Thermodynamics Miscellaneous

Thermodynamics Miscellaneous

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Thermodynamics

Thermodynamics Miscellaneous
  1. A small steam whistle (perfectly insulated and doing no shaft work) causes a drop of 0.8 kJ/kg in enthalpy of steam from entry to exit. If the kinetic energy of the steam at entry is negligible, the velocity of the steam at exit is








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    v²1
    		 + h1 =
    v²2
    22

    Heres v1 = 0, therefore
    h1 =
    v²2
    		 + h2
    2

    or v2 = √2(h1 - h2) = √2 × 0.8 × 1000 = 40 m/sec.

    Correct Option: B

    v²1
    		 + h1 =
    v²2
    22

    Heres v1 = 0, therefore
    h1 =
    v²2
    		 + h2
    2

    or v2 = √2(h1 - h2) = √2 × 0.8 × 1000 = 40 m/sec.

  1. A steam turbine receives steam steadily at 10 bar with a enthalpy of 3000 kJ/kg and discharges at 1 bar with an enthalpy of 2700 kJ/kg. The work output is 250 kJ/kg. The changes in kinetic and potential energies are negligible. The heat transfer from the turbine casing to the surroundings is equal to








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    ∆h = ∆Q – ∆W
    ∴ ∆Q = (– 300 + 250) kJ/kg = – 50 kJ/kg

    Correct Option: B


    ∆h = ∆Q – ∆W
    ∴ ∆Q = (– 300 + 250) kJ/kg = – 50 kJ/kg

  1. An engine working on air standard Otto cycle is supplied with air at 0.1 MPa and 35°C. The compression ratio is 8. The heat supplied is 500 kJ/kg. Property data for air: cp = 1.005 kJ/ kgK, cv = 0.718 kJ/kgK, R = 0.287 kJ/kgK. The maximum temperature (in K) of the cycle is ________(correct to one decimal place).








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    P1 = 0.1 MPa
    T1 = 35°C = 35 + 273 = 308 K

    Heat Supplied Qs = Cv (T3 – T2)
    Cv (T3 – T2) = 500 kJ/kg ..............(i)

    Given,
    V2
    		γ - 1=
    T1
    V1T2

    ⇒
    1
    		1.4 - 1=
    308
    		 = 707.60K
    8T2

    ⇒ T2 = 707.60 K
    After putting the value of T2 is equation(i):-
    0.718 (T3 – 707.6) = 500
    ⇒ T3 = 1403.98 K

    Correct Option: C

    P1 = 0.1 MPa
    T1 = 35°C = 35 + 273 = 308 K

    Heat Supplied Qs = Cv (T3 – T2)
    Cv (T3 – T2) = 500 kJ/kg ..............(i)

    Given,
    V2
    		γ - 1=
    T1
    V1T2

    ⇒
    1
    		1.4 - 1=
    308
    		 = 707.60K
    8T2

    ⇒ T2 = 707.60 K
    After putting the value of T2 is equation(i):-
    0.718 (T3 – 707.6) = 500
    ⇒ T3 = 1403.98 K

  1. An air-standard Diesel cycle consists of the following processes: 1-2: Air is compressed isentropically. 2-3: Heat is added at constant pressure. 3-4: Air expands isentropically to the original volume. 4-1: Heat is rejected at constant volume. If γ and T denote the specific heat ratio and temperature, respectively, the efficiency of the cycle is








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    Heat applied,Qs = cp (T3 – T2)
    Heat rejected, Qr = cV (T4 – T1)

    η = 1 -
    Qr
    		= 1 -
    1
    (T4 - T1)
    Qsγ(T3 - T2)

    Correct Option: B

    Heat applied,Qs = cp (T3 – T2)
    Heat rejected, Qr = cV (T4 – T1)

    η = 1 -
    Qr
    		= 1 -
    1
    (T4 - T1)
    Qsγ(T3 - T2)

  1. For the same values of peak pressure, peak temperature and heat rejection, the correct order of efficiencies for Otto, Duel and Diesel cycles is








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    For same values of peak pressure and temperature. Diesel cycle is most efficient and Otto cycle is least. Efficiency of dual cycle lies in between.ηDiesel > ηDual > ηOtto

    Correct Option: B

    For same values of peak pressure and temperature. Diesel cycle is most efficient and Otto cycle is least. Efficiency of dual cycle lies in between.ηDiesel > ηDual > ηOtto