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 ne

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).

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Edana Edana 5 years 2021-08-18T03:52:53+00:00 1 Answers 208 views 0

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  1. lethu
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    2021-08-18T03:54:33+00:00
    Reply

    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

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