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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.

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.

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.

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.

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.

NT

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