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

`\dot m_(cond) = 3.523\,(kg)/(s)`

Step-by-step explanation:

The heat rate received by the cooling water is:

`\dot Q_(cooling) = (200\,(kg)/(s) )\cdot (4.196\,(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\,(kJ)/(kg), h_(g) = 2591.3\,(kJ)/(kg)`

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

`\dot m_(cond) = (-8392\,kW)/(209.34\,(kJ)/(kg) - 2591.3\,(kJ)/(kg) )`

`\dot m_(cond) = 3.523\,(kg)/(s)`