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

1 MW = 106 J s–1

500-MW = 500 x 106 J s–1

1 kg of bituminous coal releases about 3.2 × 104 kJ

1 kg of bituminous coal releases about 3.2 × 107 J

To get 500 x 106 J we will need 500 x 106 J /3.2 × 107 J x 1 Kg = 15.62 Kg of bituminous coal /s

so in 1 year we will need 15.62 Kg x 60 s/m x 60 m/h x 24h/d x 365d/y = 4.92 x 108 Kg of bituminous coal /year

This is for 100% efficiency for 25% efficiency we will need 4 times this amount of coal

4.92 x 108 Kg of bituminous coal /year x 4 = 19.71 x 108 Kg of bituminous coal /year

If 1 Kg of coal generates 100 g of ash then 19.71 x 108 Kg of bituminous coal will genarate

19.71 x 108 Kg x 100 g = 19.71 x 107 Kg Kg of Ash

b) 1.9 × 1010 kJ per mole 235U

1 mole of uranium contains 4% 235U

1 mole uranium which is 238 g of U generates 1.9 × 1010 kJ x 0.04 = 7.6 x 1011 J

500-MW = 500 x 106 J s–1

500 MW /y is 500 x 106 J s–1 x 60 s/m x 60 m/h x 24h/d x 365d/y = 15.77 x 1015 J/y

So we will need 15.77 x 1015 J/y/7.6 x 1011 J x 238 = 4.9 x 106 g of U or 4937 Kg of U

This is for 100% efficiency for 25% it will be 4937 x 4 = 19751 Kg of Uranium

So this weight of radioactive wastes needs to be processed 19751 Kg

c) 1.5 kW m–2

1.5 x 103 W/m2

For 500 MW we need 500 x 106/1.5 x 103 = 3.3 x 105 m2 This is for 100% efficiency for 25% efficiency we will need 4 times area

3.3 x 105 x 4 = 1.3 x 106 m2 of collection surface

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.