Refrigeration and Air-conditioning Miscellaneous

Refrigeration and Air-conditioning Miscellaneous

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Refrigeration and Air-conditioning

Refrigeration and Air-conditioning Miscellaneous
  1. A thin layer of water in a field is formed after a farmer has watered it. The ambient air conditions are: temp. 20°C and relative humidity 5%. An extract of steam tables is given below.

    Neglecting the heat transfer between the water and the ground, the water temperature in the field after phase equilibrium is reached equals








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    Given data:
    Φ = 5% = 0.05
    Tdb = 20°C
    From table
    At 20°C, Ps = 2.34 kPa

    Φ =
    Pv
    Ps

    0.05 =
    Pv
    2.34

    Pv =.117 k Pa By Interpolation method,
    Tdb = - 15 +
    (-10 + 15
    		× (0.117 - .1)
    0.26 - 0.10

    = – 15 + 0.531 = – 14.5° C

    Correct Option: C

    Given data:
    Φ = 5% = 0.05
    Tdb = 20°C
    From table
    At 20°C, Ps = 2.34 kPa

    Φ =
    Pv
    Ps

    0.05 =
    Pv
    2.34

    Pv =.117 k Pa By Interpolation method,
    Tdb = - 15 +
    (-10 + 15
    		× (0.117 - .1)
    0.26 - 0.10

    = – 15 + 0.531 = – 14.5° C

  1. A heat engine having an efficiency of 70% is used to drive a refrigerator having a coefficient of performance of 5. The energy absorbed from low temperature reservoir by the refrigerator for each kJ of energy absorbed from high temperature source by the engine is








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    Given: ηengine = 0.7, (C.O.P)R = 5

    ηengine =
    W
    		...(i)
    Q1


    Now,
    (C.O.P)R =
    Q2
    W

    ∴ W =
    Q2
    		...(ii)
    5

    Again,
    0.7 =
    Q2
    		×
    1
    5Q1

    or
    Q2
    		= 3.5
    Q1

    Energy absorbed from low temperature reservoir by the refrigerator for each kJ of energy absorbed from high temperature source by the engine = 3.5 kJ.

    Correct Option: C

    Given: ηengine = 0.7, (C.O.P)R = 5

    ηengine =
    W
    		...(i)
    Q1


    Now,
    (C.O.P)R =
    Q2
    W

    ∴ W =
    Q2
    		...(ii)
    5

    Again,
    0.7 =
    Q2
    		×
    1
    5Q1

    or
    Q2
    		= 3.5
    Q1

    Energy absorbed from low temperature reservoir by the refrigerator for each kJ of energy absorbed from high temperature source by the engine = 3.5 kJ.

  1. The thermodynamic cycle shown in figure (T-s diagram) indicates








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    T-S diagram represent a reversed brayton cycle used in air conditioning of aeroplanes where air is used as a refrigerant.

    Correct Option: B

    T-S diagram represent a reversed brayton cycle used in air conditioning of aeroplanes where air is used as a refrigerant.

  1. The COP of a Carnot heat pump operating between 6°C and 37°C is








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    (COP)c.p. =
    T1
    		=
    310
    		= 10
    T1 - T231

    Correct Option: A

    (COP)c.p. =
    T1
    		=
    310
    		= 10
    T1 - T231

  1. A heat pump with refrigerant R22 is used for space heating between temperature limits of – 20°C and 25°C. The heat required is 200 MJ/ h. Assume specific heat of vapour at the time of discharge as 0.98 kJ/kgK. Other relevant properties are given below. The enthalpy (in kJ/ kg) of the refrigerant at isentropic compressor discharge is_________.








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    Given :T1' = 253 K
    T2' = 298 K
    S1 = S2 = S2'

    ∴ S1 = S2 = S2' + CP ln
    T2
    T2'

    1.7841= 1.7183 + 0.98 ln
    T2
    298

    ∴ T2 = 318.69 K
    ∴ Enthalpy of discharge of compressor
    ⇒ h2 = h2' + Cp (T2' – T1')
    ∴ h2 = 413.02 + 0.98 (318.69 – 298)
    ∴ h2 = 433.3 kJ/kg .

    Correct Option: A


    Given :T1' = 253 K
    T2' = 298 K
    S1 = S2 = S2'

    ∴ S1 = S2 = S2' + CP ln
    T2
    T2'

    1.7841= 1.7183 + 0.98 ln
    T2
    298

    ∴ T2 = 318.69 K
    ∴ Enthalpy of discharge of compressor
    ⇒ h2 = h2' + Cp (T2' – T1')
    ∴ h2 = 413.02 + 0.98 (318.69 – 298)
    ∴ h2 = 433.3 kJ/kg .