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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.

Respuesta :

Answer:

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

[tex]\frac{p1}{\rho g} + \frac{v1^2}{2g} = \frac{p2}{\rho g} + \frac{v2^2}{2g}[/tex]

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

so

[tex]\frac{80*10^3}{1.25*9.81} + \frac{210^2}{2(9.81)} = \frac{p2}{1.25*9.81} + \frac{60^2}{2(9.81)}[/tex]

solve we get

P2 = 105.312

so exit pressure is 105.312 kPa

and

exit temperature will be

[tex]\frac{P1}{T1} = \frac{P2}{T2}[/tex]

put here value we get

[tex]\frac{80}{278} = \frac{105.312}{T2}[/tex]

T2 = 365.96 K

so

change in entropy if exit pressure is 90 kPa

= Cp ln [tex]\frac{T2}{T1}[/tex] - R ln [tex]\frac{p2}{p1}[/tex]

= 1.005 ln [tex]\frac{365.96}{278}[/tex] - 0.287 ln [tex]\frac{90}{80}[/tex]

so change in entropy = 0.2425 KJ/kg-K

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