Find my total W error in my calculations please

[Brwmogazos]

hi guys...

i have been doing some calculations to find W, the amount of heat dissipated by a radiator.

As you all know
W = m * Cg * (Toutlet – Tinlet)

where W = Watts
m = mass flow rate, kg/s
Cg = specific heat capacity,for H20, 4186 J/(Kg°C)

So W = (lpm * 4186 * ΔT) ÷ 60
according to this link

http://thermal-management-testing.com/ThermoChill.htm

I guess the equation to calculate W is divided by 60 to find the amount of W/minutes ratio?

i have been trying to calculate the heat dissipation for my waterchill rad.
my pump is the hydor L20 which has an 700l/h rating
so m= 700l/h or 11.66l/minute

Cg is 4186 and ΔΤ is say 5C (between the cpu temp and the water temp in the intake of the radiator)

now if i calculate W i get 11.66*4186*5 =244043W

looking at the equation of W he then divides it by 60 which is 4067

So W = 4067W per minute?

Am a bit confused here...can someone enlighten me?

looking at the graphs from that site though he gets:



obviously no where near that... so what am i doing wrong here?

[pdf27]

I watt is 1 joule per second. Divide 4067 by 60 again, and you get 67.8W. This is of the right order of magnitude.
Problem is, that graph is for a 5 degree coolant-air temperature difference - not a 5 degree water temperature difference across the radiator. Hence, the two aren't directly comparable.

[Brwmogazos]

thanks mate...

what i am trying to do is compare these results with the results taken from my bios...

i know the bios isnt something accurate..i am just trying to do a comparison...

what i am thinking of is...

Calculating the W of my cpu when it is overclocked...

then substitute that W to the equation of W=*m*Cg*DT where i know the Tin due to a thermometer i have in the reservoir.

Calculate the T out from that and then compare it to my bios results...

would that be correct?i know that the pump after the reservoir will generate heat into the water before going into the radiator...but i am just trying to find a way of comparing "experimental" data from my bios with calculated data from my equations...

another thing

if i use Tin the water temp into the res...and Tout the cpu temp...would that be correct to find W for the radiator?

thanks again in advance...i really need your thoughts-opinion in both these 2 questions of mine.

thanks again for your time.

[pdf27]

would that be correct?i know that the pump after the reservoir will generate heat into the water before going into the radiator...but i am just trying to find a way of comparing "experimental" data from my bios with calculated data from my equations...
That would give you a reasonably accurate figure (probably to well within one degree, depending on the pump used) for the heat difference between the CPU heat sensor and the CPU water block outlet temperature. I'm not sure exactly what you can do with that however, unless you're comparing thermal grease or something.

if i use Tin the water temp into the res...and Tout the cpu temp...would that be correct to find W for the radiator?
No, as that assumes the outlet water temperature is going to be the same as the CPU sensor temperature. This is highly unlikely.

The problem you've got here is insufficient information. If you had a fairly accurate flow meter and a pair of accurate thermocouples (one each at inlet and outlet to the CPU block) you would be able to calculate accurately the heat dissipation of the CPU. You should be able to get these temperatures by moving the thermometer you already have around the circuit, and the flow rate is probably best got from one of the pump head curves on here - you're unlikely to get your hands on a decent calibrated flow meter easily.
You can repeat the process for the radiator to figure out how much heat you're dumping to it fairly easily. The difference (probably very little) will be what you are adding to the loop from the pump and the like.

[bigben2k]

PDF27 has it right: your temp probes should be measuring temps at the water block inlet and outlet.

What's not obvious here, is that the wattage that your going to calculate as a result, is achieved at a state of balance (i.e. the loop has been running stable temps for a while), and that wattage is what's actually dissipated by your cooling solution. If your cooling solution were more efficient for some reason (like turning up the fan), your measurements would be different.

As PDF27 also pointed out, it's not easy to measure this. We're talking about temp differences in the order of 0.1 and less, especially at higher flow rates. Because of that, it's really critical that all your measurements, including flow, are as accurate as they can be.

I'm opting to use Platinum RTD, and I'm shooting for 0.001 deg C resolution, which should give me a narrow enough margin to make a somewhat accurate measurement. I haven't picked the flow meter yet, but I'm leaning towards a magnetic unit. Anything that can give me 1% accuracy or better, but isn't outrageously expensive.


PDF27: you need to look into the WBTA (in my sig): you'd be a welcome member! (end shameless plug)

[pdf27]

Ermm... possibly not as welcome as all that.
I'm pretty much only theoretical in that I've never even seen a watercooled PC in real life. I'm just hanging around here trying to figure out how to build my next PC (watercooled, heavily silenced) after I graduate and get a job.
I'm at university finishing off my MEng in Aeronautical engineering (basically Jet Engines) so simple thermodynamics like that is a doddle. However, being as I have no experience whatever with waterblocks then I really don't have a lot to contribute. Besides, my exams start in two weeks so I should really be working.

Edit: just a thought, but for heat flows it may well be easier to measure the air flows/temperatures rather than the water flows. The temperature differences and volumes will be higher in both cases...

[MadHacker]

Originally Posted by Brwmogazos

i have been trying to calculate the heat dissipation for my waterchill rad.
my pump is the hydor L20 which has an 700l/h rating
so m= 700l/h or 11.66l/minute

so what am i doing wrong here?

OK so the math is NFG to me... (suck at it) but one thing i do know is that even though your pump does 700l/h is that what it is doing after you have your CPU block and Rad pluss all the hose in there? don't think so...

bigben2k menitoned that having acurate flow rate would be inportant to this calculation...

i could be wrong... like i said i sux at math... (1+1 = 3)

[bigben2k]

Originally Posted by pdf27

Ermm... possibly not as welcome as all that.
I'm pretty much only theoretical ...

Edit: just a thought, but for heat flows it may well be easier to measure the air flows/temperatures rather than the water flows. The temperature differences and volumes will be higher in both cases...

I'm theoretical too, but it didn't stop me from starting the WBTA...

Temp measurements of air may be tricker, because the difference (air in vs air out) may be very, very difficult to measure, especially under high airflow rates.

[OvcA]

4067 W seems to be the right number (just W, meaning J/s, not W/min) although it is obviosly wrong. I realy do not see the poin in dividing by 60. As it stands we have:

P = (m/dt)*c*dT = (700 l / h) * (1000 kg / m^3) * (4186 J / kg K) * 5 K = (700 / 3600 l/s) * (1 kg/l) * (4186 J / kg K) * 5 K = 4070 J/s = 4070 W

If we acount for the reduced flow we get a result in the order of magnatude of 10^2 which is still quite considarably off. I can't imagine a CPU producing more than 200 W of heat while being only water cooled. We than have to take into account the losses from heat transfer and secondary heat paths which means that the flow through the radiator would have to be about 35 l/h netting a whooping 95 % flow drop. A bit to much I should think.

[prandtl]

Originally Posted by OvcA

4067 W seems to be the right number (just W, meaning J/s, not W/min) although it is obviosly wrong. I realy do not see the poin in dividing by 60. As it stands we have:

P = (m/dt)*c*dT = (700 l / h) * (1000 kg / m^3) * (4186 J / kg K) * 5 K = (700 / 3600 l/s) * (1 kg/l) * (4186 J / kg K) * 5 K = 4070 J/s = 4070 W

Taking those value, 4070W is the good result. The problem here lies in the flowrate and the value of dT. No way the temperature difference between inlet and outlet is 5°C (or K).

[pdf27]

Ah, OK - his units were in such a mess that I just assumed he'd got a /60 in the wrong place.
Look back at the assumptions again - you're assuming that even at 700 l/h water flow through the water block you're going to get a 5 deg C temperature difference across the block! That's totally unphysical - 0.2 deg C is more like it. To heat that much water, you're going to need a hell of a lot of heat - 4kW isn't unrealistic.

Note: Numbers do add up right now I take the time to go over them properly - 0.19 kg/s x 4 kJ/kg x 5k.
hf for water is about 4 kJ/kg at 300k according to my steam tables anyway, so that one got looked up right. Basically, your mass flows and temperature differences are WAY out of reality.

[pdf27]

Originally Posted by bigben2k

Temp measurements of air may be tricker, because the difference (air in vs air out) may be very, very difficult to measure, especially under high airflow rates.

Not so sure about that one - you need roughly 3,500 times the volume of air as of water to get the same temperature change @ 300K. I would personally suspect (not having tried it with water) that it would be easier to duct fairly cold air over the rad to control the outlet water temperatures at lowish velocities. This would give you a reasonably large delta T to measure as well as making the flow rate easier to measure (there are plenty of air flow measurers out there for sailing and the like, but even if you use a simple pitot-static arrangement it's pretty easy to do).
Remember if you're testing water blocks then getting the peak performance from the radiator isn't all that important - you really only have to be able to attain a consistent block inlet temperature/pressure drop across block. Aside from that the block doesn't know or care about airflow over the rad.

[Brwmogazos]

Some good feedback you gave me there guys thanks...

I will leave my theroretical calculations as they are then...then discuss about the big errors in my calculations due to the flow rate loss and temp difference...

could someone give me any info about what flow rate loss i can expect? someone mentioned the real flow rate will be about 35l/m?

can someone reference that with a link or something please?

+ the temp difference wasnt right at all i know...i never expected it to be that HUGE, i will comment on that as well...
again has anyone seen any reviews with water temp measurements before and after the rad?

i bought myself an aquarium thermometer and my max water temp in the waterchill res reached 39.8C under load and the minimum was 18.2C which was when i started pooring water into the watercooling setup so its totally innacurate...

IIRC the idle water temps before my pump were about 32C i aint too sure though...

[prandtl]

Originally Posted by Brwmogazos

could someone give me any info about what flow rate loss i can expect? someone mentioned the real flow rate will be about 35l/m?

that's 2100 liter per hour
Haven't you read what we wrote yesterday? Your flow will be lower than the actual pump rating. The pump use in watercooling are velocity based pump, not volume, so the flow is not a constant vs the pressure.
Besides, just use your creativity and measure your flow yourself (foe example: use a one gallon jug and put the outlet tube of the rad in it, measure the time it takes to fill it... not really hard).

[OvcA]

The 35 l/h was just a fit for the calculations. Depending on the P/Q curve you should expect a 70% - 50% flow drop.
I imagine that you have already read the related articles here. You should also take a look at BillA's work over at overclockers.com.

On the subject of measuring flow. You should get a decent estimate (let's say a 10% error) if you make relative measurments. First measure the flow at zero head pressure. Now you have a reference point for which we also know the real flow (rated value of the pump). All subsequent measurments (at different head pressures, meaning also different flow rates) can than be express in relevance to this value.
The only problem is that in our case the flow resistance increases by the square of fluid velocity. This hols true if Reynold's number is > 1000 (between 1 in 1000 it's neither linear nor square).

Let flow be 500 l/h and hose diameter 1/2" = 12cm

v = (dV/dt) / S

v - avarage water velocitiy
S - hose cross-section
dV/dt - volume flow

Re = [ro] * v * d / [ni] = [ro] * (dV/dt) / S * d / [ni]

Re - Reynold's number
[ro] - desity of water (1000kg/m^3)
d - fluid layer thickness (equal to the hose diameter)
[ni] - viscosity of water (10^-3 Ns/m^2)

Re = (1000 kg/m^3) * (500/3600/1000 m^3/s) / ([pi] * (0.06m)^2) * 0.12 / 10^-3 = 1473

Unfortunatly the square aproximation (even worse with low flow rates becouse Re can in that case drop well below 1000) will in the end account for most of the error.

[Brwmogazos]

Originally Posted by prandtl

that's 2100 liter per hour
Haven't you read what we wrote yesterday? Your flow will be lower than the actual pump rating. The pump use in watercooling are velocity based pump, not volume, so the flow is not a constant vs the pressure.
Besides, just use your creativity and measure your flow yourself (foe example: use a one gallon jug and put the outlet tube of the rad in it, measure the time it takes to fill it... not really hard).

Easy mate...

it was a typo... i meant 35l/h not /minute...

ovca noticed straight away...

thanks for the tip...i will try and measure the flow myself.

[Brwmogazos]

Originally Posted by OvcA

The 35 l/h was just a fit for the calculations. Depending on the P/Q curve you should expect a 70% - 50% flow drop.
I imagine that you have already read the related articles here. You should also take a look at BillA's work over at overclockers.com.

On the subject of measuring flow. You should get a decent estimate (let's say a 10% error) if you make relative measurments. First measure the flow at zero head pressure. Now you have a reference point for which we also know the real flow (rated value of the pump). All subsequent measurments (at different head pressures, meaning also different flow rates) can than be express in relevance to this value.
The only problem is that in our case the flow resistance increases by the square of fluid velocity. This hols true if Reynold's number is > 1000 (between 1 in 1000 it's neither linear nor square).

Let flow be 500 l/h and hose diameter 1/2" = 12cm

v = (dV/dt) / S

v - avarage water velocitiy
S - hose cross-section
dV/dt - volume flow

Re = [ro] * v * d / [ni] = [ro] * (dV/dt) / S * d / [ni]

Re - Reynold's number
[ro] - desity of water (1000kg/m^3)
d - fluid layer thickness (equal to the hose diameter)
[ni] - viscosity of water (10^-3 Ns/m^2)

Re = (1000 kg/m^3) * (500/3600/1000 m^3/s) / ([pi] * (0.06m)^2) * 0.12 / 10^-3 = 1473

Unfortunatly the square aproximation (even worse with low flow rates becouse Re can in that case drop well below 1000) will in the end account for most of the error.

So with such big variations between the theoretical and actual flow rates and temp difference can u get the reynolds number as high as needed to prove the flow would be turbulent in any watercooling setup nowadays?

[OvcA]

Originally Posted by Brwmogazos

So with such big variations between the theoretical and actual flow rates and temp difference can u get the reynolds number as high as needed to prove the flow would be turbulent in any watercooling setup nowadays?

Re below 2000 brings about laminar flow, while above 3000 means turbolent flow is predominant. One has to take into account that there will always be a (albeit very thin) boundary layer of laminar flow becouse of surface tension. What happens between 2000 in 3000 is that laminar flow is gradualy breaking up into vortexes persisting more frequently along the sides.

Re = [ro] * v * d / [ni] holds true for flow through a hose. Things do however get far more complicated when one puts an arbitrary object (say a waterblock) into the path of the flow. Now some particles scatter (depending on the shape of the object) creating more vortexes as they collide with each other. This phenomena is used for inducing turbolent flow as can be seen in most of moderen waterblocks.

To get back to your question, I do not see any elegant way to mathematicaly prove that flow in the waterblock is turbulent. One can only prove that laminar flow is predominant in the hoses.

[Les]

OvcA
Yep, Re ~ 1450 for 500l/h in 12cm ID tube, but Re~ 14500 in 1/2"(12.7mm) ID tube.
Pointing out in case others missed the slip/typo, not trying to score points.

[pHaestus]

English units strike again, eh?

[OvcA]

Originally Posted by Les

OvcA
Yep, Re ~ 1450 for 500l/h in 12cm ID tube, but Re~ 14500 in 1/2"(12.7mm) ID tube.
Pointing out in case others missed the slip/typo, not trying to score points.

Tnx mate. That was a nasty mistake.
In light of this find you should disregard my statements about flow through the hoses being laminar. On the bright side we can atleast safely use the square aproximation for calculating flow resistance.

[Les]

Nasty no,bit silly yes.
I constantly make similar. I can only correct when am in familiar territory and can recognize the answer is not sensible