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16 September, 21:02

Air enters a nozzle steadily at 280 kpa and 77°c with a velocity of 50 m/s and exits at 85 kpa and 320 m/s. the heat losses from the nozzle to the surrounding medium at 20°c are estimated to be 3.2 kj/kg. determine (a) the exit temperature and (b) the total entropy generated for this process.

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  1. 16 September, 21:12
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    (a)

    Supposing ideal gas behavior, the specific enthalpy of air is given by

    h = cp∙T

    Specific heat capacity of dry air at 40° is [1]:

    cp = 1.005 kJ/kgK

    Solve the final temperature is

    T₂ = T₁ + (q - (1/2) ∙ (v₂² - v₁²)) / cp

    = 77°C + (3200J/kg - (1/2) ∙ ((320m/s) ² - (50m/s) ²)) / 1005J/kgK

    = 77°C - 53K

    = 24°C

    (b)

    The whole entropy change is the sum of the surrounding and entropy change of the gas.

    ∆s_sur = - q / T_sur

    = 3200J/kg / (20 + 273) K

    = 10.92J/K

    The entropy change of an ideal gas that is under change of state from

    T₁, V₁ to T₂, V₂ is to ∆S = n∙Cv∙ln (T₂/T₁) + n∙R∙ln (V₂/V₁)

    In terms of specific properties

    ∆S = m∙cp∙ln (T₂/T₁) - m∙R'∙ln (p₂/p₁) or in other words,

    ∆s = cp∙ln (T₂/T₁) - R'∙ln (p₂/p₁)

    R' is the specific gas constant of air, it is:

    R' = R/M = 287J/kgK

    The specific entropy change of the air flowing through the nozzle is:

    ∆s air = 1005kJ/kgK ∙ ln ((24 + 273) / (77+273)) - 287J/kgK∙ln (85/280)

    = 177.12J/kgK

    => total entrophy is = 10.92J/K + 177.12J/kgK = 188.04J/Kkg
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