Post Rad Chiller Concept

[Martinm210]

I'm pretty rough on the watts and thermal side of things.

But if I figured this out right, it takes about 395watts of heat energy to change a 1.5 GPM flow rate by only 1 degree celcius.

I could be wrong, but this is what I used:

1.5gallons per minute.

1.5 gallons = 5,678 milliliters = 5,678 grams

Divide by 60 for one second

= 94.6grams water / second

q=(specific heat of H20)X(grams of H20) x dT
=(4.18 J/g-K)x(94.6 grams H20)x(1K)
=395 Joules

A Joule is a watt-sec and we were measuring for one second, so that 395 watts.

And that's assuming 100% efficiency.

Probably need to go with the bigger TECs?

That seems like alot of TEC to change the water temperature though...Maybe I figured something wrong..?

Yeah I think that's wrong, I'll go digging some more...

[NickS]

I'm pretty rough on the watts and thermal side of things.

But if I figured this out right, it takes about 395watts of heat energy to change a 1.5 GPM flow rate by only 1 degree celcius.

I could be wrong, but this is what I used:

1.5gallons per minute.

1.5 gallons = 5,678 milliliters = 5,678 grams

Divide by 60 for one second

= 94.6grams water / second

q=(specific heat of H20)X(grams of H20) x dT
=(4.18 J/g-K)x(94.6 grams H20)x(1K)
=395 Joules

A Joule is a watt-sec and we were measuring for one second, so that 395 watts.

And that's assuming 100% efficiency.

Probably need to go with the bigger TECs?

That seems like alot of TEC to change the water temperature though...Maybe I figured something wrong..?

Oh god, SHUT UP!

I just took a test on q=MCdT in chem yesterday. It's 2AM saturday morning, I'm not supposed to be reading this!!!

[Martinm210]

Oh god, SHUT UP!

I just took a test on q=MCdT in chem yesterday. It's 2AM saturday morning, I'm not supposed to be reading this!!!

Well you're all fresh on this, I've been out of school for WAY too long..getting old and crusty...

I did find what you just said though on this french radiator review article:

Q = Mx Cp x dT (Equation 1)

Q = power dissipated by the radiator in W (calculate)
M = mass flow of water kg / s (measured)
Cp = heat capacity of water J / kg K, it is 4186 J / kg K here (known)
DT = temperature difference between the entrance and exit of the radiator for Water in K (1 K = 1 ° C) (measured)

So at 1.5GPM, you need 395 watts to change the temperature 1 degree...that's ALOT!

But I'm sure once you've exceed the heat dissipated by the parts getting cooled it would continue to cool down until it reached the ambient then it would start stabilizing by the cold loop radiator?