Introduction to Environmental Engineering and Science (3rd Edition) PDF

 Introduction to Environmental Engineering and Science Gilbert M. Masters Wendell P. Ela Third Edition Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk © Pe...

 Introduction to Environmental Engineering and Science Gilbert M. Masters Wendell P. Ela Third Edition Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk © Pearson Education Limited 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners. ISBN 10: 1-292-02575-1 ISBN 13: 978-1-292-02575-9 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America  P E A R S O N C U S T O M L I B R A R Y Table of Contents 1. Mass and Energy Transfer Gilbert M. Masters/Wendell P. Ela 1 2. Environmental Chemistry Gilbert M. Masters/Wendell P. Ela 47 3. Mathematics of Growth Gilbert M. Masters/Wendell P. Ela 87 4. Risk Assessment Gilbert M. Masters/Wendell P. Ela 127 5. Water Pollution Gilbert M. Masters/Wendell P. Ela 173 6. Water Quality Control Gilbert M. Masters/Wendell P. Ela 281 7. Air Pollution Gilbert M. Masters/Wendell P. Ela 367 8. Global Atmospheric Change Gilbert M. Masters/Wendell P. Ela 501 9. Solid Waste Management and Resource Recovery Gilbert M. Masters/Wendell P. Ela 601 Index 687 I This page intentionally left blank Mass and Energy Transfer 1 Introduction 2 Units of Measurement 3 Materials Balance 4 Energy Fundamentals Problems When you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meagre and unsatisfactory kind; it may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of science. —William Thomson, Lord Kelvin (1891) 1 Introduction This chapter begins with a section on units of measurement. Engineers need to be familiar with both the American units of feet, pounds, hours, and degrees Fahrenheit as well as the more recommended International System of units. Both are used in the practice of environmental engineering. Next, two fundamental topics, which should be familiar from the study of ele- mentary physics, are presented: the law of conservation of mass and the law of con- servation of energy. These laws tell us that within any environmental system, we theoretically should be able to account for the flow of energy and materials into, and out of, that system. The law of conservation of mass, besides providing an important tool for quantitatively tracking pollutants as they disperse in the environment, reminds us that pollutants have to go somewhere, and that we should be wary of approaches that merely transport them from one medium to another. From Introduction to Environmental Engineering and Science. Third Edition. Gilbert M. Masters, Wendell P. Ela. Copyright © 2008 by Pearson Education, Inc. Published by Prentice Hall. All rights reserved. 1  Mass and Energy Transfer In a similar way, the law of conservation of energy is also an essential account- ing tool with special environmental implications. When coupled with other thermo- dynamic principles, it will be useful in a number of applications, including the study of global climate change, thermal pollution, and the dispersion of air pollutants. 2 Units of Measurement In the United States, environmental quantities are measured and reported in both the U.S. Customary System (USCS) and the International System of Units (SI), so it is important to be familiar with both. Here, preference is given to SI units, although the U.S. system will be used in some circumstances. Table 1 lists conversion factors between the SI and USCS systems for some of the most basic units that will be en- countered. In the study of environmental engineering, it is common to encounter both extremely large quantities and extremely small ones. The concentration of some toxic substance may be measured in parts per billion (ppb), for example, whereas a country’s rate of energy use may be measured in thousands of billions of watts (terawatts). To describe quantities that may take on such extreme values, it is useful to have a system of prefixes that accompany the units. Some of the most important prefixes are presented in Table 2. Often, it is the concentration of some substance in air or water that is of interest. Using the metric system in either medium, concentrations may be based on mass (usually mg or g), volume (usually L or m3), or number (usually mol), which can lead to some confusion. It may be helpful to recall from chemistry that one mole of any substance has Avogadro’s number of molecules in it (6.02 * 1023 molecules/mol) and has a mass equal to its molecular weight. Liquids Concentrations of substances dissolved in water are usually expressed in terms of mass or number per unit volume of mixture. Most often the units are milligrams (mg), TABLE 1 Some Basic Units and Conversion Factors Quantity SI units SI symbol ⫻ Conversion factor ⫽ USCS units Length meter m 3.2808 ft Mass kilogram kg 2.2046 lb Temperature Celsius °C 1.8 (°C) + 32 °F Area square meter m2 10.7639 ft2 Volume cubic meter m3 35.3147 ft3 Energy kilojoule kJ 0.9478 Btu Power watt W 3.4121 Btu/hr Velocity meter/sec m/s 2.2369 mi/hr Flow rate meter3/sec m3/s 35.3147 ft3/s Density kilogram/meter3 kg/m3 0.06243 lb/ft3 2  Mass and Energy Transfer TABLE 2 Common Prefixes Quantity Prefix Symbol ⫺15 10 femto f 10⫺12 pico p 10⫺9 nano n 10⫺6 micro m 10⫺3 milli m 10⫺2 centi c 10⫺1 deci d 10 deka da 102 hecto h 103 kilo k 106 mega M 109 giga G 1012 tera T 1015 peta P 1018 exa E 1021 zetta Z 1024 yotta Y micrograms (mg), or moles (mol) of substance per liter (L) of mixture. At times, they may be expressed in grams per cubic meter (g/m3). Alternatively, concentrations in liquids are expressed as mass of substance per mass of mixture, with the most common units being parts per million (ppm) or parts per billion (ppb). To help put these units in perspective, 1 ppm is about the same as 1 drop of vermouth added to 15 gallons of gin, whereas 1 ppb is about the same as one drop of pollutant in a fairly large (70 m3) back-yard swimming pool. Since most concentrations of pollutants are very small, 1 liter of mixture has a mass that is essentially 1,000 g, so for all practical purposes, we can write 1 mg/L = 1 g/m3 = 1 ppm (by weight) (1) 3 1 mg/L = 1 mg/m = 1 ppb (by weight) (2) In unusual circumstances, the concentration of liquid wastes may be so high that the specific gravity of the mixture is affected, in which case a correction to (1) and (2) may be required: mg/L = ppm (by weight) * specific gravity of mixture (3) EXAMPLE 1 Fluoridation of Water The fluoride concentration in drinking water may be increased to help prevent tooth decay by adding sodium fluoride; however, if too much fluoride is added, it can cause discoloring (mottling) of the teeth. The optimum dose of fluoride in drinking water is about 0.053 mM (millimole/liter). If sodium fluoride (NaF) is purchased in 25 kg bags, how many gallons of drinking water would a bag treat? (Assume there is no fluoride already in the water.) 3  Mass and Energy Transfer Solution Note that the mass in the 25 kg bag is the sum of the mass of the sodium and the mass of the fluoride in the compound. The atomic weight of sodium is 23.0, and fluoride is 19.0, so the molecular weight of NaF is 42.0. The ratio of sodium to fluoride atoms in NaF is 1:1. Therefore, the mass of fluoride in the bag is 19.0 g/mol mass F = 25 kg * = 11.31 kg 42.0 g/mol Converting the molar concentration to a mass concentration, the optimum con- centration of fluoride in water is 0.053 mmol/L * 19.0 g/mol * 1,000 mg/g F = = 1.01 mg/L 1,000 mmol/mol The mass concentration of a substance in a fluid is generically m C = (4) V where m is the mass of the substance and V is the volume of the fluid. Using (4) and the results of the two calculations above, the volume of water that can be treated is 11.31 kg * 106 mg/kg V = = 2.97 * 106 gal 1.01 mg/L * 3.785 L/gal The bag would treat a day’s supply of drinking water for about 20,000 people in the United States! Gases For most air pollution work, it is customary to express pollutant concentrations in volumetric terms. For example, the concentration of a gaseous pollutant in parts per million (ppm) is the volume of pollutant per million volumes of the air mixture: 1 volume of gaseous pollutant = 1 ppm (by volume) = 1 ppmv (5) 106 volumes of air To help remind us that this fraction is based on volume, it is common to add a “v” to the ppm, giving ppmv, as suggested in (5). At times, concentrations are expressed as mass per unit volume, such as mg/m3 or mg/m3. The relationship between ppmv and mg/m3 depends on the pressure, tem- perature, and molecular weight of the pollutant. The ideal gas law helps us establish that relationship: PV = nRT (6) where P ⫽ absolute pressure (atm) V ⫽ volume (m3) n = mass (mol) 4  Mass and Energy Transfer R ⫽ ideal gas constant = 0.082056 L # atm # K - 1 # mol - 1 T = absolute temperature (K) The mass in (6) is expressed as moles of gas. Also note the temperature is expressed in kelvins (K), where K = °C + 273.15 (7) There are a number of ways to express pressure; in (6), we have used atmospheres. One atmosphere of pressure equals 101.325 kPa (Pa is the abbreviation for Pascals). One atmosphere is also equal to 14.7 pounds per square inch (psi), so 1 psi = 6.89 kPa. Finally, 100 kPa is called a bar, and 100 Pa is a millibar, which is the unit of pressure often used in meteorology. EXAMPLE 2 Volume of an Ideal Gas Find the volume that 1 mole of an ideal gas would occupy at standard tempera- ture and pressure (STP) conditions of 1 atmosphere of pressure and 0°C temper- ature. Repeat the calculation for 1 atm and 25°C. Solution Using (6) at a temperature of 0°C (273.15 K) gives 1 mol * 0.082056 L # atm # K-1 # mol-1 * 273.15 K V = = 22.414 L 1 atm and at 25°C (298.15 K) 1 mol * 0.082056 L # atm # K-1 # mol-1 * 298.15 K V = = 22.465 L 1 atm From Example 2, 1 mole of an ideal gas at 0°C and 1 atm occupies a volume of 22.414 L (22.414 * 10 - 3 m3). Thus we can write 1 m3 pollutant>106 m3 air mol wt (g/mol) mg/m3 = ppmv * * * 103 mg/g ppmv 22.414 * 10-3 m3/mol or, more simply, ppmv * mol wt mg/m3 = (at 0°C and 1 atm) (8) 22.414 Similarly, at 25°C and 1 atm, which are the conditions that are assumed when air quality standards are specified in the United States, ppmv * mol wt mg/m3 = (at 25°C and 1 atm) (9) 24.465 In general, the conversion from ppm to mg/m3 is given by ppmv * mol wt 273.15 K P(atm) mg/m3 = * * (10) 22.414 T (K) 1 atm 5  Mass and Energy Transfer EXAMPLE 3 Converting ppmv to mg/m3 The U.S. Air Quality Standard for carbon monoxide (based on an 8-hour mea- surement) is 9.0 ppmv. Express this standard as a percent by volume as well as in mg/m3 at 1 atm and 25°C. Solution Within a million volumes of this air there are 9.0 volumes of CO, no matter what the temperature or pressure (this is the advantage of the ppmv units). Hence, the percentage by volume is simply 9.0 percent CO = * 100 = 0.0009% 1 * 106 To find the concentration in mg/m3, we need the molecular weight of CO, which is 28 (the atomic weights of C and O are 12 and 16, respectively). Using (9) gives 9.0 * 28 CO = = 10.3 mg/m3 24.465 Actually, the standard for CO is usually rounded and listed as 10 mg/m3. The fact that 1 mole of every ideal gas occupies the same volume (under the same temperature and pressure condition) provides several other interpretations of volumetric concentrations expressed as ppmv. For example, 1 ppmv is 1 volume of pollutant per million volumes of air, which is equivalent to saying 1 mole of pollu- tant per million moles of air. Similarly, since each mole contains the same number of molecules, 1 ppmv also corresponds to 1 molecule of pollutant per million molecules of air. 1 mol of pollutant 1 molecule of pollutant 1 ppmv = 6 = (11) 10 mol of air 106 molecules of air 3 Materials Balance Everything has to go somewhere is a simple way to express one of the most funda- mental engineering principles. More precisely, the law of conservation of mass says that when chemical reactions take place, matter is neither created nor destroyed (though in nuclear reactions, mass can be converted to energy). What this concept allows us to do is track materials, for example pollutants, from one place to another with mass balance equations. This is one of the most widely used tools in analyzing pollutants in the environment. The first step in a mass balance analysis is to define the particular region in space that is to be analyzed. This is often called the control volume. As examples, the control volume might include anything from a glass of water or simple chemical mixing tank, to an entire coal-fired power plant, a lake, a stretch of stream, an air basin above a city, or the globe itself. By picturing an imaginary boundary around 6  Mass and Energy Transfer Control volume boundary Accumulation Inputs Outputs Reactions: Decay and generation FIGURE 1 A materials balance diagram. the region, as is suggested in Figure 1, we can then begin to quantify the flow of ma- terials across the boundary as well as the accumulation and reaction of materials within the region. A substance that enters the control volume has four possible fates. Some of it may leave the region unchanged, some of it may accumulate within the boundary, and some of it may be converted to some other substance (e.g., entering CO may be oxidized to CO2 within the region). There is also the possibility that more substance may be produced (e.g., CO may be produced by cigarette smoking within the con- trol volume of a room). Often, the conversion and production processes that may occur are lumped into a single category termed reactions. Thus, using Figure 1 as a guide, the following materials balance equation can be written for each substance of interest: a b = a b - a b + a b Accumulation Input Output Reaction (12) rate rate rate rate The reaction rate may be positive if generation of the substance is faster than its decay, or negative if it is decaying faster than it is being produced. Likewise, the ac- cumulation rate may be positive or negative. The reaction term in (12) does not imply a violation of the law of conservation of mass. Atoms are conserved, but there is no similar constraint on the chemical compounds, which may chemically change from one substance into another. It is also important to notice that each term in (12) quan- tifies a mass rate of change (e.g., mg/s, lb/hr) and not a mass. Strictly, then, it is a mass rate balance rather than a mass balance, and (12) denotes that the rate of mass accu- mulation is equal to the difference between the rate the mass enters and leaves plus the net rate that the mass reacts within the defined control volume. Frequently, (12) can be simplified. The most common simplification results when steady state or equilibrium conditions can be assumed. Equilibrium simply means that there is no accumulation of mass with time; the system has had its inputs held constant for a long enough time that any transients have had a chance to die out. Pollutant concentrations are constant. Hence the accumulation rate term in (12) is set equal to zero, and problems can usually be solved using just simple algebra. A second simplification to (12) results when a substance is conserved within the region in question, meaning there is no reaction occurring—no radioactive decay, bacterial decomposition, or chemical decay or generation. For such conserva- tive substances, the reaction rate in (12) is 0. Examples of substances that are typi- cally modeled as conservative include total dissolved solids in a body of water, heavy metals in soils, and carbon dioxide in air. Radioactive radon gas in a home or 7  Mass and Energy Transfer Stream Cs, Qs Accumulation = 0 Cm, Qm Mixture Reaction = 0 Wastes C , Q w w Q = flow rate C = concentration of pollutant FIGURE 2 A steady-state conservative system. Pollutants enter and leave the region at the same rate. decomposing organic wastes in a lake are examples of nonconservative substances. Often problems involving nonconservative substances can be simplified when the reaction rate is small enough to be ignored. Steady-State Conservative Systems The simplest systems to analyze are those in which steady state can be assumed (so the accumulation rate equals 0), and the substance in question is conservative (so the reaction rate equals 0). In these cases, (12) simplifies to the following: Input rate = Output rate (13) Consider the steady-state conservative system shown in Figure 2. The system contained within the boundaries might be a lake, a section of a free flowing stream, or the mass of air above a city. One input to the system is a stream (of water or air, for instance) with a flow rate Qs (volume/time) and pollutant concentration Cs (mass/volume). The other input is assumed to be a waste stream with flow rate Qw and pollutant concentration Cw. The output is a mixture with flow rate Qm and pollutant concentration Cm. If the pollutant is conservative, and if we assume steady state conditions, then a mass balance based on (13) allows us to write the fol- lowing: Cs Qs + Cw Qw = Cm Qm (14) The following example illustrates the use of this equation. More importantly, it also provides a general algorithm for doing mass balance problems. EXAMPLE 4 Two Polluted Streams A stream flowing at 10.0 m3/s has a tributary feeding into it with a flow of 5.0 m3/s. The stream’s concentration of chloride upstream of the junction is 20.0 mg/L, and the tributary chloride concentration is 40.0 mg/L. Treating chloride as a conser- vative substance and assuming complete mixing of the two streams, find the downstream chloride concentration. Solution The first step in solving a mass balance problem is to sketch the prob- lem, identify the “region” or control volume that we want to analyze, and label the variables as has been done in Figure 3 for this problem. 8  Mass and Energy Transfer Control volume boundary Cs = 20.0 mg/L Cm = ? Qs = 10.0 m3/s Qm = ? Cw = 40.0 mg/L Qw = 5.0 m3/s FIGURE 3 Sketch of system, variables, and quantities for a stream and tributary mixing example. Next the mass balance equation (12) is written and simplified to match the prob- lem’s conditions a b = a b - a b + a b Accumulation Input Output Reaction rate rate rate rate The simplified (12) is then written in terms of the variables in the sketch 0 = CsQs + CwQw - CmQm The next step is to rearrange the expression to solve for the variable of interest— in this case, the chloride concentration downstream of the junction, Cm. Note that since the mixture’s flow is the sum of the two stream flows, Qs + Qw can be substituted for Qm in this expression. CsQs + CwQw CsQs + CwQw Cm = = Qm Qs + Qw The final step is to substitute the appropriate values for the known quantities into the expression, which brings us to a question of units. The units given for C are mg/L, and the units for Q are m3/s. Taking the product of concentrations and flow rates yields mixed units of mg/L # m3/s, which we could simplify by applying the conversion factor of 103 L = 1 m3. However, if we did so, we should have to reapply that same conversion factor to get the mixture concentration back into the desired units of mg/L. In problems of this sort, it is much easier to simply leave the mixed units in the expression, even though they may look awkward at first, and let them work themselves out in the calculation. The downstream con- centration of chloride is thus (20.0 * 10.0 + 40.0 * 5.0) mg/L # m3/s Cm = = 26.7 mg/L (10.0 + 5.0) m3/s This stream mixing problem is relatively simple, whatever the approach used. Drawing the system, labeling the variables and parameters, writing and simplify- ing the mass balance equation, and then solving it for the variable of interest is the same approach that will be used to solve much more complex mass balance problems later in this chapter. 9  Mass and Energy Transfer Batch Systems with Nonconservative Pollutants The simplest system with a nonconservative pollutant is a batch system. By defini- tion, there is no contaminant flow into or out of a batch system, yet the contami- nants in the system undergo chemical, biological, or nuclear reactions fast enough that they must be treated as nonconservative substances. A batch system (reactor) assumes that its contents are homogeneously distributed and is often referred to as a completely mixed batch reactor (CMBR). The bacterial concentration in a closed water storage tank may be considered a nonconservative pollutant in a batch reac- tor because it will change with time even though no water is fed into or withdrawn from the tank. Similarly, the concentration of carbon dioxide in a poorly ventilated room can be modeled as a nonconservative batch system because the concentration of carbon dioxide increases as people in the room breathe. For a batch reactor, (12) simplifies to Accumulation rate = Reaction rate (15) As discussed before, the reaction rate is the sum of the rates of decay, which are negative, and the rates of generation, which are positive. Although the rates of reaction can exhibit many dependencies and complex relationships, most nuclear, chemical, and biochemical reaction rates can be approximated as either zero-, first-, or second-order reaction rates. In a zero-order reaction, the rate of reaction, r(C), of the substance is not dependent on the amount of the substance present and can be expressed as r(C) = k (generation) or r(C) = -k (decay) (16) where k is a reaction rate coefficient, which has the units of mass # volume # time - 1 -1 (e.g., mg # L-1 # s-1). The rate of evaporation of water from a bucket is a zero-order reaction because the rate of loss of the water is not dependent on the amount of water in the bucket but is only dependent on the nearly constant surface area of the water exposed to the air. Using (15) and (16), the mass balance for the zero-order reaction of a sub- stance in a batch reactor is dC V = -Vk dt The equation is written as a zero-order decay, denoted by the negative sign on the right-hand side of the equation. So that each term in the mass balance has the cor- rect units of mass/time, both the accumulation and reaction terms are multiplied by the volume of the batch reactor. Although in a batch system, the volume coefficient disappears by dividing both sides by V, it is worth remembering its presence in the initial balance equation because in other systems it may not cancel out. To solve the differential equation, the variables are separated and integrated as C t dC = -k dt (17) 3 3 C0 0 which yields C - C0 = -kt 10  Mass and Energy Transfer 200 Decay 150 k Concentration 100 50 Production −k 0 0 20 40 60 80 100 Time FIGURE 4 Concentration of a substance reacting in a batch system with zero-order kinetics. Solving for concentration gives us C = C0 - kt (18) where C0 is the initial concentration. Using (18) and its analog for a zero-order gen- eration reaction, Figure 4 shows how the concentration of a substance will change with time, if it is reacting (either being generated or destroyed) with zero-order ki- netics. For all nonconservative pollutants undergoing a reaction other than zero- order, the rate of the reaction is dependent on the concentration of the pollutant present. Although decay and generation rates may be any order, the most commonly encountered reaction rate for generation is zero-order, whereas for decay it is first- order. The first-order reaction rate is r(C) = kC (generation) or r(C) = - kC (decay) (19) where k is still a reaction rate constant, but now has the units of reciprocal time (time - 1). Radioactive decay of radon gas follows first-order decay—the mass that decays per given time is directly proportional to the mass that is originally present. Using (15) and (19), the mass balance for a pollutant undergoing first-order decay in a batch reactor is dC V = - VkC dt This equation can be integrated by separation of variables and solved similarly to (17). When solved for concentration, it yields C = C0 e - kt (20) That is, assuming a first-order reaction, the concentration of the substance in question decays exponentially. The first-order time dependence of a nonconservative pollutant’s concentration in a batch system can be seen in Figure 5. Although not nearly as common as first-order processes, sometimes a sub- stance will decay or be generated by a second-order process. For instance, hydroxyl 11  Mass and Energy Transfer 200 150 Decay Concentration 100 Production 50 0 0 20 40 60 80 100 Time FIGURE 5 Concentration of a substance reacting in a batch system with first-order kinetics. radical reactions with volatile organic pollutants is a key step in smog generation. However, if two hydroxyl radicals collide and react, they form a much less potent hydrogen peroxide molecule. This is a second-order reaction, since two hydroxyl radicals are consumed for each hydrogen peroxide produced. The second-order reaction rate is r(C) = kC2 (generation) or r(C) = - kC2 (decay) (21) where k is now a reaction rate constant with units of (volume # mass - 1 # time - 1). Again substituting (21) into (15), we have the differential equation for the second- order decay of a nonconservative substance in a batch reactor dC V = - VkC2 dt which can be integrated and then solved for the concentration to yield C0 C = (22) 1 + C0kt Figure 6 shows how the concentration of a substance changes with time if it decays or is produced by a second-order reaction in a batch reactor. Steady-State Systems with Nonconservative Pollutants If we assume that steady-state conditions prevail and treat the pollutants as noncon- servative, then (12) becomes 0 = Input rate - Output rate + Reaction rate (23) The batch reactor, which has just been discussed, can’t describe a steady-state system with a nonconservative substance because now there is input and output. Although there are an infinite number of other possible types of reactors, simply em- ploying two other types of ideal reactors allows us to model a great number of envi- ronmental processes. The type of mixing in the system distinguishes between the 12  Mass and Energy Transfer 200 150 Concentration 100 50 Production Decay 0 0 20 40 60 80 100 Time FIGURE 6 Concentration of a substance reacting in a batch system with second-order kinetics. two other ideal reactors. In the first, the substances within the system are still homogeneously mixed as in a batch reactor. Such a system is variously termed a continuously stirred tank reactor (CSTR), a perfectly mixed flow reactor, and a complete mix box model. The water in a shallow pond with an inlet and outlet is typically modeled as a CSTR as is the air within a well-ventilated room. The key concept is that the concentration, C, within the CSTR container is uniform through- out. We’ll give examples of CSTR behavior first and later discuss the other ideal reactor model that is often used, the plug flow reactor (PFR). The reaction rate term in the right-hand side of (23) can denote either a substance’s decay or generation (by making it positive or negative) and, for most en- vironmental purposes, its rate can be approximated as either zero-, first-, or second- order. Just as for the batch reactor, for a CSTR, we assume the substance is uniformly distributed throughout a volume V, so the total amount of substance is CV. The total rate of reaction of the amount of a nonconservative substance is thus d(CV)>dt = V dC>dt = Vr(C). So summarizing (16), (19), and (21), we can write the reaction rate expressions for a nonconservative substance: Zero-order, decay rate = - Vk (24) Zero-order, generation rate = Vk (25) First-order, decay rate = - VkC (26) First-order, generation rate = VkC (27) Second-order, decay rate = - VkC2 (28) Second-order, generation rate = VkC2 (29) Thus, for example, to model a CSTR containing a substance that is decaying with a second-order rate, we combine (23) with (28) to get a final, simple, and useful expression for the mass balance involving a nonconservative pollutant in a steady- state, CSTR system: Input rate = Output rate + kC2V (30) 13  EXAMPLE 5 A Polluted Lake Consider a 10 * 106 m3 lake fed by a polluted stream having a flow rate of 5.0 m3/s and pollutant concentration equal to 10.0 mg/L (Figure 7). There is also a sewage outfall that discharges 0.5 m3/s of wastewater having a pollutant concen- tration of 100 mg/L. The stream and sewage wastes have a decay rate coefficient of 0.20/day. Assuming the pollutant is completely mixed in the lake and assuming no evaporation or other water losses or gains, find the steady-state pollutant concen- tration in the lake. Solution We can conveniently use the lake in Figure 7 as our control volume. Assuming that complete and instantaneous mixing occurs in the lake—it acts as a CSTR—implies that the concentration in the lake, C, is the same as the concen- tration of the mix leaving the lake, Cm. The units (day - 1) of the reaction rate con- stant indicate this is a first-order reaction. Using (23) and (26): Input rate = Output rate + kCV (31) We can find each term as follows: There are two input sources, so the total input rate is Input rate = QsCs + QwCw The output rate is Output rate = QmCm = (Qs + Qw)C (31) then becomes QsCs + QwCw = (Qs + Qw)C + kCV And rearranging to solve for C, QsCs + QwCw C = Qs + Qw + kV 5.0 m3/s * 10.0 mg/L + 0.5 m3/s * 100.0 mg/L = 0.20/d * 10.0 * 106 m3 (5.0 + 0.5) m3/s + 24 hr/d * 3600 s/hr So, 100 C = = 3.5 mg/L 28.65 Outfall Qw = 0.5 m3/s Cw = 100.0 mg/ L Incoming Lake stream Outgoing V = 10.0 × 106 m3 Qs = 5.0 m3/s k = 0.20/day Cs = 10.0 mg/L C=? Cm = ? Qm = ? FIGURE 7 A lake with a nonconservative pollutant. 14  Mass and Energy Transfer Idealized models involving nonconservative pollutants in completely mixed, steady-state systems are used to analyze a variety of commonly encountered water pollution problems such as the one shown in the previous example. The same simple models can be applied to certain problems involving air quality, as the following example demonstrates. EXAMPLE 6 A Smoky Bar A bar with volume 500 m3 has 50 smokers in it, each smoking 2 cigarettes per hour (see Figure 8). An individual cigarette emits, among other things, about 1.4 mg of formaldehyde (HCHO). Formaldehyde converts to carbon dioxide with a reaction rate coefficient k = 0.40/hr. Fresh air enters the bar at the rate of 1,000 m3/hr, and stale air leaves at the same rate. Assuming complete mixing, estimate the steady-state concentration of formaldehyde in the air. At 25°C and 1 atm of pressure, how does the result compare with the threshold for eye irrita- tion of 0.05 ppm? Solution The bar’s building acts as a CSTR reactor, and the complete mixing inside means the concentration of formaldehyde C in the bar is the same as the concentration in the air leaving the bar. Since the formaldehyde concentration in fresh air can be considered 0, the input rate in (23) is also 0. Our mass balance equation is then Output rate = Reaction rate (32) However, both a generation term (the cigarette smoking) and a decay term (the conversion of formaldehyde to carbon dioxide) are contributing to the reaction rate. If we call the generation rate, G, we can write G = 50 smokers * 2 cigs/hr * 1.4 mg/cig = 140 mg/hr We can then express (32) in terms of the problem’s variables and (26) as QC = G - kCV so G 140 mg/hr C = = Q + kV 1,000 m3/hr + (0.40/hr) * 500 m3 = 0.117 mg/m3 Oasis Indoor concentration C V = 500 m3 Q = 1,000 m3/hr Q = 1,000 m3/hr Fresh air C=? 140 mg/hr k = 0.40/hr FIGURE 8 Tobacco smoke in a bar. 15  Mass and Energy Transfer We will use (9) to convert mg/m3 to ppm. The molecular weight of formaldehyde is 30, so C (mg/m3) * 24.465 0.117 * 24.465 HCHO = = = 0.095 ppm mol wt 30 This is nearly double the 0.05 ppm threshold for eye irritation. Besides a CSTR, the other type of ideal reactor that is often useful to model pollutants as they flow through a system is a plug flow reactor (PFR). A PFR can be visualized as a long pipe or channel in which there is no mixing of the pollutant along its length between the inlet and outlet. A PFR could also be seen as a conveyor belt carrying a single-file line of bottles in which reactions can happen within each bottle, but there is no mixing of the contents of one bottle with another. The behav- ior of a pollutant being carried in a river or in the jet stream in the Earth’s upper atmosphere could be usefully represented as a PFR system. The key difference be- tween a PFR and CSTR is that in a PFR, there is no mixing of one parcel of fluid with other parcels in front or in back of it in the control volume, whereas in a CSTR, all of the fluid in the container is continuously and completely mixed. (23) applies to both a CSTR and PFR at steady state, but for a PFR, we cannot make the simplification that the concentration everywhere in the control volume and in the fluid leaving the region is the same, as we did in the CSTR examples. The pollutant concentration in a parcel of fluid changes as the parcel progresses through the PFR. Intuitively, it can then be seen that a PFR acts like a conveyor belt of differentially thin batch reactors translating through the control volume. When the pollutant en- ters the PFR at concentration, C0, then it will take a given time, t, to move the length of the control volume and will leave the PFR with a concentration just as it would if it had been in a batch reactor for that period of time. Thus, a substance decaying with a zero-, first-, or second-order rate will leave a PFR with a concentration given by (18), (20), and (22), respectively, with the understanding that t in each equation is the residence time of the fluid in the control volume and is given by t = l>v = V>Q (33) where l is the length of the PFR, v is the fluid velocity, V is the PFR control volume, and Q is the fluid flowrate. EXAMPLE 7 Young Salmon Migration Every year, herons, seagulls, eagles, and other birds mass along a 4.75 km stretch of stream connecting a lake to the ocean to catch the fingerling salmon as they migrate downstream to the sea. The birds are efficient fishermen and will con- sume 10,000 fingerlings per kilometer of stream each hour regardless of the num- ber of the salmon in the stream. In other words, there are enough salmon; the birds are only limited by how fast they can catch and eat the fish. The stream’s average cross-sectional area is 20 m2, and the salmon move downstream with the stream’s flow rate of 700 m3/min. If there are 7 fingerlings per m3 in the water en- tering the stream, what is the concentration of salmon that reach the ocean when the birds are feeding? 16  Mass and Energy Transfer Lake Q = 700 m3/min Stream C 0 = 7 fish/m3 Q = 700 m3/min 5 km L = 4.7 C=? A = 20 m2 Ocean FIGURE 9 Birds fishing along a salmon stream. Solution First we draw a figure depicting the stream as the control volume and the variables (Figure 9). Since the birds eat the salmon at a steady rate that is not dependent on the concentration of salmon in the stream, the rate of salmon consumption is zero- order. So, 10,000 fish # km - 1 # hr - 1 k = = 0.50 fish # m - 3 # hr - 1 20 m2 * 1,000 m/km For a steady-state PFR, (23) becomes (18), C = C0 - kt The residence time, t, of the stream can be calculated using (33) as V 4.75 km * 20 m2 * 1,000 m/km t = = = 2.26 hr Q 700 m3/min * 60 min/hr and the concentration of fish reaching the ocean is C = 7 fish/m3 - 0.50 fish # m-3 # hr - 1 * 2.26 hr = 5.9 fish/m3 Step Function Response So far, we have computed steady-state concentrations in environmental systems that are contaminated with either conservative or nonconservative pollutants. Let us now extend the analysis to include conditions that are not steady state. Quite often, we will be interested in how the concentration will change with time when there is a sudden change in the amount of pollution entering the system. This is known as the step function response of the system. In Figure 10, the environmental system to be modeled has been drawn as if it were a box of volume V that has flow rate Q in and out of the box. Let us assume the contents of the box are at all times completely mixed (a CSTR model) so that the pollutant concentration C in the box is the same as the concentration leaving the box. The total mass of pollutant in the box is therefore 17  Mass and Energy Transfer Flow in Flow out Q, Ci Control volume V Q, C Concentration C Decay coefficient k d Generation coefficient k g FIGURE 10 A box model for a transient analysis. VC, and the rate of accumulation of pollutant in the box is VdC>dt. Let us designate the concentration of pollutant entering the box as Ci. We’ll also assume there are both production and decay components of the reaction rate and designate the decay coefficient kd and the generation coefficient kg. However, as is most common, the decay will be first-order, so kd’s units are time - 1, whereas the generation is zero-order, so kg’s units are mass # volume - 1 # time - 1. From (12), we can write a b = a b - a b + a b Accumulation Input Output Reaction rate rate rate rate dC V = QCi - QC - VkdC + kgV (34) dt where V⫽ box volume (m3) C = concentration in the box and exiting waste stream (g/m3) Ci = concentration of pollutants entering the box (g/m3) Q = the total flow rate in and out of the box (m3/hr) kd = decay rate coefficient (hr - 1) kg = production rate coefficient (g # m # hr - 1) The units given in the preceding list are representative of those that might be encountered; any consistent set will do. An easy way to find the steady-state solution to (34) is simply to set dC>dt = 0, which yields QCi + kgV Cq = (35) Q + kdV where Cq is the concentration in the box at time t = q. Our concern now, though, is with the concentration before it reaches steady state, so we must solve (34). Rearranging (34) gives QCi + kgV = -a + kd b # aC - b dC Q (36) dt V Q + kdV which, using (35), can be rewritten as = -a + kd b # (C - C q ) dC Q (37) dt V One way to solve this differential equation is to make a change of variable. If we let y = C - Cq (38) 18  Mass and Energy Transfer then dy dC = (39) dt dt so (37) becomes = -a + kd by dy Q (40) dt V This is a differential equation, which we can solve by separating the variables to give y t dy = - a + kd b dt Q (41) 3 V 3 y0 0 where y0 is the value of y at t = 0. Integrating gives y = y0e- Akd + V Bt Q (42) If C0 is the concentration in the box at time t = 0, then from (38) we get y0 = C0 - C q (43) Substituting (38) and (43) into (42) yields C - C q = (C0 - C q )e- Akd + V Bt Q (44) Solving for the concentration in the box, writing it as a function of time C(t), and expressing the exponential as exp( ) gives C(t) = C q + (C0 - C q )expc - akd + bt d Q (45) V Equation (45) should make some sense. At time t = 0, the exponential function equals 1 and C = C0. At t = q , the exponential term equals 0, and C = C q. Equation (45) is plotted in Figure 11. C∞ QCi + kgV Concentration C∞ = Q + kdV C0 0 Time, t FIGURE 11 Step function response for the CSTR box model. 19  Mass and Energy Transfer EXAMPLE 8 The Smoky Bar Revisited The bar in Example 6 had volume 500 m3 with fresh air entering at the rate of 1,000 m3/hr. Suppose when the bar opens at 5 P.M., the air is clean. If formalde- hyde, with decay rate kd = 0.40/hr, is emitted from cigarette smoke at the constant rate of 140 mg/hr starting at 5 P.M., what would the concentration be at 6 P.M.? Solution In this case, Q = 1,000 m3/hr, V = 500 m3, G = kg, V = 140 mg/hr, Ci = 0, and kd = 0.40/hr. The steady-state concentration is found using (35): QCi + kgV G 140 mg/hr Cq = = = Q + kdV Q + kdV 1,000 m /hr + 0.40/hr * 500 m3 3 = 0.117 mg/m3 This agrees with the result obtained in Example 6. To find the concentration at any time after 5 P.M., we can apply (45) with C0 = 0. C(t) = C q e 1 - expc - akd + bt d f Q V = 0.117{1 - exp[-(0.40 + 1,000>500)t]} at 6 P.M., t = 1 hr, so C(1 hr) = 0.11731 - exp(-2.4 * 1)4 = 0.106 mg/m3 To further demonstrate the use of (45), let us reconsider the lake analyzed in Example 5. This time we will assume that the outfall suddenly stops draining into the lake, so its contribution to the lake’s pollution stops. EXAMPLE 9 A Sudden Decrease in Pollutants Discharged into a Lake Consider the 10 * 106 m3 lake analyzed in Example 5 which, under a conditions given, was found to have a steady-state pollution concentration of 3.5 mg/L. The pollution is nonconservative with reaction-rate constant kd = 0.20/day. Suppose the condition of the lake is deemed unacceptable. To solve the problem, it is de- cided to completely divert the sewage outfall from the lake, eliminating it as a source of pollution. The incoming stream still has flow Qs = 5.0 m3/s and con- centration Cs = 10.0 mg/L. With the sewage outfall removed, the outgoing flow Q is also 5.0 m3/s. Assuming complete-mix conditions, find the concentration of pollutant in the lake one week after the diversion, and find the new final steady- state concentration. Solution For this situation, C0 = 3.5 mg/L V = 10 * 106 m3 Q = Qs = 5.0 m3/s * 3,600 s/hr * 24 hr/day = 43.2 * 104 m3/day 20  Mass and Energy Transfer Cs = 10.0 mg/L = 10.0 * 103 mg/m3 kd = 0.20/day The total rate at which pollution is entering the lake from the incoming stream is Qs Cs = 43.2 * 104 m3/day * 10.0 * 103 mg/m3 = 43.2 * 108 mg/day The steady-state concentration can be obtained from (35) QCs 43.2 * 108 mg/day Cq = = Q + kdV 43.2 * 104 m3/day + 0.20/day * 107 m3 = 1.8 * 103 mg/m3 = 1.8 mg/L Using (45), we can find the concentration in the lake one week after the drop in pollution from the outfall: C(t) = C q + (C0 - C q )expc - a kd + bt d Q V 43.2 * 104 m3/day C(7 days) = 1.8 + (3.5 - 1.8)expc - a0.2/day + b * 7 days d 10 * 106 m3 C(7 days) = 2.1 mg/L Figure 12 shows the decrease in contaminant concentration for this example. Concentration (mg/L) 3.5 2.1 1.8 0 0 7 days Time, t FIGURE 12 The contaminant concentration profile for Example 9. 4 Energy Fundamentals Just as we are able to use the law of conservation of mass to write mass balance equations that are fundamental to understanding and analyzing the flow of materi- als, we can use the first law of thermodynamics to write energy balance equations that will help us analyze energy flows. One definition of energy is that it is the capacity for doing work, where work can be described by the product of force and the displacement of an object caused by that force. A simple interpretation of the second law of thermodynamics suggests that when work is done, there will always be some inefficiency, that is, some portion of the energy put into the process will end up as waste heat. How that waste heat 21  Mass and Energy Transfer affects the environment is an important consideration in the study of environmental engineering and science. Another important term to be familiar with is power. Power is the rate of doing work. It has units of energy per unit of time. In SI units, power is given in joules per second (J/s) or kilojoules per second (kJ/s). To honor the Scottish engineer James Watt, who developed the reciprocating steam engine, the joule per second has been named the watt (1 J/s = 1 W = 3.412 Btu/hr). The First Law of Thermodynamics The first law of thermodynamics says, simply, that energy can neither be created nor destroyed. Energy may change forms in any given process, as when chemical energy in a fuel is converted to heat and electricity in a power plant or when the potential energy of water behind a dam is converted to mechanical energy as it spins a turbine in a hy- droelectric plant. No matter what is happening, the first law says we should be able to account for every bit of energy as it takes part in the process under study, so that in the end, we have just as much as we had in the beginning. With proper accounting, even nuclear reactions involving conversion of mass to energy can be treated. To apply the first law, it is necessary to define the system being studied, much as was done in the analysis of mass flows. The system (control volume) can be any- thing that we want to draw an imaginary boundary around; it can be an automobile engine, or a nuclear power plant, or a volume of gas emitted from a smokestack. Later when we explore the topic of global temperature equilibrium, the system will be the Earth itself. After a boundary has been defined, the rest of the universe be- comes the surroundings. Just because a boundary has been defined, however, does not mean that energy and/or materials cannot flow across that boundary. Systems in which both energy and matter can flow across the boundary are referred to as open systems, whereas those in which energy is allowed to flow across the boundary, but matter is not, are called closed systems. Since energy is conserved, we can write the following for whatever system we have defined: Total energy Total energy Total energy Net change P Q P Q P Q P Q crossing boundary + of mass - of mass = of energy in (46) as heat and work entering system leaving system the system For closed systems, there is no movement of mass across the boundary, so the second and third term drop out of the equation. The accumulation of energy represented by the right side of (46) may cause changes in the observable, macroscopic forms of en- ergy, such as kinetic and potential energies, or microscopic forms related to the atomic and molecular structure of the system. Those microscopic forms of energy include the kinetic energies of molecules and the energies associated with the forces acting between molecules, between atoms within molecules, and within atoms. The sum of those microscopic forms of energy is called the system’s internal energy and is represented by the symbol U. The total energy E that a substance possesses can be described then as the sum of its internal energy U, its kinetic energy KE, and its potential energy PE: E = U + KE + PE (47) 22  Mass and Energy Transfer In many applications of (46), the net energy added to a system will cause an in- crease in temperature. Waste heat from a power plant, for example, will raise the temperature of cooling water drawn into its condenser. The amount of energy needed to raise the temperature of a unit mass of a substance by 1 degree is called the specific heat. The specific heat of water is the basis for two important units of energy, namely the British thermal unit (Btu), which is defined to be the energy required to raise 1 lb of water by 1°F, and the kilocalorie, which is the energy required to raise 1 kg of water by 1°C. In the definitions just given, the assumed temperature of the water is 15°C (59°F). Since kilocalories are no longer a preferred energy unit, values of spe- cific heat in the SI system are given in kJ/kg°C, where 1 kcal/kg°C = 1 Btu/lb°F = 4.184 kJ/kg°C. For most applications, the specific heat of a liquid or solid can be treated as a simple quantity that varies slightly with temperature. For gases, on the other hand, the concept of specific heat is complicated by the fact that some of the heat energy absorbed by a gas may cause an increase in temperature, and some may cause the gas to expand, doing work on its environment. That means it takes more energy to raise the temperature of a gas that is allowed to expand than the amount needed if the gas is kept at constant volume. The specific heat at constant volume cv is used when a gas does not change volume as it is heated or cooled, or if the volume is al- lowed to vary but is brought back to its starting value at the end of the process. Similarly, the specific heat at constant pressure cp applies for systems that do not change pressure. For incompressible substances, that is, liquids and solids under the usual circumstances, cv and cp are identical. For gases, cp is greater than cv. The added complications associated with systems that change pressure and volume are most easily handled by introducing another thermodynamic property of a substance called enthalpy. The enthalpy H of a substance is defined as H = U + PV (48) where U is its internal energy, P is its pressure, and V is its volume. The enthalpy of a unit mass of a substance depends only on its temperature. It has energy units (kJ or Btu) and historically was referred to as a system’s “heat content.” Since heat is cor- rectly defined only in terms of energy transfer across a system’s boundaries, heat content is a somewhat misleading descriptor and is not used much anymore. When a process occurs without a change of volume, the relationship between internal energy and temperature change is given by ¢ U = m cv ¢T (49) The analogous equation for changes that occur under constant pressure involves enthalpy ¢ H = m cp ¢T (50) For many environmental systems, the substances being heated are solids or liquids for which cv = cp = c and ¢U = ¢H. We can write the following equation for the energy needed to raise the temperature of mass m by an amount ¢T: Change in stored energy = m c ¢T (51) Table 3 provides some examples of specific heat for several selected sub- stances. It is worth noting that water has by far the highest specific heat of the substances listed; in fact, it is higher than almost all common substances. This is one 23  Mass and Energy Transfer TABLE 3 Specific Heat Capacity c of Selected Substances (kcal/kg°C, Btu/lb°F) (kJ/kg°C) Water (15°C) 1.00 4.18 Air 0.24 1.01 Aluminum 0.22 0.92 Copper 0.09 0.39 Dry soil 0.20 0.84 Ice 0.50 2.09 Steam (100°C)a 0.48 2.01 Water vapor (20°C)a 0.45 1.88 a Constant pressure values. of water’s very unusual properties and is in large part responsible for the major ef- fect the oceans have on moderating temperature variations of coastal areas. EXAMPLE 10 A Water Heater How long would it take to heat the water in a 40-gallon electric water heater from 50°F to 140°F if the heating element delivers 5 kW? Assume all of the elec- trical energy is converted to heat in the water, neglect the energy required to raise the temperature of the tank itself, and neglect any heat losses from the tank to the environment. Solution The first thing to note is that the electric input is expressed in kilo- watts, which is a measure of the rate of energy input (i.e., power). To get total energy delivered to the water, we must multiply rate by time. Letting ¢t be the number of hours that the heating element is on gives Energy input = 5 kW * ¢t hrs = 5¢t kWhr Assuming no losses from the tank and no water withdrawn from the tank during the heating period, there is no energy output: Energy output = 0 The change in energy stored corresponds to the water warming from 50°F to 140°F. Using (51) along with the fact that water weighs 8.34 lb/gal gives Change in stored energy = m c ¢T = 40 gal * 8.34 lb/gal * 1 Btu/lb°F * (140 - 50)°F = 30 * 103 Btu Setting the energy input equal to the change in internal energy and converting units using Table 1 yields 5¢t kWhr * 3,412 Btu/kWhr = 30 * 103 Btu ¢t = 1.76 hr 24  Mass and Energy Transfer There are two key assumptions implicit in (51). First, the specific heat is assumed to be constant over the temperature range in question, although in actuality it does vary slightly. Second, (51) assumes that there is no change of phase as would occur if the substance were to freeze or melt (liquid-solid phase change) or evapo- rate or condense (liquid-gas phase change). When a substance changes phase, energy is absorbed or released without a change in temperature. The energy required to cause a phase change of a unit mass from solid to liquid (melting) at the same pressure is called the latent heat of fusion or, more correctly, the enthalpy of fusion. Similarly, the energy required to change phase from liquid to vapor at constant pressure is called the latent heat of vapor- ization or the enthalpy of vaporization. For example, 333 kJ will melt 1 kg of ice (144 Btu/lb), whereas 2,257 kJ are required to convert 1 kg of water at 100°C to steam (970 Btu/lb). When steam condenses or when water freezes, those same amounts of energy are released. When a substance changes temperature as heat is added, the process is referred to as sensible heating. When the addition of heat causes a phase change, as is the case when ice is melting or water is boiling, the addition is called latent heat. To account for the latent heat stored in a substance, we can include the following in our energy balance: Energy released or absorbed in phase change = m L (52) where m is the mass and L is the latent heat of fusion or vaporization. Figure 13 illustrates the concepts of latent heat and specific heat for water as it passes through its three phases from ice, to water, to steam. Values of specific heat, heats of vaporization and fusion, and density for water are given in Table 4 for both SI and USCS units. An additional entry has been included in the table that shows the heat of vaporization for water at 15°C. This is a useful number that can be used to estimate the amount of energy required to cause Latent heat of vaporization, 2,257 kJ 100 Boiling water Steam 2.0 kJ/°C Temperature (°C) Melting ice Water 4.18 kJ/°C latent heat of fusion 333 kJ 0 2.1 kJ/°C Ice Heat added to 1 kg of ice (kJ) FIGURE 13 Heat needed to convert 1 kg of ice to steam. To change the temperature of 1 kg of ice, 2.1 kJ/°C are needed. To completely melt that ice requires another 333 kJ (heat of fusion). Raising the temperature of that water requires 4.184 kJ/°C, and converting it to steam requires another 2,257 kJ (latent heat of vaporization). Raising the temperature of 1 kg of steam (at atmospheric pressure) requires another 2.0 kJ/°C. 25  Mass and Energy Transfer TABLE 4 Important Physical Properties of Water Property SI Units USCS Units Specific heat (15°C) 4.184 kJ/kg°C 1.00 Btu/lb°F Heat of vaporization (100°C) 2,257 kJ/kg 972 Btu/lb Heat of vaporization (15°C) 2,465 kJ/kg 1,060 Btu/lb Heat of fusion 333 kJ/kg 144 Btu/lb Density (at 4°C) 1,000 kg/m3 62.4 lb/ft3 (8.34 lb/gal) surface water on the Earth to evaporate. The value of 15°C has been picked as the starting temperature since that is approximately the current average surface temper- ature of the globe. One way to demonstrate the concept of the heat of vaporization, while at the same time introducing an important component of the global energy balance, is to estimate the energy required to power the global hydrologic cycle. EXAMPLE 11 Power for the Hydrologic Cycle Global rainfall has been estimated to average about 1 m of water per year across the entire 5.10 * 1014 m2 of the Earth’s surface. Find the energy required to cause that much water to evaporate each year. Compare this to the estimated 2007 world energy consumption of 4.7 * 1017 kJ and compare it to the average rate at which sunlight strikes the surface of the Earth, which is about 168 W/m2. Solution In Table 4, the energy required to vaporize 1 kg of 15°C water (roughly the average global temperature) is given as 2,465 kJ. The total energy required to vaporize all of that water is Increase in stored energy = 1 m/yr * 5.10 * 1014 m2 * 103 kg/m3 * 2,465 kJ/kg = 1.25 * 1021 kJ/yr This is roughly 2,700 times the 4.7 * 1017 kJ/yr of energy we use to power our society. Averaged over the globe, the energy required to power the hydrologic cycle is W 1.25 * 1024 J/yr * 1 J/s = 78.0 W/m2 365 day/yr * 24 hr/day * 3,600 s/hr * 5.10 * 1014 m2 which is equivalent to almost half of the 168 W/m2 of incoming sunlight striking the Earth’s surface. It might also be noted that the energy required to raise the water vapor high into the atmosphere after it has evaporated is negligible com- pared to the heat of vaporization (see Problem 27 at the end of this chapter). 26  Mass and Energy Transfer Many practical environmental engineering problems involve the flow of both matter and energy across system boundaries (open systems). For example, it is com- mon for a hot liquid, usually water, to be used to deliver heat to a pollution control process or, the opposite, for water to be used as a coolant to remove heat from a process. In such cases, there are energy flow rates and fluid flow rates, and (51) needs to be modified as follows: # Rate of change of stored energy = m c ¢T (53) # where m is the mass flow rate across the system boundary, given by the product of fluid flow rate and density, and ¢T is the change in temperature of the fluid that is carrying the heat to, or away from, the process. For example, if water is being used # to cool a steam power plant, then m would be the mass flow rate of coolant, and ¢T would be the increase in temperature of the cooling water as it passes through the steam plant’s condenser. Typical units for energy rates include watts, Btu/hr, or kJ/s, whereas mass flow rates might typically be in kg/s or lb/hr. The use of a local river for power plant cooling is common, and the following example illustrates the approach that can be taken to compute the increase in river temperature that results. EXAMPLE 12 Thermal Pollution of a River A coal-fired power plant converts one-third of the coal’s energy into electrical energy. The electrical power output of the plant is 1,000 MW. The other two- thirds of the energy content of the fuel is rejected to the environment as waste heat. About 15 percent of the waste heat goes up the smokestack, and the other 85 percent is taken away by cooling water that is drawn from a nearby river. The river has an upstream flow of 100.0 m3/s and a temperature of 20.0°C. a. If the cooling water is only allowed to rise in temperature by 10.0°C, what flow rate from the stream would be required? b. What would be the river temperature just after it receives the heated cool- ing water? Solution Since 1,000 MW represents one-third of the power delivered to the plant by fuel, the total rate at which energy enters the power plant is Output power 1,000 MWe Input power = = = 3,000 MWt Efficiency 1>3 Notice the subscript on the input and output power in the preceding equation. To help keep track of the various forms of energy, it is common to use MWt for thermal power and MWe for electrical power. Total losses to the cooling water and stack are therefore 3,000 MW - 1,000 MW = 2,000 MW. Of that 2,000 MW, Stack losses = 0.15 * 2,000 MWt = 300 MWt 27  Mass and Energy Transfer Electrical output Stack heat 1,000 MWe 300 MWt 33.3% efficient steam plant Coal Cooling water 1,700 MWt 3,000 MWt Qc = 40.6 m3/s Tc = 30.0 °C Qs = 100.0 m3/s Stream Qs = 100.0 m3/s Ts = 20.0 °C Ts = 24.1 °C FIGURE 14 Cooling water energy balance for the 33.3 percent efficient, 1,000 MWe power plant in Example 12. and Coolant losses = 0.85 * 2,000 MWt = 1,700 MWt a. Finding the cooling water needed to remove 1,700 MWt with a tempera- ture increase ¢T of 10.0°C will require the use of (1.53) along with the specific heat of water, 4,184 J/kg°C,given in Table 4: # Rate of change in internal energy = m c ¢T # 1,700 MWt = m kg/s * 4,184 J/kg°C * 10.0°C * 1 MW>(106 J/s) # 1,700 m = = 40.6 * 103 kg/s 4,184 * 10.0 * 10-6 or, since 1,000 kg equals 1 m3 of water, the flow rate is 40.6 m3/s. b. To find the new temperature of the river, we can use (53) with 1,700 MWt being released into the river, which again has a flow rate of 100.0 m3/s. # Rate of change in internal energy = m c ¢T 1 * 106 J/s 1,700 MW * a b MW ¢T = = 4.1°C 100.00 m3/s * 103 kg/m3 * 4,184 J/kg°C so the temperature of the river will be elevated by 4.1°C making it 24.1°C. The results of the calculations just performed are shown in Figure 14. The Second Law of Thermodynamics In Example 12, you will notice that a relatively modest fraction of the fuel energy contained in the coal actually was converted to the desired output, electrical power, and a rather large amount of the fuel energy ended up as waste heat rejected to the environment. The second law of thermodynamics says that there will always be some waste heat; that is, it is impossible to devise a machine that can convert heat 28  Mass and Energy Transfer Hot reservoir Th Qh Heat to engine Heat Work engine W Qc Waste heat Cold reservoir Tc FIGURE 15 Definition of terms for a Carnot engine. to work with 100 percent efficiency. There will always be “losses” (although, by the first law, the energy is not lost, it is merely converted into the lower quality, less use- ful form of low-temperature heat). The steam-electric plant just described is an example of a heat engine, a device studied at some length in thermodynamics. One way to view the steam plant is that it is a machine that takes heat from a high-temperature source (the burning fuel), converts some of it into work (the electrical output), and rejects the remainder into a low-temperature reservoir (the river and the atmosphere). It turns out that the maximum efficiency that our steam plant can possibly have depends on how high the source temperature is and how low the temperature is of the reservoir accepting the rejected heat. It is analogous to trying to run a turbine using water that flows from a higher elevation to a lower one. The greater the difference in elevation, the more power can be extracted. Figure 15 shows a theoretical heat engine operating between two heat reser- voirs, one at temperature Th and one at Tc. An amount of heat energy Qh is trans- ferred from the hot reservoir to the heat engine. The engine does work W and rejects an amount of waste heat Qc to the cold reservoir. The efficiency of this engine is the ratio of the work delivered by the engine to the amount of heat energy taken from the hot reservoir: W Efficiency h = (54) Qh The most efficient heat engine that could possibly operate between the two heat reservoirs is called a Carnot engine after the French engineer Sadi Carnot, who first developed the explanation in the 1820s. Analysis of Carnot engines shows that the most efficient engine possible, operating between two temperatures, Th and Tc, has an efficiency of Tc h max = 1 - (55) Th where these are absolute temperatures measured using either the Kelvin scale or Rankine scale. Conversions from Celsius to Kelvin, and Fahrenheit to Rankine are K = °C + 273.15 (56) R = °F + 459.67 (57) 29  Mass and Energy Transfer Stack gases Electricity Steam Generator Boiler Turbine Cool water in Water Fuel Cooling water Boiler Warm water out feed pump Condenser Air FIGURE 16 A fuel-fired, steam-electric power plant. One immediate observation that can be made from (55) is that the maximum possible heat engine efficiency increases as the temperature of the hot reservoir increases or the temperature of the cold reservoir decreases. In fact, since neither in- finitely hot temperatures nor absolute zero temperatures are possible, we must con- clude that no real engine has 100 percent efficiency, which is just a restatement of the second law. Equation (55) can help us understand the seemingly low efficiency of thermal power plants such as the one diagrammed in Figure 16. In this plant, fuel is burned in a firing chamber surrounded by metal tubing. Water circulating through this boil- er tubing is converted to high-pressure, high-temperature steam. During this conver- sion of chemical to thermal energy, losses on the order of 10 percent occur due to incomplete combustion and loss of heat up the smokestack. Later, we shall consider local and regional air pollution effects caused by these emissions as well as their pos- sible role in global warming. The steam produced in the boiler then enters a steam turbine, which is in some ways similar to a child’s pinwheel. The high-pressure steam expands as it passes through the turbine blades, causing a shaft that is connected to the generator to spin. Although the turbine in Figure 16 is shown as a single unit, in actuality, tur- bines have many stages with steam exiting one stage and entering another, gradually expanding and cooling as it goes. The generator converts the rotational energy of a spinning shaft into electrical power that goes out onto transmission lines for distri- bution. A well-designed turbine may have an efficiency that approaches 90 percent, whereas the generator may have a conversion efficiency even higher than that. The spent steam from the turbine undergoes a phase change back to the liquid state as it is cooled in the condenser. This phase change creates a partial vacuum that helps pull steam through the turbine, thereby increasing the turbine efficiency. The condensed steam is then pumped back to the boiler to be reheated. The heat released when the steam condenses is transferred to cooling water that circulates through the condenser. Usually, cooling water is drawn from a lake or river, heated in the condenser, and returned to that body of water, which is called once-through cooling. A more expensive approach, which has the advantage of 30  Mass and Energy Transfer requiring less water, involves the use of cooling towers that transfer the heat directly into the atmosphere rather than into a receiving body of water. In either case, the re- jected heat is released into the environment. In terms of the heat engine concept shown in Figure 15, the cold reservoir temperature is thus determined by the tem- perature of the environment. Let us estimate the maximum possible efficiency that a thermal power plant such as that diagrammed in Figure 16 can have. A reasonable estimate of Th might be the temperature of the steam from the boiler, which is typically around 600°C. For Tc, we might use an ambient temperature of about 20°C. Using these values in (55) and remembering to convert temperatures to the absolute scale, gives (20 + 273) h max = 1 - = 0.66 = 66 percent (600 + 273) New fossil fuel-fired power plants have efficiencies around 40 percent. Nuclear plants have materials constraints that force them to operate at somewhat lower tem- peratures than fossil plants, which results in efficiencies of around 33 percent. The average efficiency of all thermal plants actually in use in the United States, including new and old (less efficient) plants, fossil and nuclear, is close to 33 percent. That suggests the following convenient rule of thumb: For every 3 units of energy entering the average thermal power plant, approximately 1 unit is converted to electricity and 2 units are rejected to the environment as waste heat. The following example uses this rule of thumb for power plant efficiency com- bined with other emission factors to develop a mass and energy balance for a typical coal-fired power plant. EXAMPLE 13 Mass and Energy Balance for a Coal-Fired Power Plant Typical coal burned in power plants in the United States has an energy content of approximately 24 kJ/g and an average carbon content of about 62 percent. For almost all new coal plants, Clean Air Act emission standards limit sulfur emis- sions to 260 g of sulfur dioxide (SO2) per million kJ of heat input to the plant (130 g of elemental sulfur per 106 kJ). They also restrict particulate emissions to 13 g>106 kJ. Suppose the average plant burns fuel with 2 percent sulfur content and 10 percent unburnable minerals called ash. About 70 percent of the ash is released as fly ash, and about 30 percent settles out of the firing chamber and is collected as bottom ash. Assume this is a typical coal plant with 3 units of heat energy required to deliver 1 unit of electrical energy. a. Per kilowatt-hour of electrical energy produced, find the emissions of SO2, particulates, and carbon (assume all of the carbon in the coal is released to the atmosphere). b. How efficient must the sulfur emission control system be to meet the sulfur emission limitations? c. How efficient must the particulate control system be to meet the particulate emission limits? 31  Mass and Energy Transfer Solution a. We first need the heat input to the plant. Because 3 kWhr of heat are required for each 1 kWhr of electricity delivered, Heat input 1 kJ/s = 3 kWhr heat * * 3,600 s/hr = 10,800 kJ kWhr electricity kW The sulfur emissions are thus restricted to 130 g S S emissions = * 10,800 kJ/kWhr = 1.40 g S/kWhr 106 kJ The molecular weight of SO2 is 32 + 2 * 16 = 64, half of which is sulfur. Thus, 1.4 g of S corresponds to 2.8 g of SO2, so 2.8 g SO2/kWhr would be emitted. Particulate emissions need to be limited to: 13 g Particulate emissions = * 10,800 kJ/kWhr = 0.14 g/kWhr 106 kJ To find carbon emissions, first find the amount of coal burned per kWhr 10,800 kJ/kWhr Coal input = = 450 g coal/kWhr 24 kJ/g coal Therefore, since the coal is 62 percent carbon 0.62 g C 450 g coal Carbon emissions = * = 280 g C/kWhr g coal kWhr b. Burning 450 g coal containing 2 percent sulfur will release 0.02 * 450 = 9.0 g of S. Since the allowable emissions are 1.4 g, the removal efficiency must be