Thermodynamics Miscellaneous


  1. If VN and α are the nozzle exit velocity and nozzle angle in an impulse turbine, the optimum blade velocity is given by









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    NA

    Correct Option: D

    NA


  1. The following data pertain to a single stage impulse steam turbine: Nozzle angle = 20° Blade velocity = 200 m/s Relative steam velocity at entry = 350 m/s Blade inlet angle = 30° Blade exit angle = 25° If blade friction is neglected the work done per kg steam is









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    Given :α = 20°
    β = 30°
    x = 350 cos 30° = 303 m/s
    ∴ Vwi = 303 + Vb
    = 303 + 200 = 503 m/s

    cos 25° =
    200 + Vwo
    = 0.9063
    350

    Vwo =118 m/s
    Work done/kg of steam
    =
    (Vwi + Vwo) Vb
    kJ
    1000

    =
    (503 + 118) 200
    = 124 kJ
    1000

    Correct Option: A


    Given :α = 20°
    β = 30°
    x = 350 cos 30° = 303 m/s
    ∴ Vwi = 303 + Vb
    = 303 + 200 = 503 m/s

    cos 25° =
    200 + Vwo
    = 0.9063
    350

    Vwo =118 m/s
    Work done/kg of steam
    =
    (Vwi + Vwo) Vb
    kJ
    1000

    =
    (503 + 118) 200
    = 124 kJ
    1000



  1. A steam power plant has the boiler efficiency of 92%, turbine efficiency (mechanical) of 94%, generator efficiency of 95% and cycle efficiency of 44%. If 6% of the generated power is used to run the auxiliaries, the overall plant efficiency is









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    η0 = ηb × ηm × ηs × ηc | 1 - ηa)
    = 0.92 × 0.94 × 0.95 × 0.44r × (1 – 0.06)
    η0 = 34%

    Correct Option: A

    η0 = ηb × ηm × ηs × ηc | 1 - ηa)
    = 0.92 × 0.94 × 0.95 × 0.44r × (1 – 0.06)
    η0 = 34%


  1. For a simple compressible system, v, s, p and T are specific volume, specific entropy, pressure and temperature, respectively. As per Maxwell’s relation,δVp is equal to
    δS









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    dH = TdS + vdP

    δTS = δVP
    δPδS

    So, ⇒δVP = δTS
    δSδP

    Correct Option: B

    dH = TdS + vdP

    δTS = δVP
    δPδS

    So, ⇒δVP = δTS
    δSδP



  1. If one mole of H2 gas occupies a rigid container wi t h a capaci t y of 1000 l i t er s and t he temperature is raised from 27°C to 37°C, the change in pressure of the contained gas (round off to two decimal places), assuming ideal gas behavior, is _____ Pa. (R = 8.314 J/mol.K).









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    P1 V = nRT1
    P2 V = nRT1
    Now, Changs in pressure,
    P2 – P1 = nR (T2 - T1)
    ⇒ ∆P × 1 = 1 × 8.314(37 – 27)
    ⇒ ∆P = 83.14 Pascal

    Correct Option: A

    P1 V = nRT1
    P2 V = nRT1
    Now, Changs in pressure,
    P2 – P1 = nR (T2 - T1)
    ⇒ ∆P × 1 = 1 × 8.314(37 – 27)
    ⇒ ∆P = 83.14 Pascal