Answer:
[tex]\dot m_{cond} = 3.523\,\frac{kg}{s}[/tex]
Explanation:
The heat rate received by the cooling water is:
[tex]\dot Q_{cooling} = (200\,\frac{kg}{s} )\cdot (4.196\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (25^{\textdegree}C-15^{\textdegree}C)[/tex]
[tex]\dot Q_{cooling} = 8392\,kW[/tex]
The saturation pressure is 12.532\,kPa at 50 °C. The specific enthalpies for saturated liquid and vapor at given pressure and temperature are, respectively:
[tex]h_{f} = 209.34\,\frac{kJ}{kg}, h_{g} = 2591.3\,\frac{kJ}{kg}[/tex]
The rate of condensation of the steam in the condenser is:
[tex]\dot m_{cond} = \frac{-8392\,kW}{209.34\,\frac{kJ}{kg} - 2591.3\,\frac{kJ}{kg} }[/tex]
[tex]\dot m_{cond} = 3.523\,\frac{kg}{s}[/tex]
