A typical coal-fired power plant burns 250 metric tons of coal every hour to gen

Question

A typical coal-fired power plant burns 250 metric tons of coal every hour to generate 2.9×106 MJ of electricity. 1 metric ton= 1000 kg; 1 metric ton of coal has a volume of 1.5 m3.The heat of combustion is 28 MJ/kg. Assume that all heat is transferred from the fuel to the boiler and that all the work done in spinning the turbine is transformed into electrical energy.

A.

Suppose the coal is piled up in a 13 m × 14 m room. How tall must the pile be to operate the plant for one day?

B.

What is the power plant's thermal efficiency (expressed as a %)?

Explanation / Answer

The plant burns 250 tons per hour to produce 2.9 *106 MJ (each hour)
- 1 ton = 1000 kg
- The Volume of 1 ton is 1.5 m^3
- Heat of combustion (the energy you get from burning per kg) 28 MJ/kg

(a) The question is only asking how much space would the coal required for one day of plant operation occupy?

First, note that the plant uses 250 tons per hour * 24 hours per day coal each day.

The volume of 1 ton of coal is 1.5 m^3.

So, the volume of coal used in a day is 1.5 * 24*250 = 9000 m^3.

If we put this into a pile of dimension 13 x 14 x h meters,

we have that 13*14*h = 9000 m^3, so h = 49.45 m.

(b) Efficiency, briefly is energy output/energy input.

For this plant, the question is how much of the available chemical energy of the coal is converted to electricity?

We know that the plant outputs 2.9*1066 MJ of electricity each hour, so how much chemical energy is input?

Note that the plant uses 250 tons of coal per hour. There are 1000 kg in a ton. Each kg, when burned, gives 28 MJ of energy.

All we do now is multiply:

250 tons * 1000 kg/ton * 28 MJ/kg = 7,000,000 MJ = 7*106 MJ.

So input = 7*106 MJ., output = 2.9 *106 MJ.

The efficiency is a fraction equal to the output/input, which is:

2.9 *106 /7*106 = 27/84 = 0.41

So the plant is 41% efficient.

That is, for each 1 part of input energy, only 0.41 part of that is converted to electrical energy. T

he rest - 59% is lost as heat (mostly).

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