Answer
a)$P_{2}=90.4kPa$
b) $\Delta V=0.021m^3$
c) $\Delta H=17.363\frac{kJ}{kg}$
Work Step by Step
First we calculate the pressure exerted by the piston:
$A=\frac{\pi D^2}{4}=\frac{\pi*(0.25m)^2}{4}=0.0491m^2$
$W=mg=12kg*9.81\frac{m}{s^2}=117.72N$
$P_{1,gauss}=\frac{W}{A}=\frac{117.72N}{0.0491m^2}=2.4kPa$
$P_{1,abs}=88kPa+2.4kPa=90.4kPa$
a)$P_{2}=P_{1,abs}=90.4kPa$
From Table A-13:
1) $P_{1}=90.4kPa$ and $T_{1}=-10^{\circ}C$
As $T\gt T_{sat}$ is a superheated vapor
Interpolating:
$\upsilon_{1}=0.2418\frac{m^3}{kg}$
$h_{1}=247.77\frac{kJ}{kg}$
From Table A-13:
2) $P_{2}=90.4kPa$ and $T_{2}=15^{\circ}C$
As $T\gt T_{sat}$ is a superheated vapor
Interpolating:
$\upsilon_{2}=0.2669\frac{m^3}{kg}$
$h_{2}=268.19\frac{kJ}{kg}$
b) $\Delta V=m(\Delta \upsilon)=0.85kg*(0.2669\frac{m^3}{kg}-0.2418\frac{m^3}{kg})=0.021m^3$
c) $\Delta H=m(\Delta h)=0.85kg*(268.19\frac{kJ}{kg}-247.77\frac{kJ}{kg})=17.363\frac{kJ}{kg}$
