Power Generation MCQs Part 15

Power Generations MCQs Part 15

To supply base load of power plant, which of the following powerplant is very suitable

To supply base load of power plant, which of the following powerplant is very suitable:

1. Diesel
2. Nuclear
3. Gas
4. None of these

Explanation: Base load power plants deliver full power at all times. Coal fired power stations and nuclear power stations are particularly suitable for meeting the base load demands.

A diesel power station has fuel consumption of 0.3 per kWh. The calorific value of fuel being 9898 kcal/kg. The overall efficiency of station is

A diesel power station has fuel consumption of 0.3 per kWh. The calorific value of fuel being 9898 kcal/kg. The overall efficiency of station is:

1. 28.96%
2. 33.96%
3. 38.96%
4. 43.96%

Solution:

Heat produced by 0.3 kg of oil = 9898 * 0.3 = 2969.4 kcal

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Heat equivalent of 1 kWh = 860 kcal

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Overall efficiency = Electrical output/Heat of combustion = 860/2969.4 = 0.2896 = 28.96%

When moving from consumers to power station, the diversity factor of a power plant:

When moving from consumers to power station, the diversity factor of a power plant:

1. Increases
2. Remains the same
3. Decreases
4. None of these

Mechanical energy is supplied to a DC generator at rate of 90 J/s. The generator delivers 40 at 90 V. The efficiency of generator is:

Mechanical energy is supplied to a DC generator at rate of 90 J/s. The generator delivers 40 at 90 V. The efficiency of generator is:

1. 33784
2. 32784
3. 31784
4. 30784

Solution:

Mass of water available = 90 * 1000 = 90000 kg/sec

————

Work done/sec = W*H = 90000 * 40 = 35316 * 10³ W = 35316 kW

This is gross plant capacity.

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Firm capacity = Plant efficiency * Gross plant capacity

Firm capacity = 0.9 * 35316 = 31784 kW

A diesel power station has fuel consumption of 0.3 per kWh. The calorific value of fuel being 9800 kcal/kg. If alternator efficiency is 96%. The efficiency of engine is

A diesel power station has fuel consumption of 0.3 per kWh. The calorific value of fuel being 9800 kcal/kg. If alternator efficiency is 96%. The efficiency of engine is:

1. 23.47%
2. 30.47%
3. 37.47%
4. 44.47%

Solution:

Heat produced by 0.3 kg of oil = 9800 * 0.3 = 2940 kcal

—————

Heat equivalent of 1 kWh = 860 kcal

—————

Overall efficiency = Electrical output/Heat of combustion = 860/2940 = 0.292517006802721 = 29.25%

—————

Engine efficiency = Overall efficiency/Alternator efficiency = 29.25/0.96 = 30.47%