A 680 MW coal fired power plant is 34% efficient at producing electricity. The c

Question

A 680 MW coal fired power plant is 34% efficient at producing electricity. The coal has a heating value of 20,000 kJ/kg, an ash content of 5.5% and a sulfur content of 4.5%. Note: 1 MW = 1000 kW , and 1 kW = 1 kJ/s.

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

b. If the efficiency of ash capture is 98.6%, how much ash is emitted into the air, kg/day?

c. How much heat is emitted to the environment (kJ/day)?

d. If 82% of the heat emitted is to be removed by cooling water, calculate the flow rate of

water (in lb/day) if the water is initially at 65 oF and rises to 95 oF. Note that 1 Btu =

1.055 kJ and that the Cp for water is 1.0 Btu/(lb- oF).

Explanation / Answer

a) 680 MW = 680000kW

per day energy required = 680000*24*3600 = 58752000000kJ

let m be mass of coal required

then 34 % effieciency leads to = m*(34/100)*20000 kJ enegy

m*(34/100)*20000 = 58752000000

m = 58752000000/(0.34*20000) = 8640000 Kg of coal = 8640 tonnes

b) efficiency of ash capture = 98.6 %===> percentage of ash released = 1.4%

coal has 5.5% of ash

====> amount of ash released = (5.5/100)*(1.4/100)*8640000 = 6652.8 Kg

c) since the powerplant is 34% efficient ====> (100-34)% = 66 % of heat will be releasd

===> 66/100*58752000000 = 38776320000kJ of energy is released or 38776320MJ

d) 82% of heat realsed = 82/100*38776320000 = 31796582400kJ = 33545394432 Btu

this is equl to m*cp*(t2-t1) = m*1*(95-65) = 33545394432

m = 33545394432/30 = 1118179814.4 lb/day

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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