Calculate the BTU's put into the water from EKO40 burn today

[Northwoodsman]

I burned 98 lbs of wood today in my EKO40 and when I put the start/ending storage tank temperatures into my spreadsheet that I've been using all winter, I calculate that I put almost 200,000 more BTU's into my storage tanks than is physically possible (knowing that wood has a BTU rating of approx. 6,400 BTU/lb, if I burnt 98 lbs of it at 100% effciency I would have an output of 627,200 BTU's). In my spreadsheet I actually have a column that takes into account the BTU's being used by the house at the same time I am heating the water storage and thus since it was about 65 F today my house didn't take many BTU's from the water.

Thus, I obviously have a mathmatical error in my spreadsheet formula's.

Here is the burn data from today's burn:

I have (2) 500 gallon propane tanks that are stacked horizontally on top of each other and I charge the tanks with boiler water from the top down.

Starting tank temps: TOP OF TOP TANK.................122F
BOTTOM OF TOP TANK..........121F

TOP OF BOTTOM TANK..........120F
BOTTOM OF BOTTOM TANK...118F

After burning 98 lbs of wood in about 5.5 hours and assuming ZERO heat demand by the house,

TOP OF TOP TANK.................175F
BOTTOM OF TOP TANK..........175F

TOP OF BOTTOM TANK..........171F
BOTTOM OF BOTTOM TANK...161F

Any help with how to accurately calculate the BTU's given by the burnt wood is greatly appreciated.

Thanks,

NWM

 

[ohbie1]

I'm no expert, but let's say (to simplify the calc.)you raised your !000 gal. from 122 to 175 degrees. That's 53 x 1000 x 8.3 = 439,900 BTUs you added to your tank.....No?

 

[jebatty]

I would do the calc a little different. Tank average start temp = 120; tank average end temp = 168, delta T = 48. 8.34 x 1000 x 48 = 400,320 btu. Gross btu's from 98 lbs wood at 6400 btu/lb = 627,200. Net tank btu delivered efficiency = 400320/627200 = 64%. Lots of other variable at work, but my gut reaction, pretty good end result.

The actual end temp likely is somewhat higher, if your tanks stratify like mine does. My bottom doesn't tend to show much temp rise until most of the water above is close to boiler/tank input temp.

 

[Piker]

Don't forget that you have about 2400 lbs of steel in those tanks that is storing Btu's as well.

I still haven't found any good numbers on how many Btu's a propane tank without insulation will radiate at 165* in about a 60* enviroment. . Given the tests that I have done I came up with about 75,000 Btu's/hr on two 500 gallon tanks... but I don't have the best measurement devices in place, so I think it's more. Any thoughts?

cheers

 

[Northwoodsman]

I am working on an EKO40 system with a friend of mine and have a question regarding the size/type of coils required for the pool type storage unit we plan to use.

The facts thus far are as follows:

900 sq. ft. gagage with 11 ft ceilings = 43K BTU required to keep garage at 60 F on a zero degree day (we are planning to purchase a unit heater)

1,512 sq. ft. main level with 8 ft. ceilings = 35K BTU required to keep garage at 70 F on a zero degree day (exisiting baseboard)

1,000 sq. ft finished basement with 8 ft ceilings = 25K BTU required to keep garage at 70 F on a zero degree day (existing baseboard)

Total BTU for house w/ garage is 103K BTU


We are planning to have approx. 1,500 gallons of storage and thus need to determine the qty and size of the coils required to go into the water tank.

I have looked at the STSS web site and reviewed their coils but am not quote sure how to calculate the ones I need.

They have 2 coils shown: 120 ft coils with 26,400 BTU capacity and 180 ft. coils with 39,600 BTU capacity
(both coils are rated with the BTU's listed above at 120F average tank temp)

However, we plan to keep the water in the tank between 140-180 and thus I'm a little confussed at the BTU rating at these higher temps.

Obviously, on colder days we need to keep the temp hotter and on warmer days we can let it get to 140F.

I've contacted STSS but have not heard back yet.

Thanks in advance for your help!

NWM

 

[Northwoodsman]

Sorry about this post.

I meant to put this new topic in a new post but accidental hit the reply button.

I have since posted it as a new boiler room post. Please check the new posts if you would like to reply.

NWM

 

[DaveBP]

Any help with how to accurately calculate the BTU’s given by the burnt wood is greatly appreciated.

Knowing how much heat is stored in a tank is tough if you only have temp gages at top and bottom. With the hot water coming in at the top of the tank "laying on top" of the cooler water below it that line of separation ( the thermocline) could be anywhere between the gages. You just don't know where it is. So you can't tell how much 'hot' water is in there even when you know precisely what temperature that water is.
Jim avoided this dilemma by using a flowmeter and thermometer to get a real time measurement he could total over time in his tests earlier this year.
By merely adding the high and low temps and dividing by 2 you get the median temp but that is not likely to be the temperature of the water if it were all thoroughly mixed to the same "average" temperature. You need that true "average" temperature to make your spreadsheet come out with accurate numbers. The more gages you have vertically along your tank the better estimate you can get. You could try a bunch of these:

(broken link removed to http://www.omega.com/Temperature/pdf/RLC-80.pdf)

That's the most affordable method I've come up with so far. You would need to stick them on and cover them with some kind of insulation and whip it off and take quick readings to get the best accuracy but it could give much better results for total BTU calculations. I believe NoFossil mentioned using them. I can't imagine I won't have at least a 10-pak of them on my tanks when they go into service.

I think you need to size your heat exchangers (coils or flat plate type) to your rate of heat transfer not to the volume of the storage. The faster you need to get the heat out of your boiler and into your loads (either storage or heated space and DHW or a combo of any of these). I mean like BTUs/hour. I believe this is going to be determined mostly by the size of your boiler and how hard you are running it. And the higher the average temp of your storage the harder it is to get that heat transfer. The higher the average temperature of your storage tank the larger the heat exchanger you will need to transfer the same amount of heat. It gets tricky to calculate because there are convection currents that effect transfer of heat and your best bet may be some empirical rules of thumb. STSS might be your best bet for this if you really want to use tubing. I think you might find flat plates more efficient and cheaper to use. The bigger the better. Pay attention to the port size as well as the number of plates. You could find a lot on this forum be searching for "flat plate" or such.

Piker,
How much the tanks radiate actually will vary a lot depending on what they are painted with. White paint or aluminized paint will radiate a lot less than flat black. Most common surfaces radiate heat about the same as they absorb heat. Do you see many white solar collectors?

 

[Fred61]

Do you see many white solar collectors?

I don't see many black radiators either! Why is there a light color, aluminized paint or polished reflector behind the elements of an electric heater? I believe a "black box" is an absorber. I would need data before I believe that black radiates better.

 

[tom in maine]

Regular flat black is both highly radiative and absorbtive. Radiators are not painted black because people do not like black radiators.

That is why most wood stoves are black, because of the high emissivity of flat black finishes.

Solar collectors usually have a selective surface that suppresses radiative heat loss.

 

[jebatty]

DaveBP, good comments, especially

I think you need to size your heat exchangers (coils or flat plate type) to your rate of heat transfer not to the volume of the storage.

I can't comment on coil hx, but with plate hx reaching an approach temp of 20F is pretty easy and relatively inexpensive (boiler output 185F, plate hx output 165F), but as approach temp moves closer, the size of the hx and cost go up considerably. Still, your comment on sizing is right on, with the addition to size not only on heat transfer but also on target output temp, which further relates to flow rate. Approach tables are available to aid in sizing a plate hx.

The flowmeter tests were very instructive, both to calculate actual btu's transferred and to determine operational function of the Termovar. My current Termovar balancing valve setting allows for minimum return water temp of 150F. In actual operation, on a cold start virtually 100% of boiler output is returned until boiler output reaches about 160-165F, then the Termovar begins to open, and with cold system return (100F for example), boiler return will drop to about 150F and hold until system return rises to roughly 130-140F, and then boiler return rises to about 160F (rising faster than system return), again holding at 160F until system return is about 160F, then both rise roughly together. As mentioned in prior posts, the Termovar appears never to completely close, always allowing some direct boiler return.

 

[Gooserider]

Don't forget that you have about 2400 lbs of steel in those tanks that is storing Btu's as well.

I still haven't found any good numbers on how many Btu's a propane tank without insulation will radiate at 165* in about a 60* enviroment. . Given the tests that I have done I came up with about 75,000 Btu's/hr on two 500 gallon tanks... but I don't have the best measurement devices in place, so I think it's more. Any thoughts?

cheers

I think you need to look at the surface area of the tanks... Radiation is essentially a surface phenomenon, defined by the temperature of a surface and it's area, with some effect from the surface material - you are putting out infrared light, so it's a question of how hot the bulb is, and the size of it, but a surface has no way of "knowing" how much mass is behind it driving that emission. The biggest effect the mass would have is to change the rate at which the surface cooled as it emmitted those BTU's.

If you took 2400lbs of steel at 165* and made a solid cube out of it, you would get relatively little radiated energy out of it at any given moment, but it would stay at that temperature for a LONG time.... OTOH, if you rolled it out into a big sheet, you would get a lot more radiated heat at a single moment, but it would cool quite quickly. (The TOTAL number of BTU's emmitted would be the same in either case - but the time it took to emmit them would be different...)

I don't have the forumlas at hand, but I beleive the numbers you would need would be:

1. The temperature differential, and the two temperatures (most formulas will want this in *Kelvin)
2. The surface area of the tank
3. The rate at which the steel surface will radiate per unit area

Hope this gets you headed in the right direction...

Gooserider

 

[DaveBP]

Don’t forget that you have about 2400 lbs of steel in those tanks that is storing Btu’s as well.

Ironically, that 2400 lbs. of steel only stores as much heat as about 35 gallons of water at the same temperature change. But it's more than nothing.

The more you know about water the more amazing it seems. Great stuff to base a planet on.

 

[Piker]

Don’t forget that you have about 2400 lbs of steel in those tanks that is storing Btu’s as well.

Ironically, that 2400 lbs. of steel only stores as much heat as about 35 gallons of water at the same temperature change. But it's more than nothing.

The more you know about water the more amazing it seems. Great stuff to base a planet on.

Really? Seems like even though 2400 lbs of steel would occupy about the same volume as 35 gallons of water, it's greater density per unit of volume than water would make it be able to hold more Btu's. Apparently this isn't the case? No physicist am I.

cheers

 

[DaveBP]

Really? Seems like even though 2400 lbs of steel would occupy about the same volume as 35 gallons of water, it’s greater density per unit of volume than water would make it be able to hold more Btu’s. Apparently this isn’t the case? No physicist am I.

It's not a function of density or volume. It depends on the material itself. It's a property called specific heat.

A pound of water takes about 9 times more BTUs to raise its temperature 1 degree than a pound of steel does. Makes it a very useful heat storage and exchange medium. As I remember, ammonia gets close but water is the champ in this regard. Depends on the range of temperatures you work with. Liquid sodium is used in some nuclear reactors but that's not likely to be very useful for wood boilers. Unless NoFossil's brother is working on something he hasn't mentioned yet.

No, I'm no physicist either. Just a college dropout who used to flirt with that stuff before I got sick of homework.
I remember enough to appreciate the magic of it all but not enough to stay out of trouble.

 

[julien]

My calculate,

Your wood have probably 6050 btu/lb, at 20% moisture, 98 lbs X 6050 = 592,900 total BTU's.

http://mb-soft.com/juca/print/311.html


Average temp of storage before heating, 122 +121 + 120 + 118 / 4 = 120.25°

Average temp of storage after heating, 175 + 175 + 171 + 161 / 4 = 170.5

Delta = 50.25°


Weight a thank storage 2,400 lb + weight of boiler 1,310 lb = 3,710 pounds. Iron have 9.42 X lower specific heat than water, 3,710 / 9.42 = 394 equivalant of pounds of water.

Your 1,000 gallons storage + 25 gallons in boiler = 1025 X 8.443 lbs per gallon = 8654 pounds

+ water equivalant of iron weight system, 394, total = 9,048

9,048 pounds X delta 50.25° = total 454,662 BTU's in storage

If my 6040 BTU's for each pound of wood (at 20% moisture) is rigth your system have efficiency over 76%, and it is really excellent !

 

[jebatty]

I posted this quite awhile ago, but it may be helpful here too. See Flow Data. See my third chart.

As the chart shows, data points were every 15 minutes, a total of 18 data points. Average BTUH were 88,778; total burn time was 4.5 hours; total BTUH = 399,500. As to efficiency (BTUH per pound of wood), it all depends on the estimate of BTU's/lb of wood. If the 6400 BTU/lb is used, then total wood BTU = 460,800, and net efficiency = 399500/460800=87%. As BTU/lb of wood estimate goes up, "efficiency" goes down, and vice versa.