Thermodynamics adiabatic diffuser question

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An adiabatic diffuser is used to reduce air velocity from 220 m/s to 30 m/s, with an inlet mass flow rate of 8 kg/s at 300°C and 100 kPa, leading to an exit pressure of 125 kPa. To determine the exit area, the conservation of energy equations and the relationship between temperature, pressure, and density must be applied. The concept of adiabatic means no heat transfer occurs during the process. The outlet temperature can be calculated using conservation of energy, which then allows for the determination of outlet density. The final calculated exit area of the diffuser is 0.37 m².
Superhawkkodaka
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An adiabatic diffuser is employed to decrease the velocity of an airstream from 220 to 30 m/s. The air enters the diffuser at a rate of 8 kg/s with a temperature of 300c and a pressure of 100kpa. Determine the exit area of the diffuser if the exit pressure is 125kpa.

Answer is 0.37 m^2

Guys I am really lost with this one.. I used equations pv=mrt, m=p'v'a, v=v'a
Note (v' is velocity as in> m/s and p' is density) just to prevent confusion

What i only got are the initial values v1, a1, p'1..
 
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I can't get a2(exit area) have any ideas guys?
 
What does adiabatic mean? How can you use this information to relate the inlet and outlet properties?
 
billy_joule said:
What does adiabatic mean? How can you use this information to relate the inlet and outlet properties?
i for got to add the equation for conservation of energy on open system
q=∑(exi
billy_joule said:
What does adiabatic mean? How can you use this information to relate the inlet and outlet properties?

adiabatic mean
billy_joule said:
What does adiabatic mean? How can you use this information to relate the inlet and outlet properties?

adiabatic means heat transfer is zero (Q=0)

i forgot to add the equation for open systems
Q= ΔH + ΔPE + ΔKE + W --- in this equation Q, W, & PE is equal to zero..
and
ΔH = ΔU + Δ(PV)

this equations might be usable.. my question is how do i use these equations to determine exit area
 
You can find the outlet temp. via conservation of energy. Then you can find outlet density as you have temp. & pressure.
Then use the mass flow equation you gave in your first post to solve for area.
 
I must add that there are probably other, potentially simpler ways to solve the problem.
 
billy_joule said:
You can find the outlet temp. via conservation of energy. Then you can find outlet density as you have temp. & pressure.
Then use the mass flow equation you gave in your first post to solve for area.

conservation of energy equation? i don't know how to get the final temperature.. can you be more specific with equation for conservation of energy?
i only know this equation for conservation of energy for open system : Q= ΔH + ΔPE + ΔKE + W
i can' simplify tthe equation that i gave to input the temperature in the equation
 
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Superhawkkodaka said:
i only know this equation for conservation of energy for open system : Q= ΔH + ΔPE + ΔKE + W
You have one unknown in that equation so you should be able to solve it.
If you have no idea how to proceed it's generally best to review your textbook and/or lecture notes. We can help with specific problems but a forum is a poor format to learn course material from.
 
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