Question 6.84: Steam enters the condenser of a steam power plant at 20 kPa ......
Steam enters the condenser of a steam power plant at 20 \mathrm{kPa} and a quality of 95 percent with a mass flow rate of 20,000 \mathrm{~kg} / \mathrm{h}. It is to be cooled by water from a nearby river by circulating the water through the tubes within the condenser. To prevent thermal pollution, the river water is not allowed to experience a temperature rise above 10^{\circ} \mathrm{C}. If the steam is to leave the condenser as saturated liquid at 20 \mathrm{kPa}, determine the mass flow rate of the cooling water required.
Steam is condensed by cooling water in the condenser of a power plant. If the temperature rise of the cooling water is not to exceed 10^{\circ} \mathrm{C}, the minimum 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. 5 Liquid water is an incompressible substance with constant specific heats at room temperature.
Properties The cooling water exists as compressed liquid at both states, and its specific heat at room temperature is c=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C} (Table A-3). The enthalpies of the steam at the inlet and the exit states are (Tables A-5 and A-6)
\begin{aligned} & \begin{array}{l} P_{3}=20 \mathrm{kPa} \\ x_{3}=0.95 \end{array} \\ & \left.\begin{array}{l} P_{4}=20 \mathrm{kPa} \\ \text { sat. liquid } \end{array}\right\} h_{3}=h_{f}+x_{3} h_{f g}=251.42+0.95 \times 2357.5=2491.1 \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 steady-flow system can be expressed in the rate form as
Mass balance (for each fluid stream):
\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}_{w} \text { and } \dot{m}_{3}=\dot{m}_{4}=\dot{m}_{s} \end{aligned}
Energy balance (for the heat exchanger):
\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 }} {}^{\nearrow 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}_{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,
\dot{m}_{w}\left(h_{2}-h_{1}\right)=\dot{m}_{s}\left(h_{3}-h_{4}\right)
Solving for \dot{m}_{w} :
\dot{m}_{w}=\frac{h_{3}-h_{4}}{h_{2}-h_{1}} \dot{m}_{s} \cong \frac{h_{3}-h_{4}}{c_{p}\left(T_{2}-T_{1}\right)} \dot{m}_{s}
Substituting,
\dot{m}_{w}=\frac{(2491.1-251.42) \mathrm{kJ} / \mathrm{kg}}{\left(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(10^{\circ} \mathrm{C}\right)}(20,000 / 3600 \mathrm{~kg} / \mathrm{s})=297.7 \mathrm{~kg} / \mathrm{s}
