In message snipped-for-privacy@no-spam.zetnet.co.uk, roger ( snipped-for-privacy@no-spam.zetnet.co.uk) wrote: The message from snipped-for-privacy@meeow.co.uk (N. Thornton) contains these words:
>> Same with CH faster pump lower
>>> drop but watts shifted is the same.
>> I'm sure thats not correct. Faster flow for same power input will mean
>> rads run a little hotter, thus will dissipate more heat. Thus
>> efficiency greater. Faster flow also means lower temp water /out/ of >> the boiler.
> Faster flow with the same power output will mean the output temperature > is lower
how do you get that? Bear in mind the flowrate will affect all 4 variables at once: boiler output temp, return temp, rad temp drop AND system efficiency. And of course by implication, temp rise across the boiler.
and there will be a lesser temperature drop across the radiator. There I agree
There is another way you can look at all this too. You can model it as a series of heat transfers, from burning gas to boiler water, from boiler water to radiator, and from rad to room air. Each of these transfers has a resistance in C/watt, and decreasing one of the thermal resitances will reduce total system resistance, thus improve efficiency. Upping pump speed achieves just that, reducing thermal resistance.
NT