A typical electrical generating plant has a capacity of 500 megawatt (MW; 1 MW =

Question

A typical electrical generating plant has a capacity of 500 megawatt (MW; 1 MW = 106 J s–1 ) and an overall efficiency of about 25%.

(a) The combustion of 1 kg of bituminous coal releases about 3.2 × 104 kJ and leaves an ash residue of 100 g. What weight of coal must be used to operate a 500-MW generating plant for 1 year, and what weight of ash must be disposed of?

(b) Enriched fuel for nuclear reactors contains about 4% 235U, fission of which gives 1.9 × 1010 kJ per mole 235U. What weight of 235U is needed to operate a 500-MW power plant, assumed to have 25% efficiency, for 1 year, and what weight of fuel must be reprocessed to remove radioactive wastes?

(c) The radiation from the sun striking the earth’s surface on a sunny day corresponds to a power of 1.5 kW m–2 . How large must the collection surface be for a 500-MW solar-generating plant? (Assume that there are 6 hours of bright sun each day and that storage facilities continue to produce power at other times. The efficiency for solar-power generation would be about 25%.)

Explanation / Answer

The capacity of A electrical generating plant has a capacity of 500 megawatt

1 MW = 106 Js-1   Therefore 500 MW = 500x106 J s-1

However, though the capacity to produce energy is 500 MW, the efficiency is 25%. Which means

500x106x25/100 = 125x106 J s-1 is the actual capacity.

Also for one year it can produce: 125x106x60x60x24x365 J = 3.95x1015J /year

Now, 1 kg of bituminous coal releases about 3.2 × 104 kJ = 3.2x107 J of energy

in other wards, 3.2x107 J of energy requires 1kg of coal to be burnt

Hence,3.95x1015 J requires 3.95x1015/3.2x107 = 1.23x108 kg of coal. This indicates that the amount of coal required to generate  3.95x1015J of energy is quite huge. so also the ash produced.

Let us see how much of ash is produced

It is given that 1 kg of coal produces 100 g of ash

Therefore,  1.23x108 kg of coal produces 1.23x108 kg x 0.100 kg/1kg = 1.23x106 kg of ash

1.23x106 kg of ash must be disposed, This is quite a lot of ash

b.   235U,on fission gives 1.9 × 1010 kJ per mole

Enriched fuel for nuclear reactors contains about 4% 235U. This means 100 g of enriched nuclear fuel contains 4 g of U235

Molar mass of U = 235 g

Therefore 235 g of U gives 1.9 × 1010 kJ = 1.9 × 1013 J of energy on fission

The actual capacity of power generating station is 125x106 J s-1 = 3.95x1015J / year

1.9 × 1013 J of energy requires fission of 235 g of U

Therefore, 3.95x1015J of energy requires 3.95x1015x235/ 1.9 × 1013 = 488.55x102 g of U235

it is given that, 100 g of enriched nuclear fuel contains 4 g of U235

In other words, 4g of U235 is present in 100 g of enriched nuclear fuel

488.55x102 g of U235 is present in 488.55x102 g x 100g/4g = 122.13x104 g = 1.22x103 kg

c. The radiation from the sun striking the earth’s surface on a sunny day corresponds to a power of 1.5 kW m–2

We calculated the actual capacity of power station = 125x106 J s-1 = 125x106 J x60x60 = 4.5x1011J /hour

Solar radiation corresponds to apower of 1.5 kW m–2 = 1.5x103 J s-1

Therefor , for 1 hour  1.5x103 J x60x60 = 5.4x106 J /hour

Hence per day(6 hours of bright sun each day) 6x 5.4x106= 32.4x106 J

Given, 1.5x103 J corresponds to 1 m2 area

Therefore, 4.5x1011J of energy corresponds to  4.5x1011/1.5x103 J = 3.0 x108 m2

Hence the collection surface for a 500-MW solar-generating plant should be 3.0 x108 m2

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