This problem presents a comparison of the work required in adiabatic pumps and compressors.
A) Saturated water vapor enters a steady-state compressor at P=1 bar. Estimate the work required to compress this vapor up to P=10 bar and T=400°C.
B) Saturated liquid water at P=1 bar enters a pump. Estimate the work required to pump the liquid up to P=10 bar, and estimate the temperature of the exiting liquid.
A) Set up an energy balance around the compressor
\begin{aligned}& \frac{\mathrm{D}}{\mathrm{dt}}\left\{\mathrm{M}\left(\widehat{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2}+\mathrm{gh}\right)\right\}=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+g h_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\text {out }}+\frac{\mathrm{v}_{\text {out }}^{2}}{2}+\mathrm{gh}_{\mathrm{out}}\right)+ \\& \dot{\mathrm{W}}_{\mathrm{S}}+\dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}}\end{aligned}
Cancelling terms
0=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}
Find \widehat{\mathrm{H}}_{\mathrm{in}}
Saturated water vapor at 1 bar \rightarrow 2674.9 \frac{\mathrm{kJ}}{\mathrm{kg}}
Find \widehat{\mathrm{H}}_{\text {out }}
Water at 10 bar and 400°C \rightarrow 3264.5 \frac{\mathrm{kJ}}{\mathrm{kg}}
\begin{gathered}0=\dot{\mathrm{m}}_{\mathrm{in}}\left(2674.9 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(3264.5 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}\\\frac{\dot{W}_{s}}{\dot{m}}=\bf 589.5 \frac{k J}{kg}\end{gathered}
B) Apply equation 3.77
\begin{gathered}\frac{\dot{W}_{s}}{\dot{m}} \approx \hat{V}\left(P_{\text {out }}-P_{\text {in }}\right)\\\frac{\dot{W}_{s}}{\dot{m}}=\left(0.001043 \frac{m^{3}}{\mathrm{~kg}}\right)(10-1 \,\rm bar \left(\frac{100 \mathrm{~cm}}{m}\right)^{3}\left(\frac{8.314 \frac{\mathrm{J}}{\mathrm{mol\,K}}}{83.14 \frac{b a r \,cm^{3}}{m o l \,K}}\right)\left(\frac{1 \mathrm{~kJ}}{1000 \mathrm{~J}}\right)\\=\bf0.94 \frac{k J}{k g}\end{gathered}
The energy balance around the pump is the same as that around the compressor:
\begin{gathered}0=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}\\\frac{\dot{W}_{s}}{\dot{m}}=\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}\end{gathered}
Entering is saturated liquid at 1 bar, which has a temperature of 99.6°C :
\widehat{\mathrm{H}}_{\mathrm{out}}=417.5+0.94=418.5 \frac{\mathrm{kJ}}{\mathrm{kg}}
Exiting liquid is at 10 bar. Interpolating between the enthalpy data for 80 and 100°C at this pressure gives T=99.7°C. The temperature is practically unchanged by the compression of the liquid.