Question

In a turbine 4500 kg/min of air expands polytropically from 425 kPa & 1360K to 101 kPa. The exponent n is equal to 1.45 for the process. Calculate the heat and work steady flow. 

Answer 1

Solution;

For polytropic process:

$W=\frac{mR(T_1-T_2)}{n-1}$

R=287J/kgK

To find $T_2$ ,We use the relation;

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{n}{n-1}}$

By substitution;

$T_2=(\frac{101}{425})^{\frac{1.45}{0.45}}×1360K=13.26K$

Therefore;

$W=\frac{4500×287(1360-13.26)}{1.45-1}=64.41MW$

Heat;

$Q=W+\Delta u$

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

$c_v=\frac{R}{n-1}=\frac{287}{0.45}=637.77kJ/kgK$

$\Delta u=4500×637.77(13.26-1360)=-64.41$

$Q=W+\Delta u=0$