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Help understanding thermodynamics/BTUs

  1. Nov 13, 2012 #1
    I am trying to design a heating element to heat a greenhouse. I found a very accurate calculator that helped me find the BTU/hr I need to heat this area to my desired night time max temp, that figure is 116,556.8BTU/hr. I need to figure out the dimensions of the stove I am building so I make it large enough to heat the space I am trying to heat. I also would like to make the fuel chamber large enough for the heater to last several hours (but this is less important and easier to figure out) I would like to know how different dimensions would affect the intensity and duration of my heat. I imagine the wider the spread of a flame, the more heat is produced, where height is not so much a factor.

    Now, let's talk about the heater:

    How it works: It is very very simple. The stove works by igniting combustible wood gas that is constantly released from underneath the top lit flame. The top lit flame begins to burn the top layer of fuel (usually wood pellets or wood pieces) and then the continuous up draft air flow (provided by the fan at the bottom of this device) pushes this gas up for the flame to burn. This gas is "preferred" and the process leaves solid chunks of carbon behind. This type of stove is not regulated by thermostat. It is a live fire which is controlled by the fan. The exhaust is clean and does not require an exhaust system for leaving the greenhouse. This is why it can be so small and have a high BTU/hr rating - because it is the exhaust is most of the heat.

    Luckily, I have some head start actual figures. There is a company selling one of these such woodgas stoves and the specs are: (This is a completely cylindrical stove. The fuel chamber rests inside of an outer chamber (which in the dimensions listed below is the "total").
    Total height: 11.5"
    Total diameter: 14"
    Fuel chamber height: 8.5"
    Fuel chamber diameter: 10"
    weight: 15lbs
    material: stainless steel
    Heat output: 50,000 BTU/hr
    Power usage: 240mA @ 9v

    Final details/thoughts needed for answering this question:
    If we made fuel type and amount of air-flow into constants, we are left with diameter and depth of the cylinder to reflect the BTU/hr rating of this stove. Obviously, the stove needs some amount of depth in order to work, but after a few inches in height....I'm not sure making the fuel chamber taller (I'm using depth and height interchangeably now) would affect the heat output. As far as I can tell, it will be the diameter of this cylinder which will change the amount of heat this puts out. To calculate the BTUs of the device two types of heat must be measured; gaseous heat and radiant heat: 1. heat from the exhaust 2. the heat which is radiating from the metal fuel chamber. The example stove I have given unfortunately doesn't tell us if they took the BTU/hr measurement from the exhaust only, or if the radiant heat from the device itself was also included. Any thoughts?

    Thanks for taking a look, any feed back is appreciated.
  2. jcsd
  3. Nov 14, 2012 #2

    Philip Wood

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    I'm worried about your heat input requirement: 116,556.8BTU/hr, because 1 BThU = 1.06 kJ, so 116,556.8BTU/hr = 116,556.8 x 1.06 kW / 3600 s =34 kW. This is a very big power; is it a very big greenhouse?
  4. Nov 14, 2012 #3
    The greenhouse has 2 propane heaters in it right now on either end of the same type. They are both 75000 BTU/hr Input and 60000BTU/hr output. I should have mentioned this earlier. Last year I only used one of them and it was enough to heat the space (though I kept it at 50F during end of February and I would like to increase that to 60F next year)....so, I suppose this is telling me a great deal. Interesting though, that if the BTU/hr output of both propane heaters are added together we have 120,000BTU/hr...close to my calculation of 116,556BTU/hr. Which is probably why there are 2 of them in there in the first place (I am renting this greenhouse and the owner knows nothing about it). The greenhouse is 48' long, 20' wide and gothic style (though I calculated volume of a cylinder greenhouse this size instead because it was easier and should be close enough to give me a good figure).
  5. Nov 14, 2012 #4

    Philip Wood

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    Fascinating! Your secret is safe with me.

    Surely, rather than the diameter and height of the cylinder, it is the rate of burning of the wood-gas which primarily determines the heat output? Or are you implying that the dimensions of the cylinder control the rate of burn?
  6. Nov 16, 2012 #5
    Yes, I agree, and the rate of burn for the wood gas is controlled by the fan. And, I guess fuel type (which has it's own BTU/hr). The numbers provided by these stove makers suggest a burn rate of 6lbs of fuel/hr @ 50,000 BTU/hr. Now, it's a little strange that they have these figures without mentioning fuel type, but they do mention the voltage and milliamps on the fan that powers it - noting that with 20 milliamps less fed to the fan, the BTU/hr rating drops by half! (25,000 btu/hr) and this is also true of the fuel consumption which drops to 3 lbs per hour. I am assuming they are using wood pellets, and probably hardwood pellets to get these figures. As, this would give you the most impressive figures.
    But, let's say that I want at least 6 hrs of sleep, before I'd have to tend to this stove again. I'm betting that the figure of 6lbs is the maximum weight this stove's fuel chamber can hold (give its size and the fact that it would make no sense to make the stove sound less impressive). So, If I want this stove to burn for 6 hours instead of 1 hr, I would need to design the fuel chamber so it has a capacity for at least 36lbs of fuel. So, in making the fuel chamber bigger I can either increase the height of the cylinder, the diameter, or both. I'm betting that increasing the diameter will actually dramatically increase the BTU/hr output of the stove (and therefore fuel consumption), but increasing the height of the stove will not cause the stove to put out extra heat, but not TONS of extra heat. This is because the only extra heat you would get would be radiant heat from a greater surface area of metal (because we made it bigger)....but the amout of exhaust heat over time would be the same. Where as, in the example where the diameter increases, the exhaust heat over time actually increases significantly. I think figuring out how much radiant heat this thing puts out will be the more difficult figure to find.
  7. Nov 16, 2012 #6
    Yes, I'm certain the dimensions are controlling the rate of burn. Now it's trying to figure out the right dimensions to create the right amount of heat output over time and keeping the heat in production for 6 hours minimum.

    So, how do we calculate the radiant heat output? and how can we calculate heat output from exhaust? For a newly designed, larger version stove.
  8. Nov 16, 2012 #7

    Philip Wood

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    "Yes, I'm certain the dimensions are controlling the rate of burn."

    First step must surely, then, be to find out more about the relationship between the dimensions and the rate of burn. I've no idea how to do this, but you probably have.

    You seem to be concerned with the modes of heat flow from the stove to the room. There's some chance of estimating their relative proportions (convective to radiative) if we could estimate the temperature of the stove, but I don't see how we could do this if we don't know the rate at which heat is produced.

    I've probably missed something, but I would have thought your first priority was simply to find the rate at which your fuel was producing heat. That heat (or a good proportion of it) will find its way into the greenhouse by some means or the other. Worry later (I'd suggest) about how much transfers by convection, and how much by radiation.

    But, as I said, I've probably missed the point.
  9. Nov 16, 2012 #8

    Philip Wood

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    Going back to hash 5, increasing the surface areas won't in itself increase the output, because the surfaces will be cooler, for a given heat output from the fuel.

    What I'm interested in is your talk of the fan and the diameter of the fuel chamber. These do sound very plausible major factors in determining the gas production rate and the burn rate. That's, imo, what you need to concentrate on.

    The height of the stove will indeed affect the proportion of radiant to convective output. My first guess is that increasing the height will increase the surface area through which heat escapes, and will therefore reduce its temperature, favouring convection over radiation. There will, though be some effect on the rate of burn. I believe that the design of combustion chambers is a specialist field. Unfortunately, I am no expert, just an interested physicist. At some stage you might possibly try the engineering forum.
  10. Nov 16, 2012 #9
    The model stove puts out 50,000 BTU/hr with a 10" diameter fuel chamber and it takes one hour to burn out 6lbs of fuel, using a 240mA electric power to a small fan.

    I want 116,000BTU/hr. My guess is: 116,000 / 50,000 = 2.32 Which means the new stove needs to produce as much heat as 2.32 of the model stove. So, we could just use three of these stoves side by side to produce the heat required, that's why I think diameter is the key. Anyway, to accomplish this with one stove, the new stove will in some way need to be 232% larger than the model stove.

    I'm guessing that the majority of the BTU/hrs coming from this stove are generated by the hot exhaust gas therefore, when building the new stove I will chose to increase the diameter to meet the BTU/hr rating necessary. So, my new diameter is 10" x 2.32 = 23.2"

    I would also like the stove to burn longer before going out. I want it to last exactly 6 times longer....So, I've already messed with the diameter, now I will increase the height 6x. So, original height of 8.5" x 6 = 51"

    Our new volume = (∏r2h) 3.14 x 134.56 x 51 = 21548.4384"cubed
    Our old volume = (∏r2h) 3.14 x 25 x 8.5 = 667.2500"cubed

    Now, since I changed the height, I also increased the amount of heated metal. The metal is capturing the heat and releasing it over time at a slower rate and over a longer period of time than air. If we compare this to a fire pit, we see that when the fire goes out, the hot air is gone. But, if we stick pieces of metal into the fire, they will radiate heat after flames are gone. The fire will have still burned for the same exact amount of time, because the fire burns by fuel, and the metal isn't a fuel (until it reaches a higher temperature that it will just not reach in this stove). So, the heated metal is actually capturing then releasing heat that otherwise would have been lost? Nope, I think I just figured this out....so this heat captured/sucked in by the metal would have heated air if the metal weren't "in the way". SO, this, I suppose shorter stoves will have a higher BTU/hr output simply because they are not heating as much metal...and the metal slows down the amount of heat output/hr.

    BUT, the thing is, the depth of burning fuel at any one time is equal in both the small stove and the large stove! (Given that all other variables are now controls) I mean, if gas is burning from the top to a depth of 3" in the small stove, it is the same 3" depth in the large stove. This means....that the....volume of burning fuel, is greater in a stove with a wider diameter no matter how tall or short it is....well, almost, it must be at least 3" high...or whatever the depth of burning fuel actually is. To make a cross reference, it is the same as boiling 1 gallon of water on the stove. It will boil faster in the pot with the larger diameter because there is more volume of water in contact with the hottest part of the pan, which has also increased in surface area.

    Ok, given everything I just presented was sound....It means that if I were to redesign the model stove and only increased the diameter, I would see a 232% increase to the BTU/hr output...BUT, I wanted it to burn longer, so I made it taller. From what I pointed out a moment ago, I suspect that adding more metal will mean decreasing the BTU/hr output because more metal means more temporarily trapped heat (which I think you mentioned earlier)...a heat sink. So, I am left wondering how many BTU/hrs to subtract from what should have been 160,000 BTU/hr output. Since I made the height 6x taller, all I know is that my resistance to BTU/hr output is 6 times greater than it used to be.
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