Question

. Steam is to be condensed from saturated vapor to saturated liquid in the condenser of a steam power plant at a temperature of 50°C with cooling water from a nearby lake, which enters the tubes of the condenser at 15°C at a rate of 200 kg/s and leaves at 25°C. Determine the rate of condensation of the steam in the condenser. Assume the condenser is well insulated.

Answer 1

Answer:

$\dot m_{cond} = 3.523\,\frac{kg}{s}$

Explanation:

The heat rate received by the cooling water is:

$\dot Q_{cooling} = (200\,\frac{kg}{s} )\cdot (4.196\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (25^{\textdegree}C-15^{\textdegree}C)$

$\dot Q_{cooling} = 8392\,kW$

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:

$h_{f} = 209.34\,\frac{kJ}{kg}, h_{g} = 2591.3\,\frac{kJ}{kg}$

The rate of condensation of the steam in the condenser is:

$\dot m_{cond} = \frac{-8392\,kW}{209.34\,\frac{kJ}{kg} - 2591.3\,\frac{kJ}{kg} }$

$\dot m_{cond} = 3.523\,\frac{kg}{s}$