Discussion in 'Stove Forum' started by Twoberth, Feb 24, 2020.
Thanks for an excellent contribution, in excellent English.
I may have missed something, because I just now read through this thread, but it seems to me an important premise had not been established to be correct. That is that both the Primus 210 and Primus 1 are pumped to the same pressure -- I suspect that this is not the case once both hot and running at full blast.
As already stated the vapor pressure in the burner and in the tank, on average, will be the same since any disequilibrium will be quickly compensated for by changes in volumes. Once both stoves are hot, the vapor pressure for the hotter stove almost has to be higher (from intro chemistry: PV=nRT). As John points out, it is likely the larger burner will be able to conduct more heat to the vaporized fuel to maintain this higher pressure.
I suspect with stoves were the tank pressure could be measured, we are likely to find that the larger/hotter Primus is indeed pressurized to a greater extent. This might be a case where pure theoretical stove science is trumped by experimental stove science.
I was initially assuming that the 210 and the No.1 would have the same operating pressure.
However, I now accept that the No.1 burner probably feeds back more heat into the burner than the 210, and this results in a higher internal pressure (P3 rather than P1).
In my diagram above the operating pressure for the No.1 is shown as P3 which is greater (>) than P1 for the 210.
OK, thanks, got it. I liked you diagrams, but was not quite sure of the final conclusion.
Well, I will not be totally sure until someone measures tank pressure and vapour flow.
However, from fluid dynamics, flow through a nozzle is proportional to the square root of the pressure difference across it (Bernoulli).
So to produce the 30% increase in stove power (mass flow) between the 210 and the No.1 by pressure increase alone, the No.1 would have to operate at twice the pressure of the 210. I haven't measured it, but I don't think it does operate at twice the pressure, so that's why I think restrictive flow through the different size vaporisation pipes also plays a part.
Thanks all. the topic has maybe advanced some.
Early-on was mention btw: " ... [tank] pressure.
This can be easily quantified w/pressure gauges, makes sense to state tank pressures if we delve deeper." +1 w/myself hehehehe
Twoberth you hold out for restrictive flow of the smaller burner results in less heat output (I struggle with fwiw)
I hold out for 2 pressure zones, as per kero's link above, despite naysayers and we shall see.
We do agree increased size / capability of vapor chamber is key to increased vapor pressure = increased heat output.
I freely admit I don't yet get this entirely.It is a closed pressure system with liquid and vapor piped within (flow is directed by pressure: liquid down-up, vapor/liquid up, vapor thru a "U" and released).
SIDE NOTE: twas easy to find water expansion liquid to steam... expands to 1600 times volume!
Expansion occurs as kerosene vaporizes but how much it expands (liquid to vapor) I've not found.
Re tank to burner being: "one continuous path", "two ends of an open pipe", "vapor pressure in the burner and in the tank, on average, will be the same"
Re two ends of open pipe, the 2 ends are not equal re pressure.
1 end is positioned at bottom of tank submerged in liquid pressure the other end subject to increasing vapor pressure.
Might the column of fuel in the riser/burner perform a function with the "dynamic equalizing" piece of this? Might that level fluctuate (burner at full roar, the surface sizzling & boiling, where ever the level equalizes at)?
Further, it seems to me the column of fuel in the riser/burner, is a barrier/separation between 2 zones. 2 zones that may (or may not?) sustain equal pressure at full roar.
Consider a simple "2 open ends of pipe anology":
One can easily blow air (pressure) through a straw in bottom of an empty cup.
With cup half full, you can not easily blow air into bottom of the cup... "2 ends of open pipe" doesn't seem to account for that back pressure / barrier?
Again, we shall see, I note too: There are flow restrictions for some burner setups but as far as I know there are no fuel / vapor / flow restrictions on No.0 & No.1 size 4 tube kero roarers.
Suggestion inline with where I'm at: No.0 roarer creates less vaporization / pressure / heat output because it has smaller/less capable vapor chamber (vs larger No.1 kero roarer).
Not naysayers, just following basic physical principles.
You cannot have two gas zones at different pressures in a pipe where they are separated only by liquid. The liquid is not an immovable plug, it's a fluid and moves along the pipe to equalise the pressures.
The movement of liquid in a pipe to equalise pressures is the basic principle on which a simple manometer works.
Another way to look at it woukd be be to change each of the parameters leaving the others static.
eg. Increasing pressure in a 210 set up the vapour flow will obvioisly gradually increase, as we all understand. Take it beyond normal operating and the flame front will eventually luft away from the burner and without sufficient turbulent flow will flame out.
Ok fit larger burner, bigger distance to the flame spreader higher pressure for optimum combustion, flame front further ahead.
What's missing is an accurate fount presure between a 210 and large burner model. Bow many strokes to get optimum combustion.
I've not fitted a large burner to a 00 to comment on what the real worl pressure differential might be. I' vuessing that I'll have to pump the thing up a lot more to get a No. 2 burner to work properly even with a 0.32 jet.
One day I will fit cheap pressure gauges from bicycle foot pumps into the air screw outlets of a 210 and a No.1 and fire them both up to run normally.
Then monitor the pressure on each as I do a tea test.
at first - on diagram is some problems...
that system is designed to trasport fuel to the burner (via generator)
it is based on different pressure in tank and end of pipe
"red" is false - no transport fuel - P1 raising to the p2
"green" is true only when pipe is closed P(tank)=P(end of pipe)
Yours "F -flow" in real, is "pressure losts" thru open pipe
so - it is dynamic process
Yes, I understand. Thanks you
P6 < P5
P5 < P4
P4 = P3
In words: P6 is less than P5 that is less than P4 that equals P3
This means P5 < P3 (P5 is less than P3)
is liquid flow?
if "yes" -> it is no true
I agree. My diagrams were for a closed system (jet closed).
As soon as you open the jet you get the result you highlighted in green.
I drew the diagrams to indicate that the liquid was ‘fluid’, and that no matter how hot the top of the burner got, P2 could never be greater than P1 as the liquid front moved downwards to equalise the pressure.
Sorry for any confusion.
P3 = P5
If P5 rises, fuel will be pressed down the pipe towards the tank. This in turn will reduce P5 until it equals P3.
The same go for the vertical version in the previous post, except that P3 then must be higher in order to keep the column of fuel above the level in the tank (above P4) in the pipe from falling down to the tank. But also then, if P5 rises, it will move the fuel down the pipe that in turn reduces P5 until it again is balanced by P3 (which has not changed, and therefore P5 is back to the initial pressure).
Note that when P5 increases and epands its volume, some tiny amount of fuel is moved back to the tank. This in turn will result in the gas volume of the tank being compressed a little, whic in turn will increase P3 a little. But this is very little and out of scope of what is discussed here.
I think your diagram was quite good at explaining. No confusion, at least not for me. I just wanted to add some explanation and found it easiest to add information to your diagram. Hope that was OK for you.
if "We" need to describe that
i propouse think about diagram as "corelated two generator system"
tank - low pressure fuel pump vapourised gas powered"
vertical pipe - high pressure gas generator
generators are powered by heat
There will of course be a very slow flow of fuel from the tank towards the vaporizer. The rise fall in P4 due to this is very very smal, but you are right, it will be a tiny tiny bit lower than P3 (which in turn is reflected in P5 being a tiny tiny bit lower). This is due to "friction" between the fuel and the pipe walls.
thats "tiny tiny" - is fuel consumption, and burner power - and stove - source...
No. It's friction (or its equivalent in fluid dynamics)
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