Need a litte help remembering.

[fireview2788]

I am trying to remember from high school science. If I have wood that burns at, we'll say 28,000 BTU and add another piece that burns at 10,000 BTU I will still only have a total of 28,000 BTUs. OR if you add another piece at 28,000 BTU you will still only have 28,000 BTU. If I remember correctly if you add = or less than energy you will not increase the amount of energy.

It's been a long time since HS science and doing what I do now, I never use this stuff.

fv

 

[smokinj]

Wow not sure I want to know the answer. 28,000 stove per hour is running full tilt. Adding more wood only gets you up to that max. It would take a bigger firebox to make more btu's.

 

[Danno77]

yes and no, lol.

If you have a split that HOLDS 28,000BTU, then you add another split that HOLDS 10,000BTU, then you can have a fire putting off anywhere between 0BTU/hr and whatever the max of your stove is!

if you burnt off both splits in one instant you would be shooting off all 38,000BTUs, but if you take an hour to do it, you'd be burning 38,000BTU/hr, if you burn them both in 30 minutes, then you'd be burning 76,000BTUs, which is about what a good stove is capable off running full tilt (a good big stove, i guess)

soooo, are we assuming that this split is putting off 28,000BTUs per hour (so, maybe it's a 20lb split burning at 4lbs an hour) and then you toss another split that is burning at 2lbs per hour? Cause the answer there is that you are burning 6lbs per hour, or combining the total energy.

clear as mud? i need a shot of caffeine. BK or some other wood nerd come check these claims out so I don't confuse nobody!

 

[bpirger]

There's about 6000 BTU/pound in wood....of course this varies from specie to specie, etc. But for run of the mill maple, ash, etc. this is a good number to use. Might be a touch higher, but 6000 is easy to deal with too. So how fast you get these BTUs out depends on how fast it burns. If you can burn 1 pound per hour, you'd have 6000 BTU/HR. That's fairly slow. SO if you burn 10 pounds per hour, you'd get 60K BTU/HR, that's pretty fast!

So when you fill up your firebox with 10 pounds of wood, you have 60K BTUs you will get. If it burns in two hours, 30K BTU/HR. 3 hours is 20K BTU/HR. The total will always be 60K...the rate depends on the fire.

So adding more wood, i,e, a bigger firebox, adds more energy into the box....but the rate it is released depends on how the fire is burning....

Think of the energy in the wood as potential energy...that is exactly what it is. When it is released, as heat, you feel warm. More wood means more potential energy. So the "integral" over the burn time will always sum up all the potential energy in the wood, i.e. you will release all the potential energy into heat...but the SUM is always the same.

Make sense?

Of course, efficiency of your burn comes into play here....as does specie, wetness, etc. You will NOT get more energy from heavier (i.e. wet wood), much energy is consumed to boil the water in the wood into steam. But when the water leaves, the wood fiber remains, and that is where the stored energy is.


So adding more pounds, adds more potential energy....but how fast it is burned controls the rate of the release of the energy....

 

[Gary_602z]

You might want re-ask your question. Just being a farm boy I know that if I add more wood that I am going to get more BTU's. Whether it is usable heat or heat going up the chimney is another question. Although if you get you stove glowing you might have a little more usable? :lol:

Unless I didn't understand the question.

Gary

 

[fireview2788]

The BTU's in the original post was a hypothetical so don't get wrapped up in that. How's this: If I put one piece of ash in the stove and start the fire I will get "X" BTUs. The next time I put 2 pieces of exactly the same size and seasoning of ash (they all came from the same round) will I get more BTUs that the first fire or the same?

My memory serves me that I will get the same.


f v

 

[Backwoods Savage]

I am trying to remember from high school science. If I have wood that burns at, we'll say 28,000 BTU and add another piece that burns at 10,000 BTU I will still only have a total of 28,000 BTUs. OR if you add another piece at 28,000 BTU you will still only have 28,000 BTU. If I remember correctly if you add = or less than energy you will not increase the amount of energy.

It's been a long time since HS science and doing what I do now, I never use this stuff.

fv

Shoot, I don't even remember high school any more leave alone what I studied.

 

[wishlist]

Sheldon? Where are you? . ( big bang tv show)

 

[Danno77]

Too many variables. I understand what you are asking, but there is now way to do this with wood.

If you give the fire EXACTLY enough air to combust at 20,000BTU with one split, and then you throw another in there and you don't change the air, then I THINK you are right that you'll still get 20,000BTUs from the two.

Too many weird variables with combustion surfaces and air speed and hot spots between the multiple splits and whether the euro is up and the moon is in alignment with Jupiter can effect the ability to do this well.

 

[fireview2788]

Thank you Danno, that's what I was thinking but also knew there were variables. Now you know why I'm a cop and not a science teacher.


f v

 

[Gary_602z]

Too many variables. I understand what you are asking, but there is now way to do this with wood.

If you give the fire EXACTLY enough air to combust at 20,000BTU with one split, and then you throw another in there and you don't change the air, then I THINK you are right that you'll still get 20,000BTUs from the two.

Too many weird variables with combustion surfaces and air speed and hot spots between the multiple splits and whether the euro is up and the moon is in alignment with Jupiter can effect the ability to do this well.

So at that rate if you add 10 splits you will still only get 20,000 BTU's? The total BTU's have to increase the BTU/hr may not go up but you would have a longer burn.

Gary

 

[Danno77]

Too many variables. I understand what you are asking, but there is now way to do this with wood.

If you give the fire EXACTLY enough air to combust at 20,000BTU with one split, and then you throw another in there and you don't change the air, then I THINK you are right that you'll still get 20,000BTUs from the two.

Too many weird variables with combustion surfaces and air speed and hot spots between the multiple splits and whether the euro is up and the moon is in alignment with Jupiter can effect the ability to do this well.

So at that rate if you add 10 splits you will still only get 20,000 BTU's? The total BTU's have to increase the BTU/hr may not go up but you would have a longer burn.

Gary

Thank you. It's important to note that by adding fuel you increase the STORED energy, but not necessarily the immediate output. So when I was posting what you quoted, it was a reference to the immediate output, not the stored energy.

 

[Wood Duck]

If you add twice the wood you have twice as much potential energy in the stove, but I don't think the answer to your question is that simple. Keep in mind that what you are asking about (the heat from the stove) is really the amount of energy released into the room in a given amount of time. The heat you get into the room depends on the temperature of the stove (as long as we are talking about only your stove other variables like size, shape, air flow, etc. don't matter). The temperature of the stove depends on how much energy is released inside the stove and how much air leaves up the flue (taking heat with it), which is controlled by the amount of air entering the stove, and the rate that the wood burns (size and density of splits). Also, some of the heat is used up to evaporate water from the wood, and there is always some, so you don't feel that heat. So, if you are giving the stove only enough air to burn a certain amount of wood, and that amount is smaller than the first split, I think adding a second split won't change things much because you still only have enough air to burn the same certain amount of wood. You would get a longer burn, but not hotter. On the other hand, if you are giving the stove air in excess of the amount needed to burn the first split, then adding a second will release more heat at the same time and the stove gets hotter and you get hotter. How much hotter is not easy to figure out because the extra air you added sucks some of the heat up the flue. Finally, I'll say that in practice wood burning is highly variable and it is not easy to predict what will happen using theoretical reasoning. I think two logs is usually hotter than one.

 

[TreePointer]

Maybe this will bring back some of what you are trying to recall from years ago:

Total energy = total BTU's involved. 28K BTU + 10K BTU = 38K BTU

Law of conservation of energy = energy is neither created nor destroyed. If you put 38K BTU in the system, there are always 38K BTU to be accounted for. That is, energy may change form but there is none magically created or destroyed. Some ways of changing form: heat, light, work, sound, phase change, et al. The energy starts out as stored energy in a log, but is released in many ways.

Reaction rate = In this case, it is the rate of combustion. Classic combustion reaction: Carbon compound/wood (CHx) + O2 --> CO2 + H20 + heat, light, etc. The reactants are to the left of the arrow. The products are to the right of the arrow.

The rate at which your wood (reactant) is consumed, the rate at which 02 (reactant) is consumed, or the rate at which heat (product) is produced is a way of expressing reaction rate. A faster reaction rate means more heat produced in a shorter time. As mentioned, there are a lot of factors that go in to this, but an important one that hasn't been mentioned yet is SURFACE AREA. This reaction happens at the surface of the wood, so if you increase surface area, the reaction rate will increase. Smaller splits increase surface area.

Limiting reagents = Reagents are what you put into the reaction. If you limit one (either the wood or the O2) your reaction stops when the limited reagent is completely consumed.

Phase change = since we are talking about water, the energy required to change from liquid water to water vapor does not raise the temperature of the system. That phase change energy goes into overcoming the forces that hold the liquid water molecules close to one another. Once those forces are overcome, the water molecules can reorganize as water vapor.

 

[bogydave]

csc(x) = 1/sinh(x) = 2/( e x - e -x )/cos(x) = ( e x + e -x )/2 * sec(x) = 1/cos(x) = 2/( e x + e -x ) +

tan(x) = sinh(x)/cosh(x) = ( e x - e -x )/( e x + e -x )/cot(x) = 1/tanh(x) = ( e x + e -x)/( e x - e -x ) = (x*y*z)

(x*y*z)/arc(z) = 1/2 ln( (z+1)/(z-1) )

solve for X & Y & Z

Then X ÷ Z X Y = BTU (north of the equator)

 

[Gark]

Good gravy, Dave! That doesn't compute... It won't factor in the price-of-rice in China. I don't grok this thread.

 

[Blue Vomit]

my head hurts...

 

[Gark]

csc(x) = 1/sinh(x) = 2/( e x - e -x )/cos(x) = ( e x + e -x )/2 * sec(x) = 1/cos(x) = 2/( e x + e -x ) +

tan(x) = sinh(x)/cosh(x) = ( e x - e -x )/( e x + e -x )/cot(x) = 1/tanh(x) = ( e x + e -x)/( e x - e -x ) = (x*y*z)

(x*y*z)/arc(z) = 1/2 ln( (z+1)/(z-1) )

solve for X & Y & Z

Then X ÷ Z X Y = BTU (north of the equator)

Respectfully submit to bogydave that should a power failure or battery shortage occur, we hope he has one of these:
(or OMG does that get done BY HAND !?!?)

 

[bogydave]

csc(x) = 1/sinh(x) = 2/( e x - e -x )/cos(x) = ( e x + e -x )/2 * sec(x) = 1/cos(x) = 2/( e x + e -x ) +

tan(x) = sinh(x)/cosh(x) = ( e x - e -x )/( e x + e -x )/cot(x) = 1/tanh(x) = ( e x + e -x)/( e x - e -x ) = (x*y*z)

(x*y*z)/arc(z) = 1/2 ln( (z+1)/(z-1) )

solve for X & Y & Z

Then X ÷ Z X Y = BTU (north of the equator)

Respectfully submit to bogydave that should a power failure or battery shortage occur, we hope he has one of these:
(or OMG does that get done BY HAND !?!?)

Have my old slide rule too. (somewhere) ( Now would need the big one with large numbers)
Forgot how to use it, maybe I should practice. Found Excel, forgot all about slide rule.
∠x R² =∑ ∆ ÷ .707 = ‰ ; is the short cut

 

[Woody Stover]

csc(x) = 1/sinh(x) = 2/( e x - e -x )/cos(x) = ( e x + e -x )/2 * sec(x) = 1/cos(x) = 2/( e x + e -x ) +

tan(x) = sinh(x)/cosh(x) = ( e x - e -x )/( e x + e -x )/cot(x) = 1/tanh(x) = ( e x + e -x)/( e x - e -x ) = (x*y*z)

(x*y*z)/arc(z) = 1/2 ln( (z+1)/(z-1) )

solve for X & Y & Z

Then X ÷ Z X Y = BTU (north of the equator)

That would look really impressive if you could toss in a few deltas and thetas. :cheese:

 

[fireview2788]

I knew there had to be some science nerds out there somewhere............



f v

 

[Danno77]

By the time I finished college I was about three classes away from having a degree in Mathematics. I scored in the 99th percentile for Math on the GRE. Fast forward almost ten years...I'm lucky to complete a simple algebraic formula!

My point is: The saying "if you don't use it, you lose it" is most definitely true with math. maybe I could re-learn it faster because I once knew it, but geesh, I've forgotten more math than I currently know, lol!

 

[Gary_602z]

I have tried the “if you don’t use it, you lose it†saying on my wife and it hasn't worked yet! :lol: If you can't find it is it lost? :smirk:

Gary

 

[amateur cutter]

Yep I'm with blue vomit, my head hurts too. More wood = more heat=happy wife = happy life. Never did like math anyway, but my equation makes kids ( or @ least it used too) & I can prove it. A C

 

[bogydave]

By now
There is probably a downloadable " app" to do it anyway.