# Objective Questions on Electrical Power Generation

1. In thermal power plants the size of coal after crushing

Any material is ground into small parts its surface area is increases make it to effectively utilize.

2. Efficiency of thermal power plant can be taken as

Efficiency of thermal power plant is given by product of
a. Boiler efficiency ( about 0.75 to 0.9)
b. Efficiency of thermal cycle ( about 0.35 to 0.5)
c. Internal efficiency of turbine (0.85 to 0.94)
d. Mechanical efficiency of turbine ( about 0.99 to 0.995)
e. Efficiency of generator ( about 0.98 to 0.985).

3. Where do compounding done in thermal power plants?

There will be several problems if steam energy is converted in one step i.e., if in a single row of nozzle blade combination all the heat is dropped the steam velocity becomes high and even supersonic. Velocity of steam is proportional to square root of heat dropped in nozzle. This makes rotor speed to a value of 30,000 rpm which is too high for practical use. So it is reduced to 3000 rpm by compounding i.e., connecting turbines in 3 stages
a) High pressure.
b) Intermediate pressure.
c) Low pressure.

4. The only place where slip ring induction motors used in thermal power plants is

In coal handling plants the slip ring IM motors are used for tippling purpose i.e., to unload the coal. Because, to lift the wagon of weight of metric tons, the starting torque should be very high which can be maintained by slip ring IM.

5. The amount of ash content in Indian coal

By standards the coal is graded by following statistics

Best Worst Average
Moisture contained 17.5% 22% 20.5%
Ash 23% 35% 30%

6. To improve the efficiency of thermal power plant ________________ is placed between boiler and turbine.

Super heater is placed between boiler and turbine. A super heater is a device which raises the temperature of the steam much above the boiling point of water. In super heater, the heat is taken from flue gases and here radiant type of heat transfer method is used. It improves efficiency in thermal power plant.

7. Which of the following is not base load power plant?

Base load power plant : Low operating cost, capacity of working continuously. Pump storage plant can only supply power at peak loads not always.

8. In which power plant, the thermal efficiency is quite low?

Because a steam power station, more than 50 % of total heat of combustion is lost as heat rejected to the condenser and the loss is unavoidable as heat energy cant be convert into mechanical energy without a drop in temparature and the steam in condenser at the lower temperature.

9. Which cooling is preferred for large turbine generator ?

Hydrogen cooling is preferred because
1. It reduces windage losses to about one teeth of its value in air.
2. It provides better cooling.
3. It reduces the oxidation of the insulation and fire hazards.
4. It reduces the windage noise.

10. A steam power generation has an overall efficiency of 20 %. 0.6 Kg of coal is burnt per kWh of electrical energy generated. Calculated the calorific value of fuel?

Let x cal/Kg be the calorific value of fuel. Heat produce by 0.6 kg of coal=0.6 X Kcal. Heat equivalent of 1 KWh = 860 KCal, ηoverall = ( Electrical output in heat units ) / ( Heat of combustion ) ⇒ 0.2 = 860 / 0.6x ⇒ x = 860 / ( 0.6 × 0.2 ) = 7166.67 KCal / Kg.

11. What will be the thermal efficiency of a 240 V, 1000 W electric kettle if it brings 2 litres of water at 15°C to boiling point 15 Minutes?

Usefull heat = MST = 2 × 1 × ( 100 - 15 ) = 170 KCal, Heat produced electrically =( VIT ) / ( 4.2 × 10-3 ) KCal = 214.2857 KCal, η ( thermal efficiency ) = ( Useful heat / total heat ) = 170 / 214.2857 = 0.7933 = 79.33 %.

12. Mechanical energy is supplied to a DC generator at the rate of 4200 J/S .The generator delivers 32.2 Amp at 120 V. How much energy is cost per minute of operation?

Power lost(PL) = Pi - Po = ( 4200 - 3864 ) = 336 W, Energy lost per minute = ( 60 sec ) of operation = PLt = ( 336 × 60 ) = ( 336 × 60 ) = 20160 J.

13. A dissel power station has fuel consumption of 0.28 Kg per kWh. The calorific value of fuel is being 10,000 KCal / kg. Determine the overall efficiency?

Heat produce by 0.28 Kg of oil=10,000 × 0.28=2800 KCal, Heat produce equivalent of 1 KWh = 860 KCal, Overall efficiency = Electrical output in heats units / Heat of combustion = 860 / 2800 = 0.307 = 30.7 %.

14. Example of base load stations are

Base load stations installed of the capacity of unvarying load known as base load which is normally required for the whole day.

15. The time taken by the turbine to fall to 0 rpm speed from 3000 rpm is called

The time taken by the turbine to fall to 0 rpm speed from 3000 rpm is called coasting time (technically called) and it is generally 40 min.

16. Load frequency control is achieved by properly matching the individual machines

The load frequency control of 2 machines can be achieved by matching both turbine inputs.

17. Uses of diesel power stations

Diesel power station is an ex of peak load station which is installed to share the peak load of the base load station during peak load hours.

18. Calorific value of solid fuel is expressed as

Valorific value : Amount of heat produced by complete combustion of a unit of wt. of fuel It is expressed as Cal / gm or KCal / kg.

19. Percentage of carbon is highest in

Percentage of carbon C is highest in anthracite coal. C = 90 %, H = 3 %, O = 2 %, ash = 5 %.

20. Cheapest plant in operation and maintenance is

Hydroelectricity eliminates the flue gas emissions from fossil fuel combustion, including pollutants such as sulfur dioxide, nitric oxide, carbon monoxide, dust, and mercury in the coal. Hydroelectricity also avoids the hazards of coal mining and the indirect health effects of coal emissions. Compared to nuclear power, hydroelectricity generates no nuclear waste, has none of the dangers associated with uranium mining, nor nuclear leaks. Unlike uranium, hydroelectricity is also a renewable energy source. Compared to wind farms, hydroelectricity power plants have a more predictable load factor. If the project has a storage reservoir, it can generate power when needed. Hydroelectric plants can be easily regulated to follow variations in power demand.

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