Question

Carbon Dioxide at P1= 500 psia, V1= 1 ft^3/min, and t1=200 F are cooled at constant volume to 130 F in an internally reversible manner. For mdot=0.9 lbm/min, determine pdv and - Vdp. For nonflow process, find p2, DELTA u, DELTA h and Q. Also compute delta S.

Answer 1

Solution;

From the gas equation;

$P_1V_1=mRT_1$

For carbon dioxide ,R=1130

$V_1=\frac{mRT_1}{P_1}=\frac{0.9×1130×(659.67}{500}=1341.77ft^3/min$

Pressure after isentropic process;

$\frac{P_2}{P_1}=(\frac{T_2}{T_1})^{\frac{\gamma}{\gamma -1}}$

$P_2=500(\frac{130+459.67}{200+459.67})^{\frac{1.4}{0.4}}=337.64psia$

Volume after isentropic process;

$\frac{P_2}{P_1}=(\frac{V_1}{V_2})^{\gamma}$

$\frac{337.64}{500}=(\frac{1341.77}{V_2})^{1.4}$

$V_2=1776.13ft^3/min$

$pdV=\frac{P_1V_1-P_2V_2}{\gamma-1}$

$pdV=\frac{500×1341.77-(337.64×17776.13}{0.4}$

$PdV=177981.17Btu/min$

$VdP=rPdV$ =1.4×177981.17

$VdP=249173.63Btu/min$

$\Delta u=mC_v(T_2-T_1)$

$\Delta u=0.9×2825(130-200)$

$\Delta u=-177,975Btu/min$

$\Delta h=mC_p(T_2-T_1)$

$\Delta h=0.9×3955×(130-200)$

$\Delta h=-249165Btu/min$

$Q=\Delta h+VdP$

$Q=-249165+249173=8Btu/min$

$\Delta s=mC_vln\frac{T_2}{T_1}$

$\Delta s=0.9×2825ln\frac{130}{200}$

$\Delta s=-1095.2656Btu/min$