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

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2