Answer to Question #183769 in Mechanical Engineering for Eric Santos

Question

Assume 5 ft3/s of Hydrogen at 15 psia and 80 oF are compressed during an irreversible adiabatic  process to 90 psia and 522 oF. For a non-flow process determine a) k, b) V2, c) W, d) ΔU and e) ΔS.

Answer 1

a. $\frac{T_2}{T_1}=(\frac{P_2}{P_1})^{\frac{k-1}{k}}$

$\frac{545.372}{299.817}=(\frac{90}{15})^{\frac{k-1}{k}}$

$\ln{1.819}=({\frac{k-1}{k}} )\ln{6}$

$1.1935k=1.79$

$k=1.5$

b. $P_1V_1^k=P_2V_2^k$

$15(5)^{1.5}=90(V_2)^{1.5}$

$V_2=1.5142 In^3/s$

c. $\triangle w = \frac{P_1V_1-P_2V_2}{k-1}= \frac{15 \times 1728 \times 5 - 90 \times 1782 \times 1.5142}{1.5-1}=-211776.76 lb-inch/s$

d. $\triangle \theta = \triangle w +\triangle u$

But $\triangle \theta =0$

So, $\triangle w = -\triangle u$

$\triangle u=211776.76 lb-inch/s$

e. $\triangle s=\frac{Q}{T}=\frac{-211776.76}{245.555}=-862.44124 lb-inch /sK$