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

c)as speed changes, wavelength and frequency change.

everything changes!

increasing transnational kinetic energy. because the transnational kinetic energy formula consists of only two variables, mass and velocity, increasing one of those properties is the only way to increase an object's transnational kinetic energy. increases to mass and velocity, however, do not have the same impact.