A) Set up an energy balance around the Flash chamber
d d t { M ( U ^ + v 2 2 + g h ) } = m ˙ i n ( H ^ i n + v in 2 2 + g h i n ) − m ˙ o u t ( H ^ o u t + v o u t 2 2 + g h o u t ) + W ˙ S + W ˙ E C + Q ˙ \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} d t d { M ( U + 2 v 2 + g h ) } = m ˙ i n ( H i n + 2 v in 2 + g h i n ) − m ˙ o u t ( H o u t + 2 v o u t 2 + g h o u t ) + W ˙ S + W ˙ E C + Q ˙
Cancelling terms
0 = m ˙ i n ( H ^ i n ) − m ˙ o u t ( H ^ o u t ) m ˙ i n ( H ^ i n ) = m ˙ o u t ( H ^ o u t ) \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} 0 = m ˙ i n ( H i n ) − m ˙ o u t ( H o u t ) m ˙ i n ( H i n ) = m ˙ o u t ( H o u t )
Find H ^ i n \widehat{\mathrm{H}}_{\mathrm{in}} H i n
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) :
1 2 Enthalpy (y) 852.3 ? ? ? 852.3 Pressure (x) 15.5 25 50 \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} Enthalpy (y) Pressure (x) 1 8 5 2 . 3 1 5 . 5 ? ? ? 2 5 2 8 5 2 . 3 5 0
y = ( y 2 − y 1 ) ( x − x 1 ) ( x 2 − x 1 ) + y 1 y = ( 853.7 − 852.3 ) ( 25 − 15.5 ) ( 50 − 15.5 ) + 852.3 = 852.7 k J k g ( H ^ i n ) = ( H ^ o u t ) = 852.7 k J k g \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} y = ( x 2 − x 1 ) ( y 2 − y 1 ) ( x − x 1 ) + y 1 y = ( 5 0 − 1 5 . 5 ) ( 8 5 3 . 7 − 8 5 2 . 3 ) ( 2 5 − 1 5 . 5 ) + 8 5 2 . 3 = 8 5 2 . 7 k g k J ( H i n ) = ( H o u t ) = 8 5 2 . 7 k g k J
Water at 1 bar and 852.7 k J k g → 852.7 \frac{\mathrm{kJ}}{\mathrm{kg}} \rightarrow 8 5 2 . 7 k g k J → Mixture of saturated liquid and saturated vapor
1 2 Percent Vapor (q) 0 ? ? ? 100 Enthalpy (x) 417.5 852.7 2674.9 \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} Percent Vapor (q) Enthalpy (x) 1 0 4 1 7 . 5 ? ? ? 8 5 2 . 7 2 1 0 0 2 6 7 4 . 9
q = ( q 2 − q 1 ) ( x − x 1 ) ( x 2 − x 1 ) + q 1 y = ( 100 − 0 ) ( 852.7 − 417.5 ) ( 2674.9 − 417.5 ) + 0 = 19.28 % v a p o r \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} q = ( x 2 − x 1 ) ( q 2 − q 1 ) ( x − x 1 ) + q 1 y = ( 2 6 7 4 . 9 − 4 1 7 . 5 ) ( 1 0 0 − 0 ) ( 8 5 2 . 7 − 4 1 7 . 5 ) + 0 = 1 9 . 2 8 % v a p o r
B) Set up an energy balance around the Flash chamber
d d t { M ( U ^ + v 2 2 + g h ) } = m ˙ i n ( H ^ i n + v in 2 2 + g h i n ) − m ˙ o u t ( H ^ o u t + v o u t 2 2 + g h o u t ) + W ˙ S + W ˙ E C + Q ˙ \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} d t d { M ( U + 2 v 2 + g h ) } = m ˙ i n ( H i n + 2 v in 2 + g h i n ) − m ˙ o u t ( H o u t + 2 v o u t 2 + g h o u t ) + W ˙ S + W ˙ E C + Q ˙
Cancelling terms
0 = m ˙ i n ( H ^ i n ) − m ˙ o u t ( H ^ out ) + Q ˙ m ˙ o u t ( H ^ o u t ) − m ˙ i n ( H ^ i n ) = Q ˙ \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} 0 = m ˙ i n ( H i n ) − m ˙ o u t ( H out ) + Q ˙ m ˙ o u t ( H o u t ) − m ˙ i n ( H i n ) = Q ˙
Find H ^ out \widehat{\mathrm{H}}_{\text {out }} H out
Water/steam at q=0.5 and P=1 bar →
1 2 Enthalpy (y) 417.5 ? ? ? 2674.9 Vapor Fraction (q) 0 50 100 \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} Enthalpy (y) Vapor Fraction (q) 1 4 1 7 . 5 0 ? ? ? 5 0 2 2 6 7 4 . 9 1 0 0
y = ( y 2 − y 1 ) ( q − q 1 ) ( q 2 − q 1 ) + y 1 y = ( 2674.9 − 417.5 ) ( 50 − 0 ) ( 100 − 0 ) + 417.5 = 1546.2 k J k g 5 k g s e c ( 1546.2 k J k g ) − 5 k g s e c ( 852.7 k J k g ) = Q ˙ Q ˙ = 3467.5 k J s e c \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} y = ( q 2 − q 1 ) ( y 2 − y 1 ) ( q − q 1 ) + y 1 y = ( 1 0 0 − 0 ) ( 2 6 7 4 . 9 − 4 1 7 . 5 ) ( 5 0 − 0 ) + 4 1 7 . 5 = 1 5 4 6 . 2 k g k J 5 s e c k g ( 1 5 4 6 . 2 k g k J ) − 5 s e c k g ( 8 5 2 . 7 k g k J ) = Q ˙ Q ˙ = 3 4 6 7 . 5 s e c k J