Am 08.12.2011 17:20, schrieb Robert Leguillon:
Unfortunately, it's not quite that simple for two reasons:
1) the secondary flow rate was much higher than the primary, moving the
equilibrium point closer to the hot side
2) the primary flow rate is unknown, and quite possible variable, moving the
equilibrium point back and forth
3) the primary flow is sometimes steam, sometimes water, sometimes both. If the
steam were to immediately condense in the brass fitting, it would impart the
same energy as water at hundreds of degrees celsius, driving the equilibrium
closer to the cold side.
Yes this is true. If the thermal resistance against the massflow is not
symmetric, then there is no precise symmetry.
But we have seen hot water outflow before. Also air bubbles can make
problems. if the heat exchanger is partially filled with air, the
thermal coupling increases. So we have other unknown parameters discovered.
This arrangement is not good enough to do an industrial test for a gas
boiler.
Its therefore a waste of time to calculate this precisely, too much
unknown factors.
These problems can be easily avoided. Fit 30 cm of copper pipe to the
heat exchanger or insert a piece of copper pipe into the hose at a
reasonable distance and measure the temperature there. Thermal
insulation can be used to avoid heat loss, but because the absolute
temperature was not much above ambient, not much loss is expected.
Anyway, thermal isolation is cheap and would eliminate the influence of
ambient air.
Peter
Date: Thu, 8 Dec 2011 17:09:53 +0100
From: [email protected]
To: [email protected]
Subject: Re: [Vo]:Will tests surface mounted thermocouples on pipe
----- Original Nachricht ----
Von: Jed Rothwell<[email protected]>
An: [email protected]
Datum: 08.12.2011 17:00
Betreff: Re: [Vo]:Will tests surface mounted thermocouples on pipe
[email protected] wrote:
How can you say this is incorrect? Do you know everything, great master?
I can say that because Houkes knows what he is doing, other experts
agree with him, and it has been my experience that the water temperature
in a pipe dominates the surface temperature even when there is another
pipe or hot body nearby. As for example, in a calorimeter where the
inlet and outlet sensors are close, and both under insulation. Or in the
tests I did last night. Air temperature and heat conducted by the pipe
do not play much of a role.
There is symmetry, and so the temperature distribution must be
symmetrical.
This is EASY to see.
Evidently not.
If your experts dont see this simple fact, then they are not experts but buggy
calculation machines.
I have calculated many linear networks, by hand, 35 years ago, when computers
could not do this.
I know how to simplify a linear network.
best, Peter
