§ME 231 Homework 5

Sep 25, 2025

§3.68

Determine the specific volume, in m3/kg, of Refrigerant 134a at 16 bar, 100°C, usinga. Table A-12.b. Figure A-1.c. the ideal gas equation of state.
P=16bar P=16bar
T=100°C=373K T=100\degree C=373K

§a.

From table A-12 where P=16barP=16bar and T=100°CT=100\degree C:
v=0.01601m3/kg v=\boxed{0.01601m^3/kg}

§b.

From table A-1:
TC=374K T_C=374K
PC=40.7bar P_C=40.7bar
M=102.03kgkmol=0.10203kgmol M=102.03\frac{kg}{kmol}=0.10203\frac{kg}{mol}
Rˉ=8.3145Jmol∗K \bar{R}=8.3145\frac{J}{mol*K}
R=RˉM=8.3145Jmol∗K0.10203kgmol=81.491Jkg∗K R=\frac{\bar{R}}{M}=\frac{8.3145\frac{J}{mol*K}}{0.10203\frac{kg}{mol}}=81.491\frac{J}{kg*K}
Reduced temperature and pressure can now be acquired to locate ZZ:
TR=TTC=373K374K=0.9973≈1.0 T_R=\frac{T}{T_C}=\frac{373K}{374K}=0.9973\approx1.0
PR=PPC=16bar40.7bar=0.3931≈0.4 P_R=\frac{P}{P_C}=\frac{16bar}{40.7bar}=0.3931\approx0.4
Figure A-1 from Fundamentals of Engineering Thermodynamics
Z≈0.86 Z\approx0.86
Pv=ZRT Pv=ZRT
v=ZRTP=0.86∗81.491Jkg∗K∗373K16bar=0.01634m3/kg v=\frac{ZRT}{P}=\frac{0.86 * 81.491\frac{J}{kg*K} * 373K}{16bar}=\boxed{0.01634m^3/kg}

§c.

With an ideal gas, compressibility effects are ignored:
Z=1 Z=1
Pv=ZRT=RT Pv=ZRT=RT
v=RTP=81.491Jkg∗K∗373K16bar=0.01900m3/kg v=\frac{RT}{P}=\frac{81.491\frac{J}{kg*K} * 373K}{16bar}=\boxed{0.01900m^3/kg}

§3.79

As shown in Fig. P3.79, 2 kg of oxygen fills the cylinder of a piston–cylinder assembly. The initial volume and pressure are 2 m3 and 1 bar, respectively. Heat transfer to the oxygen occurs at constant pressure until the volume is doubled. Determine the heat transfer for the process, in kJ, assuming the specific heat ratio is constant, k = 1.35. Kinetic and potential energy effects can be ignored.
Figure Fig. P3.79 from Fundamentals of Engineering Thermodynamics
m=2kg m=2kg
V1=2m3 V_1=2m^3
P1=1bar P_1=1bar
V2=2V1=4m3 V_2=2V_1=4m^3
P2=P1=1bar P_2=P_1=1bar
k=1.35 k=1.35
W=∫V1V2PdV W=\int_{V_1}^{V_2}PdV
=P∫V1V2dV =P\int_{V_1}^{V_2}dV
=P(V2−V1) =P(V_2-V_1)
=1bar∗(4m3−2m3) =1bar*(4m^3-2m^3)
=200kJ =200kJ
M=32.00kg/kmol M=32.00kg/kmol
Rˉ=8.3145Jmol∗K \bar{R}=8.3145\frac{J}{mol*K}
R=RˉM R=\frac{\bar{R}}{M}
=8.3145Jmol∗K32.00kgkmol =\frac{8.3145\frac{J}{mol*K}}{32.00\frac{kg}{kmol}}
=259.83Jkg∗K =259.83\frac{J}{kg*K}
PV=mRT PV=mRT
T=PVmR T=\frac{PV}{mR}
T1=P1V1mR T_1=\frac{P_1V_1}{mR}
=1bar∗2m32kg∗259.83Jkg∗K =\frac{1bar * 2m^3}{2kg * 259.83\frac{J}{kg*K}}
=384.87K =384.87K
T2=P2V2mR T_2=\frac{P_2V_2}{mR}
=1bar∗4m32kg∗259.83Jkg∗K =\frac{1bar * 4m^3}{2kg * 259.83\frac{J}{kg*K}}
=769.73K =769.73K
From table A-20 where k=1.35≈1.350k=1.35\approx1.350:
cv=0.743kJkg∗K c_v=0.743\frac{kJ}{kg*K}
ΔU=mcv(T2−T1) \Delta U=mc_v(T_2-T_1)
=2kg∗0.743kJkg∗K(769.73K−384.87K) =2kg * 0.743\frac{kJ}{kg*K}(769.73K - 384.87K)
=571.9kJ =571.9kJ
ΔU=Q−W \Delta U = Q-W
Q=ΔU+W Q=\Delta U+W
=571.9kJ+200kJ=771.9kJ =571.9kJ + 200kJ=\boxed{771.9kJ}

§3.98

Air undergoes a polytropic process in a piston–cylinder assembly from p1 = 1 bar, T1 = 295 K to p2 = 7 bar. The air is modeled as an ideal gas and kinetic and potential energy effects are negligible. For a polytropic exponent of 1.6, determine the work and heat transfer, each in kJ per kg of air,a. assuming constant cυ evaluated at 300 K.b. assuming variable specific heats.
P1=1bar P_1=1bar
T1=295K T_1=295K
P2=7bar P_2=7bar
n=1.6 n=1.6
PVn=C PV^n=C

§a.

From table A-20 for air where T=300KT=300K:
cv=0.718kJkg∗K c_v=0.718\frac{kJ}{kg*K}
P1P2=(T1T2)nn−1 \frac{P_1}{P_2}=\left(\frac{T_1}{T_2}\right)^\frac{n}{n-1}
(P1P2)n−1n=T1T2 \left(\frac{P_1}{P_2}\right)^\frac{n-1}{n}=\frac{T_1}{T_2}
T2=T1(P1P2)n−1n T_2=\frac{T_1}{\left(\frac{P_1}{P_2}\right)^\frac{n-1}{n}}
=295K(1bar7bar)1.6−11.6 =\frac{295K}{\left(\frac{1bar}{7bar}\right)^\frac{1.6 - 1}{1.6}}
=612.0K =612.0K
From table A-1:
M=28.97kg/kmol M=28.97kg/kmol
Rˉ=8.3145Jmol∗K \bar{R}=8.3145\frac{J}{mol*K}
R=RˉM=8.3145Jmol∗K28.97kgkmol R=\frac{\bar{R}}{M}=\frac{8.3145\frac{J}{mol*K}}{28.97\frac{kg}{kmol}}
=287Jkg∗K =287\frac{J}{kg*K}
Pv=RT Pv=RT
v=RTP v=\frac{RT}{P}
v1=RT1P1=287Jkg∗K∗295K1bar v_1=\frac{RT_1}{P_1}=\frac{287\frac{J}{kg*K} * 295K}{1bar}
=0.8467m3/kg =0.8467m^3/kg
v2=RT2P2=287Jkg∗K∗612.0K7bar v_2=\frac{RT_2}{P_2}=\frac{287\frac{J}{kg*K} * 612.0K}{7bar}
=0.25092m3/kg =0.25092m^3/kg
W=P2V2−P1V11−n W=\frac{P_2V_2-P_1V_1}{1-n}
w=P2v2−P1v11−n w=\frac{P_2v_2-P_1v_1}{1-n}
=7bar∗0.25092m3/kg−1bar∗0.8467m3/kg1−1.6 =\frac{7bar * 0.25092m^3/kg - 1bar * 0.8467m^3/kg}{1 - 1.6}
=−151.6kJ/kg =\boxed{-151.6kJ/kg}
ΔU=mcv(T2−T1) \Delta U=mc_v(T_2-T_1)
Δu=cv(T2−T1) \Delta u=c_v(T_2-T_1)
=0.718kJkg∗K∗(612.0K−295K) =0.718\frac{kJ}{kg*K} * (612.0K - 295K)
=227.61kJ/kg =227.61kJ/kg
ΔU=Q−W \Delta U=Q-W
Δu=q−w \Delta u=q-w
q=Δu+w q=\Delta u+w
=227.61kJ/kg−151.6kJ/kg =227.61kJ/kg - 151.6kJ/kg
=76kJ/kg =\boxed{76kJ/kg}

§b.

A change in how cvc_v is interpreted only affects the value of Δu\Delta u; hence, ww can be borrowed from the previous section:
w=−151.6kJ/kg w=\boxed{-151.6kJ/kg}
From table A-22 where T=295K≈300KT=295K\approx300K:
u1=142.56kJ/kg u_1=142.56kJ/kg
From table A-22 where T=612.0K≈610KT=612.0K\approx610K:
u2=442.42kJ/kg u_2=442.42kJ/kg
Δu=u2−u1 \Delta u=u_2-u_1
=442.42kJ/kg−142.56kJ/kg =442.42kJ/kg - 142.56kJ/kg
=299.9kJ/kg =299.9kJ/kg
Δu\Delta u from the line above is about as far from the value computed in part a as I would have reasonably expected with a fixed specific heat.
q=Δu+w q=\Delta u+w
=299.9kJ/kg−151.6kJ/kg =299.9kJ/kg - 151.6kJ/kg
=148.3kJ/kg =\boxed{148.3kJ/kg}
Though qq is off by a good amount, it is only fair to evaluate the discrepancy for Δu\Delta u as qq is an additive function of Δu\Delta u and our fixed ww. Δu\Delta u is off by 24%24\% which "sounds about right" with such a bold assumption of a fixed cvc_v which itself varies by 23%23\% between the two temperatures. I needed to calculate the error percentages for the sake of my own sanity lol; feeling better to see the error percentage and difference in cvc_v to be so close in percentage.