Velocity exiting Heat Exchanger

[jdawg]

Homework Statement

Cold air enters a heat exchanger at 400 [K] and exits it at 1200 [K]. At the 0.025 [m²] inlet, the pressure is measured to be 1.0 [MPa] and the velocity is measured to be 3 [m/s]. The area of the exit is also 0.025 [m²] and the pressure here is measured to be 300 [kPa]. What is the velocity exiting the heat exchanger? Use conservation of mass principles.

Homework Equations

The Attempt at a Solution

So the mass flow rate going in equals the mass flow rate going out. m1=m2

I used this formula to calculate the mass flow rate:
m1=ρ_airA1v1
m=(1.225)(0.025)(3)=0.091875 kg/s

Since its an ideal gas, I substituted the mass flow rate equation into the ideal gas equation:

P₂ =(m/(A₂v₂))RT₂

I found R to be 0.287 kJ/(kg*K)

v₂=[(0.091874 kg/s)(0.287 kJ/(kg*K))(1200K)]/[(300000 N/m²)(0.025 m²)]

I got v₂= 4.2189 m/s as my velocity which is incorrect since my answer can only be 20,30,27,or 3.3333 m/s.

I can't see what I did wrong.

 

[SteamKing]

Homework Statement

Cold air enters a heat exchanger at 400 [K] and exits it at 1200 [K]. At the 0.025 [m²] inlet, the pressure is measured to be 1.0 [MPa] and the velocity is measured to be 3 [m/s]. The area of the exit is also 0.025 [m²] and the pressure here is measured to be 300 [kPa]. What is the velocity exiting the heat exchanger? Use conservation of mass principles.

Homework Equations

The Attempt at a Solution

So the mass flow rate going in equals the mass flow rate going out. m1=m2

I used this formula to calculate the mass flow rate:
m1=ρ_airA1v1
m=(1.225)(0.025)(3)=0.091875 kg/s

In your calculation of the inlet mass flow, you have apparently used the density of air at normal atmospheric pressure rather than a pressure of 1.0 MPa.

 

[jdawg]

Perfect, thanks!