Calculate BTU’s Produced By a Pellet Stove

[whitsett2014]

Wow, 306 BTU/hr is about the same heat as a 100W incandescent bulb emits?

Not really sure earlll,
Remember though, this was only for 3 of the 'cooler' surfaces that I measured and determined to be worth calculating. Also I'm not guaranteeing that these numbers are 100% accurate. It seems to be fairly accurate, though there is really nothing to check it against.

 

[whitsett2014]

the Advance may not be the best choice for this test...the actual amount of pellets being fed is semi related to the feed adjuster knob and the temperature being read at the ESP probe. The stove will manipulate the actual on/off times of the auger in relationship to the exhaust temp....

Of course, I could be way off base, and in that case/either way...you are doing an excellent job. I am intersted in what you end up with. Super LEEDS platinum ultra is cool./ do they have that yet? ;-)

Delta-T,
The first point you make is taken out of the equation I think, by the fact that we are doing the calculation in the dead of winter (stated earlier in the post) in which case, we assume the stove will never reach it's set thermostat temperature (and never actually does to my knowledge).

As for your question, I'm fairly new to the company (started my co-op here Jan. 2, 2011) and as far as I know, we are only LEED Platinum. We were the first LEED Gold certified office building (before they invented Platinum) in Ohio. We are currently striving to achieve NZE(net zero energy) VERY soon.

Regards,
Whitsett

 

[jtakeman]

Wow, 306 BTU/hr is about the same heat as a 100W incandescent bulb emits?

Not really sure earlll,
Remember though, this was only for 3 of the 'cooler' surfaces that I measured and determined to be worth calculating. Also I'm not guaranteeing that these numbers are 100% accurate. It seems to be fairly accurate, though there is really nothing to check it against.

It has to due with surface area. You would get burnt from the 100 watt bulb. But you can touch the side of the stove. me thinx?

 

[whitsett2014]

Wow, 306 BTU/hr is about the same heat as a 100W incandescent bulb emits?

Not really sure earlll,
Remember though, this was only for 3 of the 'cooler' surfaces that I measured and determined to be worth calculating. Also I'm not guaranteeing that these numbers are 100% accurate. It seems to be fairly accurate, though there is really nothing to check it against.

It has to due with surface area. You would get burnt from the 100 watt bulb. But you can touch the side of the stove. me thinx?

Indeed you can.

 

[tjnamtiw]

One last time.............. you are totally ignoring the hottest surfaces of the stove which are behind the trim panels on the sides and back. Since you've only been there a few weeks, maybe you should spend some quality time up close and personal with the stoves learning how they work and are constructed. To assume a wood burning (pellet form) space heater gives off only 306 btu/hr of radiant heat is ludicrous.

 

[whitsett2014]

To assume a wood burning (pellet form) space heater gives off only 306 btu/hr of radiant heat is ludicrous.

I did not make that assumption. You only said that I did. I stated that this was only for the three surfaces on the front, foot of the stove under the glass. Never once did I say that this stove in its entirety only gives off 305 btu/hr of radiant heat. I have only done this calculation for these three surfaces because as of right now I don't have a temperature sensor that is able to measure the temp. of the glass on the front, and I also don't have the means (or time) to take the whole stove apart to try to determine if there is actually anything behind what you are saying (which I have also already stated). Again, I'm not guaranteeing the accuracy of these calculations either. I have taken a Physics course in thermodynamics and that is about all that I am going off of.

Regards,
Whitsett

 

[whitsett2014]

Something to add here...
After putting some more thought into tj's comment about the stove being double walled, and thus concealing most of it's thermal radiation from me, I have a hypothesis... The blower is located inside of this double wall, so I think that it is fairly safe to assume that the heat created from thermal radiation is being blown out through the vents by the blower. It also has a 'Combustion Air intake Back draft Damper' (the purpose of this is to get rid of the harmful emissions caused by combustion out of the ventilation leading outside) which I assume is connected directly to the burn box. This really only makes sense because if you put the blower directly from the burn box you would be blowing all of the emission from the combustion into your house/facility. However I don't know for a fact that this is how the blower and back draft damper are set up, I'm only speculating (and it makes sense to me ).

Conclusion- It is safe to assume that the majority of the radiant heat is being distributed through the blower out of the vents. (open for discussion of course)

Regards,
Whitsett

 

[tjnamtiw]

To assume a wood burning (pellet form) space heater gives off only 306 btu/hr of radiant heat is ludicrous.

I did not make that assumption. You only said that I did. I stated that this was only for the three surfaces on the front, foot of the stove under the glass. Never once did I say that this stove in its entirety only gives off 305 btu/hr of radiant heat. I have only done this calculation for these three surfaces because as of right now I don't have a temperature sensor that is able to measure the temp. of the glass on the front, and I also don't have the means (or time) to take the whole stove apart to try to determine if there is actually anything behind what you are saying (which I have also already stated). Again, I'm not guaranteeing the accuracy of these calculations either. I have taken a Physics course in thermodynamics and that is about all that I am going off of.

Regards,
Whitsett

Sorry if I got your back up but you DID say "Infrared temperature gun says otherwise
nothing notable as far as surface temperature change is concerned about the top, back, or sides. Less than 15 degrees C, most cases less than 10." in an earlier post and I was getting frustrated that you were ignoring most of the surface area of the stove. If you study the stoves a little, you will see that the sides and back can be exposed quite easily with snap latches and a few screws. Then you will see what I mean. Check out the owner's manual. I'm a Physics major from a loooooong time ago and I just gave away my thermodynamics text books last fall. :cheese:

 

[kofkorn]

Just a real quick thought...

Radiant heat only applies to surfaces that are directly exposed to the ambient environment. If you need to remove a panel to expose it, the radiant heat provided by the inside surface only affects the panel in front of it. If you think of the stove as a system, to get the proper numbers for radiant heat, you should only evaluate the exterior surfaces.

Now the inside panels may be providing additional CONVECTION heat that is not be accounted for and is escaping through vented holes. Figuring out this additional heat is not fun at all. But I don't think that Whitsett should be removing covers during his RADIANT calculations.

Thoughts are appreciated.

 

[tjnamtiw]

To help you better understand what I am talking about, I've taken a picture of the side of my Quadrafire Castile insert with the decorative side panel installed and then removed. The temperature of the outer panel is 80 degrees top to bottom (heavy cast piece). The inner panel that you see is the outer section of the double walled burn chamber. The surface temperature measured with a contact pyrometer is from 130 to 155 F from bottom to top. The upper switch in the picture is a snap switch that turns on the room circulating fan (convection fan) at 110 degrees. If these panels didn't get hot, the stoves would never work. I relocated the switch and replaced it with an adjustable one so I had some control over when the fan starts and stops. This was suggested by another poster, B-Mod.
While this is a different stove than yours, the principles of operation are the same. Judging from your comments above about the combustion blower, you need some guidance there as well. The combustion blower actually SUCKS air from the combustion chamber and blows the hot air out the flue. The air is made up by air coming into the combustion air intake and flowing into a chamber below the fire pot. The fire pot has holes in it that allow the air to flow with velocity into the combustion chamber. That's what creates the lively flame action. The combustion air flow and convection air flow paths are, HOPEFULLY, separated.

 

[tjnamtiw]

Just a real quick thought...

Radiant heat only applies to surfaces that are directly exposed to the ambient environment. If you need to remove a panel to expose it, the radiant heat provided by the inside surface only affects the panel in front of it. If you think of the stove as a system, to get the proper numbers for radiant heat, you should only evaluate the exterior surfaces.

Now the inside panels may be providing additional CONVECTION heat that is not be accounted for and is escaping through vented holes. Figuring out this additional heat is not fun at all. But I don't think that Whitsett should be removing covers during his RADIANT calculations.

Thoughts are appreciated.

He could quite easily shut down the stove, remove the panels, measure the surfaces, refire the stove, take his readings and reverse the process. Of course, for safety and liability reasons, AND if it's a union shop, this could be done by a maintenance man. Surely management would give him that support.
Yes, there are quite a few slots to allow the radiant heat to escape into the environment.
One thing no one has addressed is whether the stoves have OAK'S or not. That would also have an impact on the net heat gain.

 

[kofkorn]

There's a bit of a mix-up. Its not an issue about whether he is able to remove panels or not. Its about the type of heat being generated. Back to the text books, there are three types of heat: Conduction, Convection, and Radiation. Stoves use Conduction to move the heat from the inside of the firebox to the outside of the fire box. From there, Convection and Radiation take over.

Radiated heat is a line of sight energy wave only. If there are any surfaces in front of the radiating surface, they absorb the radiation and repeat the conduction process. Radiated heat does not escape through vents, Convection heat does. If he has to remove panels, then he is exposing additional radiating surfaces that wouldn't normally be exposed during typical operation. To calculate the actual Radiated heat provided by the stove, only the exterior panels should be evaluated.

I'm not ignoring the heat through the vents, but this is not radiated heat. This additional convection heat being lost through the vent holes is significant, but also difficult to calculate as the air flow rate is not known. However, it cannot be calculated using the radiated energy equations. Additional convection equations need to be evaluated.

 

[whitsett2014]

Hmmm...
The question to answer all questions is, where is the distribution blower getting the air from? If it is inside the stove, but outside of the burn box, the radiation of the burn box need not be taken into account because (for the most part) the radiated heat is being blown out of the vent. I realize that some of it would escape through other holes, but this is small enough (I think) to ignore, and even if it is considerable, much to difficult to calculate as kofkorn stated.

So,
Combustion blower sucks air (and harmful combustion emissions) from the burn box where the actual fire is located. Heat is lost out of the flue. Negative pressure is created inside the burn box. (the negative pressure part is in the manual)
The fire box is made of metal and releases thermal radiation to the air inside the outer shell of the stove.
The Distribution Blower intake is outside of the firebox (to prevent high concentrations of harmful combustion emissions) but inside of the outer shell of the stove.
Therefore I think that the thermal radiation on the inside of the stove can be (mostly) neglected because it is largely the same air that is being blown out by the distribution blower.
The heat that is not being blown out by the distribution blower is being absorbed by the outer shell (which I have measured temperatures) and then further radiated.
OR is being simply transferred through convection, which makes for a tedious calculation of which I have no interest in.

If my assumptions about the locations of the intake/output of the combustion and distribution blowers are correct, this seems to be an accurate deduction.

 

[tjnamtiw]

semantics :coolgrin: Heat is heat and it is contributing to the environment around the stove. If you measure the surface area and know it's STEADY STATE temperature, then can't you assume it is losing as much heat energy to the surroundings as it is absorbing to maintain that temperature? He can do what ever he wants but if he wants an accurate number then you have to take into account ALL TYPES of heat generation, don't you. He clearly doesn't know yet how the stove operates so wouldn't that be the first thing to learn to better understand how the heat is generated? Jumping feet first into a theoretical analysis without knowing how a machine operates is NOT the way to start.

 

[whitsett2014]

There's a bit of a mix-up. Its not an issue about whether he is able to remove panels or not. Its about the type of heat being generated. Back to the text books, there are three types of heat: Conduction, Convection, and Radiation. Stoves use Conduction to move the heat from the inside of the firebox to the outside of the fire box. From there, Convection and Radiation take over.

Radiated heat is a line of sight energy wave only. If there are any surfaces in front of the radiating surface, they absorb the radiation and repeat the conduction process. Radiated heat does not escape through vents, Convection heat does. If he has to remove panels, then he is exposing additional radiating surfaces that wouldn't normally be exposed during typical operation. To calculate the actual Radiated heat provided by the stove, only the exterior panels should be evaluated.

I'm not ignoring the heat through the vents, but this is not radiated heat. This additional convection heat being lost through the vent holes is significant, but also difficult to calculate as the air flow rate is not known. However, it cannot be calculated using the radiated energy equations. Additional convection equations need to be evaluated.

I agree completely with this, and it is exactly what I have been trying to say. Put very nicely korn.

 

[jtakeman]

You guys are beating on this pretty hard. You will "note" he is only guessing at the input BTU's of the test pellets. So the route is an unknown. You can beat the last BTU's out of it. But without a given input your end result is still just a wild stab at it.

A bit of overkill here?

 

[whitsett2014]

He clearly doesn't know yet how the stove operates so wouldn't that be the first thing to learn to better understand how the heat is generated? Jumping feet first into a theoretical analysis without knowing how a machine operates is NOT the way to start.

My assumptions are sound, and I agree with you that all forms of heat transfer should be considered. However this does not change the facts that korn stated about thermal radiation. The convection heat transfer is going to be much too tedious to calculate, and I'm certain that I do not have the proper tools to do so.

 

[kofkorn]

A bit of overkill here?

I don't disagree, it's all in how far he want's to go with it. At this point, I feel that the accuracy concerns will far outweigh any results we get out of it. It's too easy to keep going without the big step back once in a while.

 

[whitsett2014]

You guys are beating on this pretty hard. You will "note" he is only guessing at the input BTU's of the test pellets. So the route is an unknown. You can beat the last BTU's out of it. But without a given input your end result is still just a wild stab at it.

A bit of overkill here?

If I'm measuring the output temperature of the air from the blower, flow rate etc. The input BTU's don't need to be taken to account. Nor do corrections for efficiencies. I'm not trying to calculate the efficiency of the stove, just the BTU output (which is more than likely going to be a rough calculation).

We know the temp of the air and how much air it is outputting compared to the temp of the ambient air, and this will give us a rough estimate of the BTU's it is outputting.

 

[jtakeman]

You guys are beating on this pretty hard. You will "note" he is only guessing at the input BTU's of the test pellets. So the route is an unknown. You can beat the last BTU's out of it. But without a given input your end result is still just a wild stab at it.

A bit of overkill here?

If I'm measuring the output temperature of the air from the blower, flow rate etc. The input BTU's don't need to be taken to account. Nor do corrections for efficiencies. I'm not trying to calculate the efficiency of the stove, just the BTU output (which is more than likely going to be a rough calculation).

We know the temp of the air and how much air it is outputting compared to the temp of the ambient air, and this will give us a rough estimate of the BTU's it is outputting.

What is the route of you equation?

 

[whitsett2014]

See page 2 for sample calculations.

 

[tjnamtiw]

That's enough for me!! I'm off to help someone who wants and needs help........................... Forget about my comments on an OAK also because that might as well be ignored too.

Have fun.

 

[whitsett2014]

Thanks for everyone who had input on the subject, it really was a big help.
I think I have everything that I need now.

Regards,
Whitsett

 

[jtakeman]

Some how I feel like a used ?????. Got what he wanted and he's out of here.

We want to see the end result when you get it. I can use it to compare with what I have been doing. The full equation wouldn't hurt for us loser's either.

 

[BDPVT]

Whitsett, I don't want the rain on your parade but in my opinion what you are attempting cannot be done through contrived calculations without significant data collection to support your rather vague, and at times, incorrect assumptions. It seems your calculations will be a wild guess at best. Design an experiment that answers the basic question. If you need to measure the BTU output of the stove into the conditioned space then that is what you should do. There are so many variables involved I won't even get into it. My head is starting to hurt. Needless to say, you have only begun to scratch the surface. I don't mean to sound so negative and hope you are successful. Good luck!