§ME 231 Homework 11

Sep 25, 2025

§6.99

Oxygen (O2) at 25°C, 100 kPa enters a compressor operating at steady state and exits at 260°C, 650 kPa. Stray heat transfer and kinetic and potential energy effects are negligible. Modeling the oxygen as an ideal gas with k = 1.379, determine the isentropic compressor efficiency and the work in kJ per kg of oxygen flowing.
Oxygen \text{Oxygen}
T1=25°C=298.2K T_1=25\degree C=298.2K
p1=100kPa p_1=100kPa
m˙=const \dot{m}=\text{const}
T2=260°C=533.2K T_2=260\degree C=533.2K
p2=650kPa p_2=650kPa
Q=KE=PE=0 Q=KE=PE=0
Ideal Gas \text{Ideal Gas}
k=1.379 k=1.379
Rˉ=8.3145kJkmol∗K \bar{R}=8.3145\frac{kJ}{kmol*K}
From table A-1 for oxygen:
M=32.00kgkmol M=32.00\frac{kg}{kmol}
R=RˉM=8.3145kJkmol∗K32.00kg/kmol=0.25983kJkg∗K R=\frac{\bar{R}}{M}=\frac{8.3145\frac{kJ}{kmol*K}}{32.00kg/kmol}=0.25983\frac{kJ}{kg*K}
cp=kRk−1=1.379∗0.25983kJkg∗K1.379−1=0.9454kJkg∗K c_p=\frac{kR}{k-1}=\frac{1.379*0.25983\frac{kJ}{kg*K}}{1.379-1}=0.9454\frac{kJ}{kg*K}
h2−h1=cp(T2−T1)=0.9454kJkg∗K(533.2K−298.2K)=222.17kJ/kg h_2-h_1=c_p(T_2-T_1)=0.9454\frac{kJ}{kg*K}(533.2K-298.2K)=222.17kJ/kg
h1−h2=−(h2−h1)=−222.17kJ/kg h_1-h_2=-(h_2-h_1)=-222.17kJ/kg
From table A-23 for oxygen at T=298.2K≈298KT=298.2K\approx 298K:
h1=8682kJkmol÷32.00kgkmol=271.313kJkg h_1=8682\frac{kJ}{kmol} \div 32.00\frac{kg}{kmol} = 271.313\frac{kJ}{kg}
From table A-23 for oxygen at T=533.2K≈530KT=533.2K\approx 530K:
h2s=15708kJkmol÷32.00kgkmol=490.875kJkg h_{2s}=15708\frac{kJ}{kmol} \div 32.00\frac{kg}{kmol} = 490.875\frac{kJ}{kg}
ηc=h1−h2sh1−h2=271.313kJkg−490.875kJkg−222.17kJ/kg=0.9883=98.83% \eta_c=\frac{h_1-h_{2s}}{h_1-h_2}=\frac{271.313\frac{kJ}{kg}-490.875\frac{kJ}{kg}}{-222.17kJ/kg}=0.9883=\boxed{98.83\%}
0=Q˙−W˙s+m˙[h1−h2s+v⃗12−v⃗222+g(z1−z2)] 0=\cancel{\dot{Q}}-\dot{W}_s+\dot{m}\left[h_1-h_{2s}+\cancel{\frac{\vec{v}_1^2-\vec{v}_2^2}{2}}+\cancel{g(z_1-z_2)}\right]
0=−W˙s+m˙(h1−h2s) 0=-\dot{W}_s+\dot{m}(h_1-h_{2s})
W˙s=m˙(h1−h2s) \dot{W}_s=\dot{m}(h_1-h_{2s})
W˙sm˙=ws=h1−h2s=271.313kJkg−490.875kJkg=−219.562kJkg \frac{\dot{W}_s}{\dot{m}}=w_s=h_1-h_{2s}=271.313\frac{kJ}{kg} - 490.875\frac{kJ}{kg}=-219.562\frac{kJ}{kg}
w=ηcws=0.9883∗−219.562kJkg=−216.99kJkg w=\eta_c w_s=0.9883*-219.562\frac{kJ}{kg}=\boxed{-216.99\frac{kJ}{kg}}

§6.108

Air as an ideal gas enters a diffuser operating at steady state at 4 bar, 290 K with a velocity of 512 m/s. The exit velocity is 110 m/s. For adiabatic operation with no internal irreversibilities, determine the exit temperature, in K, and the exit pressure, in bar,a. for k = 1.4.b. using data from Table A-22.
Air \text{Air}
Ideal Gas \text{Ideal Gas}
Diffuser \text{Diffuser}
m˙=const \dot{m}=\text{const}
p1=4bar p_1=4bar
T1=290K T_1=290K
v⃗1=512m/s \vec{v}_1=512m/s
v⃗2=110m/s \vec{v}_2=110m/s
Q=0 Q=0
σ=0 \sigma=0

§a.

k=1.4 k=1.4
R=287Jkg∗K R=287\frac{J}{kg*K}
cp=kRk−1=1.4∗287Jkg∗K1.4−1=1.005kJkg∗K c_p=\frac{kR}{k-1}=\frac{1.4 * 287\frac{J}{kg*K}}{1.4 - 1}=1.005\frac{kJ}{kg*K}
0=Q˙−W˙+m˙[h1−h2+v⃗12−v⃗222+g(z1−z2)] 0=\cancel{\dot{Q}}-\cancel{\dot{W}}+\dot{m}\left[h_1-h_2+\frac{\vec{v}_1^2-\vec{v}_2^2}{2}+\cancel{g(z_1-z_2)}\right]
0=m˙[h1−h2+v⃗12−v⃗222] 0=\dot{m}\left[h_1-h_2+\frac{\vec{v}_1^2-\vec{v}_2^2}{2}\right]
0=h1−h2+v⃗12−v⃗222 0=h_1-h_2+\frac{\vec{v}_1^2-\vec{v}_2^2}{2}
0=cp(T1−T2)+v⃗12−v⃗222 0=c_p(T_1-T_2)+\frac{\vec{v}_1^2-\vec{v}_2^2}{2}
0=T1−T2+v⃗12−v⃗222cp 0=T_1-T_2+\frac{\vec{v}_1^2-\vec{v}_2^2}{2 c_p}
T2=T1+v⃗12−v⃗222cp=290K+(512m/s)2−(110m/s)22∗1.005kJkg∗K=414.4K T_2=T_1+\frac{\vec{v}_1^2-\vec{v}_2^2}{2 c_p}=290K + \frac{(512m/s)^2-(110m/s)^2}{2 * 1.005\frac{kJ}{kg*K}}=\boxed{414.4K}
T2T1=(p2p1)k−1k \frac{T_2}{T_1}=\left(\frac{p_2}{p_1}\right)^\frac{k-1}{k}
(T2T1)kk−1=p2p1 \left(\frac{T_2}{T_1}\right)^\frac{k}{k-1}=\frac{p_2}{p_1}
p2=p1(T2T1)kk−1=4bar∗(414.4K290K)1.41.4−1=13.95bar p_2=p_1\left(\frac{T_2}{T_1}\right)^\frac{k}{k-1}=4bar * \left(\frac{414.4K}{290K}\right)^\frac{1.4}{1.4-1}=\boxed{13.95bar}

§b.

From table A-22 at T=290KT=290K:
h1=290.16kJkg h_1=290.16\frac{kJ}{kg}
s1o=1.66802kJkg∗K s^o_1=1.66802\frac{kJ}{kg*K}
h2=h1+v⃗12−v⃗222 h_2=h_1+\frac{\vec{v}_1^2-\vec{v}_2^2}{2}
h2=290.16kJkg+(512m/s)2−(110m/s)22=415.2kJ/kg h_2=290.16\frac{kJ}{kg} + \frac{(512m/s)^2-(110m/s)^2}{2}=415.2kJ/kg
Excerpt from table A-22 around h=415.2kJ/kgh=415.2kJ/kg:
TT hh s°s\degree
410K410K 411.12kJkg411.12\frac{kJ}{kg} 2.01699kJkg∗K2.01699\frac{kJ}{kg*K}
420K420K 421.26kJkg421.26\frac{kJ}{kg} 2.04142kJkg∗K2.04142\frac{kJ}{kg*K}
T2=410+(420−410)415.2−411.12421.26−411.12=414.02K T_2=410+(420-410)\frac{415.2-411.12}{421.26-411.12}=\boxed{414.02K}
s2o=2.01699+(2.04142−2.01699)415.2−411.12421.26−411.12=2.0268kJkg∗K s^o_2=2.01699+(2.04142-2.01699)\frac{415.2-411.12}{421.26-411.12}=2.0268\frac{kJ}{kg*K}
0=s2o−s1o−Rln⁡p2p1 0=s^o_2-s^o_1-R \ln \frac{p_2}{p_1}
−(s2o−s1o)=−Rln⁡p2p1 -(s^o_2-s^o_1)=-R \ln \frac{p_2}{p_1}
s2o−s1o=Rln⁡p2p1 s^o_2-s^o_1=R \ln \frac{p_2}{p_1}
s2o−s1oR=ln⁡p2p1 \frac{s^o_2-s^o_1}{R}=\ln \frac{p_2}{p_1}
exp⁡s2o−s1oR=p2p1 \exp \frac{s^o_2-s^o_1}{R} = \frac{p_2}{p_1}
p2=p1exp⁡s2o−s1oR=4bar∗exp⁡2.0268kJkg∗K−1.66802kJkg∗K287Jkg∗K=13.96bar p_2 = p_1 \exp \frac{s^o_2-s^o_1}{R} = 4bar * \exp \frac{2.0268\frac{kJ}{kg*K} - 1.66802\frac{kJ}{kg*K}}{287\frac{J}{kg*K}}=\boxed{13.96bar}

§6.115

Carbon dioxide (CO2) expands isothermally at steady state with no irreversibilities through a turbine from 10 bar, 500 K to 2 bar. Assuming the ideal gas model and neglecting kinetic and potential energy effects, determine the heat transfer and work, each in kJ per kg of carbon dioxide flowing.
Carbon Dioxide \text{Carbon Dioxide}
T1=T2=500K≡T T_1=T_2=500K \equiv T
m˙=const \dot{m}=\text{const}
σ=0 \sigma=0
Turbine \text{Turbine}
p1=10bar p_1=10bar
p2=2bar p_2=2bar
Ideal Gas \text{Ideal Gas}
KE=PE=0 KE=PE=0
R=0.1889kJkg∗K R=0.1889\frac{kJ}{kg*K}
Δs=s2−s1=s2o−s1o−Rln⁡p2p1 \Delta s = s_2 - s_1 = s^o_2 - s^o_1 - R \ln \frac{p_2}{p_1}
s2−s1=−Rln⁡p2p1=−0.1889kJkg∗K∗ln⁡2bar10bar=0.304kJkg∗K s_2 - s_1 = - R \ln \frac{p_2}{p_1} = - 0.1889\frac{kJ}{kg*K} * \ln \frac{2bar}{10bar}=0.304\frac{kJ}{kg*K}
0=Q˙T+m˙(s1−s2)+σ˙ 0=\frac{\dot{Q}}{T}+\dot{m}(s_1-s_2)+\cancel{\dot{\sigma}}
−m˙(s1−s2)=Q˙T -\dot{m}(s_1-s_2)=\frac{\dot{Q}}{T}
m˙(s2−s1)=Q˙T \dot{m}(s_2-s_1)=\frac{\dot{Q}}{T}
s2−s1=qT s_2-s_1=\frac{q}{T}
q=T(s2−s1)=500K∗0.304kJkg∗K=152kJkg q=T(s_2-s_1)=500K * 0.304\frac{kJ}{kg*K}=\boxed{152\frac{kJ}{kg}}
0=Q˙−W˙+m˙[h1−h2+v⃗12−v⃗222+g(z1−z2)] 0=\dot{Q}-\dot{W}+\dot{m}\left[h_1-h_2+\cancel{\frac{\vec{v}_1^2-\vec{v}_2^2}{2}}+\cancel{g(z_1-z_2)}\right]
0=Q˙−W˙+m˙[h1−h2] 0=\dot{Q}-\dot{W}+\dot{m}[h_1-h_2]
Enthalpy is a function of temperature and also stays constant:
0=Q˙−W˙+m˙[h1−h1] 0=\dot{Q}-\dot{W}+\cancel{\dot{m}[h_1-h_1]}
0=Q˙−W˙ 0=\dot{Q}-\dot{W}
W˙=Q˙ \dot{W}=\dot{Q}
w=q=152kJkg w=q=\boxed{152\frac{kJ}{kg}}