Thermodynamics

Thermodynamics 21. Thermodynamics 21.1. FIRST LAW OF THERMODYNAMICS Converting internal energy to mechanical energy is much more difficult than the reverse, and perfect efficiency is impossible. A heat  engine is a device or system that can perform this conversion; the human body and the earth’s atmosphere are heat engines, as are gasoline and diesel motors, aircraft jet engines, and steam turbines. All heat engines operate by absorbing heat from a reservoir of some kind at a high temperature, performing work, and then giving off heat to a reservoir of some kind at a lower temperature (Fig. 21-1).

 Figure 21-1. Two general principles apply to all heat engines. The The first  first law of thermodynamics is an expression of the principle of conservation of  energy. According to this law, in any process that a system of some kind (such as a heat engine) undergoes, we have

Here Q is the net heat added to the system during the process; if the system gives off heat, Q is negative. When the internal energy of the system U increases, U increases, ΔU  ΔU is is positive; when U decreases, U decreases, ΔU  ΔU is is negative. The net work done by the system during the process is W ; if work  is done on the system, W is W is negative. If the system is a heat engine that operates in a cycle, energy may be stored and released from storage, but the engine does not undergo a net change in its internal energy during each cycle. In this case

The net heat input is the amount of heat Q 1 the engine takes in from the high-temperature reservoir minus the amount of heat Q 2 the engine gives off to the low-temperature reservoir, as in Fig. 21-1, so that

21.2. WORK DONE BY AND ON A GAS The work output of most heat engines is produced by an expanding gas. If the volume of the gas changes from V 1 to V 2 at the constant  pressure p  pressure p,, the work done is

If the gas is compressed rather than expanded, V 2 is less than V 1 and W is W is negative. This means that work is done on the gas during a compression. In the p-V  the p-V (pressure-volume) (pressure-volume) diagram of Fig. of  Fig. 21-2 the expansion of a gas at constant pressure appears as a horizontal line from V 1 to V 2. The area under the line is equal to p to  p((V 2 − V 1) and so equals the work W  work  W done done in the expansion. If the gas pressure varies during the expansion, the expansion appears as a curved line on a  p-V diagram,  p-V diagram, as in Fig. 21-3. 21-3. We can imagine the region under the curve as divided into thin strips, each corresponding to a small expansion at a different constant pressure so that the total area under the curve equals the work done in this situation also.

 Figure 21-2.

 Figure 21-3. Three important kinds of expansion and comparison that can occur in a gas are as follows:

1. An isobaric process is one that takes place at constant pressure. pressure . 2. An isothermal  isothermal process process is one that takes place at constant temperature. temperature . The expansions and compressions of a gas in a container that is surrounded by a constant-temperature heat reservoir are approximately isothermal. 3. An adiabatic process is one that takes place in a system so isolated from its surroundings that heat neither enters nor leaves the system during the process . Most rapid thermodynamic processes are approximately adiabatic because heat transfer takes time and a rapid process may be completed before much heat has passed through the walls of the system.

21.2.1. SOLVED PROBLEM 21.1

Show that the work done by a gas expanding at the constant pressure  p from V 1 to V 2 is given by W = W = p(  p(V 2 − V 1). Figure 21-4 shows a gas-filled cylinder with a movable piston. The cross-sectional area of the cylinder is  A,  A, and the gas pressure is p is  p.. Since p Since p = F / A,  A, the force the gas exerts on the piston is The work done by the gas in moving the piston through the distance Δ s is

But A But A Δ s is the change ΔV  ΔV = = V 2 − V 1 in the volume of the gas, so

 Figure 21-4. 21.2.2. SOLVED PROBLEM 21.2

A gas expands by 1.2 L at a constant pressure of 2.5 bar. During the expansion 500J of heat is added. Find the change in the internal energy of the gas. From Q = ΔU  ΔU + + W we W we have

Here W = W = p(  p(V 2 − V 1) since the expansion is isobaric. Since  p = (2.5 bar) (10 5 Pa/bar) = 2. 2.5 × 10 5 Pa and V 2 − V 1= (1.2L)/(10 3 L/m 3) −3 3 = 1. 1.2 × 10 m , we have

21.2.3. SOLVED PROBLEM 21.3

A sample of gas expands from V 1 to V 2. Is the work done by the gas greatest when the expansion is ( a) isobaric, (b (b) isothermal, or (c (c) adiabatic? How does the temperature vary during each expansion?

(a) At the constant pressure p pressure  p the work done is p is p((V 2 − V 1) and is the greatest of the three expansions. The temperature must increase during the expansion in order to maintain the pressure constant despite the increase in volume. (b) Since pV/T  Since pV/T = = constant, during an expansion at constant temperature the pressure must drop as V increases, V increases, and the work done is accordingly less than in (a ( a). (c) In an adiabatic expansion the temperature must drop since all the work done is at the expense of the internal energy of the gas. The final pressure is therefore lower than in ( a) or (b (b), and the least amount of work is done. See Fig. 21-5. 21-5.

 Figure 21-5 . 21.2.4. SOLVED PROBLEM 21.4 At 100ºC and atmospheric pressure the heat of vaporization of steam is  L v = 2260 kJ/kg, the density of water is 10 3kg/m 3, and the density of steam is 0.6 kg/m 3. What proportion of  L  L v represents work done to expand water into steam against the pressure of the atmosphere? The volumes of 1 kg of water and 1 kg of steam at 100ºC and atmospheric pressure are, respectively,

Atmospheric pressure is p is  p = 1. 1.013 × 10 5 Pa. Hence the work done in the expansion is

This is

of the heat of vaporization of water. The remainder of the heat of vaporization goes into pulling the water molecules apart to create a gas from a liquid and so becomes internal energy of the s team. 21.2.5. SOLVED PROBLEM 21.5

Show that the power output of each cylinder of a reciprocating engine of any kind (steam, gasoline, diesel) is given by the formula P  formula P c = pLAn,  pLAn, where

In general, P  general,  P = = Fs/t . Here F  Here F is is the force exerted on the piston during each power stroke by the pressure  p,  p, so since p since p = F/A, F = F = pA.  pA. The distance traveled by the piston per power stroke is  L,  L, and the distance it covers per second is therefore  s/t =  s/t = Ln.  Ln. Hence

21.2.6. SOLVED PROBLEM 21.6

The four-cylinder, four-stroke diesel engine of a car develops 60 kW at 2600 rev/min. The pistons of this engine are 100 mm in diameter, and they travel 130 mm. Find the average pressure on the pistons during each power stroke. The area of each piston is

In a four-stroke engine a power stroke occurs in each cylinder once every 2 rev, so there are 1300 power strokes per minute or  1300/60 = 21.7 strokes per second. Because the engine has four cylinders, its total power output is  P =  P = 4 P c= 4 pLAn and

21.3. SECOND LAW OF THERMODYNAMICS Internal energy resides in the kinetic energies of randomly moving atoms and molecules, whereas the output of a heat engine appears in the ordered motions of a piston or a wheel. Since all physical systems in the universe tend to go in the opposite direction, from order to disorder, no heat engine can completely convert heat to mechanical energy or, in general, to work. This fundamental principle leads to the  the  second law of thermodynamics: thermodynamics: It is impossible to construct a continuously operating engine that takes heat from a source and  performs an exactly equivalent amount of work. Because some of the heat input to a heat engine must be wasted and because heat flows from a hot reservoir to a cold one, every heat engine must have a low-temperature reservoir for exhaust heat to go to as well as a high-temperature reservoir from which the input heat is to come, as in Fig. 21-1. 21-1.

21.4. CARNOT ENGINE An ideal heat engine is one in which every process that occurs is reversible without any loss of energy. Such an engine is not subject to such practical mechanisms of energy loss as friction and heat conduction to the outside world. An example of an ideal heat engine is the imaginary Carnot engine which consists of a cylinder filled with an ideal gas that has a movable piston at one end. Figure 21-6 shows the four stages in the operation of a Carnot engine. The engine does work during the two expansions, and work is done on the engine during the two compressions; the net work done per cycle is the area enclosed by the curve.

 Figure 21-6.. (Top portion from Modern Technical Physics, 6th Ed., Arthur Beiser, ©1992. Reprinted by permission of Pearson  Education, Inc.) A Carnot engine is the most efficient engine that can operate between the temperatures T 1 and T 2 at which heat is absorbed and exhausted. If heat Q 1 is absorbed at the absolute temperature T 1 and heat Q 2 is given off at the absolute temperature T 2, then Q 1/Q 2 = T  /T   1/T 2 in such an engine. Its efficiency is therefore

The smaller the ratio between T 2 and T 1, the more efficient the engine. Because no reservoir can exist at a temperature of 0 K or 0ºR, which is absolute zero, no heat engine can be 100 percent efficient. 21.4.1. SOLVED PROBLEM 21.7

The Carnot engine uses only isothermal and adiabatic processes in its operating cycle. Why? An engine has maximum efficiency when all the processes that occur in its it s operation are reversible without wi thout the performance of work, since any other processes must necessarily involve the waste of energy. Heat flow from a hot reservoir to a cooler one is not reversible in this sense because the natural direction of heat flow is from hot to cold (in fact, this is an alternate statement of the second law of  thermodynamics). However, in an isothermal process, the heat flow occurs at a constant temperature, so the process can be reversed

without any work being lost. An adiabatic process is also reversible in the same sense because no heat enters or leaves a system during such a process. Hence an engine that uses only isothermal and adiabatic processes is the most efficient possible. 21.4.2. SOLVED PROBLEM 21.8

A 1-MW (10 6-W) generating plant has an overall efficiency of 40 percent. How much fuel oil whose heat of combustion is 45 MJ/kg does the plant burn each day? In 1 day the plant produces

of electric energy. Since its efficiency is 0.4 and Eff = work output/heat input, we have

The mass of fuel required to supply this amount of heat is

21.4.3. SOLVED PROBLEM 21.9

A Carnot engine absorbs 1 MJ of heat from a reservoir at 300ºC and exhausts heat to a reservoir at 150ºC. Find the work it does. The intake and exhaust temperatures are, respectively, T 1 = 300ºC + 273 = 573 K and T 2 = 150ºC + 273 = 423 K. The engine efficiency is

and so the work done is

21.4.4. SOLVED PROBLEM 21.10

Three designs are proposed for an engine that is to operate between 500 and 300 K. Design A is claimed to produce 750 J of work per  kilojoule of heat input, B is claimed to produce 500 J, and C is claimed to produce 250 J. Which design would you choose? The efficiency of an ideal engine operating between T 1 = 500 K and T 2 = 300 K is

The claimed efficiencies of the proposed engines are

Both A and B claim efficiencies greater than that of an ideal engine and hence could not possibly work as stated. Design C is therefore the only possible choice. 21.4.5. SOLVED PROBLEM 21.11

A steam engine is being planned that is to use steam at 400ºF and whose efficiency is to be 20 percent. Find the maximum temperature at which the spent steam can emerge. The intake temperature is T 1 = 400ºF + 460º = 860ºR. We proceed as follows:

The maximum exhaust temperature is therefore T 2 = 688ºR − 460º = 228ºF.

21.5. INTERNAL COMBUSTION ENGINES An internal combustion engine is relatively efficient because it generates the input heat within the engine itself. The operating cycle of a four-stroke gasoline engine is shown in Fig. 21-7. 21-7. In the intake stroke, a mixture of gasoline vapor and air from the carburetor is sucked into the cylinder through the intake valve as the piston moves downward. In the compression stroke, both valves are closed and the fuelair mixture is compressed to 1/7 or 1/8 of its original volume. At the top of the stroke the spark plug is fired, which ignites the fuel-air  mixture. Gases from the burning f uel expand and force the piston downward in the power stroke. Finally the piston moves upward again to push the waste gases out through the exhaust valve.

 Figure 21-7.. (From Konrad B. Krauskopf and Arthur Beiser, The Physical Universe, 10th Ed., ©2003, The McGraw-Hill Companies.  Reproduced with permission of The McGraw-Hill Companies.)

In a diesel engine, only air is drawn into the cylinder in the intake stroke. At the end of the compression stroke, diesel fuel is injected into the cylinder and is ignited by the high temperature of the compressed air. No spark plug is needed. Diesel engines are more efficient but also heavier and more expensive than gasoline engines. 21.5.1. SOLVED PROBLEM 21.12

Describe the operating cycle of a gasoline engine with the help of a  p p-V diagram. V diagram. In Fig. 21-8 the events that occur during each cycle of the engine are as follows: ab: ab: intake of fuel-air mixture bc: bc: compression stroke cd : ignition of fuel-air mixture; mixture; the heat Q 1 is released into the cylinder  de: de: power stroke e: exhaust valve opens, which starts the release of the waste waste heat Q 2

ba: ba: exhaust stroke

 Figure 21-8.. (From Modern Technical Physics, 6th Ed., Arthur Beiser, ©1992. Reprinted by permission of Pearson Education, Inc.)

21.6. REFRIGERATION refrigerator is a heat engine that operates backward to extract heat from a low-temperature reservoir and transfer it to a highA refrigerator is temperature reservoir (Fig. ( Fig. 21-9). 21-9). Because the natural tendency of heat is to flow from a hot region to a cold one, energy must be  provided to a refrigerator to reverse the flow, and this energy adds to the heat exhausted by the refrigerator. Figure 21-10 shows a refrigerator whose working substance is an easily liquified gas called a refrigerant . The operation of this refrigeration system proceeds as follows: 1. The compressor , usually driven by an electric motor (or by the car engine in the case of an automotive air  conditioner), brings the refrigerant to a high pressure, which raises its temperature as well. 2. The hot refrigerant passes through the condenser , an array of thin tubes that give off heat from the refrigerant to the atmosphere. The condenser is on the back of most household refrigerators. As it cools, the refrigerant becomes a liquid under high pressure.

 Figure 21-9.

 Figure 21-10.. (From Modern Technical Physics, 6th Ed., Arthur Beiser, ©1992. Reprinted by permission of Pearson Education, Inc.) valve , from which it emerges at a lower pressure and 3. The liquid refrigerant now goes into the expansion valve, temperature. 4. In the evaporator the evaporator the cool liquid refrigerant absorbs heat from the storage chamber and vaporizes. Farther along in the evaporator the refrigerant vapor absorbs more heat and becomes warmer. The warm vapor then goes back to the

compressor to start another cycle. In Step 4 of this cycle, heat is extracted from the storage chamber by the refrigerant. In Step 1, work is done on the refrigerant by the compressor. In Step 2, heat from the refrigerant leaves the system. A refrigerator might remove two or more times as much heat from its storage chamber as the amount of work done. If a refrigerator absorbs heat Q 1 at the absolute temperature T 1 and ejects heat Q 2 at the absolute temperature T 2, itscoefficient itscoefficient of   performance (CP) is given by

In an ideal refrigerator, which is a Carnot engine run backward, Q  /Q  1/Q 2 = T  /T  1 2 and

21.6.1. SOLVED PROBLEM 21.13

In an effort to cool a kitchen during the summer, the refrigerator door is left open and the kitchen’s door and windows are closed. What will happen? Since no refrigerator can be completely efficient, more heat is exhausted by the refrigerator into the kitchen than is extracted from the kitchen. The net effect, then, is to increase the kitchen’s temperature. 21.6.2. SOLVED PROBLEM 21.14

A 1-kW refrigerator whose coefficient of performance is 2.0 takes heat from a freezer compartment at −20ºC and exhausts it at 40ºC. How does its CP compare with that of an ideal refrigerator? Here T 1 = −20ºC + 273 = 253 K and T 2 = 40ºC + 273 = 313 K. The CP of an ideal refrigerator operating between these temperatures is

The actual refrigerator is 2. 2 .0/4. 0/4.2 = 0. 0.48 = 48 percent as efficient as the ideal refrigerator. 21.6.3. SOLVED PROBLEM 21.15

At what rate does the refrigerator of  Prob. 21.14 remove heat from the freezer compartment? Since P  Since P = = W/t and W/t and Q 1 = (CP)(W  (CP)(W ), ),

For every joule of energy supplied to the refrigerator, it removes 2 J of heat from the freezer compartment. 21.6.4. SOLVED PROBLEM 21.16

A refrigerator which is half as efficient as an ideal refrigerator extracts heat from a storage chamber at 0ºF and exhausts it at 100ºF. How many foot-pounds of work per Btu extracted does this refrigerator require? We begin by finding the coefficient of performance of the ideal refrigerator. Since

Here Q 1 = 1 Btu and, since 1 Btu = 778 ft·lb,

This refrigerator is half as efficient as an ideal refrigerator, so the required work per Btu is twice as great, or 278 ft·lb. 21.6.5. SOLVED PROBLEM 21.17

The British unit of refrigeration capacity is the ton, ton, which is that rate of heat extraction that can freeze 1 ton of water at 32ºF to ice at 32ºF per day. Since the heat of fusion of water is 144 Btu/lb,

The refrigerator of Prob. of  Prob. 21.16 has a capacity of 2 tons. How much power is required to operate its compressor? Since the refrigerator requires 278 ft·lb of work per Btu of heat extracted,

In terms of horsepower,

21.7.   Multiple-Choice Questions

21.1. A heat engine takes in heat at one temperature and turns

(a) all of it into work  (b) some of it into work and rejects the rest at a lower temperature (c) some of it into work and rejects the rest at the same temperature (d ) some of it into work and rejects the rest at a higher temperature 21.2. A process that can be reversed without energy input from an outside source is one that takes place at constant

(a) pressure (b) density (c) velocity (d ) temperature 21.3. In an adiabatic process in a system,

(a) its temperature stays the same (b) its pressure stays the same (c) no heat enters or leaves it (d ) no work is done by or on it 21.4. Without work being done on it, a gas cannot be

(a) compressed isobarically (b) compressed isothermally (c) compressed adiabatically

(d ) expanded adiabatically 21.5. To be completely efficient (which is impossible), the exhaust temperature of a frictionless heat engine would have to be

(a) 0 K  (b) 273 K  (c) less than its intake temperature (d ) the same as its intake temperature 21.6. The Carnot cycle does not include an

(a) isobaric expansion (b) isothermal expansion (c) adiabatic expansion (d ) adiabatic compression 21.7. The efficiency of a Carnot engine operating between the absolute temperatures T 1 and T 2 is

(a) equal to T 2/T 1 (b) 100% (c) the maximum possible between these temperatures (d ) the same as that of an actual engine operating between these temperatures 21.8. Which of the following engines is the most efficient?

(a) gasoline engine (b) diesel engine

(c) steam engine (d ) Carnot engine 21.9. The fuel in a diesel engine is ignited by

(a) a spark plug (b) the hot air into which it is injected (c) being compressed until it is hot enough (d ) exhaust gases left over from the previous cycle 21.10. The heat a refrigerator absorbs from its contents is

(a) less than it gives off  (b) the same amount it gives off  (c) more than it gives off  (d ) any of the above, depending on its design 21.11. Four kilojoules of heat is given off by a gas when it is compressed from 0.08 to 0.05 m 3 under a pressure of  200 kPa. The internal energy of the gas

(a) decreases (b) is unchanged (c) increases (d ) any of the above, depending on the initial and final temperatures 21.12. A Carnot engine that absorbs heat at 300ºC and exhausts heat at 100ºC has an efficiency of 

(a) 33% (b) 35% (c) 65% (d ) 67% 21.13. If a Carnot engine absorbs 10 kJ of heat per cycle when it operates between 500 and 400 K, the work it does  per cycle is

(a) 2 kJ (b) 2.5 kJ (c) 8 kJ (d ) 10 kJ 21.14. To have an efficiency of 40 percent, a heat engine that exhausts heat at 350 K must absorb heat at no less than

(a) 210 K  (b) 583 K  (c) 875 K  (d ) 1038 K  21.15. The coefficient of performance of a refrigerator that gives off 3 J of heat for every joule of mechanical energy input is

(a) 1 (b) 2

(c) 3 (d ) 4

21.8. Supplementary Problems 21.1. Why is it impossible for a ship to use the internal energy of seawater to operate its engine? 21.2. One kilojoule of heat is added to a gas as it expands from 8 to 10 L at a constant pressure of 2 bar. What is the change in the internal energy of the gas? 21.3. A 2400-hp diesel locomotive burns 160 gal/h of fuel. If the heat of combustion of the diesel oil used is 1 .2 × 10 5Btu/gal, find the efficiency of the engine. 21.4. Steam enters a turbine engine at 550ºC and emerges at 90ºC. The engine has an actual overall efficiency of 35  percent. What percentage of its ideal efficiency is this? 21.5. The six-cylinder, four-cycle gasoline engine of a car has pistons 3.4 in. in diameter whose stroke (length of  travel) is 4 in. If the mean effective pressure on the pistons during the power stroke is 70 lb/in. 2, find the number of  horsepower developed by the engine when it operates at 2000 rev/min. 21.6. A four-cylinder, four-stroke diesel engine develops 100 kW at 66 rev/s. The pistons of the engine are 108 mm in diameter and their travel is 127 mm. Find the average pressure on the pistons during the power stroke. 21.7. A Carnot engine absorbs 200 kJ of h eat at 500 K and exhausts 150 kJ. What is its exhaust temperature? 21.8. A Carnot engine absorbs 1 MJ of heat at 327ºC and exhausts heat to a reservoir at 127ºC. How much work does it do? 21.9. A Carnot engine whose efficiency is 35 percent absorbs heat at 500ºC. What must its intake temperature be if its efficiency is to be 50 percent with the same exhaust temperature? 21.10. A Carnot engine absorbs 500 Btu of heat at 500ºF and performs 1 × 10 temperature?

5

ft·lb of work. What is its exhaust

21.11. Three designs for a refrigerator to operate between −20 and 30ºC are proposed. Design A is claimed to need 100 J of work per kilojoule of heat extracted, design B to need 200 J, and design C to need 300 J. Which design

would you choose and why? 21.12. Describe the operating cycle of a four-stroke diesel engine with the help of a  p p-V diagram. V diagram. 21.13. An ideal refrigerator extracts heat from a freezer at −20ºC and exhausts it at 50ºC. How many joules of heat are extracted per joule of work input? 21.14. How much work must an ideal refrigerator perform to make 1 kg of ice at −10ºC from 1 kg of water at 20ºC when 20ºC is also its exhaust temperature? 21.15. The energy efficiency ratio (EER) of a refrigerator or air conditioner is the ratio between the heat in Btu extracted per hour and the power in watts the machine uses. Find the relationship between EER and coefficient of   performance. 21.16. What horsepower rating is needed for the motor of a 5-ton air conditioner whose coefficient of performance is 2.5?

21.9. Answers to Multiple-Choice Questions 21.1. (b) 21.2. (d ) 21.3. (c) 21.4. (a), (b (b), (c (c) 21.5. (a) 21.6. (a) 21.7. (c) 21.8. (d ) 21.9. (b) 21.10. (a)

21.11. (c) 21.12. (b) 21.13. (a) 21.14. (b) 21.15. (b)

21.10. Answers to Supplementary Problems 21.1. There would be no suitable low-temperature reservoir to absorb the waste heat from the engine. 21.2. The internal energy increases by 600 J.

 Figure 21-11. 21.3. 32 percent 21.4. 63 percent 21.5. 38.2 hp 21.6. 651 kPa

21.7. 375 K  21.8. 0.33 MJ 21.9. 732ºC 21.10. 253ºF 21.11. A is more efficient than a Carnot refrigerator, and B has the same efficiency, so both are impossible as  practical refrigerators. C is less efficient than a Carnot refrigerator and so might work as claimed. 21.12. In Fig. 21-11 the events that occur during each cycle of the engine are as follows: ab, ab, intake of air; bc, bc, compression stroke; cd , fuel injection and burning; ce, ce, power stroke; e, exhaust valve opens; ba, ba, exhaust stroke. 21.13. 4.6 J 21.14. 50 kJ 21.15. EER = 3.42(CP) 21.16. 9.43 hp