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.

(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