Question

 If 10 kg/min of air are compressed isothermally from P1=96kPa and V1=7.65m^3/min to P3=620kPa, find the work change of entropy and the heat for: a) nonflow process and b) steady flow process with v1=15m/s and v2=60m/s.

Answer 1

Solution;

Given;

$\dot{m}=10kg/min$

$P_1=96kPa$

$V_1=7.65m^3/min$

$P_2=620kPa$

$v_1=15m/s$

$v_2=60m/s$

For nonflow;

Work;

$W=nRT_1ln(\frac{P_1}{P_2})$

$n=0.3452$

$R=8.314J/mil.K$

$T_1=\frac{PV}{n}=\frac{96×0.1275}{0.3452}=34.46K$

$W=0.3452×8.314×34.46ln(\frac{96}{620})=-189.83kJ$

$Q=U+W$

But U=0

Hence;

$Q=W=-189.83kJ$

Entropy;

$S=\frac{Q}{T_1}=\frac{-189.83}{34.46}=-5.509kJ$

For steady flow;

$Q=W+m(\frac{v_1^2-v_2^2}{2})$

$Q=-189.83+10(\frac{15^2-60^2}{2000})=-206.705kJ$

$S=\frac{-206.705}{34.46}=-6.0kJ$