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.

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

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

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

so

solve we get

P2 = 105.312

so exit pressure is 105.312 kPa

and

exit temperature will be

put here value we get

T2 = 365.96 K

so

change in entropy if exit pressure is 90 kPa

= Cp ln - R ln

= 1.005 ln - 0.287 ln

so change in entropy = 0.2425 KJ/kg-K

answer:

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