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toothpaste666
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Homework Statement
At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are 718∘C and 427∘C, and of the second 419∘C and 270∘C.
A) If the heat of combustion of coal is 2.8×107J/kg, at what rate must coal be burned if the plant is to put out 950 MW of power? Assume the efficiency of the engines is 65 % of the ideal (Carnot) efficiency.
B)Water is used to cool the power plant. If the water temperature is allowed to increase by no more than 5.5 C∘ , estimate how much water must pass through the plant per hour.
Homework Equations
e = 1-QL/QH
eideal = 1-TL/TH
e = W/QH
The Attempt at a Solution
for part A
second engine
e2 = W/(QH2) = (950000000J/s)/(QH2)
e2ideal =1-(TL2/TH2) = 1-(427+273)/(718+273) = 1 - 543/692 = .215
e2 = .65(e2ideal) = .65(.215) = .140
.140 =(950000000J/s)/(QH2)
QH2 = (950000000J/s)/(.140) = 6785714286 J/s
QH2 = QL1
e1ideal = 1 - 700/991 = .294
e1 = .65(.294) = .191
.191 = 1 - (QL1)/(QH1)
(QL1)/(QH1) = 1 - .191 = .809
QL1 = (.809)(QH1)
so QH2 = .809(QH1)
QH2 = 6785714286 J/s
.809(QH1) = 6785714286 J/s
QH1 = 8387780329 J/s - heat input of first engine
(8387780329 J/s) (kg/(2.8x10^2 J))
= 299.6 kg/s
but it came up wrong. I can't figure out what I did wrong