Solving Flow Process Problem: PCB Dissapates 20W

[Lancelot59]

I'm attempting to solve the following problem:

A hollow PCB dissapates 20W into air which enteres at 32 degrees C, 1atm, and a volume flow rate of 800cm³/s. Find the average temperature the air leaves at.

I started by first simplifying the second law energy balance equation:

$$\frac{dE}{dt}+\Sigma_{out}m^{\cdot}_{e}(h_{e}+ \frac{1}{2} v_{e}^{2}+gz_{e})- \Sigma_{in}m^{\cdot}_{i}(h_{i}+\frac{1}{2}v_{i}^{2}+gz_{i})=Q^{\cdot}-W^{\cdot}$$

Since this is a steady state problem, energy rate of change is 0, and there are no effects from kinetic or potential energy. There is also no work being done, and the mass flow rates are equal. So I believe this simplifies to:

$$m^{\cdot}((h_{e})-h_{i})=Q^{\cdot}$$

From here I'm not sure what I can do to solve the problem. Was this the correct approach to begin with?

 

[Lancelot59]

Update: I tried running the ideal gas law using the volume flow rate, and after sorting the units I believe I can get the mass flow rate from that. What to do after that however, I'm not sure.

 

[LawrenceC]

What is a PCB? Can it be considered a vessel of some sort that is imparting the air flow with 20 watts? If it is, you have the correct equation. Q is energy per unit time. If you assume KE changes are negligible and express enthalpy as specific heat times temperature, the solution is at hand. Be sure your units agree when plugging in the numbers.

 

[Lancelot59]

In this instance PCB = Printed Circuit Board. I did manage to sort everything out. Thanks.