Answer to Question #151908 in Molecular Physics | Thermodynamics for mikee

Question

Air flows steadily at a rate of 0.85 kg/s through an air compressor, entering at 10 m/s speed, 150 KPa pressure and 1.35 m3/kg specific volume and leaving at 8 m/s and 0.53 m3/kg.

Determine the value of leaving pressure and the change in specific enthalpy if the internal energy of the air entering is 105 KJ/kg greater than that of the air leaving. Cooling water in the compressor jackets add heat to the air at the rate of 60KW and the work done on the system is 106.58 kW

Answer 1

Given:

$m=0.85[\frac{kg}{s}]$

$v_1=10[\frac{m}{s}]$

$p_1=150*10^3Pa$

$V_1=1.35[\frac{m^3}{kg}]$

$v_2=8[\frac{m}{s}]$

$V_2=0.53[\frac{m^3}{kg}]$

$(U_1-U_2)=105[\frac{kJ}{kg}]$

Solution:

$Q_l=60[\frac{kJ}{s}]=\frac{60}{0.85}=70.6[\frac{kJ}{kg}]$

$Q-W=\Delta\;h+\Delta\;KE+\;\Delta\;PE$

$\Delta\;KE=\frac{V_2^2-V_1^2}{2}=\frac{8^2-10^2}{2}=-18[\frac{J}{kg}]=-18*10^{-3}[\frac{kJ}{kg}]$

$\Delta\;PE=0$

$70.6-106.58=\Delta\;h-18*10^{-3}$

$\Delta\;h=-36[\frac{kJ}{kg}]$

$\Delta\;h=\Delta(U+pV)=(U_2-U_1)+(p_2V_2-p_1V_1)$

$p_2=\frac{-36-105+150*10^3\cdot1.35}{0.53}=381809\;Pa=381.8\;kPa$