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.
The message from snipped-for-privacy@meeow.co.uk (N. Thornton) contains these words:
You can't get a quart out of a pint pot. The above are all variables but if you start with a steady state situation and then turn the pump speed up the output temperature will go down and even when the new equilibrium is reached with a higher return temperature the output temperature would still be lower as the radiator at the other end of the equation is still putting out the same amount of heat so should have the same average temperature as before.
It is a matter of necessity rather than implication. Increasing the flow of water through the boiler must decrease the temperature rise across it.
I may be making a mistake answering this have wined and dined very well this evening but as I see it once equilibrium is achieved the average temperature in the boiler and the average temperature in the radiator should not change over a wide range of pump speeds provided the heat input is constant.
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:
/out/
temperature
equilibrium
example at end will hopefully show that heat input to the water will not be the same, with the same rate of gas burn. Thus we know rads will not output the same amount of heat, nor have same average temp.
improve
I would suggest we know that is not the case by looking at extreme examples to see in what way pumping speed affects things.
Extreme case 1: the water goes so fast that the entire system is at the same water temp. Output, return and rad all at same temp. Impractically fast flow of course. This gives us min ave boiler temp, max ave rad temp, and max efficiency - at least if you ignore the infinite pump power. :)
Extreme case 2: the pump runs so slow that by the time the water reaches the rad it has cooled down. In this example the return temp is cold, the rads are cold, and the boiler output temp astronomical. (We'd best ignore boiling here, since we dont have to deal with boiling in a properly functioning CH system.) With very high bioler output temp, our efficiency is going to be crap. Temp drop across boiler heat exchanger is relatively low, therefore far less heat flows from gas to water.
This view makes it easy to see what pump speed does. And I suggest it is easy to see that efficiency will not be the same in both cases. And since we know return and output temps affect efficiency in real world systems, and that adjusting flowrate will alter these temps, it is easy to see flowrate will afect efficiency. And thus the assumption above, that heat input is the same, is false.
It is necesary to think of _all_ the variables at once imho. I think what is happening in this thread is people thinking of some of them, but ignoring others temporarily. But of course I could be wrong, totally crazy, or a short pygmy with a long beard.
The message from snipped-for-privacy@meeow.co.uk (N. Thornton) contains these words:
I usually edit posts such that at least the first line of my latest input is on the initial screen. However in this particular case I didn't want to delete too much hence this totally unnecessary paragraph. :-)
AFAICT the examples, as well as being impractical, do no such thing
You might but it is knowledge I would dispute.
Above (and below) you refer to all sorts of factors that have some bearing on a heating system but ISTM that here you are ignoring the one that actually determines the return temperature - the need to dissipate the heat load. A radiator with all its surface at the same temperature is not going to give out a materially different amount of heat to when it merely has the same average temperature and if the average temperature in the radiator doesn't change it is hardly likely the average temperature in the boiler will either.
In practical terms the highest flow rate will be at maximum pump speed and I see no reason why my ideas would not work at that limit.
Right, so we lag the pipes thus disposing of that red herring. The practical limit for the slow pump speed is governed by the need to avoid boiling. Anything slower than that should play by the rules.
Is boiler efficiency to be directly related to output temperature? Up to now I have taken it to be average temperature as a likely candidate but what actually determines the temperature difference at the flame/heat exchanger interface? The higher the pump speed the lower the output temperature but the lower the pump speed the lower the input temperature. Which one of these has the bigger effect, or do they just cancel each other out?
If we assume that the efficiency is governed by that well worn formula (t1 -t2)/t1 the temperature difference in question is not the difference between flow and return but the reduction in temperature of the flue gases. I don't know offhand what the gas combustion temperature should be but I wouldn't be at all surprised if it wasn't of the order of 400 to 500 degrees C in which case it would take about a 7 degree difference in t2 to make a 1% difference in efficiency and I suspect that any difference in t2 would be at least an order of magnitude less for any real changes in flow and return temperatures that changing pump speed could generate.
snipped-for-privacy@meeow.co.uk (N. Thornton) wrote in news: snipped-for-privacy@posting.google.com:
I wish you would use a proper news server; I'm sure there's something interesting and informative to be got out of this thread, except it isn't a thread.
A lot of us, as you've been told, use news.individual.net ....
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