This has actually been bothering me for a while. Most of us would be aware that lantern outputs would largely depend on the fuel flowrates through the orifices of their jets or gas tips, which in turn, are are mainly dependent on both the cross-sectional areas of the orifices and the operating pressures of the lanterns. However, I have not really come across any posts or articles on the internet being written about anyone actually calculating the fuel flowrates for such an operating lantern. That's to say, done so without actually measuring the said flowrates or fuel consumption on the operating lantern itself. In this post, I would finally make an attempt to theoretically calculate the fuel flowrates(both volumetric and mass) of a real lantern example, which in this case, the Petromax 500HK lanterns (namely, the models no.829 and 523). The ultimate objective would of course, be able to obtain a fairly accurate calculation which, would not deviate too far off the actual measured output of a controlled lantern test rig or setup. In such a setup, the fount pressure of the operating lantern would essentially been regulated, kept or maintained steady at a specific value, for instance, 2 bar or 2000000 Pa over the measuring period. Expectedly, this would have be a little technical in nature. It seems that this is certainly not as simple as I first thought. Certainly not for anyone, such as myself who, does not actually have any backgound or any appreciable knowledge in the field of Fluid Mechanics. Maths and Physics aren't my favourite subjects either . Please feel free to comment and any inputs would be welcomed. Anyways, here it goes. I'd start off with using the following equation for volumetric flowate of a fluid, (in this case, kerosene, which I'd consider a liquid and supposedly, incompressible) through an orifice. *(Note: this would be the standard fluid flowrate formula through an orifice, which had been derived by other people with profound knowledge, from a combination of Bernoulli's equation and the continuity equation. I'd rather not go through the derivation details here):- Volumetric flowrate, Q = Cd×A2×√(2×(ΔP)÷(ρ×(1-(A2÷A1)^2)) where, ΔP is the pressure difference across the jet or gas tip orifice (difference between the jet's upstream pressure, P1 and the downstream pressure, P2). Therefore, the volumetric flowrate, Q = Cd×A2×√(2×(P1-P2)÷(ρ×(1-(A2÷A1)^2)) where, *With reference to the following generalities: Based on the above, the orifice discharge coefficient, Cd ranges between 0.80 (with 60°angle of approach) to 0.82 (with 19° angle of approach). and the cross sectional drawing of the Optimus lantern jet: For this, it is assumed: Cd = 0.8158536585 (with 55°÷2= 27.5° angle of approach for the Petromax/Optimus jet or gas tip). **since they are pretty much similar in design, size and geometry. A2 = Cross sectional area of the gas tip, jet or nozzle orifice. A2 = π (r2)², where r2 is radius of orifice. A2 = π (D2÷2)², where D2 is 0.24mm÷1000mm/m A2 = π×(0.24÷1,000÷2)^(2) A2 = 4.52389342E−8 m² A1 = Cross sectional area of the internal fuel path immediately upstream or just before the jet orifice. A1 = π (r1)², where r1 is the radius of the area, A1. A1 = π (D1÷2)², where D1 is 3.1mm ÷1000 mm/m = 0.0031 m. A1 = π (D1/2)² A1 = π×(0.0031÷2)^(2) = 0.0000075477 m². P1 taken as 200000 Pa or kg/ms² (2 bar). P2 taken as 101325 Pa or kg/ms² (atmospheric pressure at the jet orifice's outlet). ρ is the density of liquid kerosene @15° or 20°C, which in general, usually ranges between 780 to 810 kg/m³. Assumed: 800kg/m³ Here, I would be using the SI Units in the calculations. Therefore, substituting the values into the equation would give Q = 0.8158536585×4.52389342E−8×√(2×(200,000−101,325)÷(800×(1−(4.52389342E−8÷0.0000075477)^(2)))) = 5.79703608E−7 m³/s or 2.0869329888 litre per hour. Range: (5.87088665E−7 to 5.76114077E−7 m³/s) or 2.113519194 to 2.0740106772 litres per hour depending on the typical density range between 780 to 810kg/m³ for liquid kerosene. The mass flowrate, m can then be calculated using: m(kg/s) = Q(m³/s) × ρ(kg/m³) Example: m = 5.79703608E−7 m³/s × 800kg/m³ m = 0.0004608913 kg/s or 0.0004608913 kg/s × 3600s/h = 1.65920868 kg/h. **(1 liter of liquid kerosene@15° or 20°C is taken as 0.8kg). To anyone who's is familiar with the operation of a Petromax 500HK, this is obviously inconsistent, illogical or incorrect, as it would imply that a properly operating Petromax 500HK(at advertised rated output), would consume the entire 1 litre kerosene filling within:- (1 litre ÷ 2.113519194 litre/hour) to (1 litre ÷ 2.0740106772 litre/hour) ~ 0.473 h or 28.39 minutes to 0.482 h or 28.93 minutes! The above calculated results also appear to be COMPLETELY OFF THE MARK , as compared with the technical data stated in Hytta's website for the 500HK lantern: From the above data, the 500HK lantern should by default, only consume 1 litre of kerosene in 8 hours. I have no idea on how did Hytta came up with the stated values. I'd suppose they were actual measured values from a number of controlled test samples which, had been averaged out to the nearest acceptable values. I'd therefore assume the values given by Hytta are pretty much VALID. As of such, the aforementioned flowrate equation and calculated results would obviously NOT BE APPLICABLE on a real operating Petromax 500HK! ***Perhaps the above calculated figures might more closely correspond to the actual flowrates ONLY when the lantern is NOT being lighted up AND when the discharged fuel remains a LIQUID. For anyone who has actually performed the cold orifice fuel flow tests on an unlighted lantern, it should be fairly obvious that the flowrates would appear to be significantly higher than those when the lantern is in actual steady-state operation. Under such cold flow tests, you might have noticed that the fount pressures would drop fairly quickly...a lot quicker than that of the same lantern that is in operation at full blast. So,...back to the drawing board, basics. This obviously needs to be corrected and revised .
I never actually tried to calculate the fuel flow of a lantern. Far to complicated for me. The "eight hours for one liter of fuel" is advertising talk. It is true but... - on the lowest possible pressure with a reduced light output. The 500HK of light output on the other hand again is true but... - with the lantern running on maximum pressure which then will the not be working for 8 hours... I think @ludwig might be the one to ask for exact calculations. Erik
@MYN Extraordinary effort, and I suspect your calculations aren’t wrong. For a pressurised tank of fuel, emptying one litre of it through the jet as liquid fuel in half-an-hour seems plausible. As you suggest, the rate of depletion of the fuel when vapourised will be less, much less evidently. That’s surely explained by the expansion of the vapourised gas in the generator (vapourising tube) creating back-pressure, slowing down considerably the passage of liquid fuel through the generator.
@MYN it's kerosene gas passing through the nozzle. Not liquid kero. Therefore the calculated volume might be correct. But the equivalent liquid mass or volume is certainly not.
These matters are covered by Herman Lahde in his treatise, “Light and Heat from Hydrocarbon Appliances” (2000). There is a chapter on these theoretical aspects and some tables. Tony
He (@ludwig ) already did, in German however: Source: petromax.nl/PX_tabel.html#BRD To do: translate it in English.... Edit: Under the typical operating conditions of pressure kerosene lamps, the vapor jet in the nozzle is at the speed of sound. Higher speeds can only be achieved with special nozzle geometries (Laval nozzles), which PM nozzles certainly do not have. Since the kerosene vapor cannot escape faster than the speed of sound, the amount of vapor escaping per unit of time should be approximately proportional to the nozzle cross-section, i.e., diameter^2. Approximately, since the speed of sound is only reached at a certain distance from the nozzle wall; closer to the wall, it is certainly lower. Also, smaller gas lamps generally have lower efficiency than larger ones, meaning they require more fuel per lumen produced. Therefore, a bore of 0.15 mm is more appropriate for the 150 series, and 0.30-0.31 mm for the 800 series. The situation is quite different at low flow velocities, where the flow is far from the speed of sound, e.g., with mixing tube or clay burner cross-sections (or with low-pressure spirit nozzles). Here, the volumetric efficiency (i.e., what passes through the cross-section in question at the same pressure difference) increases with diameter^5 (in words: five). Ludwig.
@bp4willi is right. The calculation must be done with the vapor density at nozzle temperature. Furthermore, I would assume, the "coefficient of discharge" is dependent on the viscosity of the gas (resp. vapor), also at nozzle temperature (Viscosities of gases rise with temperature). I made similar "calculative estimations" for my alcohol lamps, to get some information on energy and impulse of the vapor jet, which determines air intake. Ludwig
Thank you so much for all the responses. What a wealth of knowledge and information from the members here. @Erik Leger , I'm actually not too sure about the figures given at the Hytta site. The 500HK and other designated outputs are not specific for a particular brand but encompasses the Petromax, Aida and Standard. I have no idea if they were originally provided by the manufacturers such as E&G or independently generated by Hytta on a later date. I'd doubt Hytta gains anything in advertising for the companies who are no longer in such business. From the specification table, the 500 or other HK values might merely be the size class of the lanterns rather than the actual candle power rating. There isn't anything stated on regard to the luminous outputs of the lanterns in the specs. Instead, I can see the calorific outputs being provided in kcal/std. The values are logical if we consider the energy density of kerosene would fall somewhere around 35MJ/litre. That'd be about 8365.2kcal/litre. If the 500HK consumes 1 litre per 8-hour period, then the calorific output for a perfectly operating lantern(with 100% energy conversion efficiency) would be approximately 8365.2kcal/l ÷ 8h/l = 1,045.65kcal/h or roughly~1050kcal/h assuming std means stunde or hour. @presscall , It might possibly be sufficiently valid for purely liquid kerosene at room temperatures but unless I happen to find another validation, I'd still be keeping my fingers crossed about it. @bp4willi , I believe you are absolutely correct about the fuel passing through the orifice being in vapour form rather than liquid. That would explain the stark differences in the earlier calculated output and the figures provided at Hytta's site. @Tony Press , I'd certainly love to have Dr. Hermann Lahde's book. Unfortunately, it isn't available in my location and if I'm not wrong, it had not actually been officially published during his time. In the 2nd page above, it was indeed stated that the cp outputs are not valid. I'd suppose the "1 litre burns hours" in column 3 of Table 11 would still be valid, despite that info being used in advertising. Nevertheless, there is also a mention about that bearing no relation to the maximum power outputs. @WimVe , An excellent piece of translated info that I certainly do not have. Speaking of sonic velocities, I did came across mentions about the importance of MACH numbers in relation to choked flows when doing an internet search about flowrate calculations. However, myself having no prior background in thermodynamics, might need some further reading and time in order to grasp and understand all that . @ludwig , Thank you for chiming in. There is certainly a lot technicalities and governing principles involved if we are to calculate the related parameters and variables in such gas pressure appliances. Especially so if we were to expect the calculations to be anywhere reflective of the devices' actual operating conditions. In my prior post, I merely assumed the orifice discharge coefficient could likely fall between 0.8 to 0.9. It might not be correct. Yes, the actual value of the discharge coefficient is dependent on many factors that I might not be able to ascertain without actually carrying out an experimental measurement without appropriate instuments/setups. I would not be able know for sure about some important related parameters like fluid frictional factors, the actual gas expansion factors, Reynolds numbers in relation to the viscosities, etc. Getting acquainted into such details would seem like a mountain of a task for me .
Based on my current understanding (or probably misunderstanding) of the subject matter, here's my second spontaneous attempt to calculate the flowrate :- So it did not took too long before I realize that the previous flowrate equation needs to be expanded so that it would become more or less valid for gaseous or vaporized substances. In this case, the fluid(vaporized kerosene) would be regarded as a gas, compressible and therefore be subjected to appreciable expansion and contraction in instantaneous volumes. This makes the calculations and considerations a lot more complicated than that for non-compressible fluids. Yes, we would actually be dealing with hot, gaseous kerosene instead of a cold liquid, unlike what I've posted before. The jet orifice would actually be discharging hot kerosene in the gaseous phase on a real operating lantern. Vaporized kerosene at elevated temperatures would also have much lower densities than liquid kerosene at room temperatures. Although essential, I'd rather not to delve into the details of thermodynamics, thermophysics, derivations and such here. These are basically heavy, mind-taxing subjects which are, I think, best left to the experts. So, I'd just use the relation for mass flowrate of compressible fluid through the orifice this time. Among the reasons to initially calculate the mass instead of volumetric flowrate are: 1. Difficulty in ascertaining the actual expansion factor of the vaporized kerosene with some variables being rather ambiguous at present. 2. It is more convenient to use mass flows instead of volumetric flows for energy-related calculations. Therefore, the mass flowrate(in kg/s) of vaporized kerosene through an orifice would in general, be expressed by the following relation instead:- ṁ = Cd×A2×P1×√((2×MW)÷(R×T1))×(γ÷(γ-1))×((P2÷P1)^(2÷γ)-(P2÷P1)^((γ+1)÷γ)) where, Cd is the orifice discharge coefficient, assumed 0.8158536585 as explained in the first post. A2 is the cross-sectional area of the orifice in m². As before, for a 0.24mm orifice of the Petromax 500HK or similar lanterns, A2 is 4.52389342E−8 m². P1 is the upstream pressure of the fuel at the point just before the jet orifice, in Pa or kg/m.s². Note: In this case, it is assumed that this pressure would eventually be in equilibrium with the fount pressure. This is despite the fact that vaporized kerosene is expected to expand in volume and might probably cause a pressure rise within the vaporizer and fount. However, I suppose this would not become significant in a properly working lantern, due to the fuel isn't really confined in a totally enclosed volume. The jet orifice is an open end where the mounting vapour pressure within the vaporizer would quickly diminish and equalize to that of the atmosphere at the orifice outlet. As of such, I would ignore the slight back-pressure created by the vapor pressure within the jet or vaporizer. Moreover, it won't be easy to estimate the exact values of such back pressures without a proper measuring setup. Therefore, in this case, P1 is assumed to be equal to the fount pressure which in this case, a value of 2 bar or 200000 kg/m.s² is used. P2 is the downstream pressure of the vaporized fuel at the outlet of the jet orifice, in Pa or kg/m.s². This is essentially the pressure of the gaseous or vaporized fuel as it exits the orifice, which, in an open-ended path, equal to the atmospheric pressure of 101325 kg/m.s². The atmospheric pressure would remain pretty much unchanged, regardless of the ambient temperatures. MW is the average molecular weight for kerosene in kg/mol. Note: It is certainly not easy to find a well-validated figure for a substance which is composed of varying amounts of many different hydrocarbons with different individual molecular mass. Here, it is taken as approximately 170g/mol or 0.17kg/mol. R is the universal gas constant taken as R = 8.31446261815324 kg.m²/s².K.mol. T1 is the temperature of the vaporized fuel in Kelvin or K. Note: I do not have a proper setup to actually measure and validate the temperature of the fuel at the vaporizer. If the vaporizer is a closed end, I'd believe the fuel might attain critical or supercritical temperatures beyond 400°C, where the internal pressure might rise beyond that of the atmospheric pressure. This is reinforced by earlier observations that the Preston loop might occasionally get red-hot during operation. For dull red-hot colour temperatures of blackbodies such as brass, steels, etc, the estimated temperature would be in excess of 600°C. The following shows the phase diagram typical of similar kerosene substances: However, I'd assume the fuel temperature would somehow be be limited by the highest or end boiling point of the heaviest fraction in a typical kerosene at atmospheric pressure for an open-ended path. I'd assume this to be 300°C or 573.15K for a kerosene with a typical boiling range between around 150 to 300°C. Therefore, T1 = 573.15 K in this case. γ is the specific heat ratio of vaporized kerosene @T1 = 573.15K. Note: This value represents the ratio of the specific heat capacity at constant pressure (Cp) to the specific heat capacity at constant volume (Cv). (γ = Cp/Cv). In general this value would slightly decrease as the temperature increases. The temperature dependence would appear something like this: The exact validated figure at the specific temperature(573.15K or some other figures) for vaporized kerosene is hard to find due to scarcity of such data. They are usually only found from specialized research papers, publications and journals. Usually, such info are used in the field of jet and rocket propulsion systems. In a rough sense, this might be infringing into the realms of 'rocket science'. From some sources it would usually range between 1.1 to 1.35 for kerosene or similar multi-hydrocarbon mixtures' vapour at approximately 573.15K. It is certainly different from the values for liquid kerosene at lower temperature ranges. In this case, I'd arbitrarily use γ = 1.2. Again, by substituting all the assumed values of the constants and variables into the new equation for mass flowrate, we'd get:- ṁ = 0.8158536585×4.52389342E−8×200,000×√(((2×0.17)÷(8.3144626182×573.15))×(1.2÷(1.2−1))×((101,325÷200,000)^(2÷1.2)−(101,325÷200,000)^((1.2+1)÷1.2))). = 0.0000283667 kg/s. or 0.0000283667kg/s × 3600s/h = 0.10212012 kg/h or = 0.81696096 kg per 8 hours. Note: the above mass flowrate is approximately for gaseous or vaporized kerosene with unknown actual volumes. Nevertheless, we are now considering the mass, which would be more reflective of the actual quantity of kerosene flowing through the orifice. We can therefore disregard the actual volumes occupied by the vaporized kerosene in the calculation. (The mass would still be consistent, regardless of the volume). Therefore, if we now take the typical density range of liquid kerosene @15 or 20°C as 780 to 810 kg/m³, we can then calculate the volumetric consumption of liquid kerosene of the 500HK lantern as follows:- Q = 0.0000283667 kg/s ÷ (780 to 810 kg/m³) = 3.63675641E−8 to 3.50206173E−8 m³/s. or Q = 0.10212012 kg/h ÷ (780 to 810 kg/m³) = 0.0001309232 to 0.0001260742 m³/h or 0.1309232 to 0.1260742 l/h or 1.0473856 to 1.0085936 litres per 8 hours. Therefore, these calculated figures would imply that a properly working 500HK lantern operating at its default, norminal or rated output would take: 1 litre ÷ (0.1309232 l/h to 0.1260742 l/h) = 7.6380656751 to 7.9318369659 hours or about 7 hours 38.28min. to 7 hours 55.91min. to fully consume a single 1-litre fount filling. This appears to be very much closer to the 8-hour lantern 'burn period per filling' stated in the specifications on Hytta's website. This time, I'd regard the calculated figures to be not too far off the mark and probably valid.?? It is also important to be aware that the above calculated range of volumetric flowrates, Q would not actually correspond to the volumetric flowrates of vaporized kerosene @583.15K having an unknown density range that is passing through the jet orifice but rather, they are more reflective of the lantern's volumetric consumption rate of liquid kerosene with a density range of 780 to 810kg/m³. As I've not carried out any controlled-setup measurements on the actual lantern to verify the calculated figures, I still cannot be absolutely certain if they are consistent or not. Perhaps members here might actually attempt such a measurement someday. Again, opinions and inputs are always welcomed. It would certainly be great if anyone could share a more appropriate formula or equation that could more accurately reflect the actual measured flowrates of a real operating lantern. The actual physically measured values would undeniably, still be the absolute or ultimate reference figures for accuracy.
@MYN The book Light and Heat from Hydrocarbon Appliances was privately published in hard copy (year 2000), and is available on the Bay of Evil and from some lamp-related sites in the USA. Best regards Tony
@Tony Press , Thank you for the info. I've never purchased anything from Ebay before. I might make an attempt someday. Its a pity that the book isn't more widely available. I would have bought it without hesitation if it is conveniently available at the local bookstores. I'm aware it was/is available from the Imperial Lighting Company in USA.
@bp4willi , Thank you. I can't be certain if my calculations are correct or not. The assumptions that I made was merely based on my limited current understanding of the subject matter. There could be things that I've ignorantly left out or not aware of. I'm not even absolutely sure if that is the correct mass flowrate formula to be used. Therefore, I cannot yet tell if the nearly-consistent results to Hytta's data was a mere coincidence or are actually correct. There's bound to be an even more accurate relationship somewhere out there.
Hi @MYN I am not sure what is more crazy - these calculations or your acetone lamp Anyway, from my memories of Fluid Mechanics from University I remember that the discharge coefficient for particular orifice shape was important. But you know that already. In my opinion discharge will be different depending on not only shape of the orifice, but also its diameter. And I am not writing about simple orifice area calculations. What I mean is that, for particular shape of the orifice, the discharge coefficient will impact gas/fluid discharge in different way for jets of 0,01mm or 0,15mm or 0,25mm or 0,3mm diameter. There should also be some other factor taken into consideration for particular jet diameter size, IMHO I would guess that - the larger jet diameter the less important orifice shape (angle of approach) is. good luck, Piotrek
Hello Myn, you can still buy this book on the internet at Imperiallightningco, but also on ebay.com. The seller is the same! The shipping cost to Malaysia would be $42.86. The book costs ‘only’ $43.95. Greetings Joerg
Yes, Wim. I do remember that. The situation is that I've not been able to conveniently order anything by mail. Would try work something out to address this.
Hi @Piotrek , You're probably right. I do have some tendencies to do or attempt seemingly crazy stuffs. You are certainly correct in implying that the discharge coefficient of a nozzle or jet orifice is also influenced in some ways by their diametres. What I've came across in some literatures was there were mentions about the coefficient values being proportional to the diametres in general, although not necessarily in a linear fashion. Since I have not yet found sufficient info on this with orifices sizes in the range that matches those of most liquid-fueled lanterns, I had no choice but to ignore this factor in the previous calculation. The angle of approach within a jet orifice is merely a single factor which affects the coefficient. I only used that to substantiate my previous assumptions out of convenience since, I happened to have at least one reference about that on hand. In fact by basic definition, its actually the ratio of the actual flowrate over the theoretical flowrate. As mentioned earlier, I would not be able to know the actual flowrate due to insufficiencies on my measuring means. Besides that I was also unable to incorporate other revelant factors such as Reynolds numbers, gas expansion factors, etc. due to inadequate expertise . The calculation was actually attempted with no prior knowledge on the actual flowrates of the devices. It was almost entirely a theoretical approach.
Thank you @Jörg Wekenmann for shedding light on its availability at present. I really need to address my existing incapacities on mail orders.
@MYN - I respect your efforts no matter how much theoretical. Curiosity in math and physics shall be valued these days. People do not want to learn even a multiplication table. Maybe try to "discuss" these calculations with some Artificial Intelligence (AL) software. AL should be able gather and analyse some huge theoretical data to help you.
By the way, I have not yet been able to find any post here or elsewhere about someone calculating the fuel flowrate of a lantern in a similar approach. That's to say, by means of some mathematical expression which contains at least two of the most well-known variables that affects the flowrate, namely the orifice cross sectional area and the operating pressure. In particular, using the figures that are relevant to the specifications of real lantern models as examples or reference. It would certainly be enlightening if anyone could share those posts.
I do find A.I. to be helpful at times, especially saving me a fair amount of time in search of references. Sometimes, the references are a little overwhelming or too much for me to digest in short time. It might not provide a direct answer. At times, the answers provided were not consistent or incorrect when I re-checked them. I suppose A.I. would be sourcing most or all info from what's available on the internet. A.I. might not be too helpful when it comes to the technical stuffs that are specific to classic pressure lamps or stoves. Probably because the details of these devices were never published or uploaded into the Web in the first place. For similar gaseous or vaporized fuel calculations, it had actually led me to some advanced research work in jet/rocket propulsion systems instead.