About Us

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

The message
from (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. :-)

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.

AFAICT the examples, as well as being impractical, do no such thing

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
thes 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.

You might but it is knowledge I would dispute.

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.

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.

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.

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?

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 affect efficiency. And thus the assumption
above, that heat input is the same, is false.

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.

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.

Yes.:-)

--
Roger

(N. Thornton) wrote in
om:

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

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 ....

mike

Thread Tools	Search this Thread
Show Printable Version	Search this Thread: Advanced Search
Display Modes
Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
CH pump - fast or slow?	N. Thornton	UK diy	1	December 30th 04 11:42 PM
Water Pump / Pressure Tank Problem !!!!!!	James Nipper	Home Ownership	6	June 28th 04 02:13 AM
Low Water Pressure - pump stops at 20 psi - need advice Please	Davis Leeman	Home Repair	2	March 14th 04 02:57 PM
Need advice with rural water well pump design	Gene	Home Repair	20	December 2nd 03 02:58 PM
Backup Sump Pump Horror Story	Harvey Krodin	Home Repair	10	June 27th 03 02:59 PM

Menu

About Us