Question

A steam power Plant has coal Consumption of 165 Tons Per Hour. Calorific Value of Coal is

3500 kcal / kg. If the power generation is 250 MW, Find overall efficiency of the plant. Use

relation 1 kcal = 4.18 kJ.​

Answer 1

Given:-

consumption of coal, $\dot{m}=165ton/hr$

Calorific value of fuel, $CV_{f}=3500kcal/kg$

Output power, P=250 MW

To Find:-

overall efficiency of plant

Explanation:-

$\text{Since, 1kcal=4.18kJ}\Rightarrow CV_{f}=3500\times4.18=14630kJ/kg\\\\\dot{m}=165ton/hr=\dfrac{165\times1000}{3600}=45.833kg/s\\\\\Rightarrow\text{Amount of heat added in the plant, }Q=\dot{m}\times CV_{f}=\dfrac{45.833\times14630}{1000}=670.536MW\\\\\text{Now, }\\\\\text{Overall efficiency of the plant, }\eta=\dfrac{P}{Q}=\dfrac{250}{670.536}=0.3728$