Water at P=25 bar and T=200°C enters a steady-state flash chamber with a flow rate of 5 kg/sec. The liquid and vapor streams exiting the flash chamber are both at P=1 bar.
A) If the flash chamber is adiabatic, what fraction of the entering water leaves the flash as vapor?
B) How much heat must be added to or removed from the flash chamber in order to vaporize half of the entering water?
A) Set up an energy balance around the Flash chamber
\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}+\mathrm{gh}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}+g h_{\mathrm{out}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}+\dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}}\end{aligned}
Cancelling terms
\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{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)=\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)\end{gathered}
Find \widehat{\mathrm{H}}_{\mathrm{in}}
Liquid water at 25 bar and 200°C → Interpolation Needed
Interpolate between data at 50 bar and saturated liquid at 200°C (15.5 bar) :
\begin{array}{|c|c|c|c|} \hline & 1 & & 2\\\hline \text{Enthalpy (y) } & 852.3 & ??? & 852.3 \\\hline \text{Pressure (x) }& 15.5 & 25 & 50\\\hline\end{array}
\begin{gathered}y=\frac{\left(y_{2}-y_{1}\right)\left(x-x_{1}\right)}{\left(x_{2}-x_{1}\right)}+y_{1}\\y=\frac{(853.7-852.3)(25-15.5)}{(50-15.5)}\\\qquad\qquad +852.3=852.7 \frac{\mathrm{kJ}}{\mathrm{kg}}\\\left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)=\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)=852.7 \frac{\mathrm{kJ}}{\mathrm{kg}}\end{gathered}
Water at 1 bar and 852.7 \frac{\mathrm{kJ}}{\mathrm{kg}} \rightarrow Mixture of saturated liquid and saturated vapor
\begin{array}{|c|c|c|c|} \hline & 1 & & 2\\\hline \text{Percent Vapor (q) } & 0 & ??? & 100 \\\hline \text{Enthalpy (x) }& 417.5 & 852.7 & 2674.9\\\hline\end{array}
\begin{gathered}q=\frac{\left(q_{2}-q_{1}\right)\left(x-x_{1}\right)}{\left(x_{2}-x_{1}\right)}+q_{1}\\y=\frac{(100-0)(852.7-417.5)}{(2674.9-417.5)}+0=\mathbf{1 9 . 2 8} \% \bf vapor\end{gathered}
B) Set up an energy balance around the Flash chamber
\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}+\mathrm{gh}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}+g h_{\mathrm{out}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}+\dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}}\end{aligned}
Cancelling terms
\begin{gathered}0=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\text {out }}\right)+\dot{\mathrm{Q}}\\\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)-\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)=\dot{\mathrm{Q}}\end{gathered}
Find \widehat{\mathrm{H}}_{\text {out }}
Water/steam at q=0.5 and P=1 bar →
\begin{array} {|c|c|c|c|}\hline & 1 & & 2\\\hline \text{Enthalpy (y)} & 417.5 & ??? & 2674.9 \\\hline \text{Vapor Fraction (q) }& 0 & 50 & 100 \\\hline\end{array}
\begin{aligned}& y=\frac{\left(y_{2}-y_{1}\right)\left(q-q_{1}\right)}{\left(q_{2}-q_{1}\right)}+y_{1} \\& y=\frac{(2674.9-417.5)(50-0)}{(100-0)} \\& \qquad\qquad +417.5 \\& \qquad\qquad =1546.2 \frac{\mathrm{kJ}}{\mathrm{kg}}\\ & 5 \frac{\mathrm{kg}}{\mathrm{sec}}\left(1546.2 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)-5 \frac{\mathrm{kg}}{\mathrm{sec}}\left(852.7 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)=\dot{\mathrm{Q}}\\ & \dot{\mathrm{Q}}=\bf 3467.5 \frac{\mathbf{kJ}}{\mathbf{sec}}\end{aligned}