Question

A heat engine is operating on a Carnot cycle and has a thermal efficiency of 47 percent. The waste heat from this engine is rejected to a nearby lake at 60 deg. F at a rate of 800 Btu/min. Determine a) the power output of the ending (Approx 17 hp) and b) the temperature of the source (Approx 1000 R).

Answer 1

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