Answer
a) $w=15.10\ kJ/kg$
b) $w=15.14\ kJ/kg$
c) $w=15.10\ kJ/kg$
Work Step by Step
From tables A-4 to A-6:
Inlet ($P_1=10\ kPa, x_1=0$): $v_1=0.001010\ m³/kg,\ h_1=191.81\ kJ/kg,\ s_1=0.6492\ kJ/kg.K$
Outlet ($P_2=15\ MPa, s_2=s_1$): $v_2=0.001004\ m³/kg,\ h_2=206.90\ kJ/kg$
a) From the energy balance:
$\dot{W}=\dot{m}(h_2-h_1)$
$w=h_2-h_1=15.10\ kJ/kg$
b) With constant specific volume (of the inlet):
$w=v_1(P_2-P_1)=15.14\ kJ/kg$
c) With average specific volume:
$v_{avg}=\frac{v_1+v_2}2=0.001007\ m^3/kg$
$w=v_{avg}(P_2-P_1)=15.10\ kJ/kg$
