Thermal Physics: Computer cooling

[StillAnotherDave]

Homework Statement

Estimate the inside temperature

Relevant Equations

W=Qh-Ql

Hello folks,
I have a bit of a time sensitive question, if anyone happens to read this soon.

A computer box consumes 80W power constantly. Outside temperature T= 290 Kelvin. Ambient pressure 10^5.

If all the heat generated is expelled at 5 litres per second, estimate the inside temperature?

I don't expect an answer, but could someone kindly guide me to which formulae I should be applying.

 

[willem2]

The heat is expellled by taking in air at 290 and expelling it at the temperature of the inside. You can use conservation of energy. Power in is power out.

(This might not not be too accurate. If the airflow passes a heat sink where most of the power is dissipated just before exiting, the outflow temperature of the air might be higher than inside temperature, but we have no details on the inside, so I don't think we can do better)

 

[StillAnotherDave]

By conservation of energy here you mean W=p1v1 - p2v2?

The question doesn't give the pressure inside the box. Are we assuming its the same as outside or does it need to be worked out? I'm wondering the dE=CvdT needs to be factored in, for a diatomic gas.

Would v1 be 5 litres? If so, what is v2?

 

[Delta2]

I think you should use the formula $Q=mC\Delta T$, ($\Delta T$ being our unknown which we seek to find).

Divide both sides by the time $\Delta t$ will give you at the left hand side the power of 80 watt (assuming all the power goes to heat) and at the right hand side you will get $\frac{m}{\Delta t}C\Delta T$, where $\frac{m}{\Delta t}$ is the mass flow rate which we are not given, we are given the volume flow rate instead which is 5litres/sec.

 

[StillAnotherDave]

I think you should use the formula $Q=mC\Delta T$, ($\Delta T$ being our unknown which we seek to find).

Divide both sides by the time $\Delta t$ will give you at the left hand side the power of 80 watt (assuming all the power goes to heat) and at the right hand side you will get $\frac{m}{\Delta t}C\Delta T$, where $\frac{m}{\Delta t}$ is the mass flow rate which we are not given, we are given the volume flow rate instead which is 5litres/sec.

Thanks. I approached it along these lines but failed to make the flow rate conversion!

 

[Delta2]

To be honest I don't know how we can make the flow rate conversion, we are not given the density of air and also where we use the ambient pressure is another question...

 

[Delta2]

The only thing i can think of is from a data table to find the density of air at a temperature of 290K (17C) and pressure of 10^5 Pa, and use that to do the flow rate conversion, assuming that the density of air remains constant.

Anyway if @Chestermiller can come here and enlighten us.

 

[Chestermiller]

The only thing i can think of is from a data table to find the density of air at a temperature of 290K (17C) and pressure of 10^5 Pa, and use that to do the flow rate conversion, assuming that the density of air remains constant.

Anyway if @Chestermiller can come here and enlighten us.

In my judgment, this is a problem in the application of the open system (control volume) version of the first law of thermodynamics. In this case, the computer box internals is the control volume (operating at steady state), and the 1st law applied to this system reads $$0=Q-\dot{m}\Delta h$$where Q is the heat flow from the electronics to the control volume, $\dot{m}$ is the molar flow rate of air passing through the control volume and $\Delta h=C_p\Delta T$ is the molar enthalpy change of the air, with Cp representing the molar heat capacity at constant pressure of air.

Treating air as an ideal gas, Cp is 3.5R and $\rho=\frac{P}{RT}$ is the molar density, P is the air pressure and T is the entering air temperature. So the molar flow rate is $\dot{m}=\rho \dot{V}$, where $\dot{V}$ is the volumetric flow rate of the entering air.

 

[Delta2]

@Chestermiller Q in the above equation is the rate of heat flow right i.e. it has dimension of energy/time=power right?

 

[Chestermiller]

@Chestermiller Q in the above equation is the rate of heat flow right i.e. it has dimension of energy/time=power right?

Yes. I should have used $\dot{Q}$.

 

[Delta2]

@Chestermiller is your solution essentially different than mine? I can see that the central equation is almost the same except that it uses molar flow rate instead of mass flow rate and $C_p$ (molar heat capacity) instead of C(specific heat).

 

[Chestermiller]

@Chestermiller is your solution essentially different than mine? I can see that the central equation is almost the same except that it uses molar flow rate instead of mass flow rate and $C_p$ (molar heat capacity) instead of C(specific heat).

I wanted to present it in a formal way, showing how it would be done formally using the open system steady state first law of thermo. The only thing I added to what you had is how to get the heat capacity and density for an ideal gas (without having to search for data).