Overclock.net › Forums › Cooling › Specialized Cooling › Peltiers / TEC › Quad TEC Water Chiller Project by Ultrasonic2's
New Posts  All Forums:Forum Nav:

# Quad TEC Water Chiller Project by Ultrasonic2's - Page 27

Quote:
 Originally Posted by Scarlet Infidel Watts is Joules per Second
OK....

Q = mcT

Q = (1000g)(4.18J/g-C)(1 C)

Q = 4180 J

So you need 4180 J of heat to raise 1 litre water 1Âº

.....1 joule = 0.00027777777778 Watt-hour. so 0.00027777777778 *4180 = 1.16w which is correct for my formula.
Yep, I always agreed with the maths. Just making sure it was clear to everyone that power(watts) = energy(joules) / time(seconds)
Quote:
 Originally Posted by zipdogso With my calculations earlier I said irrespective of everything water takes 1.16w to heat 1Âº in one hour and since it has been pretty much ascertained that a closed loop is generally in temp equilibrium taking the volume of the whole loop would seem fair. Now i don't think thats the case as flow has made a fair difference here...then it dawned on me the calculation is NOT volume of the loop per se but is actually the flow rate per second. So you use exactly the same calculation using the flow rate in litres per second instead of the actual volume of the loop. The calculations should then be right...I did wonder why originally it didn't all seem to fit as correctly as I thought it should. EDIT For those of you how have seen the calculation - flow rate litres per second x 4190 joules (thermal capacity of water) = watts/Kelvin. My calculation is the same but with the physical time component added. For those who want a better explaination - It is the same because 4190 joules = 1.16w and 1 kelvin degree is the same size as one celcius degree (so Kelvin is often used to indicate a difference between two celcius temps.) BUT 1ÂºK is not the same temerature as 1ÂºC - 1ÂºK = -272ÂºC they have completely different scales. Flow rate in litres per second gives the weight per litre because 1kg = 1 Litre. Although the calcs say flow rate litres per second you are actually using the weight of the water (in grams.) to match the thermal capacity calculation = 4.193 Kj/kgÂºK which in grams is 4190j/gÂºK (it is rounded down a bit.)
id like to say these calculations are Irrelevant there a time based calculation on how long it will take to reduce X volume of water at Y rate(watts) ..

How is that of any use to this discussion ?

im dealing with the thermal resistance between the hot spot ( the TEC ) and the cold spot (the water entering the chillier). The calculations you've been posting have no relevance to C/W
Edited by Ultrasonic2 (muffy) - 3/28/09 at 1:26pm
 My System (13 items)
 My System (13 items)
Impressive stuff! REP+
 Home Workstation v1.0 (15 items)
CPUMotherboardGraphicsRAM
AMD FX-8320 3.6GHz 1.152v Gigabyte 990FXA-UD5 EVGA GTX 960 SSC 4GB 8GB G. Skill DDR3-1333 CL9 1.5v
Hard DriveOptical DriveCoolingOS
WD 1TB SATA3 + 4TB Storage External USB Thermalright Archon Win 7 Pro x64
MonitorKeyboardPowerCase
Samsung 27 inch LED 1920x1080 HD Logitech K120 Seasonic X-760 Modular Fractal Define R4
MouseAudio
Logitech Altec Lansing 2.1
 Home Workstation v1.0 (15 items)
CPUMotherboardGraphicsRAM
AMD FX-8320 3.6GHz 1.152v Gigabyte 990FXA-UD5 EVGA GTX 960 SSC 4GB 8GB G. Skill DDR3-1333 CL9 1.5v
Hard DriveOptical DriveCoolingOS
WD 1TB SATA3 + 4TB Storage External USB Thermalright Archon Win 7 Pro x64
MonitorKeyboardPowerCase
Samsung 27 inch LED 1920x1080 HD Logitech K120 Seasonic X-760 Modular Fractal Define R4
MouseAudio
Logitech Altec Lansing 2.1
stick around rick, these engineers are awesome!
Quote:
 Originally Posted by Ultrasonic2 (muffy) id like to say these calculations are Irrelevant there a time based calculation on how long it will take to reduce X volume of water at Y rate(watts) .. How is that of any use to this discussion ? im dealing with the thermal resistance between the hot spot ( the TEC ) and the cold spot (the water entering the chillier). The calculations you've been posting have no relevance to C/W
Thermal resistance of a heatsink is calculation of the the missing thermal heat transfer and is quoted a ÂºC/W which is watts lost per ÂºC.

In my calculations I said:-

Quote:
 Originally Posted by zipdogso so Volume (Ltrs.) x 1.16 x ÂºC change divided by the TEC operating wattage is how long it will take to cool that amount of water. If you had a 4 litre loop and you wanted a 5ÂºC drop with a 250w operating point on the TEC :- 4 x 1.16 x 5 -------------- = 0.0928 hours = 0.0928 x 60 = 5.5 minutes 250 This of course is calculated assuming no losses and constant value of heat from TEC and the water is not being heated....if you timed it and it actually took 8 minutes :- Volume (Ltrs.) x 1.16 x ÂºC change divided by the time in hours is the true wattage transferred to the water :- 4 x 1.16 x 5 ----------------- = 174.04....call it 175w (8/60) = 0.1333 So your chiller is leaking 75w somewhere...get the neoprene out....
In the calculation shown above the chiller lost 75w over 5ÂºC....

hence the thermal resistance of that chiller is 5/75 = 0.06 ÂºC/W

Now what were you saying about my calculations not being relevant....

I wont mention calculations anymore I'll let you get on with COOLflow or what ever it is....
Edited by zipdogso - 3/28/09 at 5:33pm
Sorry zipdogso, unless you have missed out loads of detail I think you are greatly simplifying the issue here.

If you want to explain it all to me in detail, with all the equations, assumptions and units then I will be happy to listen. PM me or start a new thread, I don't think this one needs more cluttering.
Quote:
 Originally Posted by zipdogso Thermal resistance of a heatsink is calculation of the the missing thermal heat transfer and is quoted a ÂºC/W which is watts lost per ÂºC. In my calculations I said:- In the calculation shown above the chiller lost 75w over 5ÂºC.... hence the thermal resistance of that chiller is 5/75 = 0.06 ÂºC/W Now what were you saying about my calculations not being relevant.... I wont mention calculations anymore I'll let you get on with COOLflow or what ever it is....
As long as you keep quoting a volume of water as part of your equation and time then your calculations are irrelevant.. Just like you posted your equations are based on x volume of water and time. It's useful in finding out how long it will take to cool a load of X watts to a temp of Y though.

THIS has nothing to do with what we're testing.

Your saying the chillier is "Leaking 75 watts" where does that come form ? There no leakage to the environment in the simulations. all of the 800watts is being transferred into the water in every model.

zipdogso don't run away im happy to debate this
 My System (13 items)
 My System (13 items)
Quote:
 Originally Posted by Ultrasonic2 (muffy) There no leakage to the environment in the simulations. all of the 800watts is being transferred into the water in every model.
Well in that case your coolflow doesn't work too well.... it is impossible for 100% off your 800w to be transferred to the water in reallife...heat in a copper bar migrates to the coldest point one of these coldest points is the outside edge from where heat radiates into the atmosphere you can see it on your simulations in the center of the bar at the supposed hot point the edges are blue indicating low temp...why...because it is radiating into the atmosphere..what is ?...some of your 800w....so it's impossible in reallife for you to transfer your entire 800w to the water...it doesn't happen ...it wont happen ..never mind what your simulation says.

Quote:
 Originally Posted by Ultrasonic2 (muffy) Your saying the chillier is "Leaking 75 watts" where does that come form ?
In my hypothetical example the "leaking 75w" is the thermal heat transfer losses through the mass of the copper and the covers in fact all the materials of the chiller...some of which you can get back by insulating the chiller. anyway the figures themselves are not real examples I just picked them out the air and was surprised they loked so cute....
velocity of the flow is very important to in cooling a load

Here is another test everything is the same as the last test but i reduced the depth of the channels from 3mm to 1.5mm and kept the base plate at the standard 3mm(same as last one)

Now this reduction in depth means for the same volume of water to pass through the chillier it has to be travelling faster. the deduction in temps it due to the increased scavenging affect because the water is moving faster.

Solid Temperature [Â°C] 24.1619 29.4251 28.2041

With the channels at 3mm the average was 34.9 now it's 28.2
 My System (13 items)
 My System (13 items)
New Posts  All Forums:Forum Nav:
Return Home
Back to Forum: Peltiers / TEC
• Quad TEC Water Chiller Project by Ultrasonic2's