Thermodynamics Miscellaneous

Thermodynamics Miscellaneous

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Thermodynamics

Thermodynamics Miscellaneous
  1. An ideal gas of mass m and temperature T1 undergoes a reversible isothermal process from an initial pressure P1 to final pressure P2. The heat loss during the process is Q. The entropy change ∆S of the gas is








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    For clausius theorem, ∮
    δQ
    		< 0 (for irreversible Heat engine).
    T

    Correct Option: B

    For clausius theorem, ∮
    δQ
    		< 0 (for irreversible Heat engine).
    T

Direction: In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K. The following are the properties and relations pertaining to air:
Specific heat at constant pressure, cp = 1.005 kJ/kgK;
Specific heat at constant volume, cv = 0.718 kJ/kgK;
Characteristic gas constant, R = 0.287 kJ/kgK.
Enthalpy, h = cpT. .
Internal energy, u = cvT.

  1. If the pressure at station Q is 50 kPa, the change in entropy (sQ – sp) in kJ/kgK is








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    ∆s =
    δq
    		for reversible process
    T


    δQ = δW = P1V1ln
    V2
    V1

    ∆s =
    δq
    		= mRln
    P1
    Tp2

    ∆s = mR ln
    P1
    P2

    Correct Option: C

    ∆s =
    δq
    		for reversible process
    T


    δQ = δW = P1V1ln
    V2
    V1

    ∆s =
    δq
    		= mRln
    P1
    Tp2

    ∆s = mR ln
    P1
    P2

  1. If the air has to flow from station P to station Q, the maximum possible value of pressure in kPa at station Q is close to








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    SQ - Sp = CvlnT2+ RlnV2
    T1V1

    From perfect gas law,
    V2
    		=
    P1

    T2

    V1P2T1

    =
    150 × 300
    		= 2.57
    300 × 350

    ∴ SQ – Sp = – 0.1107 + 0.287 ln 2.57 = 0.16 kJ/kgK

    Correct Option: B

    SQ - Sp = CvlnT2+ RlnV2
    T1V1

    From perfect gas law,
    V2
    		=
    P1

    T2

    V1P2T1

    =
    150 × 300
    		= 2.57
    300 × 350

    ∴ SQ – Sp = – 0.1107 + 0.287 ln 2.57 = 0.16 kJ/kgK

  1. One kilogram of water at room temperature is brought into contact with a high temperature thermal reservoir. The entropy change of the universe is








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    P2 = 87.4 kPa
    P2 = 87kPa

    Correct Option: D


    P2 = 87.4 kPa
    P2 = 87kPa

  1. A rope-brake dynamometer attached to the crank shaft of an I.C. engine measures a brake power of 10 kW when the speed of rotation of the shaft is 400 rad/s. The shaft torque (in Nm) sensed by the dynamometer is _____.








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    P = Tω

    ∴ T =
    P
    		 =
    10000
    		= 25 N-m
    ω400

    Correct Option: A

    P = Tω

    ∴ T =
    P
    		 =
    10000
    		= 25 N-m
    ω400