Question 6.82: Refrigerant-134a at 700 kPa, 70°C, and 8 kg/min is cooled by......
Refrigerant-134a at 700 \mathrm{kPa}, 70^{\circ} \mathrm{C}, and 8 \mathrm{~kg} / \mathrm{min} is cooled by water in a condenser until it exists as a saturated liquid at the same pressure. The cooling water enters the condenser at 300 \mathrm{kPa} and 15^{\circ} \mathrm{C} and leaves at 25^{\circ} \mathrm{C} at the same pressure. Determine the mass flow rate of the cooling water required to cool the refrigerant.
Refrigerant-134a is condensed in a water-cooled condenser. The mass flow rate of the cooling water required 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. \mathrm{3} There are no work interactions. \mathrm{4} Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid.
Properties The enthalpies of R-134a at the inlet and the exit states are (Tables A-11 through A-13)
\begin{aligned} & \left.\begin{array}{l} P_{3}=700 \mathrm{kPa} \\ T_{3}=70^{\circ} \mathrm{C} \end{array}\right\} h_{3}=308.33 \mathrm{~kJ} / \mathrm{kg} \\ & \left.\begin{array}{l} P_{4}=700 \mathrm{kPa} \\ \text { sat. liquid } \end{array}\right\} h_{4}=h_{f @ 700 \mathrm{kPa}}=88.82 \mathrm{~kJ} / \mathrm{kg} \end{aligned}
Water exists as compressed liquid at both states, and thus (Table A-4)
\begin{aligned} & h_{1} \cong h_{f @ 15^{\circ} \mathrm{C}}=62.98 \mathrm{~kJ} / \mathrm{kg} \\ & h_{2} \cong h_{f @ 25^{\circ} \mathrm{C}}=104.83 \mathrm{~kJ} / \mathrm{kg} \end{aligned}
Analysis We take the heat exchanger as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as
Mass balance (for each fluid stream):
\dot{m}_{\mathrm{in}}-\dot{m}_{\mathrm{out}}=\Delta \dot{m}_{\mathrm{system}}{ }^{\nearrow 0 \text { (steady) }}=0 \rightarrow \dot{m}_{\mathrm{in}}=\dot{m}_{\mathrm{out}} \rightarrow \dot{m}_{1}=\dot{m}_{2}=\dot{m}_{w} \text { and } \dot{m}_{3}=\dot{m}_{4}=\dot{m}_{R}
Energy balance (for the heat exchanger):
\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 }} {}^{\nearrow 0 \text { (steady) }}}_{\begin{array}{c} \text { Rate of change in internal, kinetic, } \\ \text { potential, etc. energies } \end{array}}=0
\begin{aligned} \dot{E}_{\text {in }} & =\dot{E}_{\text {out }} \\ \dot{m}_1 h_1+\dot{m}_3 h_3 & =\dot{m}_2 h_2+\dot{m}_4 h_4 \quad(\text { since } \dot{Q}=\dot{W}=\Delta \mathrm{ke} \cong \Delta \mathrm{pe} \cong 0) \end{aligned}
Combining the two, \quad \dot{m}_{w}\left(h_{2}-h_{1}\right)=\dot{m}_{R}\left(h_{3}-h_{4}\right)
Solving for \dot{m}_{w}: \quad \quad \dot{m}_{w}=\frac{h_{3}-h_{4}}{h_{2}-h_{1}} \dot{m}_{R}
Substituting,
\dot{m}_{w}=\frac{(308.33-88.82) \mathrm{kJ} / \mathrm{kg}}{(104.83-62.98) \mathrm{kJ} / \mathrm{kg}}(8 \mathrm{~kg} / \mathrm{min})=\mathrm{4 2 . 0} \mathrm{k g} / \mathrm{min}
