§ME 231 Quiz 4

Sep 25, 2025
A rigid tank has 5 kg air inside. The initial state is T1=500 K, P1=300 kPa. After a process, the air's final state is T2=300 K. Air can be treated with constant k=1.4. T0 =300 K, p0=100 kPa. What is the exergy change of the system from the initial state to the final one.
Air \text{Air}
V=const V=\text{const}
m=5kg m=5kg
T1=500K T_1=500K
P1=300kPa P_1=300kPa
T2=300K T_2=300K
k=1.4 k=1.4
T0=300K T_0=300K
P0=100kPa P_0=100kPa
From table A-22 at T=500KT=500K:
u1=359.49kJkg u_1=359.49\frac{kJ}{kg}
From table A-22 at T=300KT=300K:
u2=214.07kJkg u_2=214.07\frac{kJ}{kg}
From table A-1 for air:
M=28.97kgkmol M=28.97\frac{kg}{kmol}
Rˉ=8.314kJkmol∗K \bar{R}=8.314\frac{kJ}{kmol*K}
R=RˉM=8.314kJkmol∗K28.97kgkmol=0.2870kJkg∗K R=\frac{\bar{R}}{M}=\frac{8.314\frac{kJ}{kmol*K}}{28.97\frac{kg}{kmol}}=0.2870\frac{kJ}{kg*K}
Assuming ideal gas:
cv=Rk−1=0.2870kJkg∗K1.4−1=0.7175kJkg∗K c_v=\frac{R}{k-1}=\frac{0.2870\frac{kJ}{kg*K}}{1.4-1}=0.7175\frac{kJ}{kg*K}
Simplification of exergy change equation assuming negligible potential and kinetic energy:
ΔE=ΔU+P0ΔV−T0ΔS+ΔKE+ΔPE \Delta E=\Delta U + P_0 \Delta V - T_0 \Delta S + \Delta KE + \Delta PE
ΔE=mΔu+P0ΔV−T0mΔs+ΔKE+ΔPE \Delta E=m\Delta u + \cancel{P_0 \Delta V} - T_0 m \Delta s + \cancel{\Delta KE} + \cancel{\Delta PE}
ΔE=mΔu−T0mΔs \Delta E=m\Delta u - T_0 m \Delta s
ΔE=m(u2−u1)−T0m(s2−s1) \Delta E = m (u_2 - u_1) - T_0 m (s_2 - s_1)
ΔE=m(u2−u1)−T0m(cvln⁡T2T1+Rln⁡V2V1) \Delta E = m (u_2 - u_1) - T_0 m (c_v \ln \frac{T_2}{T_1} + R \ln \frac{V_2}{V_1})
ΔE=m(u2−u1)−T0m(cvln⁡T2T1+Rln⁡1) \Delta E = m (u_2 - u_1) - T_0 m (c_v \ln \frac{T_2}{T_1} + \cancel{R \ln 1})
ΔE=m(u2−u1)−T0mcvln⁡T2T1 \Delta E = m (u_2 - u_1) - T_0 m c_v \ln \frac{T_2}{T_1}
ΔE=5kg∗(214.07kJkg−359.49kJkg)−300K∗5kg∗0.7175kJkg∗K∗ln⁡300K500K=−177.3kJ \Delta E = 5kg * (214.07\frac{kJ}{kg} - 359.49\frac{kJ}{kg}) - 300K * 5kg * 0.7175\frac{kJ}{kg*K} * \ln \frac{300K}{500K}=\boxed{-177.3kJ}