a wide range of machine parameters and load conditions. Selection and ordering information, as well as engineering information can be found in the SIMOREG 6RA70 DC MASTER catalog, available from you local Siemens sales representative.
SIMOREG drives are designed for connection to a three-phase AC supply. They, in turn, supply the armature and field of variable-speed DC motors. SIMOREG drives can be selected for connection to 230, 400, 460, 575, 690, and 830 VAC, making them suitable for global use.
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Siemens SIMOREG DC MASTER 6RA70 drives are available up to 1000 HP at 500 VDC in standard model drives. In addition, drives can be paralleled, extending the range up to 6000 HP. Siemens SIMOREG drives have a wide range of microprocessor-controlled internal parameters to control DC motor operation. It is beyond the scope of this course to cover all of the parameters in detail, however; many concepts common to most applications and drives will be covered later in the course.
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Power Modules
The SIMOREG 6RA70 is available in a power module and base drive panels. The power module contains the control electronics and power components necessary to control drive operation and the associated DC motor.
Base Drive Panels
The base drive panel consists of the power module mounted on a base panel with line fuses, control transformer, and contactor. This design allows for easy mounting and connection of power cables.
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High Horsepower Designs
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High horsepower designs are also available with ratings up to 14,000 amps. These drives have input ratings up to 700 VAC and can operate motors with armature ratings up to 750 VDC. For additional information on high horsepower design SIMOREG 6RA70 DC MASTER drives, contact your Siemens sales representative.
Converting AC to DC
Thyristor
A primary function of a DC drive, such as the SIMOREG 6RA70 DC MASTER, is to convert AC voltage into a variable DC voltage. It is necessary to vary to DC voltage in order to control the speed of a DC motor. A thyristor is one type of device commonly used to convert AC to DC. A thyristor consists of an anode, cathode, and a gate.
Gate Current
A thyristor acts as a switch. Initially, a thyristor will conduct (switch on) when the anode is positive with respect to the cathode and a positive gate current is present. The amount of gate current required to switch on a thyristor varies. Smaller devices require only a few milliamps; however, larger devices such as required in the motor circuit of a DC drive may require several hundred milliamps.
Holding Current
Holding current refers to the amount of current flowing from anode to cathode to keep the thyristor turned on. The gate current may be removed once the thyristor has switched on. The thyristor will continue to conduct as long as the anode remains sufficiently positive with respect to the cathode to allow sufficient holding current to flow. Like gate current, the amount of holding current varies from device to device. Smaller devices may require only a few milliamps and larger devices may require a few hundred milliamps.
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The thyristor will switch off when the anode is no longer positive with respect to the cathode.
AC to DC Conversion
The thyristor provides a convenient method of converting AC voltage to a variable DC voltage for use in controlling the speed of a DC motor. In this example the gate is momentarily applied when AC input voltage is at the top of the sinewave. The thyristor will conduct until the input’s sinewave crosses zero. At this point the anode is no longer positive with respect to the cathode and the thyristor shuts off. The result is a half-wave rectified DC.
The amount of rectified DC voltage can be controlled by timing the input to the gate. Applying current on the gate at the beginning of the sinewave results in a higher average voltage applied to the motor. Applying current on the gate later in the sinewave results in a lower average voltage applied to the motor.
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DC Drive Converter
The output of one thyristor is not smooth enough to control the voltage of industrial motors. Six thyristors are connected together to make a 3Ø bridge rectifier.
Gating Angle
As we have learned, the gating angle of a thyristor in relationship to the AC supply voltage, determines how much rectified DC voltage is available. However, the negative and positive value of the AC sine wave must be considered when working with a fully-controlled 3Ø rectifier. A simple formula can be used to calculate the amount of rectified DC voltage in a 3Ø bridge. Converted DC voltage (VDC) is equal to 1.35 times the RMS value of input voltage (VRMS) times the cosine of the phase angle (cosα). VDC = 1.35 x VRMS x cosα The value of DC voltage that can be obtained from a 460 VAC input is -621 VDC to +621 VDC. The following table shows sample values of rectified DC voltage available from 0° to 180°. It is important to note that voltage applied to the armature should not exceed the rated value of the DC motor.
Volts RMS 460 VAC 460 VAC 460 VAC 460 VAC 460 VAC 460 VAC 460 VAC
α 0 30 60 90 120 150 180
Cosine 1.00 0.87 0.50 0.00 -0.50 -0.87 -1.00
Formula VDC = 460 x 1.35 x 1 VDC = 460 x 1.35 x 0.87 VDC = 460 x 1.35 x 0.50 VDC = 460 x 1.35 x 0 VDC = 460 x 1.35 x (-0.50) VDC = 460 x 1.35 x (-0.87) VDC = 460 x 1.35 x (- 1)
VDC 621 538 310.5 0 -310.5 -538 -621
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The following illustration approximates the output waveform of a fully controlled thyristor bridge rectifier for 0°, 60°, and 90°. The DC value is indicated by the heavy horizontal line. It is important to note that when thyristors are gated at 90° the DC voltage is equal to zero. This is because thyristors conduct for the same amount of time in the positive and negative bridge. The net result is 0 VDC. DC voltage will increase in the negative direction as the gating angle (α) is increased from 90° to a maximum of 180°.
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Review 4 1.
An increase of torque causes a corresponding ____________ in horsepower a. increase b. decrease
2.
Typically, DC motor armature voltage is either rated for ____________ VDC or ____________ VDC.
3.
Identify the following insulation classes.
4.
The SIMOREG 6RA70 DC MASTER ____________ drive consists of the power module mounted on a panel with line fuses, control transformer, and a contactor.
5.
A thyristor is one type of device commonly used to convert ____________ . a. DC to AC b. AC to DC
6.
The approximate converted DC voltage of a six-pulse converter when the thyristors are gated at 30° is ____________ VDC.
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Basic Drive Operation
Controlling a DC Motor
A thyristor bridge is a technique commonly used to control the speed of a DC motor by varying the DC voltage. Examples of how a DC rectifier bridge operates are given on the next few pages. Voltage values given in these examples are used for explanation only. The actual values for a given load, speed, and motor vary.
It is important to note that the voltage applied to a DC motor be no greater than the rated nameplate. Armature windings are commonly wound for 500 VDC. The control logic in the drive must be adjusted to limit available DC voltage to 0 - 500 VDC. Likewise, the shunt field must be limited to the motor’s nameplate value.
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Basic Operation
A DC drive supplies voltage to the motor to operate at a desired speed. The motor draws current from this power source in proportion to the torque (load) applied to the motor shaft.
100% Speed, 0% Load
In this example an unloaded motor connected to a DC drive is being operated at 100% speed. The amount of armature current (Ia) and unloaded motor needs to operate is negligible. For the purpose of explanation a value of 0 amps is used. The DC drive will supply only the voltage required to operate the motor at 100% speed. We have already learned the amount of voltage is controlled by the gating angle (COSα) of the thyristors. In this example 450 VDC is sufficient. The motor accelerates until CEMF reaches a value of Va - IaRa. Remember that Va = IaRa + CEMF. In this example IaRa is 0, therefore CEMF will be approximately 450 VDC.
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100% Speed, 100% Load
A fully loaded motor requires 100% of rated armature current at 100% speed. Current flowing through the armature circuit will cause a voltage drop across the armature resistance (Ra). Full voltage (500 VDC) must be applied to a fully loaded motor to operate at 100% speed. To accomplish this, thyristors are gated earlier in the sine wave (36.37°). The DC drive will supply the voltage required to operate the motor at 100% speed. The motor accelerates until CEMF reaches a value of Va - IaRa. Remember that Va = IaRa + CEMF. In this example armature current (Ia) is 100% and Ra will drop some amount of voltage. If we assume that current and resistance is such that Ra drops 50 VDC, CEMF will be 450 VDC.
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1 Quad, 4 Quad
Up to this point we have only looked at a drive in singlequadrant operation. A single-quadrant DC drive will have six thyristors. In the speed-torque chart there are four quadrants of operation according to direction of rotation and direction of torque. A four-quadrant DC drive will have twelve thyristors.
Single-Quadrant Operation
Single-quadrant drives only operate in quadrant I. Motor torque (M) is developed in the forward or clockwise (CW) direction to drive the motor at the desired speed (N). This is similar to driving a car forward on a flat surface from standstill to a desired speed. It takes more forward or motoring torque to accelerate the car from zero to the desired speed. Once the car is at desired speed your foot can be let off the accelerator a little. When the car comes to an incline a little more gas, controlled by the accelerator, maintains speed. To slow or stop a motor in single-quadrant operation the drive lets the motor coast.
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Changing Direction of a DC Motor
There are two ways to change the direction a DC motor rotates. 1. Reverse Armature Polarity 2. Reverse Field Polarity
Reversing in SingleQuadrant Operation
Field contactor reverse kits can be used to provide bidirectional rotation from a single-quadrant drive. To turn the motor in the forward direction the “F” contacts are closed, applying DC voltage in one polarity across the shunt field. Simply reversing the polarity of the field, by opening the “F” contacts and closing the “R” contacts, will reverse direction of a DC motor. It is important to note that field reversal will only work when a quick reversal is not required. The field circuit is inductive and must be brought to 0 current before opening the contacts.
Stopping a Motor
48
Stopping a motor in single-quadrant operation can be done by simply removing voltage to the motor and allowing the motor to coast to a stop. Alternatively, voltage can be reduced gradually until the motor is at a stop. The amount of time required to stop a motor depends on the inertia of the motor and connected load. The more inertia the longer the time.
Dynamic Braking
Dynamic braking is often used on single quadrant drives as a means of stopping a motor quickly. Dynamic braking is not recommended for continuous or repetitive operation. Dynamic braking kits for use with Siemens SIMOREG® drives are typically designed to stop a load operating at base speed a maximum of three consecutive times. After three consecutive stops a waiting period of 15 minutes is required. Dynamic braking develops stopping torque by using a contact (MAUX) to connect a resistor (Rdb) across the armature terminals after the drive controller turns off power to the motor. The field remains energized to supply stopping torque. This is because motor torque (M) depends on armature current (Ia) and field flux (Φ). Armature current (Ia) reverses direction as the motor now acts like a generator. A reversal in armature current (Ia) results in a reversal of torque applied to the motor. Torque, now applied in the opposite direction, acts as a brake to the motor. Stored energy in the rotating motor is applied across the resistor and converted to heat. The resistor is sized to allow 150% current flow initially. Armature voltage decreases as the motor slows down, producing less current through the resistors. The motor is finally stopped due to frictional torque of the connected load.
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Four-Quadrant Operation
The dynamics of certain loads require four-quadrant operation. If motor voltage is suddenly reduced, negative torque is developed in the motor due to the inertia of the connected load. The motor acts like a generator by converting mechanical power from the shaft into electrical power which is returned to the drive. This is similar to driving a car downhill. The car’s engine will act as a brake. Braking occurs in quadrants II and IV.
Regen
In order for a drive to operate in all four quadrants a means must exist to deal with the electrical energy returned by the motor. Electrical energy returned by the motor tends to drive the DC voltage up, resulting in excess voltage that can cause damage. One method of getting four-quadrant operation from a DC drive is to add a second bridge connected in reverse of the main bridge. The main bridge drives the motor. The second bridge returns excess energy from the motor to the AC line. This process is commonly referred to as regen. This configuration is also referred to as a 4-Quad design.
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Motoring
The motor receives power from the incoming line. In this example the motor is operating at full speed (500 VDC).
100% Speed, -100% Load
When the motor is required to stop quickly, the motoring bridge shuts off and the regen bridge turns on. Due to the initial inertia of the connected load the motor acts like a generator, converting mechanical power at the shaft into electrical power which is returned to the AC line. The IaRa voltage drop (-50 VDC) is of opposite polarity then when the drive was supplying motoring power. The control logic is gating thyristors in the regen bridge at an angle of 130° and the resultant DC voltage on the bridge is 400 VDC, in the opposite polarity. Because the regen bridge is of opposite polarity, the voltage applied to the motor acts like an electrical brake for the connected load.
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Regen vs. Dynamic Braking
Regen and dynamic braking provide the same amount of braking power to slow a motor from maximum speed in field weakening to base speed. This is because field strength increases until the motor reaches base speed. However, from base speed to stop, regen is capable of slowing a motor at a faster rate. In addition, regen can develop torque at zero speed to bring the motor to a complete stop. Another advantage of regen is that regen braking is not limited in duty cycle and cool-down periods. Applications that require frequent braking or have overhauling loads should consider four quadrant operation with regen braking.
Reversing
52
A four-quadrant drive can easily reverse the direction of rotation of a DC motor simply by applying armature voltage in the opposite polarity. This is accomplished by using what was the regen bridge to motor. The bridge that was used to drive the motor in the forward direction becomes the regen bridge.
Review 5 1.
When torque is developed in the forward direction and the armature is turning in the forward direction, the motor is operating in quadrant ____________ .
2.
When the armature is turning in the forward direction but torque is developed in the reverse direction, the motor is operating in quadrant ____________ .
3.
The direction of rotation of a DC motor, operated from a 6-pulse converter, can be reversed by reversing the polarity of the DC voltage applied to the ____________ field.
4.
____________ ____________ is a method used to stop a motor quickly by applying a resistor to the armature.
5.
Which of the following is an advantage of a 4-quad converter? a. Instead of being dissipated in heat, excess energy is returned to the supply line. b. From base speed to zero speed a 4-quad converter will stop a motor faster than a 1-quad converter. c. A 4-quad converter can reverse motor direction by simply applying voltage in the opposite polarity across the armature. d. all of the above.
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SIMOREG 6RA70 DC MASTER Electronics
Up to this point we have looked at the power components of a DC Drive necessary to control the speed of a DC motor. The actual control of these components is accomplished with electronic hardware and technology software. Speed Control with CEMF Feedback
Speed control is one mode of operation. The drive will attempt to maintain a constant speed regardless of the load’s torque. A speed reference is input into a ramp function generator which applies reference voltage to the speed controller over a specified period of time. This allows a smoother acceleration of the motor and connected load. The output of the speed controller is routed to the firing circuit, which controls the amount of voltage applied to the armature. You will recall that Va (applied voltage) = IaRa + CEMF. IaRa is proportional to load and is generally 10% of nameplate armature voltage at 100% load. Therefore, as load torque/ current varies between 0 and 100%, IaRa varies from 0 to 50 VDC for a 500 VDC armature. Va and Ia are constantly monitored. Ra is measured during the comissioning and tuning of the drive. Because Va, Ia, and Ra are known values, CEMF (Ea) can be precisely calculated. CEMF is proportional to speed and the speed controller uses this value to calculate actual speed. Speed control with CEMF feedback can only be used on applications where the motor operates between zero and base speed. CEMF feedback provides approximately 2-5% speed regulation.
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Speed Control with Tach Feedback
A tachometer can be used when a more accurate measurement of speed is needed, or when the motor will be operated above base speed. A measurement of actual speed is returned to the speed controller. The speed controller will make armature voltage adjustments to maintain constant speed with variations in load. If, for example, load is suddenly increased the motor will slow, reducing speed feedback. The speed controller will output a higher signal to the current controller, which will increase the firing angle of the firing circuit. The resulting increased armature voltage applies more torque to the motor to offset the increased load. Motor speed will increase until it is equal with the speed reference setpoint. When the motor is rotating faster than desired speed armature voltage is reduced. In a four-quad drive DC armature voltage could momentarily be reversed to slow the motor at a faster rate to the desired speed. Several tachs can be used with the SIMOREG 6RA70. DC tachs can provide approximately 0.10 to 2% regulation. Digital (pulse) tachs can provide approximately 0.10 to 0.25% regulation. These values vary depending on the tach and the operating conditions.
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Current Measurement
The drive monitors current, which is summed with the speed control signal at the current controller. The drive acts to maintain current at or below rated current by reducing armature voltage if necessary. This results in a corresponding reduction in speed until the cause of the overcurrent is removed.
Torque Control
Some applications require the motor to operate with a specific torque regardless of speed. The outer loop (speed feedback) is removed and a torque reference is input. The current controller is effectively a torque controller because torque is directly proportional to current.
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Tuning the Drive
A feature of the SIMOREG 6RA70 DC MASTER is the ability to self tune for a given motor and associated load. An improperly tuned control may result in an excessive speed overshoot when changing from one speed to another. Oscillations can occur which contribute to system instability.
A properly tuned drive will have an initial overshoot of approximately 43% and settle into a new speed quickly. This provides a stable system with quick response.
The SIMOREG 6RA70 DC MASTER has three self-tuning routines to match the performance of the drive to the controlled motor and associated load. •
Armature Tuning tunes the drive to the motor characteristics
•
Speed Tuning tunes the drive to the connected load
•
CEMF Tuning tunes the drive for field weakening
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CUD1 Board
The CUD1 board is the main control board for the SIMOREG 6RA70. This board contains the necessary software and hardware interfaces for operating the drive in speed or torque control. It has input and output connections for wiring the control devices of various functions such as start/stop pushbuttons and speed potentiometer. The CUD1 board has comprehensive diagnostics for troubleshooting. CUD1 also contains the necessary software for self-tuning. Programmable binary outputs, used to indicate the condition of the drive, are available on X171. Binary inputs are also available to start and stop the drive on X171. In addition, there are two programmable binary inputs for such functions as reverse and jog. The 6RA70 accepts analog inputs for speed control on X174. Programmable analog outputs on X175 provide meter indication of various drive parameters such as current and voltage. A motor temperature switch can be connected to X174 and is used to stop the drive if the motor becomes overheated. Connections are also available on X173 for a digital tach.
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Typical Connections
The following diagram shows a typical connection used to operate the drive. A normally open (NO) contact is used to start and stop the drive.
Alternately, pushbuttons can be used to start and stop the drive.
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Programming and Operating Sources
SIMOREG 6RA70 drives can be programmed and operated from various sources, such as the PMU, OP1S, or other SIMATIC® HMI device such as the TP170A, TP170B, OP27, or MP370. In addition to these, various methods of serial communication is available through RS232 or RS485 connections. These will be discussed later in this section with the option boards. The PMU can be used alone or with the OP1S. The OP1S can be mounted directly on the PMU or up to 200 meters away with an external power supply. Parameters, such as ramp times, minimum and maximum speed, and modes of operation are easily set. The changeover key (“P”) toggles the display between a parameter number and the value of the parameter. The up and down pushbuttons scroll through parameters and are used to select a parameter value, once the “P” key sets the parameter. The OP1S has a numbered key pad for direct entry.
SIMATIC HMI Devices
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Another, more robust option, is a SIMATIC HMI device such as the TP170A. The TP170A uses a touch-sensitive screen for control and monitoring. It is powered from the drive and standard PROFIBUS connections.
CUD2 Expansion Board
The CUD2 is typically selected when additional inputs and outputs (I/O) are required. CUD2 I/O is selectable. An advantage to the CUD2 expansion board is that it mounts directly on the CUD1 and requires no additional hardware. The CUD2 provides four optically isolated binary inputs, four selectable binary inputs to ground, two analog inputs, one analog input for motor temperature evaluation, two binary outputs, and one serial interface. In addition to the expanded I/O, the CUD2 provides a parallel interface for paralleling up to six power modules.
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EB1 and EB2 Expansion Boards
EB1 and EB2 are half-sized expansion boards that provide a number of additional I/O possibilities. EB1 has three binary inputs and four bidirectional binary I/O. Bidirectional I/O can be configured as a binary input or output. One of the analog inputs is used as a voltage or current reference input. Two of the analog inputs can also be configured as binary inputs. EB2 has two binary inputs, one analog input, one analog output, and four relay contacts. Three of the contacts are normally open (NO) and one of the contacts can be configured as normally open (NO) or normally closed (NC).
T400 Technology Board
I/O Isolated Binary Inputs
CUD2 4
EB1 0
EB2 0
Binary Inputs
4
3
2
Bidirectional Binary I/O
0
4
0
Analog Inputs
2
3
1
Analog Outputs
2
2
1
Relay (Binary) Outputs
2
0
4
Serial Interface
1
0
0
Parallel Converter Interface
1
0
0
The T400 is an option board that is used to provide specialized features for applications, such as winders, tension control, position control, and hoisting gear. In addition to applying built-in technology functions, users familiar with the Siemens PLC software SIMATIC STEP-7 can also implement their own process functions.
To implement the various control functions required by specific applications the T400 has two analog outputs, five analog inputs, two binary outputs, eight binary inputs, four bidirectional binary inputs/outputs, two incremental encoder inputs, and two serial interfaces. 62
Communications
One of the strong points of the SIMOREG 6RA70 is its serial interface capabilities, which makes it easy to integrate the drive with other automation components. Communication options are available for PROFIBUS-DP, SIMOLINK®, CAN, and DeviceNet communications.
SLB
The SLB communication board is used for peer-to-peer communication with other Siemens drives via SIMOLINK. SIMOLINK is a high speed fiber optic ring bus that allows various data to be passed from one drive to the next. Communication is not limited to the SIMOREG 6RA70. SIMOLINK can also communicate between Siemens AC drives such as the MASTERDRIVE MC and MASTERDRIVE VC.
CBP2
PROFIBUS-DP is an open bus standard for a wide range of applications in various manufacturing and automation applications. Siemens DC drives can easily communicate with other control devices such as programmable logic controllers (PLCs) and personal computers (PCs) through the PROFIBUSDP communication system and other various protocols. The CBP2 board is required to communicate via PROFIBUS-DP.
63
CBC
ISO is a federation of standards organizations from over 100 countries that develops voluntary standards for business, science, and technology. The official name is Organization Internationale de Normalisation, also known in the United States as the International Organization for Standardization. The CBC communication board is used to communicate with CAN protocol, which is an ISO standard (ISO 11898) for serial data communications. CAN protocol was initially developed in 1986 for the automotive industry. Today communication with CAN protocol can also be found in other industrial automation applications. One device, such as a PLC or computer, acts as a master. SIMOREG drives equipped with CBC boards and other controllable devices configured for CAN act as slaves. CAN uses a simple twisted pair of wires for transmission of control and parameter value data between SIMOREG drives with CBC boards.
64
CBD
The CBD communication board is used to communicate with DeviceNet. DeviceNet is another communication protocol that was developed based on the CAN technology. DeviceNet provides a low-level network for DeviceNet enabled devices such as sensors, motor starters, and drives to communicate with higher-level devices such as computers and PLCs. DeviceNet can read the state of devices, such as on/off, as well as start and stop motors (motor starters). SIMOREG 6RA70 DC MASTERs equipped with a CBD board can be added to a DeviceNet network. A DeviceNet enabled master device can control the operation, such as start, stop, accel, and decel.
SBP
Digital tachometers (encoders) can be used to measure the actual speed of a motor. The SBP encoder board can be also be used to monitor an external encoder, such as might be connected to the driven machine.
65
Electronics Box
The electronics box contains the CUD1 board (main control board) and option boards. The CUD1 board is plugged into slot 1.
Mounting Option Boards
There are several option boards available, which will be discussed later in this section. Option boards are automatically recognized by the drive. Up to six boards can be installed in the electronics box. A Local Bus Adapter (LBA) is required if mounting positions 2 or 3 are needed. In addition, adapter boards (ADB) are necessary for slots D, E, F, and G when utilizing the half-size option boards.
66
There are a few rules that must be followed when mounting option boards: •
Option boards may be plugged into positions 2 or 3, however, position 2 must be filled first.
•
When used, a technology board (T400) is always installed in position 2.
•
If a communication board (CBP2, CBC. or CBD) is used with a technology board the communication board is placed in slot G.
•
It is unnecessary and not possible to use expansion boards EB1 and EB2 in conjunction with the technology board T400. T400 has its own expanded inputs and outputs (I/O).
•
It is unnecessary and not possible to use the pulse encoder board (SBP) or the SIMOLINK communication board (SLB) in conjunction with T400. T400 has provision to connect an encoder.
•
A maximum of two supplementary boards of the same type may be used in one drive. For example, no more than two communication boards or two expansion boards can be used.
The following chart shows the mounting positions for CUD1 and option boards. Board
LBA
ADB
Location 1
CUD1 CUD2 CBP2 CBC CBD SLB SBP T400 EB1 EB2
No No Yes Yes Yes Yes Yes Yes Yes Yes
No No Yes Yes Yes Yes Yes No Yes Yes
Yes Yes No No No No No No No No
Location 2 D E No No No No Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes
Location 3 F G No No No No Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes No Yes Yes Yes Yes
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Review 6 1.
____________ is the designation of the main electronic control board in the SIMOREG 6RA70 DC MASTER.
2.
A ____________ ____________ ____________ is required when mounting option boards in the electronics box.
3.
Position ____________ must be filled first when installing option boards.
4.
____________ tuning tunes a drive to the motor characteristics.
5.
Technology board T400 can be installed in location ____________ .
6.
An advantage of the CUD2 expansion board is that it mounts directly on ____________ and requires no additional hardware.
7.
____________ expansion board has the most bidirectional binary I/O. a. CUD2 b. EB1 c. EB2
8.
____________ is used to communicate with PROFIBUS-DP.
9.
____________ is used to communicate with other Siemens drives via SIMOLINK.
10. A second digital tachometer is connected to the drive through an ____________ board when T400 is not used.
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Parameters and Function Blocks
The SIMOREG 6RA70 DC MASTER features an extensive parameter set that can easily be adapted to almost any drive task, from simple to complex. A wide scope of parameters include: • • • • • •
Acceleration/Deceleration Control Automatic Restart Function Field Reversal Various Arithmetic and Boolean Logic Operations Technology Controllers Velocity/Speed and Diameter Calculators
In addition, the SIMOREG 6RA70 DC MASTER has extensive status indicators and display parameters for monitoring. The SIMOREG 6RA70 DC MASTER also supports a large database of faults and alarms. This provides the operator with a clear indication of what may be needed to correct the problem. There are numerous parameters within the SIMOREG 6RA70 DC MASTER. It is beyond the scope of this course to cover these in any detail. However, it is important to understand how parameters and function blocks work together. Parameters
Parameter values are used to provide settings to the drive. In the Siemens SIMOREG 6RA70 DC MASTER each parameter is clearly designated by an assigned number. Parameters are differentiated according to their function: • •
Function Parameters (can be read and written) Visualization Parameters (can only be read)
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Function Parameters
Acceleration or deceleration times are examples of function parameters. A feature of DC drives is the ability to increase or decrease the armature voltage gradually. This accelerates and decelerates the motor smoothly with less stress on the motor and connected load. Parameters P303 and P304 work together to instruct the SIMOREG 6RA70 DC MASTER how much acceleration/ deceleration time is needed from 0 to 100% speed. P303 and P304 can be set to any value between 0.0 to 650 seconds. If P303 were set to 20.00, for example, the drive would take 20 seconds to accelerate the motor from 0 to 100% speed. Acceleration and deceleration time is linear which means the time speed curve can be accurately tracked. The motor would be at 25% speed after 5 seconds and 50% speed after 10 seconds.
Rounding is a feature that can be added to the acceleration/ deceleration curve. This feature smooths the transition between starting and finishing a ramp.
70
Visualization Parameters
Visualization parameters are used for visualizing internal quantities. These parameters are only displayed and cannot be changed by the operator. Visualization parameters are distinguished by a lower case “r” . Parameter r038, for example, displays the value of voltage output to the motor.
Function Blocks
A function block consists of several parameters grouped together to perform a specific task. The following function block represents one example of how a proportional/integral (PI) controller can be used in speed control of a SIMOREG 6RA70 DC MASTER.
Function Parameters
The response of a function block is determined by function parameters. Proportional gain and integral time, for example, determine the response of a PI-controller. Each parameter has a name, identifying number, value range, and a factory setting. Function parameters can be indexed.
71
Indexing and Data Sets
In many applications it may be desirable to configure the SIMOREG 6RA70 DC MASTER for variations in operation. For example, there may be a situation in an application where it is desirable to have different acceleration times. Indexed parameters can have up to four different values stored with them. Each value stored is part of a data set. Parameter P303, acceleration time, is an example of an indexed parameter. P303 can have four different acceleration times stored. P303 could, for example, have the following values: P303.1 = 0.50 P303.2 = 1.00 P303.3 = 3.00 P303.4 = 8.00 If data set 1 is active, the acceleration time is 0.50 seconds. If data set 2 is active, the acceleration time is 1.00 second. Data sets are operator selected and can be changed at any time.
PI-Controller
PI-controllers are commonly used in drive technology. In our example the desired speed and actual speed are input to a summation point. The two signals are opposite in polarity. When the actual speed is equal to the desired speed the deviation, which is input into the PI-controller, is zero (0). Whenever desired speed and actual speed are different there is a deviation. Changes in load on the motor, for example, can affect motor speed. A sudden increase in load would cause the motor to slow down. This would decrease the feedback from actual speed and the deviation would become more positive. It is also possible that the application may require the motor to slow down or speed up. Until the motor reaches the new desired speed there will be a deviation.
72
The PI-controller’s job is to make speed corrections quickly a minimal amount of overshoot and oscillation. Parameter P225 (gain) and parameter P226 (time) are used to tune the PIcontroller’s performance. The end result should be a fast response time with about a 43% initial overshoot. The motor should then settle in to the new desired speed.
Connectors and Binectors
Connectors and binectors are elements used to exchange signals between function blocks. Connectors are used to store analog values. Analog values are stored in the form that is represented by 16 bit or 32 bit words. Binectors are used to store binary (digital) information.
73
Connectors and binectors are identified by a name and number. Connectors with 16 bit resolution are identified with a “K”. Connectors with 32 bit resolution are identified with a “KK” . Binectors are identified with a “B”.
BICO
74
BICO is the term used to describe the method of connecting function blocks together. This is performed with BInectors and COnnectors. A connection between two function blocks consists of a connector or binector and a BICO parameter. With BICO parameters you can determine the sources of the input signals of a function block. This allows the user to “softwire” function blocks to meet specific application requirements.
Engineering Tools
There are several engineering tools available optionally. These tools aid in programming, operating, troubleshooting, and managing SIMOREG 6RA70 DC MASTER drives. SIMOVIS
SIMOVIS® can be used to aid start-up, setting and saving parameters, and as a diagnostic tool for troubleshooting. A feature of SIMOVIS is the graphics display. With this feature oscilloscope functions can be displayed on a computer screen.
75
QuickStart
QuickStart is another tool useful in setting up SIMOREG DC drives. A feature of this tool is the ability to communicate with specific drives in a network of drives. A Wizard driven menu leads operators through a simplified start-up procedure step-bystep.
Drive ES
Drive ES is used to integrate Siemens drives with the SIMATIC automation world. There are three Drive ES packages available.
Package Drive ES Basic
Drive ES Graphic
Description Structured similar to SIMOVIS allowing commissioning, parameter handling, oscilloscope readout, and fault evaluation. Based on STEP 7 for integration into SIMATIC. Provides graphic configuring of BICO function blocks. Requires Drive ES Basic and a SIMATIC programming tool called SIMATIC CFC.
Drive ES SIMATIC Provides function blocks and examples of SIMATIC projects. Requires Drive ES Basic.
76
Applications
When applying a DC drive and motor to an application it is necessary to know the horsepower, torque, and speed characteristics of the load. The following chart shows typical characteristics of various loads.
Loads generally fall into one of three categories:
Category
Description
Constant Torque
The load is essentially the same throughout the speed range. Hoisting gear and belt conveyors are examples.
Variable Torque
The load increases as speed increases. Pumps and fans are examples.
Constant The load decreases as speed increases. Winders Horsepower and rotary cutting machines are examples.
77
Application Examples
The Siemens SIMOREG 6RA70 DC MASTER drives are designed to handle the most challenging applications. The following examples are just some of applications the SIMOREG can be used on. Winders/Coilers
78
DC motors offer superior characteristics at low speed for winder and coiler operation and performance. In winder applications maintaining tension at standstill is a very important operation. DC motors offer a wide speed range at rated torque. On many winder applications that run in an extended speed range a smaller horsepower DC motor could do the same job as a larger horsepower AC motor.
Marine Applications
DC drives offer several advantages in marine applications. Compact sizing is one of the biggest advantages. DC drives also adapt well from generator supplies such as found in the marine industry.
Crane/Hoist
DC offers several advantages in applications that operate at low speed, such as cranes and hoists. Advantages include low speed accuracy, short-time overload capability, size, torque proving control, and load sharing.
79
Mining/Drilling
DC is often preferred in the high horsepower applications required in the mining and drilling industry. DC drives offer advantages in size and cost. They are rugged, dependable, and proven in the industry.
Extruding
Extruding is a price competitive industry. DC offers economical solutions in the 60 to 1000 HP range which is commonly used in extruding applications.
80
Selecting a Siemens DC Drive
The following flow diagram, along with the selection charts, will help you select the right DC drive for your application.
81
The following tables provide catalog numbers for SIMOREG DC drives up to 1000 HP. For larger drives consult your Siemens sales representative. Power Module Single Quad, Non-Regen
82
Horsepower 240 VDC 500 VDC 3 7.5 7.5 15 15 30 25 60 40 75 60 125 75 150 125 250 150 300 250 500 700 1000
Rated Armature (Amps DC) 15 30 60 100 140 210 255 430 510 850 1180 1660
Catalog Number 6RA7018-6FS22-0Z+X01 6RA7025-6FS22-0Z+X01 6RA7028-6FS22-0Z+X01 6RA7031-6FS22-0Z+X01 6RA7075-6FS22-0Z+X01 6RA7078-6FS22-0Z+X01 6RA7082-6FS22-0Z+X01 6RA7085-6FS22-0Z+X01 6RA7087-6FS22-0Z+X01 6RA7091-6FS22-0Z+X01 6RA7093-4GS22-0Z+X01 6RA7095-4GS22-0Z+X01
Power Module Four Quad, Regen
Base Drive Signle Quad, Non-Regen
Base Drive Four Quad, Regen
Horsepower 240 VDC 500 VDC 3 7.5 7.5 15 15 30 25 60 40 75 60 125 75 150 125 250 150 300 250 500 700 1000
Rated Armature (Amps DC) 15 30 60 100 140 210 255 430 510 850 1180 1660
Horsepower 240 VDC 500 VDC 3 7.5 7.5 15 15 30 25 60 40 75 60 125 75 150 125 250 150 300 250 500 700 1000
Rated Armature (Amps DC) 15 30 60 100 140 210 255 430 510 850 1180 1660
Horsepower 240 VDC 500 VDC 3 7.5 7.5 15 15 30 25 60 40 75 60 125 75 150 125 250 150 300 250 500 700 1000
Rated Armature (Amps DC) 15 30 60 100 140 210 255 430 510 850 1180 1660
Catalog Number 6RA7018-6FV62-0Z+X01 6RA7025-6FV62-0Z+X01 6RA7028-6FV62-0Z+X01 6RA7031-6FV62-0Z+X01 6RA7075-6FV62-0Z+X01 6RA7078-6FV62-0Z+X01 6RA7082-6FV62-0Z+X01 6RA7085-6FV62-0Z+X01 6RA7087-6FV62-0Z+X01 6RA7091-6FV62-0Z+X01 6RA7093-4GV62-0Z+X01 6RA7095-4GV62-0Z+X01
Catalog Number 6RA7013-2FS22-0 6RA7018-2FS22-0 6RA7025-2FS22-0 6RA7030-2FS22-0 6RA7072-2FS22-0 6RA7075-2FS22-0 6RA7077-2FS22-0 6RA7082-2FS22-0 6RA7083-2FS22-0 6RA7087-2FS22-0 6RA7091-2FS22-0 6RA7094-2FS22-0
Catalog Number 6RA7013-2FV62-0 6RA7018-2FV62-0 6RA7025-2FV62-0 6RA7030-2FV62-0 6RA7072-2FV62-0 6RA7075-2FV62-0 6RA7077-2FV62-0 6RA7082-2FV62-0 6RA7083-2FV62-0 6RA7087-2FV62-0 6RA7091-2FV62-0 6RA7094-2FV62-0
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Review 7
84
1.
Parameters that can be read only are referred to as ____________ parameters.
2.
A function block consists of several ____________ grouped together to perform a specific task.
3.
____________ is the term used to describe the method of connecting function blocks together.
4.
Winders are examples of ____________ ____________ applications.
5.
Identify the category of the following speed, torque, and horsepower graphs.
Review Answers
Review 1
1) SIMOREG; 2) 15; 3) 15; 4) b. Dividing Distance by Time; 5) force; 6) external force.
Review 2
1) electromagnet; 2) current; 3) CEMF; 4) a. Permanent Magnet, b. Compound, c. Shunt, d. Series.
Review 3
1) a. increase; 2) c. not turning; 3) shunt; 4) b. torque; 5 flux.
Review 4
1) a. increase; 2) 250 VDC or 500 VDC; 3) a. A, b. B, c. F, d. H; 4) base; 5) b. AC to DC; 6) 538.
Review 5
1) I; 2) II; 3) shunt; 4) Dynamic Braking; 5) d. all of the above.
Review 6
1) CUD1; 2) Local Bus Adapter; 3) 2; 4) Armature; 5) 2; 6) CUD1; 7) b. EB1; 8) CBP2; 9) SLB; 10) SBP.
Review 7
1) visualization; 2) parameters; 3) BICO; 4) constant horsepower; 5) a. Constant Torque, Constant Horsepower, Variable Torque.
85
Final Exam
The final exam is intended to be a learning tool. The book may be used during the exam. A tear-out answer sheet is provided. After completing the test, mail the answer sheet in for grading. A grade of 70% or better is passing. Upon successful completion of the test a certificate will be issued. Those receiving a score of less than 70% will be provided a second test. Questions
1.
The type of DC motor best suited for use with DC drives is the ____________ wound motor. a. b. c. d.
2.
____________ is the trade name for Siemens DC drives. a. b.
3.
c. d.
SIMOVIS SIMOLINK
armature voltage CEMF field current (If) armature resistance (Ra)
A decrease in field flux strength (Φ) causes a/an ____________ . a. b. c. d.
86
SIMOREG SIMOVERT
The base speed of a motor is an indication of how fast the motor will turn with rated ____________ and rated load (amps) at rated flux (Φ). a. b. c. d.
4.
series shunt compound series or shunt
decrease in armature voltage increase in armature voltage increase in motor torque decrease in motor torque
5.
____________ is the voltage induced into an armature conductor of a DC motor in opposition to applied voltage. a. b. c. d.
6.
____________ current refers to the minimum amount of current flowing from anode to cathode to keep a thyristor turned on. a. d. c. d.
7.
0 310.5 538 621
A 1-quad drive uses ____________ thyristors to convert AC to a variable voltage DC. a. b.
9.
Armature current Gating current Holding current Field current
The value of rectified DC voltage obtained from a 460 VAC 3Ø source when the thyristors are gated at 60° is ____________ VDC. a. b. c. d.
8.
Armature voltage Field voltage CEMF EMF
4 6
c. d.
8 12
____________ is a method sometimes used on 1-quad drives as a means of stopping a motor quickly by converting mechanical energy to heat. a. b. c. d.
Regen Field reversal Dynamic Braking Armature reversal
87
10.
Which of the following is not an advantage of regen over dynamic braking? a. b. c. d.
11.
____________ tuning tunes the 6RA70 drive to the motor characteristics. a. b.
12.
CUD1 CUD2
c. d.
SLB CBP2
ADB Adapter Board LBA Local Bus Adapter ADB and LBA No additional hardware
D E
c. d.
F G
The command to start the 6RA70 drive is received on ____________ of CUD1. a. b. c. d.
88
CEMF Field
If a communication board (CBP2, CBC, or CBD) is used with a technology board, the communication board must be placed in slot ____________ . a. b.
15.
c. d.
CUD2 requires ____________ to install in the 6RA70. a. b. c. d.
14.
Speed Armature
____________ is the main control board in the 6RA70 which controls drive operation. a. b.
13.
Regen brakes faster from max speed to base speed Regen brakes faster from base speed to stop Regen is not limited to duty cycle and cooldown periods Regen can develop torque at zero speed
Terminal 106 of XS Terminal 4 of X174 Terminal 37 of X171 Terminal 14 of X175
16.
Hoisting gear is an example of a ____________ load. a. b. c. d.
17.
____________ is used to communicate with PROFIBUS-DP. a. b.
18.
c. d.
CBD SBP
SIMATIC SIMOVIS
c. d.
BICO QuickStart
____________ is an example of a visualization parameter. a. b.
20.
CBC CBP2
____________ is the term used to describe the method of connecting function blocks together. a. b.
19.
constant torque variable torque constant horsepower constant speed
P225 r038
c. d.
K165 B0205
The correct catalog number for a SIMOREG 6RA70 DC MASTER, base drive, four quad, to be used with armature amps rated for 100 amps is ____________ . a. b. c. d.
6RA7031-6FS22-0Z+X01 6RA7031-6FV62-0Z+X01 6RA7030-2FS22-0 6RA7030-2FV62-0
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Notes
90
Notes
91
Notes
92
Table of Contents
Introduction ..............................................................................2 Electron Theory .........................................................................4 Conductors, Insulators and Semiconductors ............................5 Electric Charges ........................................................................7 Current......................................................................................9 Voltage .................................................................................... 11 Resistance .............................................................................. 13 Simple Electric Circuit ............................................................. 15 Ohm’s Law ............................................................................. 16 DC Series Circuit .................................................................... 18 DC Parallel Circuit ...................................................................23 Series-Parallel Circuits ............................................................30 Power......................................................................................34 Magnetism .............................................................................37 Electromagnetism ..................................................................39 Introduction to AC ...................................................................42 AC Generators ........................................................................44 Frequency ...............................................................................47 Voltage and Current ................................................................48 Inductance ..............................................................................51 Capacitance ............................................................................56 Inductive and Capacitive Reactance .......................................61 Series R-L-C Circuit .................................................................67 Parallel R-L-C Circuit ................................................................69 Power and Power Factor in an AC Circuit ................................ 71 Transformers ...........................................................................75 Three-Phase Transformers ......................................................80 Review Answers .....................................................................83 Final Exam ..............................................................................84 1
Introduction
Welcome to the first course in the STEP series, Siemens Technical Education Program designed to prepare our distributors to sell Siemens Energy & Automation products more effectively. This course covers Basics of Electricity and is designed to prepare you for subsequent courses on Siemens Energy & Automation products. Upon completion of Basics of Electricity you will be able to:
2
•
Explain the difference between conductors and insulators
•
Use Ohm’s Law to calculate current, voltage, and resistance
•
Calculate equivalent resistance for series, parallel, or series-parallel circuits
•
Calculate voltage drop across a resistor
•
Calculate power given other basic values
•
Identify factors that determine the strength and polarity of a current-carrying coil’s magnetic field
•
Determine peak, instantaneous, and effective values of an AC sine wave
•
Identify factors that effect inductive reactance and capacitive reactance in an AC circuit
•
Calculate total impedance of an AC circuit
•
Explain the difference between real power and apparent power in an AC circuit
•
Calculate primary and secondary voltages of single-phase and three-phase transformers
•
Calculate kVA of a transformer
The objectives listed above may sound strange to you. You may also wonder why you would need to know these things to sell Siemens Energy & Automation products. Developing a basic knowledge of electrical concepts, however, will help you to better understand customer applications. In addition, you will be better able to describe products to customers and determine important differences between products. If you are an employee of a Siemens Energy & Automation authorized distributor, fill out the final exam tear-out card and mail in the card. We will mail you a certificate of completion if you score a passing grade. Good luck with your efforts.
3
Electron Theory
Elements of an Atom
All matter is composed of molecules which are made up of a combination of atoms. Atoms have a nucleus with electrons orbiting around it. The nucleus is composed of protons and neutrons (not shown). Most atoms have an equal number of electrons and protons. Electrons have a negative charge (-). Protons have a positive charge (+). Neutrons are neutral. The negative charge of the electrons is balanced by the positive charge of the protons. Electrons are bound in their orbit by the attraction of the protons. These are referred to as bound electrons. Electron Proton Nucleus
Free Electrons
4
Electrons in the outer band can become free of their orbit by the application of some external force such as movement through a magnetic field, friction, or chemical action. These are referred to as free electrons. A free electron leaves a void which can be filled by an electron forced out of orbit from another atom. As free electrons move from one atom to the next an electron flow is produced. This is the basis of electricity.
Conductors, Insulators and Semiconductors
Conductors
An electric current is produced when free electrons move from one atom to the next. Materials that permit many electrons to move freely are called conductors. Copper, silver, aluminum, zinc, brass, and iron are considered good contors. Copper is the most common maal used for contors and is relatively insive.
Insulators
Materials that allow few free electrons are called insulators. Materials such as plastic, rubber, glass, mica, and ceramic are good insulators.
An electric cable is one example of how conductors and insulators are used. Electrons flow along a copper conductor to provide energy to an electric device such as a radio, lamp, or a motor. An insulator around the outside of the copper conductor is provided to keep electrons in the conductor. Rubber Insulator Copper Conductor
5
Semiconductors
Semiconductor materials, such as silicon, can be used to manufacture devices that have characteristics of both conductors and insulators. Many semiconductor devices will act like a conductor when an external force is applied in one direction. When the external force is applied in the opposite direction, the semiconductor device will act like an insulator. This principle is the basis for transitors, diodes, and other solidstate electronic devices.
Transistor
Diode
Review 1 1.
List the three basic elements of an atom and state the charge of each (positive, negative, or neutral). Element
Charge
____________ ____________ ____________ ____________ ____________ ____________ 2.
An electron forced out of orbit by an external force is called a ____________ ____________ .
3.
Conductors allow ____________ free electrons to flow when an external electric force is applied.
4.
Which of the following materials are good conductors? a. copper b. plastic c. silver d. rubber
5.
6
e. f. g. h.
aluminum glass iron mica
Semiconductor devices can be manufactured to allow ____________ electrons to flow in one direction and ___ _________ electrons to flow in the opposite direction.
Electric Charges
Neutral State of an Atom
Elements are often identified by the number of electrons in orbit around the nucleus of the atoms making up the element and by the number of protons in the nucleus. A hydrogen atom, for example, has only one electron and one proton. An aluminum atom (illustrated) has 13 electrons and 13 protons. An atom with an equal number of electrons and protons is said to be electrically neutral.
Outer Band
Positive and Negative Charges
Electrons in the outer band of an atom are easily displaced by the application of some external force. Electrons which are forced out of their orbits can result in a lack of electrons where they leave and an excess of electrons where they come to rest. The lack of electrons is called a positive charge because there are more protons than electrons. The excess of electrons has a negative charge. A positive or negative charge is caused by an absence or excess of electrons. The number of protons remains constant.
Neutral Charge
Negative Charge
Positive Charge
7
Attraction and Repulsion of Electric Charges
The old saying, “opposites attract,” is true when dealing with electric charges. Charged bodies have an invisible electric field around them. When two like-charged bodies are brought together, their electric field will work to repel them. When two unlike-charged bodies are brought together, their electric field will work to attract them. The electric field around a charged body is represented by invisible lines of force. The invisible lines of force represent an invisible electrical field that causes the attraction and repulsion. Lines of force are shown leaving a body with a positive charge and entering a body with a negative charge. Unlike Charges Attract
Coulomb’s Law
8
Like Charges Repel
During the 18th century a French scientist, Charles A. Coulomb, studied fields of force that surround charged bodies. Coulomb discovered that charged bodies attract or repel each other with a force that is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them. Today we call this Coulomb’s Law of Charges. Simply put, the force of attraction or repulsion depends on the strength of the charged bodies, and the distance between them.
Current
Electricity is the flow of free electrons in a conductor from one atom to the next atom in the same general direction. This flow of electrons is referred to as current and is designated by the symbol “I”. Electrons move through a conductor at different rates and electric current has different values. Current is determined by the number of electrons that pass through a cross-section of a conductor in one second. We must remember that atoms are very small. It takes about 1,000,000,000,000,000,000,000,000 atoms to fill one cubic centimeter of a copper conductor. This number can be simplified using mathematical exponents. Instead of writing 24 zeros after the number 1, write 1024. Trying to measure even small values of current would result in unimaginably large numbers. For this reason current is measured in amperes which is abbreviated “amps”. The letter “A” is the symbol for amps. A current of one amp means that in one second about 6.24 x 1018 electrons move through a cross-section of conductor. These numbers are given for information only and you do not need to be concerned with them. It is important, however, that the concept of current flow be unstood.
Units of Measurement
The following chart reflects special prefixes that are used when dealing with very small or large values of current: Prefix
Symbol
Decimal
1 kiloampere 1 milliampere 1 microampere
1 kA 1 mA 1 mA
1000 A 1/1000 A 1/1,000,000 A
9
Direction of Current Flow
Some authorities distinguish between electron flow and current flow. Conventional current flow theory ignores the flow of electrons and states that current flows from positive to negative. To avoid confusion, this book will use the electron flow concept which states that electrons flow from negative to positive.
_
+ Electron Flow
10
_
+ Conventional Current Flow
Voltage
Electricity can be compared with water flowing through a pipe. A force is required to get water to flow through a pipe. This force comes from either a water pump or gravity. Voltage is the force that is applied to a conductor that causes electric current to flow. Water Flow Through a Pipe
Current Flow Through a Conductor
Electrons are negative and are attracted by positive charges. They will always be attracted from a source having an excess of electrons, thus having a negative charge, to a source having a deficiency of electrons which has a positive charge. The force required to make electicity flow through a conductor is called a difference in potential, electromotive force (emf), or more simply referred to as voltage. voltage is designated by the letter “E”, or the letter “V”. The unit of measurement for voltage is volts which is also designated by the letter “V”.
11
Voltage Sources
An electrical voltage can be generated in various ways. A battery uses an electrochemical process. A car’s alternator and a power plant generator utilizes a magnetic induction process. All voltage sources share the characteristic of an excess of electrons at one terminal and a shortage at the other terminal. This results in a difference of potential between the two terminals. Shortage of Electrons Excess of Electrons
_
+
Batter y
Voltage Circuit Symbol
The terminals of a battery are indicated symbolically on an electrical drawing by two lines. The longer line indicates the positive terminal. The shorter line indicates the negative terminal.
+ _
Units of Measurement
12
The following chart reflects special prefixes that are used when dealing with very small or large values of voltage: Prefix
Symbol
Decimal
1 kilovolt 1 millivolt 1 microvolt
1 kV 1 mV 1 mV
1000 V 1/1000 V 1/1,000,000 V
Resistance
A third factor that plays a role in an electrical circuit is resistance. All material impedes the flow of electrical current to some extent. The amount of resistance depends upon composition, length, cross-section and temperature of the resistive material. As a rule of thumb, resistance of a conductor increases with an increase of length or a decrease of crosssection. Resistance is designated by the symbol “R”. The unit of measurement for resistance is ohms (Ω). Resistance Circuit Symbols
Resistance is usually indicated symbolically on an electrical drawing by one of two ways. An unfilled rectangle is commonly used. A zigzag line may also be used.
Resistance can be in the form of various components. A resistor may be placed in the circuit, or the circuit might contain other devices that have resistance. Units of Measurement
The following chart reflects special prefixes that are commonly used when dealing with values of resistance: Prefix
Symbol
Decimal
1 kilohm 1 megohm
1 kΩ 1 MΩ
1000 Ω 1,000,000 Ω
13
Review 2 1.
Elements are identified by the number of ____________ in orbit around the nucleus.
2.
A material that has an excess of electrons is said to have a ____________ charge.
3.
A material that has a deficiency of electrons is said to have a ____________ charge.
4.
Like charges ____________ and unlike charges ____________ .
5.
The force that is applied to a conductor to cause current flow is ____________ .
6.
Electrons move from ____________ . a. positive to negative b. negative to positive
7.
With an increase of length or a decrease of crosssection of a conductor, resistance will ____________ . a. increase b. decrease
14
Simple Electric Circuit
An Electric Circuit
A fundamental relationship exists between current, voltage, and resistance. A simple electric circuit consists of a voltage source, some type of load, and a conductor to allow electrons to flow between the voltage source and the load. In the following circuit a battery provides the voltage source, electrical wire is used for the conductor, and a light provides the resistance. An additional component has been added to this circuit, a switch. There must be a complete path for current to flow. If the switch is open, the path is incomplete and the light will not illuminate. Closing the switch completes the path, allowing electrons to leave the negative terminal and flow through the light to the positive terminal.
+
An Electrical Circuit Schematic
Switch
_
_
+
The following schematic is a representation of an electrical circuit, consisting of a battery, a resistor, a voltmeter and an ammeter. The ammeter, connected in series with the circuit, will show how much current flows in the circuit. The voltmeter, connected across the voltage source, will show the value of voltage supplied from the battery. Before an analysis can be made of a circuit, we need to understand Ohm’s Law. +
+ _
_ A
+ V
R
_
15
Ohm’s Law
George Simon Ohm and Ohm’s Law
The relationship between current, voltage and resistance was studied by the 19th century German mathematician, George Simon Ohm. Ohm formulated a law which states that current varies directly with voltage and inversely with resistance. From this law the following formula is derived: I=
E R
or Current =
Voltage Resistance
Ohm’s Law is the basic formula used in all electrical circuits. Electrical designers must decide how much voltage is needed for a given load, such as computers, clocks, lamps and motors. Decisions must be made concerning the relationship of current, voltage and resistance. All electrical design and analysis begins with Ohm’s Law. There are three mathematical ways to express Ohm’s Law. Which of the formulas is used depends on what facts are known before starting and what facts need to be known. I=
Ohm’s Law Triangle
16
E R
E=IxR
R=
E I
There is an easy way to remember which formula to use. By arranging current, voltage and resistance in a triangle, one can quickly determine the correct formula.
Using the Triangle
To use the triangle, cover the value you want to calculate. The remaining letters make up the formula.
I=
E R
E=IxR
R=
E I
Ohm’s Law can only give the correct answer when the correct values are used. Remember the following three rules: • • • Examples of Solving Ohm’s Law
Current is always expressed in amperes or amps Voltage is always expressed in volts Resistance is always expressed in ohms
Using the simple circuit below, assume that the voltage supplied by the battery is 10 volts, and the resistance is 5 Ω. +
_ A
+
+ _
V
R
_
To find how much current is flowing through the circuit, cover the “I” in the triangle and use the resulting equation. I=
E R
I=
10 Volts 5Ω
I = 2 Amps
Using the same circuit, assume the ammeter reads 200 mA and the resistance is known to be 10 Ω. To solve for voltage, cover the “E” in the triangle and use the resulting equation. E=IxR
E = 0.2 x 10
E = 2 Volts
Remember to use the correct decimal equivalent when dealing with numbers that are preceded with milli (m), micro (µ) or kilo (k). In this example had 200 been used instead of converting the value to 0.2, the wrong answer of 2000 volts would have been calculated.
17
DC Series Circuit
Resistance in a Series Circuit
A series circuit is formed when any number of resistors are connected end-to-end so that there is only one path for current to flow. The resistors can be actual resistors or other devices that have resistance. The following illustration shows four resistors connected end-to-end. There is one path of current flow from the negative terminal of the battery through R4, R3, R2, R1 returning to the positive terminal. R1
R3
R2
R4
+ _
Formula for Series Resistance
The values of resistance add in a series circuit. If a 4 Ω resistor is placed in series with a 6 Ω resistor, the total value will be 10 Ω. This is true when other types of resistive devices are placed in series. The mathematical formula for resistance in series is: Rt = R1 + R2 + R3 + R4 + R5 11 KΩ
2 KΩ
2 KΩ
100 Ω
1 KΩ
R1
R2
R3
R4
R5
+ _
Rt = R1 + R2 + R3 + R4 + R5 Rt = 11,000 + 2,000 + 2,000 + 100 + 1,000 Rt = 16,100 Ω
18
Current in a Series Circuit
The equation for total resistance in a series circuit allows us to simplify a circuit. Using Ohm’s Law, the value of current can be calculated. Current is the same anywhere it is measured in a series circuit. E R 12 I= 10 I=
I = 1.2 Amps
+ _
5Ω
1Ω
2Ω
2Ω
10 Ω
R1
R2
R3
R4
Rt + _
12 Volts
Equivalent Circuit
Original Circuit
Voltage in a Series Circuit
12 Volts
Voltage can be measured across each of the resistors in a circuit. The voltage across a resistor is referred to as a volt age drop. A German physicist, Kirchhoff, formulated a law which states the sum of the voltage drops across the resistances of a closed circuit equals the total voltage applied to the circuit. In the following illustration, four equal value resistors of 1.5 Ω each have been placed in series with a 12 volt battery. Ohm’s Law can be applied to show that each resistor will “drop” an equal amount of voltage. 12 V
3V
3V
3V
3V
1.5 Ω
1.5 Ω
1.5 Ω
1.5 Ω
R1
R2
R3
R4
12 Volt Battery +
_
19
First, solve for total resistance: Rt = R1 + R2 + R3 + R4 Rt = 1.5 + 1.5 + 1.5 + 1.5 Rt = 6 Ω
Second, solve for current: I= E R I = 12 6 I = 2 Amps
Third, solve for voltage across any resistor: E=IxR E = 2 x 1.5 E = 3 Volts
If voltage were measured across any single resistor, the meter would read three volts. If voltage were read across a combination of R3 and R4 the meter would read six volts. If voltage were read across a combination of R2, R3, and R4 the meter would read nine volts. If the voltage drops of all four resistors were added together the sum would be 12 volts, the original supply voltage of the battery. Voltage Division in a Series Circuit
It is often desirable to use a voltage potential that is lower than the supply voltage. To do this, a voltage divider, similar to the one illustrated, can be used. The battery represents Ein which in this case is 50 volts. The desired voltage is represented by Eout, which mathematically works out to be 40 volts. To calculate this voltage, first solve for total resistance. Rt = R1 + R2 Rt = 5 + 20 Rt = 25 Ω
20
Second, solve for current: Ein Rt 50 I= 25 I=
I = 2 Amps
Finally, solve for voltage: Eout = I x R2 Eout = 2 x 20 Eout = 40 Volts
+ _
Ein
R1
5Ω
R2
20 Ω
Eout 40 Volts
21
Review 3 1.
The basic Ohm’s Law formula is ____________ .
2.
When solving circuit problems; current must always be expressed in ____________ , voltage must always be expressed in ____________ and resistance must always be expressed in ____________ .
3.
The total current of a simple circuit with a voltage supply of 12 volts and a resistance of 24 Ω is ____________ amps.
4.
What is the total resistance of a series circuit with the following values: R1=10 Ω, R2=15 Ω, and R3=20 Ω? ____________ Ω.
5.
What is total current of a series circuit that has a 120 volt supply and 60 Ω resistance?
6.
In the following circuit the voltage dropped across R1 is ____________ volts and R2 is ____________ volts.
+ _
7.
1.5 Ω
R1
R2
12 Volts
In the following circuit voltage dropped across R1 is ____________ volts, and R2 is ____________ volts.
+ _
22
1.5 Ω
5Ω
20 Ω
R1
R2
100 Volts
DC Parallel Circuit
Resistance in a Parallel Circuit
A parallel circuit is formed when two or more resistances are placed in a circuit side-by-side so that current can flow through more than one path. The illustration shows two resistors placed side-by-side. There are two paths of current flow. One path is from the negative terminal of the battery through R1 returning to the positive terminal. The second path is from the negative terminal of the battery through R2 returning to the positive terminal of the battery.
+ _
Formula for Equal Value Resistors in a Parallel Circuit
R1
R2
To determine the total resistance when resistors are of equal value in a parallel circuit, use the following formula: Rt =
Value of any one Resistor Number of Resistors
In the following illustration there are three 15 Ω resistors. The total resistance is: Rt =
Value of any one Resistor Number of Resistor
15 Rt = 3 Rt = 5 Ω
+ _
R1
R1
R1
15 Ω
15 Ω
15 Ω
23
Formula for Unequal There are two formulas to determine total resistance for Resistors in a Parallel Circuit unequal value resistors in a parallel circuit. The first formula is used when there are three or more resistors. The formula can be extended for any number of resistors. 1 = 1 + 1 + 1 Rt R1 R2 R3
In the following illustration there are three resistors, each of different value. The total resistance is: 1 = 1 + 1 + 1 Rt R1 R2 R3 1 = Rt
Insert Value of the Resistors
1 = 4 + 2 + 1 Rt 20 20 20
Find Lowest Common Denominator
1 = 7 Rt 20
Add the Numerators
Rt = 20 1 7
Invert Both Sides of the Equation
Rt = 2.86 Ω
Divide
+ _
24
1 + 1 + 1 10 5 20
R1
R2
R3
5Ω
10 Ω
20 Ω
The second formula is used when there are only two resistors. R1 x R2 Rt = R1 + R2
In the following illustration there are two resistors, each of different value. The total resistance is: R1 x R2 Rt = R1 + R2 5 x 10 5 + 10
Rt =
50 Rt = 15 Rt =
3.33 Ω
+ _
Voltage in a Parallel Circuit
R1 5Ω
R2 10 Ω
When resistors are placed in parallel across a voltage source, the voltage is the same across each resistor. In the following illustration three resistors are placed in parallel across a 12 volt battery. Each resistor has 12 volts available to it. 12 Volt Battery + _
R1
12 V
R2
12 V
R3
12 V
25
Current in a Parallel Circuit
Current flowing through a parallel circuit divides and flows through each branch of the circuit. It
+ _
R1
R2
I1
I2
R3 I3
It
Total current in a parallel circuit is equal to the sum of the current in each branch. The following formula applies to current in a parallel circuit. It = I 1 + I 2 + I 3
Current Flow with Equal Value Resistors in a Parallel Circuit
When equal resistances are placed in a parallel circuit, opposition to current flow is the same in each branch. In the following circuit R1 and R2 are of equal value. If total current (It) is 10 amps, then 5 amps would flow through R1 and 5 amps would flow through R2. It = 10 Amps
+ _
R1 I1 = 5 Amps
R2 I2 = 5 Amps
It = 10 Amps It = I 1 + I 2 It = 5 Amps + 5 Amps It = 10 Amps
26
Current Flow with Unequal Value Resistors in a Parallel Circuit
When unequal value resistors are placed in a parallel circuit, opposition to current flow is not the same in every circuit branch. Current is greater through the path of least resistance. In the following circuit R1 is 40 Ω and R2 is 20 Ω. Small values of resistance means less opposition to current flow. More current will flow through R2 than R1.
12 Volts + _
R1 40 Ω I1 = 0.3 Amps
R2 20 Ω I2 = 0.6 Amps
It = 0.9 Amps
Using Ohm’s Law, the total current for each circuit can be calculated. I1 =
E R1
I1 = 120 Volts 40 Ω I1 = 0.3 Amps
I2 =
E R2
I2 = 120 Volts 20 Ω I2 = 0.6 Amps
It = I 1 + I 2 It = 0.3 Amps + 0.6 Amps It = 0.9 Amps
27
Total current can also be calculated by first calculating total resistance, then applying the formula for Ohm’s Law. R1 x R2 Rt = R1 + R2 40 Ω x 20 Ω Rt = 40 Ω + 20 Ω 800 Ω Rt = 60 Ω Rt = 13.333 Ω
It = E Rt It = 12 Volts 13.333 Ω It = 0.9 Amps
28
Review 4 1.
The total resistance of a parallel circuit that has four 20 Ω resistors is ____________ Ω.
2.
Rt for the following circuit is ____________ Ω.
+ _
3.
R2
R3
10 Ω
20 Ω
30 Ω
Rt for the following circuit is ____________ Ω.
+ _
4.
R1
R1 5Ω
R2 10 Ω
Voltage available at R2 in the following circuit is ____________ volts. 12 Volts
+ _
R1 5Ω
R2 10 Ω
5.
In a parallel circuit with two resistors of equal value and a total current flow of 12 amps, the value of current through each resistor is ____________ amps.
6.
In the following circuit current flow through R1 is _____ _______ amps, and R2 is ____________ amps. 24 Volts + _
R1 10 Ω
R2 10 Ω
29
Series-Parallel Circuits
Series-parallel circuits are also known as compound circuits. At least three resistors are required to form a series-parallel circuit. The following illustrations show two ways a series-parallel combination could be found. Parallel Branches + _
Parallel Branches Devices in Series + _
Simplifying a Series-Parallel
The formulas required for solving current, voltage and resistance problems have already been defined. To solve a series-parallel circuit, reduce the compound circuits to equivalent simple circuits. In the following illustration R1 and R2 are parallel with each other. R3 is in series with the parallel circuit of R1 and R2. R1 10 Ω R3 10 Ω + _
30
R2 10 Ω
First, use the formula to determine total resistance of a parallel circuit to find the total resistance of R1 and R2. When the resistors in a parallel circuit are equal, the following formula is used: R =
Value of any One Resistor Number of Resistors
R = 10 Ω 2 R = 5Ω
Second, redraw the circuit showing the equivalent values. The result is a simple series circuit which uses already learned equations and methods of problem solving. R3 10Ω
R3 5Ω
+ _
Simplifying a Series-Parallel Circuit to a Parallel Circuit
In the following illustration R1 and R2 are in series with each other. R3 is in parallel with the series circuit of R1 and R2.
R1 10Ω + _
R3 20Ω R2 10Ω
First, use the formula to determine total resistance of a series circuit to find the total resistance of R1 and R2. The following formula is used: R = R1 + R 2 R = 10 Ω + 10 Ω R = 20 Ω
31
Second, redraw the circuit showing the equivalent values. The result is a simple parallel circuit which uses already learned equations and methods of problem solving.
+ _
+ _
32
R = 20 Ω
R3 = 20 Ω
Rt = 10 Ω
Review 5 1.
Calculate equivalent resistance for R1 and R2 and total resistance for the entire circuit. R1 20 Ω R3 10 Ω
+ _
R2 30 Ω
R1/R2 equivalent resistance = ____________ Ω Total resistance = ____________ Ω 2.
Calculate equivalent resistance for R1 and R2 and total resistance for the entire circuit.
R1 30 Ω + _
R3 20 Ω R2 10 Ω
R1/R2 equivalent resistance = ____________ Ω Total resistance = ____________ Ω
33
Power
Work
Whenever a force of any kind causes motion, work is accomplished. In the illustration below work is done when a mechanical force is used to lift a weight. If a force were exerted without causing motion, then no work is done.
Electric Power
In an electrical circuit, voltage applied to a conductor will cause electrons to flow. Voltage is the force and electron flow is the motion. The rate at which work is done is called power and is represented by the symbol “P”. Power is measured in watts and is represented by the symbol “W”. The watt is defined as the rate work is done in a circuit when 1 amp flows with 1 volt applied.
Power Formulas
Power consumed in a resistor depends on the amount of current that passes through the resistor for a given voltage. This is expressed as voltage times current. P=ExI or P = EI
Power can also be calculated by substituting other components of Ohm’s Law. P = I2R and P=
34
E2 R
Solving a Power Problem
In the following illustration power can be calculated using any of the power formulas. I = 2 Amps + _
12 Volts
R=6Ω
P = EI P = 12 Volts x 2 Amps P = 24 Watts P = I2 R P = (2 Amps)2 x 6 Ω P = 24 Watts P=
E2 R
P=
(12 Volts)2 6Ω
P=
144 6
P = 24 Watts
Power Rating of Equipment
Electrical equipment is rated in watts. This rating is an indication of the rate at which electrical equipment converts electrical energy into other forms of energy, such as heat or light. A common household lamp may be rated for 120 volts and 100 watts. Using Ohm’s Law, the rated value of resistance of the lamp can be calculated. P=
E2 R
2 which can be transposed to R = E P
R=
(120 Volts)2 100 Watts
R = 144 Ω
35
Using the basic Ohm’s Law formula, the amount of current flow for the 120 volt, 100 watt lamp can be calculated. I= I=
E R 120 Volts 144 Ω
I = 0.833 Amps
A lamp rated for 120 volts and 75 watts has a resistance of 192 Ω and a current of 0.625 amps would flow if the lamp had the rated voltage applied to it. R=
E2 P
R=
(120 Volts)2 75 Watts
R = 192 Ω I=
E2 P
I=
120 Volts 192 Ω
I = 0.625 Amps
It can be seen that the 100 watt lamp converts energy faster than the 75 watt lamp. The 100 watt lamp will give off more light and heat. Heat
36
Current flow through a resistive material causes heat. An electrical component can be damaged if the temperature is too high. For this reason, electrical equipment is often rated for a maximum wattage. The higher the wattage rating, the more heat the equipment can dissipate.
Magnetism
The principles of magnetism are an integral part of electricity. Electromagnets are used in some direct current circuits. Alternating current cannot be understood without first understanding magnetism. Types of Magnets
The three most common forms of magnets are the horse-shoe, bar and compass needle.
All magnets have two characteristics. They attract and hold iron. If free to move, like the compass needle, the magnet will assume roughly a north-south position. Magnetic Lines of Flux
Every magnet has two poles, one north pole and one south pole. Invisible magnetic lines of flux leave the north pole and enter the south pole. While the lines of flux are invisible, the effects of magnetic fields can be made visible. When a sheet of paper is placed on a magnet and iron filings loosely scattered over it, the filings will arrange themselves along the invisible lines of flux.
37
By drawing lines the way the iron filings have arranged themselves, the following picture is obtained. Broken lines indicate the paths of magnetic flux lines. The field lines exist outside and inside the magnet. The magnetic lines of flux always form closed loops. Magnetic lines of flux leave the north pole and enter the south pole, returning to the north pole through the magnet.
Interaction between Two Magnets
38
When two magnets are brought together, the magnetic flux field around the magnet causes some form of interaction. Two unlike poles brought together cause the magnets to attract each other. Two like poles brought together cause the magnets to repel each other.
Electromagnetism
Left-Hand Rule for Conductors
An electromagnetic field is a magnetic field generated by current flow in a conductor. Whenever current flows a magnetic field exists around the conductor. Every electric current generates a magnetic field. A definite relationship exists between the direction of current flow and the direction of the magnetic field. The left-hand rule for conductors demonstrates this relationship. If a current-carrying conductor is grasped with the left hand with the thumb pointing in the direction of electron flow, the fingers will point in the direction of the magnetic lines of flux.
Current-Carrying Coil
A coil of wire carrying a current, acts like a magnet. Individual loops of wire act as small magnets. The individual fields add together to form one magnet. The strength of the field can be increased by adding more turns to the coil. The strength can also be increased by increasing the current.
39
Left-Hand Rule for Coils
A left-hand rule exists for coils to determine the direction of the magnetic field. The fingers of the left hand are wrapped around the coil in the direction of electron flow. The thumb points to the north pole of the coil.
Electromagnets
An electromagnet is composed of a coil of wire wound around a core. The core is usually a soft iron which conducts magnetic lines of force with relative ease. When current is passed through the coil, the core becomes magnetized. The ability to control the strength and direction of the magnetic force makes electromagnets useful. As with permanent magnets, opposite poles attract. An electromagnet can be made to control the strength of its field which controls the strength of the magnetic poles. A large variety of electrical devices such as motors, circuit breakers, contactors, relays and motor starters use electromagnetic principles.
40
Review 6 1.
The rate at which work is done is called ___________ .
2.
The basic formula for power is ____________ .
3.
In a circuit with a 12 volt supply and 4 Ω resistance the power consumed is ____________ watts.
4.
The two characteristics of all magnets are; they attract and hold ____________ , and if free to move will assume roughly a ____________ position.
5.
Lines of flux always leave the ____________ pole and enter the ____________ pole.
6.
The left-hand rule for conductors states that when the ___________ hand is placed on a current-carrying conductor with the ____________ pointing in the direction of electron flow, the fingers will point in the direction of ____________ .
41
Introduction to AC
The supply of current for electrical devices may come from a direct current source (DC), or an alternating current source (AC). In direct current electricity, electrons flow continuously in one direction from the source of power through a conductor to a load and back to the source of power. Voltage in direct current remains constant. DC power sources include batteries and DC generators. In allowing current an AC generator is used to make electrons flow first in one direction then in another. Another name for an AC generator is an alternator. The AC generator reverses terminal polarity many times a second. Electrons will flow through a conductor from the negative terminal to the positive terminal, first in one direction then another.
42
AC Sine Wave
Alternating voltage and current vary continuously. The graphic representation for AC is a sine wave. A sine wave can represent current or voltage. There are two axes. The vertical axis represents the direction and magnitude of current or voltage. The horizontal axis represents time. + Direction
0
Time
- Direction
When the waveform is above the time axis, current is flowing in one direction. This is referred to as the positive direction. When the waveform is below the time axis, current is flowing in the opposite direction. This is referred to as the negative direction. A sine wave moves through a complete rotation of 360 degrees, which is referred to as one cycle. Alternating current goes through many of these cycles each second. The unit of measurement of cycles per second is hertz. In the United States alternating current is usually generated at 60 hertz. Single-Phase and Three-Phase AC Power
Alternating current is divided into single-phase and three-phase types. Single-phase power is used for small electrical demands such as found in the home. Three-phase power is used where large blocks of power are required, such as found in commercial applications and industrial plants. Single-phase power is shown in the above illustration. Three-phase power, as shown in the following illustration, is a continuous series of three overlapping AC cycles. Each wave represents a phase, and is offset by 120 electrical degrees. Phase 1 Phase 2 Phase 3 +
0
-
43
AC Generators
Basic Generator
A basic generator consists of a magnetic field, an armature, slip rings, brushes and a resistive load. The magnetic field is usually an electromagnet. An armature is any number of conductive wires wound in loops which rotates through the magnetic field. For simplicity, one loop is shown. When a conductor is moved through a magnetic field, a voltage is induced in the conductor. As the armature rotates through the magnetic field, a voltage is generated in the armature which causes current to flow. Slip rings are attached to the armature and rotate with it. Carbon brushes ride against the slip rings to conduct current from the armature to a resistive load.
Pole Piece
Magnetic Field Armature
Brush
R1
Slip Ring
Basic Generator Operation
An armature rotates through the magnetic field. At an initial position of zero degrees, the armature conductors are moving parallel to the magnetic field and not cutting through any magnetic lines of flux. No voltage is induced.
R1
44
Generator Operation from Zero to 90 Degrees
The armature rotates from zero to 90 degrees. The conductors cut through more and more lines of flux, building up to a maximum induced voltage in the positive direction. 90 Degrees
R1
Generator Operation from 90 to 180 Degrees
The armature continues to rotate from 90 to 180 degrees, cutting less lines of flux. The induced voltage decreases from a maximum positive value to zero.
S
180 Degrees
R1
Generator Operation from
The armature continues to rotate from 180 degrees to 270 degrees. The conductors cut more and more lines of flux, but in the opposite direction. voltage is induced in the negative direction building up to a maximum at 270 degrees.
270 Degrees
R1
45
Generator Operation from 270 to 360 Degrees
The armature continues to rotate from 270 to 360 degrees. Induced voltage decreases from a maximum negative value to zero. This completes one cycle. The armature will continue to rotate at a constant speed. The cycle will continuously repeat as long as the armature rotates. 360 Degrees
S
One Revolution R1
46
Frequency
The number of cycles per second made by voltage induced in the armature is the frequency of the generator. If the armature rotates at a speed of 60 revolutions per second, the generated voltage will be 60 cycles per second. The accepted term for cycles per second is hertz. The standard frequency in the United States is 60 hertz. The following illustration shows 15 cycles in 1/4 second which is equivalent to 60 cycles in one second.
1/4 Second
Four-Pole AC Generator
The frequency is the same as the number of rotations per second if the magnetic field is produced by only two poles. An increase in the number of poles, would cause an increase in the number of cycles completed in a revolution. A two-pole generator would complete one cycle per revolution and a fourpole generator would complete two cycles per revolution. An AC generator produces one cycle per revolution for each pair of poles.
One Revolution R1
47
Voltage and Current
Peak Value
The sine wave illustrates how voltage and current in an AC circuit rises and falls with time. The peak value of a sine wave occurs twice each cycle, once at the positive maximum value and once at the negative maximum value. Peak Value
+
0
Time
-
Peak-to-Peak Value
Peak Value
The value of the voltage or current between the peak positive and peak negative values is called the peak-to-peak value. +
0
Peak-to-Peak Value
Time
-
Instantaneous Value
The instantaneous value is the value at any one particular time. It can be in the range of anywhere from zero to the peak value. +
0
-
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Instantaneous Value
Time
Calculating Instantaneous Voltage
The voltage waveform produced as the armature rotates through 360 degrees rotation is called a sine wave because instantaneous voltage is related to the trigonometric function called sine (sin θ = sine of the angle). The sine curve represents a graph of the following equation: e = Epeak x sin θ
Instantaneous voltage is equal to the peak voltage times the sine of the angle of the generator armature. The sine value is obtained from trigonometric tables. The following table reflects a few angles and their sine value. Angle
Sin
θ
Angle
Sin
θ
30 Degrees
0.5
210 Degrees
-0.5
60 Degrees
0.866
240 Degrees
-0.866
90 Degrees
1
270 Degrees
-1
120 Degrees
0.866
300 Degrees
-0.866
150 Degrees
0.5
330 Degrees
-0.5
180 Degrees
0
360 Degrees
0
The following example illustrates instantaneous values at 90, 150, and 240 degrees. The peak voltage is equal to 100 volts. By substituting the sine at the instantaneous angle value, the instantaneous voltage can be calculated. 90° = +100 Volts
+
150° = +50 Volts 0 240° = -86.6 Volts -
Any instantaneous value can be calculated. For example: 240° e = 100 x -0.866 e = -86.6 volts
49
Effective Value of an AC Sine Wave
Alternating voltage and current are constantly changing values. A method of translating the varying values into an equivalent constant value is needed. The effective value of voltage and current is the common method of expressing the value of AC. This is also known as the RMS (root-mean-square) value. If the voltage in the average home is said to be 120 volts, this is the RMS value. The effective value figures out to be 0.707 times the peak value. + Peak Value 169.7 Volts 0
-
169.7 Vpeak x 0.707 = 120 Vrms
The effective value of AC is defined in terms of an equivalent heating effect when compared to DC. One RMS ampere of current flowing through a resistance will produce heat at the same rate as a DC ampere. For purpose of circuit design, the peak value may also be needed. For example, insulation must be designed to withstand the peak value, not just the effective value. It may be that only the effective value is known. To calculate the peak value, multiply the effective value by 1.41. For example, if the effective value is 100 volts, the peak value is 141 volts. Review 7
50
1.
The graphic representation of AC voltage or current values over a period of time is a ____________ ____________ .
2.
Each phase of three phase AC power is offset by ____________ degrees.
3.
An AC generator produces ____________ cycle per revolution for each pair of poles.
4.
What is the instantaneous voltage at 240 degrees for a peak voltage of 150 volts?
5.
What is the effective voltage for a peak voltage of 150 volts?
Inductance
The circuits studied to this point have been resistive. Resistance and voltage are not the only circuit properties that effect current flow, however. Inductance is the property of an electric circuit that opposes any change in electric current. Resistance opposes current flow, inductance opposes change in current flow. Inductance is designated by the letter “L”. . The unit of measurement for inductance is the henry (h). Current Flow and Field Strength
Current flow produces a magnetic field in a conductor. The amount of current determines the strength of the magnetic field. As current flow increases, field strength increases, and as current flow decreases, field strength decreases. 0 Degrees No Current
30 Degrees Increasing Current
90 Degrees Maximum Current
Any change in current causes a corresponding change in the magnetic field surrounding the conductor. Current is constant in DC, except when the circuit is turned on and off, or when there is a load change. Current is constantly changing in AC, so inductance is a continual factor. A change in the magnetic field surrounding the conductor induces a voltage in the conductor. This self-induced voltage opposes the change in current. This is known as counter emf. This opposition causes a delay in the time it takes current to attain its new steady value. If current increases, inductance tries to hold it down. If current decreases, inductance tries to hold it up. Inductance is somewhat like mechanical inertia which must be overcome to get a mechanical object moving or to stop a mechanical object from moving. A vehicle, for example, takes a few moments to accelerate to a desired speed, or decelerate to a stop.
51
Inductors
Inductance is usually indicated symbolically on an electrical drawing by one of two ways. A curled line or a filled rectangle can be used.
Inductors are coils of wire. They may be wrapped around a core. The inductance of a coil is determined by the number of turns in the coil, the spacing between the turns, the coil diameter, the core material, the number of layers of windings, the type of winding, and the shape of the coil. Examples of inductors are transformers, chokes, and motors. Simple Inductive Circuit
In a resistive circuit, current change is considered instantaneous. If an inductor is used, the current does not change as quickly. In the following circuit, initially the switch is open and there is no current flow. When the switch is closed, current will rise rapidly at first, then more slowly as the maximum value is approached. For the purpose of explanation, a DC circuit is used.
+ _
R1 L1
Inductive Time Constant
52
The time required for the current to rise to its maximum value is determined by the ratio of inductance (in henrys) to resistance (in ohms). This ratio is called the time constant of the inductive circuit. A time constant is the time (in seconds) required for the circuit current to rise to 63.2% of its maximum value. When the switch is closed in the previous circuit, current will begin to flow. During the first time constant current rises to 63.2% of its maximum value. During the second time constant, current rises to 63.2% of the remaining 36.8%, or a total of 86.4%. It takes about five time constants for current to reach its maximum value.
100.0% 98.1% 94.9% 86.4%
63.2%
First Time Constant
Second Time Constant
Third Time Constant
Fourth Time Constant
Fifth Time Constant
Similarly, when the switch is opened, it will take five time constants for current to reach zero. It can be seen that inductance is an important factor in AC circuits. If the frequency is 60 hertz, current will rise and fall from its peak value to zero 120 times a second. 100.0%
First Time Constant
Second Time Constant
Third Time Constant
Fourth Time Constant
Fifth Time Constant
36.8% 13.6% 5.1% 1.9% 0%
Calculating the Time Constant of an Inductive Circuit
The time constant is designated by the symbol “τ”. To determine the time constant of an inductive circuit use one of the following formulas: τ (in seconds) =
L (henrys) R (ohms)
τ (in milliseconds) =
L (millihenrys) R (ohms)
τ (in microseconds) =
L (microhenrys) R (ohms)
53
In the following illustration, L1 is equal to 15 millihenrys and R1 is equal to 5 Ω. When the switch is closed, it will take 3 milliseconds for current to rise from zero to 63.2% of its maximum value.
+ _
R1 5 Ω L1 15 mh
τ=
15 mh 5Ω
τ = 3 milliseconds
Formula for Series Inductors The same rules for calculating total resistance can be applied. In the following circuit, an AC generator is used to supply electrical power to four inductors. There will always be some amount of resitance and inductance in any circuit. The electrical wire used in the circuit and the inductors both have some resistance and inductance. Total inductance is calculated using the following formula: Lt = L 1 + L 2 + L 3
R1
2 mh
2 mh
1 mh
1 mh
L1
L2
L3
L4
AC Generator
Lt = L1 + L2 + L3 + L4 Lt = 2 mh + 2 mh + 1 mh + 1 mh Lt = 6 mh
54
Formula for Parallel Inductors
In the following circuit, an AC generator is used to supply electrical power to three inductors. Total inductance is calculated using the following formula: 1 = 1 + 1 + 1 Lt L1 L2 L3
R1
1 = Lt
L1
L2
L3
5 mh
10 mh
20 mh
1 + 1 + 1 5 10 20
1 = 7 Lt 20 Lt = 2.86 mh
55
Capacitance
Capacitance and Capacitors
Capacitance is a measure of a circuit’s ability to store an elec cal charge. A device manufactured to have a specific amount of capacitance is called a capacitor. A capacitor is made up of a pair of conductive plates separated by a thin layer of insulating material. Another name for the insulating material is dielectric material. When a voltage is applied to the plates, electrons are forced onto one plate. That plate has an excess of electrons while the other plate has a deficiency of electrons. The plate with an excess of electrons is negatively charged. The plate with a deficiency of electrons is positively charged. Negative Plate Dielectric Material Positive Plate
Direct current cannot flow through the dielectric material because it is an insulator; however it can be used to charge a capacitor. Capacitors have a capacity to hold a specific quantity of electrons. The capacitance of a capacitor depends on the area of the plates, the distance between the plates, and the material of the dielectric. The unit of measurement for capacitance is farads (F). Capacitors usually are rated in µF (microfarads), or pF (picofarads). Capacitor Circuit Symbols
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Capacitance is usually indicated symbolically on an electrical drawing by a combination of a straight line with a curved line, or two straight lines.
Simple Capacitive Circuit
In a resistive circuit, voltage change is considered instantaneous. If a capacitor is used, the voltage across the capacitor does not change as quickly. In the following circuit, initially the switch is open and no voltage is applied to the capacitor. When the switch is closed, voltage across the capacitor will rise rapidly at first, then more slowly as the maximum value is approached. For the purpose of explanation, a DC circuit is used.
+ _
R1 C1
Capacitive Time Constant
The time required for voltage to rise to its maximum value in a circuit containing capacitance is determined by the product of capacitance, in farads, times resistance, in ohms. This product is the time constant of a capacitive circuit. The time constant gives the time in seconds required for voltage across the capacitor to reach 63.2% of its maximum value. When the switch is closed in the previous circuit, voltage will be applied. During the first time constant, voltage will rise to 63.2% of its maximum value. During the second time constant, voltage will rise to 63.2% of the remaining 36.8%, or a total of 86.4%. It takes about five time constants for voltage across the capacitor to reach its maximum value. 100.0% 98.1% 94.9% 86.4%
63.2%
First Time Constant
Second Time Constant
Third Time Constant
Fourth Time Constant
Fifth Time Constant
57
Similarly, during this same time, it will take five time constants for current through the resistor to reach zero. 100.0%
First Time Constant
Second Time Constant
Third Time Constant
Fourth Time Constant
Fifth Time Constant
36.8% 13.6% 5.1% 1.9% 0%
Calculating the Time Constant of a Capacitive Circuit
To determine the time constant of a capacitive circuit, use one of the following formulas: τ (in seconds) = R (megohms) x C (microfarads) τ (in microseconds) = R (megohms) x C (picofarads) τ (in microseconds) = R (ohms) x C (microfarads)
In the following illustration, C1 is equal to 2 µF, and R1 is equal to 10 Ω. When the switch is closed, it will take 20 microseconds for voltage across the capacitor to rise from zero to 63.2% of its maximum value. It will take five time constants, 100 microseconds for this voltage to rise to its maximum value.
+ _
R1 10 Ω C1 2µF
V τ = RC τ = 2µF x 10 Ω τ = 20 microseconds
58
Formula for Series Capacitors
Connecting capacitors in series decreases total capacitance. The effect is like increasing the space between the plates. The formula for series capacitors is similar to the formula for parallel resistors. In the following circuit, an AC generator supplies electrical power to three capacitors. Total capacotance is calculated using the following formula: 1 = 1 + 1 + 1 Ct C1 C2 C3
R1
5 µF
10 µF
20 µF
C1
C2
C3
1 = 1 + 1 + 1 Ct 5 10 20 1 = 7 Ct 20 Ct = 2.86 µF
Formula for Parallel Capacitors
In the following circuit, an AC generator is used to supply electrical power to three capacitors. Total capacitance is calculated using the following formula: Ct = C1 + C2 + C3
R1 C1
C2
C3
5 µF
10 µF
20 µF
Ct = 5 µF + 10 µF + 20 µF Ct = 35 µF
59
Review 8 1.
The total inductance for this circuit is ___________ .
R1
2.
4 mh
2 mh
L1
L2
3 mh
1 mh
L3
L4
The total inductance for this circuit is ____________ . R1 L1 5 mh
3.
L3 10 mh
The total capacitance for this circuit is ____________ .
R1
4.
L2 10 mh
5 µF
10 µF
10 µF
C1
C2
C3
The total capacitance for this circuit is ____________ . R1
60
C1
C2
C3
5 µF
10 µF
10 µF
Inductive and Capacitive Reactance
In a purely resistive AC circuit, opposition to current flow is called resistance. In an AC circuit containing only inductance, capacitance, or both, opposition to current flow is called reactance. Total opposition to current flow in an AC circuit that contains both reactance and resistance is called impedance, designated by the symbol “Z”. Reactance and impedance are expressed in ohms. Inductive Reactance
Inductance only affects current flow when the current is changing. Inductance produces a self-induced voltage (counter emf) that opposes changes in current. In an AC circuit, current is changing constantly. Inductance in an AC circuit, therefore, causes a continual opposition. This opposition to current flow is called inductive reactance and is designated by the symbol XL. Inductive reactance is dependent on the amount of inductance and frequency. If frequency is low, current has more time to reach a higher value before the polarity of the sine wave reverses. If frequency is high, current has less time to reach a higher value. In the following illustration, voltage remains constant. Current rises to a higher value at a lower frequency than a higher frequency. +
+ Current
Current
0
0
_
_ Low Frequency
High Frequency
The formula for inductive reactance is: XL = 2πfL XL = 2 x 3.14 x frequency x inductance
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In a 60 hertz, 10 volt circuit containing a 10 mh inductor, the inductive reactance would be: XL = 2πfL XL = 2 x 3.14 x 60 x 0.10 XL = 3.768 Ω
Once inductive reactance is known, Ohm’s Law can be used to calculate reactive current. I=
E Z
I=
10 3.768
I = 2.65 Amps
Phase Relationship between Current and Voltage in an Inductive Circuit
Current does not rise at the same time as the source voltage in an inductive circuit. Current is delayed depending on the amount of inductance. In a purely resistive circuit, current and voltage rise and fall at the same time. They are said to be “in phase.” In this circuit there is no inductance. Resistance and impedance are the same. +
Voltage Current
0
_
In a purely inductive circuit, current lags behind voltage by 90 degrees. Current and voltage are said to be “out of phase”. In this circuit, impedance and inductive reactance are the same. 90 Degrees +
Voltage Current
0
_
62
All inductive circuits have some amount of resistance. AC current will lag somewhere between a purely resistive circuit, and a purely inductive circuit. The exact amount of lag depends on the ratio of resistance and inductive reactance. The more resistive a circuit is, the closer it is to being in phase. The more inductive a circuit is, the more out of phase it is. In the following illustration, resistance and inductive reactance are equal. Current lags voltage by 45 degrees. 45 Degrees +
XL = 10 Ω Voltage Current
0 R = 10 Ω _
Calculating Impedance in an Inductive Circuit
When working with a circuit containing elements of inductance, capacitance, and resistance, impedance must be calculated. Because electrical concepts deal with trigonometric functions, this is not a simple matter of subtraction and addition. The following formula is used to calculate impedance in an inductive circuit: Z = R2 + XL2
In the circuit illustrated above, resistance and inductive reactance are each 10 ohms. Impedance is 14.1421 ohms. A simple application of Ohm’s Law can be used to find total circuit current. Z = 102 + 102 Z=
200
Z = 14.1421 Ω
Vectors
Another way to represent this is with a vector. A vector is a graphic represention of a quantity that has direction and magnitude. A vector on a map might indicate that one city is 50 miles southwest from another. The magnitude is 50 miles and the direction is southwest. Vectors are also used to show electrical relationships. As mentioned earlier, impedance (Z) is the total opposition to current flow in an AC circuit containing reactance, inductance, and capacitance.
63
14 .14 2
1
Ω
The following vector illustrates the relationship between reactance and inductive reactance of a circuit containing equal values of each. The angle between the vectors is the phase angle represented by the symbol θ. When inductive reactance is equal to resistance the resultant angle is 45 degrees. It is this angle that determines how much current will lag voltage.
Z
=
XL = 10 Ω
θ R = 10 Ω
Capacitive Reactance
Capacitance also opposes AC current flow. Capacitive reactance is designated by the symbol XC.The larger the capacitor, the smaller the capacitive reactance. Current flow in a capacitive AC circuit is also dependent on frequency. The following formula is used to calculate capacitive reactance. 1 XC = 2πfC
Capacitive reactance is equal to 1 divided by 2 times pi, times the frequency, times the capacitance. The capacitive reactance for a 60 hertz circuit with a 10 microfarad capacitor is: 1 XC = 2πfC 1 XC = 2 x 3.14 x 60 x 0.000010 XC = 265.39 Ω
Once capacitive reactance is known, Ohm’s Law can be used to calculate reactive current. I=
E Z
I=
10 265.39
I = 0.0376 Amps
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Phase Relationship between The phase relationship between current and voltage are Current and Voltage opposite to the phase relationship of an inductive circuit. In a purely capacitive circuit, current leads voltage by 90 degrees. 90 Degrees +
Voltage Current
0
_
All capacitive circuits have some amount of resistance. AC current will lead somewhere between a purely resistive circuit and a purely capacitive circuit. The exact amount of lead depends on the ratio of resistance and capacitive reactance. The more resistive a circuit is, the closer it is to being in phase. The more capacitive a circuit is, the more out of phase it is. In the following illustration, resistance and capacitive reactance are equal. Current leads voltage by 45 degrees. 45 Degrees
XC = 10 Ω
+ Voltage Current 0 R = 10 Ω _
Calculating Impedance in a Capacitive Circuit
The following formula is used to calculate impedance in a capacitive circuit: Z = R2 + XC2
In the circuit illustrated above, resistance and capacitive re tance are each 10 ohms. Impedance is 14.1421 ohms. Z = 102 + 102 Z=
200
Z = 14.1421 Ω
65
The following vector illustrates the relationship between resistance and capacitive reactance of a circuit containing equal values of each. The angle between the vectors is the phase angle represented by the symbol θ. When capacitive reactance is equal to resistance the resultant angle is -45 degrees. It is this angle that determines how much current will lead voltage. R = 10 Ω
θ Z = 2 .14 14
XC = 10 Ω
1 Ω
Review 9 1.
In a circuit containing inductance, capacitance, or both, opposition to current flow is called ____________ .
2.
Total opposition to current flow in a circuit that contains both reactance and resistance is called ____________ .
3.
In a 50 hertz circuit, containing a 10 mh inductor, the inductive reactance is ____________ ohms.
4.
In a purely inductive circuit, ____________ a. current and voltage are in phase b. current leads voltage by 90 degrees c. current lags voltage by 90 degrees
5.
In a purely capacitive circuit, ____________ a. current and voltage are in phase b. current leads voltage by 90 degrees c. current lags voltage by 90 degrees
66
6.
In a 50 hertz circuit, containing a 10 microfarad capacitor, the capacitive reactance is ____________ ohms.
7.
In a circuit with 5 Ω resistance, and 10 Ω inductive reactance, impedance is ____________ ohms.
8.
In a circuit with 5 Ω resistance, and 4 Ω capacitive reactance, impedance is ____________ ohms.
Series R-L-C Circuit
Circuits often contain elements of resistance, inductance, and capacitance. In an inductive AC circuit, current lags voltage by 90 degrees. In a AC capacitive circuit, current leads voltage by 90 degrees. It can be seen that inductance and capacitance are 180 degrees apart. Since they are 180 degrees apart, one element will cancel out all or part of the other element.
XL
XC
R
An AC circuit is: • • • Calculating Total Impedance in a Series R-L-C Circuit
Resistive if XL and XC are equal Inductive if XL is greater than XC Capacitive if XC is greater than XL
The following formula is used to calculate total impedance of a circuit containing resistance, capacitance, and inductance: Z = R2 + (XL - XC)2
In the case where inductive reactance is greater than capacitive reactance, subtracting XC from XL results in a positive number. The positive phase angle is an indicator that the net circuit reactance is inductive, and current lags voltage. In the case where capacitive reactance is greater than inductive reactance, subtracting XC from XL results in a negative number. The negative phase angle is an indicator that the net circuit reactance is capacitive and current leads voltage. In either case, the value squared will result in positive number.
67
Calculating Reactance and Impedance in a Series R-L-C Circuit
In the following 120 volt, 60 hertz circuit, resistance is 1000 Ω, inductance is 5 mh, and capacitance is 2 µF. To calculate total impedance, first calculate the value of XL and XC, then impedance can be calculated.
R = 1000 Ω
L = 5 mh
C = 2 µF
XL = 2πfL XL = 6.28 x 60 x 0.005 XL = 1.884 Ω 1 XC = 2πfC 1 XC = 6.28 x 60 x 0.000002 XC = 1,327 Ω
Z = R2 + (XL - XC)2 Z = 10002 + (1.884 - 1,327)2 Z = 1,000,000 + ( - 1,325.116)2 Z = 1,000,000 + 1,755,932.41 Z = 2,755,932.41 Z = 1,660.1 Ω
Calculating Circuit Current in a Series R-L-C Circuit
Ohm’s Law can be applied to calculate total circuit current. I= E Z I=
120 1,660.1
I = 0.072 Amps
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Parallel R-L-C Circuit
Calculating Impedance in a Parallel R-L-C Circuit
Total impedance (Zt) can be calculated in a parallel R-L-C circuit if values of resistance and reactance are known. One method of calculating impedance involves first calculating total current, then using the following formula: E Zt = t It
Total current is the vector sum of current flowing through the resistance plus, the difference between inductive current and capacitive current. This is expressed in the following formula: It =
IR2 + (IC - IL)2
In the following 120 volt, 60 hertz circuit, capacitive reactance has been calculated to be 25 Ω and inductive reactance 50 Ω. Resistance is 1000 Ω. A simple application of Ohm’s Law will find the branch currents. Remember, voltage is constant throughout a parallel circuit.
R = 1000 Ω
XL = 50 Ω
XC = 25 Ω
IR =
E R
E IL = XL
E IC = XC
IR =
120 1000
120 IL = 50
120 IL = 25
IL = 2.4 Amps
IL = 4.8 Amps
IR = 0.12 Amps
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Once the branch currents are known, total current can be calculated. It =
IR2 + (IC - IL)2
It =
0.122 + (4.8 - 2.4)2
It =
0.0144 + 5.76
It =
5.7744
It = 2.4 Amps
Impedance is now found with an application of Ohm’s Law. E Zt = t It Zt =
120 2.4
Zt = 50 Ω
70
Power and Power Factor in an AC Circuit
Power consumed by a resistor is dissipated in heat and not returned to the source. This is true power. True power is the rate at which energy is used. Current in an AC circuit rises to peak values and diminishes to zero many times a second. The energy stored in the magnetic field of an inductor, or plates of a capacitor, is returned to the source when current changes direction. Although reactive components do not consume energy, they do increase the amount of energy that must be generated to do the same amount of work. The rate at which this non-working energy must be generated is called reactive power. Power in an AC circuit is the vector sum of true power and reactive power. This is called apparent power. True power is equal to apparent power in a purely resistive circuit because voltage and current are in phase. voltage and current are also in phase in a circuit containing equal values of inductive reactance and capacitive reactance. If voltage and current are 90 degrees out of phase, as would be in a purely capacitive or purely inductive circuit, the average value of true power is equal to zero. There are high positive and negative peak values of power, but when added together the result is zero. True Power and Apparent Power Formulas
The formula for apparent power is: P = EI
Apparent power is measured in volt-amps (VA). True power is calculated from another trigonometric function, the cosine of the phase angle (cos θ). The formula for true power is: P = EI cos θ
True power is measured in watts.
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In a purely resistive circuit, current and voltage are in phase. There is a zero degree angle displacement between current and voltage. The cosine of zero is one. Multiplying a value by one does not change the value. In a purely resistive circuit the cosine of the angle is ignored. In a purely reactive circuit, either inductive or capacitive, current and voltage are 90 degrees out of phase. The cosine of 90 degrees is zero. Multiplying a value times zero results in a zero product. No power is consumed in a purely reactive circuit. Calculating Apparent Power in a simple R-L-C Circuit
In the following 120 volt circuit, current is equal to 84.9 milliamps. Inductive reactance is 100 Ω and capacitive reactance is 1100 Ω. The phase angle is -45 degrees. By referring to a trigonometric table, the cosine of -45 degrees is found to be .7071.
R = 1000 Ω XL = 100 Ω XC = 1100 Ω 120 VAC
The apparent power consumed by the circuit is: P = EI P = 120 x 0.0849 P = 10.2 VA
The true power consumed by the circuit is: P = EI cos θ P = 120 x 0.0849 x 0.7071 P = 7.2 Watts
Another formula for true power is: P = I2R P = 0.08492 x 1000 P = 7.2 Watts
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Power Factor
Power factor is the ratio of true power to apparent power in an AC circuit. Power factor is expressed in the following formula: PF =
True Power Apparent Power
Power factor can also be expressed using the formulas for true power and apparent power. The value of EI cancels out because it is the same in the numerator and denominator. Power factor is the cosine of the angle. PF =
EI cos θ EI
PF = cos θ
In a purely resistive circuit, where current and voltage are in phase, there is no angle of displacement between current and voltage. The cosine of a zero degree angle is one. The power factor is one. This means that all energy delivered by the source is consumed by the circuit and dissipated in the form of heat. In a purely reactive circuit, voltage and current are 90 degrees apart. The cosine of a 90 degree angle is zero. The power factor is zero. This means the circuit returns all energy it receives from the source to the source. In a circuit where reactance and resistance are equal, voltage and current are displaced by 45 degrees. The cosine of a 45 degree angle is .7071. The power factor is .7071. This means the circuit uses approximately 70% of the energy supplied by the source and returns approximately 30%.
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Review 10 1.
An AC circuit is ____________ if inductive reactance and capacitive reactance are equal.
2.
A series AC circuit is ____________ if there is more inductive reactance than capacitive reactance.
3.
A series AC circuit is ____________ if there is more capacitive reactance than inductive reactance.
4.
In a 120 VAC, 60 hertz series circuit, with 1000 Ω of resistance, 10 mh of inductance and 4 µF of capacitance, impedance is ____________ Ω and current is ____________ amps.
5.
In the illustrated circuit,
R = 1000 Ω XL = 200 Ω XC = 1000 Ω 120 VAC 60 Hz
It is __________ amps, and impedance is __________ Ω.
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6.
True power is measured in ____________ .
7.
A circuit with 0.2 amps flowing through 100 Ω of resistance, is consuming ____________ watts.
Transformers
Mutual Induction
Transformers are electromagnetic devices that transfer electrical energy from one circuit to another by mutual induction. Mutual induction is the coupling of inductances by their mutual magnetic fields. In a single-phase transformer there are two coils, a primary and a secondary coil. The following circuit illustrates mutual induction. The AC generator provides electrical power to the primary coil. The magnetic field produced by the primary induces a voltage into the secondary coil, which supplies power to a load.
Primary Coil
Secondary Coil
Transformers are used to step a voltage up to a higher level, or down to a lower level. Transformers are used extensively in power distribution systems, allowing power companies to transfer electrical energy many miles. Power generators typically generate high voltages. This voltage varies, depending on the generator, but a typical voltage might be 15 KV. The voltage is stepped up through a transformer to higher levels for transmission to substations. Typical voltages range from 115 KV to 765 KV. The electrical power is received at substation transformers many miles away where it is stepped down. Typical voltage might be 34 KV or 69 KV. From here, electrical power is fed to a distribution substation. It can also be fed directly to factory locations. If the power is fed to a factory, transformers at the factory site reduce the voltage to usable levels. The power fed to a distribution substation is reduced by transformers at the substation for factory and home use.
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Coefficient of Coupling
Mutual inductance between two coils depends on their flux linkage. Maximum coupling occurs when all the lines of flux from the primary coil cut through the secondary winding. The amount of coupling which takes place is referred to as coefficient of coupling. To maximize coefficient of coupling, both coils are often wound on an iron core which is used to provide a path for the lines of flux. The following discussion of step-up and step-down transformers applies to transers with an iron core.
Lines of Flux Confined to Iron Core
Lines of Flux that don’t Couple
Voltage, Current, and the Number of Turns in a Coil
There is a direct relationship between voltage, impedance, current, and the number of coil turns in a transformer. This relationship can be used to find either primary or secondary voltage, current, and the number of turns in each coil. It is the number of turns which determine if a transformer is a step up or step down transformer. The following “rules-of-thumb” apply to transformers: •
If the primary coil has fewer turns than the secondary coil, it is a step-up transformer.
•
If the primary coil has more turns than the secondary coil, it is a step-down transformer.
When the number of turns on the primary and seconday coils of a transformer are equal, input voltage, impedance, and current are equal to output voltage, impedance, and current.
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Step-Up Transformer
A step-up transformer is used when it is desirable to step voltage up in value. The following circuit illustrates a stepup transformer. The primary coil has fewer turns than the secondary coil. The number of turns in a transformer is given as a ratio. When the primary has fewer turns than the secondary, voltage and impedance are stepped up. In the circuit illustrated, voltage is stepped up from 120 VAC to 240 VAC. Because impedance is also stepped up, current is stepped down from 10 amps to 5 amps. 1:2
Primary Coil 900 Turns
Secondary Coil 1800 Turns
120 VAC Supply 10 Amps
Step-Down Transformer
240 VAC 5 Amps
A step-down transformer is used when it is desirable to step voltage down in value. The following circuit illustrates a step-down transformer. The primary coil has more turns than the secondary coil. The step-down ratio is 2:1. voltage and impedance are stepped down, current is stepped up. 2:1
Primary Coil 1800 Turns
240 VAC Supply 5 Amps
Seconday Coil 900 Turns
120 VAC Out 10 Amps
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Single-Phase Transformer
120 or 240 VAC single-phase transformers are used to supply lighting, receptacle, and small appliance loads. A transformer with a 240 VAC secondary can be used to supply 240 VAC to larger appliances such as stoves, air conditioners and heaters. A 240 VAC secondary can be tapped in the center to provide two sources of 120 VAC power. Primary
Primary
120 VAC
120 VAC
240 VAC
Secondary
Formulas for Calculating the Number of Primary and Secondary Turns of a Transformer
Ground
Secondary
There are a number of useful formulas for calculating, voltage, current, and the number of turns between the primary and secondary of a transformer. These formulas can be used with either step-up or step-down transformers. The following legend applies to the transformer formulas: ES EP IS IP NS NP
= = = = = =
secondary voltage primary voltage secondary current primary current turns in the secondary coil turns in the primary coil
To find voltage: ES =
EP x IP IS
EP =
ES x IS IP
IP =
ES x IS EP
To find current: IS =
EP x IP ES
To find the number of coil turns: NS =
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120 VAC
ES x NP EP
NP =
EP x NS ES
Using the values for the step-down transformer in the example of the previous page, the secondary voltage can be verified. ES =
EP x IP IS
ES =
240 Volts x 5 Amps 10 Amps
ES =
1200 10
ES = 120 Volts
Transformer Ratings
Transformers are rated in kVA (kilovolt-amps). This rating is used rather than watts because loads are not purely resistive. Only resistive loads are measured in watts. The kVA rating determines the current a transformer can deliver to its load without overheating. Given volts and amps, kVA can be calculated. Given kVA and volts, amps can be calculated. kVA =
Volts x Amps 1000
Amps =
kVA x 1000 Volts
Using the illustrated step-down transformer, the kVA rating can be calculated. The kVA rating of a transformer is the same for both the primary and the secondary. Primary kVA =
240 x 5 1000
Primary kVA = 1.2 kVA Secondary kVA =
120 x 10 1000
Secondary kVA = 1.2 kVA
Transformer Losses
Most of the electrical energy provided to the primary of a transformer is transferred to the secondary. Some energy, however, is lost in heat in the wiring or the core. Some losses in the core can be reduced by building the core of a number of flat sections called laminations.
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Three-Phase Transformers
Delta Connections
Three-phase transformers are used when three-phase power is required for larger loads such as industrial motors. There are two basic three-phase transformer connections, delta and wye. Delta transformers are used where the distance from the supply to the load is short. A delta is like three singlephase transformers connected together. The secondary of a delta transformer is illustrated below. For simplicity, only the secondary of a three-phase transformer is shown. The voltages shown on the illustration are secondary voltages available to the load. Delta transformers are schematically drawn in a triangle. The voltages across each winding of the delta triangle represents one phase of a three phase system. The voltage is always the same between any two wires. A single phase (L1 to L2) can be used to supply single phase loads. All three phases are used to supply three phase loads. L1
480 Volts
480 Volts
480 Volts
L2 480 Volts 480 Volts
L3
L1 to L2 = 480 volts L2 to L3 = 480 volts L1 to L3 = 480 volts
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Balanced Delta Current
When current is the same in all three coils, it is said to be balanced. In each phase, current has two paths to follow. For example, current flowing from L1 to the connection point at the top of the delta can flow down through one coil to L2, and down through another coil to L3. When current is balanced, coil current is 58% of the line current measured on each phase. If the line current is 50 amps on each phase, coil current would be 29 amps.
29 Amps
50 Amps
L1
50 Amps
L3
29 Amps
50 Amps
L2
29 Amps
Unbalanced Delta Current
When current is different in all three coils, it is unbalanced. The following diagram depicts an unbalanced system. 43.6 Amps Coil A 30 Amps
L1
Coil B 20 Amps
26.4 Amps
L2
Coil C 10 Amps
36 Amps
L3
Though current is usually measured with an ammeter, line cur rent of an unbalanced delta transformer can be calculated with the following formulas: IL1 =
IA2 + IB2 + (IA x IB)
IL2 =
IB2 + IC2 + (IB x IC)
IL3 =
IA2 + IC2 + (IA x IC)
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Wye Connections
The wye connection is also known as a star connection. Three transformers are connected to form a “Y” shape. The wye transformer secondary, (shown below) has four leads, three phase connectors, and one neutral. The voltage across any phase (line-to-neutral) will always be less than the line-to-line voltage. The line-to-line voltage is 1.732 times the line-to-neutral voltage. In the circuit below, line-to-neutral voltage is 277 volts. Line-to-line voltage will be 480 volts (277 x 1.732). L1
480 Volts 277 Volts L2
277 Volts
N
480 Volts
480 Volts
277 Volts
L3
Review 11
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1.
If the primary of a transformer
