Question

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is 120 m/s. For adiabatic operation with no internal irreversibility, determine the exit temperature, in K, and the exit pressure, in bar, for (a) Constant specific heats with k

Answer 1

Step-by-step explanation:

Calculating the exit temperature for K = 1.4

The value of
`c_p` is determined via the expression:

`c_p = (KR)/(K_1)`

where ;

R = universal gas constant =
`(8.314 \ J)/(28.97 \ kg.K)`

k = constant = 1.4

`c_p = (1.4((8.314)/(28.97) ))/(1.4 -1)`

`c_p= 1.004 \ kJ/kg.K`

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

`0=(h_1-h_2)+((v_1^2-v_2^2))/(2)` ------ equation(1)

we can rewrite the above equation as :

`0 = c_p(T_1-T_2)+ ((v_1^2-v_2^2))/(2)`

`T_2 =T_1+ ((v_1^2-v_2^2))/(2 c_p)`

where:

`T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K`

`T_2= 280+(((510)^2-(120)^2))/(2(1.004)) *(1)/(10^3)`

`T_2 = 402.36 \ K`

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation:
`(T_2)/(T_1) = ((P_2)/(P_1))^(k)/(k-1)`

`P_2=P_1((T_2)/(T_1))^(k)/(k-1)`

`P_2 = 5 ((402.36)/(280) )^(1.4)/(1.4-1)`

`P_2 = 17.79 \ bar`

Therefore, the exit pressure is 17.79 bar