Quote:
Originally Posted by Scarlet Infidel
This is good. It seems flow rate has a huge impact on temperatures here. Could you take the pressure readings like you were showing before and then use these to reduce the flow on the more restrictive versions?

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.)
Edited by zipdogso  3/28/09 at 10:23am