alk

Enthalpy, simply put, is the internal energy of the fluid in your loop PLUS the flow work. If you have a glass of water, at some temperature x, then it has an internal energy. Now lets say your watercooling loop is flowing and at some point in the loop the water is at temperature x (Same temp as before). Yes, the internal energy is the same, BUT the water is flowing! So what we get is something along the lines of H = U + VP where H is the enthalpy, U is internal energy, V is the volume, and P is the pressure.

Now thats all very well, but out system is flowing. How do we have a defined volume or internal energy at that point? Well we have to use specific quantities, signified by lower case quantites; h, u, v. A specific quntity is a quantity/unit mass. Eg. h = J/kg

The other thing to note is that in a steady flow system such as your WC loop, the mass flow rate will determine the properties of your loop.

Now, what I should have mentioned a wayyyyy back at the start is the Steady Flow Energy Equation. This is:

Q-W=dH (+dKe+dPe)

Heat input - Work input = Change in Enthalpy (plus change in kinetic energy plus change in potential energy)

Ke and Pe are usually pretty small in compraison to the enthalpy, so we can disregard them in most cases (Some exceptions are in high speed turbines and compressors)

Often we hear of people talking about "Heat Dump" from their pumps. This is where the inefficiencies in the pump generate friction in the propellor bearins which heats the water and also increase the enthalpy of the water through the flowwork.

We also need to consider what a pump is. Sounds obvious right? Well, a pump, by defenition, is a device that moves a fluid from a region of low pressure to a region of higher pressure by adding energy to the fluid.

In your WC loop you have restrictions - waterblocks & radiators. These act as non-isentropic throttling valves (this means there is heat added to the fluid during the throttling process). The pressure of the fluid out of the block will be less than the pressure into the block. If we consider the waterblocks, and radiator in a loop as throttling valves we can look specifically at the pump and the difference in pressure between the inlet, and exit pressures.

So lets paint a scenario. A laing D5 vario pump is operating at its maximum setting, consuming 23W. It is connected to a D-Tek Fuzion WB.

The water inlet temperature has just exited the radiator (we will assume the radiator induces no flow restrictions) and is at 25 degrees C.

So we know that:

Pump Power= 23W
Tin = 25 degC
vwater = 1/rhowater = 0.001 m^3/kg

Heres a flowchart for the D-tek Fuzion:


For this we will assume that the the D5 can pump 3gpm through the block.

This translates to a pressure drop accross the block of 4.4 PSI.

I like working in SI units, so this means our values are:

dP = 30.34 kPa
m. = 0.1894 kg/s = 189.4 g/s

Our enthalpy can be calculated using:

dh ~= v*dP = -Specific Pump work
= 0.001(30.34)
= 0.03034 kJ/kg = 30.34 J/kg

dH = m. * dh
=0.1894 * 30.34
=5.7464 W = Pump work.

Since we know how much energy we are putting into the pump (23W) we can calculate the heat dump:

Heat dump = Input Power - Pump Work
= 23 - 5.75
= 17.25 W

Heat dump = m.*Cp*dT
17.25 = 189.4*4.2*dT
dT = 0.02169 degC

T2 = T1 + dT
T2=25.02 deg C

As you can see, even with the pumps MASSIVE inefficiency (Nth = Wout/Qin= 5.75/23 = 25% efficient) there is still only a marginal increase in temperature.

Obviously, this only takes into account a loop with a D-Tek Fuzion restricting it. As we add more restrictions, our dP increases and our pump becomes more efficient - in other words : more restrictions = less heat dump in a frictionless environment (thanks Martinm210!). If we consider flow friction, then the increased friction between the pump propellor blades will add more heat. IE. More of the pump work is transfered into heat instead of work. In this case we assumed that dH = - Pump Work!

Using the methods I used in this post, you can also work out how much heat your WB is cooling and also how much heat your radiator is disipating.

Hope this post was insightful for you enthusiasts!

Martinm210

Thanks for the post, it will take me some time to absorb this though..

Been over 10 years now since my college days...

The heat dump is less as restriction increases still doesn't click for me though, I always understood there was a portion of pump heat dump not necessarily caused by the motor efficiency, but instead by the impeller/water friction and internal losses. The same should hold true throughout the system. Pressure drop is itself a good measurement of energy lost for a given flow rate, so it would only make sense to me that the engery loss has been converted to heat.

It also wouldn't make sense that pumping into an extremely restrictive system would create less heat dump. Imagine a closed valve directly at the pump outlet and the internal water friction between the impeller and water within...that can't be an efficient condition...

Anyhow, just trying to get my hands around this...I never did quite understand heat dump, but I have read through most of Cathar's comments.

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rottenotto

Hence, as the passage becomes more restrictive, the increase in friction adds more heat........... that's the way I read it, anyway.

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Wankerfx

I thought the enthalpy equation was:

q=mcΔT

q is the amount of heat in J/KJ
m is the mass
c is the specific heat capacity of the object (J/g.ºC or KJ/Kg.ºC)
ΔT is the temperature difference

It looks like I have a few more things to learn after Grade 12.

Awesome informations by the way.

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alk

well, no and yes. The block is made of copper, induces a restriction, and the heat added is the heat from your CPU/GPU, etc.

alk

Well you see enthalpy can be an increase in flow work or internal energy. If the water gets hotter due to friction, this is an increase in enthalpy. Similarly, in a less restrictive loop, and increase in flow work is also an increase in the enthalpy. So you see, in both scenarios, the enthalpy increases. The thing that will happen in a restricitve loop is that the flow rate will drop, which will of course reduce your flow work. I made a big assumption saying that the change in enthalpy was the same as the pump work!

Nice post tho! made a few corrections here and there!

ira-k

Well that gave me a head-ache......Thanks for the work....Don't for get you have to take laminar and turbulent flow into account in a WC'ing loop to for heat dissipation....

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Martinm210

FYI

Here is some info from an XS member on heat dump calculations regarding pressure drop:

I have to admit, even though I've been accused of having a mental condition in liking this stuff too much, I still do not completely understand how heat dump works in a water cooling loop.

Maybe when I start getting a full night's sleep, one of these days it'll all make sense to me. For now I keep thinking heat dump "For the system" is the same (or nearly insignificant) no matter the restriction or somehow related directly to the pump efficiency curve regardless of restriction.

PUMP ONLY
Obviously in a zero pressure or restrictionless condition, the pump itself would transfer that work energy in to kinetic energy (Water Flow Rate) which would be it's most efficient and least amount of pump heat dump (From water friction, perhaps likely not from a motor RPM state). The same would hold true at zero flow rate with a closed valve at the outlet. If water is not moving at all, all of that pump work energy has to go somewhere, and it must be lost in the form of heat and your net result would be the most heat dump (Water/impeller friction).

BLOCK ONLY
While a block does not add energy, all that restriction and what we commonly measure in terms of pressure drop is nothing more than a measurement of energy loss relative to flow rate. I suppose you could consider in this effect that pressure drop is also a good measurement of water block heat dump. Energy is lost, so it must be in the form of heat..

SYSTEM AS A WHOLE
So in the end, I think what we're most interested in is the net result of the whole system that is introduced. So in a more free flowing system, the pump curve would be riding on the right side where the added pressure energy by the pump is lower, but the added flow rate energy is higher. And because it's a closed loop system it will find a fixed flow rate, one where the energy lost due to friction is equal to the energy that can be added by the pump. So I'm wondering if it would make sense that less restrictive loops would have less heat dump by the pump and more heat dumped by the restriction and pressure drops throughout the system?

In the end it would seem to make sense that on a closed loop system it's a bit of a wash regarding "Heat dump of the system". Heat will dump either at the pump or in the system restrictions if the energy input was equal. Although I have to think under a more restrictive system, than the heat generated by the motor that is dumped will increase as restriction is increased.

Perhaps it all depends on the pumps efficiency curve, and the net result could actually go either way depending where you are on that efficiency curve?

Sorry for talking out loud, these are just my random thoughts in trying to understand this better...

Something keeps nagging at my side that the pump efficiency curve is key to understanding how "system heat dump" changes regarding restriction and everything else is moot because it's a closed loop.

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