At a steam power plant, steam engines work in pairs, the heat output of the firs

Question

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 730 C and 430 C, and of the second 405 C and 290 C

Part 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 970 MW of power? Assume the efficiency of the engines is 65% of the ideal (Carnot) efficiency.

Express your answer using two significant figures

Part B

Water is used to cool the power plant. If the water temperature is allowed to increase by no more than 4.5C, estimate how much water must pass through the plant per hour.

Express your answer using two significant figures.

Explanation / Answer

(a) the coal rate:
Note that the sum of the work of each engine equals 970MW, so:
W1 + W2 = 970000 kW -- eq 1
& thermal efficiency of heat engines is work output divided by heat input:
e = W/Qa

so we can rewrite equation 1:
e1*Qa1 + e2*Qa2 = 970000 kW
since, Qa1-Qr1 = Qa2 ("output of heat from one being the approximate input of the second") and Qa1-Qr1 = eff1*Qa1, so:

e1*Qa1 + e2*eff1*Qa1 = 970000 kW -- eq 2

the thermal effficiency of heat engine 1 is:
e1 = 0.65*eff1 carnot
e1 = 0.65*[(730-430)/(730+273)] = 0.6*0.299 = 0.194

likewise, the thermal efficiency of heat engine 2 is:
e2 = 0.65*[(730-430)/(405+273)] = 0.288

substituting values to eq 2:
(0.194 + 0.288*0.299)*mf*28,000 = 970000
mf = 123.68 kg/sec

B) Cooling water must remove heat rejected from the plant

Energy in= Energy out

QL= mw*cp* delta T

8798.13 MW = mw *4186*4.5

mw= 8798.17 *10^6/4186*4.5 = 467066 kg/s =467066/3600 kg/h= 129.74 kg/h

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