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