Question 6.E.79: Water at 50°F and 50 psia is heated in a chamber by mixing i......
Water at 50^{\circ} \mathrm{F} and 50 psia is heated in a chamber by mixing it with saturated water vapor at 50 psia. If both streams enter the mixing chamber at the same mass flow rate, determine the temperature and the quality of the exiting stream.
Liquid water is heated in a chamber by mixing it with saturated water vapor. If both streams enter at the same rate, the temperature and quality (if saturated) of the exit stream is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible.
Properties From steam tables (Tables A-5E through A-6E),
\begin{aligned} & h_{1} \cong h_{f @ 50^{\circ} \mathrm{F}}=18.07 \mathrm{Btu} / \mathrm{lbm} \\ & h_{2}=h_{g @ 50 \mathrm{psia}}=1174.2 \mathrm{Btu} / \mathrm{lbm} \end{aligned}
Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as Mass balance:
\begin{aligned} \dot{m}_{\mathrm{in}}-\dot{m}_{\text {out }} & =\Delta \dot{m}_{\text {system }}{ }^{\nearrow 0 \text { (steady) }}=0 \\ \dot{m}_{\text {in }} & =\dot{m}_{\text {out }} \\ \dot{m}_{1}+\dot{m}_{2} & =\dot{m}_{3}=2 \dot{m} \\ \dot{m}_{1} & =\dot{m}_{2}=\dot{m} \end{aligned}
Energy balance:
\begin{aligned} \underbrace{\dot{E}_{\text {in }}-\dot{E}_{\text {out }}}_{\begin{array}{c} \text { Rate of net energy transfer } \\ \text { by heat, work, and mass } \end{array}} & =\underbrace{\Delta \dot{E}_{\text {system }} {}^{\swarrow 0 \text { (steady) }}}_{\begin{array}{c} \text { Rate of change in internal, kinetic, } \\ \text { potential, etc. energies } \end{array}}=0 \\ \dot{E}_{\text {in }} & =\dot{E}_{\text {out }} \\ \dot{m}_{1} h_{1}+\dot{m}_{2} h_{2} & =\dot{m}_{3} h_{3} \quad(\text { since } \dot{Q}=\dot{W}=\Delta \mathrm{ke} \cong \Delta \mathrm{pe} \cong 0) \end{aligned}
Combining the two gives
\dot{m} h_{1}+\dot{m} h_{2}=2 \dot{m} h_{3} \text { or } h_{3}=\left(h_{1}+h_{2}\right) / 2
Substituting,
h_{3}=(18.07+1174.2) / 2=596.16 \mathrm{Btu} / \mathrm{lbm}
At 50 \mathrm{psia}, h_{f}=250.21 \mathrm{Btu} / \mathrm{lbm} and h_{g}=1174.2 \mathrm{Btu} / \mathrm{lbm}. Thus the exit stream is a saturated mixture since h_{f}<h_{3}<h_{g}. Therefore,
T_{3}=T_{\text {sat } @ 50 \text { psia }}=\mathrm{2 8 0 . 9 9}^{\circ} \mathrm{F}
and
x_{3}=\frac{h_{3}-h_{f}}{h_{f g}}=\frac{596.16-250.21}{924.03}=\mathrm{0 . 3 7 4}
