A building is maintained at a temperature T by means of an ideal heat pump which uses a river at temperature T0 as a source of heat. The hea

Question

A building is maintained at a temperature T by means of an ideal heat pump which uses a river at temperature T0 as a source of heat. The heat pump consumes power W, and the building loses heat to its surroundings at a rate α(T−T0), where α is a positive constant. Show that T is given by T=T0+W2α(1+1+4αT0/W‾‾‾‾‾‾‾‾‾‾‾√).

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Khải Quang 4 years 2021-08-12T06:47:17+00:00 2 Answers 177 views 0

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  1. Tryphena
    0
    2021-08-12T06:48:24+00:00
    Reply

    Answer:

    T = (T_o + W/2α (1 + √(4αT_o/(W) + 1)] has been proved!

    Explanation:

    We know that efficiency of pump is given by;

    η = (Q/t)/P

    Where;

    P = the power consumed in the house

    Q/t = the rejected power.

    Now, P can also be expressed as W/t

    Likewise, Q/t can also be expressed as; α(T – T_o)

    From the question, pump consumed power of W. Thus, P = W in this case.

    So, efficiency can be expressed as;

    η = (α(T – T_o))/(W) – – – – (1)

    Now, for the carnot engine of the pump, efficiency is expressed as;

    η = T/(T – T_o) – – – – (2)

    Equating equation 1 and 2,we have;

    (α(T – T_o))/(W) = T/(T – T_o)

    α(T – T_o)² = WT

    Expanding, we have;

    T² – (2T•T_o) + T_o² = WT/α

    T² – (2T•T_o + WT/α) + T_o² = 0

    T² -T(2T_o + W/α) + T_o² = 0

    To find T, let’s use quadratic formula which is;

    x = [-b ± √(b² – 4ac)]/2a

    T = -(-(2T_o + W/α)) ± √(-(2T_o + W/α)² – (4•1•T_o²)]/(2•1)

    T = (2T_o + W/α)) ± √((-2T_o – W/α)² – (4T_o²)]/(2)

    T = (2T_o + W/α)) ± √((4T_o² + 2T_o•W/α + (W/α)² – 4T_o²)]/(2)

    T = (2T_o + W/α)) ± √(2T_o•W/α + (W/α)²)]/(2)

    T = (T_o + W/2α)) ± (1/2)√(2T_o•W/α + (W/α)²)]

    T = (T_o + W/2α)) ± (W/2α)√(4T_o/(W/α) + 1)]

    T = (T_o + W/2α)) ± (W/2α)√(4αT_o/(W) + 1)]

    Now, let’s arrange to correspond with what’s in the question. And it means we will make use of the positive sign where we have ±.

    Thus;

    T = (T_o + W/2α (1 + 1√(4αT_o/(W) + 1)]

    T = (T_o + W/2α (1 + √(4αT_o/(W) + 1)]

  2. mit
    0
    2021-08-12T06:49:07+00:00
    Reply

    Answer:

    We want to prove that

    Show that T is given by T=To+W/2α[1+√(1+4αTo/W)]

    Explanation:

    The efficiency of the pump is given as

    n = Q/t / P

    Where Q/t is the rejected power

    And P is the consumed power to cool the house

    Given that,

    The at rejected or loss is α(T−To)

    So, Q/t = α(T−To)

    Pump Consumed =W

    Then, power consumed P=W/t

    Then,

    n = Q/t /P

    n = α(T−To) / W . Equation 1

    Generally,

    The efficiency of a carnot engine is given as

    n = TH/(TH —TC)

    Where TC is cold temperature

    TC=To

    And TH is high temperature

    TH=T

    n= T/ (T—To) Equation 2

    Equation the two equation we have,

    α(T−To) / W = T/ (T—To)

    Cross multiply

    α(T−To) × (T-To) =W× T

    α(T² — 2TTo + To²)=WT

    Divide both sides by α

    T² — 2TTo + To²=WT/α

    Rearrange

    T² — 2TTo + To²—WT/α=0

    T² — 2TTo —WT/α + To²=0

    T² — (2To + W/α)T + To²=0

    This form a quadratic equation

    So let use formula method.

    T = [- b ± √(b²-4ac) ]/ 2a

    Where a=1, b= — (2To + W/α), c= To²

    Then, applying the formula

    T = [- b ± √(b²-4ac) ]/ 2a

    T = [- -(2To + W/α) ± √(-(2To +W/α))² – 4×1×To²)]/ 2×1

    T = [(2To + W/α) ± √4To²+4ToW/α+ W²/α² – 4To²]/ 2

    T = [(2To+W/α) ±√(4ToWα+W²/α²)]/2

    T=[(2To+W/α)±√(W²/α²(4Toα/W+1)]/2

    T=[(2To+W/α)±W/α√(4Toα/W+1)]/2

    Divide through by 2

    T=To+W/2α±W/2α√(4Toα/W+1)

    T=To+W/2α(1 ±√(4Toα/W+1)

    Rearranging to conform to what we want to proof

    T=To+W/2α(1 ±√(1 + 4αTo/W)

    This is the required proof..

    T = To + W/2α(1 + √(1 + 4αTo/W)

    OR

    T=To+W/2α(1 —√(1 + 4αTo/W)

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