head and flow

[Nick C]

could someone explain for me the differnece and importance of head and flow? I sometimes think I've got the terms correct then someone throws me a curve...

and also, what are the main types of tubing,a nd where can I find out pros, cons, etc about them? I'll search in the forums here too.
thanx

[pHaestus]

Flow rate is in units of volume fluid moved per unit time. Head is usually used when referring to total pumping head, which is a measure of the total pressure that a pump must develop to produce a certain water velocity. Total head is the sum of the pressure from height differences, the pressure from friction losses in the loop, and the velocity head. You see "head loss" used over "pressure drop" by engineers and mfgrs because the numbers for head loss are not dependent on the identity of the fluid being pumped. For our purposes, pressure drop and head loss are interchangable I think.

So a pump's performance is quantified by a P-Q curve where the pump's total head is plotted vs. flow rate to yield how pump performance decreases as they are faced with more restrictive (higher head loss) discharge situations.

To further complicate things, flow rate (Q) is not what really is relevant to cooling; fluid velocity (Q/area) is what carries through in all heat transfer and turbulence calculations.

[murray13]

pHaestus said: ...flow rate (Q) is not what really is relevant to cooling; fluid velocity (Q/area) is what carries through in all heat transfer and turbulence calculations.

While that is true it is not totallly complete. Velocity is related to the rate that heat is transfered into a liquid, (how fast the WB will give up it's heat to the water). How much heat is transfered is related to the volume of water flow.

Think of it like this: if you increase the velocity you lower the deg C/watt, if you increase the flow you increase how many watts will be transfered.

One tiny little channel might get a very fast velocity but wont have that much flow so it will transfer heat very efficiently but cant move a whole lot of it. Conversly if you have one very big channel it's velocity will be small but it's flow will be big giving you a very inefficient transfer but it will move lots of watts of heat.

This is the biggest problem WB designers have in creating a high performance WB. High flow is easy to get all by itself. High velocity can be created (with loss of head). But doing both at the same time is a very tricky problem.

I know I went a little off topic but maybe this will help some of those out there who havent read up on fluid/thermodynamics.

Have a great day!

[BillA]

Originally posted by murray13
. . . . How much heat is transfered is related to the volume of water flow.
. . . .

wrong
gonna have to do some more reading there murray13

[myv65]

To expand on Bill's statement a little, how much heat is transferred is comparitively constant for a given system. This doesn't take much thought to deduce. Just ask yourself a question, "Does the rate of water flow have anything to do with the amount of energy my CPU is using?" The answer is a definite "No".

What does change is the total power put into the water, as creating higher flow requires more pump power. So long as the pump uses less than 1/3 the power of the chips being cooled, this is generally of no consequence. When folks start putting 200 watt Little Giant pumps or the large Iwaki pumps in there, it's another matter.

I think what you were trying to get at is that 1/2 the flow volume moving twice as fast would be more efficient than one unit volume moving at one unit velocity. Generally speaking this is a true statement. Note that creating these circumstances means changing the flow pathways, so construction of those pathways also plays a part. ie, you can make a pathway half as large and double the flow speed, but if the pathway doesn't work well with the die it's no good.

[murray13]

myv65 said: ....What does change is the total power put into the water....

And since water has a limited amount of thermal energy it can absorb per mass you need more mass to move more heat (without a phase change).

And doesn't the heat trasnfer formula use mass, specific heat and delta T???

q = m(DT)Cp

Where q = heat transferred, m = mass, DT = the change in temperature and Cp = the specific heat

[pHaestus]

I believe you are confusing concepts of kinetics with thermodynamics. With a larger volume of water, it will take longer for temperature to rise (dT/dt). With a constant heat source, however, the 2 different volumes will come to the same equilibrium temperature at steady state.

[myv65]

Originally posted by murray13
And since water has a limited amount of thermal energy it can absorb per mass you need more mass to move more heat (without a phase change).

Yeah, but look at the differentials that you see in water cooling. 23 gph carries a 75 watts heat load with a delta-T of 1°C.

If you ignore the energy of the pump (and some people continue to argue that pump energy is irrelevant, sheesh), you would simply have a 1/x relationship between flow rate and delta-T, essentially independent of actual water temperature. What would change versus "heat transfer efficiency" is the bulk water temperature. As flow velocity drops and "h" tanks along with it, the bulk water temp must go up to maintain the heat transfer.

[murray13]

I think I know where everyone is going, for a given WB if you increase velocity, flow must also follow. Giving a rise in pressure (head).

What I originally said is true. To calculate heat transfer you must know the mass. Velocity alone has no units of mass.

[myv65]

Actually, what you originally said was vague. And to calculate heat transfer in this instance, you do not need to know mass. There are many ways to calculate heat transfer. One option is to measure mass flow rate and delta-T and multiply by the specific heat. Another is to measure energy consumption directly. Given a choice between the two, you'd quickly find it's a lot easier to accurately measure electrical usage than the required properties to determine heat transfer via the water directly.

Irrespective of both velocity and mass flow rate, if a cooling system is able to operate in a steady-state condition then it is capable of removing all heat from the heat source. I believe this is likely the main reason why Bill stated you were incorrect. Specifically referring to your prior post, how much water flows has virtually zero bearing on the amount of heat transferred in a water cooled PC.

[pHaestus]

forced convective heat transfer is related to turbulence and surface area. Turbulence is directly related to water velocity, not flow rate, by the appearance of Reynolds number in calculation.

I think the term I am digging for is called the h coefficient (encompasses the S.A. and Re and some more terms) but I dont have any of that in front of me at the moment. I am betting that Dave or Bill know it much better.

Anyway flow rate is useful because it can be measured anywhere in the loop (constant) much more easily than velocity at a lot of points. Plus I assume water velocity can be estimated at any point with cross sectional area.

[Since87]

Originally posted by pHaestus

Anyway flow rate is useful because it can be measured anywhere in the loop (constant) much more easily than velocity at a lot of points. Plus I assume water velocity can be estimated at any point with cross sectional area.

Only the average velocity through a plane surface through which the whole flow passes can be calculated easily.

Areas through which only a part of the flow passes can have dramatically different different average velocities. (Consider eddies.)

An interesting image to consider. (Provided by roscal in this OCAU thread.)



Edit: Upon thinking about it more, I don't think it is necessarily easy to determine the average velocity even through an area which all the flow passes through. Considering the image above, it appears that it could be very difficult to predict.

[murray13]

Boy have I sure got a lot of people involved by not being able to chose the correct way of explaining myself. I understand what you guys are saying.

I was only looking at the WB itself disreguarding everything else. From the point of designing one that you can't test. (what I'm in the process of doing) I have all the formulas, done all the math, and understand what I'm trying to say. I just didn't do a very good job of it.

Sorry to get everyone mixed up in this misunderstanding. Thats one of the reasons I don't post very much. Never been very good at relating what I am thinking.

[bigben2k]

Murray13: Oh that's all right, I think we all learn a little bit more each time! I'm working on that PSU array today, got anymore input?

[Alchemy]

Originally posted by pHaestus
forced convective heat transfer is related to turbulence and surface area. Turbulence is directly related to water velocity, not flow rate, by the appearance of Reynolds number in calculation.

I think the term I am digging for is called the h coefficient (encompasses the S.A. and Re and some more terms) but I dont have any of that in front of me at the moment. I am betting that Dave or Bill know it much better.

Heat transfer problems like these will often begin by calculating the convective heat transfer coefficient h on all surfaces where there is significant heat transfer and the conductive heat transfer coefficients k through all solids where there is conductive heat transfer. The areas to which k and h apply must also be known. In cases like these where there can be a long chain of k's and h's - such as the radiator where there is convective transfer from the water to the tubes, conductive transfer across the tubes, conductive transfer along the fins, and convective transfer across the fins and tubes to air - this approach becomes very calculation intensive and error propogation becomes a major problem.

Anyway, if these things are done well, you can find the difference between the temperature of the CPU and the temperature of the air through the radiator by this relationship:

heat = UA (Tcpu - Tair), where

1/(UA) = 1/(kA,block) + 1/(hA,block) +1/(hA,water-radiator) +1/(kA,radiator tubes) +1/(kA, radiator fins) + 1/(hA, radiator fins/tubes-air)

That's the theory, but since most of these coefficients and areas are incredibly hard to determine, a technique called the Wilson method exploits these sorts of relationships to determine the coefficients.

Oh, also:

You see "head loss" used over "pressure drop" by engineers and mfgrs because the numbers for head loss are not dependent on the identity of the fluid being pumped. For our purposes, pressure drop and head loss are interchangable I think.

Head loss in a centrifugal pump is highly dependent on what sort of fluid is pumped. Head loss numbers are always for water unless otherwise stated.

Alchemy

[redleader]

And since water has a limited amount of thermal energy it can absorb per mass you need more mass to move more heat (without a phase change).

How is there an upper limit? Even if you dissallow a phase change you could, in theory, rachet up the pressure as far as is needed for infinate thermal energy to pass through a given amount of water. I'm sure eventually you'd hit practical limitations in preventing a phase change, but thats beside the point.

[Les]

Some sums.
Calculators* used for Jet sums, unfortunately, employ a "Head Loss" proportional to "flow"^1.75 power relationship.Unfortunately, because analysises of Billa data appear to show a quadratic relationship(eg http://forum.oc-forums.com/vb/showth...6&pagenumber=1 ).


http://www.jr001b4751.pwp.blueyonder.co.uk/ChJet1.jpg



* http://www.aps.anl.gov/asd/me/

Flomerics Jet http://www.coolingzone.com/Content/...las/fcalc10.htm
Kryotherm Channel http://www.kryotherm.ru/soft.htm

Edit: Have removed "Heat Tranfer Coeff"v "Pressure Drop"graph and reposted (below) using a quadratic relationship between PD an Flow.Have left link to graph.

[Les]

Have edited previous post.
Finally found a calculator for Nozzles http://www.pressure-drop.com/index.html . It has "Pressure Drop" proportional to "Flow" squared. Kryotherm also uses a quadratic relationship So PD conversion of the velocity and flow calculations for Jet Impingement and Channel cooling may have a little more meaning.
New plots for "h" v PD :


Think it only confuses further but............

[bigben2k]

Les,


I have to admit, I have a hard time interpreting this data.

Perhaps you could give us a brief interpretation, and what you conclude?

Otherwise, thanks for another great link!!!

[Les]

Ben
Would probably take me 10years to consolidate my understanding of Fluid Dynamics and Heat Transfer to produce brief description and conclusions.That is one of the reasons for presenting graphically.
The graphs are intended to maybe highlight areas worthy of further investigation.
Would treat all with caution (particularly the "h" v PD graph) and as indicators of the "maybe possible".

[bigben2k]

Well, I wasn't calling for an essay, but more of a description of what you graphed.

I had to look over it several times to notice that you used 5mm for Cathar's block, when he's reduced it to 3mm, if in fact it refers to the opening width of the nozzling plate.

Perhaps you could describe the seven different parameters you used, as in: what are they?

[Les]

"Mini - Cathar 8Ch(1x5mm) in 30x17mm Area" was intended to be self descrcriptive.
It refers to a Mini-channeled wb, of which the Cathar protype was an example, with eight 1mm wide and 5mm deep channels in an area 30mm long and 17mm wide.Being 17mm wide the fins would be 1mm thick. The calculated h here refers to flow through channel cooling as calculated by Kryotherm and expressed as W/m*m*c [1/(C/W) divide by the bp area(0.017m x 0.03m), where C/W is the Thermal Resistance of the wb with a 0.1mm thick bp].