Answer:
Explanation:
Given that,
Efficiency of Carnot engine is 47%
η =47%=0.47
The wasted heat is at temp 60°F
TL=60°F
Rate of heat wasted is 800Btu/min
Therefore, rate of heat loss QL is
QL' = 800×60 =48000
The power output is determined from rate of heat obtained from the source and rate of wasted heat.
Therefore,
W' = QH' - QL'
Note QH' = QL' / (1-η)
W' = QL' / (1-η) - QL'
W'=QL' η / (1-η)
W'= 48000×0.47/(1-0.47)
W'=42566.0377 BTU
1 btu per hour (btu/h) = 0.00039 horsepower (hp)
Then, 42566.0377×0.00039
W'=16.6hp
Which is approximately 17hp
b. Temperature at source
Using ratio of wanted heat to temp
Then,
TH / TL = QH' / QL'
TH = TL ( QH' / QL')
Since, QH' = QL' / (1-η)
Then, TH= TL( QL' /QL' (1-η))
TH=TL/(1-η)
TL=60°F, let convert to rankine
°R=°F+459.67
TL=60+459.67
TL=519.67R
TH=519.67/(1-0.47)
TH=980.51R
Which is approximately 1000R
