Thermodynamics - Steady State Nozzle, find area of inlet/exit

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leafjerky
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Homework Statement


In a jet engine, a flow of air at 1000 K, 200 kPa, and 40 m/s enters a nozzle, where the air exits at 500 m/s and 90 kPa. What is the exit temperature, inlet area, and exit area, assuming no heat loss?

Homework Equations


min = mout = m
where m = mass air flow
dE/dt cv = Qcv - Wcv + Σmin(h+ (Vin/2)2 + gz) - Σmout(h+ (Vout/2)2 + gz)

exit area:
m = ρAV
where m = mass air flow, ρ = density, A = area, V = velocity

The Attempt at a Solution


I've already solved for the exit temperature to be 890 K. I just need to know how to find the exit/inlet areas.
 
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leafjerky said:

Homework Statement


In a jet engine, a flow of air at 1000 K, 200 kPa, and 40 m/s enters a nozzle, where the air exits at 500 m/s and 90 kPa. What is the exit temperature, inlet area, and exit area, assuming no heat loss?

Homework Equations


min = mout = m
where m = mass air flow
dE/dt cv = Qcv - Wcv + Σmin(h+ (Vin/2)2 + gz) - Σmout(h+ (Vout/2)2 + gz)

exit area:
m = ρAV
where m = mass air flow, ρ = density, A = area, V = velocity

The Attempt at a Solution


I've already solved for the exit temperature to be 890 K. I just need to know how to find the exit/inlet areas.
If there is no mass addition between the inlet and the exit, then doesn't continuity apply?

I don't know if you will be able to find the individual areas, but you should be able to calculate an area ratio between the inlet and exit.
 
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SteamKing said:
If there is no mass addition between the inlet and the exit, then doesn't continuity apply?

I don't know if you will be able to find the individual areas, but you should be able to calculate an area ratio between the inlet and exit.

In our problems with nozzles, I believe we are assuming the following:
  • closed system
  • steady state (de/dt = 0)
  • W = 0
  • Q = 0
  • PE = 0
I have emailed my teacher regarding the problem with the areas. I figured she must have left m out on accident. I just wanted to come on here and make sure. Thanks for your help as always SteamKing