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 770 ∘C and 460 ∘C, and of the second 435 ∘C and 280 ∘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 860 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 4.5∘C, estimate how much water must pass through the plant per hour.

Answer 1

  (a) the coal rate:

Note that the sum of the work of each engine equals 860MW,

so:
W1 + W2 = 860000 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 = 860000 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 = 860000 kW -- eq 2

the thermal effficiency of heat engine 1 is:
e1 = 0.60*eff1 carnot
e1 = 0.60*[(770 - 460)/(770+273)] = 0.6*0.244 = 0.178

likewise, the thermal efficiency of heat engine 2 is:
e2 = 0.60*[(435-280)/(435+273)] = 0.131

substituting values to eq 2:
(0.178+ 0.131*0.244)*mf*28,000 = 860000
mf =146.28 kg/sec

b)

assuming an electric consumption of 13,500 kW-hr per capita or 1.5 kW/person for continuously operated power plant, then the number of person that a 860000 kW power plant can serve is:
n = 860000/1.5 =573333.33 persons

this would be 6 to 7 towns for a860,000 plus/minus population per town.