Answer
a) See answer below.
b) $Q_{1-2}=5641\ kJ$ (in)
c) $W_{b,2-3}=1291\ kJ$ (out)
Work Step by Step
From tables A-4 to A-6:
1: ($x_1=0.5, T_1=100°C$): $h_1=1547.4\ kJ/kg$
2: ($x_2=1, T_2=100°C$): $h_2=2675.6\ kJ/kg,\ u_2=2506.0\ kJ/kg,\ s_2=7.3542\ kJ/kg.K$
3: ($P_3=15\ kPa, s_3=s_2$): $u_3=2247.9\ kJ/kg$
a) 1-2: Starts below the saturation curve, moving horizontally (constant T) to it (increase in s).
2-3: Starts at the saturation curve, moving vertically (constant s) away from it (decrease in T).
b) From the energy balance:
$Q-W_b=\Delta U\rightarrow Q=\Delta H$
$Q_{1-2}=m(h_2-h_1),\ m=5\ kg$
$Q_{1-2}=5641\ kJ$ (in)
c) From the energy balance:
$-W_{b,2-3}=\Delta U$
$-W_{b,2-3}=m(u_3-u_2)$
$W_{b,2-3}=1291\ kJ$ (out)
