# Power Generation MCQs Part 15

Power Generations MCQs Part 15

Contents

## 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%