Question

A steam power plant has coal consumption of 16200 Kg/hr with calorific value of coal as 17793.9 kJ/kg. If the speed of steam turbine is 1000 rpm and generated torque is 477464.8293Nm.​

Answer 1

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Answer 2

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A Steam Power Plant has Coal Consumption of 16200 Kg/hr with Calorific Value of Coal as 17793.9 kJ/kg. If the speed of steam turbine is 1000 rpm and generated torque is 477464.8293 Nm. Find the input Power

Concept:

• Power and Energy
• Efficiency

Given:

• Coal consumption = 16200 Kg/hr
• Calorific value of coal = 17793.9 kJ/kg
• Angular velocity of steam turbine = 1000rpm =
• Generated Torque = 477464.8293 Nm

Find:

• Input Power
• Output Power
• Efficiency

Solution:

Input power = coal consumption x calorific value

Input power = 16200kg/h x 17793.9kJ/kg = 2.88*10⁸kJ/h = 80072kJ/s

The input power is 80072000W.

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