Simple Solutions to Energy Calculations

SIMPLE SOLUTIONS TO

ENERGY CALCULATIONS FOURTH EDITION

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SIMPLE SOLUTIONS TO

ENERGY CALCULATIONS FOURTH EDITION

RICHARD VAILLENCOURT

Library of Congress Cataloging-in-Publication Data Vaillencourt, Richard, 1951Simple solutions to energy calculations / Richard Vaillencourt. -- 4th ed. p. cm. Includes index. ISBN 0-88173-575-2 (alk. paper) -- ISBN 0-88173-576-0 (electronic) -- ISBN 1-4200-7327-3 (distributor Taylor & Francis : alk. paper) 1. Energy conservation. 2. Energy auditing--Mathematics. 3. Energy auditing. 4. Engineering mathematics--Formulae. I. Title. TJ163.3.V35 2008 696--dc22 2007037433 Simple solutions to energy calculations 4th edition / Richard Vaillencourt. ©2008 by The Fairmont Press. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the publisher. Published by The Fairmont Press, Inc. 700 Indian Trail Lilburn, GA 30047 tel: 770-925-9388; fax: 770-381-9865 http://www.fairmontpress.com Distributed by Taylor & Francis Ltd. 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487, USA E-mail: [email protected] Distributed by Taylor & Francis Ltd. 23-25 Blades Court Deodar Road London SW15 2NU, UK E-mail: [email protected]

Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 0-88173-575-2 (The Fairmont Press, Inc.) 1-4200-7327-3 (Taylor & Francis Ltd.) While every effort is made to provide dependable information, the publisher, authors, and editors cannot be held responsible for any errors or omissions.

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DEDICATION This book is dedicated to Mr. Harold Gorman, PE— the man who taught me how to think like an engineer. For me, he is truly “the leader of the band.”

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TABLE

OF

CONTENTS

Introduction ............................................................................ ix Chapter 1:

The Walk Through Audit................................1

Chapter 2:

Simple Thermodynamics...............................25

Chapter 3:

Lighting ............................................................47

Chapter 4:

Pumps ...............................................................79

Chapter 5:

Fans .................................................................117

Chapter 6:

Motors .............................................................127

Chapter 7:

Insulation........................................................135

Chapter 8:

Fuel Switching...............................................147

Chapter 9:

Heat Recovery ...............................................157

Chapter 10: Heating, Venting, and Air Conditioning ..........................................167 Chapter 11: Summary of Calculations............................191 Chapter 12: Energy Myths and Magic: An Examination of Conservation Measures and Calculations .........................199 Richard’s Retrofit Rules .......................................................211 Index .......................................................................................217

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INTRODUCTION Most energy conservation projects are implemented only after passing through several levels of analysis and decision making. It is obvious that it is not cost effective to spend a lot of time on the initial levels in this process. That does not mean that the information gathered and methods of analysis do not require accuracy and technical correctness. On the contrary, there is no time that it is acceptable to use inaccurate or technically flawed data or methods. On the other hand, it is foolhardy to waste too much time trying to nail things down to the last decimal place when an analysis based on the information that you already have would indicate that the measure is economically unacceptable, or so good that it’s a sure thing. Since I must admit that I am undeniably slow and lazy, I wanted some tools to quickly and easily answer the question: “Is this project worth spending time and money to get detailed answers?” I sure didn’t want to pay someone to tell me that my idea was ridiculous. Worse yet, I didn’t want my boss to think that I couldn’t figure it out myself. Some energy evaluations are quite easy. Some others are extremely difficult, but mostly unnecessary at this stage. The engineers who write the energy auditing handbooks seldom take the trouble to show you how easy the easy evaluations are. They seem to desire to impress you with their understanding of the difficult ones by going into great detail whether you want to or not. To solve my dilemma, I developed several “plug ix

and chug” computer spreadsheets for various energy evaluations that seemed to occur over and over. Lighting, occupancy sensors, variable speed drives for fans and pumps, etc., are some of the areas that are common to any facility. My simple solutions are presented here, hopefully, to make your life easier. Perhaps you need to do this type of analysis often enough that you can develop your own spreadsheet specific to your needs. I offer them to all the other slow and lazy engineers out there so that you can stay under cover. I also admit to having a high degree of anxiety about going public with my methods. I am sure that there are many very smart engineers who will attack this book on the grounds that it commits several serious errors. First, its goal is to answer only one question. Namely, “Should I drop this project or look deeper?” If you asked them (at $200/hour) they would feel obligated to deliver several pages of calculations to justify their fee. Second, they will complain that I made almost no attempt to justify, derive, or demonstrate with rigid completeness the validity of my methods. I told you I was lazy. If you use them and they seem to work and you want to know why—go for it! The proof is left to the student. Finally, they will say that the solutions are simple because I left the complicated things out. You bet I did! I told you I was slow. That is why I wrote this book. I believe that you can grasp the significance of a project’s worth and outline its limits and feasibility without being forced to work through the intricate, unnecessary, and mostly irrelevant engineering gynmastics so faithfully performed by impractical engineers. Proficiency at doing x

the unnecessary does not increase productivity. The only warning is to remind you, again, that the purpose of this approach is to provide a decision making tool to help you to decide whether to invest real time and money into developing the details of a project. Don’t bet your job on the exact value of the savings calculated from these equations. Your recommendation should be: 1) no, this won’t pay; or, 2) yes, this looks good so far and should be the target of an in-depth study.

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CHAPTER 1

THE WALK THROUGH AUDIT INTRODUCTION The saying goes that a journey of a thousand miles starts with the first step. The same applies to the journey to a completed Energy Conservation Measure (ECM). The first step is to have an idea. There are many ways to get any particular ECM idea. You can read about someone else doing it. Or you can have a vendor (who is trying to sell something) point it out to you. Or you can have your wife (spouse) say, “Hey! Look, stupid! If you put a variable speed drive on your cooling tower fan you can save a lot of money.” Most of the time your ideas will come from walking through the facility with your eyes and ears and mind open. This is called a “Walk Through Audit.” Why is it called an audit? Why not a “Survey” or “Analysis” or some other engineering term? Because the federal government created this term. Spoken like true accountants, they were hoping to create more accountants out of engineers. The original thrust of the activity that they were trying to define (and therefore regulate) involved balancing (auditing) the energy books. As accountants, they feel that it is important to identify the energy (cash) flow. They can’t help it. Their need to balance the books is either implanted 1

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into their psyche in college or is the genetic defect that caused them to become accountants. Whatever the cause, the demand for a one page, all encompassing view is in sharp contrast to the engineering approach of solving a large problem by solving all its smaller parts first. There is an argument that says that without the “whole picture” you will run the danger of wasting your opportunities. This is absurd. If the evaluation indicates that you will save 10,000 Btu or kWh with a two-year payback, that is real energy and real savings. It does not matter if the 10,000 units were a small part of a major consumer or the entire consumption of a smaller consumer. The dollars saved and the dollars spent are absolute numbers. If it is reliable and cost effective, do it. Stop wasting time trying to find “The Perfect Project.” Think of the money you saved by not spending the time on the overview. Anyway, using the accounting term makes it easier to get them to read the report. Since the accountants control the money it’s better to humor them. Most walk through audits are not formal affairs. The really good plant engineers are doing it when they walk through the plant to do their normal work. They are always looking for a better way to get something done. Not just energy, but: storage, material handling, purchasing supplies, water, sewage, pollution control, hazardous waste management, maintenance, installations, demolitions, etc., etc., etc. This informal, but very effective walk through audit can produce some very significant results because a good plant engineer knows where the skeletons are buried.

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However, bear in mind that this productivity can be improved by a formal walk through audit. If you take the time to concentrate only on energy utilization and to record your observations you will start to see the individual trees in the forest in front of you. Be careful. Sometimes the forest turns out to be a murky swamp. This productivity can be amplified if the walk through audit is performed by an experienced energy professional. Why? Because the intimate knowledge of the facility in the brain of the plant engineer can be tapped by another engineer and combined with his specialized training and experience to develop a larger list of potential projects than either engineer can do by himself. A walk through energy audit consists of two distinct operations. The first is to determine with as much accuracy as is possible exactly what the existing conditions are. This is accomplished during the walk-through stage. This is most difficult and most important. If you incorrectly establish that a certain piece of equipment utilizes several times more energy than it really uses, you will just as incorrectly calculate a larger savings than is possible. The extreme case (that I have seen too often in reports) is that you will calculate energy savings that exceed the annual consumption of the entire facility. The second operation is defining and evaluating (ECM). That will be addressed later in this book. The walk through audit report is meant to be a first cut report only. The idea is to identify measures that appear to be worth a detailed evaluation. Many of these measures will not pass a second cut. Quite often due to feasibility or product quality conflicts, not economics.

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Your job is to point out the opportunity, ballpark the potential, and make them tell you “why not.” You have accomplished your goal. The measure has not been overlooked. It has been evaluated and discarded for (hopefully) logical reasons. The limitations of a walk-through audit should be clearly understood. You cannot completely understand any complicated processes and even most simple ones as they apply in any specific application. However, the process auxiliaries will, in most cases, be standard applications of pumps, fans, cooling (air and water), lighting, etc. The goal is to suggest ways to provide these functions in such a way that the process is “unaware” of a change, yet the operating expense is reduced and reliability is unchanged, or improved. This approach will not produce detailed plans and specifications. The goal is to point out the possibilities with enough detail to get a feel for the difficulties, and cost and savings estimates reasonable enough to indicate which ideas deserve an in-depth study. The best and most often used tool you can have is common sense. You don’t need a college degree, but a certificate from the school of hard knocks will help. You must always ask yourself two questions: 1) Will there really be a savings? and, 2) Will many people complain about the change? A dictionary definition of the word “engineering” is: “1. the science of making practical application of pure sciences.” Similarly, the word “art” has, as two of its meanings: “1. the principles governing any craft or skill. 2. skilled workmanship or execution.” We, as energy en-

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gineers, must strive to employ the practical application of our training and experience, with a major emphasis on being practical. Why? Because in the world of building managers and production managers energy conservation has been a synonym for suffering. As energy engineers we really haven’t done much to dispel these fears. In many cases we have earned the reputation for going off half-cocked. A reputation for squeezing off the flow of Btus with little regard for anything else. To counter this we must continuously learn the “principles governing our craft.” And since many of these principles affect non-engineering facts of life, we must apply them skillfully. We must know when they are important enough to push for and when to back off. This is art. The elements of style, finesse, subtlety, etc., must be constantly balanced with facts and reality. Engineering realities, financial realities, cultural realities, and the most frustrating of all: human nature. Have you ever wondered why a production supervisor is not pleased to see you? And if not openly hostile, he is at best uncooperative. That’s because Energy Management means changing his procedures, retraining his men, and giving him more paperwork. He is a successful production supervisor because he produces. Why should he go out of his way to do something that will only make you look good? And give him more headaches. What’s in it for him? It’s all right to tell him about making the company more competitive or profitable, and I’m sure he agrees with those goals. But my experience has shown me that

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in reality his next raise or promotion depends entirely upon whether he gets the product out on time without mistakes. What does that have to do with energy management? Nothing. That’s your problem, not his. Your next raise or promotion depends on how much you reduce the company’s energy bills. Nobody comes to you and asks you why orders are not being filled on time. “We should not expect to utilize in practice all the motive power of combustibles. The attempts made to attain this result would be far more hurtful than useful if they caused other important considerations to be neglected. The economy of the combustible is only one of the conditions to be fulfilled in heat engines. In many cases it is only secondary. It should often give precedence to safety, to strength to the durability of the engine, to the small space which it must occupy, to small cost of installation, etc. Sadi Carnot—1824 To put it in another perspective, let’s look at the two ways of making a baby. By far, the most efficient and reliable method is to mix egg and sperm in a test tube then implant the resulting gamete in the “appropriate environment” for development. This is much more reliable than the random biological chance of the egg and sperm meeting while coincidentally already in the “appropriate environment” at the same time. It certainly uses a lot less energy on everybody’s part. And, if you throw in the cost of several dates, an engagement ring, the wedding ring,

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the wedding, the reception and the honeymoon (I realize that this is the old-fashioned approach), the test tube method is much more cost effective. Personally, I refuse to give up the biological method. As a matter of fact, I have been known to expend all that time and energy in a biological attempt that I sincerely hoped would fail! So the first question: “Would there be any real savings?” is answered with a resounding, “NO!” Laboratories can rebate, government can mandate and give tax credits, and I still will want to use the biological approach. Salesmen can offer me trips to a tropical paradise as an incentive, and that will only confuse me because I would want to go there only in the hope of increasing my biological attempts. So the answer to the second question, “Will many people complain about the change?” is affirmative, and usually unprintable. What does this have to do with a walk through audit? Everything. To get the plant engineer on your side you must state clearly, and more importantly, you must believe that efficiency is not the only parameter that merits your attention. You only have a few hours to spend with the plant engineer. You need him to keep you out of trouble. You need him on your side. You need to convince him that you are not blind to his needs and are not going to make a name for yourself without any care for his problems. By far, the hardest thing to overcome when dealing with a client plant engineer is that he is, in many cases, defensive. This is completely understandable. First: He lives, breathes, eats and sleeps with his facility. Whether he wants them to, or not, the plant personnel will find him when they need him.

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Simple Solutions to Energy Calculations

Second: As I mentioned above, he is always looking for a better way to get something done. Most of the time some “bean counter” is pushing an idea that he read about in Popular Science. Having been there, I know how you can start with a chip on your shoulder. All it takes is for someone to come in off the street, make a proposal to some production manager (that is the same thing you have been talking about for two years) that gets approved in two weeks, and you can get bitter. Third: There is no way that anyone off the street can know his plant and processes better than him. Fourth: He has the attitude: If there was a better way to do it, I would have done it already. These attitudes are very difficult to avoid developing if you are the plant engineer, and very difficult to overcome if you are the one coming in “ off the street.” As a conservation engineer, remember that it is his plant and he has dealt with a lot of “snake oil salesmen”: They waste his time and annoy the daylights out of him. Admit immediately that you know he knows more than you about his plant. But try and point out that he shouldn’t be expected to know everything about the latest technology. Maybe you do have something to offer. Together, you might be able to sort out what will, or will not, work for him. The hardest thing for an energy auditor to deal with is the wrong information. There is no way (see #3, above) that you can tell that it’s wrong. If you’re smart, you will take a long time to describe what you think will happen, or needs to happen if a certain ECM is going to work. Hopefully, in this discussion, if it is a real dialogue,

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enough information will be passed between the two engineers so that both the auditor and the plant engineer know enough to keep each other out of trouble. That is the essence of this last discussion. KEEP EACH OTHER OUT OF TROUBLE. If this attitude is made clear to the customer you will be seen as an asset. Your goals are the same as his. You are there to present options. To improve the profitability and the productivity. You are really there to help him. If you’re a vendor and you feel that the client really doesn’t know the answers to important questions, don’t stop asking and hope everything will turn out right. If it’s important enough, an obvious wrong answer might lead you to the conclusion that the best course of action is to walk away from the project. Give the plant engineer full power of veto! You may think that you know what will work but the plant engineer knows what will be accepted. If he indicates that a certain measure is not worth it to him, by all means try to find out why and even try to change his mind. But if he still insists—drop it! I can’t say this strongly enough! You don’t have to know or accept his reasons. You don’t have to agree with him. You do have to give him that power. It may be that he has been trying to get funding for the exact same ECM for the last five years and is just plain tired of fighting (and losing) over it. Or, it may be that he thinks that it is absolutely brilliant. So brilliant that he wants you to leave it out so that he can suggest it after you’re gone. It doesn’t matter. If he wants it left out you won’t be able to get around him anyway so why make an enemy? If

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you push the plant engineer unnecessarily I guarantee the project is doomed to failure. If you prove to him that you are on his side and he has veto power, you will get the first projects implemented. Now you can build from this success to push for the ones you held back before, if you still want to. It is my recommendation that you purchase, develop, copy, or steal an audit checklist form. You only have a short time to perform the audit and usually no access for a second visit to get what you missed. Trust me. You will not get everything. But with a good form to prompt you, you won’t forget the obvious things. Find a form and always be ready to modify it. It doesn’t matter how many pages you carry. Obviously many of the pages will not apply to the plant that you are doing. If you have sections designed for a sewage treatment plant you won’t use them in a grocery store. But the fact that you note “N/A” where appropriate under the “gas fuel” section will give you the confidence that you did not forget it entirely. The last page should be a fist of questions left unanswered when you left, but promised to be answered by plant personnel. A copy of this page with a delivery date should be left with the person at the plant responsible for getting the answers to you. Keep this list short! Usually this consists of energy consumption records and costs, specific equipment operating times and strategies, horsepower of inaccessible motors, etc. Try very hard to keep it to information that you would like to have. And also try to limit the questions to ones that the plant engineer can delegate and

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nag someone else about. If he has to bug the accounting department for the fuel and electric bills he will enjoy it so much that you can be sure it will happen. It is rare for him to have the accountants owe him some information. The momentary feeling of power is intoxicating. If there is a piece of information that you feel that you need to do your work, try very hard to get it while you are there. Many times the best intentions of the plant personnel are forgotten when they are faced with putting out daily fires. If the information is necessary for a complete and accurate report, you may be “on hold” for a long time. It is best to be ready to make an assumption (spelled “g-u-e-s-s”) if the information is not forthcoming. At this stage assumptions are acceptable if they are clearly identified in the report as your best guess. It is much better to produce a list of ECM with a potential, but known flaw in it, than to wait indefinitely for details. I suggest that you avoid making recommendations for production processes unless you are specifically asked. It doesn’t matter that you were the Corporate Energy Manager for a textile company for over five years. All that means is that you can easily recognize textile machinery and its basic function. I know that this is a difficult restraint because the really big savings are in production conservation measures, not to mention the most fun. But you have to walk before you can run. Do a really good job on the easy stuff and you might get the chance to have some real fun. Screw up on a lighting measure and there will be a shoot-on-site order put out on you. I usually ask if there is any “pet project” that they would like my help in getting evaluated and approved.

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Simple Solutions to Energy Calculations

After about two years as a Corporate Energy Manager I had pretty much accomplished the easy stuff. The “no brainer” projects with major savings and almost no capital. Now I wanted to tackle some of the more interesting ideas that were a little more risky and certainly more expensive. At the very least, they required a lot of thinking. It was very frustrating to send report after report to what I came to call “the Black Hole”: my boss’ desk. Nothing ever came back from there. Not even acknowledgment that he received them. It got to the point where I was asking a vendor that I trusted to take my evaluations and recommendations and submit them on his letterhead. Quite often that worked. If the Outside Consultant Syndrome exists at your plant, you might as well use it! So, as an Outside Consultant, I offer to evaluate the energy part of some in house pet project. It is important to emphasize that the production part is still the responsibility of the plant personnel. Your role is to determine how much energy, or energy dollars, can be saved if this project will work without affecting production or quality. What you should normally look at is all the auxiliary equipment and functions that support the process. Specifically, cooling water, heating method, exhaust, fume/dust collection, etc. The goal is to provide all these services for the least cost such that the process never experiences a change in its ability to function. I used the term “least cost” on purpose. In order to be useful to the plant, your goal must be to save money. Certainly you are concentrating on the energy consumption. But the reason you are looking at the energy consumption

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is to reduce its cost to the operation as a whole. If energy didn’t cost anything, you wouldn’t be reading this book. There would be no need to conserve. As soon as it became expensive, the term Energy Manager was invented. My own experiences with failed conservation measures taught me an important lesson: Most of the failures occurred when I went for every last Btu. That’s when more than one person noticed that I changed something. That meant that they could talk to each other and complain as a group. Actually, I have to admit that I have gone just a bit too far once or twice. That’s when performance fell into the “marginal” or “unsatisfactory” category (not enough Btuh in or out to meet the peak conditions, etc.). The surprise was that the entire project was considered a failure and discarded! I was never asked to just back off a little. It always went back to what was there before. The goal of the previous discussion is to help an outside consultant develop a style and attitude that will encourage cooperation and support from inside the plant. Such support will improve the accuracy and the volume of information that you can acquire in the short time provided. In house engineers can use many of the elements on colleagues and plant personnel. I know that it takes a lot of courage and personal restraint to go to a production supervisor and let him know that you are looking at changing something in his little kingdom. He will invariably start off with, “You want to do what?!” followed by either a loud laugh or a loud groan. Strangling him will only give you a momentary pleasure. And it will look bad on your resume.

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His second reaction will be, “You can’t do that because (fill in the blank) 1. 2. 3. 4. 5. 6. 7.

it’ll never work we tried that before you came and it didn’t work it’s too complicated for the operators it’ll ruin quality the tenants/customers won’t like it it’ll ruin our sales the boss/owner won’t like it

Any one of the above is a valid reason to walk away from a project. The problem is that they are said without regard for the truth only to get you to walk away. They are thrown at you like hand grenades that you have to throw back (prove that they aren’t true) before they explode. The good news is that, like a hand grenade, if they are thrown back fast enough they explode on the other guy. Once he has been blown up by his own grenade he has to cooperate with you or be identified as an obstructionist. Or worse in corporate circles, he’ll be labeled: “Not A Team Player.” At all times, whenever possible, for as long as possible, talk to the front line people. The people who run the machines, open and close the valves, turn the lights on and off, set the thermostats, change the light bulbs, etc., etc. You will certainly run the risk of meeting the plant Bullwinkle (Mr. Know-It-All) and the plant Archie Bunker (everything’s being done wrong). But the vast majority of these people know what’s going on for real. They also know when something can be improved, and when it can’t. They will be very happy to tell you exactly what’s

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on their mind. Most of the time it’s information that is worth taking the time to get. As you perform your calculations, you start to walk a fine line between engineering and opinion. The final numbers are only as good as your assumptions. The best approach is to take the attitude: “This is what I thought I saw and what I thought I heard. If these assumptions are reasonably correct then the report is reasonably correct.” Making your assumptions is like playing “The Price Is Right.” You should always be trying to get as close as possible to the actual facts without going over. The best assumptions are the ones that are honest and slightly conservative. The idea is to keep from overstating the possible savings while at the same time avoid grossly understating the savings to such a point that a promising opportunity is discarded. If at all possible, get the plant engineer or the front line operator to make the assumptions for you. Most of the time they will have a better idea about reality as it exists in their world than you will. More importantly, they won’t be able to accuse you of influencing the outcome in your favor. (I have never been able to figure out how I would benefit from suggesting an ECM, but I have been accused of somehow acting in my own self-interest.) By all means try to keep them honest. If they are making assumptions that are outrageous gently bring them back to reality. I have seen attempts to justify an ECM by assuming the annual run time was 9,000 hours per year. There is no way to assume that there are more than 8,760 hours unless it’s leap year or you are a lawyer billing a client.

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UTILITY COSTS There is a lot of wasted effort put into determining the “correct” utility costs to use in a calculation. Much of this is the result of confusing what is needed for energy management and what is needed to evaluate a retrofit project. The fine details of ON PEAK, OFF PEAK, SHOULDER, etc. rate structures are necessary to understand the effects of day to day operations. When you are an “in house” energy manager it should be your job to know these details and know them well. But to evaluate a retrofit project identified in a walk through audit all you need is an understanding of the seasonal rates and whether the project has a seasonal operating schedule. After that, all you need to determine is the facility’s average incremental costs for demand and consumption. I know this is heresy. It always amazes me that the cost of oil and natural gas can fluctuate every two weeks and it is perfectly acceptable to use the annual average cost. It is even acceptable to use an annual weighted average when there is dual fuel capability that allows the facility to switch to the cheapest fuel several times a year. But when it comes to electricity suddenly it becomes necessary to do an hourly analysis. Not only is there this double standard, but what does the extra effort get you or the customer? If an evaluation is performed using the billing charges for peak demand and the average kWh costs, and done a second time paying careful attention to the various kWh adjustments and time of day variations, the differences will not be significant enough to mention. I have made

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this comparison and found the differences to be less than 5%. And who knows which one is right?

CALCULATIONS The energy cost calculations need to be done before any measure can be evaluated. These are usually simple averages. If you are contemplating a measure that truly is used seasonally, such as most air conditioning at a school, it will become necessary to calculate the seasonal energy costs. The method is the same, you just have to perform the calculations for discrete periods of the year. In all cases, you start with at least twelve consecutive months of energy bills.

Incremental Electrical Demand (W) Charges

Most rates today are not declining block rates. Therefore, the incremental demand charge is the charge shown on the bill divided by the peak demand shown on the bill. Otherwise, if you have a copy of the utility rate structure, the demand charge will be clearly shown there. But beware, rates change very often these days. If you don’t have an up to date copy of the rate you will probably be calculating lower costs. Most utilities only charge for the peak demand that occurs during their peak usage period. If you are evaluating a measure that will reduce the demand during their off peak usage period, i.e., nighttime parking lot lights, it will not reduce the customer’s demand charges. The cus-

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tomer probably isn’t paying any demand charges for the use of those fixtures. Always be careful that you are not claiming a reduction in a charge that doesn’t exist. Some people might think that you don’t know what you are doing. Others that believed your numbers will be disappointed when the savings don’t show up.

Incremental Electrical Energy (kWh) Charges

Once you have determined the proper demand charge and the demand billed for a certain month, it is easy to determine the incremental cost per kWh. Simply subtract the demand charge from the total charges. The remaining dollars can be divided by the total kWh to give you a cost per kWh. This cost will automatically include all surcharges, credits, taxes, etc. that are attributed to the energy consumption. Perform this calculation for every month and average the twelve values and you will have a good historical annual cost per kWh for your savings calculations. The computer spreadsheet on the following pages shows you one format for an approach that requires only data entry from the electric bills and rates.

Incremental Fuel Costs

Fuel units of consumption should be converted to Btu, or therms or MMBtu (million Btu). This will allow all sources of fuel to be averaged to develop a cost for the heat value purchased. The total cost per Btu will reflect the real costs relevant to the way the facility historically purchases fuel. If you also determine the various losses in

BILLING DATA

MONTH/YR. —————— JAN. ‘93 FEB. ‘93 MAR. ‘93 APR. ‘93 MAY. ‘93 JUN. ‘93 JUL. ‘93 AUG. ‘93 SEPT. ‘93 OCT. ‘93 NOV. ‘93 DEC. ‘93

DEMAND RATE ($) ———— $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00

SUMMARY DATA

0.0 $0 0

kW kWh

DEMAND COST ($) ———— $0 $0 $0 $0 $0 $0 $0 $0 $0 $0 $0 $0

TOTAL USAGE (kWh) ——— 0 0 0 0 0 0 0 0 0 0 0 0

TOTAL COST ($) ——— $0 $0 $0 $0 $0 $0 $0 $0 $0 $0 $0 $0

TOTAL BILL GROSS AVG. $/kWh INCREMENTAL $ PER kWh

kWh COST ($)/kWh ———— $0.0000 $0.0000 $0.0000 $0.0000 $0.0000 $0.0000 $0.0000 $0.0000 $0.0000 $0.0000 $0.0000 $0.0000 $0 $0.0000 $0.0000

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ANNUAL PEAK DEMAND DEMAND COST kWh USAGE

BILLED DEMAND (kW) ———— 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

The Walk Through Audit

CUSTOMER NAME

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Simple Solutions to Energy Calculations

EXAMPLES

AND

CALCULATIONS

ABC, INC. BILLING DATA DEMAND BILLED DEMAND RATE DEMAND COST ($)

(kWh)

TOTAL COST

MONTH/YR.

($)

——— 1,251.2 1,209.6 1,152.0 1,180.8 1,337.6

——— $11,261 $10,886 $10,368 $10,627 $12,038

——— 393,600 499,200 374,400 435,200 460,800

——— $36,900 $43,345 $30,563 $37,109 $43,951

———— $0.0651 $0.0650 $0.0539 $0.0609 $0.0693

JUN. JUL. AUG. SEP. OCT. NOV. DEC.

$9.00 $9.00 $9.00 $9.00 $9.00 $9.00 $9.00

1,513.6 1,632.0 1,561.6 1,638.4 1,465.6 1,264.0 1,289.6

$13,622 $14,688 $14,054 $14,746 $13,190 $11,376 $11,606

534,400 412,800 534,400 566,400 502,400 473,600 460,800

$45,315 $38,636 $46,120 $49,372 $45,402 $42,966 $40,407

$0.0593 $0.0580 $0.0600 $0.0611 $0.0641 $0.0667 $0.0625

SUMMARY DATA ANNUAL PEAK DEMAND DEMAND COST kWh USAGE TOTAL BILL GROSS AVG. $/kWh INCREMENTAL $ PER kWh (a) (b) (c)

1,638.4 $148,464 5,648,000 $500,086 $0.0885 $0.0622

kW kWh (b) (c)

(Total Cost – Demand Cost) ÷ Total kWh Total Cost ÷ Total kWh Average Monthly “kWh Cost” {Σ Col. (a) ÷ 12 = $/kWh}

($)

(a) kWh COST

—————— ———— JAN. ‘92 $9.00 FEB. ‘92 $9.00 MAR. ‘92 $9.00 APR. ‘92 $9.00 MAY. ‘92 $9.00 ‘92 ‘92 ‘92 ‘92 ‘92 ‘92 ‘92

(kW)

TOTAL USAGE

($)/kWh

The Walk Through Audit

21

combustion of each fuel and losses in the distribution of the thermal fluid system (the “end use efficiency”), you can develop an understanding of which fuel, including electricity, is the best choice for the specific application. The following computer spreadsheets show you one format for an approach that requires only data entry from the fuel bills.

ABC, INC. #6 FUEL OIL MONTH/YR.

JAN. FEB. MAR. APR. MAY. JUN. JUL. AUG. SEP. OCT. NOV. DEC.

‘91 ‘91 ‘91 ‘91 ‘91 ‘91 ‘91 ‘91 ‘91 ‘91 ‘91 ‘91

TOTALS

GALLONS DELIVERED

TOTAL COST

0 0 0 0 0 0 0 0 0 0 0 0

$0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00

0

$0.00

(b) 0.15 MMBtu/Gal COST PER MMBtu ($/MMBtu)

$0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00

AVERAGE

(c)

$0.00

22

Simple Solutions to Energy Calculations

NATURAL GAS

MONTH/YR.

————— JAN. ‘91 FEB. ‘91 MAR. ‘91 APR. ‘91 MAY. ‘91 JUN. ‘91 JUL. ‘91 AUG. ‘91 SEP. ‘91 OCT. ‘91 NOV. ‘91 DEC. ‘91 TOTALS

USAGE (THERMS)

TOTAL COST

0

$0.00

———— 0 0 0 0 0 0 0 0 0 0 0 0

(b) 0.1 MMBtu/Therm COST PER MMBtu ($/MMBtu)

——— ————— $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 AVERAGE

(c)

$0.00

The Walk Through Audit

23

TOTAL FUEL; GAS & OIL

MONTH/YR.

————— JAN. ‘91 FEB. ‘91 MAR. ‘91 APR. ‘91 MAY. ‘91 JUN. ‘91 JUL. ‘91 AUG. ‘91 SEP. ‘91 OCT. ‘91 NOV. ‘91 DEC. ‘91 TOTALS

(b)

(a)

COST PER

OIL (GAL)

GAS

TOTAL

TOTAL

MMBtu

DELIVERED

(THERMS)

THERMS

COST

($/MMBtu)

————— 0 0 0 0 0 0 0 0 0 0 0 0 0.0

———— 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

———— 0 0 0 0 0 0 0 0 0 0 0 0

——— $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00

0.0

0

$0.00

———— $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00 $0.00

AVERAGE

(a) (Gallons × MMBtu/Gal × (1 Therm/0.1 MMBtu) + (Natural Gas Therms) (b) Total Cost ÷ (total Therms × (1 MMBtu/10 Therms)) (c) Average Monthly Fuel Costs {ΣCol. (b) ÷ 12 = $/MMBtu)

(c) $0.00

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CHAPTER 2

SIMPLE THERMODYNAMICS INTRODUCTION As far as using energy is concerned, thermodynamics can be boiled down to one simple rule. The neat thing about thermodynamics is that this rule cannot be broken. It cannot be negotiated. It cannot be bent. But once you understand it, energy conservation becomes relatively easy to spot. Here it is: ENERGY ALWAYS MOVES FROM HOT TO COLD You make something hot by putting something hotter next to it. The energy will move from the warm object to the cool object making the colder object warmer. You make something cold by putting something even colder next to it. It is the same thing as above: one object is warmer than the other. The colder object, in effect, sucks the energy from the warmer object, making the warmer object colder. The inside walls of your refrigerator are the coldest part. They suck the heat out of the air that’s in the refrigerator. The air is next to the bottle of milk that you put in the refrigerator and it sucks the heat out of the milk. The inside walls don’t get warmer because the energy that is sucked into them is constantly being sucked away into 25

26

Simple Solutions to Energy Calculations

the compressor and pushed out into the kitchen by way of the condenser coils on the back of the refrigerator. Notice that like water in two connected tanks, the energy tries to reach an equal level in the two objects. Also notice that the heat from the warm objects placed in the refrigerator does not disappear. It is simply (well, not so simply) moved from inside the refrigerator to outside and dumped into the kitchen. If your kitchen is air conditioned, then the heat will be picked up by the evaporator coil of your kitchen air conditioner and transferred to the outside world using the same thermodynamic rule. The following is an important corollary to the first rule: The greater the temperature difference (∆T) between the hot and cold sides of the transfer, the faster the energy moves from the hot to the cold. Looking at it from a more important angle: THE CLOSER TOGETHER BOTH TEMPERATURES ARE, THE LOWER THE ENERGY TRANSFER.

This is the blueprint for how to control the use of energy in your building. When looking at heating, ventilating, and air conditioning (HVAC), if you want to minimize the amount of energy to maintain the space temperature, try to keep the space temperature as close to the outdoor temperature as you can get away with. If you keep the inside temperature equal to the outside temperature, your heating and cooling energy consumption will actually disappear. (Unfortunately, so will your job.)

Simple Thermodynamics

27

That is why resetting thermostats closer to the outside temperature when no one is in the building (to complain) saves energy (a strategy called “night setback”). On the other hand, if you let the space temperature on the inside become significantly warmer in the winter (or colder in the summer) than the outside air temperature happens to be (either through calibration errors, or unrestrained occupant control, etc.), then the energy transfer through the walls, windows, and roof to the rest of the world is significantly greater. And remember, all that energy came through a meter with a bill attached. This is because all thermostats and other controls are trying to maintain equilibrium. That is, they are continually adjusting the energy input under their control such that it exactly equals the energy that is being lost from the space. Equilibrium means that the temperature in the space (or whatever you are trying to measure) does not change. When equilibrium is lost, the temperature will change. If the rate of energy (Btus per hour, or Btuh) leaving the space through conduction, convection, ventilation, etc. is greater than the Btuh that is being added, the temperature will drop. Conversely, if the Btuh entering the space through conduction, convection, ventilation, and internal heat gains is greater than the Btuh being removed by the cooling system, the temperature will rise. The rest of this chapter explores how this applies to various individual pieces of equipment or controls in your building operations or production equipment. Since we are focusing on energy reduction strategies, we will always be looking for the ∆T and how to make it small. This is your first step to reduce energy use! Certainly en-

28

Simple Solutions to Energy Calculations

ergy will be used intelligently when you employ the most efficient equipment to provide the equilibrium energy, but the cheapest Btu is the one that you don’t use.

REFRIGERATION In order to understand this simple law as it applies to refrigeration equipment, I feel that I need to provide a simple overview of the refrigeration cycle, the component equipment, and how it operates. First, you start with a tank of liquid refrigerant under pressure. A refrigerant is simply a liquid that has a low boiling point. Ammonia boils at about –25°F at atmospheric pressure. Refrigerant R-12 boils at about –20°F at atmospheric pressure. Therefore, if you have R-12 at about 100°F, the pressure has to be above 118 psig for it to be a liquid. It cannot exist as a liquid at 100°F if the pressure is below 118 psig. If you start with a tank of R-12 that is above 118 psig and less than 100°F, then allow the pressure to fall below 118 psig, the liquid will immediately evaporate (boil, flash to vapor, etc.). It will do so by extracting the energy needed to change phase from liquid to vapor (the latent heat of vaporization) from its surroundings. (See Figure 2-1) If you take a pipe (P) and connect this tank (A) of 100°F liquid R-12 at 118 psig to a finned tube radiator coil (C) at approximately 35 psig through a pressure-reducing, thermal expansion valve (B), it will immediately evaporate in the radiator coil at a temperature of 45°F. It will suck the necessary 100 Btu/lb from the air

Simple Thermodynamics

29

around the fins. The fins will eventually start to approach a temperature of 45°F. If you are blowing warm air from your room at 80°F through the fins, it will emerge from the other side at close to 55°F.

Figure 2-1. Refrigeration Cycle 1 Once the R-12 vapor has been evaporated, it must be discharged to make room for more liquid to be boiled (D). When the tank is empty, simply replace it with a full tank and keep going. However, if you want to limit your refrigerant expense, avoid the heavy fines plus jail time for willful discharge of a greenhouse gas, and avoid being labeled a heartless capitalist by your children for ruining their world, then we need to somehow close this cycle by capturing the Freon vapor and making it a high-pressure liquid again. This is done by adding a compressor and a condenser

30

Simple Solutions to Energy Calculations

to the system (see Figure 2-2). The compressor (E) takes the R-12 vapor from (D) and raises the pressure to the 118 psig we started with. It is still in a vapor form because it still contains the latent heat of vaporization extracted from the air stream that was cooled. This high-pressure vapor is delivered to a heat exchanger (G) that removes the absorbed heat of vaporization and thus condenses the R-12 back into a liquid at 118 psig where it is collected in the tank. The process is now continuous and good for the environment and (very important) legal. A very important fact is that temperature and pressure are inseparable. As pressure goes up, temperature goes up. As pressure goes down, temperature goes down. So what? Since you get nothing for nothing, the pertinent energy question is: where is energy being pumped into the system? Your cost is the energy to run the compressor. The compressor is doing all the work (aside from peripheral fans or pumps) to make this refrigeration cycle work. A simple fact is that the amount of energy required by any vapor compressor is directly related to the difference between the inlet pressure and the outlet pressure (∆P). Therefore, the amount of energy required by a refrigeration compressor is directly related to the difference between the evaporator coil pressure and the condenser coil pressure. The closer these two pressures are to each other, the smaller the ∆P, the lower the compressor horsepower. Since these pressures are related to their temperature, the closer you can keep the evaporator temperature to the

Simple Thermodynamics

31

condenser temperature, the lower the ∆P on the compressor resulting in a lower kW that your refrigeration equipment will run at. This goes for window air conditioners, packaged rooftop units, chillers of all sizes, etc. What does this mean in real life? Direct Expansion (D/X) A D/X coil is an evaporator that is in the media being cooled, i.e., the supply air stream to your room. Standard design values have the leaving air temperature (LAT) from the D/X coil at 55°F. The thermal expansion valve, the coil, the total quantity of refrigerant are all designed to produce this by ensuring that the internal pressure in the D/X coil will result in a 45°F temperature inside the coil. You are pretty much stuck with this. The only place that normally has some flexibility is the condenser. An air-cooled condenser is usually designed to operate in an environment with an entering air temperature up to 105°F. On a hot day, on a black roof, the air temperature at the air-cooled condenser can be close to that limit. If the condenser is cooled by outside air that starts at 90°F, the temperature inside the condenser must be higher than 90°F for the heat flow to be from inside the condenser coil to the outside air (see Rule #1). Therefore, the condenser pressure will be much higher if the surrounding air temperature is 90°F than when the air temperature is at 70°F. For instance: at 100°F, R-12 is at about 115 psig and at 45°F it is at 35 psig. Therefore the ∆P is 80 psi (115 – 35 = 80). At 70°F, R-12 is at about 70 psig and the resulting ∆P is lowered to 35 psi. Therefore, operating with a condenser

32

Simple Solutions to Energy Calculations

temperature of 100° will result in about a 23% increase in horsepower over operating the condenser at 70°. One way that a condenser gets hotter than it has to be is for the fins to be covered with dirt. Dirt is a great insulator. More importantly, the dirt clogs the fins greatly restricting the air flow which greatly reduces the heat transfer capability of the condenser coil. The temperature (pressure) then builds up in the condenser and the horsepower increases. This is easily fixed by cleaning the coils.

Chillers

Air-cooled chillers are susceptible to the same cleaning and ambient air temperature issues as with D/X systems. Water-cooled chillers are much more efficient because the condenser cooling water temperature can be maintained at a constant 85° temperature regardless of the ambient outside air dry-bulb temperature. Often it is possible to adjust your cooling tower control strategy to deliver colder water to the condenser to reduce the chiller horsepower requirements even further. But remember that you can have too much of a good thing. Most chillers have low-temperature safety controls to shut the chiller down if the condenser water falls below 70°F. The standard design for chillers is for the chilled water temperature (leaving water temperature or LWT) to be 45°F with a return water temperature (entering water temperature or EWT) of 55°F. The condenser design is usually 95°F LWT and 85°F EWT. The difference with chillers is that the evaporator

Simple Thermodynamics

33

temperature can also be adjusted. If the load on the chiller is relatively light and the air-handling units do not have to dehumidify (such as is the case in the winter), then the LWT can be allowed to go up. It is not unheard of for the LWT to be raised to 50°F, or even 55°F. This will increase the evaporator pressure which is on the suction side of the compressor and thus lower the ∆P. Obviously, if the evaporator temperature can be raised at the same time that the condenser temperature is lowered, the horsepower requirement will go down dramatically. In conclusion: with a refrigeration system, the important temperature difference is between the evaporator and the condenser. Keep those two temperatures close together and the energy consumption will be as low as it can be.

Boilers

When I speak of boilers, I am simply referring to your primary heat generator. Whether it is producing steam, hot water, or hot air, the same rules about the temperature difference apply. Whether it is using oil, gas, or electricity, the total consumption will be lowest when you find the primary temperature difference and keep it as low as possible. Energy efficiency is defined by the amount of usable Btus that you get from of the total Btus of fuel that you put in. (Remember: 100% is impossible except in rare, specific cases that only look like 100%. Such cases are beyond the scope of this book). For a boiler, the usable Btus are the ones that are transferred to the water. The “unusable” Btus

34

Simple Solutions to Energy Calculations

are the ones that go up the stack or are radiated through the walls of the boiler. I put the word unusable in quotes because while you may not get those Btus back, some of them are necessary to the operation of the boiler. With a heating system there are two conflicting interests. The first is to get the heat from the heating system into the room as fast as possible. The second is to minimize the amount of energy that is lost in the combustion process, the boiler standby losses, and the distribution system. These two interests are in conflict because of that thermodynamic corollary that says: The greater the ∆T, the faster heat is transferred. Inside a boiler the primary ∆T is between the fireside and the waterside. The goal is to have a high ∆T between the fireside and the waterside to rapidly move the energy from the products of combustion into the water where you want it. However, for the boiler as a whole, the primary ∆T is between the inlet combustion air temperature and the stack temperature (the temperature at the location where the flue gases leave the boiler). At the chimney end, you want the temperature of the flue gasses going up the chimney to be as close to the inlet combustion air temperature as you can (with certain restrictions). In the extreme case, if the flue gas temperature was the same as the inlet combustion air temperature, that would mean that all the energy from the fuel had been extracted and there would be no excess energy going up the stack. While this sounds like a desirable goal, it is not possible to achieve, and if you are using fuel oil, it would be damaging to the boiler and chimney. It is not possible because of that pesky first law. You

Simple Thermodynamics

35

need a temperature difference to cause heat to flow. If the flue gases are the same temperature as the water there can be no heat transfer. If the flue gases are at a lower temperature than the water then the water will be heating the flue gases. So what temperature do you look for? The industry standard rule of thumb is 150° above the process temperature. If you have a process steam temperature of 300° (53 psig), then the stack temperature “target” would be 450°. If your combustion air enters the boiler at 80°, then you would have a net stack temperature of 450° – 80° = 370°. If your steam temperature is 246° (13 psig), then the stack temperature target would be 396° and your net stack temperature would be 316°, or 15% lower. In addition, your boiler wall temperature would be 54° lower (18%). Therefore, lower process (steam, water, etc.) temperature means lower stack temperature means higher combustion efficiency and lower radiation losses. Please remember that this is just a target number. Less than 150° ∆T will mean higher combustion efficiency. More than 150° ∆T will mean lower combustion efficiency, but that may be the best that you boiler can achieve because of the way it was built, etc. But if your stack temperature appears to be slowly rising, then you have some operating problems that are lowering your boiler efficiency, but can be corrected. Once again, you can have too much of a good thing! If you are burning oil, you must ensure that your stack gases never drop below the acid dewpoint. If that occurs

36

Simple Solutions to Energy Calculations

anywhere inside your system, hot sulfuric acid will condense on the metal surfaces. This is not a good thing! Meanwhile, back inside the room, the radiators were designed for a particular ∆T and will operate with much slower heat transfer rates if you lower the water temperature closer to the room air temperature. This means that the response time to a thermostat’s request for heat will be slower. So when it comes to the ∆T between the radiator fins and the room air, a higher ∆T means that the radiator will work much better. However, the higher temperature is obtained by having the boiler and the piping system at a high temperature. The higher temperature increases the energy losses before the useful energy gets to the room. Temperature setback controls based on the outside air temperature work on this angle. When it is warm outside, the required rate that heat must be transferred into the room is considerably lower than the system was designed to handle. Even though lowering the temperature in the system will hinder its performance at the end-use, (i.e., how fast it can deliver the heat energy into the room) because the heat load requirements of the room are lower, the sluggish response may be all that is required. Meanwhile, because the boiler and distribution system are at a significantly lower temperature, the heat losses are also significantly lower. However, it must be considered that the lower performance capability at the end-use (radiators) may cause any particular room to have too slow a response to a local increase in the heating requirement. In most cases, this is

Simple Thermodynamics

37

not a problem, but be aware of it so that you can recognize it in case it becomes a problem.

Outside Air/Ventilation

Outside air is the most expensive air that you can use at your facility. It is expensive because it is rarely at the temperature that you want it to be. Therefore, you must apply heating or cooling energy to make it the correct temperature. Of course when it is at the right temperature, it is the least expensive air that you can use for exactly the same reason: you do not have to add energy to change its temperature. Unfortunately, outside air is very desirable when you consider worker productivity and indoor air quality which is why there are several regulations dictating the required volumes of outside air that you must bring in to your facility. The trick for intelligent use of energy is to know when to maximize, or minimize the volume of outside air depending on how close it is to the temperature that you need. So the important question is: what temperature do you need? The easiest answer is: the discharge air temperature setpoint of your air-handling unit because that is the ∆T that the heating and cooling system is working on. So if your DAT is 55oF to provide space cooling, then whenever the outside air is 55oF or less, you can use it for “free” cooling. This is the classic Economizer Cycle. Temperatures lower than 55° can be used because in most cases you can blend very cold outside air with the warm (+/– 78o) return air to achieve the required DAT.

38

Simple Solutions to Energy Calculations

But what about heating? It is very unlikely that there will be a time when outside air temperature can provide “free” heating. But where else can you find “free” heat? One common (and commonly overlooked) place is in your exhaust air. The air that is being exhausted from your building is the temperature of your return air; the upper 70s. If outside air is colder than the exhaust, then you have a ∆T that can be used to minimize the energy lost with the exhaust air. With a heat exchanger of various possible designs, the cold outside air can be preheated by the energy in the exhaust. Remember, you already paid to heat that air that you are throwing out. It may be cost-effective to recycle some of that heat to lower your cost to heat the outside air that is needed to replace the exhaust. This is an important point: every cubic foot of exhaust air will be replaced with outside air. You can only exhaust the volume of air that you return to the building. One way or another, the exhaust volume will get back into your building and become a load on your heating or cooling equipment. If you completely seal up a room and start an exhaust fan, the fan will eventually stop exhausting air. Even though it will still be running, the high negative pressure on the room side of the fan will eventually be so large that the fan will not be able to overcome the ∆P and air will stop moving. So pay attention to your exhausts. The only way that you can beat the laws of thermodynamics is by refusing to play. Shut off your exhaust fans when not needed. Reduce your exhaust volumes to the lowest possible unless the make-up air is the right temperature. Whenever possible,

Simple Thermodynamics

39

recycle the energy that you paid for in the exhaust stream and use it to pre-condition the make-up air.

Demand Controlled Ventilation

A common area of excess exhaust is one that is built into the system at the design stage. If there is a room built for a large number of people, then the design of the airhandling unit must accommodate the maximum number of occupants. For example: an auditorium at a high school may be designed to handle up to 300 occupants. Therefore the outside air capacity designed and built into the AHU must be large enough to meet the code requirements for 300 people. But the auditorium is actually fully occupied for maybe six times a year for four hours. Those twenty-four hours are the only times that the ventilation and exhaust volumes need to be at their design levels. However, since the AHU cannot tell how many people are in the room, whenever the AHU is operating, it probably is operating at its maximum ventilation design air volumes. By putting in a sensor that can determine approximately how much ventilation air is actually required, the volumes can be reduced to match the need, not the full design. This is called demand controlled ventilation (DCV). The sensor is usually a CO2 sensor. Regulations for design volumes are based on keeping the level of CO2 below 1,000 parts per million (PPM). CO2 is used as an indirect indicator of the number of people in the room, i.e. more people in the room make more CO2. The control strategy is to adjust the outside air dampers in response to the CO2 level in the room to maintain a

40

Simple Solutions to Energy Calculations

maximum setpoint. As the CO2 level increases, the outside air dampers open more, and vice versa. This strategy also works for gymnasiums, conference rooms, etc.

Pumps and Fans

Can you apply these thermodynamic principles to mechanical devices such as pumps and fans? Well, not directly. The connection that can be used it the fact that pumps and fans are used to transfer energy from one location to another. It should be clear that the greater the amount of energy that needs to be transferred, the larger the pump or fan has to be. (As well as the pipes and ducts, etc.) Therefore, if you can control your temperature differences to keep them at a minimum, then you can design your energy distribution system to operate at the same minimum. Unfortunately, in the real world designers first must design for the worst condition. Second; they must design for the lowest first cost. Finally; they must design to cover their butts. (Oops! Did I say that out loud?) The result is that pumps, fans, and distribution systems are usually oversized even for the peak condition, and definitely oversized for over 90% of the hours they are in use. This is not necessarily evil. Many times I have laid out a piping plan then calculated the expected pressure drop (at peak gpm) and used that value to design the pump station. When I inspected the final installation, I was convinced that the plumber was paid by the fitting. The extra fitting losses put the pump performance in jeopardy.

Simple Thermodynamics

41

Therefore, I now design with a healthy “safety factor.” The “safety” that I am talking about is the “safety” of my PE license. Obviously, if there is extra horsepower installed to cover the peak, the actual extra horsepower at the minimum load is quite large. With a VFD you can cause the operating pump and fan horsepower to closely match the actual load at any given time. In order to take advantage and evaluate this, you must identify the real thermodynamic loads, install sensors to measure it, and install control equipment to respond to the sensors and adjust your distribution horsepower. Knowing how to identify the minimum thermodynamic requirement for a space or process, i.e., what is the lowest ∆T that will meet my goals and keep me out of trouble, is the first step. And remember: this value could change throughout the day, month, and year. By investigating, recognizing, and fully understanding all the subtle, and not so subtle, nuances that come with your design problem you can identify the lowest ∆T necessary to satisfy your peak operating conditions. Armed with that information you can design for the lowest first cost installation by not grossly oversizing your equipment. After that, maintaining the lowest ∆T even as that value changes, with the necessary control equipment will ensure that your thermal energy consumption is as low as it possibly can be. And, finally, installing the necessary equipment to vary the energy used by your distribution system to match the now varying distribution requirements will complete the cost-saving trifecta.

42

Simple Solutions to Energy Calculations

Insulation

Wait a minute! Insulation is completely passive. Insulation does not “use” energy. The point is that you need to look for ∆Ts that control energy consumption wherever you can find them. If you want to stop losing energy from one location to another you must keep the temperature of both spaces equal to each other. Insulation can change the ∆T between the inside and outside temperatures. Insulation effectively keeps the surface temperature of an object very close to the temperature of the objects next to it. The surface temperature on each side of a piece of insulation is very close to the temperature of the object on that side. In Figure 2-3 the ∆T between the surface temperature of the pipe (which is reasonable to assume to be the same as the water) and the air is 70° and the heat loss is 95 Btuh/lineal foot. By adding 1” of pipe insulation (Figure 2-4), the outside surface temperature of the insulation is reduced to 87°. Therefore the effective ∆T between the pipe/insulation “system” and the air is reduced by 90% to only 7°. This is the ∆T that is ultimately driving the heat transfer from the hot metal of the pipe to the ambient air through conduction and convection. Therefore, the heat loss is re-

Figure 2-3. HW Pipe w/o Insulation

Simple Thermodynamics

43

duced to only 12 Btuh/lineal foot. Remember: insulation does not stop heat transfer. As long as there is a ∆T, heat will move through the material. If the water isn’t flowing through the pipe in the example above, then the water inside the pipe will eventually (given enough time) become the same temperature as the air around the insulation.

Figure 2-4. HW Pipe w/1” Thick Insulation Insulation can only slow the rate of heat transfer. But it slows it down a lot! This is a very good thing because it means that to maintain the temperature of the water inside the pipe you only have to add a little heat all the time instead of a lot of heat all the time.

Industrial Processes

Understanding these thermodynamic principles can mean very large savings in industrial process systems. One example: I have observed many cases where pro-

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cess cooling water systems use chilled water (45°F) when the actual cooling water temperature requirement is 70° or 80°F. A common example of this is a water-cooled air compressor. Most often this is done because the standard design for a cooling tower is to deliver water at 85°F on its worst design day. If the compressor needs 70° water, then the cooling tower will not deliver the necessary cooling under those conditions. But those conditions will occur for only a small fraction of the year. For most of the year, free cooling from a tower will be more than adequate. That extra 25°F ∆T developed from mechanical refrigeration increases the cost. The fact that you are using a refrigeration compressor to do the cooling increases the operating cost even more. And ironically, it usually is provided by a chiller that already uses a cooling tower! (Not to mention the fact that 45° cooling water can cause excess condensation inside the air compressor, damaging the internal components.) Finally, operating the compressed air cylinders at cooler temperatures will increase the total Btus that are rejected to the cooling water because the ∆T is greater. This further increases the load on the cooling system and increases the operating costs. So the first step is to lower the ∆T across the compressor cooling system by increasing the entering water temperature, and there is a bonus. By moving the temperature of the water entering the air compressor cooling system closer to the temperature of the water leaving the system it becomes possible to gain the savings of turning off your mechanical refrigeration equipment. Installing a separate cooling tower, or simply a sepa-

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rate loop from the existing chiller’s cooling tower to the air compressor, will significantly reduce the operating cost for this cooling water system. During the few hours that this will not satisfy the cooling requirement, chilled water can then be used. Even so, you should look to reduce the ∆T in this system by using the cooling tower to cool the process water as low as possible, then use a heat exchanger and the chilled water to cool it the rest of the way, but only to the required temperature (70° in this example).

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CHAPTER 3

LIGHTING INTRODUCTION I do not believe that I am a lighting expert. First as Plant Engineer, then as Corporate Energy Manager I have been exposed to many different options promoted by vendors. My decision as to whether to recommend implementation had to be based upon my own assessment of the validity. To be real honest, I had two main questions: 1) Would there really be a savings? 2) Will many people complain about the change? Now, as an energy conservation consultant, I have made many lighting retrofit recommendations. I found that to perform with integrity I had to answer the same two main questions: 1) Would there really be a savings? 2) Will many people complain about the change? Much to my dismay, I found myself evaluating the claims of my competitors. This happens when you lose a bid. It also happens when a customer trusts you enough to let you know that your competitor has proposed to “save more energy for less money” but the customer doesn’t believe it. He is faced with explaining to his boss why one is real and the other isn’t. In tough economic times this can be very difficult. Let’s not mince words, Plant Managers and Accountants 47

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want the least cost option—period. If you want to spend more money you are the one they question, not the nice vendor offering a cheaper deal. Every piece of technology has its place and can perform “as advertised”: The vendor that thinks that his one product will be the best in all cases needs to be discovered and discarded. Use him where it’s best. Use another where they are best. Or find a engineer that is not limited in his choices and force him to convince you that his choices are what’s best for you. New lighting products are capable of producing the best light possible for the energy consumed. But how much conservation is too much? When can a customer tell if the savings are too good to be true? The customer must rely upon the expertise of the vendor or engineer making the recommendations. Unfortunately, too often, the person making the recommendations has his own interests at heart and is willing to sacrifice the long term interests of the customer. The driving force for all energy conservation decisions is payback period… period. This necessarily brings up the subject of watts saved and burn time. Anyone who has tried to get an accurate determination of burn time knows that this is very difficult to get. The best solution is a negotiated settlement. The area of reduced watts is the real test of the vendor’s integrity. Anyone can promise a watt reduction. The question is whether the customer can be convinced that he will be satisfied with the light levels after the job is finished. The vendor is armed with a confusing bag of terminology and subjective reasoning to boggle the most analytical mind. If the customer knew

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what the vendor was talking about then the customer probably would have done that retrofit long before the vendor showed on the scene. Now mix in the confusing array of Utility Rebate Programs. Many of these programs are as good as they should be. They promote conservation and allow building owners to upgrade their facilities with the best technology. Unfortunately this means that they are considered money in the bank to vendors. Often the payment comes directly from the utility upon completion of the job. Certainly this accomplishes the favorable goal of removing the economic obstacle to conservation. However, it also removes the connection between customer satisfaction and payment. It is in the vendor’s best interest to maximize watts saved in order to assure that the annual savings are high enough to cover the net cost after the rebate. In other words, make the deal so sweet that it can’t be refused.

DEFINITIONS A wise (and remarkably unpretentious) college professor once told me that the only difference between a student and a Ph.D. is vocabulary. In order to make themselves indispensable, the typical lighting salesman will use a lot of lighting vocabulary. If you don’t understand what he just told you, don’t panic. My experience has been that he didn’t understand what he just told you. The really good lighting vendors truly appreciate an informed customer. It will make their job that much easier if they teach you the vocabulary, and they know it.

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So don’t be afraid to let them know that you don’t understand what they are trying to tell you. If they are not happy to take the time to teach you, tell them that there are others who would love to have the chance. Here are some of the frequently used terms. You will see how difficult they can make them, and how easy they really are to understand.

Candela

One candela is the luminous intensity radiating from an area of 1/600,000 square meter of a blackbody radiator elevated to the temperature of solidifying platinum at a pressure of 101,325 Newtons per square meter. Which means absolutely nothing to me. This is what I quote to people who think that the metric system is always the easiest system to work with. Fortunately, one candela also happens to be approximately the light from one candle.

Lumen

A lumen is equal to the luminous flux through a unitsolid angle (steradian) from a uniform point source of one candela, it is also equal to the flux striking a unit surface all points of which are at a unit distance from a uniform point source of one candela. Otherwise defined as a unit of luminous flux (light). For the present, just understand that this is a small amount of light. A 60 watt incandescent is rated to generate 890 lumens, a single 40 watt fluorescent lamp is rated at 3,050 lumens.

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Foot-candle

A foot-candle is a unit of measure defining the amount of light at a specific location. Its units of measure are: lumens per square foot. As you can see, we have built up from candela to lumens to foot-candle. Again, the definition may be intimidating as you work your way from candela to arrive at foot-candles. Don’t worry, only in the rarest circumstance will you run across candela in retrofit work. The real world concept of foot-candle is self-explanatory. Much like the lumen, it relates to a candle at a one foot distance. It is the amount of light at a specific location equal to the amount of light that would be there if the source was a candle held one foot away. Thus 50 footcandles is the amount of light that would be there if the source was 50 individual candles held one foot away.

Relative Light Output (RLO)

Relative Light Output is a new term that has emerged as more fixtures are being left in place and upgraded for efficiency. RLO is the heart of the retrofit conservation measure. It is the place for unscrupulous vendors to hide. And it is the place to expose them. Basically it is a percentage calculation comparing the amount of light from a given retrofit measure to the amount of light before any changes were made. Unfortunately, now it gets vague. The question of how much light was there before changes were made brings up some new definitions that are relatively easy to understand, but are virtually impossible to quantify.

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Therefore the vendor can make the result almost anything he wants. Fortunately, even though you can’t pin it down, you can recognize when it’s outrageous.

Initial

Whether this term is used with lumens or foot-candles, it is referring to a new system, or one that has been cleaned and relamped with new lamps.

Depreciation

Light sources, i.e., lamps and fixtures, do not work like new all their lives. Lamps lose their brightness and fixtures lose their ability to reflect or transmit light. For lamps, the strict term is Lamp Lumen Depreciation, sometimes abbreviated as LD. For fixtures it’s Lumen Dirt Depreciation (DD), which means exactly what it says. As the fixture gets dirtier, fewer lumens bounce off as more are absorbed. These are all expressed as a percentage of the initial values.

Maintained

As it applies to lighting, this term refers to the level of light (percentage of initial) that will be experienced after several months have passed. The LD and DD can occur rapidly at first then level off for a long period of time. Thus the light level will remain relatively stable, or will be maintained unless the conditions change. Their importance to our discussion lies in the fact that they are necessary percentages to calculate the condition of the existing system being evaluated for retrofit. This is

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where the reflector saves 50% of your electricity and gives you “the same light as before.” The reflector vendors will tell you this, but they won’t tell you that they mean the minute before you made the change and that they have assumed 80% LD and 80% DD. So the real question is what is the light level that is required? You know what IES standards are. You know what certain other industry standards are. You might even know what light level they really need. Plant personnel will most often insist that more is better.

Coefficient of Utilization (CU)

The percentage of light from the lamps that escapes from the fixture that actually reaches the work surface. CU is a function of the fixture efficiency and the room characteristics. Unfortunately, we don’t like to see light bulbs. We enclose them and cover them and diffuse them. Every time that there is something between what you want to see and the light bulb, you lose some useful light. Please remember that this is necessary. Glare from bright sources can be so distracting that you cannot perform the task at hand. Fortunately, for most retrofit options the CU is not changed and becomes a moot point. Only when you change something about the fixture or the room do you change the CU.

Color Temperature

Color temperature and the concept of cool and warm colors is a combination of the arts and sciences. Unfortunately that means that their usage is contradictory. The

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color temperature is the temperature, in degrees Kelvin, that a perfect (blackbody) radiator would have to be to appear the same color. A glowing ember is a good example. When it is just “red hot” the temperature is about 1800° K. A blue flame is closer to 6000° K. The contradiction comes in calling the lower temperature, red, a “warm” color and the higher temperature, blue, a “cool” color. So why do you need to hear this? All you need to remember is that if you want to keep the same color rendition you need to start with the same color temperature. Two hundred degrees Kelvin either way will not make a significant difference.

Color Rendering Index

The term Color Rendering describes the effect a light source has on the appearance of colored objects. A light source has an absolute effect on how colors are perceived. If the light source produces light without a certain wavelength, blue for example, then blue will not be seen in any objects. The blue objects will appear to be black or gray. The Color Rendering Index (CRI) is a rating from 1 to 100 that attempts to quantify the effect by comparing the way eight sample colors appear under the light source in question to the way they appear under an ideal light source (a blackbody radiator). There are three things worth noting: 1) the CRI does not give you any idea whether some colors were better than others; 2) the comparison is only valid for comparing light sources operating at the same color temperature; and 3) I don’t know who is looking at these colors and decides how good they are.

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Unfortunately, most clients will not really understand the effects of light source and color. But, often they will insist that color rendition is important. Most of the time what they really mean is, “Don’t change the colors that we are used to seeing.” If you don’t want the colors to change try to use a new light source that is close to the same color temperature and the same, or higher, CRI.

Efficacy

Sometimes it is useful to know how well a light source converts electricity into light. This is calculated by dividing the lumen output by the watts consumed. The resulting quantity of lumens per watt is the called the efficacy of the light source.

Ballast Factor (BF)

The ballast factor is another measure of a system’s performance. Unfortunately the ballast manufacturers chose a term that isn’t descriptive so that there is no clue as to what it means. If they called it RLO you might not need them to explain. Basically, they compare the light output of the fluorescent lamp and ballast combination in question to the theoretical light output from a mythical perfect ballast. Thus a lamp and ballast combination with a ballast factor of 0.88 means that the light output will be equal to 88% of the rated lamp lumens in the lamp catalog.

Lamp Identification Nomenclature

If you want to confuse and intimidate the novice lighting customer you will rattle off the term “F40T12/CW/RS/

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WM” instead of saying “four-foot fluorescent lamp”: I admit that there is a lot more information in the first method, and it takes less time to say, so it is worth learning what it means. This is what my professor was talking about when he mentioned vocabulary making you smarter. All lamps (the industry correct term for a light bulb is “lamp”) are identified by a letter and number system. The letter defines the shape of the lamp. Sometimes the letter is descriptive, sometimes it is arbitrary. The number defines the physical size of the largest diameter of the lamp. The “T” in the example above stands for “Tubular,” the “12” stands for the number of 1/8” in the diameter. I have no idea why it is or was important to know the diameter in eighths of an inch, but that’s the way it is. So the lamp is a tubular lamp, 1-1/2” in diameter. The rest of the nomenclature is specific to identifying it as a reduced wattage, cool white, fluorescent lamp.

LIGHTING RETROFIT MEASURES Lighting retrofit projects are relatively easy and can produce significant savings. With a little bit of information you can easily make a project go smoothly and profitably. When certain unscrupulous vendors sense that you have money to spend they will try to convince you that their product is the only thing that will work. Remember, when the only tool you have to sell is a hammer, everything looks like a nail. What follows is some help in understanding how things really work.

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This is an excellent conservation measure. The efficacy of fluorescent lamps is so much greater than incandescent that the savings potential can be as high as 87%! So what are the problems? The first, and often underlying problem, is size. The standard 60W A19 incandescent lamp has a maximum overall length (MOL) of 4-7/16” and a major diameter of 2-3/8”. The lumen for lumen compact fluorescent is a 13W double or quad tube. The quad tube (or double twin), with a screw-in adapter has a MOL, depending on the manufacturer, closer to 6” with the major diameter around 2-3/4”. The problem with the diameter is where it occurs. In an A19 lamp it is at the opposite end from the screw base. In the compact fluorescent and adapter it is right at the base. Most fixtures accommodate the A19 by virtually recessing the socket into the fixture body behind a reflector. In order to get a screw-in adapter to fit you usually need a socket extender close to 1-1/4” long. The reason that I called this an underlying problem is that most of the time in a commercial/industrial setting the incandescent lamp is 75 or 100 watts. The lumen for lumen compact fluorescent for the 75W lamp is a 20W fluorescent with an MOL of 8-3/16”. I don’t know of a compact fluorescent that will replace the 1750 lumens of a 100 watt A19. The whole reason for going over this is to enlighten you about the real and often encountered obstacles to utilizing this legitimate technological improvement. What you will often see is that the vendor will recommend 13W quads with adapters for all your hihats and porcelain

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sockets regardless of the existing wattage because that is pretty much the only one that will fit. This is fine if you are used to 60W incandescent lamps. Where it becomes a gray area is with decorative lighting, i.e., hihats and wall washers. In those cases there will be no loss of “productivity” or inconvenience if there is a reduction in the light level. Realistically, in many cases the only person who will get upset is the architect who is spending your money to make his “statement.” If you want to bite the bullet and replace the entire fixture there is another set of circumstances that are ultimately beneficial. The fixture manufacturers have gotten their acts together and designed new fixtures with the proper ventilation and optical reflectors that result in optimal output. Quite often these fixtures have a considerably higher CU than the existing incandescent fixtures and therefore require less lumens to produce equal RLO. In these situations you can reduce watts even further by reducing the lumen input. That is what I mean about an informed decision to conserve. Only you can decide if the two questions (Will there really be a savings? Will many people complain about the change?) have been answered correctly. As a final note on the subject, there have been recent studies on the effect of heat build-up on the light output from a compact fluorescent. Compact fluorescents have the same problems with lamp wall temperature (LWT) as all other fluorescents. Studies have shown that the LWT rises above design temperature when a compact fluorescent is enclosed in a hihat fixture or an integral reflector/lens. This has the overall effect of reducing the light output by as

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much as 20%. This, of course, makes the 13W compact fluorescent less than equal to a 60W incandescent lamp.

Fluorescent Current Limiters

These devices are connected to the ballast of a fluorescent fixture to reduce the lamp current. This is similar to what an electronic ballast does only it doesn’t change the frequency. Therefore, whatever the reduction in power consumption there is a corresponding reduction in light output. However, with new lamps and a cleaned fixture, the RLO immediately after retrofit will very likely be acceptable. Is this a trick? Well yes and no. The lower lamp current will reduce the rate of lumen depreciation, therefore the light level will remain high longer. Many devices also clean up the sine wave of the lamp current, thus reducing the crest factor which has a large, positive impact on the lamp life and LD. It is a trick only when you are not told of the light loss or the reasons why the RLO should be acceptable. One problem to mention: stroboscopic effect. My experience has been that reduced wattage lamps are very susceptible to exhibiting the sixty cycle flickering called stroboscopic effect. Very definitely eight-foot slimline reduced wattage lamps are good candidates and any lamp that is exposed to cool air that will lower the LWT. This is easily seen in surface mounted strip fixtures that are located next to air conditioning diffusers. Another problem: old ballasts. My experience has been that old ballasts (12 years or more) are susceptible to failure in the first month of use. I don’t know why, though I have some guesses that it has to do with the

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ballast capacitor and resonance. One last problem: low power factor. Measurements that I made on one brand indicated that they would contribute up to 50 VARs (lagging) per unit. This doesn’t sound like much but we installed close to 14,000 of them in one plant so it made a big dent in their electric bill where they were being penalized $10/month for every KVAR. This problem has been addressed by the industry and has been corrected by many of the manufacturers.

Fluorescent Fixture Reflectors

The way reflectors are generally sold they are the embodiment of the phrase, “it’s all done with mirrors.” I mean that literally and in the sense that it’s mostly hype. A new, baked white enamel fixture will have a reflectance factor between 82% and 85%. The best mirror reflector will have a reflectance factor of 96%. With only 14% improvement in reflectance, how can you get 50% more light out of the fixture. Admittedly, that 14% improvement represents a 17% increase over the original. And, it is true that the shape of the reflector is designed to get the light out of the fixture in one bounce. But both those conditions combine to produce an increase in the CU of the fixture of approximately 20%. Don’t get me wrong. I think that 20% is a significant achievement. But it is continuously oversold because it is also relatively costly and can be difficult to install. Reflectors can be a useful tool in your kit when used properly. However; claims made by reflector salesmen ask you to believe that 50% of the light output from a fluorescent lamp is absorbed by the fixture. This is not so.

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This kind of wild claim is easy to debunk when you know the simple dynamics of light reflections and the formulas for calculating fixture output. We have already discussed LD and DD. Common assumptions are that at the end of a fluorescent lamp’s useful life the LD is 70%. The amount of DD is dependent upon the environment. In an office you might expect a range of 85% to 95%. The vendor selling “the same light as before” is counting on large percentages of LD and DD. Percentages closer to 70% LD and 70% DD. Well, so what? The retrofit is complete and lo and behold the light level in the space is even a little better! The man is a hero. Better light for half the cost! But remember, you didn’t change the location that caused the dirt and you didn’t change the policy of relamping on burn out, so the same factors exist that made the 70% LD and 70% DD assumptions correct. Therefore, in about a year, the LD and DD will return and you will be back to 50% light output. But there’s one big problem. You started with 50% light output! Fifty percent of 50% is only 25%. The only way out for a plant engineer is to maintain your system at its peak performance until you can get your resume back out on the street. One more trick to expose. That first 100 hours (3 weeks at one shift) that the new lamps were in place, and the evaluation was made, is only the lamp burn-in period. New lamps experience a 10% lumen depreciation in the first 100 hours of use; then they level off. This is so well known that the lumen rating of a lamp that is published in the catalog and used in all foot-candle calculations is really the rated output after 100 hours. The new lamps gave the vendor a nice 10% edge against complaints of low light level.

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So how do you expose this? Ask him to work out the following equation for RLO: % of existing initial lumens × % increase in CU (or % increase in reflectance) ———————————————————————— RLO The typical numbers should be like this: 0.5 ×—— 1.2

(1/2 the lamps) (or % increase in reflectance)

0.6.

The typical “real world” situation may have numbers that calculate out a little differently: Before retrofit 1.0 × 0.7 × 0.7 ×—— 1.0

(All the existing lamps) (LD) (DD) (existing CU)

0.49

49% of the initial foot-candles. After retrofit 0.5 × 1.1 × 1.0 ×—— 1.2

(1/2 the lamps) (LD of new lamp) (DD of clean fixture) (CU increase due to reflector)

0.66

66% of the initial foot-candles.

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Which is a 35% increase over the existing conditions just prior to the retrofit. That sounds great! Half the watts with 35% more light. But look ahead ten or twelve months: One year later 0.5 × 0.7 × 0.7 ×—— 1.2

(LD) (DD) (CU increase)

0.29

29% of the initial foot-candles. Which is a 41% decrease in the amount of light present just prior to the retrofit. Let me stress that I have seen many applications that can suffer a 41% light loss without any loss in productivity or safety. In hallways, for instance, this enables the retrofit to proceed without complaints. As time passes the light loss is gradual and unneeded, therefore unnoticed. But if the vendor is claiming “the same light as before” because you feel that you need the same light as before, be aware of how he will prove it. Another “proof” will be the light meter test. The vendor will measure the light under a fixture before and after a reflector retrofit and, once again, it’s at least the same if not better! Above I showed you how that can happen if there is poor maintenance and low LD and DD. But this will be the case, or very close, on even a well maintained system. The reason is simple. Just look at any of those wonderful line drawings that reflector manufacturers include in

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their literature. There they show the rays of light radiating from the tube, hitting the reflector and bouncing down in one bounce. This is exactly what is going on and why the CU is increased. The reflector is designed to get the light out of the fixture in one bounce and it does it by directing the light downward. Therefore, although the total lumens are less, they are concentrated into a smaller square foot area so the lumens per square foot (foot-candles) remains very much the same as before—under the fixture. The best light meter test should include before and after readings taken carefully under the fixture and in between fixtures. Again, I have recommended reflectors where this is a desirable effect, i.e., hallways, etc. But do it on purpose, not because you were tricked. Lest I be accused of indiscriminate reflector bashing, let me identify a retrofit option that is not a hallway and is perfect for reflectors. A fixture that is a three lamp fixture with an eighteen cell deep parabolic lens is an extremely difficult fixture to retrofit any other way. If you just remove a lamp the row of cells below it will be very noticeably dark. However, a reflector and lamp socket relocation specifically designed for the fixture, can prove to be effective. The light is redistributed to all three cell rows from two lamp locations. The RLO numbers work out like this: 0.667 (3 lamps to 2 lamps) × 1.2 (CU increase) —— 0.8 80% of the initial light output remains.

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It is clear that if you want to factor even just a small amount of depreciation into the existing systems light level at the time of retrofit you will have equal or better light in—immediately after the retrofit. Even better is the case where there are reduced wattage (34W) fluorescent lamps and reduced wattage ballasts in the existing system. This system starts with a ballast factor of 0.88. If the retrofit is to a two lamp T8 system (BF of 0.95), then the RLO numbers are these: 0.667 (3 lamps to 2 lamps) × 1.14 (lumen increase due to T8 and higher BF) × 1.2 —— (CU increase) 0.91 91% RLO without hedging over depreciation.

Electronic Ballasts

I happen to be a true believer in electronic ballasts. The theory behind the use of high frequency to maintain the phosphorescent excitation at a higher level than at 60 hertz makes sense to me. I believe that it was a wise move on the part of the ballast manufacturers to choose the design goal of maintaining the existing light output and reduce the watts. The industry has certainly gone through some turbulent times. Starting with the original design flaws that caused a high failure rate, overcoming the bad press from that, then dealing with demands for product that have been unbelievably high. Long lead times are a serious problem and frustration.

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Dealing with these issues I have met and worked with some very nice, very dedicated people who manufacture and distribute this product. Then there are the others. The notion that one electronic ballast can operate almost any type of lamp (T8 or T12) is, as most tricks are, only partially correct. Fluorescent lamps are designed to operate at a specific lamp current. The original purpose of a ballast was to limit this current to the design level and avoid the tendency for it to “run away,” over pressure the tube, and cause damage. Most T12 lamps operate at 425ma (slimline) or 430ma (4’ or shorter rapid start). The 4’ T8 lamps all operate at 265ma. In order to accommodate both, the manufacturers who try, designed the ballast to operate at a lamp current somewhere in between. As a result, the T8 lamps are overdriven and the T12 lamps are under-driven. The operating watts are a reflection of the lamp current. A two lamp T8 system driven by one of these electronic ballasts will use 76 watts. Only four watts less than a wattmiser lamp on a standard ballast. I have never heard this information “volunteered” by a salesman. They have told me when I insisted. To be fair, there is a considerable increase in light output over a wattmiser system. But over driving an expensive lamp to get more light is not sound engineering practice.

HIGH INTENSITY DISCHARGE LIGHTING Although fluorescent lighting is technically high intensity discharge (HID) lighting, the term refers to

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Mercury Vapor (MV), Metal Halide (MH), High Pressure Sodium (HPS), and Low Pressure Sodium (LPS) light. Of those four types, LPS is the most efficient. Unfortunately, it has the color rendition that is the least acceptable. Although HPS light has a color rendition that may not be wholeheartedly liked, it is widely accepted. Especially for outdoor use. While MV and MH have the color rendering characteristics similar to fluorescent lamps, MV fixtures are not very efficient and MH fixtures are equal or slightly less efficient than fluorescent fixtures. A further drawback for MH fixtures is that the lamp life in wattages below 400 Watts is half that of HPS, MV and fluorescent lamps. The problem is that fluorescents are pretty much limited to indoor use (poor low temperature performance) and fixture light output under 40,000 lumens. MH fixtures can have a light output up to 110,000 lumens. Plus the MH lamp is a virtual point light source that allows reflectors and lenses to direct the light. Without a doubt, wherever there is a MV light it should be replaced. You should always try to install HPS lighting. Outdoor lighting should be easy to get approved with HPS. This is the highest efficiency and the longest lamp life. The failure mode is also self-correcting. The failure mode of an MV lamp is to gradually become dimmer and dimmer until the light output is nonexistent. (I have seen this in a dead storage warehouse. I thought all the lights were turned off until I saw one giving off a dull glow.) What this means is that unless you group lamp, you really don’t relamp until it is painfully obvious. Especially an outdoor fixture that is in the back

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lot and 35 feet in the air. The failure mode of a HPS fixture is to blink off and on until the neighbor’s complain. What this means is that there is a real potential for reducing the lumens as well as the watts if you change from MV to HPS. Why? Because the existing condition for years has probably been lamps with an average lumen depreciation of 70% or morel If this overall light output has been acceptable, then you can replace the MV fixtures with HPS fixtures and not size them lumen for lumen. You can at least downsize to 90% lumens or even 80% lumens without causing complaints about the light levels. (Complaints about the color are another subject.) I, personally, do not recommend low pressure sodium lights for anything but the most basic security applications. It is my belief that it takes more than just lumens to allow effective “seeing.” The monochromatic light grossly distorts all colors such that contrast and perception are greatly impaired. While objects become visible, they are difficult to identify without concentrating on them. This may be fine when the goal is to illuminate objects for safety and security. But if the tasks require identification and recognition, even if it’s not too critical, the task becomes very difficult.

Occupancy Sensors

I have recommended and installed hundreds of occupancy sensors. But evaluating the savings is a guess at best. Why? Because the savings are based upon correcting human errors. There is no easy way to accurately determine the level of human error for each office. All you can

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do is apply an assumed burn time percentage reduction. This is dangerous at best. Your safest approach is to get the facility’s personnel to make the guess for you. When calculating the savings, don’t double dip. Base the kWh reduction on reduced burn time for the wattage of the fixture after it is retrofit to the lowest watts. You may be surprised how poor the payback is under those conditions.

CALCULATIONS All calculations are derived from the same goals. You must always remember: the first goal is to define the existing conditions. The second goal is to define the proposed conditions. The third, and final goal is to subtract the second from the first. Defining the existing conditions is probably the most critical part. The calculations for a lighting retrofit are the easiest to handle in this business. The wattage before and after can be selected from a table of values for various fixture types. Since many utilities now have some kind of lighting rebate program, they have developed these tables to standardize the savings calculations. If you can’t get such a table from your utility, then I suggest that you use the ANSI values from a ballast catalogue. It is important to use the fixture wattages since that will include all parasitic ballast consumption. The only other parameter required is burn time. The following information must be gathered or guessed in order to arrive at an answer to the questions

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of economics: Define the existing conditions 1. Fixture watts before retrofit. 2. Quantity of fixtures. 3. Hours of operation (burn time). Define the proposed conditions 4. Fixture watts after retrofit. 5. Hours of operation (burn time). Define the economics 6. Cost to retrofit. 7. Incremental cost of electricity per kWh. 8. Incremental cost of electricity per kW, demand. These are all reasonably self-explanatory. The only one to comment on is fixture quantity. The accuracy of the count should be inversely proportional to the quantity of the fixtures. It will make very little difference in your decision if the count you use is 3,500 instead of the actual 3,347, even though the error is greater than 150 fixtures. But if you are using 35 fixtures when the actual count is 25, you are in for a surprise. The real goal is to determine the magnitude of the range of cost and savings. When you have an idea of how much money you are going to ask the accountants for, you know just how much time you need to spend answering their questions about your initial analysis. The following computer spreadsheet shows you one format for an approach that requires only data entry from the walk through survey.

Lighting

71

LIGHTING RETROFIT ECM I.D. EXISTING: {Description}

PROPOSED: {Description}

QTY: FIXTURE WATTS: TOTAL BURN TIME: ENERGY COSTS: INCREMENTAL kW: INCREMENTAL kWh:

0 0 0

$0:00 $0.0000

ENERGY COST SAVINGS: kWh SAVINGS: ANNUAL kW SAVINGS: TOTAL SAVINGS:

$0 $0 $0

PAYBACK PERIOD:

0.0

QTY: FIXTURE WATTS: TOTAL BURN TIME:

0 0 0

ENERGY CALCULATIONS: kW SAVED: 0.00 kWh SAVED: 0 ESTIMATED INSTALLED COSTS: MATERIAL: LABOR: CONTINGENCY: TOTAL/FIXTURE: TOTAL ECM:

$0.00 $0.00 15% $0.00 $0

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Simple Solutions to Energy Calculations

{Example and Equations}

LIGHTING RETROFIT OFFICE FLUORESCENTS EXISTING: QTY:

4’-3LT12ES/ES*

278

FIXTURE WATTS:

124

TOTAL BURN TIME:

4431

ENERGY COSTS:

INCREMENTAL kW:

$9.00

INCREMENTAL kWh:

$0.0565

ENERGY

COST SAVINGS:

QTY.

4’-3LT8-EB/LP** 278

FIXTURE WATTS:

78

TOTAL BURN TIME:

4431

ENERGY CALCULATIONS:

kW SAVED:

(a)

12.79

kWh SAVED:

(b)

56,672

ESTIMATED

kWh SAVINGS:

(c) $3,202

TOTAL SAVINGS:

(e)

ANNUAL kW SAVINGS: (d)

PAYBACK PERIOD:

PROPOSED:

$898

$4100

4.1

INSTALLED COSTS:

MATERIAL: LABOR:

CONTINGENCY:

$32.50 $20.00 15%

TOTAL/FIXTURE:

(f) $60.38

TOTAL ECM:

(g) $16786

* {4’; 3-Lamp; T12; reduced wattage lamps/reduced wattage ballast}

** {4’; 3-Lamp; T8; electronic ballast/low power design}

(a) {[(Exist Qty.) × (Exist. Watts)]—[(Prop. Qty. × (Prop. Watts)]} + 1,000

(b) (c) (d) (e) (f) (g)

Exist. Qty. × Exist. Watts × Exist. Hrs. – Prop. Qty. × Prop. Watts × Prop. Hrs

kWh Saved × Incremental $/kWh kW Saved × Incremental $/kW × 12 Months kWh $ Saved + kW $ Saved (Mat’l $ + Lab. $) × (1 + Contingency) Total $/Fixt. × Prop. Qty.

+ 1,000

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73

The following spreadsheet offers an approach to Occupancy Sensors that can be used for the entire facility before or after the lighting retrofit.

OCCUPANCY SENSOR ECM I.D. PROPOSED: SENSOR QTY:

0

EXISTING BURN TIME: % BURN TIME REDUCTION:

0% 0%

DESCRIPTION: FIXTURE QTY: FIXTURE WATTS:

{Description} 0 0

TOTAL kW:

0.00

DESCRIPTION: FIXTURE QTY: FIXTURE WATTS:

{Description} 0 0

TOTAL kW:

0.00

DESCRIPTION: FIXTURE QTY: FIXTURE WATTS:

{Description} 0 0

TOTAL kW:

0.00

DESCRIPTION: FIXTURE QTY: FIXTURE WATTS:

{Description} 0 0

TOTAL kW: TOTAL SENSOR kW:

0.00 0.00

ENERGY CALCULATIONS: INCREMENTAL kWh:

ESTIMATED INSTALLED COSTS: $0.0000

kWh REDUCTION: kWh SAVINGS:

0 $0

PAYBACK PERIOD:

ERR

MATERIAL: LABOR: MISC.: CONTINGENCY: TOTAL SENSOR: TOTAL ECM:

$0.00 $0.00 $0.00 15% $0.00 $0

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Simple Solutions to Energy Calculations

{Example and Equations}

OCCUPANCY SENSOR ADMIN. OFFICES PROPOSED: SENSOR QTY:

30

EXISTING BURN TIME:

2860

% BURN TIME REDUCTION:

25%

DESCRIPTION: FIXTURE QTY: FIXTURE WATTS:

4’-4LT8-EB 50 106

TOTAL kW:

(a)

5.30

DESCRIPTION: FIXTURE QTY: FIXTURE WATTS:

4’-2LT8-EB 25 62

TOTAL kW:

(a)

1.55

DESCRIPTION: {Description} FIXTURE QTY: 0 FIXTURE WATTS: 0

TOTAL kW:

(a)

0.00

DESCRIPTION: {Description} FIXTURE QTY: 0 FIXTURE WATTS: 0

TOTAL kW:

(a)

0.00

TOTAL SENSOR kW:

6.85

ENERGY CALCULATIONS:

ESTIMATED INSTALLED COSTS:

INCREMENTAL $/kWh: $0.0650

MATERIAL: LABOR: MISC. CONTINGENCY: TOTAL SENSOR: TOTAL ECM:

kWh REDUCTION: kWh SAVINGS:

(b) 4,898 (c) $318

PAYBACK PERIOD:

7.04

(a) (b) (c)

[(Fixture Qty.) × (Fixture Watts)] ÷ 1,000 Σ(a) * Burn Time * % Burn Time Reduction kWh Saved × Incremental $/kWh

$50.00 $15.00 $0.00 15% $74.75 $2,242

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75

ANECDOTES Richard’s Retrofit Rules Lighting Rule #1: “You can never save more energy than shutting it off.” Lighting Rule #2: “There is at least one person who won’t like it.” Lighting Rule #3: “Always retrofit lighting at night.” Lighting Rule #4: “The occupancy sensor will turn the lights off when the company president is in the bathroom.” I recently completed a lighting retrofit project for a major office building, forty stories of luxury office space. Being very careful, I chose to play it safe and replace the incandescent hihat fixtures with new compact fluorescent hihat fixtures. To be certain of success, I chose replacement fixtures with equal, or greater, lumen output. Fortunately, I insisted on installing a sample set of fixtures in the pickiest tenant’s space. It was a lawyer’s office. Four floors of the forty, 10% of the rentable space, controlled by this one law firm. They had clout and I knew it, so I wanted them to sign off on the selection. Essentially, the building manager agreed that if they were with us, the other tenants would not complain. The test area was one of the break rooms; six fixtures, each with two 13W compact fluorescent lamps, in a 10-by12 windowless room. The only other light came from the fully lit soda machine door and the snack vending machine.

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Simple Solutions to Energy Calculations

I thought it looked great. The firm’s Facility Coordinator thought that it was acceptable. I still waited a week before I ordered 1,960 fixtures to complete the project. I should have waited two weeks. “Marilyn thinks it’s too harsh and too bright.” I quickly put the 700 fixtures for their floors on hold. I never found out what Marilyn had on everybody, but they were all worried about her feelings. She didn’t have a corner office, but maybe there was some significance to the fact that her office was between two corner offices. After two more tries I got the best response from Marilyn that any retrofit engineer can get: “I couldn’t find the fixtures that were retrofit.” I now ordered 700 fixtures with two 9W compact fluorescent lamps each. Installing fixtures at a rate of 40 fixtures a night, we had 240 installed before the Facility Coordinator asked us to stop because the Art Committee (I’m not making this up) was “afraid that the paintings on the walls would look different.” We installed several wall washers so that they could see the effect. Fortunately, they couldn’t agree whether the change was better (more light on the paintings) or worse (slightly different color rendition). We were allowed to proceed. Approximately 200 fixtures later, someone noticed that the brass trim was not as wide as the old fixture. Once again installation stopped. My first response to the building manager was to advise him to take a hard line. Since they had over six weeks to notice the difference and had given the OK to proceed, they should bear the cost to remove and return the narrow trim and the increased cost of the wide

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77

trim. My second response was to advise him to stall for a week. They would have to go back to work sometime and stop looking at the ceiling. Fortunately, faced with a $12,000 charge and a workforce wasting far too much time looking at the ceiling, Marilyn said that the trim was OK. The installation finally went full speed ahead. I learned the lesson about installing lighting retrofit measures at night when I was a corporate energy manager. I was implementing a fluorescent lighting retrofit project for the home office facility. To avoid interfering with personnel I arranged for the contractor to work nights after everyone went home. One morning, a few days into the project, I was confronted in my office by a lynch mob. “We cannot work under these conditions! We are all getting terrible headaches!” Suddenly I felt very, very lucky. I was lucky because I hadn’t retrofit their area yet! Not only did it shut them down, it shut down anyone who had a legitimate complaint! No one wanted to take the chance of looking that foolish again. I was lucky because, while it was obvious in this case that their complaints were completely unfounded, if I had retrofit their area I wouldn’t have an argument to counter with. It would have been a power struggle just to complete the project. Another story involves the fact that I strongly believe that you should always try to live with retrofit measures that you are recommending for other people. So I installed an occupancy sensor in my office/study at home. With only about 150 watts of various incandescent lights it certainly

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Simple Solutions to Energy Calculations

isn’t cost effective, but it provided me with invaluable experience of what it is like to live with one. I really had no trouble putting up with the lights going off while I was working at the computer, or reading technical manuals, or talking on the phone. I increased the time delay and reduced the number of these annoyances (and reduced the savings potential). I got used to waving my arms or rocking in my chair to put the lights back on (and reduced the life of the lamps). I even put up with the faint hum on the telephone. I found that I could eliminate the hum by shutting the lamps off by their switches during the day (and again reduced the savings potential). But there was one thing that I could not tolerate. One night, in bed, snuggling with my wife, I noticed that she was tense and unresponsive. When I asked what was wrong, she answered: “I hate that damn switch in your study!” (I’m not making this up.) “When I go in there during the day it turns the lights on when I don’t want them on. When I’m talking on the phone at night it turns the lights off when I don’t want them off. I have to flap my arms like some stupid bird to get the lights to work! That switch controls me! I want to control it!” It took me less than ten minutes to put the standard light switch back in the study. It may be “my” office, but this house is her domain. And I will not tolerate anything that interferes with my “snuggling” with my wife. Don’t forget: the customer probably got to be president of the company because he is a control junkie. If he doesn’t have control of everything in his environment he is extremely uncomfortable.

CHAPTER 4

PUMPS INTRODUCTION Most inexperienced energy engineers discover the Affinity Laws and go crazy. The potential for energy savings is almost unbelievable. In the simplest situations, the savings are absolutely true. The true test is to be able to recognize when things are not simple and the evaluation needs to be modified. Rest assured, the modifications all follow the Affinity Laws, the calculations are simple, and the savings are usually considerable. But the savings potential will be less, and in some cases too small to consider when certain easily recognized situations exist. At the time of this writing, the hottest item in “pumping energy conservation” is variable speed drives (VSD). As always, a good engineer will recognize those situations where it will be more cost effective to downsize the pump or change the operating strategy. But these situations will be discovered while you are evaluating the potential for a VSD application. Pumps that are never required to run at full speed are a dead give away for downsizing. Remember, you can always put a VSD on a downsized pump to optimize the retrofit costs and the savings. 79

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Simple Solutions to Energy Calculations

DEFINITIONS Affinity Law

I had to look up the word “affinity” in the dictionary to get an idea why it was chosen for the formulas defining the relationships between speed ratios, head ratios and horsepower ratios for centrifugal pump and fan applications. I found out that the word “affinity” means “relationship.” It is my feeling that a deliberate effort is made every now and then to name useful and clever formulas with words that tend to hide the connection to their use in the real world. Possibly it is just a result of engineers taking a cue from doctors and lawyers. Anyway, the ratio equations are straightforward and simple to manipulate to determine any useful information with a few known facts:

N = rpm

N 1 F1 N 1 = ; N 2 F2 N 2 F = Flow

2

=

H1 N1 ; H2 N2

H = head

3

=

HP1 HP2

HP = horsepower

Head

With respect to fluid flow, this term is synonymous with the word “Pressure.” By far it most often refers to pressure given in units of Feet of Water Column, other times it may be Pounds per Square Inch.

Static Head

The static head is the pressure at the pump caused totally by the highest elevation of the water above the

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81

pump (which is why the units often are in feet). It is the pressure developed or required even when there is zero flow in the system (static conditions). In a closed loop the static head at the outlet of the pump must be the same as the static head at the inlet to the pump. Therefore the static head that the pump must overcome to move water in a closed loop is zero. However, even in a closed loop the minimum pressure required by the system header piping to provide the necessary flow through all connected devices should also be considered to be the static head of the system. If, for example, you are supplying cooling water to several parallel fan coils designed for operation with a 30 psi inlet pressure, the static (or zero flow) head that the pump must deliver into the piping system is the 30 psi (approximately 70 feet). Even though you could certainly let the header pressure go to zero when no flow is required, for all practical purposes it is better to maintain the required pressure at all times. Why? Because 1) there will be no complaints if the system is maintained such that it is capable of responding instantly to a call for flow, and 2) in most cases in the real world the probability of all connected devices being at zero flow is very small.

Friction Head

The friction head is the pressure losses due solely to the friction of the water moving against the pipe and fittings. (If you have started with units of feet, remember to use the same units in all your calculations.)

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Simple Solutions to Energy Calculations

Pressure Drop (PD)

The pressure drop is the change in pressure between the inlet of something and the outlet of something. In the simplest system of a closed loop without any equipment, the pressure drop from the beginning to the end is the friction head. In the next simplest system of a fluid transfer arrangement, like from a sump to the top of a cooling tower, the pressure drop from the beginning to the end is the friction head plus the static head. All pieces of equipment that require cooling or heating water will specify what the pressure drop through the equipment is expected to be when you operate it at its rated heat exchange level. This is the friction head that needs to be overcome in order to push water through the equipment at the flow rate for which it is designed. If equipment is in parallel, the equipment with the highest PD is the pump controlling factor. If the equipment is in series, the sum of each PD is the pump controlling factor.

Total Dynamic Head (TDH)

The TDH is the highest pressure that the piping and equipment system will impose on the pump. This is simply a sum of the static head, the friction head and the controlling pressure drop of the various pieces of equipment.

WHAT

TO

LOOK

FOR

Centrifugal pumps used to circulate cooling (or heating) water to HVAC or process applications offer

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83

major savings potential. Look for throttling valves. Not just at the outlet from the pump, but at the various end use equipment. Many process cooling water loops are supplied by a pump that runs constantly pressurizing a plant-wide header with various machine taps ending in a valve (solenoid or manual) at each machine. This is schematically equivalent to having a throttling valve on the pump outlet that is actuated in discrete steps.

Each of those machine valves varies the flow requirement from the pump by an incremental amount. I usually recommend a variable speed drive (VSD) controlled by a pressure sensor in the header. As a valve opens, the flow increases and the pressure drops. The pressure sensor will send a signal to speed up the pump to maintain the pressure setpoint. As a valve closes, the flow will decrease and the pressure will rise, slowing the pump down. This measure should also be evaluated for HVAC chilled water loops with two-way, modulating valves at each coil. If the existing system uses three-way modulating bypass valves, the savings will be optimized if the valves are replaced with two-way valves or modified with

84

Simple Solutions to Energy Calculations

the bypass isolated and capped. Cooling towers transfer heat from the inside of a facility to the outside. When used with refrigeration equipment the heat is transferred from the condenser side of refrigeration machines to the outside air. When used with process equipment the heat is transferred from a process heat exchanger or cooling jacket to the outside air. Installing a VSD controlled by return water temperature will ensure that the gpm (mass flow rate) is matched to the Btu rejection requirements. (Caution! Check with the equipment manufacturer to determine the minimum flow rate required to maintain turbulent flow through the condenser and evaporator tubes.) Another common application is a boiler feed water pump that runs continuously while the actual inlet flow to the boiler is controlled by a modulating (throttling) float valve. The calculations must take into account the fact that the pump must provide a minimum pressure higher than the boiler pressure. This means that there is a minimum horsepower requirement even if the flow is negligible. Therein lies the difficulty. There are many applications where the minimum speed of the pump will be dictated by the minimum TDH. The trick is to recognize that there is a minimum TDH and then determine what it is. All pumps are selected to operate at a certain maximum design flow rate, FD and maximum design total dynamic head; HD (D=Design). HD is the total of static and friction head. (Figure 4-1). For each system there may be a specific minimum static head requirement, HMin, If there is no minimum static head, such as a closed loop for hot water baseboard

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85

heat, then the system curve will start at zero head at zero flow. The friction head, however, varies as the square of the flow rate (Figure 4-2).

Figure 4-1

Figure 4-2

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Simple Solutions to Energy Calculations

However, most systems have a static head to overcome before any flow can be started. Since the static head will remain constant for all flows, the system curve does not start at zero H because of the static head requirement (Figure 4-3). To use a VSD means that you are developing the “family” of pump curves, i.e., a curve for each rpm. The minimum rpm curve (rpm 4) is the one closest to the minimum static head (HMin) at zero flow (Figure 4-4). That minimum rpm curve can be found by the Affinity Laws. The same laws identify the minimum horsepower requirement. Potential energy savings will be available between (HMin) and HD.

Figure 4-3

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87

Figure 4-4

N min Fmin N min = ; ND FD ND N min ND

2

2

=

H min N min ; HD ND

= %SPEED

%SPEED

min

=

2 min

=

3

H min HD

H min HD

Therefore: HPmin = HPD *

=

H min HD

3

HPmin HPD

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Simple Solutions to Energy Calculations

In the real world, the amount of pump horsepower savings is determined usually by the amount of friction head variation caused by the variation in flow rates. The greater the friction losses at high flow as compared to low flow the greater the horsepower savings available. The best situation is one that will also have a reduced pressure requirement. However, this is also a situation that should be evaluated for downsizing the pump first. The application of a VSD to a smaller pump results in an optimization of cost and savings.

THINGS

TO

LOOK OUT

FOR

ARE

1.

Constant high flow requirements: If there is no variation in flow and the pump is maxed out, there is no place to save energy. If, however, there is no variation in flow and the pump is grossly under loaded, first try to change the pump to the right size. If operating personnel can’t part with their existing equipment (see Pump Retrofit Rule #2), then the installation of a VSD will save a lot of money.

2.

Non-variable fluid transfer systems: In many process pump applications the goal is to move a quantity of fluid from one place to another as fast as possible. The size of the pump and motor are determined by first cost economics. If they could afford to buy a 100 horsepower pump and reduce the transfer time by 50% they probably would. There is no potential for savings because the process requires the short burst

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89

of consumption and then the pump is shut off (see Pump Retrofit Rule #1). This is the situation with municipal water supply systems. The district water distribution and booster pumps are generally controlled by a float switch in the storage tank. As the float drops to the low level setting, the pump is started and runs flat out until the float high level is satisfied and the pump is shut off. It may be argued that an analog signal from the float could be used to vary the speed of the pump to match the flow demanded from the system. But remember that the static head in these storage tank systems is usually quite high requiring a relatively high minimum speed. In addition, I have learned that the action of the large changes in water level has the desired effect of breaking the ice that forms on the surface of the water. 3.

Positive displacement pumps: The use of a VSD on a positive displacement pump is not a problem. The only reason to be aware of such pumps is that they do not necessarily follow the affinity laws. The affinity laws apply to centrifugal loads. Centrifugal loads exhibit the characteristic of requiring variable torque with the variable flow. Positive displacement loads require constant torque over the total range of flows. The horsepower to speed relationship can be a straight line at worst or a square law relationship. Either way saves energy, but the calculated savings will be less than with a centrifugal load that follows the cubed relationship.

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Simple Solutions to Energy Calculations

Typical examples of positive displacement loads are piston type pumps for highly viscous fluids and mechanical aerators (paddle wheels) for aeration lagoons.

CALCULATIONS Remember, you have two goals: 1.

Correctly define the existing conditions in terms of total energy consumption.

2.

Estimate the total energy consumption that will occur after the project is implemented.

The first goal is simply the horsepower of the pump motor times the hours it runs. The second goal is simply the horsepower of the pump motor at reduced speeds (flows) times the hours that it runs at those reduced speeds. If you have the system design documents with system and pump curves, or a table giving the gpm that the system requires at the TDH it was designed for, then your problem is very simple. The minimum and design pressure will be known so that you can calculate the minimum pump speed for the system, i.e., where the system curve intersects the zero pressure axis. The Affinity Laws will define the speed percentage for the flow portion of the system curve. Remember:

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91

N1 F = 1; & N2 F2

N = rpm

H min HD

= % Min Speed

F = flow

H = head

Therefore, the equation giving the speed percentage for any flow is: % Speech 1 = Min% + N1 = ND

F1 * 100% – Min% = FD

H min F + 1 * 1– HD FD

H min ND

Finally, the horsepower required for any flow, F1 is the Affinity Law:

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Simple Solutions to Energy Calculations

N1 3 *= HPD = HP1 = ND

N = rpm

H min F + 1 * 1– HD FD

H min ND

3

*HPD

HP = horsepower

Is that simple, or what? This lends itself nicely to setting up a computer spreadsheet to calculate the horsepower, and kW, for various flow rates and durations. Determining the various flow rates (gpm) and times that are required to meet the system’s functional needs is what you get paid to do. If you didn’t think that there would be any variations then you wouldn’t be evaluating a VSD. Unfortunately, for the majority of cases that I have had to deal with, the system design information is long gone. When you run into this situation: don’t panic. The calculations for these evaluations may, at first, seem anything but simple. Indeed, the concepts behind the equations and their derivations can be intimidating. However, you don’t have to derive the formulas every time you use them. And, while it is always better to understand the finer points of an internal combustion engine*, you don’t need to be a mechanic to drive a car. Certainly driving the car without knowing the rules of the road can be very dangerous, and just driving around without knowing the route can land you someplace that you don’t want to be. *Even with degrees in electrical and mechanical engineering, I firmly believe that internal combustion engines operate on the principles of black magic. As evidence I offer the following observation: anyone who works on them must resort to chanting magical words, usually containing four letters and delivered at high volume.

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93

But going slow at first, in familiar territory, by yourself, can be very productive and give you valuable experience before you have to deal with a super highway in rush hour. All you need are three equations. To find the horsepower at any flow you simply have to fill in the unknowns in the following equation: cy)

(1) Horsepower = (GPM*TDH) ÷ (3960 * Pump Efficien-

To determine the TDH you can use the next two equations: (2) TDH = Static Head (ft.) + Friction Head (ft.) (3) Friction Head (ft.) = (0.001246*GPM2*Equivalent Lineal Feet of Pipe) ÷ (Pipe Diameter in inches)5

(Note: the constant 0.001246 is the factor for galvanized or black iron (rough) pipe. For brass, copper, or plastic [smooth] pipe, the factor is 0.000623.) Combined the horsepower equation looks like this: Horsepower =

GPM * 0.001246 * GPM 2 * Equivalent Lineal Feet of Pipe ÷ Pipe Diameter in inches

3690 * Pump Efficiency

5

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Simple Solutions to Energy Calculations

The object is to input all known parameters and then manipulate the unknown parameters in the horsepower equation (design gpm and TDH) to arrive at an equation that results in a calculated full load horsepower consistent with the nameplate on the motor and the selected motor load factor. From that point on you can vary the gpm in the equation to determine the horsepower at the different flows. The following information must be gathered (or guessed) in order to arrive at an answer to the question of TDH: 1.

Static Head This parameter should be the easiest to determine. If there are no gages, the facility personnel should know the pressure requirements and/or the physical elevation changes of the systems involved. If available, use the original construction plans and specifications. If all else fails, look through equipment specifications for similar equipment. You can at least base your assumptions on realistic values from someone’s published data. 2.

Equivalent Lineal Feet of Pipe This is an undeniable wild guess. If the system is a plant wide cooling loop, use the perimeter of the area served. In any case, don’t forget to figure on a round trip for a circulating system. A good estimation of the effect of valves and fittings is to take the estimated distance and multiply it by 2.5 to get the equivalent lineal feet. Get a feel for the complexity of the run. At a city water supply facility it was more appropriate to use a

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95

1.5 multiplier because the pumps served several miles of relatively straight pipe. This parameter is, however, a value that can, and should, be “tweaked” to establish the estimated system curve. Once all the other data are satisfied, making realistic changes in this value can bring the horsepower calculated at your peak flow gpm in the spreadsheet to match the installed motor. 3.

Average Pipe Diameter Always start off with the header diameter at the pump outlet. This will be most accurate for short runs such as boiler feed water systems. If the resulting peak horsepower in the spreadsheet is unrealistically low, adjust this value downward to a realistic effective average diameter that will reflect the effects of longer branch systems with smaller diameter pipes in distant branches. If you followed closely, you will note that you have solved the two steps. The energy before was the pump horsepower running at a constant load all the time. The energy after is the pump horsepower running at various gpm for various periods. Just run the horsepower equation for the various flow rates and you will know how much energy it will take to run the system with a VSD. Subtract the after energy from the before energy and you know how much energy you will save. I told you it was simple. The following information must be gathered or guessed in order to arrive at an answer to the questions of economies:

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Simple Solutions to Energy Calculations

1.

Active Pump Horsepower The key word is ”active.” The pump has to be running to save energy. This may sound obvious to some, but if you find a duplex pumping system that is alternated, both pumps need to be controlled but only one uses energy at a time. 2.

Motor Efficiency If the efficiency isn’t stamped on the nameplate then make a logical assumption (guess). 3.

Assumed Load Factor No motor is loaded to 100%. If a design engineer calculates the horsepower required to be 4.5 horsepower he won’t select a 5 horsepower motor (if he’s smart). Why? Because he knows that the chances of his design information being that correct are slim. He should select a 7.5 horsepower motor. If you come across a motor that is fully loaded, you have also come across a situation where there is no potential for conservation. If you assume the existing condition utilizes the full nameplate horsepower you will be making the all too common error of overstating the savings potential. If I don’t have anything else to base my assumptions on, I use a load factor of 80%. 4. 5.

Pump Run Time

Assumed Normal “OFF” Time These may sound like a redundant way to say the same thing. What I am addressing is the situation where

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97

the controls are automatic. Many times all the facility personnel will be able to tell you is that the pumps are “enabled” during certain time periods. You will have to develop some level of understanding about the control strategy so that you can arrive at an assumed percentage of the enabled ON time that the pumps will be automatically turned OFF. 6.

Maximum gpm Most of the time this will be given. If you find that you have to guess, you can start with the formula: GPM = 15*D2 D = Pipe Diameter, inches 15 = Flow velocity factor, {15 = avg; 10 = conservative; 20 = borderline} Again, if available, use the original construction plans and specifications. 7.

Pump Efficiency For gpm ranges less than 100 gpm use 70% to 75%, for 100 to 500 gpm use 75% to 80%; for large pumps: 85%. 8.

Operating Scenario; Percent Time at Percent Flow I usually use four steps for percent flow: 100%, 85%, 75%, & 65% (each for an equal 25% of the time) for circulating pumps that are not used in comfort heating or cooling. As stated in previous chapters, try to get the facility personnel to make these assumptions for you. They

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Simple Solutions to Energy Calculations

should have a better feel for what’s right and wrong. (This may sound like it contradicts Retrofit Rule #3. It does. But unless you have better data than they do, let them make the guess.) Don’t forget to check the minimum speed requirement for the system parameters. You don’t want to look like you don’t know what you’re doing. Especially if you don’t. The following spreadsheet offers an approach that can be used for circulating pumps. For pumps that are used in comfort cooling applications, utilize the local area temperature history through its bin temperature observation data. First you must make an assumption of the percentage of the building cooling load that is caused by the internal heat gains (people, equipment, etc.) and the remaining percentage is assumed caused by the building envelope heat gains. By assuming that the highest outside temperature represents 100% load, the lower temperatures represent a lower percentage load and thus a corresponding lower flow requirement. This percentage flow requirement would then be cubed when applying the Affinity Laws. % Load at Bin Temp. = (% Internal load influence) +

Bin Temp. – Design IAT Max. Bin Temp. – Design IAT

* % Building envelope influence

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99

CIRCULATING PUMP VARIABLE SPEED CONTROLLERS CIRCULATING PUMP IDENTIFICATION BASE ASSUMPTIONS: OPERATING ASSUMPTIONS: ACTIVE PUMP HORSEPOWER: 0 00% HOURS @ 100% FLOW ASSUMED EFFICIENCY: 0% 00% HOURS @ 0% FLOW ASSUMED LOAD FACTOR: 0% 00% HOURS @ 0% FLOW BURN TIME: 0 00% HOURS @ 0% FLOW ASSUMED NORMAL “OFF” TIME: 0% NET HOURS: 0 INCREMENTAL COST/kWh: $0.0000 ENERGY CALCULATIONS: ACTIVE kW: 0.0 kWh REDUCTION: 0 ENERGY SAVINGS: $0 PAYBACK PERIOD: ERR

COST ASSUMPTIONS: QTY: 0 HP: COST. $0 REBATE: $0 NET COST$0

SYSTEM ASSUMPTIONS:

EQUIV. LIN. FT.

0

AVG. PIPE DIA. (IN.)

0

STATIC HEAD; FT. DESIGN GPM:

PUMP EFFICIENCY MOTOR EFF.

(a) (b)

0

(a)

0

(b)

00%

00%

If there are no plans or specifications to get this from, make reasonable assumptions. Same as above. Or use the equation given before: GPM = 15*D2

Be creative and play with all the various values in the “Systems Assumptions” that are not specifically known so you can observe the effect on the maximum horsepower and TDH. You will begin to see the effects of variations on the system, and, quite often, it will point out how oversized a pump really is.

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Simple Solutions to Energy Calculations

WATER PUMP LOAD DATA: CIRCULATING PUMP IDENTIFICATION (n) GPM —— (1) 3.4 (2) 6.7 — — — — (14) 46.9 (15) 50.3

(o) FT. TDH ——— 288.69 288.90 — — 302.67 304.74

(p) (q) (r) (s) % MOTOR PSI MOTOR SPEED HP TDH kW ——— ——— ——— ——— 97% 4.76 125 4.1 97% 4.76 125 4.1 — — — — — — — — 100% 5.11 131 4.4 100% 5.16 132 4.4

(n)

(b) is inserted into the last row then divided into 15 ascending steps (arbitrarily)

(o)

{[0.001246 * (n)2 * Equiv. Lineal Feet] ÷ (Pipe Diameter in inches)5} + Static Head See Standard Plumbing Engineering Design; Pg. 142—reference above equation

(p)

p x=

a Col. o

15

+

n x * 1– n 15

a Col. o

15

(q)15 (GPM * TDH ÷ (3960 * Pump Efficiency) = ((n)15 * (o)15) ÷ (3960 * Pump Eff.) (q)x Percent Speed3 * HPmax = (p)x3 * (q)15 (r)x

TDHx * 0.4331=(o)x * 0.4331

(S)x (Motor HPx * 0.746) ÷ Motor Efficiency = ((q) x * 0. 746) ÷ Motor Efficiency

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101

This equation assumes that the system was designed reasonably correct, i.e., that the highest outside temperature applies the maximum load: 100%. The temperature bins below the highest bin represent a decreased load due to a lower ∆t applied to the building envelope. The total load percentage is the internal loads percentage plus the reduced external loads percentage. The following spreadsheet offers an approach that can be used for a chilled water circulating loop based upon the local temperature history. CIRCULATING PUMP VARIABLE SPEED CONTROLLERS CHILLED WATER CIRCULATING LOOP BASE ASSUMPTIONS: ACTIVE PUMP HORSEPOWER:

0

ASSUMED EFFICIENCY:

00%

ASSUMED LOAD FACTOR:

00%

BURN TIME:

0

ASSUMED NORMAL “OFF” TIME:

0%

NET HOURS:

0

ASSUMED BASELOAD SPEED:

00%

% HOURS @ BASELOAD SPEED:

00%

NET HOURS @ BASELOAD SPEED:

0

INCREMENTAL COST/kWh:

$0.0000

ENERGY CALCULATIONS: ACTIVE kW:

COST ASSUMPTIONS: 0.0

kWh REDUCTION:

0

ENERGY SAVINGS:

$0

PAYBACK PERIOD:

ERR

QTY:

0

HP: COST:

$0

REBATE:

$0

NET COST:

$0

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Simple Solutions to Energy Calculations

CHILLED WATER LOOP DATA BASELOAD ASSUMPTIONS:

BUILDING ENVELOPE INFLUENCE:

00%

INTERNAL LOADS INFLUENCE:

ANNUAL COOLING SEASON HOURS:

00% 0

COOLING LOAD DATA (CLOSEST CITY) START MONTH TO END MONTH HOURS PER DAY DESIGN IAT: ACCUM. BIN

TEMP ——

97.5 92.5 87.5 82.5 77.5

BIN

HOURS ———

7 34 122 249 396

BIN

HOURS ———

7 41 163 412 808

00

% LOAD ———

100% 93% 86% 79% 73%

% ANNUAL HOURS COOLING: % ANNUAL HOURS @ BASELOAD:

ANNUAL % TIME ————

0.19% 0.94% 3.39% 6.92% 11.00%

00% 00%

VSD

kWh ——

0 0 0 0 0

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103

(EXAMPLE

AND

CALCULATIONS)

CIRCULATING PUMP VARIABLE SPEED CONTROLLERS CHILLED WATER CIRCULATING LOOP BASE ASSUMPTIONS: ACTIVE PUMP HORSEPOWER: MOTOR EFFICIENCY: ASSUMED LOAD FACTOR: RUN TIME: ASSUMED NORMAL “OFF” TIME: NET HOURS: ASSUMED BASELOAD SPEED: % HOURS @ BASELOAD SPEED: NET HOURS @ BASELOAD SPEED: INCREMENTAL COST/kWh:

(a) (b) (c) (d) (e) (f)

15 89% 80% 3600 0% 3600 65% 78% 2792 $0.0675

(c) (d) (e)

Annual hours that the chilled water system is “enabled” Assumed % that controls may be satisfied causing the system to shut off, usually 0% (b) * (a) Automatically input from “Internal Loads Influence” cell Automatically input from bin data table; “% Hours At Base Load”

(f)

(e) * (c)

(a) (b)

ENERGY CALCULATIONS:

ACTIVE kW: kWh REDUCTION: ENERGY SAVINGS: PAYBACK PERIOD: (g)

(h)

(i)

(g) 10.1 (h) 24,588 (i) $1,660 3.37

COST ASSUMPTIONS:

QTY: HP: COST. REBATE: NET COST

1 15 $7,590 $2,000 $5,590

(Active HP * Load Factor * 0. 746) ÷ Motor Efficiency [(g) * (c)] [(Sum of Col. (o)) + ((g) * (d)3 * (ƒ))] {existing kWh} – [{kWh under VSD control} + {kWh @ minimum, baseload speed}] (h) * Incremental $/kWh

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Simple Solutions to Energy Calculations

CHILLED WATER LOOP DATA BASELOAD ASSUMPTIONS: BUILDING ENVELOPE INFLUENCE: INTERNAL LOADS INFLUENCE: ANNUAL COOLING SEASON HOURS:

(k) (l) (a)

35% 65% 3600

(k)

Assumed percentage of the cooling load caused by heat gains through the building envelope and from infiltration, etc. This is the part of the load that responds to bin temperatures

(l)

(1 – (k)) Assumed percentage of the cooling load caused by heat gains from occupants and equipment. This is the part of the load that represents the base load on the system.

COOLING LOAD DATA {ALBANY, NY} MAY 1 TO SEPTEMBER 30 24 HOURS PER DAY DESIGN IAT 72

BIN TEMP ——— 97.5 92.5 87.5 82.5 77.5

BIN HOURS ———— 7 34 122 249 396

ACCUM. (n) (o) BIN (m) ANNUAL VSD HOURS % LOAD % TIME kWh ———— ———— ———— ——— 7 100% 0.19% 70 41 93% 0.94% 276 163 86% 3.39% 788 412 79% 6.92% 1,254 808 73% 11.00% 1,521

% ANNUAL HOURS COOLING: 22.4% % ANNUAL HOURS @ BASELOAD:

(p) 77.6%

(q)

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105

(m) (l) + {[(Bin Temp. – Design IAT) ÷ (Max. Bin Temp. – Design IAT)] * (k)} Assumes that the system was designed reasonably correct, i.e., that the highest temperature applies the maximum load: 100%. The temperature bins below the highest bin represent a decreased load due to a lower ∆t applied to the building envelope. The total load percentage is the internal loads percentage plus the reduced external loads percentage. (n) {Bin Hours ÷ (a)} * 100 (o) (m)3 * Bin Hours * (g) {At last! The application of the Affinity Law.} (p) [Total Accumulated Bin Hours ÷ (a)] * 100 (q) [100-(p)]

Evaluating the energy savings for boiler feed water pumps, basically the only application for a VSD exists when the pump runs continuously against a modulating valve that controls the flow. Because the static head is usually high (150 to 250 feet) the minimum speed to overcome the boiler pressure is usually higher than 90%. So where are the savings? Even 95% cubed results in a 14% savings. Since these are typically long run time situations (all winter for heating or all year for process applications) the savings are often sufficient to justify the implementation. In addition, very often the pumps are so oversized that the flow control valve is wasting a lot of energy by throttling, and the output pressure of the pump

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Simple Solutions to Energy Calculations

is very much higher than needed. There are reasons for over sizing boiler feed water pumps. First, having each pump sized to serve all the boilers in a multiple boiler installation adds redundancy. Second, if the pump is sized only to handle the equivalent gpm of the steam flow it will never be able to raise the boiler water level during periods of peak usage. Finally, if it can only just keep up, it will not be able to handle any emergency requirement to raise the water level in the boiler in a hurry. This is one application where I don’t recommend downsizing. If you don’t change the ability of the pump to deliver whatever peak flow it was designed to produce, you can’t be sued for screwing things up. (Successfully sued, that is. You can always be sued.) Typically when you apply the water horsepower formula you will be amazed when you calculate a required horsepower 1/2 to 1/3 of the connected horsepower. This is the horsepower that will be delivered by a VSD. The following spreadsheet offers an approach to calculating the savings with a boiler feedwater pump.

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107

BOILER FEED WATER PUMP BASE ASSUMPTIONS:

ACTIVE PUMP HORSEPOWER: ASSUMED EFFICIENCY:

ASSUMED LOAD FACTOR: BURN TIME:

0

0%

0%

0

ASSUMED NORMAL “OFF” TIME: 0% NET HOURS:

INCREMENTAL COST/kWh:

OPERATING ASSUMPTIONS: 00% HOURS @

100%

FLOW

00% HOURS @

0%

FLOW

00% HOURS @

00% HOURS @

0%

0%

0

$0.0000

ENERGY CALCULATIONS: ACTIVE kW: 0.0 kWh REDUCTION: 0 ENERGY SAVINGS: $0 PAYBACK PERIOD: ERR

COST ASSUMPTIONS: QTY: 0 HP: COST $0 REBATE: $0 NET COST: $0

SYSTEM ASSUMPTIONS: EQUIV. LIN. FT. 0 STATIC HEAD; FT. 0 BOILER PRESSURE (PSIG): 0 AVG. PIPE DIA. (IN.) 0 BOILER HP 0 PPH;STEAM 0 GPM RETURN RATE 0 GPM @ 2.5 SF: 0 PUMP EFFICIENCY 00% MOTOR EFF. 00%

FLOW

FLOW

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Simple Solutions to Energy Calculations

WATER PUMP LOAD DATA: BOILER FEED WATER PUMP GPM 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

FT. TDH 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

% SPEED 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

MOTOR HP 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

PSI TDH T 0 0 0 0 0 0 0 0 0 0 0 0 0 0

MOTOR kW 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

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109

(Example and Calculations) CIRCULATING PUMP VARIABLE SPEED CONTROLLERS BOILER FEED WATER PUMP (d) BASE ASSUMPTIONS:

ACTIVE PUMP HORSEPOWER: ASSUMED EFFICIENCY:

ASSUMED LOAD FACTOR: RUN TIME:

ASSUMED NORMAL “OFF” TIME:

NET HOURS:

INCREMENTAL COST/kWh:

(a) (b) (c) (d) (e)

(a) (b) (c)

10

87% 80%

8760

(e)

OPERATING ASSUMPTIONS: (1) 25% HRS @ 100%

FLOW

(3) 25% HRS @ 50%

FLOW

(2) 25% HRS @ 75% (4) 25% HRS @ 25%

FLOW FLOW

0%

8760 $0.0000

Annual hours that the boiler is active. Assumed % that controls may be satisfied causing the system to shut off; usually 0% (b) * (a) Operating assumptions defining the amount of time that each flow step will experience Operating assumptions defining the variations in flow in discrete steps

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Simple Solutions to Energy Calculations ENERGY CALCULATIONS: COST ASSUMPTIONS:

ACTIVE kW: kWh REDUCTION: ENERGY SAVINGS: PAYBACK PERIOD: (f)

(f) (g) (h)

6.9 23,017 $1554 3.5

QTY: HP: COST. REBATE: NET COST

1 10 $7,245 $1,800 $5,445

(Active HP*Load Factor*0. 746) ÷ Motor Efficiency {Existing kWh} – {Step 1 kWh – Step 2 kWh – Step 3 kWh – Step 4 kWh} ((f)*(c)) - [(d1)*(c)*@VLOOKUP{[(m)*(E1)], “Load Data Table,” (s)}] –

(g)

[(d2)*(c)*@VLOOKUP {[(m)*(E2)], “Load Data Table,” (s)}] – [(d3)*(c)*@VLOOKUP {[(m)*(E3)], “Load Data Table,” (s)}] – [(d2)*(c)*@VLOOKUP {[(m)*(E4)], “Load Data Table,” (s)}] –

(h)

(g)*Incremental $/kWh SYSTEM ASSUMPTIONS:

EQUIV. LIN. FT. STATIC HEAD; FT. BOILER PRESSURE (PSIG): AVG. PIPE DIA. (IN.) BOILER HP PPH;STEAM GPM RETURN RATE GPM @2.5 SF: PUMP EFFICIENCY MOTOR EFF. (u) (t) (i) (k) (l) (m)

500 289 125 2.5 300 10,056 20 50 75% 87%

Manual Data Entry (u) ÷ 0.4331 Manual Data Entry, or: PPH ÷ 33.52 (i) * 33.52 (k) ÷ 500 (l) * 2.5

(t) (u) (i) (k) (l) (m)

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111

WATER PUMP LOAD DATA: BOILER FEED WATER PUMP (n)

(1) (2) — — (14) (15)

GPM ——— 3.4 6.7 — — 46.9 50.3

(o) FT. TDH ——— 288.69 288.90 — — 302.67 304.74

(p) (q) (r) (s) % MOTOR PSI MOTOR SPEED HP TDH kW ———— ——— ——— ———— 97% 4.76 125 4.1 97% 4.76 125 4.1 — — — — — — — — 100% 5.11 131 4.4 100% 5.16 132 4.4

(n)

(m) is inserted into the last row then divided into 15 ascending steps (arbitrarily)

(o)

{[ 0.001246*(n)2 * Equiv. Lineal Feet] ÷ (Pipe Diameter in inches)5} + Static Head See Standard Plumbing Engineering Design, Pg. 142—reference above equation

(p)

p x=

t

Col. o

15

+

n x * 1– n 15

t

Col. o

15

(q) 15 (GPM * TDM ÷ (3960 * Pump Efficiency) = ((n) 15 *(o) 15 ) ÷ (3960*Pump Eff.) (q)x Percent Speed3, HPmax = (p)x3 * (q)15 (r)

TDHx * 0.4331 = (o)x*0.4331

(s)x (Motor HPx * 0.746) ÷ Motor Efficiency = ((q)x * 0.746) ÷ Motor Efficiency

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Simple Solutions to Energy Calculations

ANECDOTES Richard’s Retrofit Rules Pumping Rule #1: “You can never save more energy than shutting it off.” Pumping Rule #2: “You will never get a production manager to accept downsizing.” Pumping Rule #3: “No one in production knows the flow requirements.” The very first VSD installation I designed involved a 50 horsepower pump in a textile dyeing application. This pump was attached to a 6” diameter pipe with two short elbows immediately connected to the outlet that fed into a pneumatic butterfly valve. The operator made sure that the valve was closed during start up, then he would gradually open the valve until the flow in the open tank was visually satisfying to him. The pump would then run for two to four hours at a time with fifteen minute breaks while the quality was checked. The total daily run time was close to 22 hours. Because I could visually see that the butterfly valve was only open a fraction I knew that there was potential here. I installed pressure gages on both sides of the valve. The pressure on the inlet side of the valve was 60 psig and the pressure on the outlet side of the valve was only 5 psig. This confirmed that there was a major energy loss

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113

across that valve. In order to quantify the savings I needed to find out what the flow rate was. See rule #3. Not even the dye house Chemist who is responsible for designing the chemical formulas knew how much water was going through the fabric. Fortunately, the equipment manufacturer was willing to make a guess at 750 gpm. The chemist thought that sounded OK, so that’s what we used. Running through the standard pump formulas, I calculated that the required pump brake horsepower for that system without a valve was five horsepower. My first reaction was that I made some stupid mistake. Since I didn’t want anyone to know that I was stupid I spent the next day studying the application and pump formulas in general. I couldn’t get away from the five horsepower answer. Finally, I called the equipment manufacturer and asked them where I was going wrong. Here I was saying that there is only a five horsepower load and there they were using 50 horsepower on all their equipment. They were very cooperative and reassuring. They agreed that five horsepower probably was all that was needed in our application of their equipment, but they sell their equipment to all kinds of dyeing operations. Some of the other operations will need that 50 horsepower. Since you are paying for it, they keep their design and inventory standardized at 50 horsepower. Armed with new self-confidence, I recommended that the 50 horsepower motors be removed and replaced with 10 horsepower motors. (Why go too far out on a limb when you don’t have to?) Then we can install the VSD equipment at the much cheaper 10 horsepower level. Not a chance.

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Simple Solutions to Energy Calculations

The production supervisors got sweaty palms and the production manager broke out in a rash. “But we may need that 50 horsepower someday!” Although my conversations with the equipment manufacturer outlining the different dyeing operations that would require 50 horsepower had led me to conclude that we would never, never, never need 50 horsepower, I explained that the motors could easily be reinstalled when the time called for it. And that the difference in the cost to install a 10 horsepower VSD vs. a 50 horsepower VSD was about $6,000 for each one of the five pumps. Money was no object when it came to production reliability, translated: their peace of mind. The 50 horsepower units were installed. They still saved a lot of money. They even improved quality with better control of the flow and pressure. But the cost to implement could have been much cheaper. One thing I’d like to caution against: resist the temptation to install one VSD on a system with two or more pumps discharging into the same header. The seductive idea is to run the VSD pump until the flow demand is greater than one pump’s flow, but less than both. The theory is that you can run one pump at full speed and the other one at increasing speeds to meet the increasing flow requirements up to two full pumps. This sounds like a good way to get the benefit without buying two systems. Believe me, if I thought it would work I would be recommending it loud and long. Unfortunately, it doesn’t pass the Affinity Law test. Imagine that each pump has a check valve installed

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115

at its outlet. If one pump is running at full speed, it will be putting out full pressure into the header. If the second pump is running at, say, 50% speed to give a total system flow of 150%, the outlet pressure must be 0.502, or 25%. Obviously, if the pressure from one pump is four times higher than the pressure from the other pump, one check valve will be open but the other will be closed. These systems will function and may give the impression that all is well. To put it simply, the feedback signal controlling the speed of the second, variable speed pump, whether it is sensing flow or pressure, cannot be satisfied until the second pump has an impact on the system. This will cause the feedback signal to keep calling for increased speed until the discharge pressures match and the “check valve” opens. The flow to the system will be satisfied, but the energy savings will be negligible, if any.

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CHAPTER 5

FANS INTRODUCTION The evaluation of fan conservation measures is very similar to pump measures. This is so because most fans are centrifugal pumps that pump air. Therefore the affinity laws apply the same way as for pumps. The differences lie in the fact that most fan systems do not have a minimum static head. Many building exhaust fan systems are grossly oversized and over-operated. They are designed, rightly so, for the highest occupancy conditions and run that way continuously for 24 hours a day. A VSD will allow plant personnel to match the exhaust quantities to the exhaust requirements without making major modifications. The best part of this is that you can reduce the exhaust by duty cycling without completely stopping it. A duty cycle that runs the exhaust fan at 50% speed, therefore 50% flow, for five minutes out of every fifteen minutes will reduce the horsepower per hour to 71%. The amount of exhausted air will be reduced by 17% thereby reducing the load on the HVAC system. This will not add any excess strain or wear on the equipment and, more importantly, no one will notice. 117

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Simple Solutions to Energy Calculations

Cooling tower fans provide additional air circulation through cascading water to increase the cooling effect of conduction, convection and evaporation. As the air temperature drops (specifically the wet bulb) less air flow will be required to provide the same cooling effect. A VSD controlled by discharge water temperature will allow the fan speed and energy to automatically match the cooling requirements. (If the cooling tower pump VSD application is also implemented, the controls should be designed such that the pump is brought up to full speed before the fan is enabled to start at slow speed.) Many variable air volume (VAV) systems were originally designed without cfm control devices. Second generation VAV systems utilized pressure relief and bypass dampers. Third generation systems utilize variable inlet vanes to control fan output. All these systems can benefit from the installation of a VSD and removal of the other flow control methods. Specifically, the conversion of a variable inlet vane flow control system to a VSD flow control system is usually very simple to implement. In most cases the existing flow control sensors and feedback hardware are directly compatible with the controls for the VSD. Induced and forced draft fans for larger boilers, rotary kilns, process furnaces, etc., generally control the flow of air through vanes or dampers to control the amount of combustion air as needed to match the firing rate of the equipment. Whether inlet vanes or discharge dampers are used, significant savings can be realized if the flow is controlled by a VSD. In many cases, especially the larger

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119

pieces of equipment, the discharge pressure is relatively high due to large static pressure losses through the firing chamber, heat exchanger, and various pollution control devices. This will limit the minimum speed to whatever is necessary to overcome those pressure requirements.

THINGS

TO

LOOK

OUT FOR

ARE

1.

Dust collection system fans. These systems require a minimum cfm to maintain carrying velocity throughout the duct system. Long duct lengths with high flow volumes require a relatively high static pressure. High static pressure fans often are not centrifugal fans. Certain fan speeds may result in resonant frequency vibration amplification (this can be avoided with VSD options).

2.

Fume Hood exhaust fans. Generally, there is little variation allowed in a single fan and hood system. Variations are possible in a laboratory with multiple fume hoods served by a single fan. However, the system is there for health reasons. I maintain a personal policy that life safety and health systems are not worth the liability to change them for energy savings only. If there is a need for a new design or redesign then energy efficiency can be designed into the system with the help and guidance of the health safety experts that set the requirements.

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Simple Solutions to Energy Calculations

DEFINITIONS {As you will see, most of these terms are repeated in different sections. I have chosen to repeat them because I felt that each section should be able to stand on its own as a reference. (Not because I needed to fill space in the book.)}

Affinity Law

I had to look up the word “affinity” in the dictionary to get an idea why it was chosen for the formulas defining the relationships between speed ratios, head ratios and horsepower ratios for centrifugal pump and fan applications. I found out that the word “affinity” means “relationship.” It is my feeling that a deliberate effort is made every now and then to name useful and clever formulas with words that tend to hide the connection to their use in the real world. Possibly it is just a result of engineers taking a cue from doctors and lawyers. Anyway, the ratio equations are straightforward and simple to manipulate to determine any useful information with a few known facts: N 1 F1 N 1 = ; N 2 F2 N 2

N = rpm

F = flow

2

=

H1 N1 ; H2 N2

H = head

3

=

HP1 HP2

HP = horsepower

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121

Pressure

With respect to air flow, this term is synonymous with the word “Head.” By far in air flow situations it most often refers to pressure given in units of Inches of Water Column, other times it may be Pounds per Square Inch.

Static Pressure

The static pressure is the pressure at the fan that is developed or required even when there is zero flow in the system (static conditions).

Velocity Pressure

The velocity pressure is the pressure losses due solely to the friction of the air moving against the ductwork and fittings.

Pressure Drop (PD)

The pressure drop is the change in pressure between the inlet of something and the outlet of something. In an air handling system, the pressure drop from the beginning to the end is the velocity pressure. All pieces of equipment will specify what the pressure drop through the equipment is expected to be when you operate it at its rated capacity. This is the friction head that needs to be overcome in order to push air through the equipment at the flow rate for which it is designed. If the equipment is in parallel, the equipment with the highest PD is the fan controlling factor. If the equipment is in series, the sum of each PD is the fan controlling factor.

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Simple Solutions to Energy Calculations

Total Pressure (TP)

The TP is the highest pressure that the ductwork and equipment system will impose on the fan. This is simply a sum of the static pressure and velocity pressure.

CALCULATIONS Remember, you have two goals: 1.

Correctly define the existing conditions in terms of total energy consumption

2.

Estimate the total energy consumption that will occur after the project is implemented.

The first goal is simply the HP of the fan motor times the hours it runs. The second goal is simply the HP of the fan motor at reduced speeds (flows) times the hours that it runs at those reduced speeds. To find the horsepower at any flow you simply have to fill in the unknowns in the following equation: Horsepower =

CFM * Pressure (In. W.C.) —————————————— 6356 * Fan Eff.

Notice the similarity to the water horsepower equation. Basically, they are both the flow times the pressure, divided by efficiency and a constant for units conversion.

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123

Determining the various flow rates (cfm) that are required to meet the system’s functional needs is what you get paid to do. If you didn’t think that there would be any variations then you wouldn’t be evaluating a VSD. However, applying the Affinity Law ratios will simplify the calculation when the horsepower and flow at full speed is known. N 1 F1 N 1 = ; N 2 F2 N 2

N = rpm

F = flow

2

=

P1 N 1 ; P2 N 2

P = pressure F1 FD

F = flow

3

3

=

HP1 HP2

HP = horsepower

* HPD = HP1

HP = horsepower

There are really only a few applications that I have run into that require the inclusion of the minimum speed formula. However, one of them comes up rather a lot. The feedback controls for the fan speed of a variable air volume system operate on the concept of maintaining a constant duct pressure at a specific design point in the duct system (rule of thumb is to place the sensor 2/3 of the way down stream from the fan outlet). At first this may sound like a no win situation because if the pressure remains constant then the pressure ratio equals one. Therefore the speed ratio must also equal one (the square

124

Simple Solutions to Energy Calculations

root of one). If the speed cannot change for the system to work, then there cannot be any horsepower savings. Where are the savings? Remember that the pressure must remain constant 2/3 of the distance down stream from the fan outlet. The fan must present a discharge pressure that is higher than the set point pressure in order to overcome friction losses along the way. In addition, branches and zone discharges both upstream and down stream will affect the pressure at the set point. Therefore the fan must increase and decrease its discharge pressure as the flow varies in order to maintain the set point down stream. But there is a minimum pressure that the fan must produce at zero flow to produce the pressure at the set point. This will result in the need for a minimum speed and horsepower requirement just as calculated for a pump.

% Speed Min =

H min HD

The HMin in this situation is the pressure set point, based on the assumption that with zero flow the pressure down stream will be equal to the fan discharge pressure. The HD is the set point pressure plus the friction losses at design flow. Hopefully, since VAV systems are relatively new, the design conditions will be easily available.

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ANECDOTES Richard’s Retrofit Rules Fan Rule #1: “You can never save any more energy than shutting it off. But when you do, someone will complain that it’s too (stuffy, cold, hot, quiet, etc.).” Fan Rule #2: “It is always a surprise to find out that fan noise is more important than the actual movement of air.” Fan Rule #3: “Employees always believe that more exhaust is better.” While performing an energy audit at a facility that made light aggregates for cinder blocks by heating shale in a rotary kiln that was 12 feet in diameter and about 150 feet long, I observed that they had already installed a VSD on a 400 horsepower forced draft fan. This eliminated the use of a throttling damper to control the firing rate of the kiln. While I was there a second kiln was down for repairs and upgrade. Facility personnel reported that the energy savings calculations may have justified the first installation, but they hadn’t bothered to make actual measurements because their bill went down and process control was improved. Therefore they had ordered a second VSD for the second kiln.

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CHAPTER 6

HIGH EFFICIENCY MOTORS INTRODUCTION Replacing standard efficiency motors with high efficiency motors has become as commonplace as replacing incandescent fixtures with fluorescent. Most utility companies offer very attractive incentives. When performing a utility subsidized energy audit most utilities even require that an energy auditor must include such an evaluation in their reports. The reality is that quite often the predicted savings do not occur. I happen to believe that it is economically smart to implement a policy to use high efficiency motors as replacements and specify them for new equipment. However, it is rare to find a case where replacement of a working motor is economically sound. The contemporary wisdom defining how to evaluate the savings consistently predicts higher savings than actually is achieved. As it turns out, the reasons for this failure are not that difficult to understand. 127

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DEFINITIONS High Efficiency

I figured that I would start with the hardest first. Or maybe this is the easiest. Since every manufacturer has their own definition of “high efficiency” it generally means “something higher than our standard efficiency motor.” Certainly, NEMA has come out with standards (or should I say; ever changing standards) that define minimum efficiencies at full load horsepower to qualify as high efficiency. These are dutifully stamped on the nameplates of all high efficiency motors. But the best you can say is that if you ever operate that motor at full load, it should work that efficiently. Next come the utility companies. They usually set their own minimum standard efficiency for their incentive programs. These are quite often higher standards than NEMA. Therefore they usually call them “Premium Efficiency” standards.

Synchronous Speed

The synchronous speed is the ideal maximum rpm attainable with the number of magnetic poles in the motor windings and the frequency of the power supply. The higher the number of poles, the lower the synchronous rpm. The synchronous speed formula is: Hertz * 120 ————— # of Poles

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Slip

Slip is the difference between the synchronous rpm and the actual rpm of a motor. The higher the efficiency, the lower the slip.

What to Look For Any motor can be a candidate for replacement with a higher efficiency model on burn out. The best indicator for economic viability is long run hours.

What to Look Out For Motors coupled directly to pumps and fans. These can certainly result in savings, but the savings will be considerably less than expected because the rpm will be slightly higher. In fact, under certain conditions, the energy consumed can be significantly higher than with the standard efficiency motor.

Cal cul at ions I promised you simple solutions and this equation is the simplest. To determine the theoretical savings from motor efficiency improvement the formula is as follows:

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kW Saved = 0.746 * hp *

1 1 – Eff. Motor 1 Eff. Motor 2

One important thing to remember is that the horsepower in this equation should be the actual horsepower load on the motor. The equation defining the real, not the theoretical, savings includes the percent load on the motor. Otherwise known as the load factor:

kW Saved = 0.746 * hp * LF *

1 1 – Eff. Motor 1 Eff. Motor 2

Therein lies this measure’s “Catch 22.” If the horsepower load on the motor is less than 50% of its nameplate then all bets are off as to the motor efficiency. If the motor is loaded up to 100% then it should be evaluated to be sure it is all right to operate that way for long periods. Otherwise it may be smart to install a larger motor. What you need in most cases is the value of the “part load” efficiency which is usually given for 25%, 50%, and 75% loading. That is, when you can get such information. The second problem arises when the motor is direct-coupled to a centrifugal load. If you remember the Affinity Law from previous chapters, you learned how a small reduction in rpm can have a major effect on the horsepower. Well, not surprisingly, it applies to increases in motor rpm as well. The increased motor efficiency necessarily results in a reduction in the motor slip. Therefore the rpm increases slightly as the efficiency increases.

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An increase of 10 rpm on a motor that normally ran at 1750 rpm means the speed ratio of before and after will be 1.006: 1760 —— 1750

= 1.006

Which may not seem significant. However, if the motor is driving a centrifugal load, then the Affinity Law causes the horsepower to increase to 102%. 1760 1750

3

= 1.006 3 = 1.02

When the high efficiency motor is expected to save only 3% to 5% of the full load horsepower, adding another 2% to the load can eat away more than half of the savings. The theoretical savings formula should be modified as follows to reflect the differences in horsepower before and after. kW Saved = 0.746 * hp 1 * 1 1 – 0.746 * hp 2 Eff. Motor 2 Eff. Motor 1

The distinguished engineer Mr. Konstantin Lobodovsky has shown that in certain applications, the conversion to high efficiency motors actually uses more energy than the previous motor in that application. Finally, once you have the correct kW reduction, all you have to do is multiply it by the annual hours that the motor runs to get the total kWh reduction.

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ANECDOTES Richard’s Retrofit Rules Motor Rule #1: “You can never save any more energy than shutting it off.” Motor Rule #2: “No one in the plant knows how much the motor is loaded.” Motor Rule #3: “If you burn your hand when you touch it, it is at least 100% loaded.” Being the project manager of a large motor retrofit job I was expected to have all the answers. With the risk of sounding defensive, the motor evaluation was required by the local utility as part of a major energy conservation report and the evaluation was performed by someone else utilizing the standard savings formula. As work progressed, the very savvy operations and maintenance personnel asked some hard questions. Questions like: “If this is supposed to save energy, why are my motor ampere readings higher than with the standard motor?” “If this is supposed to save energy, why are the motor temperature readings higher than before? And isn’t the higher temperature going to adversely effect the life of the motor?” Of course I was afraid to answer those questions the way that I really felt. The utility company would have

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choked and then my boss would have choked… me. So I deferred to the motor manufacturer’s representative. The end result was that some of the motors were removed as defective and replaced. In most cases the replacement motors produced amp readings lower than before, if only slightly. Other motors were replaced with heavier duty motors to lower operating temperature. The representative also said that the higher temperatures were the result of better heat transfer characteristics in the design of the motor housing that allow the cooling fan to be downsized, thus reducing the energy losses called “windage” and would not be detrimental to these motors. Ultimately, the customer was satisfied because he managed to get all new motors for a very low net cost after the utility rebate. I was satisfied because ultimately I didn’t have to answer any of those questions. However, the bottom line should be emphasized. The majority of the motors at that site were not direct-coupled. The specifications called for tachometer readings before replacement, and sheave adjustments to maintain the same output rpm. Therefore the majority of the motors showed lower amperes and temperatures.

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CHAPTER 7

INSULATION INTRODUCTION The evaluation of conservation measures involving insulation starts with your ability to feel warmth and cold. Honestly, some of the best measures that I have identified in practice came to me when I came close to burning myself. It’s pretty safe to say that there will be some savings potential if you can feel the heat even before you touch it. Pipes are easy. Especially `hot pipes. The evaluation relies upon determining the heat loss from conduction, convection, and radiation. At first this may seem complicated. If you were required to determine the various heat transfer factors and coefficients, etc. ad infinitum ad nauseam it would be very complicated. But would I be talking about this if I didn’t have an easy way out? Fortunately for me, Owens Corning, Manville, Dow, and everyone else trying to sell insulation have done all the work. They all publish tables showing the total heat loss from various diameter pipes. You get the heat loss in Btuh per lineal foot for horizontal and vertical orientations, with wind, without wind, at all temperatures, with weatherproof jackets, without jackets, and most important—bare pipe. 135

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Now here’s the part where I cheat: If I have observed a 30” diameter condensate return tank that is bare steel holding 200°F water, I don’t have to perform the heat loss calculations for a bare cylinder if I have a table that gives me the values for a 30” diameter pipe. Honestly, the various laws of physics really can’t tell whether the object is a pipe or a tank. Yes, the pipe doesn’t have any ends. Therefore your analysis is conservative at worst. If the payback calculation is only marginal then you can add the heat loss from the ends. My experience is that this conservative approach shows payback periods in months. That is the essence of this book. The fast analysis that tells you whether it’s worth going deeper. When the surface is clearly not a pipe, even when you stand back and squint your eyes, then it must be a flat surface. Fear not! The same sources have developed heat loss tables for vertical and horizontal surfaces. There you can get the Btuh per square foot for the tank ends that were missing from the pipe heat loss table. Yes, most tank ends are curved, not flat. So what? Again the result is that your analysis is conservative.

DEFINITIONS {As you will see, most of these terms are repeated in different sections. I have chosen to repeat them because I felt that each section should be able to stand on its own as a reference. (Not because I needed to fill space in the book.)}

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Btu

This stands for British Thermal Unit. A Btu is defined as the amount of heat that it takes to increase the temperature of one pound of water by one degree Fahrenheit. Do not confuse Btu with temperature! The complete combustion of a standard wooden kitchen match will release approximately one Btu. The temperature of the flame will be over 1200°F. To get a perspective on quantities of Btus, one gallon of gasoline contains approximately 128,000 Btus.

Btuh

Simply means Btu per hour.

MBH

Simply means 1,000 Btu per hour.

MMBtu and MMBtuh

Means 1,000,000 Btu and Btuh.

Conduction

The transmission of heat by molecular vibrations from one part of a body to another part of the same body or to another body in physical contact with it, without appreciable displacement of the particles of the body. The key idea here is physical contact.

R-factor; Thermal Resistance

Resistance is a property of a material that determines the amount of heat that will flow through a unit area given a difference in temperature.

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The R-factor is the measure of the ability of a material to resist letting heat pass through it to a colder area and so it is otherwise called the thermal resistance. The higher the R-factor the better it is. The units of measure for R-factor are ∆°F/Btuh/sq.ft.

U-factor; Over-all Coefficient of Heat Transfer

Time rate of transfer of heat by conduction, across unit area for unit difference of temperature. The U-factor is the measure of the ability of a material to let heat pass through it to a colder area and so it is otherwise called the thermal conductivity. It is the mathematical inverse of the sum of all the R-factors. The lower the U-factor the better it is. The units of measure for U-factor are Btuh/sq.ft./°F(∆t).

End-use efficiency

Sources of thermal energy generally have to deal with two areas where heat is lost before it arrives at the “end-use” There are combustion losses, i.e., heat lost up the stack. Then there are transmission losses, i.e., heat lost to the areas that the pipes or ducts pass through before they reach the end-use. The end-use efficiency is used to determine the amount of fuel that needs to be burned at the boiler in order for there to be the required amount of Btus at the end-use. The formula for calculating end-use efficiency is as follows: End-use efficiency = Combustion efficiency * transmission efficiency

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Remember that heat lost by space heating pipes into the space that is to be heated is not really lost. All you lose is control.

CALCULATIONS The best situation for cost effective insulation projects is when the heat source is electricity: the most expensive source of Btu’s known to man. (Actually, the most expensive source of Btus known to man so far is atomic energy. But we use that to make electricity, so by the time we get it, electricity is more expensive.) Electricity has the advantage of being 100% efficient at its end use. Therefore the conversion from units of Btu to units of kWh is very simple: 3,413 Btu = 1 kWh Once you have determined the total Btu saved for the year: Btuh * Annual hours at the elevated temperature = Annual Btu loss You simply divide the annual Btu saved by 3,413 to get the annual kWh saved. Other situations deal with fuel energy conservation. The heat lost by steam distribution pipes and fittings should not be overlooked. As a matter of fact, this loss must be calculated and steam traps sized correctly to

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transport the condensate away from the supply pipes or there will be serious consequences to the system. To evaluate the heat loss, and cost of these heat losses, refer to the product literature as mentioned above. By subtracting the Btuh loss after the new insulation from the Btuh loss before the new insulation you will know how many Btuh will be saved. Remember, the most important information that you are getting from these tables is the heat loss for the bare surface. This is the information that is pure physics without any egocentric sales promises. The heat loss from their insulation may be subject to suspicion, but the loss from a bare surface is so high that once you become aware of it, to do nothing will be inexcusable (even for an accountant). For example: the heat loss from the bare condensate tank described above is 1,857 Btuh per lineal foot. With 1.5” of fiberglass insulation with a bright metal jacket the heat loss drops to 198 Btuh per lineal foot. Therefore the Btuh saved is 1,659 Btuh per lineal foot. Assuming a five-foot-long tank that is hot for 3,600 hours during the winter, the annual Btu loss is 29,862,000 Btu per year. How much will this save? The answer lies in the cost of the fuel and how long the system operates at elevated temperatures. Suppose natural gas costs $5.00 per MMBtu. This is the price as it passed through the gas meter. The combustion efficiency should be no lower than 80% and the distribution efficiency should be around 85%. Therefore the end-use efficiency will be 68%:

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End-Use efficiency = 0.8 * 0.85 = 0.68 This will be used to determine how many Btus had to be consumed at the burner in order to have those Btus to waste at the condensate tank. (End-Use Btu) ÷ (End-Use Efficiency) 29,862,000 ÷ 0.68 = 43,924,706 Btu So the cost of the 1,659 Btuh per lineal foot that was lost before insulating the tank was: 43,914,706 * $5.00 = $219.57 per year 1,000,000

The heat and cooling energy lost through the building envelope presents a very frustrating set of circumstances to me. First, it is usually a significant amount of energy and, more importantly, money. However, the money that would need to be spent to reduce the energy losses is typically 10 to 12 times greater than the savings. Unless, of course, you involve an architect in the project. Then there is no limit to the amount of money it will cost. At any rate, you’re welcome to do the analysis. The heat loss from conduction through a wall is calculated one of two ways: Q=

Ti – To * A; & Q = U * A * Ti – To R

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Q Q Ti To U R A

= = = = = = =

Btuh Heat loss, Btuh Indoor air temperature, °F Outdoor air temperature; °F Heat loss coefficient Thermal resistance Square foot of envelope

When trying to calculate the Btuh saved with increased insulation, you can subtract the U-factors and multiply the answer times the area and the ∆T to get the savings. The trick is that you can’t just subtract the R-factors. You must subtract the inverse of the R-factors. ∆Q = ∆T * A * 1 – 1 ; & ∆Q = U 1 – U 2 * A * ∆T R1 R2

∆Q ∆T U R A

= = = = =

Btuh saved (Outside Air Temp. – Inside Air Temp.) Heat loss coefficient Thermal resistance Square foot of envelope

I will talk briefly about infiltration. The first thing to say is that nobody can calculate the heat losses due to infiltration with any certainty. That is why the ASHRAE manual is so important. Because there is no way to determine the actual annual infiltration losses, when there is any dispute about the actual savings you can always point to the manual and say that you followed industry standards. There is always comfort in knowing that you used the most complicated method to arrive at an arbitrary answer.

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So, since I believe that whatever answer I come up with is only a rough approximation, I use the simplest method that I know of to make my first cut. If the measure looks like it will be marginal but still desirable, I am forced to proceed with the ASHRAE methods to cover my tail. The simplest method uses the total linear feet of perimeter around the windows and doors and applies a standard 0.25 cfm per foot to arrive at a total infiltration cfm. Then you simply apply the standard Btu equations for cfm to arrive at the heat loss. The trick in this method is to use only half of the total lineal feet of perimeter of all the windows. Why? Because the quantity of air that comes in must cause an equal quantity of air to be forced out. So while half the cracks are letting air into the building the other half are letting air out of the building. The equation is as follows: Total perimeter lineal feet of crack * 0.25 = cfm 2

CFM * 1.08 * ∆T = Btuh That’s it. If you want to calculate the infiltration through an active door, go directly to ASHRAE. There you will find impressive tables with factors for various levels of activity through the door. If you’re forced to guess, you might as well guess with the best.

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ANECDOTES Richard’s Retrofit Rules Insulation Rule #1: “If you burn your hand on the equipment you can probably save some energy. Insulation Rule #2: “If the insulation is in the way of production or maintenance workers it won’t last long enough to have a payback.” Insulation Rule #3: “If it’s already covered with asbestos, the payback period will be measured in decades.” Insulation Rule #4: “Sometimes you have to point out the safety features of adding insulation, not the energy saved.” When I first started doing walk through audits I found myself in plastic molding facilities looking at extruder barrels, dies, and molds that were kept at temperatures from 150°F up to 700°F by electric heat. In none of the first three facilities that I went through did I see insulation on these pieces of equipment. It didn’t take a genius to see, and feel, the potential energy savings. The heat radiating from these parts was awful and in some cases dangerous. Some of the locations were in air conditioned rooms! But being a little cautious, I felt that it would be better to first ask why they weren’t insulated. It just seems smarter to first assume that there is a good reason that isn’t obvious to an outsider. They

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may think a question is dumb, but they will think that the person is dumb if he makes a recommendation in writing that they all know is dumb. The answers were varied, but mostly consisted of “I don’t know” and, “They never are insulated.” With some further investigation, it finally came down to two reasons: 1.

The extruder barrels only use the electric heaters during the start up of the process run. After the extruder is running up to speed the heat of the friction and the shear forces of the plastic forced through the barrel at high pressures is more than enough to maintain the temperature. In fact, after start up, the barrel has to be cooled with water to keep the temperature from getting too high.

2.

They tried insulation many years ago when the only high temperature insulation available was the rigid blocks of extremely brittle material that needed to be custom formed to the equipment. Since the dies, and barrels, etc. require frequent maintenance or production changes, it did not take long for the insulation to be wrecked by excessive handling. At the worst it was a maintenance nightmare. So the practice was abandoned.

It was easy then to recommend the use of custommade blanket type insulation that can withstand high temperatures, is reliable, and is easy to remove and install. Now, I find that many insulation contractors and specialty

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manufacturers are marketing insulation blankets to these customers. The only thing that has come to light as a drawback is that many of the older machines were outfitted with electric heater bands that have relatively low temperature electrical insulation on the wires. When these are open to the air, the temperature does not get above their rated design. But when they are covered with insulation, the heat is trapped inside and the electrical insulation can fail.

CHAPTER 8

FUEL SWITCHING INTRODUCTION Fuel switching measures are, for me, some of the most practical approaches to energy management. The real conservation that is desired, for all intents and purposes, is the conservation of money. When you do an evaluation that seeks to identify the equipment and fuel that will accomplish the same results with the least expense you can get creative. In the right situation, with the proper in house support, you can even do some experimenting. Primarily we are talking about conversion of electric heat to oil or gas. But remember, even though electricity is very expensive, there are times that it is still the best choice. If the utilization is intermittent, if there are great difficulties in getting the necessary pipes and heat exchangers to a remote location, if there are maintenance or contamination problems to worry about, if very high temperatures are required, etc., all of these reasons, and many more, are reason enough to use this expensive source of Btus. But do it on purpose! Use it after careful consideration of all the alternatives. Then you know that you have done your job the right way. 147

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DEFINITIONS {As you will see, most of these terms are repeated in different sections. I have chosen to repeat them because I felt that each section should be able to stand on its own as a reference. (Not because I needed to fill space in the book.)}

Btu

This stands for British Thermal Unit. A Btu is defined as the amount of heat that it takes to increase the temperature of one pound of water by one degree Fahrenheit. Do not confuse Btu with temperature! The complete combustion of a standard wooden kitchen match will release approximately one Btu. The temperature of the flame will be over 1200°F. To get a perspective on quantities of Btus, one gallon of gasoline contains approximately 128,000 Btus.

Btuh

Simply means Btu per hour.

MBH

Simply means 1,000 Btu per hour.

MMBtu and MMBtuh

Means 1,000,000 Btu and Btuh.

Radiation

The emission and propagation of energy through space or through a material medium in the form of waves. If it’s glowing, it’s radiating.

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Strictly speaking, it doesn’t have to be glowing to radiate energy, but in the context of this chapter we are talking about infrared radiation and mostly where the source is hot enough to be radiating some portion of its energy in the visible, red, wavelength.

IR Radiation

Infrared radiation. This is sometimes used to describe a “new” technology in high temperature ovens, etc. The truth is that all ovens with heating elements inside them transfer heat through the radiation of infrared energy. What they are trying to get across is the idea that the equipment has been specifically designed to improve the radiation transfer characteristics of the system. Another myth is that “Infrared heaters do not heat the air so you automatically save energy.” Of course they heat the air. They heat whatever object that the infrared radiation is pointed at and then that object heats the air. But how are these heaters controlled? If they are controlled by a thermostat in the room, as most of them are, then the heaters will stay ON until the air in the room is warm enough to satisfy the thermostat setting. The potential for savings comes from the fact that when you are in a room full of warm objects, the air can be kept much cooler than normal and you will still feel very comfortable.

End-use Efficiency

Sources of thermal energy generally have to deal with two areas where heat is lost before it arrives at the “end-use.” There are combustion losses, i.e., heat lost up

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the stack. Then there are transmission losses, i.e., heat lost to the areas that the pipes or ducts pass through before they reach the end-use. The end-use efficiency is used to determine the amount of fuel that needs to be burned at the boiler in order for there to be the required amount of Btus at the end-use. The formula for calculating end-use efficiency is as follows: End-Use efficiency = Combustion efficiency ÷ transmission efficiency Remember that heat lost by space heating pipes into the space that is to be heated is not really lost. All you lose is control.

Common Conversion Factors #6 Fuel Oil #4 Fuel Oil #2 Fuel Oil Gasoline Coal Natural Gas Propane Wood

153,000 Btu/gal 145,000 Btu/gal 138,000 Btu/gal 128,000 Btu/gal 22,000,000 Btu/ton +/- 1,000 Btu/ft3 95,000 Btu/gal +/- 8,500 Btu/lb.

CALCULATIONS As stated above, the best situation for cost effective fuel switching projects is when the heat source is electric-

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ity: the most expensive source of Btu’s known to man. (Actually, the most expensive source of Btus known to man so far is atomic energy. But we use that to make electricity, so by the time we get it, electricity is more expensive.) Electricity has the advantage of being 100% efficient at its end use. Therefore the conversion from units of Btu to units of kWh is very simple: 3,413 Btu = 1 kWh or: 3,413 Btuh = 1 kW One of the most useful calculations that you can perform allows you to compare all fuel sources by their cost per useful heat content. This is their cost per MMBtu adjusted for their end-use efficiency. $/Unit ————————————— (Btu/Unit) * (End-use Eff.)

*1,000,000 = $/MMBtu

So if electricity costs $0.075/kWh at 100% end-use efficiency and natural gas costs $0.42/CCF (CCF = 100 cubic feet) at 68% end-use efficiency, then the comparison results as follows: $0.075 Electricity: —————— 3,413 * 100%

*1,000,000 = $21.98/MMBtu

$0.42 Natural Gas: ———————— *1,000,000 = $6.18/MMBtu (100 * 1,000) * 68%

Finding the existing peak consumption for the equipment under investigation is usually pretty easy. Most of

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the time there is a nameplate that tells you the capacity in Btuh, MBH, gallons per hour, cubic feet per hour, or connected electrical load in Watts or kW. With this number known you can size the peak fuel consumption rate required by the replacement equipment. This may be several times higher than the running load for the process, but it is more than likely the rate of heat delivery required for a reasonable start up period. Therefore, it now becomes necessary to evaluate the running consumption in order to correctly calculate the potential for savings. This is extremely difficult to nail down. If it is an electrically heated device, the use of a recording power meter will give you accurate results for the processes completed over the period of the recording. Whatever the method you use to arrive at an annual consumption value, remember to adjust it to reflect possible differences in end-use efficiency between fuel sources and distribution methods. In many cases when you are comparing similar fuel sources, i.e., natural gas to propane or #4 fuel oil to #2 fuel oil, the annual consumption in Btus will be identical. The operating savings will be the net cost of the existing fuel source savings minus the cost to purchase the proposed fuel. Annual Btu ————————————— Existing End-use Eff.

* Existing $/Btu –

Annual Btu Annual ————————————— * Proposed $/Btu = Operating Proposed End-use Eff. Savings

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That’s all there is to it. The really hard part is to calculate the cost to install the new equipment. You must take into consideration all the necessary appurtenances to make the conversion complete. A chimney is not usually easy to install as a retrofit item. Again, this is where creativity can blossom. Remember to let your imagination run past the obvious difficulties. For instance, if they are using electric heat because there is a restriction against an open flame in an explosion proof area, consider installing the burner equipment outside of the restricted area. You can then pipe the heat into the restricted area with some kind of thermal fluid, i.e., steam, hot oil, hot water, etc. Don’t forget that you have to include the cost of the energy for the pump when you calculate the operating costs.

ANECDOTES Richard’s Retrofit Rules Fuel Switching Rule #1: “The production managers think that using oil or gas fuel sources will contaminate the product.” Fuel Switching Rule #2: “There is no easy way to install a chimney.” Fuel Switching Rule #3: “Transporting heat through pipes or ducts is much more difficult, and messy, than using wires.”

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Fuel Switching Rule #4: “Infrared is defined by the surface temperature of the infrared source, not the fuel.” My first fuel switching project involved converting steam boilers that burned #4 fuel oil to “dual fuel” burners. This maintained the #4 fuel burning capability and added the ability to burn natural gas. The natural gas could be purchased very cheaply at interruptible rates. This is actually a very simple project requiring the purchase and installation of packaged components and the extension of a gas line into the building. However, politically I opened a hornet’s nest. I found out that my boss had removed dual fuel burners and removed the gas pipe about five years before I was hired. Unfortunately I found out after I had officially made the proposal. His decision came under the category: “It seemed like a good idea at the time.” The burners needed extensive repairs and the gas pipe was in the way of an expansion project. And it was about ten months before the first oil embargo. Gas was expensive and oil was very cheap. As you can see, timing is everything. The best answer for a fuel switching project is to leave the existing equipment in place for a back up, if at all possible. Even if it adds some extra cost to the project. It will help production managers to sign on to the project if they think that they can say “I told you so!” and get back to their old ways fast. And you may need that back door. Another project involved fuel switching at the start of the design stage. The company was planning an expansion of both the building and production equipment. The heart of the project was the purchase and installation of a

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new, larger, faster, high temperature tenter frame (continuous process fabric curing/drying oven). The current available steam pressure of 100 psi could only produce temperatures of about 340°F. The tenter frame required 400°F. Therefore, the original plans included the typical electrically heated equipment. This would add a connected load of 700 kW which would require a serious increase in the size of the electrical service equipment. After considerable investigation, it was decided to install a natural gas fired hot oil heater. This would produce oil at 600°F and circulate it through the tenter frame. The additional installation costs were easily offset by the reduction in the electrical service requirements. And the operating savings were significant, even taking into account the operation of the 30 horsepower circulation pumps.

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CHAPTER 9

HEAT RECOVERY INTRODUCTION The evaluation of heat recovery conservation measures relies heavily on two things: 1) having a large enough temperature difference between the heat source and the heat sink, and 2) having someplace that you can use the heat if it can be recovered. This is one of those areas in energy conservation engineering where you can really get your creative juices flowing. Just remember that, while it may have been exciting to be a Pioneer, a lot of Pioneers were killed by Indians. But at this stage of the engineering trail you should be free to explore. The best way to cover your trail is to start off by announcing that you are “only in the brainstorming stage.” Then if they start shooting flaming arrows, or just burst out laughing, you can retreat with honor. Or just run. Heat transfer is a one way street. You can only go from hot to cold. The way you make something hot is to place it next to something hotter. The way you make something cold is to place it next to something colder (that way you make the something colder become hotter and 157

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your object, which is hotter, becomes colder—clear?). In all cases, if the two objects are left together long enough they will both become the same temperature somewhere between the two original temperatures. This is one of the laws of thermodynamics. I didn’t bother to look up which law it is because, honestly, who cares? It is important to remember that this law exists because it should force you to remember the golden rule: Do Unto Others, Then Leave! In order for there to be a worthwhile project to recover heat (or cold) you must take that energy away immediately. Otherwise your temperature difference driving the heat transfer will eventually disappear. The two things for an ideal situation are a reasonably steady flow of wasted energy and a reasonably steady need for the recovered energy. This can be overcome to some degree with adequate storage capacity for either side of the transfer. But storage costs capital and loses heat (cold) to standby losses. High temperature exhaust streams can produce hot water. Process cooling is often accomplished by running city water through a heat exchanger, cooling the process fluid and heating the city water. The more audits you do, the more you’ll be surprised to see this clean, hot water dumped down the drain. The first question is whether the plant can use that much hot water. If not, look into a cooling tower and water recirculation system. This will use more energy, but will save major amounts of money in water and sewer charges. (Remember, you are there to help save money, not just energy.)

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DEFINITIONS Delta T

No, this is not the name of a Mississippi river boat. The temperature difference between two items or between two locations on the same item is called the Delta T because the Greek letter ∆ is used to indicate the mathematical difference between two numbers.

Approach Temperature

This is really the Approach Temperature Difference. This is the smallest ∆T between the fluid stream that you are transferring from and the fluid stream that you are transferring to. In essence, it is the closest number of degrees that the target fluid can approach the source fluid to maintain heat transfer. It is a function of the heat transfer characteristics of the heat exchanger under evaluation. Specifically it allows you to define the lowest (or highest) temperature that the target fluid can achieve. In a cooling tower, it is the ∆T between the ambient air wet bulb temperature and the leaving water temperature. In a heat exchanger it is the expected difference between the entering source fluid temperature and the outlet target fluid temperature.

Specific Heat

This is simply the number of Btus it takes to raise a pound of a substance one degree Fahrenheit. The specific heat of water is 1.0.

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THINGS

TO

LOOK

FOR

1.

Stacks above the roof and drainage pits in the floor. I like to look at a building as an energy conversion box. The energy comes in through pipes and wires and leaves through the building envelope. If it doesn’t leave with the product through the loading dock, then it leaves through the walls, out the exhausts, and down the drain.

2.

Water vapor in the air. Open tanks are very difficult to do anything with, because, in most cases, if they could be closed they would have been closed already. But a fog in the air could also come from a hot water drain line (or steam leaks).

3.

Listen for water splashing or air moving through a duct. Touch (carefully) the pipes and the ducts. If they are hot or cold find out more.

THINGS

TO

LOOK

OUT FOR

1.

No place to use the recovered heat. It does no one any good to recover heat from one location only to send down the drain at another location.

2.

Contaminated fluid streams. Many exhaust streams contain contaminants that are down right scary. There are many reliable approaches to avoiding cross

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contamination. But remember, no savings are worth being responsible for a health disaster. Secondly, there are contaminants that will stay a vapor at elevated temperatures. You must be aware of the dew point of these vapors. Some of the contaminants will form corrosive liquids when they condense out of the vapor stage. Others may become pools of flammable liquid that can spontaneously ignite.

CALCULATIONS One of the first calculations should be to determine how much energy is available for recovery. The actual formula depends on the medium that is carrying the energy.

Air

Btuh = 1.08 * CFM * ∆T

{The constant, 1.08, is the conversion factor taking the density of air times the specific heat of air times 60 (min./hr). In the real world the density and specific heat will vary depending on the actual conditions. However, this variation is so slight that it will be negligible for all but the most extreme cases.}

Water

Btuh = 500 * GPM * ∆T

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{The constant, 500, is the conversion factor taking the density of water lbs./gal) times the specific heat (1.0) times 60 (min./hr).}

Other Liquids (Oil, Glycol, Etc.) Btuh = GPM * (Lbs./Gal.) * Specific Heat * ∆T * 60 min/hr The questions that you have to answer are specific to the situation: what is the GPM or CFM and what is the ∆T? But as I said before: that’s what you get paid for.

ANECDOTES Richard’s Retrofit Rules Heat Recovery Rule #1: “The heat source will always be the furthest distance it can be from where you can use it.” Heat Recovery Rule #2: “The waste heat stream with the most energy is the one with the most dangerous chemicals.” Heat Recovery Rule #3: “If the project goes smoothly during installation, the output temperatures will be lower than expected.” Heat Recovery Rule #4: “If you find a large source of wasted energy you will not be able to find a matching use for it.”

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Heat Recovery Rule #5: “Production managers do not trust ‘second hand’ energy.” My first heat recovery project was in a textile dyehouse. My first month on the job as Corporate Energy Manager I observed clean, hot water going down the drain. Not just a little: 75 gpm at 140°, from three dye machines at a time. This was the cooling water discharge from a shell and tube heat exchanger that cooled the dye bath water from 205° to 100°. This was clean, fresh water. I then found out about a storage tank, piping, and pumping system that had been installed to store this clean, hot water for use when the process called for hot water. This whole system had been installed, used for a while, then abandoned in place. The reason, as told to me by the dyehouse manager, was that the hot water made the pipes rust more than the cold water. Which made no sense at all to me. The reason that they believed this was that several dyelots had several yards of fabric ruined by rust, and, of course, “That never happened until we started using the recovered hot water.” Fortunately, the position of Energy Manager was new to this company and I was the first one. This meant that politically I was favored, at least for a while. Therefore, I was able to get them to reactivate the recovery system. This went on for about two weeks until the dyehouse manager came into my office with about ten yards of fabric to show me the rust that was ruining his product. He unfolded the fabric and exposed a large collection of flakes of rust, both tiny and up to 1/4” across. I picked

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up the largest flake and remarked that it was incredibly flat to have come from the inside of a pipe. I then turned it over and discovered that one side was painted gray. I then pointed out that I was pretty sure that the inside of the water pipes had not been painted recently. Finally, after he told me which dye machine this had come from, I reminded him that that machine was the only machine that was not connected to the recovered hot water system. This is the source of Rule #5. All this evidence must have been there all those earlier times that rust showed up. But their prejudice against second hand heat blinded them. So where did the rust come from? It took some observation time, and a serendipitous coincidence of being in the right place at the right time, but I finally solved the mystery. The dye baths involved were open tanks that the fabric sat in and the dye was recirculated through the tank. One day I stopped to ponder the rust question at a tank that was full of clean, standing water and no fabric. I observed tiny flakes of rust floating on the surface. I was boggled by the thought that rust introduced through a submerged fill pipe would float to the surface. So I bent over and touched one of them which broke the surface tension of the water and the flake sank to the bottom. Just as I was beginning to realize the truth, a large flake that was about 1-1/2” across went splook into the water in front of my face and drifted to the bottom like a leaf in a light breeze. I looked up and observed the two-ton hoist on a monorail above all five tanks for loading and unloading

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the 2,000 pound (dry weight) dyelots. This monorail was painted gray. I emphasize “was” because the effect of the clouds of steam from the dye baths had rusted most of the paint and as a dyelot was moved with the hoist the monorail would flex and more rust and paint would fall into the dye baths. The solution was to institute an annual preventive maintenance schedule of scraping and repainting the monorail. A second heat recovery project involved recovering hot water from the exhaust of a tenter frame (an oven for drying fabric and/or curing the chemicals added for various reasons). This oven routinely operated at 400°F with about 12,000 cfm exhaust. By installing a water coil in the exhaust I was able to reclaim close to 80 gpm of 120° water for use in the dyehouse. All went very well until about three or four weeks passed. On Monday morning, at about 9:00 am, the duct caught fire just down stream from the coil. Because it was contained in the ductwork it wasn’t too serious. The fire department put it out reasonably quickly and there was no collateral damage. In about three days the duct was replaced and the operation was back in service. Until another three or four weeks passed and the duct caught on fire again in the morning. What was happening? The chemicals from all the different dyes, and fabric treatments, etc. were condensing in the ductwork then cooling down further when the oven was off during third shift. This chemical soup was turning into a witches brew that would spontaneously ignite

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when it was reheated enough times. Certainly the potential for heat recovery was greatly increased during the fire, but the city fire department did not appreciate their role in this heat recovery project. They made it clear that they would no longer provide the water supply without some hefty labor charges. And that, of course, made the payback period much too high. Factory Mutual is still scratching their heads over that one.

CHAPTER 10

HEATING, VENTILATING, AIR CONDITIONING

AND

INTRODUCTION HVAC projects are a pain in the neck! If you ever thought that you knew what you were doing as an engineer, take on a heating or cooling project in an office building with more than two employees. As a matter of fact, I recommend that all energy engineers be required to work on the HVAC system in their own office before they go out in the field. Within two weeks you will be happy to spend whatever energy it takes to stop the complaints. You will never again make fun of the designer that oversized everything in order to get an instantaneous response. You will understand why the unit was left on overnight so that the president’s secretary was comfortable the instant that she arrived. And you will never, never suggest that the electric heaters under the desks should be removed and banned. Why? Because life is too short to make that many enemies, especially if they know where you work. All the careful formulas, calculations, measurements, 167

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etc., miss the fact that your goal is to create a level of comfort for a human being. That isn’t so bad, except that I lied. You don’t have to create a level of comfort for a human being, you must satisfy many human beings simultaneously! I’m sorry to inform you that the best that you can hope for is failure. Your real goal is to avoid a lynch mob. Why do I paint such a depressing picture? Because no matter how sophisticated, how expensive, how well maintained a system that you design, you have absolutely no control over the primary mechanism for maintaining human comfort: human metabolism. You cannot make up for rapid and unexpected changes brought about by food (hot or cold), emotions (glad, mad, sad, or horny), rapid changes in activity (lunchtime volleyball followed by sitting at the computer), etc. The bad news is that there are no simple solutions to HVAC calculations. The good news is that some are less complicated than others. There are some that argue that you give up a degree of accuracy by using less complicated methods. In this case I feel that you give up a false sense of increased accuracy and you save a major amount of time. The best news is that the situations that you will deal with in evaluating an energy conservation measure are almost exclusively retrofit situations where you have only existing equipment to upgrade and replace. Generally speaking, by the time you’re called in to survey the equipment, it will be in such bad shape and so badly controlled (if at all), that anything will be an improvement.

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DEFINITIONS R-factor; Thermal Resistance Resistance is a property of a material that determines the amount of heat that will flow through a unit area given a difference in temperature. The R-factor is the measure of the ability of a material to resist letting heat pass through it to a colder area and so it is otherwise called the thermal resistance. The higher the R-factor the better it is. The units of measure for R-factor are ∆°F/Btuh/sq.ft.

U-factor; Over-all Coefficient of Heat Transfer Time rate of transfer of heat by conduction, across unit area for unit difference of temperature. The U-factor is the measure of the ability of a material to let heat pass through it to a colder area and so it is otherwise called the thermal conductivity. It is the mathematical inverse of the sum of all the R-factors. The lower the U-factor the better it is. The units of measure for U-factor are Btuh/sq.ft./°F(∆t).

Degree Day The concept of a Degree Day is based on the assumption that all buildings require heat input when the outside air temperature is 65°F, or lower. To calculate a Degree Day you must know the high and low temperature for the day. Then you subtract the average of the high and low temperature from 65. If the high is 50° and the low is 30°

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Simple Solutions to Energy Calculations

then the number of degree days for that day will be 25.

Bin Temperature

The historical temperature data collected by the NOAA are collected into groups of five degrees, i.e., 0 to 4 degrees; 5 to 9 degrees; -5 to -9 degrees; 30 to 34 degrees; etc. These are called temperature bins, or five-degree bins. This is a rather loose adaptation from the heat load calculations that pictured the heat loss graph with five degree sections, called “bins” that you “poured” energy into to fill them. By calculating the Heat Loss (Gain) Factor; which is the summary of all the transmission formulas without the ∆T term which gives you a value of Btuh/∆T, you then plot the total heat loss/gain for any given OAT. If the design IAT is 72°, and the heat loss factor (HLF) is 500 MBh/∆T, then the heat loss at 0° is 36.0 MMBtuh. At 5° the heat loss is 33.5 MMBtuh.

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The average heat loss between 36.0 MMBtuh and 33.5 MMBtuh is multiplied by the number of hours (called bin hours) that this temperature is experienced throughout the year to determine the total consumption for that period of hours in MMBtu. The easiest way to calculate the average heat loss factor is to use the temperature at center of the bin, which is 2.5°. This results in a heat loss per hour of 34.75 MMBtuh. If the total bin hours are 10 hours, then the total bin heat loss is 347.5 MMBtu. (In other words it takes 347.5 MMBtu to “fill” this bin.) This is a large improvement over Degree Days.

Ventilation

Outdoor air that is intentionally brought into the building to freshen the indoor air.

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Infiltration

Outdoor air that leaks into the building uncontrolled and usually unwanted.

EWT

Entering Water Temperature: The temperature of the water entering a heat exchange device. One device’s EWT is another device’s LWT. Normal centrifugal chiller condenser temperatures discharge water at approximately 105°F. This would be the EWT for the cooling tower serving that condenser.

LWT

Leaving Water Temperature: The temperature of the water leaving a heat exchange device. One device’s LWT is another device’s EWT. Normal centrifugal chiller condenser temperatures are designed for an EWT of approximately 86°F. This would be the LWT for the cooling tower. In general, if the condenser water is lower, the chiller efficiency goes higher. However, again in general, most chillers require that the condenser water temperature stay above 70°F for proper operation.

THINGS

TO

LOOK

FOR

The simplest savings can be found by looking at holes and controls. Holes in the building envelope and insulation that allow infiltration and excessive conduction losses not only

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waste heating and cooling energy, they cause some of the worst complaints. The cold draft on the back of the neck, cold feet, etc. are the most difficult to overcome usually requiring the application of brute force in the form of energy from the heating system. Damper controls that are not functioning, or damper actuators that are not connected, are another big energy waste in air handling systems that I have found. The outside air damper frozen in the 100% open position wastes an incredible amount of heating and cooling energy. Caution: Outside air dampers that are frozen closed are generally out of code compliance and will add energy load in both summer and winter when you correct the problem. Since larger dampers control larger volumes of air, a broken damper means major energy costs or savings! A larger damper also requires stronger actuating systems. These stronger requirements often are only marginally met and so they fail if there is only a slight increase in friction or a decrease in pneumatic pressure, etc. Dampers require constant supervision of the human variety. I have seen too many computer based energy management systems (EMS) that have dutifully reported that the damper had been told to move, but the reality upon visual inspection was that the damper was rusted open for months. The discharge air sensor was completely satisfied because the heating and cooling coils were adding enough energy to correct the problem. Therefore, no complaints—no problem! Hot and chilled water valves that are stuck or not seated completely cause another major waste condition where the preheat and heating coils are warming 45°F

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Simple Solutions to Energy Calculations

outside air to 70° so that the cooling coil can cool it back down to 55°. All the while the discharge air sensor is completely satisfied and the energy management computer is happy because it thinks that it’s doing a wonderful job. After all, it told the outside air dampers to open for economizer, and the hot water valves to close, and the discharge air sensor is reading 55° like it is supposed to. Everything is right in its world.

CALCULATIONS Existing building simulation software was developed originally to assist in the design of new systems for new buildings. Its adaptation to energy conservation requires the same level of detail regarding building materials, orientation, physical equipment, occupancy, etc. to allow the complicated and interdependent algorithms to develop an answer. This is extremely time consuming, especially since it has to be partially repeated one, or more, times to evaluate various options. Unfortunately, in my opinion, the principal value of such an effort is the impression on the customer when you deliver the reams of printout that such an exercise will produce. With all that time and data sheets and calculations and hourly graphs, etc. ad infinitum, it becomes extremely difficult for someone to criticize the results. This takes advantage of the newest GIGO computer law: Garbage In—Gospel Out. The baseline model of the existing conditions is usu-

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ally compared to the last year’s actual energy consumption. If the model’s results are significantly different from the billing history, certain operating parameters and assumptions are adjusted until the results are acceptable. In simple terms: the Garbage In is modified until the Garbage Out is visually pleasing to the observer. Likewise, the verification process for energy savings always includes the comparison of the energy consumption before the retrofit and after the retrofit. Unless the weather data are continuously updated, the program will use weather data that are a composite of multiple years’ data. If either of the years under scrutiny were extreme in comparison to the composite, then the program results are less trustworthy. This is compounded by the calculate-and-compare method of establishing the baseline. If you modify your input such that your output matches an extremely cold weather year, then your model will incorrectly evaluate the response to an average or extremely warm weather year. My first point is that these problems cannot be avoided without a major investment in time and software. My second point is that the improvement over simpler methods is not significant enough to be economically justifiable. In an article in Strategic Planning for Energy and the Environment*, James Waltz relates some of his experiences making accurate predictions with the use of building modeling software. He states: *”Practical Experience in Achieving High Levels of Accuracy in Energy Simulations of Existing Buildings“; James P. Waltz, PE, Strategic Planning for Energy and the Environment, 1995, Vol. 15, No. 2

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A high level of accuracy can only be achieved through optimization of three factors: 1.

An intimate understanding of the simulation tool

2.

An intimate understanding of the building to be simulated

3.

Careful analysis and critique of output data

It is certainly possible to acquire and maintain the first factor. Anyone can become familiar with the modeling software that they use. In fact, as a manager of an ESCO engineering office, I assigned a specific individual to become our “Modeling Guru.” Mostly because I didn’t want to invest all the necessary engineering man-hours for the rest of the office to get that familiar. Not only that, but the thought of having several computers tied up while the program ran its course was unacceptable. Factors number two and three are the areas where I take considerable exception to the value of the results. Especially number three: “Careful analysis and critique of output data.” I understand that when Mr. Waltz talks about this “careful analysis” that he is referring to the person with the “intimate understanding of the simulation tool” and the “intimate understanding of the building” making the analysis. Unfortunately the analysis and critique never stops there. I have observed on more than one occasion the dedicated and extensive observations and careful data

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entry of an experienced engineer become corrupted by an engineering review process where the reviewing authority rejected the results because they were not close enough to his experience. In other words, the model indicated that some portion, say the ventilation heat load, was too low or too high, in his opinion. Remember, the reviewer had never left his office. It got so bad that the engineers writing the report got to know what the reviewers expected and made their adjustments before sending the report. Factor number two, an intimate understanding of the building, is very difficult to obtain in less than a week for a good sized building. Certainly you can only observe one season. If the building has been modified and added to by several different architects over the last 20 years, or the assignment is a college campus with multiple buildings, it is impossible! I would love to finally get an assignment to do a study of a college campus with multiple (20 or more) buildings that didn’t have to be completed in two to three months. In my world if you don’t get them an answer in less than three months you’ve lost their attention and the opportunity. It’s not right. It’s just reality! So if there is no time to become intimately familiar with the building, and the engineering review process can throw your careful work out the window, what can you do? As in all cases, the more information, the better. Unfortunately you seldom have enough factual HVAC load data. And too much analysis with too few facts is just a waste of time. Predicting future HVAC loads is inherently less reliable than predicting the weather. Not only do you have to deal

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with the unreliable and sweeping weather generalizations based on historical averages of data from locations remote from the actual building site, you also have to deal with the unreliable occupancy predictions based on unsubstantiated, subjective, data usually from someone’s personal memory. Then throw in generalizations about the building envelope transmission, infiltration, thermal storage, orientation, solar gains, etc., and the Farmer’s Almanac starts to look like a NASA publication in comparison. Simple HVAC modeling based on connected capacity, reasonable operating assumptions, and bin temperature data can produce sensible and defensible results. The best part is that once you admit that it is only your best “guess,” you are open to making reasonable modifications when suggested by personnel with a more intimate knowledge of the building’s operation and operating history. For instance: if the boiler operator tells you that he absolutely needs to run his backup boiler on the coldest day of the year, any model that does not include that consumption, whether the heat loss calculations indicate that it is required or not, will be hopelessly flawed. So how do we proceed? First you must determine the installed capacity of the HVAC equipment. If the chiller can only produce 100 tons of chilled water, any analysis that requires more than 100 tons will not give you an accurate prediction. Second, discuss with the operating personnel whether the installed equipment meets the building’s needs on the peak day. Here you must use your judgment. If you don’t think that the source is reliable keep asking people until you have at least shown “due diligence.” If there is no

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information, I settle on 80% of the installed active capacity (exclude any clearly redundant equipment). This is what I assume to be the design building load, since a reasonable engineer would include a 20% to 25% safety factor in their original design. This percentage is adjusted as information is received and evaluated. It is assumed that internal loads are fixed by occupancy (as does most modeling software), i.e., the number of people remains constant, their activity and electrical loads also remain constant, etc. For cooling loads I start at 65% internal load. This means that only 35% of the cooling load is varied as the weather, i.e., outdoor air temperature, changes. This is obviously only a guess. In the past it has proved to be a successful starting point in my iterations toward the final calculations. At 65% internal load you are assuming that the minimum load at the design outdoor air temperature will be 65%. This at least keeps the calculations conservative. The internal loads for the heating calculations are assumed to be 0%. This is chosen because the equipment will be designed for the coldest temperatures which usually occur at night when the building is most likely unoccupied. Again, this means that your calculations will start conservative. You can, however, include the internal gains if you can isolate the bin temperature data that apply to occupied periods. This will increase the accuracy of your analysis by lowering your occupied consumption but maintaining the peak load requirements at unoccupied times. Taking a look at the various heat loss equations for the building envelope conduction, ventilation, and infitration:

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Conduction Ti – To * A; & Q = U * A * Ti – To R

Q=

Q Ti To U R A

= = = = = =

Heat loss; Btuh Indoor air temperature; °F Outdoor air temperature; °F Heat Loss coefficient Thermal resistance Square foot of envelope

Ventilation and Infiltration CFM * 1.08 * (Ti – To) = Btuh The comparison of the heat loss at the outdoor air temperature To2 with respect to the outdoor temperature To1 is as follows for conduction: CFM * 1.08 * (Ti – To) = Btuh And for ventilation and infiltration: Q2 = U * A * (Ti – To2) ——————————— Q1 = U * A * (Ti – To1) Both of these become: Q2 = (Ti – To2) ——————————— Q1 = (Ti – To1)

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Assuming To1 = the design outdoor temperature, TD; and To2 = the bin temperature in question, TB the relationship becomes: QB = (Ti – TB) ———————— QD = (∆TD) Therefore: (Ti – TB) QB = QD ————— (∆TD) If the design indoor air temperature (IAT) is 72°F and the design outdoor air temperature OAT is 10°F, then the design ∆TD is (72 – 10) = 62°. If the bin temperature is 35°F, then the heat loss load percentage is: (72 – 35) ———— (62)

= 0.597 = 59.7%

The OAT cooling load calculations require the inclusion of the base load percentage as follows: Bin T – IAT * 1 – % Base Load + % Base Load Design ∆T

If the design IAT is 72°F and the design OAT is 97°F, then the design ∆T is (97 – 72) = 25°. The Base Load is assumed to be 65%.

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Simple Solutions to Energy Calculations

If the bin temperature is 85°F, then the percentage cooling load on the installed equipment is: 85 – 72 * 1 – 0.65 + 0.65 = 0.832 = 83.2% 25

Finally, when dealing with the heating load, don’t forget to include the combustion efficiency and standby losses of the boilers in your calculations. If you are looking only at controls, then the efficiency calculations can be included at any point in the calculations. If you are also looking at a burner or boiler retrofit or replacement, then the best method is to apply the existing fuel-to-steam efficiency to your building load calculations to model the actual heat input requirements at the burner. Then do a separate “after” calculation with the new efficiency to calculate the input Btu’s required with the new equipment. This also tends to force you to clearly show all your parameters on your spreadsheet which makes it much easier to review and troubleshoot. An example of such an approach for a heating load is shown below. The spreadsheet calculates the estimated heat loss based upon the installed capacity and assumed safety factors. The cut off at 67.5°F reflects the change over as the outside air rises above the IAT setpoint. The “Design Peak Load” is determined by taking the boiler input Btuh and applying the Design Boiler Efficiency to determine the available output Btuh. Then the 80% Design Load Factor is applied to determine the output

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Btuh that most likely was calculated when the installation was designed. The % Load times the Design Peak Load is the Required Btuh. The Required Btuh times the Annual Hours for that Bin Temperature divided by 1,000,000 is the Existing MMBtu. BASE ASSUMPTIONS ————————————————————— Largest Boiler Btuh; input Design Boiler Efficiency Design Load Factor Design Peak Load; Btuh BIN

ANNUAL

———

————

TEMP

67.5 62.5 57.5 52.5 47.5 42.5 37.5 32.5 27.5 22.5 17.5

HOURS

755 718

8,380,000 80% 80% 5,363,200

%

REQUIRED EXISTING

———

————— —————

LOAD

5.8%

423

21.5%

1,151,263

755

31.9%

1,712,854

655

26.7%

780 937 625 402 247

233

589,671

16.2%

651

308,875

MMBtu

11.0%

675 656

Btuh

37.2% 42.4% 47.6% 52.9% 58.1%

870,467

1,432,059

588 938

1,115

1,993,650

1,555

2,555,242

1,597

2,274,446 2,836,038 3,116,834

2,131 1,140 770

12.5 172 63.4% 3,397,629 584 ————————————————————————————

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The “Required Btuh” at 67.5° Bin Temp. is simply: 5.759% × 5,363,200 Btuh = 308,875. The “Existing MMBtu” for that bin temperature is 308,875 Btuh × 755 Annual Hours/1,000,000 = 233.2 MMBtu. On the following page is a sample spreadsheet showing the variation in building cooling load as the outside air temperature changes. The cut off at 62.5° reflects the field survey data information that the controls are set such that the chiller does not come on until the outside air rises above the 60°. Therefore, it doesn’t matter what any envelope heat transfer model comes up with for a load, if the chiller is off there is no consumption. The “Design Peak Load” is determined by taking the chiller peak tons and applying the 80% Design Load Factor to determine the required output tons that most likely was calculated when the installation was designed. If you have more detailed information about the facility and can determine the actual peak tons from that information, by all means go with what you feel is more accurate. The % Load times the Design Peak Load is the Existing Load {Tons}. The Existing Load {Ton-Hrs} is the Existing Load times the Annual Hours for that Bin Temperature. The Ton-Hrs can be converted to kWh if you know the efficiency of the cooling system. (You can even get fancier if you know the part-load efficiencies. A simple “Lookup” command in a spreadsheet can apply the appropriate part-load efficiency to the percentage of peak load for each bin temperature.) Looking back at the heating consumption model, there is a reality check that I use to keep my feet on the ground. I compare the modeled total annual consump-

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BASE ASSUMPTIONS CHILLER PLANT CAPACITY; TONS:

650

ASSUMED DESIGN LOAD FACTOR:

80%

DESIGN PEAK LOAD; TONS:

520

ASSUMED BASE LOAD; %:

65%

DESIGN IAT:

72

DESIGN PEAK ∆T:

25.5

Weather Data: (Syracuse/Hancock IAP)

MCWB HRS. HRS)

——

74 73 71 68 66 63 60 56 52 47

EXISTING

EXISTING

BIN

BIN

ANNUAL

%

LOAD

LOAD

TEMP

TEMP

HOURS

LOAD

(TONS)

( T O N S-

——

——

——— ———

———

————

97.5 92.5 87.5 82.5 77.5 72.5 67.5 62.5 57.5 52.5

95/99 90/94 85/89 80/84 75/79 70/74 65/69 60/64 55/59 50/54

171.5 165.5 158.5 150.5 143.5 135.5 127.5 118.5

100.0% 93.1% 86.3% 79.4% 72.5% 65.7% 58.8% 52.0%

520.0 484.3 448.6 412.9 377.3 341.6 305.9 270.2

89,180.0 80,153.9 71,107.5 62,147.6 54,136.1 46,282.5 39,000.0 32,018.2

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Simple Solutions to Energy Calculations

Design Boiler Efficiency Design Load Factor Design Peak Load; Btuh REQUIRED Btuh ——————

308,875 589,671 870,467 1,151,263 1,432,059 1,712,854 1,993,650 2,274,446 2,555,242 2,836,038 3,116,834 3,397,629

EXISTING MMBtu —————

80% 80% 5,363,200

MMBtu

233 HISTORICAL TOTAL 423 CONSUMPTION: 588 10,500 755 938 MODEL VS HISTORICAL 1,115 113% 1,555 2,131 1,597 1,140 770 584 TOTAL 11830.0 ————————————————————————————

tion to the historical annual consumption. But you must be careful to be sure that the historical consumption just includes heating loads. If the comparison indicates that the model is within 15% of the historical, I consider the model to be valid. Again, this is the “calculate and compare” approach, but I just check to make sure that I am not so far off that I look like a complete idiot. Remember, your bin data are an historical average. In the Northeast, the past several

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187

winters have been either above average cold or above average warm. Therefore, I think that 15% plus or minus puts me in the ballpark for reliability of potential savings predictability. The % Load column is a useful number for calculating the load on a variable speed drive for the heating or cooling system pumps. The idea behind the installation of a VSD for the pump in such a system is that the Btuh required by the system is directly related to the mass flow of the transfer fluid (assuming constant temperature of the transfer fluid). If you only need 25% of the peak Btuh then you only need 25% of the peak gpm. (See the chapter on Pumps to determine how to translate that 25% gpm requirement to pump rpm.) You must remember that this does not take into account humidity. Which is why it works best for heating in the northern climates. The shortcoming of all methods is that the historical data, due to its complexity, gives priority to dry bulb temperatures and only averages wet bulb temperatures. The bin data mention only the MCWB (Mean Coincident Wet Bulb) for each bin temperature for the month. I have certainly used this information to determine the latent load for each bin temperature and bin hours. This is truly reaching beyond “simple.” Even so, while the method is better than nothing, it is even further from reality than the sensible load predictions. One final comment. This approach is meant as a solution to the problem of lack of available time to do it the best way. If you have the time and the tools, by all means, use them! But if you want to optimize accuracy and time spent, this method seems to work.

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Simple Solutions to Energy Calculations

ANECDOTES A large university campus utilized large central air handling units in each of its 12 main buildings. The heating and cooling was supplied from a central chilled water loop and a central high temperature hot water loop. When I say large air handling units, in this case I mean 50,000 to 70,000 cfm. These units were often served from the same fresh air intake system, providing up to 150,000 cfm of outside air. This system amounted to some very large concrete, subterranean rooms located on the other side of the outside air dampers. Upon actual inspection of these dampers, it was found that over 75% were not working at all and were stuck 100% open. It was clear that maintenance personnel had not been near these dampers in a very long time. In almost all fresh air intake rooms there were lounge chairs and couches. In one room there was an extension cord, a refrigerator, a microwave, and lots of empty beer cans. I’m just glad the resident wasn’t there when I was. Another facility with large (50,000 cfm) air handling units in an electronic microchip operation required precise temperature and humidity controls. What they termed “photospec” air. This was a state-of-the-art facility so there were EMS connections and readouts everywhere. I spent several hours in the control room paging through the computer screens with their wonderful graphics showing me the temperatures inside the air handling unit from the outside air intake to the discharge air plenum. It also showed the position of all the dampers and the hot water, chilled water, and humidifier valves.

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It was late spring in northern Vermont. (Late spring in northern Vermont is followed immediately by early fall.) The high internal loads required cooling. The outside air temperature was 48°F and the economizer control had the outside air damper full open. The EMS indicated that the valves for the hot water preheat and main heating coils were 100% closed. However, since the discharge air temperature downstream of these coils read 70°F, the chilled water coil downstream from the fan was full open so that the discharge air temperature was at the desired setpoint of (you guessed it!) 48°F. The hot water valve for the heating coil was stuck open. If you do the math they lost 1.73 MMBtuh heating the outside air then lost another 1.73 MMBtuh cooling it back down to where it was when they started.

Richard’s Retrofit Rules HVAC Rule #1: There are only two kinds of damper linkages: Those that are broken and those that are about to break. The larger the damper, the more likely it is that the actuator system will be broken. HVAC Rule #2: The more money that you spend on an energy management computer, the more money you will need to spend on the operating personnel. (Otherwise, the only thing that they will be able to do is OVERRIDE.) HVAC Rule #3: There is no such thing as a “temporary” override.

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Simple Solutions to Energy Calculations

HVAC Rule #4: Unit ventilators are the noisiest, most uncontrollable, and hardest to maintain way to heat and cool a room. (That is why they are the most common. People are so used to being uncomfortable while the equipment is still making so much noise that nobody is ever really sure if it is broken.) HVAC Rule #5: Air source heat pumps north of 400 N. Latitude are just plain stupid!

CHAPTER 11

SUMMARY

OF

CALCULATIONS

LIGHTING Relative Light Output (RLO) X.X xX.X xX.X xX.X xX.X

———

X.XX

(% lumens vs. original) (% change in Ballast Factor) (Lumen Depreciation at time of evaluation) (Dirt Depreciation at time of evaluation) (% change in Coefficient of Utilization after retrofit) (% RLO)

PUMPING Centrifugal Pump Affinity Laws N 1 F1 = N 2 F2 N1 N2

2

=

191

H1 H2

192

Simple Solutions to Energy Calculations

N 1 F1 = N 2 F2 N1 N2

N = rpm;

2

H1 H2

=

H = Head (Pressure);

HP = Horsepower

% Minimum pump speed (rpm): %N min

H min HD

N min = ND

% Pump speed (rpm) at any flow; F1:

%N

N1 = ND

H min + HD

F1 * 1– FD

H min HD

Horsepower at any flow; F1: HP1 = HPD *

H min F + 1 * 1– F HD D

Standard Pump Formulas: GPM * TDH HP = ———————— 3960 * Pump Eff.

H min HD

3

Summary of Calculations

D=

193

GPM K

GPM = D 2 * K

D = Pipe diameter; inches K = 10 (good) 4 fps K = 15 (avg.) 6 fps

K = 20 (poor) 8 fps TDH = Static Head + Friction Head

Friction Head =

0.001246 * GPM 2 * L D5

D

= Pipe Diameter, inches

L

= Equivalent length of pipe

0.001246

= factor for galvanized or black iron pipe (rough)

0.000623 H(ft)

= factor for copper or plastic pipe (smooth) = PSI ÷ 0.4331 PPH GPM = —— 500

PPH

= Pounds per Hour (of water; either as steam or liquid)

194

Simple Solutions to Energy Calculations

FANS See PUMPING for Affinity Law Formulas Standard Fan Formulas Horsepower =

CFM * Pressure (In. W.C.) ———————————— 6356 * Fan Eff.

MOTORS Basic Motor Formula HP kW = 0.74 —— Eff.

Motor Efficiency Saving kW Saved = 0.746 *

HP1 HP2 – EffMotor1 EffMotor2

HP1 = existing motor load HP2 = motor load for new motor

Summary of Calculations

195

INSULATION Heat Loss through Walls (Conduction) Q=

Q Ti To U R A

Ti – To = A; & Q = U * A * Ti – To R

= = = = = =

Heat loss, Btuh Indoor air temperature, °F Outdoor air temperature; °F Heat Loss coefficient Thermal resistance Square foot of envelope

Energy Savings from Increased Insulation ∆Q = ∆T * A *

Temp.)

1 – 1 ; & ∆Q = U – U * A * ∆T i o R1 R2

∆Q ∆T

= =

Btuh saved (Outside Air Temp. – Inside Air

U R A

= = =

Heat Loss coefficient Thermal resistance Square foot of envelope

Infiltration Heat Loss Total perimeter lineal feet of crack * 0.25 = CFM 2

196

Simple Solutions to Energy Calculations

CFM * 1.08 * ∆T = Btuh

Heat Loss from Cylinders (like pipes) 2 * π * k * L * To – Tin Ro In R in

k = L = To = Tin = ro = rin =

conductivity, Btu/hr.ft. °F, or Btu/hr. ft2 °F/ft. length of cylinder temperature outside of cylinder temperature inside of cylinder outer radius of cylinder inner radius of cylinder

FUEL SWITCHING Unit Cost for Fuel Source $/Unit ———————————— (Btu/Unit) * (end-use Eff.)

* 1,000,000 = $/MMBtu

Annual Cost Savings (MMBtu1 * $/MMBtu1) – (MMBtu2 * $/MMBtu2) MMBtu1 = Annual consumption before switch MMBtu2 = Annual consumption after switch (may be equal to MMBtu1)

Summary of Calculations

197

HEAT RECOVERY Energy Available in an Air Stream Btuh = CFM * 1.08 * ∆T

Energy Available in a Water Stream Btuh = GPM * 500 * ∆T

Energy Available in a Liquid Stream Btuh = GPM * Lbs/Gal * Specific Heat * ∆T * 60 min/hr

HVAC Conduction Q=

Q Ti To U R A

= = = = = =

Ti – To * A; & Q = U * A * Ti – To R

Heat loss; Btuh Indoor air temperature, °F Outdoor air temperature; °F Heat loss coefficient Thermal resistance Square foot of envelope

198

Simple Solutions to Energy Calculations

Ventilation and Infiltration CFM * 1.08 * (Ti – To) = Btuh

Estimated Heat Loss At Any Bin Temperature QB = QD *

Ti – TB ∆T D

Estimated Heating Load (% of peak load) At Any Bin Temperature: Bin T – IAT Design ∆T

= % Load

Estimated Cooling Load (% of peak load) At Any Bin Temperature: Bin T – IAT * 1 – % Base Load + % Base Load = % Load Design ∆T

CHAPTER 12

ENERGY MYTHS

AND

MAGIC:

An Examination of Conservation Measures and Calculations

“ONE

MAN’S MAGIC

IS ANOTHER MAN’S ENGINEERING.”* Practical engineering not only looks at the pure science, it looks at how it is applied, where it is applied, and why it is applicable in the specific situation in question. This is important! Magic can turn into Myth by myth-application (sorry, couldn’t resist). The sad truth is that once a customer has been burned by an over zealous claim most other legitimate opportunities at that location are lost due to the mistrust that is generated. My experience has been that this industry is plagued with two kinds of myths. The first myth deals with energy conservation measures that are flawed but look like they must work. The second of the myths deals with conservation measures that are absolutely true. They are true engineering magic. The myth is not whether the specific measure will produce energy savings. The myth is that anyone can calculate what those savings will be. As an *The Notebooks of Lazarus Long, by Robert A. Heinlein 199

200

Simple Solutions to Energy Calculations

engineer, I can easily guarantee that the sun will come up in the morning. I cannot guarantee that you will be able to see it! I would like to address individual energy saving strategies as they are presented in the marketplace and discuss their actual operation and performance in the real world. Not only is it important to determine if the savings will be real, it is just as important to be able to calculate the predicted savings. If my income is predicated upon the customer seeing the sun in the morning, I will have a huge budget to ensure that my client can get above the clouds.

EXAMPLE #1 To start with a simple example; consider the energy savings claims from reducing the temperature of domestic hot water. This is a standard recommendation suggested in almost every energy conservation manual. Let’s explore how this will reduce the energy consumption and how to calculate the savings. In the Energy Conservation Workbook for Multifamily Housing distributed by the New York State Energy Office (may it rest in peace), measure number 301: “Reduce water temperature: For every 10°F, that water temperature is reduced, approximately 1.7 million Btu’s will be saved per apartment per year.” Who knows where that number came from? (By reverse engineering, 1.7 MMBtu at a 10°F ∆T means that 56 gallons per day was the base number.) In the Energy Workbook for Office Buildings distrib-

Energy Myths and Magic

201

uted by the Missouri Department of Natural Resources; Division of Energy, conservation measure number F1: “Hot water should be stored and distributed at the lowest useful temperature… Normally, 110°F water is adequate for domestic use.” They go on to give the following formula to calculate the savings: Q = Annual Gallons * 8.35 * (T – 110) This equation is certainly correct, but what about the real world application? Suppose that 100° is the desired temperature at the faucet. This will ultimately be determined by the hands of the user. If the water temperature is 140° the user will be scalded and will add cold water to achieve the desired 100°. Let’s assume that the faucet in question delivers two gallons per minute of water. If the hot water is 140° and the cold water is 60°, then to achieve 100° on your skin the ratio of hot to cold must be 50%/50%. If 50% of the two gpm is hot water, then the consumption of 140° water will be one gpm. One gpm at an 80°F ∆T (60° heated to 140°) will consume 40 MBh. To lower the hot water temperature to 110° would change the hot to cold ratio to 80%/20%. If 80% of the two gpm is hot water then the consumption of 110° water will be 1.6 gpm. This increased consumption; 1.6 gpm at a 50°F ∆T (60° heated to 110°) will consume the same 40 MBh. There will be no energy savings from the temperature reduction.

202

Simple Solutions to Energy Calculations

The calculation method in the NY State Energy Office manual predicts annual savings of 5.1 MMBtu per apartment. The Missouri manual comes up with the same number if you allow 56 gpd for 365 days per year. Certainly, the heat loss in the distribution piping and storage tank will be lower if the water temperature is lower, but that is seldom calculated because it is usually considered insignificant. (Or too difficult.)

Control and Thermodynamic Continuity

In order to properly evaluate the effect of any specific conservation measure you must determine and follow an unbroken thermodynamic and control connection from the energy source to the end-use. What I mean is that the end-use will absorb all the energy that it’s given only if it needs it to achieve equilibrium. The energy that it needs is the responsibility of thermodynamic system. The energy that is “given” is the responsibility of the control system and equipment. I have tried to write that many different ways and I fear that I do not have the skill to say it clearly and concisely. Therefore, let me give an example that I hope is better at getting the point across. Take a simple room. The heat is lost through the envelope. A hot water radiator is in the room, backed up by an oil burner. There is no thermostat or control valve on the radiator. The radiator is sufficiently sized for the area, with a safety factor, but it is not an infinite source of Btu’s. At some point the Btu’s discharged into the room will raise the IAT until the heat loss driven by the OAT

Energy Myths and Magic

203

and ∆T exactly equals the radiator capacity. That is the thermodynamic connection. If you add a thermostat and a control valve, the energy from the oil burner is still connected thermodynamically. The heat is still being lost at a rate determined by the ∆T between the IAT and OAT. But the thermodynamic connection has an added component: the thermostat. If the heat loss is sufficiently greater than the heat input, the IAT will drop. This will cause the thermostat to increase the heat input until it is sufficiently greater than the heat loss, causing the IAT to rise. This is the thermodynamic connection between the room and the thermostat. The thermostat responds and affects the thermodynamic connection between the room and the oil burner. This is the control connection from the end-use to the energy source. It doesn’t matter how large the oil burner is. The control connection will only cause it to send out energy at a rate that satisfies the thermodynamic connection between the room and the thermostat. If this makes sense, then you will understand the next energy myth. Because it also doesn’t matter how hot the hot water is (as long as it is hot enough to meet the requirements for the radiator to work.)

EXAMPLE #2 tem.

Water temperature reset for a hot water heating sys-

The energy magic is that when the outside air temperature is warmer, the IAT/OAT ∆T is lower than the

204

Simple Solutions to Energy Calculations

equipment was designed to handle. Therefore the quantity of Btu’s needed for equilibrium is lower, so the water temperature delivered by the boiler is lowered. The theory is that this will prevent overheating of the space. This is energy magic when the system is a simple, single-zone, circulating loop for many rooms, with a single thermostat (like a residential application). In this case all areas that are not near the thermostat and have a lower heat loss will be overheated. This not only wastes energy by making spaces warmer than needed, it will increase the IAT/OAT ∆T, this will cause the heat loss to increase, and even more energy will be wasted. However, this type of control is regularly installed on commercial systems with a single loop and multiple zone thermostats. This turns the magic into a myth. The zone thermostats are all that should be needed to avoid the overheating of the individual zones. It doesn’t matter how hot the water is, when the total heat transfer is sufficient to satisfy the thermostat, the thermostat will stop the flow of heat. The major effect of lower water temperature will be to slow down the response time of heating equipment allowing the space to be calling for heat longer. Finally, even in cases where it is the correct thing to do, calculations can only provide a dismally poor approximation of how much energy this will save. The energy savings must be calculated by determining the nature of the thermodynamic connection back to the energy source. That is a moving target based on the heat loss characteristics of the room, the varying OAT, and the actions of the occupants.

Energy Myths and Magic

205

Thermodynamically, why not just assume that the boiler runs continuously instead of just enough to meet the design heat loss calculation? Because the room’s occupants will do something to relieve their discomfort. The only thing that they can do is increase the room’s heat loss such that it reaches equilibrium at a lower IAT. That’s engineering language to say that they open the windows. My experience has been that standard calculation methods assume an IAT that is a fixed number of degrees above the design IAT. The excess energy to meet that IAT is calculated for the various bin temperatures and the savings are totaled. That will certainly give you an answer. How close is it? That’s the myth.

EXAMPLE #3 Consider the energy savings claims from Optimal Start options in energy management systems. The theory stems from the idea that a building that is being maintained at occupied temperatures when it is empty is certainly wasting energy. Let’s explore how these control strategies reduce the energy consumption for the customer. First, you start with the assumption that the HVAC system always comes on too early in the morning. The optimal start option will look at the outside temperature and the inside temperature to determine the correct starting time such that the interior of the building will be at the correct temperature (warm or cool) by the time the first occupant arrives. Therefore, the energy that is wasted

206

Simple Solutions to Energy Calculations

in a building at occupied design temperatures with no occupants is saved. So how are these savings calculated? The first question is how long is the building at design temperature without any occupants? That is impossible to answer. If the required warm-up/cool-down time was the same every day, there wouldn’t be any need for the optimal start option. So the assumption is that the building operators err on the side of caution and set the scheduled start time to cover the worst condition with a safety factor. Some engineers assume that the building “averages” one hour per day of excess temperature control. So what does this mean to the system’s energy consumption? For both heating and cooling, the energy to bring the building back to setpoint temperature must be expended whether it’s on time, early, or late. So the energy saved is the fan and pump energy and the energy to overcome heat gains or losses to maintain the design temperature. For heating, the energy to maintain the design temperature at start-up occurs at the coldest time of the day. In addition, there are no internal gains from lights, people, or equipment. It is, in fact, at or near the peak load conditions depending on the outside air temperature on any particular morning. So savings have the potential for being significant. However, there should be zero outside air brought in until the building is occupied. In addition, on a very cold morning, the required warm-up time may be such that the previously fixed schedule start time would have left the facility uncomfortably cold by the time the first occupant

Energy Myths and Magic

207

arrived. In that case, more energy is used with the optimal start than would have been used without it. For cooling, the energy to maintain the design temperature also occurs at the coolest time of the day. It also occurs when there are no internal gains. It should, in fact, be minimal and possibly zero some of the time since the outside air will be below the design setpoint causing a conductive heat loss and the HVAC unit to operate on economizer. Therefore, the only savings come from fan energy. Fan energy savings calculations should be the easiest, assuming that there are no variable speed drives or VAV options. With VSD and VAV the energy will be highest at start up and taper off to minimum as the design temperature is approached. Even so, there is still the question of hours use during optimal start and unnecessary hours use without optimal start. Whether the magnitude of the watts or Btuh is significant or insignificant, the total annual energy wasted is directly a function of how long the waste is going on. This is virtually impossible to determine with any accuracy. So “everybody” accepts a value of one hour per day since no one can prove or disprove it.

CONCLUSION “WHAT

ARE THE FACTS?

WHAT ARE THE FACTS?

AGAIN

AND AGAIN AND AGAIN—

SHUN WISHFUL THINKING, IGNORE DIVINE REVELATION, FORGET WHAT “THE STARS FORETELL” AVOID OPINION, CARE NOT WHAT THE NEIGHBORS THINK, NEVER MIND THE UNGUESS-

208 ABLE

Simple Solutions to Energy Calculations

“VERDICT

OF HISTORY”—WHAT ARE THE FACTS, AND TO HOW

MANY DECIMAL PLACES?

YOU

PILOT ALWAYS INTO AN UNKNOWN

FUTURE, FACTS ARE YOUR SINGLE CLUE.

GET

THE FACTS!”*

It is my sincere wish that energy engineers will continue to look at what they are doing to make sure that it makes sense. It is my hope that these few examples will be adequate to show the need of finding the thermodynamic and control continuity from the energy source to the enduse. Other examples could have been chosen; such as: •

Infrared heaters: If they save energy because they “only heat objects, not the air” (which is true), that defines the thermodynamic continuity. But they are controlled by thermostats that respond to air temperature. Therefore, until the air is heated to satisfy the thermostat, the energy source will stay ON.

•

DDC controls: Energy savings are claimed from the reduction in the temperature overshoot in a zone. However, if the energy source is staying ON longer to cause the overshoot, doesn’t it stay OFF longer while the overshoot becomes undershoot? If you shorten the ON time don’t you also shorten the OFF time? Is it possible that the net result will be approximately the same energy but greater comfort?

•

Variable speed drives: If you apply the affinity laws to the flow and horsepower, you must apply the affinity

*The Notebooks of Lazarus Long, by Robert A. Heinlein

Energy Myths and Magic

209

laws to the pressure. This is often forgotten. When the pump must deliver a certain minimum pressure to the system before flow can even occur, i.e., if it must overcome the lift to the cooling tower on the roof, there will be some minimum speed that must be met before any flow can occur. In other words, if you want 25% flow and the output pressure from the pump won’t be greater than the lift to the tower unless the pump is at 60% rpm, there will be no flow in the system until the pump speed is greater than 60%. There is not enough time to go into so many of these calculations. (I have discussed this concept of minimum speed and how to calculate its value in previous papers.) That is not to say that the measures cannot be saving energy. The conclusion I want to convey is that we must examine the myths and carefully turn them into magic with sound practice. Or discard them for the fantasies that they are. I would be happy to hear from other engineers regarding methods of analysis and calculations for myths or magic. I am always looking for better ways to get the right answers. My address is: Richard R. Vaillencourt P.O. Box 459 Canterbury, CT 06331 Or call me at (860) 546-1124.

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RICHARD’S RETROFIT RULES LIGHTING Lighting Rule #1: “You can never save more energy than shutting it off.” Lighting Rule #2: “There is at least one person who won’t like it.” Lighting Rule #3: “Always retrofit lighting at night.” Lighting Rule #4: “The occupancy sensor will turn the lights off when the company president is in the bathroom.”

PUMPING Pumping Rule #1: “You can never save any more energy than shutting it off.” Pumping Rule #2: “You will never get a production manager to accept downsizing.” Pumping Rule #3: “No one in production knows the flow requirements.” 211

212

Simple Solutions to Energy Calculations

FANS Fan Rule #1: “You can never save any more energy than shutting it off. But when you do, someone will complain that it’s too (stuffy, cold, hot, quiet, etc.).” Fan Rule #2: “It is always a surprise to find out that the fan noise is more important than the actual movement of air.” Fan Rule #3: “Employees always believe that more exhaust is better.”

MOTORS Motor Rule #1: “You can never save any more energy than shutting it off.” Motor Rule #2: “No one in the plant knows how much the motor is loaded.” Motor Rule #3: “If you burn your hand when you touch it, it is at least 100% loaded.”

INSULATION Insulation Rule #1: “If you burn your hand on the equipment you can probably save some energy.”

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213

Insulation Rule #2: “If the insulation is in the way of production or maintenance workers it won’t last long enough to have a payback.” Insulation Rule #3: “If it’s already covered with asbestos, the payback period will be measured in decades.” Insulation Rule #4: “Sometimes you have to point out the safety features of adding insulation, not the energy saved.”

FUEL SWITCHING Fuel Switching Rule #1: “The production managers think that using oil or gas fuel sources will contaminate the product.” Fuel Switching Rule #2: “There is no easy way to install a chimney.” Fuel Switching Rule #3: “Transporting heat through pipes or ducts is much more difficult, and messy, than using wires.” Fuel Switching Rule #4: “Infrared” is defined by the surface temperature of the infrared source, not the fuel.”

HEAT RECOVERY Heat Recovery Rule #1: “The heat source will always be the furthest distance it can be from where you can use it.”

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Simple Solutions to Energy Calculations

Heat Recovery Rule #2: “The waste heat stream with the most energy is the one with the most dangerous chemicals.” Heat Recovery Rule #3: “If the project goes smoothly during installation the output temperatures will be lower than expected.” Heat Recovery Rule #4: “If you find a large source of wasted energy you will not be able to find a matching use for it.” Heat Recovery Rule #5: “Production managers do not trust “second hand” energy.”

HVAC HVAC Rule #1: “There are only two kinds of damper linkages. Those that are broken and those that are about to break. The larger the damper, the more likely it is that the actuator system will be broken.” HVAC Rule #2: “The more money that you spend on an energy management computer, the more money you will need to spend on the operating personnel. (Otherwise, the only thing that they will be able to do is OVERRIDE.)” HVAC Rule #3: “There is no such thing as a ‘temporary override.’”

Richard’s Retrofit Rules

215

HVAC Rule #4: “Unit ventilators are the noisiest, most uncontrollable, and hardest to maintain way to heat and cool a room. (That is why they are the most common. People are so used to being uncomfortable while the equipment is still making so much noise that nobody is ever really sure if it is broken.)” HVAC Rule #5: “Air source heat pumps north of 40° N. Latitude are just plain stupid!”

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INDEX A Affinity Laws 79, 80, 90, 98, 120, 123, 130 approach temperature 159 assumptions 11, 15 audit assumptions 15 checklist 10 goal 4 limitations 4 walk through 1 average pipe diameter 95 B bin temperature 170 Btu 137, 148 Btuh 137, 148 C calculations air heat recovery 161 building envelope 141 conduction heat transfer 180 cost per MMBtu 151 energy costs 17 fans 122 fan horsepower 122 fuel switching 151, 196 heat recovery 197 high efficiency motors 131 infiltration 142, 180 insulation 139 liquids heat recovery 162 load by bin temperatures 102

maximum gpm 97 minimum horsepower 86 minimum pump speed 84 minimum speed 124 motors 129 PUMP 88 pump horsepower 96 TDH 84 variable speed pump 115 ventilation 180 water heat recovery 161, 162 Carnot, Sadi 6 conduction 137 control and thermodynamic continuity 202 cooling tower 84 D dampers 173 Degree Day 169 Delta T 159 design peak load 182, 183

E ECM 3 electric heat conversion 147 EMS 173 end-use efficiency 138, 150 Energy Conservation Measure 3 energy management systems 173 217

218

Simple Solutions to Energy Calculations

equivalent lineal feet of pipe 94 failures 13 energy management systems 173 EWT 172

insulation 135, 195 flat surface 136 pipe 135 internal loads cooling 179 heating 179

F fans cooling tower 118 dust collection 119 duty cycling 117 exhaust 117 forced draft 118 fume exhaust 119 fume hood 119 induced draft 118 VAV 118 friction head 81, 82 fuel switching 147

L lighting 191 LWT 172

H head 80 heat gain factor 170 heat loss factor 170 heat recovery 157 exhaust 158 process cooling 158 storage 158 HLF 170 horsepower active pump 96 HVAC 167, 197 safety factor 178 I infiltration 172 infrared 149

M MBH 137, 148 MCWB 187 mean coincident wet bulb 187 MMBtu 137 MMBtuh 148 motors 127, 194 direct-coupled 129 efficiency 96 high efficiency 128, 129 load factor 130 part load efficiency 130 premium efficiency 128 slip 129 O over-all coefficient of heat transfer 138, 169 P plant engineer relationship 7 pressure 121 drop 82, 121 PUMP boiler feed water 84, 106, 107 centrifugal 82

Index constant high flow 88 efficiency 97 load factor 96 minimum requirement 84 positive displacement 89 transfer 88 pumping 191 R R-factor 138, 169 radiation 149 S safety factors 179 simple thermodynamics 25 simulation software 174 specific heat 159 spreadsheet chilled water pump 103 circulating pump 99 combined gas & oil costs 23 cooling load 185 electric costs 19, 20 fuel oil costs 23 lighting retrofit 71 natural gas & oil costs 22 occupancy sensor retrofit 73, 74

219 static head 80, 94 static pressure 122 synchronous speed 128 T TDH 82 thermal conversion factors 150 thermal resistance 137, 169 total dynamic head 82 total pressure 122 U U-factor 138, 169 utility costs 16 demand charges 17 electricity spreadsheet 19, 20 fuel spreadsheet 23 incremental fuel 23 incremental kWh 18 V valves chilled water 173 hot water 173 variable speed drives 79 velocity pressure 122 ventilation 171 VSD 79