Answer
a) $w=853.4\ kJ/kg$ (in)
b) $w=1422.3\ kJ/kg$ (in)
Work Step by Step
For isentropic processes on ideal gases:
$\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^\dfrac{k-1}{k}$
Given $P_2=450\ kPa,\ P_1=90\ kPa,\ T_1=303\ K,\ k=1.667$
$T_2=576.9\ K$
a)
From the energy balance:
$-W=\Delta U = mc_v(T_2-T_1)$
$w=-c_v(T_2-T_1)$
With $c_v=3.1156\ kJ/kg.K$
$w=-853.4\ kJ/kg$ (in)
b) From the energy balance:
$\dot{W}_s+\dot{m}h_1=\dot{m}h_2$
$w=c_p(T_2-T_1)$
With $c_p=5.1926\ kJ/kg.K$
$w=1422.3\ kJ/kg$ (in)
