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Unread 11-12-2002, 11:22 AM   #28
8-Ball
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Join Date: Feb 2002
Location: Oxford University, UK
Posts: 452
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Shaft01,

did you read my post?

If not, then please do and read the linked thread aswell.

If you have and don't get it, let me try and explain.


If you consider a single unit of water, 1ml for example, travelling through a radiator at X ml/m(millilitres per minute), it could transfer Y Joules of het to the air flowing through the radiator, agreed.

Now consider the same 1ml of water travelling through the same radiator, but this time at X/2 ml/m , ie half the flow rate. We can "assume" that this will now be able to transfer 2Y Joules of heat to the air, as it is in the radiator for twice the length of time.

Agreed

(NB I put "assume" in inverted commas as the actual heat transfered would be a little below 2Y. This is because heat flux is a function of temperature gradient. After losing Y Joules of heat in the first half of the radiator, the temp of the water will have dropped, lowering the temperature gradient, therefore lowering the flux of heat from the water to the air.)

Now here's the key part, taking each case in turn, how much heat will be dissipated for a constant flow of water.

Case 1

X ml of water per minute.
Each ml of water transfers Y Joules of heat to the air

Total Heat transferred per minute = X x Y = XY

Case 2

X/2 ml of water per minute.
Each ml of water transfers 2Y Joules of heat to the air

Total Heat transferred per minute = X/2 x 2Y = XY (2's cancel)

So would you agree that in a world where variation of temp gradient does not change the heat flux, and turbulence does not exist, both of these cases would have the same heat output.

Now, we all know that turbulence does play a part, and will allow the faster flowing water to dissipate a little more heat due to the break down of the planar flow at the interface.

Therefore, case 1 will actually be able to throw out X x (Y plus a little bit)

And,

As I said before, temp gradient does affect the heat flux so in fact, case 2 should dissipate X x (Y minus a little bit.

Ergo, by this logic, the argument of "keeping the water in the radiator for longer allows it to give out more heat" is not valid.

Someone please tell me if I'm spouting a load of horse sh*t

8ball
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