A 36% efficient coal fired power plant is producing 790 MW of electricity. The c

Question

A 36% efficient coal fired power plant is producing 790 MW of electricity. The coal has a heating value of 19,000 kJ/kg, an ash content of 7.1% and a sulfur content of 5.2%. Note: 1 MW = 1,000,000 W; 1 W = 1 J/s.

a.       What is the daily input rate of coal (kg/day)?


b.      If the efficiency of ash capture is 98.8%, how much ash is emitted into the air; give your answer in units of kg/day and in units of g/kwh of electricity produced?

c.       The efficiency for SO2removal is 94%. Calculate the emission rate of SO2, kg/day. Atomic weights are: S=32, O=16

Explanation / Answer

a.

790 MW x (10^6 W/1 MW) x (1 (J/s)/1W) = 790,000,000 J/s output

Ideal output = actual output / efficiency = 2.2 x 10^9 J/s

2.2 x 10^9 J/s x (3600 s/1 h) x (24 h/1 day) x (1 kJ/1000 J) x (1 kg/19,000 kJ) = 1.0 x 10^7 kg/day of coal

b. 7.1% ash

= 1.0 x 10^7 kg/day x 0.071 = 7.1 x 10^5 kg/day

7.1 x 10^5 kg/day x (1 day/790,000,000J) x (3,600,000 J/kWh) x (1000g/1kg) = 3.2 x 10^6 g/kWh

98.8% captured, thus only 1.2 % released.

Multiply 0.012 to both of these answers

8.5 x 10^3 kg/day

3.9 x 10^4 g/kWh

c.

1.0 x 10^7 kg/day coal x (5.2 % sulfur) x (1 mol sulfur/32 g sulfur) x (1 mol SO2/1 mol sulfur) x (64 g SO2/1 mol SO2) x (6% uncaptured)

= 6.2 x 10^4 kg/day

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