SSC JE Mechanical Engineering 27 Oct 2020 Shift 2 Papers

Question : 101 of 200

Marks: +1, -0

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?

2 kg/s
0.002 kg/s
6.22 kg/s
0.00622 kg/s

Validate

Solution:

Given
Heat supplied in boiler

= Q_S = 3500

kJ/kgHeat rejected in condenser

= Q_R = 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

⇒m=1200/600=2

kg/s

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