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Replacing liquid with air in a liquid-cooled heat sink

Replacing liquid with air in a liquid-cooled heat sink

Replacing liquid with air in a liquid-cooled heat sink

(OP)
Folks-
I soon have to deal with a design where we'll try to use a liquid-cooled heat sink to cool another design having a lower heat load using air at a higher mass flow rate.

If a detailed analysis was done on the liquid-cooled heat exchanger with heat load Q1 using a liquid having thermal diffusivity alpha1, mass flowrate Vdot1, drawing liquid in at Tin1, and exhausting fluid at Tout1, is it approximately true that:

Tout2-Tin2
=(Tout1-Tin1)[(Q2)(alpha1)(Vdot1)]/[(Q1)(alpha1)(Vdot1)]

where
Tout2= the outlet (exhaust) air temperature
Q2= the heat load with the air-cooled setup
alpha2 = the thermal diffusivity of air,
Vdot2 = the mass flowrate of the the air, and
Tin2 = the inlet temperature of the air

????
Or am I oversimplifying things if radiation and natural convection are ignored?



Tunalover

RE: Replacing liquid with air in a liquid-cooled heat sink

tunalover,

Looks like you have an error in your formula:

Fluid temperature rise = Q / Cp Vdot
where Cp = specific heat

Warning - This only gives the temperature rise of the fluid.  It's of little value in determining the heat sink temperature, which is usually the more important value.  I'm afraid that solution is much more complicated.

ko  (www.ecooling.biz)

RE: Replacing liquid with air in a liquid-cooled heat sink

(OP)
I meant
Tout2-Tin2
=(Tout1-Tin1)[(Q2)(alpha2)(Vdot2)]/[(Q1)(alpha1)(Vdot1)]

Is that better?

Tunalover

RE: Replacing liquid with air in a liquid-cooled heat sink

The liquid dT doesn't help you predict the air dT since, among other things, the flow is different. Also, the energy equation uses specific heat, not thermal diffusity:  
Q = Mdot (Cp) dT

Using your terms
Tout2 - Tout1 = Q1 / (Mdot1 Cp1)

I'm using Mdot for mass flow to avoid confusion with volume flow.

Hope this helps

ko  (www.ecooling.biz)

RE: Replacing liquid with air in a liquid-cooled heat sink

>>????
>>Or am I oversimplifying things if radiation and natural >>convection are ignored?

  In forced flow natural convection is not really a concern unless you were dealing creeping flows (really slow moving flows). Otherwise, the heat transfer due to the forced convection is >> any natural convective heat transfer.
  Radiation does not need to be considered in cases were the temperature difference between the object radiating and the the object recieving radiation is large.
   

RE: Replacing liquid with air in a liquid-cooled heat sink

Edit: Radiation does not need to be considered in small temperature difference cases.

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