Question

The power developed by a turbine in a certain steam power plant is 1206 kW. The heat supplied to boiler is 3500 kJ/kg. The heat rejected by steam to cooling water is 2900 kJ/kg. the feed pump work required to condensate back into the boiler is 6 kW. What will be mass flow rate of cycle? 

1. 2 kg/s
2. 0.002 kg/s
3. 6.22 kg/s
4. 0.00622 kg/s

Answer 1

Correct Answer - Option 1 : 2 kg/s

Concept:

In a steam power plant, The energy balance is given as

Heat supplied in boiler - Heat rejected in condenser = Turbine work - Pump work

Calculation:

Given 

Heat supplied in boiler = QS = 3500 kJ/kg

Heat rejected in condenser = QR = 2900 kJ/kg

Turbine work = 1206 kW

Pump work = 6 kW

Net work output = 3500 - 2900 = 600 kJ/kg

Now W  = 600 × m

Here W = 1206 - 6 = 1200 kJ/s

\(\Rightarrow m = \frac {1200}{600} = 2\ kg/s \)