Homework Chapter 22 Solutions! ! 22.5!

22.7!

22.11!

22.25!

22.27!

22.30!

22.40!

22.57!

22.63!

22.71!

!

page �1

Problem 22.5!A particular heat engine has a mechanical power output of 5 kW and an efficiency of 25%. The engine expels 8000 J of exhaust energy in each cycle.!(a)!Find the energy taken in during each cycle.!(b)!Find the time interval for each cycle.!

Solution!(a)!The efficiency being 25% means the following.!

� !

� !

The first law say the following.!� !

Together, we have the following.!� !

(b)!For the power, we need the total work done by the engine.!� !

The mechanical power by the engine through each cycle is!

� !

!

e =

Wtotal by

Qin

= 0.25

Wtotal by = 0.25Qin

ΔEint = Qin +Qout +Wtotal on = Qin − 8000 J − 0.25Qin = 0

0.75Qin = 8000 J ⇒ Qin = 10,667 J

Wtotal by = 0.25Qin = 0.25 ⋅10,667 J = 2,667 J

power =

Wtotal by

cycle=

2,667 Jcycle

=5,000 J

s ⇒ cycle = 0.533 s

page �2

Problem 22.7!Suppose a heat engine is connected to two energy reservoirs, one a pool of molten aluminum (660 °C) and the other a block of solid mercury (–38.9 °C). The engine runs by freezing 1 g of aluminum and melting 15 g of mercury during each cycle. The heat of fusion of aluminum is 3.97 x 105 J/kg. The heat of fusion of mercury is 1.18 x 104 J/kg.!What is the efficiency of the engine?!

Solution!Here is a diagram of the cycle.!

� !The efficiency definition is this.!

� !

The heat in is!

� !

The heat out is!

� !

The total work done by the engine is!� !

The efficiency is!

� !

!

Qin Qout

Wtotal

freezing aluminum

melting mercury

e =

Wtotal by

Qin

Qin = Lf ,AlmAl = (3.97×105 J

K)(0.001 kg) = 397 J

Q

out= L

f ,Hgm

Hg= −(1.18×104

J

K)(0.015 kg) = −177 J

397 J −177 J +Wtotal on = 0 ⇒ Wtotal on = −Wtotal by = 220 J

e =220 J397 J

= 0.55416

page �3

Problem 22.11!A freezer has a coefficient of performance of 6.3. It is advertised as using electricity at a rate of 457 kWh/yr.!(a)!On average, how much energy does it use in a single day?!(b)!On average, how much energy does it remove from the refrigerator in a single day?!(c)!What is the maximum mass of water at 20 °C could the freezer freeze in a single day?!

Solution!(a)!Here is the diagram for the cycle.!

� !The work done on the freezer is powered by the electricity. Over 1 day, the energy used by the freezer is!

� !

(b)!The definition of the COP is!

� !

The work done on the freezer in a single day is 4.5043 x 106 J. The heat pulled from the freezer into the engine is!

� !

(c)!The heat pulled in depends on the mass of the water. For a mass of water, m, the heat required to be removed is!

� !

Using the per day values above,!

� !

!

Qin Qout

Wtotal

freezer at 0 °C

room at ? °C

Wtotal on

day=

457 kW hyr

1 yr365.25 day

3600 s1 h

1000 Js

1 kW=

4.5043×106 Jday

e =

Qin

Wtotal on

= 6.3

Qin = 6.3Wtotal on = (6.3)(4.5043×106 J) = 2.8377×107 J

Qin = mcΔT + mLf = m(4186 J

kg⋅°C⋅ 20 °C + 3.33×105 J

kg) = m(4.1672×105 J

kg)

2.8377×107 J = m(4.1672×105 J

kg) ⇒ m = 68.097 kg

page �4

Problem 22.25!An ideal gas is taken through a carnot cycle. The isothermal expansion occurs at 250 °C and the isothermal compression takes place at 50 °C. The gas takes in 1.2 x 103 J of energy from the hot reservoir during the isothermal expansion.!(a)!Find the energy expelled to the cold reservoir in each cycle.!(b)!Find the total work done by the engine in each cycle.!

Solution!(a)!The efficiency of this engine is!

� !

The energy going into the engine is 1200 J. According to the first law,!� !

� !

The work done by the engine is!� !

!

e = 1−

Tc

Th

= 1−323 K523 K

= 0.38241 =Wtotal by

Qin

⇒ Wtotal by = 0.38241Qin

Qin +Qout +Wtotal on = 0

Qin +Qout − 0.38241Qin = 0 ⇒ Qout = −0.6176Qin = −0.6176(1200 J) = 741.11 J

Wtotal by = 0.38241Qin = 0.38241(1200 J) = 458.89 J

page �5

Problem 22.27!Argon enters a turbine at a rate of 80 kg/min, at a temperature of 800 °C, at a pressure of 1.5 MPa. It expands adiabatically as it pushes on the turbine blades and exists at a pressure of 300 kPa.!(a)!Calculate the temperature at exit.!(b)!Calculate the maximum power output of the turing turbine.!(c)!The turbine is one component of a model closed-cycle gas turbine engine. Calculate the

maximum efficiency of the engine.!

Solution!Here is the diagram of the engine.!

� !(a)!Let’s look at the values per minute. The process of the engine is adiabatic. At the initial state, !

� !

The pressure and the temperature are known. The amount of argon is 80 kg. This is!

� !

The initial volume then is!

� !

At the final state, !� !

� !

� !

The relationship between the volume and temperature for the process is!

� !

Argon is monoatomic, so f = 3 and gamma = 5/3.!

� !

From above, !

� !

!

Qin Qout

Wtotal

argon at 800 °C at 1.5 MPa

argon at ? °C

at 300 kPa

PiVi = nRTi

80 kg Ar

1 mol40 g

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

1000 g1 kg

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ = 2000 mol

(1.5×106 Pa)Vi = (2000 mol)(8.314 J

mol ⋅K)(1073 K) ⇒ Vi = 11.895 m3

PfVf = nRTf

(300 kPa)Vf = (2000 mol)(8.314 J

mol ⋅K)Tf

Vf = (5.5427×10−2 m3

K)Tf

TiVi

γ−1 = TfVfγ−1

(1073 K)(11.895 m3)2/3 = 5591.3 m2K = TfVf

2/3

Vf

2/3 = ((5.5427×10−2 m3

K)Tf )

2/3 = (0.14537 m2

K 2/3)Tf2/3

page �6

Using this,!

� !

� !

(b)!Continuing to look at the values per minute, power output of the engine is just the work done by the engine. The work done is just the change in the internal energy, and the internal energy depends only on the temperature change.!

� !

� !

So the power output is!

� !

(c)!The efficiency of the turbine is!

� !

The total thermal energy put into the turbine that could be extracted for work is!

� !

� !

!

5591.3 m2K = Tf (0.14537 m2

K 2/3)Tf2/3 = (0.14537 m2

K 2/3)Tf5/3

Tf

5/3 = 3.8462×104 K 5/3 ⇒ Tf = 563.66 K

Wtotal on = ΔEint =

32nRΔT =

32(2000 mol)(8.314 J

mol ⋅K)(563.66 K −1073 K)

Wtotal by = 1.2704×107 J

Pout =

Wtotal by

Δt=

1.2704×107 Jmin

1 min60 s

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ = 2.1173×105 W = 211.73 kW

e =

Wtotal by

Qin

=1.2704×107 J

Qin

Eint =

32nRTin =

32(2000 mol)(8.314 J

mol ⋅K)(1073 K) = 2.6763×107 J

e =

1.2704×107 J

2.6763×107 J= 0.47469

page �7

Problem 22.30!At point A in a carnot cycle, 2.34 mol of a monoatomic ideal gas has a pressure of 1400 kPa, a volume of 10 L, and a temperature of 720 K. The gas expands isothermally to point B and then expands adiabatically to point C where its volume is 24 L. An isothermal compression brings it to point D where its volume is 15 L. An adiabatic process returns the gas to point A.!(a)!Determine all of the unknown pressures, volumes, and temperatures as you fill in the following

table.!(b)!Find the energy added by heat, the work done by the engine, and the change in the internal

energy for each of the steps AB, BC, CD, and DA.!(c)!Calculate the efficiency of the cycle.!(d) Show that efficiency is equal to the carnot inefficiency.!

Solution!(a)!Here is the table. The value of gamma is 5/3.!

The easy ones to do are the ones defined by the processes. !

Points A and B are isothermally related. Points C and D are related isothermally.!� !

From the ideal gas model,!� !

� !

� !

For the adiabatic processes, we have!

� !

� !

� !

� !

P V n TA 1.4x106 Pa 10x10-3 m3 2.34 mol 720 KB 2.34 molC 24x10-3 m3 2.34 molD 15x10-3 m3 2.34 mol

P V n TA 1.4x106 Pa 10x10-3 m3 2.34 mol 720 KB 2.34 mol 720 KC 24x10-3 m3 2.34 mol TCD

D 15x10-3 m3 2.34 mol TCD

TA = TB and TC = TD ≡TCD

PBVB = nRTB ⇒ PBVB = (2.34 mol)R(720 K)

PCVC = nRTC ⇒ PC (24×10−3 m3) = (2.34 mol)RTCD

PDVD = nRTD ⇒ PD(15×10−3 m3) = (2.34 mol)RTCD

PBVBγ = PCVC

γ ⇒ PBVB5/3 = PC (24×10−3 m3)5/3

TBVBγ−1 = TCVC

γ−1 ⇒ (720 K)VB2/3 = TCD(24×10−3 m3)2/3

PDVDγ = PAVA

γ ⇒ PD(15×10−3m3)5/3 = (1.4×106 Pa)(10×10−3 m3)5/3

TDVDγ−1 = TAVA

γ−1 ⇒ TCD(15×10−3 m3)2/3 = (720 K)(10×10−3 m3)2/3

page �8

The easy one to solve first is TCD.!

� !

The pressure at C and D are!

� !

� !

Finally, !� !

� !

Their ratio is!

� !

� !

� !

!

TCD =

(720 K)(10×10−3 m3)2/3

(15×10−3 m3)2/3= 549.46 K

P V n TA 1.4x106 Pa 10x10-3 m3 2.34 mol 720 KB 2.34 mol 720 KC 4.45x105 Pa 24x10-3 m3 2.34 mol 549 KD 15x10-3 m3 2.34 mol 549 K

PC =

(2.34 mol)R(549.46 K)

24×10−3 m3= 4.4540×105 Pa

PD =

(2.34 mol)R(549.46 K)

15×10−3 m3= 7.1264×105 Pa

P V n TA 1.4x106 Pa 10x10-3 m3 2.34 mol 720 KB 2.34 mol 720 KC 4.45x105 Pa 24x10-3 m3 2.34 mol 549 KD 7.13x105 Pa 15x10-3 m3 2.34 mol 549 K

PBVB = (2.34 mol)R(720 K)

PBVB5/3 = (4.4540×105 Pa)(24×10−3 m3)5/3

PBVB5/3

PBVB

=(4.4540×105 Pa)(24×10−3 m3)5/3

(2.34 mol)R(720 K) ⇒ VB

2/3 = 6.3496×10−2 m2

VB = 1.6000×10−2 m3 = 16×10−3 m3

PB =

(2.34 mol)R(720 K)

1.6000×10−2 m3= 8.7547×105 Pa

P V n TA 1.4x106 Pa 10x10-3 m3 2.34 mol 720 KB 8.75x105 Pa 16x10-3 m3 2.34 mol 720 KC 4.45x105 Pa 24x10-3 m3 2.34 mol 549 KD 7.13x105 Pa 15x10-3 m3 2.34 mol 549 K

page �9

(b)!Here are the energies.!

The work done in process AB is!

� !

The change in the internal energy in process BC is!

� !

The work done in process AB is!

� !

(c)!The efficiency is!

� !

(d)!The carnot efficiency is!

�

∆Eint Q WAB 0 JBC 0 JCD 0 JDA 0 J

WAB = −nRTAB ⋅ ln

VB

VA

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

= −(2.34 mol)R(720 K) ⋅ ln16×10−3 m3

10×10−3 m3

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

= −6583.5 J

∆Eint Q WAB 0 J +6584 J –6584 JBC 0 JCD 0 JDA 0 J

ΔEBC =

32nRΔTBC =

32(2.34 mol)R(549.46 K − 720 K) = −4976.7 J

∆Eint Q WAB 0 J +6584 J –6584 JBC –4977 J 0 J –4977 JCD 0 JDA +4977 J 0 J +4977 J

WCD = −nRTCD ⋅ ln

VD

VC

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

= −(2.34 mol)R(549.46 K) ⋅ ln15×10−3 m3

24×10−3 m3

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

= 5024.2 J

∆Eint Q WAB 0 J +6584 J –6584 JBC –4977 J 0 J –4977 JCD 0 J –5024 J +5024 JDA +4977 J 0 J +4977 J

e =Wtotal by

Qin

=1560 J6584 J

= 0.23694

ecarnot = 1−Tc

Th

= 1−549.46 K720 K

= 0.23686

page �10

Problem 22.40!A one more sample of hydrogen gas is contained in the left side of the container. It has equal volumes on the left and right. The right side is evacuated. When the valve is opened, the gas streams into the right side. Assume that the hydrogen behaves like an ideal gas.!(a)!What is the entropy change of the gas?!(b)!Does the temperature of the gas change?!

Solution!(b)!When the gas expands, the number of moles is constant, the volume doubles and the pressure

halves. The temperature does not change.!� !

� !

� !

(a)!Because the change in the temperature is zero, the change in the internal energy is zero. This means the heat is equal to the work.!

� !The actual path is not known as it is not quasi-static and reversible, however. This means we have to take a reversible path to calculate the entropy change. Let’s take the isothermal path.!The change in the entropy is!

� !

� !

!

PiVi = nRTi

PfVf =

12Pi ⋅ 2Vi = PiVi = nRTf

Tf = Ti

Q = W

ΔS =

dQT∫ = −

1T

dWisothermal∫ =1T

P(V )dV∫ =1T

nRTV

dV∫ = nRdVV∫ = nR ⋅ ln(V )

ΔS = nR ⋅ ln

Vf

Vi

⎛

⎝⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟= nR ⋅ ln

2Vi

Vi

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

= nR ln(2) = (1 mol)(8.314 Jmol ⋅K

)ln(2) = 5.7628 JK

page �11

Problem 22.57!A fixed this quantity of inert gas moves cyclically between the cylinders, expanding in the hot one and contracting in the cold one. Figure P.22.57 represents a model for its thermodynamic cycle. Consider n moles of an ideal monoatomic gas being taken once through the cycle, consisting of two isothermal processes at temperatures 3Ti and Ti and two constant-volume processes. Let us find the efficiency of this engine.!(a)!Find the energy transferred by heat into the gas during the

isovolumetric process AB.!(b)!Find the energy transferred by heat into the gas during the

isothermal process BC.!(c)!Find the energy transferred by heat into the gas during the

isovolumetric process CD.!(d) Find the energy transferred by heat into the gas during the isothermal process DA.!(e) Identify which of the results from parts (a) through (d) are positive and evaluate the energy

input to the engine by heat.!(f)! From the first law of thermodynamics, find the work done by the engine.!(g)!From the results of parts (e) and (f), evaluate the efficiency of the engine.!

Solution!(a)!Isovolumetric means that the work is zero. The heat is just the change in the internal energy.

For a monoatomic gas,!

� !

(b)!Isothermal means that the change in the internal energy is zero. The heat is just the work.!

� !

(c)!The heat is the the internal energy change.!

� !

(d)!The heat is the work.!

� !

(e)!The total heat in is!� !

(f)! The total work done by the engine is!� !

!

QAB = ΔEAB =

32nRΔTAB =

32nR(3T

i−Ti) = 3 ⋅nRTi

QBC = −WBC = nRTBC ⋅ ln

VC

VB

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

= nR(3Ti) ⋅ ln 2 = 3 ln2 ⋅nRTi

QCD = ΔECD =

32nRΔTCD =

32nR(Ti − 3Ti) = −3 ⋅nRTi

QDA = −WDA = nRTDA ⋅ ln

VA

VD

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

= nR(Ti) ⋅ ln(1 / 2) = − ln 2 ⋅nRTi

Qin = QAB +QBC = 3 ⋅nRTi + 3 ln2 ⋅nRTi = 3(1 + ln 2) ⋅nRTi

Wtotal by engine = −WBC −WDA = 3 ln2 ⋅nRTi − ln 2 ⋅nRTi = 2 ln2 ⋅nRTi

page �12

(g)!The efficiency is!

� !

! e =

2 ln2 ⋅nRTi

3(1 + ln 2) ⋅nRTi

=2 ln2

3(1 + ln 2)= 0.27292

page �13

Problem 22.63!A power plant, having a carnot efficiency, produces 1 GW of electrical power from turbines that take in steam at 500 K and reject water at 300 K into a flowing river. The water downstream is 6 K warmer due to the output of the power plant. Determine the flow rate of the river.!

Solution!Here is a schematic of the engine.!

� !This engine has the carnot efficiency so that the efficiency is!

� !

This means the heat in in one second is or the power in is!

� !

� !

The heat output is 1.5 GW. This heat goes into the river. In one second, 1.5 GJ of heat goes into the river. This heat can heat up this much mass by 6 K.!

� !

The rate of water flow is!

� !

!

steam!500 K engine water!

300 K

work!1 GJ/s

e = 1−

Tc

Th

= 1−300 K500 K

= 0.4

e = 0.4 =

Wtotal by engine

Qin

=109 JQin

⇒ Qin = 2.5×109 J

Pin = 2.5 GW

Q = 1.5×109 J = mcΔT = m(4186 J

kg⋅°C)(6 K) ⇒ m = 5.9723×104 kg

dmdt

= 5.9723×104 kgs

page �14

Problem 22.71!A 1 mol sample of an ideal gas γ=1.4 is carried through the carnot cycle described in Figure 22.10. At point A, the pressure is 25 atm and the temperature is 600 K. At point C, the pressure is 1 atm and the temperature is 400 K.!(a)!Determine the pressure and volume at points A, B, C, and D.!(b)!Calculate the net work done per cycle.!

Solution!(a)!Here is the schematic of the cycle.!

� !The volume at point A is!

� !

The volume at point C is!

� !

At point B,!

� !

� !

At point D, !

� !

� !

(b)!The work in process AB is!

� !

!

A!25 atm!600 K

C!1 atm!400 K

B!600 K

D!400 K

VA =

nRTA

PA

=(1 mol)(8.314 J

mol ⋅K)(600 K)

(25 atm)(101,325 Paatm

)= 1.9693×10−3 m3

VC =

nRTC

PC

=(1 mol)(8.314 J

mol ⋅K)(400 K)

(1 atm)(101,325 Paatm

)= 32.821×10−3 m3

TBVBγ−1 = TCVC

γ−1 ⇒ (600 K)VB0.4 = (400 K)(32.821×10−3 m3)0.4 ⇒ VB = 11.910×10−3 m3

PB =

nRTB

VB

=(1 mol)(8.314 J

mol ⋅K)(600 K)

(11.910×10−3 m3)= 4.1884×105 Pa = 4.1336 atm

TDVDγ−1 = TAVA

γ−1 ⇒ (400 K)VD0.4 = (600 K)(1.9693×10−3 m3)0.4 ⇒ VD = 5.4268×10−3 m3

PD =

nRTD

VD

=(1 mol)(8.314 J

mol ⋅K)(400 K)

(5.4268×10−3 m3)= 6.1281×105 Pa = 6.0480 atm

WAB = −nRTAB ⋅ ln

VB

VA

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

= −(1 mol)(8.314 Jmol ⋅K

)(600 K) ⋅ ln11.910×10−3 m3

1.9693×10−3 m3

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

= −8977.6 J

page �15

The work in process BC is!

� !

The work in process CD is!

� !

The work in process DA is!

� !

The total work done by the gas is 2998.6 J

WBC =

52nRΔTBC =

52(1 mol)(8.314 J

mol ⋅K)(400 K − 600 K) = −4157.0 J

WCD = −nRTCD ⋅ ln

VD

VC

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

= −(1 mol)(8.314 Jmol ⋅K

)(400 K) ⋅ ln5.4268×10−3 m3

32.821×10−3 m3

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

= 5979.0 J

WDA =

52nRΔTDA =

52(1 mol)(8.314 J

mol ⋅K)(600 K − 400 K) = 4157.0 J

page �16