Air at 80 kPa and 400 K enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the airstream is decreased from 230 to 30 m/s as it passes through the diffuser. Find (a) the exit temperature of the air and (b) the exit area of the diffuser.

a)  425.6 K = 152.6 degree Celsius

b)  678.6 square centimeter

Explanation:

Initial enthalpy = Final enthalpy

m(h1 + v1^2/2) = m(h2 + v2^2/2)

(h1 + v1^2/2) = (h2 + v2^2/2)

h 2 = (h1 + v1^2/2) - (v2^2/2)

h 2 = 400.98 + (230^2 – 30^2)/2 * 10^-3

= 426.98 KJ/Kg

T2 = T1’ + (T2’ -T1’)(h2-h1’)/( h2’-h1’)

T2 = 420 + (430-420)(431.43-421.26)(426.98-421.26)/(431.43 – 421.26)

T2 = 425.6 K = 152.6 degree Celsius

Area = mRT2/P2V2

Area = (600/60*60*1) *0.287*425.6/(100*30) *10^4

Area = 678.6 square centimeter

c. magnitude and direction

a is the best answer your welcome