§ME 231 Homework 8

Sep 25, 2025

§5.14

A power cycle receives QH by heat transfer from a hot reservoir at TH = 1200 K and rejects energy QC by heat transfer to a cold reservoir at TC = 400 K. For each of the following cases, determine whether the cycle operates reversibly, operates irreversibly, or is impossible.a. QH = 900 kJ, Wcycle = 450 kJb. QH = 900 kJ, QC = 300 kJc. Wcycle = 600 kJ, QC = 400 kJd. η = 75%
TH=1200K T_H=1200K
TC=400K T_C=400K
ηreversible=TH−TCTH=0.667 \eta_{reversible}=\frac{T_H-T_C}{T_H}=0.667
βmax=TCTH−TC=400K1200K−400K=0.500 \beta_{max}=\frac{T_C}{T_H-T_C}=\frac{400K}{1200K-400K}=0.500

§a.

QH=900kJ Q_H=900kJ
Wcycle=450kJ W_{cycle}=450kJ
η=WQH=450K900K=0.5<ηreversible=0.667  ⟹  Irreversible \eta=\frac{W}{Q_H}=\frac{450K}{900K}=0.5<\eta_{reversible}=0.667\implies\boxed{\text{Irreversible}}

§b.

QH=900kJ Q_H=900kJ
QC=300kJ Q_C=300kJ
W=QH−QC=900kJ−300kJ=600kJ W=Q_H-Q_C=900kJ-300kJ=600kJ
β=QCW=300kJ600kJ=0.500=βmax  ⟹  Reversible \beta=\frac{Q_C}{W}=\frac{300kJ}{600kJ}=0.500=\beta_{max}\implies\boxed{\text{Reversible}}

§c.

Wcycle=600kJ W_{cycle}=600kJ
QC=400kJ Q_C=400kJ
QH=QC+Wcycle=400kJ+600kJ=1000kJ Q_H=Q_C+W_{cycle}=400kJ+600kJ=1000kJ
η=WcycleQH=600kJ1000kJ=0.600<ηreversible=0.667  ⟹  Irreversible \eta=\frac{W_{cycle}}{Q_H}=\frac{600kJ}{1000kJ}=0.600<\eta_{reversible}=0.667\implies\boxed{\text{Irreversible}}

§d.

η=75%=0.75>ηreversible=0.667  ⟹  Impossible \eta=75\%=0.75>\eta_{reversible}=0.667\implies\boxed{\text{Impossible}}

§5.19

At a particular location, magma exists several kilometers below Earth's surface at a temperature of 1100°C, while the average temperature of the atmosphere at the surface is 15°C. An inventor claims to have devised a power cycle operating between these temperatures having a thermal efficiency of 79%. Investigate this claim.
TH=1100°C=1373K T_H=1100\degree C=1373K
TC=15°C=288.2K T_C=15\degree C=288.2K
ηclaimed=75%=0.75 \eta_{claimed}=75\%=0.75
ηreversible=TH−TCTH=1373K−288.2K1373K=0.7901>ηclaimed  ⟹  Irreversible, Possible \eta_{reversible}=\frac{T_H-T_C}{T_H}=\frac{1373K-288.2K}{1373K}=0.7901>\eta_{claimed}\implies\boxed{\text{Irreversible, Possible}}

§5.31

At steady state, a reversible refrigeration cycle discharges energy at the rate 𝑄H to a hot reservoir at temperature TH, while receiving energy at the rate 𝑄C from a cold reservoir at temperature TC.a. If TH = 13°C and TC = 2°C, determine the coefficient of performance.b. If 𝑄H = 10.5 kW, 𝑄C = 8.75 kW, and TC = 0°C, determine TH, in °C.c. If the coefficient of performance is 10 and TH = 27°C, determine TC, in °C.

§a.

TH=13°C=286.2K T_H=13\degree C=286.2K
TC=2°C=275.2K T_C=2\degree C=275.2K
β=TCTH−TC=275.2K286.2K−275.2K=25 \beta=\frac{T_C}{T_H-T_C}=\frac{275.2K}{286.2K-275.2K}=\boxed{25}

§b.

QH=10.5kW Q_H=10.5kW
QC=8.75kW Q_C=8.75kW
TC=0°C=273.2K T_C=0\degree C=273.2K
β=TCTH−TC=QCQH−QC \beta=\frac{T_C}{T_H-T_C}=\frac{Q_C}{Q_H-Q_C}
TC(QH−QC)TH−TC=QC \frac{T_C(Q_H-Q_C)}{T_H-T_C}=Q_C
TC(QH−QC)=QC(TH−TC) T_C(Q_H-Q_C)=Q_C(T_H-T_C)
TC(QH−QC)=QCTH−QCTC T_C(Q_H-Q_C)=Q_CT_H-Q_CT_C
TC(QH−QC)+QCTC=QCTH T_C(Q_H-Q_C)+Q_CT_C=Q_CT_H
TH=TC(QH−QC)+QCTCQC T_H=\frac{T_C(Q_H-Q_C)+Q_CT_C}{Q_C}
TH=TCQH−QCQC+TC=273.2K∗10.5kW−8.75kW8.75kW+273.2K=327.8K=54.65°C T_H=T_C\frac{Q_H-Q_C}{Q_C}+T_C=273.2K*\frac{10.5kW-8.75kW}{8.75kW}+273.2K=327.8K=\boxed{54.65\degree C}

§c.

β=10 \beta=10
TH=27°C=300.2K T_H=27\degree C=300.2K
β=TCTH−TC \beta=\frac{T_C}{T_H-T_C}
β(TH−TC)=TC \beta(T_H-T_C)=T_C
βTH−βTC=TC \beta T_H-\beta T_C=T_C
βTH=TC+βTC \beta T_H=T_C+\beta T_C
βTH=TC(1+β) \beta T_H=T_C(1+\beta)
TC=βTH1+β=10∗300.2K1+10=272.9K=−0.25°C T_C=\frac{\beta T_H}{1+\beta}=\frac{10 * 300.2K}{1 + 10}=272.9K=\boxed{-0.25\degree C}

§5.59

At steady state, a thermodynamic cycle operating between hot and cold reservoirs at 1000 K and 500 K, respectively, receives energy by heat transfer from the hot reservoir at a rate of 1500 kW, discharges energy by heat transfer to the cold reservoir, and develops power at a rate of (a) 1000 kW, (b) 750 kW, (c) 0 kW. For each case, apply Eq. 5.13 on a time-rate basis to determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
TH=1000K T_H=1000K
TC=500K T_C=500K
QH=1500kW Q_H=1500kW

§a.

QC=1000kW Q_C=1000kW
QHTH−QCTC=−σcycle \frac{Q_H}{T_H}-\frac{Q_C}{T_C}=-\sigma_{cycle}
σcycle=QCTC−QHTH=1000kW500K−1500kW1000K=0.5>0  ⟹  Irreversible \sigma_{cycle}=\frac{Q_C}{T_C}-\frac{Q_H}{T_H}=\frac{1000kW}{500K}-\frac{1500kW}{1000K}=0.5>0\implies\boxed{\text{Irreversible}}

§b.

QC=750kW Q_C=750kW
σcycle=QCTC−QHTH=750kW500K−1500kW1000K=0  ⟹  Reversible \sigma_{cycle}=\frac{Q_C}{T_C}-\frac{Q_H}{T_H}=\frac{750kW}{500K}-\frac{1500kW}{1000K}=0\implies\boxed{\text{Reversible}}

§b.

QC=0kW Q_C=0kW
σcycle=QCTC−QHTH=0kW500K−1500kW1000K=−1500kW1000K<0  ⟹  Impossible \sigma_{cycle}=\frac{Q_C}{T_C}-\frac{Q_H}{T_H}=\frac{0kW}{500K}-\frac{1500kW}{1000K}=-\frac{1500kW}{1000K}<0\implies\boxed{\text{Impossible}}