Question

During a steady flow process, the pressure of the working substance drops from 180 psia to 30 psia, the speed increases from 210 fps to 1050 fps, the internal energy of the open system decreases 30 Btu/lb, and the specific volume increases from 2 to 10 ft³/lb. Heat loss is 20 Btu/lb. Determine the horsepower for 12 lb/min mass flow.

Answer 1

$P_1+K_1+W_1+U_1+Q=P_2+K_2+W_2+U_2+W,$

$K_1=\frac{v_1^2}{2k}=\frac{210^2}{2\cdot32.174\cdot778}=0.88~\frac{Btu}{lb},$

$K_2=\frac{v_2^2}{2k}=\frac{1050^2}{2\cdot32.174\cdot778}=22.02~\frac{Btu}{lb},$

$W_1=p_1V_1=\frac{180\cdot 144\cdot2}{778}=66.63~\frac{Btu}{lb},$

$W_2=p_2V_2=\frac{30\cdot 144\cdot10}{778}=55.53~\frac{Btu}{lb},$

$\Delta U=U_2-U_1=30~\frac{Btu}{lb},$

$Q=-20~\frac{Btu}{lb},$

$0.88+66.63-20=22.02+55.53-30-W,$

$W=0.04~\frac{Btu}{lb},$

$W=\frac{0.04\cdot12}{42.4}=0.011~hp.$