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In message , roger
) wrote:
The message
from (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

The message
from (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.

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.

And of course by implication, temp rise across the
boiler.

It is a matter of necessity rather than implication. Increasing the flow
of water through the boiler must decrease the temperature rise across
it.

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.

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.

--
Roger

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