Question 6.80: A stream of refrigerant-134a at 1 MPa and 12°C is mixed with......
A stream of refrigerant-134a at 1 \mathrm{MPa} and 12^{\circ} \mathrm{C} is mixed with another stream at 1 \mathrm{MPa} and 60^{\circ} \mathrm{C}. If the mass flow rate of the cold stream is twice that of the hot one, determine the temperature and the quality of the exit stream.
Two streams of refrigerant-134a are mixed in a chamber. If the cold stream enters at twice the rate of the hot stream, the temperature and quality (if saturated) of the exit stream are 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. \mathrm{3} There are no work interactions. \mathrm{4} The device is adiabatic and thus heat transfer is negligible.
Properties From R-134a tables (Tables A-11 through A-13),
\begin{aligned} & h_{1} \cong h_{f @ 12^{\circ} \mathrm{C}}=68.18 \mathrm{~kJ} / \mathrm{kg} \\ & h_{2}=h_{@ 1 \mathrm{MPa}, 60^{\circ} \mathrm{C}}=293.38 \mathrm{~kJ} / \mathrm{kg} \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}=3 \dot{m}_{2} \text { since } \dot{m}_{1}=2 \dot{m}_{2} \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 \quad 2 \dot{m}_{2} h_{1}+\dot{m}_{2} h_{2}=3 \dot{m}_{2} h_{3} or h_{3}=\left(2 h_{1}+h_{2}\right) / 3
Substituting,
h_{3}=(2 \times 68.18+293.38) / 3=143.25 \mathrm{~kJ} / \mathrm{kg}
At 1 \mathrm{MPa}, h_{f}=107.32 \mathrm{~kJ} / \mathrm{kg} and h_{g}=270.99 \mathrm{~kJ} / \mathrm{kg}. Thus the exit stream is a saturated mixture since h_{f}<h_{3}<h_{g}. Therefore,
T_{3}=T_{\text {sat } @ 1 \mathrm{MPa}}=39.37^{\circ} \mathrm{C}
and
x_{3}=\frac{h_{3}-h_{f}}{h_{f g}}=\frac{143.25-107.32}{163.67}=\mathrm{0 . 2 2 0}
