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A steam power station has an overall efficiency of 25%, and 0.5 kg of coal is burnt per kWh of electrical energy generated. Determine the calorific value of fuel.

(Take heat equivalent of 1 kWh as 860 kcal)


1. 1720 kcal/kg
2. 6880 kcal/kg
3. 3400 kcal/kg
4. 4650 kcal/kg

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Correct Answer - Option 2 : 6880 kcal/kg

Concept:

Calorific values of fuel: The calorific value of fuel is the quantity of heat produced by its combustion at constant pressure and under normal conditions. Calorific value mention in kcal / kg.

The overall efficiency of the steam power station: The overall efficiency of the steam power station is defined as the ratio of the power available at the generator terminal to the rate of energy released by the combustion of fuel. It is given by,

\({\eta _{overall}} = \frac{{EO}}{{HC}}\)

Where EO is the Electrical output in the heat unit.

HC is Heat combustion.

Calculation: 

Given: Overall efficiency = 25 % , 0.5 kg coal burnt.

Let x cal /kg be the calorific value of the fuel.

Heat equivalent of 1 kWh = 860 kcal

\({\eta _{overall}} = \frac{{EO}}{{HC}}\)

\(0.25 = \frac{{860}}{{0.5x}}\)

\(x = \frac{{860000}}{{25 \times 5}}\)

\(x = 6880\;kcal/kg\)

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