Question

Air at a velocity of 210 m/s decelerates through a thermally insulated diffuser to a velocity of 60 m/s. The temperature and pressure at the inlet are 278 K and 80 kPa. Assuming one-dimensional and steady flow, calculate: • The exit temperature. • The change in entropy if the exit pressure is 90 kPa. • The exit pressure assuming no dissipative phenomena occur.

Answer 1

exit velocity is 60 m/s

exit pressure is 105.312 kPa

change in entropy = 0.2425 KJ/kg-K

exit temperature = 365.96 K

Step-by-step explanation:

given data

velocity v1 = 210 m/s

velocity v2 = 60 m/s

temperature t1 = 278 K

pressure p1 = 80 kPa

to find out

The exit temperature and change in entropy and exit pressure assuming no dissipative phenomena

solution

we know here at inlet

velocity is 210 m/s

and at exit velocity is 60 m/s

and

by the bernoullis equation exit pressure will be

`(p1)/(\rho g) + (v1^2)/(2g) = (p2)/(\rho g) + (v2^2)/(2g)`

and here put all value and we know ρ = ρ air = 1.25 kg/m³

so

`(80*10^3)/(1.25*9.81) + (210^2)/(2(9.81)) = (p2)/(1.25*9.81) + (60^2)/(2(9.81))`

solve we get

P2 = 105.312

so exit pressure is 105.312 kPa

and

exit temperature will be

`(P1)/(T1) = (P2)/(T2)`

put here value we get

`(80)/(278) = (105.312)/(T2)`

T2 = 365.96 K

so

change in entropy if exit pressure is 90 kPa

= Cp ln
`(T2)/(T1)` - R ln
`(p2)/(p1)`

= 1.005 ln
`(365.96)/(278)` - 0.287 ln
`(90)/(80)`

so change in entropy = 0.2425 KJ/kg-K