5. Consider a 700-MW, 28 percent efficient coal-fired power plant that uses cool

Question

5. Consider a 700-MW, 28 percent efficient coal-fired power plant that uses cooling water withdrawn from a nearby river (with an upstream flow of 120-m3/s and temperature 20 C) to take care of waste heat. The heat content of the coal is 12,000 Btu/lb, and the sulfur content is 1.8% by mass. (a) How much electricity (kWhr/yr) would the plant produce each year? (b) How many pounds per hour of coal would need to be burned at the plant? (c) If the cooling water is only allowed to rise in temperature by 15 °C, what flow rate (in m'/s) from the stream would be required? (d) What would be the river temperature just after it receives the heated cooling water? (e) Estimate the hourly SO2 emissions (in kg/h) from the plant assuming that all of the sulfur is oxidized to So2 during combustion. Report on the required efficiency of the SO2 scrubber, if the plant is only allowed to emit the legal limit of 0.6 lb SO2 per million Btu of heat input. (f)

Explanation / Answer

a) The electrical energy produced by 700MW power plant = 700MJ/s (million joule per second)

b) Since the efficiency of the plant is 28%, the heat energy required to be produced to generate 700MJ/s of electrical energy is = 700/ 0.28

= 2500MJ/s

Consumption of coal in producing 2500MJ/s of heat energy is = 2500 MJ/s / calorific value of coal in MJ/kg

= 2500 / 21

= 119.05 Kg/s

1 pound = 0.4536 Kg

119.05 Kg = 119.05/0.4536 = 262.45 pounds

c) using the relation , V is directly proportional to T

thus, V1/ V2 = T1 / T2

120 / V2 = 20/ 35

V2 = 210 m3/sec
When water temperature is allowed to rise in 15 degrees then the velocity is 210m3/sec

d) If the temperature of discharge is significantly warmer than the natural water then negative effects may occur. There are several significant including diminished dissolved oxygen levels, fish kills and influxes of invasive species.

e) Sulphur content of consumed coal i.e., Sulphur produced per second is = 1.8 % of mass

= 1.8% * 119.05

= 2.1429 Kg/s

Now, S+O2 = SO2

the molecular mass of both S and O2 is 32, and they combine on a one to one basis :

2.15Kg of S + 2.15 Kg of O2 = 4.3 kg of SO2 / sec

Emission of sulphurdioxide in Kg/sec = 4.3 Kg/s

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