Calculate BTU’s Produced By a Pellet Stove

whitsett2014 said:

Lets say that the stove is set at rate 3 (on a scale of 1-6) and 6 is the maximum feed rate. That puts me at 3/6*5.7lb/hr=2.85lb/hr. Over the course of the day the stove runs at a constant speed for 9 hours. It would burn 25.65lb of pellets.

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This is a bad assumption. The settings do not necessarily work out to be an even division of the max rate. Based on your assumption, at setting 1, there would likely not be enough pellets feeding to keep the fire going. My stove has a max rating of ~62,000 BTU. It has 9 feed settings. On feed setting 1 the stove is rated to put out ~20,000 BTU. And even then, I wouldn't assume that you can simply divide that range into equal bits. You would need to run a few bags through to determine your feed rate.

Whitsett,
The pellet stove efficiency rating published by their manufacturers is a combined rating which includes combustion efficiency, electrical efficiency and heat transfer efficiency.
Now pellet stoves mostly burn at over 98% combustion efficiency, and the electrical efficiency is about 99%. Let's analyse a typical pellet stove efficiency.
Assume, for the sake of argument, that the heat transfer efficiency of a pellet stove is 60%. If you add up 98%, 99% and say 60%, you get 257. Divide that figure by three to get the overall efficiency and it comes out at 85.7%.
78% is the EPA's, (Environmental Protection Agency), assumed default efficiency for a pellet stove's overall performance. It's clear then, that the heat transfer efficiency of pellet stoves is often less than 60%.

Combustion Efficiency And BTU./Hr.
The published Btu. rating of a pellet stove relates to its combustion efficiency. This is a measure of the heat produced from burning fuel. It does not directly relate to heat available to the home as some of it disappears up the flue.
One pound of hardwood pellets will produce around 8,200 btu. Softwood pellets slightly more. As combustion efficiency is so good in pellet stoves, this figure is very close to the actual heat 'input' of the stove. So take a stove with a published rating of 40,000 btu./hr. Divide 40,000 by 8,200 and you see that at that output, the stove is burning 4.8 lbs. of pellets an hour.
This 'heat input rating' is the main figure we have for assessing the capability of a pellet stove.

EPA Approval And Pellet Stove Exemption
As well as assuming an efficiency of 78% for pellet stoves, the EPA also stipulates that they produce particulate emissions of less than 2.5 grams per hour to be approved. Tests for this must be carried out at an independent testing laboratory such as Omni.
To be exempt, a pellet stove must have an air to fuel combustion ratio of more than 35:1.

Heat Exchanger Efficiency
From the above heat transfer efficiency figures, it's clear that there must be a wide range of heat exchanger efficiencies in pellet stoves. To be effective a heat exchanger must have a large surface area, and the stove must direct hot air evenly over that area.
Broadly speaking, the heavier the heat exchanger, the more effective it is, so as a rule if comparing pellet stoves of a given heat input, the heavier stove is likely to perform the best. Two provisos to that: Some heat exchangers are made of aluminium which is lighter than steel, and don't compare a cast iron stove by weight with one fabricated from steel. Cast iron bodies are thicker and heavier than steel so compare like with like.
Bottom line? Pellet stove efficiency is only as good as the efficiency of its heat exchange system

It may seem so, but you can find many people on this site that have taken the rated information to determine the amount of time that a single bag will burn, and found out that the stove feeds very differently than the rating. Plus the stoves don't feed based on weight. They use an auger which feeds based on volume. The pellet lengths vary significantly from manufacturer to manufacturer and this make for a very different pellet density, which dramatically affects the Lb/hr feed rate.

Check out J-Takeman's pellet comparison testing which is all done using the same stove and settings.

https://www.hearth.com/econtent/index.php/forums/viewthread/60581/

The burn times vary from ~20 hrs to 27 1/2 hrs for a 40lb bag.

I suggest strongly that you measure your consumption and not calculate it.

kofkorn said:

It may seem so, but you can find many people on this site that have taken the rated information to determine the amount of time that a single bag will burn, and found out that the stove feeds very differently than the rating. Plus the stoves don't feed based on weight. They use an auger which feeds based on volume. The pellet lengths vary significantly from manufacturer to manufacturer and this make for a very different pellet density, which dramatically affects the Lb/hr feed rate.

Check out J-Takeman's pellet comparison testing which is all done using the same stove and settings.

https://www.hearth.com/econtent/index.php/forums/viewthread/60581/

The burn times vary from ~20 hrs to 27 1/2 hrs for a 40lb bag.
I suggest strongly that you measure your consumption and not calculate it.

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Actually, That's for both bags. LG's were the only ones from 1 bag of fuel. I do agree, You have to measure the stoves consumption to eliminate any variables. Stoves do not feed by specific weight. Just volume. And hour to hour they will very slightly because stepper motors do stahl slightly. Taking an average of multiple hours of running should help.

Sorry J, you're right I forgot that piece of info. I wish I could get those temps you measured for 24 hrs out of one bag

whitsett2014 said:

I appreciate all of the useful information that everyone is providing!

However I would like to redirect this forum to the original question now, How to measure and/or calculate the instantaneous output of a pellet stove.
The answer yet alludes me.

Never give up,
Never surrender.
Whitsett

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Are you trying to actually measure output? Way to many variables to control in an uncontroled setting IMHO. How precise does this evaluation need to be? I suggest you calculate/ estimate based on info supplied by the manufacturer and the suggestions already made here. I do not think you are going to do better than that. Hope it works for you.

whitsett2014 said:

How to measure and/or calculate the instantaneous output of a pellet stove.

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Quick answer?? VERY CAREFULLY

As you can see by all of the information given so far, this is not a simple thing and not feasible without significant testing equipment and controlled space. Too many variables. All you can do is to make some assumptions, lock some of the variables and state your results with a low tolerance of accuracy. Other than that, you'll be chasing your tail for a very long time.

whitsett2014 said:

BDPVT said:

Whitsett,
It's clear then, that the heat transfer efficiency of pellet stoves is often less than 60%.
Bottom line? Pellet stove efficiency is only as good as the efficiency of its heat exchange system

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BDPVT,
Hey, thanks for the post! I had a couple of questions raised as a result...
1) Are the EPA efficiency numbers available for the public?
2) Given this overall percentage, could you take the expected output (by measuring the flow rate and gathering the input data) and multiply it by the average efficiency to find the instantaneous output of the stove? Of course this number would be strictly theoretical... I will need something more concrete.
For question two would you use only the transfer efficiency, or the total average efficiency?

To answer your question, they need to be reasonably accurate. Maximum +- a couple thousand BTU's hopefully.. This goal seems to be slipping through my fingers rather quickly without thousands of dollars of equipment though.

Thanks,
Whitsett

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Answer to question 1) Yes, if the stove is certified it should be listed on EPA website. Table will show heat output BTU/HR. You may also get this info from the manufacturer.
Answer to question 2) You should be able to back into this number, but you are correct, it is theoretical. I would use transfer efficiency or total BTU output if possible. Problem is most stoves use the EPA 78% Default Efficiency Rate so you don't know for sure how they really compare.

your Tfinal should be in the 150F - 250F range (65C - 120C) . 100F is really low.

using your numbers assuming Density of Air = 1.29 Kg/m3:

(1.005 kJ/(kg* Deg K)) * (1.29 kg / m3) = 1.2964 kJ/(m3* deg K)

Prev ans * 3.822774 m3 / min = 4.931 kJ/(m * deg K)

Assuming output temp ~= 250F temp differential is 103 K

Prev Ans * 103 Deg K = 507.93 kJ / min ~= 480 BTU / min

Prev Ans * 60 = 28800 BTU / hr

Prev Ans / .78 = 36923 BTU / hr (you divide by the exchanger efficiency because you are measuring the output temp of the room blower, not the temp difference of the combustion chamber. So the efficiency increases to get the stove rating. I hope this make sense.)

This seems much closer. When you factor in the remainder of the heat going out the exhaust and what not, you should be really close.

After all my experiments with improving the heat transfer efficiency of the Quad's tubular heat exchanger, if your stoves have tubes also then I can tell you that each tube has a different exit temperature as well as a different flow rate. That should make life interesting. Even if it doesn't have tubes, you can pretty well bet that there is inconsistent temps across the air exit.

No matter how you cut it this whole thread borders on the "Dilbert Principle".

Ro3bert said:

No matter how you cut it this whole thread borders on the "Dilbert Principle".

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Depends there Ro3bert.

The co-op does well maybe a new slot, and more $s , does bad, maybe an assisted exit exit at next review time. Who knows.

Maybe the boss is looking for a more direct measurement method such as placing the stove inside a big heavy metal box on a concrete floor with a controlled electric heater inside an identical box next to it and taking thermal images until the electrically heated box and the box with the stove in reach the same temperature.

Who knows.

Maybe the boss is looking for some get up and go or the ability to slice and dice a problem.

oooooooooooooooooh, now THAT would be the best way to measure it, Smokey!! Put a know # of btu's into the chamber and measure the heat rise. then put the stove in there exhausted outside and measure again. You're not as dumb as everyone said you were!! :bug:

Ro3bert said:

No matter how you cut it this whole thread borders on the "Dilbert Principle".

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I won't deny it. It's fun to exercise some of that thermodynamics theory I paid so much for.

tjnamtiw said:

oooooooooooooooooh, now THAT would be the best way to measure it, Smokey!! Put a know # of btu's into the chamber and measure the heat rise. then put the stove in there exhausted outside and measure again. You're not as dumb as everyone said you were!! :bug:

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I'm also very good at stomping out burning fire dance performers.

ha ha ha. thinking about it, you idea would work even better if you got the stove 'steady state' and then rolled it into the insulated chamber and recorded the heat rise over time. Then replace the stove with electric resistance heaters that are 100% efficient. Let the chamber come to ambient temp again and then adjust the electric heaters to duplicate the heat rise of the stove. Then you would absolutely have the total heat given off by the pellet stove nailed down. From that you would get the btu's per hour total output from the blowers, radiant heat and even the motors. Weigh the pellets in the hopper before and after the test to see how many pounds you used and you have the input btu's using the supplier's rating.

kofkorn said:

your Tfinal should be in the 150F - 250F range (65C - 120C) . 100F is really low.

using your numbers assuming Density of Air = 1.29 Kg/m3:

(1.005 kJ/(kg* Deg K)) * (1.29 kg / m3) = 1.2964 kJ/(m3* deg K)

Prev ans * 3.822774 m3 / min = 4.931 kJ/(m * deg K)

Assuming output temp ~= 250F temp differential is 103 K

Prev Ans * 103 Deg K = 507.93 kJ / min ~= 480 BTU / min

Prev Ans * 60 = 28800 BTU / hr

Prev Ans / .78 = 36923 BTU / hr (you divide by the exchanger efficiency because you are measuring the output temp of the room blower, not the temp difference of the combustion chamber. So the efficiency increases to get the stove rating. I hope this make sense.)

This seems much closer. When you factor in the remainder of the heat going out the exhaust and what not, you should be really close.

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knock it off.

Can some explain how the EPA comes up with 85% efficiency from an appliance that only has a heat exchanger efficiency of 60%? Are we 100% sure that is what they can claim in that case?

I mean, I don't have a hard time believing the stove is wasting that much heat but for the manufacturer to be able to claim 80% efficiency in such a case is horribly misleading to the consumer.

Checkthisout said:

Can some explain how the EPA comes up with 85% efficiency from an appliance that only has a heat exchanger efficiency of 60%? Are we 100% sure that is what they can claim in that case?

I mean, I don't have a hard time believing the stove is wasting that much heat but for the manufacturer to be able to claim 80% efficiency in such a case is horribly misleading to the consumer.

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Simple,

The EPA just adds the universal government fudge factor.

Do you think they actually have anything remotely resembling a clue?