Most 1 - 2 pint kerosene stoves operate at roughly the same pressure. This is controlled initially by pumping, and maintained by thermal feedback into the burner and tank. If these 1-2 pinters are all fitted with the same diameter (0.32mm) jet, then they should all have the same mass flow of vaporised fuel out of the jet. If the burner is correctly designed to provide sufficient air to burn all the fuel correctly, then they should all have the same power output and the same fuel consumption. But they don’t. What am I missing?
@Twoberth The parameters of ‘roughly’ probably explains much of the variation. I guess you’ve been looking at Stove Reference Library catalogue listings for individual stove fuel consumption to conclude there’s a difference in outputs from ‘same jet size’ burners?
Yes, and other online sources. For example, quoted fuel consumption rates for Primus 210 and 1 are 154gm/hr and 223gm/hr. Corresponding power is quoted as 1.9 and 2.8 KW respectively. This is 30% difference, so it must be from tank pressure difference. I am surprised.
210 burner is smaller (maybe 30% smaller than in No. 1/5?) so maybe it will create more restriction for a flow than bigger burner (i.e. pressure loss in/over the burner if bigger). So Amount of fuel that flows through the burner (even with same tank pressure) is smaller.
Yes that’s a possibility, but burner pipe internal diameters even on a 210 are huge compared to the jet.
Smaller v. larger roarer burner introduces another factor too though and that’s the size of the fuel vapourisation chamber the burner tubes feed into and in turn the area of the flame front (circumference at burner head). The greater capacity for fuel to be vapourised in the larger chamber of a No.1 burner probably increases the vapour pressure at the jet. My reasoning is that the pocket of vapourised fuel between the pressurised liquid fuel is in a greater volume of enclosure but the size/area of the flame impacting on it is much greater than the proportional difference between the vapourising chamber volumes. Greater distance between the jet and the vaporising chamber on the larger roarer burner enables more air to mix with the more intense stream of vapourised fuel, creating a greater heat output. ... I think! John
At some point one would think we have brassies pretty much figured out but every week it seems something new comes up (this topic as suggested is new to me). Twoberth, I accept the premise: 1 - 2 pint kerosene stoves operate at roughly the same [tank] pressure. This can be easily quantified w/pressure gauges, makes sense to state tank pressures if we delve deeper. I welcome a link to the KW comparisons. 1 pinter w/1 pint burner (ie 210) = 1.9 KW 2 pinter w/2 pint burner (No.1) = 2.8 KW1880 boil tests will reflect the same differences. A related point: my impression KW / BTU rates are determined by the parts from the top of the liquid fuel upwards. So not stove size.... might we? discuss outputs of 2 or 3 roarer burner sizes (vs stove sizes). Maybe, while we're at it, add the 2 size lipsticks into the mix? I note: the No.1 "J" model types, 1 pint burners on 2 pint tanks, provides only longer run time (output / boil time determined by the 1 pint burner size regardless of stove/tank size). ---------------------------------------- John, you have me convinced. A related point, pressure is non-directional, pressure does not flow (current related thought being, liquid vs vapor pressure factors into pre1880 vapor chamber design). My impression of John's thought is: *tank pressure or tank size or vapor flow restriction (below jet)… none of these are factors. Also meaning, these can all be the same at their maximum effect and will not increase/reduce a burner's heat output. John makes sense and if I follow, the size of vapor chamber (and vaporizing surface within) is a determining factor for max heat output. Pressure in vapor chamber is higher than tank pressure. With normal burner function a larger vapor chamber will produce higher vapor pressures increasing velocity/flow released from jet. ---------------------------------------- Confession, if it sounds like I too have this figured out... I "think" John has it explained correctly but I do not entirely understand the functionality/process (entirely, how it works). The "not completely understanding" brassies is part of the attraction, the jaw drop reaction seeing one light for the 1st time and resign to thinking... will you look at that, how the heck.... this is some kind of magic! In many places our brassies are a complete unknown. In places that are familiar, imo, most amaze and admit "I don't know how it works, I just know it works!" It's magical. @presscall
Liquid fuel is propelled into vapor chamber (below jet) by pressure in the tank. Vaporized fuel is propelled into mixing chamber (above jet) by pressure in the vapor chamber. Easy enough butFWIW I remain stumped. kerophile thank you. The link IDs 2 pressure zones: the design and operation of a pressure stove is a dynamic balance between the pressure in the tank and the pressure in the fuel vapour “bubble” under the jet. further, at operating temp: the 2 establish "dynamic equilibrium". I suppose it is the self-maintaining / dynamic equilibrium that boggles my mind. Once higher pressure vaporized fuel is established in vapor chamber, fuel from tank (liquid fuel at lower pressure) continues to release into vapor chamber... I can only "guess" liquid vs vapor pressure has a role? My above mention of pressure is non-directional is, I don't entirely understand why, once vapor chamber has higher pressure, why liquid fuel does not stop or even reverse flow? Because there is no "flow" per se. There is flow rate and flow direction, and flow direction, as I understand it is entirely pressure driven/directed (higher pressure releases into lower pressure). IF pressure is equal, I would think pressure release does not occur, flow (from tank) stops... but it does not stop, it continues. It is continuous? It is freakin magic and I'm obviously missing a piece of science behind it.
Hi @OMC , It is indeed possible for the dynamic equilibrium to be disturbed in the reverse direction. Stoves with burners running at high power, and causing high thermal feed-back to supply pipework and tank, can be heard to “burp” and bubble as fuel vapour bubbles are forced into the tank, where they condense back to the liquid phase. Best Regards, Kerophile.
ah hah. thanks again Such a burp can help restore equilibrium, granted not an optimal / ideal situ but stove burp is my something new learned today. I am only now thinking, myself and fellow stovies report hearing fuel boiling/bubbling in tank… this may be burping as well? I was unaware of burps, so that had not occurred to me in the past. 8R/99 copies and Ben's (boiling / burping?) vesta come to mind, @z1ulike . So next newbie that presents us with too much thermal feed back performance issue we can tell him "That baby needs to be burped!" ? … i'll get my coat.
Interesting comments from you all. The vapour pressure at the jet can't be sustained at a higher level than the rest of the system, as it's a continuous path from tank to jet and the pressure will be equalized at all points along the path. In dynamic equilibrium, if the pressure at the jet momentarily rises above that of the rest of the system, then the vapour bubble will push the liquid front backwards to re-equalise the pressure. In extreme cases, as George points out, this bubble can be forced back into the tank and condensed. @Afterburner I think you are nearer the mark, and I think the boundary layer set up by the flow has a more significant effect in the smaller 210 tubes, and for the same tank pressure, restricts the liquid flow more than in the larger No. 1 burner. @OMC The power and consumption data came from George here. You are right, it is the burner size that matters not the stove size. A 210 burner will have the same max. power output on a 210 tank as on a No.1 tank. @kerophile George, thanks for the link back to you post from Grigoriy's original, This is the one we discussed by PM earlier. These four 'horsemen' of stove power I have no problem with. However, in the case of these two properly designed and sized burners, the last two 'horsemen' cannot be responsible for the difference in power between the burners, as they have both been specifically designed to entrain sufficient air, and to maintain vaporisation over a range of tank pressures. I have never had to throttle either stove back because it couldn't burn all the fuel properly, or had to re-pump a hot stove because it couldn't maintain vaporisation. Interesting stuff! As George says, 'Aren't stoves fascinating'.
A vapour bubble percolating back to the tank is a feature of some gravity-fed alcohol stoves, specifically one in my experience, a Bukta. I’ve never encountered that extreme on a pumped fuel stove and I’m inclined to think that the designers anticipated that ‘vapour bubble’ phenomenon to create an effective ‘non return’ fluid dynamic Where those peaks of vapour pressure disperse out of the jet and not against the flow. In a self-pressurising naptha stove the vapour bubble is no doubt responsible for the characteristic ‘ram jet’ sound, the ‘back pressure’ stopped short of the tank in part by the restriction presented by the wick.
Re "...The vapour pressure at the jet can't be sustained at a higher level than the rest of the system, as it's a continuous path from tank to jet and the pressure will be equalized at all points along the path." Caveat: I've made it clear I don't entirely get this yet but is "continuous path" a best description? reminder: I have inserted vapor vs liquid pressure above. The link IDs 2 pressure zones, and makes sense (vs 1 continuous vapor path). At play here, it seems to me is the (fluctuating) column of fuel in the rizer. That is liquid fuel, a significant resistance / barrier between tank pressure zone and vapor pressure zone. An overly pressurized vapor chamber can burp excess vapor pressure back into tank (bubbles up through fuel). To do so requires enough vapor pressure to displace the entire column of fuel in rizer back down to bottom of tank (below, maybe way below the level the fuel gravitates to). IF on rare occasion that happens, then immediately fuel re-enters the rizer. The impression I continue to get is as John and others suggest, at full roar the vapor pressure below jet would be higher than the tank pressure. All that said, I myself don't fully understand it, I am inclined to accept 2 pressure zones. *Higher/increased vapor pressure = increased velocity = increased heat output. *burners of same design, a larger vapor chamber = increased vapor pressure.
Forgive the simple diagrams, as this is what I see in my head. I tend to see things in simple high school pictures, but I have difficulty in putting these pictures into words. This is what I meant by equalising the pressure by pushing the liquid front backwards. P is pressure and F is flow. P2>P3>P1 and F2>F3>F1. That's what I see in my head. Couple that with the restricted flow in the smaller 210 burner and I think I understand the power output differences.
My thoughts: For a stove under normal working conditions, the gas volume (air+vaporized fuel) in the stove's tank is much larger than the volume of gas in the vaporizer. If, for some reason, the pressure of the gas in the vaporizer is, say doubled, it will certainly push back fuel towards the tank since you cannot have different pressure on the two ends of an open pipe containing fluid without the fluid moving towards the lower pressure end. When the "gas bubble" in the vaporizer pushes fuel back towards the tank, it will at the same time increase its volume. If it, say doubles in volume, its pressure will be halved. The pressure in the tank will however be almost the same since the amount of fuel pushed back into the tank will be small in relation to the total volume of gas in the tank. This minute amount of pressed back fuel will only shrink the gas volume in the tank very little. The result is only a very small increase in the pressure in the tank. This means that the pressure difference very soon will equalize in the vaporizer and the tank, and fuel flow towards the tank will stop. This "negative feedback" stabilizes the pressure in the system. Also, when fuel is pressed back towards the tank, there will be less contact between fuel and hot methal in the vaporizer that is hot enough to vaporize the fuel, therby reducing vaporization which in turn reduces pressure buildup in the vaporizer. This is also a "negative feedback" that contributes to stabilization. This is of course a rather simplified description, but hope this in some way could contribute to the discussion. If not, print it on some cheap paper, curl it up and throw it in the waste basket. (Please be patient with my English. It is not my native language)
