What volume of water is there in your loop?

[fornowagain]

My loop is currently running at 1.5GPM. Anything under 1 GPM, I would consider marginally slow tbh. You'll see resistance or temperature vs flow plots start to curve upwards below one GPM, especially impingement blocks.

As a worked example its supposed to help someone do their own calcs and I prefer whole numbers for clarity in the base equation, and I don't use '1' unless its deliberate or I'm not paying attention lol. I also try and keep the derivative values high enough to avoid 0.0x numbers as these always cause problems. The 4GPM value was supposed to be very high to make a point.

I apologise for biting its not like me, I did remove it. Anyway just for you.

Example.

A computer WC rig with a flow of 0.75GPM & 7/16" bore. The water moves at 0.49m/s. A 250W heat load. Water at 22C has a density of 997 kg/m3 giving a mass flow rate of 0.047 kg/s, specific heat capacity for water is 4186 J/(Kg°C)

250 = 4186 x 0.0474 x dT

Therefore dT = 1.26°C

If you doubled the flow to 1.5GPM? Fluid velocity increases to 0.975m/s, mdot (mass flow rate) increases to 0.094 kg/s.

Therefore dT = 0.63°C

So for the 250w going into the cpu block, say with a coolant inlet temp of 20C, the outlet is 21.26C. If you seriously increased flow by doubling it you only get a marginal improvement of 20.63C

Might as well make use of the second example, if you look at the values you can see that as the flow rates increase you see diminishing returns pretty quickly. If you drop below the 1 GPM mark, temp differences across radiators/blocks becomes more significant.

 

[pastymuncher]

Just under 2L of Feser in mine. Aquaextreme 50z pump, Aquaextreme MP-05 Pro Limited Edition CPU block, Aquaextreme MP-1 GPU block, XSPC 250mm passive res, Thermochill PA120.3 (new version) about 10 feet of 7/16" tubing and 2x six foot lengths of 15mm copper pipe. Radiator is in a rad box mounted on a windowsill outside my computer room.

 

[JonJ678]

Useful equations once again, thank you. I think you missed the point slightly though, or at least missed my point.

I'm with you on mass of water affecting time to equilibrium in the obvious way, and on radiated heat from the reservoir being potentially significant. Say, if the reservoir is large and finned or hanging out the window in winter.

I'm also in agreement regarding flow rate, as I think Paul is. However, what effect does a reservoir of various sizes have on flow rate?

In a closed loop, as long as the peripheral resistance is low enough, I'd expect the pump to start the water moving slowly then swiftly accelerate it up to the equilibrium mass flow rate. However, if you put say a swimming pool partway down the loop, the water entering the pump is essentially stationary and that returning to the pool loses all its kinetic energy to the damping effect of the pool.

Is it reasonable then that any reservoir will impose resistance to flow, and larger/worse designed ones will have a bigger effect? The xspc reservoir top (laing ddc) for example has a tube running through the centre aiming the incoming water at the pump inlet. This reduces the drag from the reservoir liquid, and perhaps explains the measurable performance increase over separate reservoirs and tops.


Shadowscotland is of the opinion that since pressure in the reservoir is uniform, as long as it is full it makes no difference to flow rate. However that a bubble will reduce flow rate, hence uses a completely full reservoir. Whereas I'm working on the basis that when spraying a hose into a swimming pool, you can feel the pressure from the hose a short distance below the surface but it swiftly disappears with depth, and pumps are likely to perform better when fed with flowing water than when being forced to accelerate it continually.

I would be interested in peoples views on this, and hope I have expressed myself clearly enough. This looks offtopic I know, but at least its my thread I'm derailing

 

[fornowagain]

Yes that's correct. There are losses from using a reservoir. Quite simply from the conservation of energy and loss of fluid momentum. Probably the best way to calculative would be velocity triangles. Assume the inlet velocity has reduced to zero at pump inlet then the entire vector is across the pump blade. If any momentum is preserved say from the outlet jet facing the pump inlet then a velocity vector becomes a component to the final annular momentum. Does that make sense? Its a little difficult to picture without a diagram. So yeah design can have a factor.

Remember also if you stall a pump its not about fluid velocity, but pressure. Water flowing through a pipe, has 3 types of energy: Kinetic energy, potential energy from gravity and energy by virtue of pressure. Energy must be conserved, the fluid energy can't change as it flows through the hose (ignoring frictional losses, as important as they are). If the loop is horizontal, the fluids potential energy from gravity doesn't change. If the incompressible fluid enters a narrow section the fluid must increase velocity to avoid slowing the fluid behind. This increased velocity has an increase in kinetic energy, but the total energy can't increase so the potential energy from pressure decreases and visa versa. Bernoulli. Now if the loop rises upwards the fluids potential energy from gravity rises, being incompressible the velocity remains the same so the pressure must drop. So one form of energy can change into another, but the total sum at any point remains the same. Excluding losses.

 

[shadowscotland]

thanks fornowagain - that's very kind of you to run the numbers for me
@ jon - with a fully filled res I was working on the coolant being a non-compressable liquid - aka one unit in a full res = one unit out. (sealed loop)

and pumps are likely to perform better when fed with flowing water than when being forced to accelerate it continually.

There are losses from using a reservoir

Completely agree

but a compensating reservoir also has gains.
Tall res with bubble - flow into top of res is due to pump - flow out of bottom is due to gravity (ideally these should be equal)

Edit: please shoot me down if i'm taking bull

 

[JonJ678]

Reassured to hear it, thank you. Sad to say I'm not doing well at the calculative side of things, but shall improve with time. For now I'm happy knowing I was on the right track.

@pastymuncher, good effort on the exterior radiator. Do you hit issues when summer is hot or winter below ambient? Unsure when feser freezes

 

[pastymuncher]

@pastymuncher, good effort on the exterior radiator. Do you hit issues when summer is hot or winter below ambient? Unsure when feser freezes

Don't know as i only started using the Feser a couple of months ago when i fitted the new Thermochill. For the past four years i just used distilled water and zerex and had no problems. Winter temps were in single figures on the cpu!! No problems with condensation either. I think i may go back to distilled water again when the Feser needs replacing. Bugger this changing liquid every year lark. Previously i had run the loop for 3.5 years without draining!! Its a lot cheaper too. The rad is'nt actually outside. It's in a radbox on the inside windowsill. I just open the window to let it suck the air through the box. The window side has a large filtered cutout to allow cold air in.

 

[fornowagain]

Tall res with bubble - flow into top of res is due to pump - flow out of bottom is due to gravity (ideally these should be equal)

Edit: please shoot me down if i'm taking bull

If I undersaynd what you're saying, think about this: A res, one loop connection at the top and one at the bottom. The loop is filled and no air in the loop. Is there any flow in the loop?

Consider the actual inlet point at the bottom. You're correct there is head pressure from the column of water at that point, but there is also an equal and opposite column of water from the pipe. So no flow.

If you add a pump the overall energy increase across the impeller from inlet to outlet and you get flow or pressure.

 

[shadowscotland]

With the res fully filled it as just like tubing in a loop - flow in = flow out.
The loop works is just like a sealed loop with a t-line.

Without the pump the loop is in static equalibrum - with the pump the loop is in moving equalibrum (and why my old 3w pump would work a 6 block loop).

The charged water molecules from a unbroken chain around the whole loop - one unit out of the pump = one unit into the pump.

Res's are great for removing bubble - but why let them reduce flow for the other 363 days of the year?

If I undersaynd what you're saying, think about this: A res, one loop connection at the top and one at the bottom. The loop is filled and no air in the loop. Is there any flow in the loop? .

No

Consider the actual inlet point at the bottom. You're correct there is head pressure from the column of water at that point, but there is also an equal and opposite column of water from the pipe. So no flow..

Agreed

If you add a pump the overall energy increase across the impeller from inlet to outlet and you get flow or pressure.

I don't get the 'and you get flow or pressure' bit. If you mean what I said above - then we agree

Edit: If we are all agreed on the sealed loop approach great.
The bubble being helpful/neutral/hinderance is another matter.
I know people like martin (liquid labs etc) and skinnie have looking into this and measured flowrate drops due to a res with a bubble.
So I'm happy to just accept they testing without the maths (but it would be good to know both)

 

[fornowagain]

Your comment about flow from gravity and gains is what interests me. There is no flow in the loop open or closed res unless there is a differential supplied by a pump (other than convective means), gravity potentials cancel. The pump imparts energy, but not necessary flow. If the loop is sufficiently restrictive or indeed blocked the pump will create no flow and simply raise the line pressure. If you mean an air gap, then the loop is still flow in = flow out, it cannot be otherwise in a closed system. With this air gap, a compressible void, you would see greater turbulent flow conditions (momentum losses amongst others, I'll have to think on that one). The bleed system would tend towards more laminar (really only true if the res is big enough to have static areas) conditions I propose (for reference look up Reynolds, Darcy–Weisbach, Energy and Hydraulic grade lines), but the underlying energy conservation remains. If the res is open then of course line pressure drops and with it energy, the pressure drop to atmospheric would be a gradient from pump outlet to discharge point.

Res's are great for removing bubble - but why let them reduce flow for the other 363 days of the year?

The losses are fairly small, similar to a 90 degree elbow or an extra Tee. Certainly not in the region of losses from a block. For me this loss is nothing with my flow rates and worth the advantages of not having to top up very often and as you say makes bleeding easy.

 

[DIABLO]

In my h50 a max of 500ml i would say.