A steam power station has an overall efficiency of 25%, and 0.5 kg of coal is burnt per kw of electrical energy generated. determine the calorific value of fuel. (take heat equivalent of 1kw as 860 kcal)

To determine the calorific value of the fuel, we can use the following formula:

$$\text{Calorific Value} (\text{CV}) = \frac{\text{Heat Produced by Coal}}{\text{Mass of Coal Burnt}}$$

Where:

• The overall efficiency of the power station is 25%.
• 1 kW of electrical energy is generated by burning 0.5 kg of coal.
• Heat equivalent of 1 kW = 860 kcal.

Now, the total energy input required to generate 1 kW of electrical energy is the energy output divided by the efficiency:

$$\text{Total Energy Input} = \frac{\text{Energy Output}}{\text{Efficiency}}$$

Given that:

• Energy Output = 860 kcal (heat equivalent for 1 kW of electrical energy)
• Efficiency = 25% = 0.25

So, the total energy input is:

$$\text{Total Energy Input} = \frac{860}{0.25} = 3440 \, \text{kcal}$$

This total energy input is the heat produced by burning 0.5 kg of coal. Now, we can find the calorific value of the fuel:

$$\text{Calorific Value} = \frac{3440}{0.5} = 6880 \, \text{kcal/kg}$$

Thus, the calorific value of the coal is 6880 kcal/kg.