At a steam power plant, steam engines work in pairs, the heat output of the firs
ID: 1528212 • Letter: A
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 740 C and 450 C, and of the second 425 C and 250 C.
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 990 MW of power? Assume the efficiency of the engines is 65% of the ideal (Carnot) efficiency.
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.
Explanation / Answer
Operating temperatures of first=7500C and 4500C
Operating temperatures of second=4250C and 2700C
(a) Coal rate,
Sum of the work of each engine equals 990 MW,
=> W1 + W2 = 990000 kW ----------1
Thermal efficiency of heat engines is the ratio of work output and heat input:
e = W / Q
equation 1 can be rewrite as:
e1*Q1 + e2*Q2 = 990000 kW
since, Q1-Qr1 = Q2
Q1-Qr1 = eff1*Q1
Since Output of heat from one being is the input of the second
e1*Q1 + e2*eff1*Q1 = 990000 kW ---------------2
Thermal effficiency of heat engine 1 is:
e1 = 0.65*eff1Carnot
=>e1 = 0.65*[(750-450)/(750+273)] = 0.65*0.293 = 0.190
Thermal effficiency of heat engine 2 is:
e2 = 0.65*eff2Carnot
e2 = 0.65*[(425-270)/(425+273)] = 0.144
substituting values to eq 2:
(0.190 + 0.144*0.293)*mf*28,000 = 990000kW
6504.16mf = 990000kW
mf=152.275 kg/sec
b)mc(deltaT)=q
q=mcT.
"Q" stands for heat, usually given in Joules.
"m" is the mass of the given substance.
"c" is the specific heat capacity of that substance, and
"T" is the change in temperature (initial temperature minus final temperature) in degrees Celsius
m*4186*4.5=q
q=990000/0.144
e2=0.144
q=6875MW=6.875 x109 W
m=6.875 x109 W / (4186 J/kg *4.5C)
m=3.64973x 105 kg/sec
Volume of water that has to pass through the planet
V= m/
= 3.64973x 105 kg/sec/ 1000 kg/m3
=364.973 m3 /sec
=(364.973 m3 /sec) x (3600sec/hr)
= 13.139 x 105 m3/ hr
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