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?

This question was previously asked in

SSC JE ME Previous Paper 8 (Held on: 27 October 2020 Evening)

Option 1 :

2 kg/s

ST 1: Logical Reasoning

1293

20 Questions
20 Marks
20 Mins

__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

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