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# Formula for figuring BTU's

Jim Davis_7 Member Posts:

**67** The formula for output is CFM X Delta T X .99 for hot air. This tells you how many btu's are being transferred into the air. What we don't know is the actual btus of the fuel. It can vary 100 btus per cu.ft. or more for natural gas or 2000 btus or more per gallon of oil etc. Stack loss does not truly evaluate anything accurate. All we can calculate is how many btus we are getting out of a certain quantity of fuel. Our job is to transfer the maximum amount the equipment is rated to deliver.

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6I have a new power burner that cannot be calculated by the amount of fuel beacause it is super efficient. The cold air intake is 60 degrees, hot air out is 130 degrees. It is 3800 CFMs. The stack temp is 440 dgrees. I would like to know a formula for figuring the BTUs without using the fuel consumption, but rather by the heat transfer from the inlet to outlet. Thanks for any help. Gene

712Are you burning gas or oil? If gas, just watch the gas meter needle for a timed period.

2,3983,800 CFM x (130-60) x 1.085 = 288,610 BTUH at sea level.

This is uncorrected for air density at temperature but close enough for government work.

This is output of course and is approximate. Clocking your meter or noting your oil consumption for input is up to you.

-Ernie White, my Dad

306287,280 btuh going into the 3800 cfm. What you don't know and can't figure out with the info you have is how much is going up the stack.

6I am burning number 2 heating oil but it can't be calculated by the size of the nozzle but rather by the heat exchange.

thanks Gene

306I'm trying to remember exactliy how we used to do it. When I used to service Torrid-heat and Timken burners we had a calibrated glass column and a stop watch. We would disconnect the oil line and hook up the column full of oil, start the burner and time it. We had a chart that read in btuh input. The input was adjustable.

6If I calculate the BTUs this way, do I need to calculate the BTU loss in the stack- If so- do you know how to figure that?

According to a friend of mine, the same furnace with a 1.1 nozzle and 120-140 PSI delivered from the pump with the same heat temps in and out and the same CFMs is 110 BTUs, measured by the fuel consumption. Is this correct or is my friend wrong? Could you give me the calculation for that?

Thanks again, Gene

546size and the pressure it's delivered at? There's a valid formula for it as there is for measuring output.

Are you sure you have exactly 3800CFM? Have you measured it off the heat exchanger?

2,440calculated is output at the equipment. Naturally that presumes that your temperature measurement across the HEX is correct (easy enough to measure) and that your aiflow measurement is correct (an entirely different matter without the right tools).

Given the parameters described by your friend, I think it impossible to have what he says he measured. One or more of his variables is off.

Taking a closer look, say he is burning 1.1 GPH (154 MBH input) and at 80% combustion efficiency so a gross output of 123.2 MBH. (It would seem that the oil consumption is more easily verified than airflow.) At a 70 degree rise he is probably moving about 1622 CFM -more or less

306I am wondering about the origin of the multipliers you used. 1.085 and .99. I had memorized 1.08 many years ago, when I saw Brad use 1.085 I thought what the heck I'm fitter and he's an engineer what do I know. I figured I probably rounded it off. I am not near my library so I can't check, however if I remember standard air is .075 lbs/cu.ft. times 60 min. in an hour times the specific heat of air .24 equals 1.08 even.

Waz up.

67Working with the guru of air flow "Rob Falke" we actually had a discussion on this last year. When doing the heating calculations in my classes I told my students that no measurement is exact so for head work it probably doesn't make much difference if we use the 1.08 or not. Well Rob said thats not correct until he checked the Psychiometric chart and found that the actual weight of air at 120-140 degrees is not .075 lbs./cu.ft but closer to .069lbs/cu.ft. When he did the calculations the multiplier came out to .99 or .98 instead of 1.08. This then explained how some of our Air Balance contractors were coming up with efficiencies over 100%. Choosing to use numbers that I could calculate in my head versus using a calculator in the end turned out to be more correct. I have often stated in the past I have stumbled by accident just as often as on purpose to dicover solutions to problems.

1,935don't you need to:

CFM x 60(to get btuh) x air weight/cubic foot x btu(cubic foot air)x delta T

hows all that come out to 1.(something)?

thats what I used for ACH heatloss.

54667The formula is CFM X Delta T X 1.08 = Btu,s for Air

GPM X 8.33 X Delta T X 60 = BTU's for Water

The 1.08 multiplier was used for air conditionin but never adjusted for heating. When factoring in the hooter air temperatures of heating it is no longer l.08 but instead .98 or .99.

2,440illustrated, with warmer air comes less density (less mass flow) hence less heat carried per unit of air. Warmer air carries less heat per cubic foot and CFM is what it is about. It takes more cubic feet of air per pound.

At 70 degrees, dry air has about 13.35 cubic feet per pound. At 80 degrees this rises to 13.6, at 100 degrees it rises to 14.12 (more or less!). At 130 degrees it rises to 14.9.

The carrying capacity is based on the specific heat of air and pounds per hour transported at a certain temperature difference. The 1.085 factor I use is for standard air, 70 degrees and at sea level. Colorado air is laid back and does not work as hard so you need more of it.

306Brad, I think you are a little too dense:) The reciprocal of 13.35 cu.ft./lb is .0749063 lbs./cu.ft.x 60 x .24=1.0786507. Don't reply to this I'm just screwin around. Thanks for the replys.

546> heating. When factoring in the hooter air

> temperatures of heating it is no longer l.08 but

> instead .98 or .99.

I'm wondering, Jim, if this applies to JUST Hooter air or all air? Speaking of Hooter's I really enjoy their, errr, food.

24When using any of the air formulas it is important to understand how to correct for changes in the air density if the air being measured is not standard air. The air constants apply to standard air at 70° F and 14.7 Pisa, (29.92hg.) If air being measured is outside of these parameters, it may require that the constant be recalculated. For most situations the standard air formulas can be used, but if precise measurements are desired, adjustments to the constants should be made. Remember, fans are doing work; they are moving in reality pounds of air. The amount of air they will move in CFM remains constant with a variable mass flow rate, so the cubic feet of air they will move over any given time period will remain the same. The difference is in the density of the air or the number or the pounds per cubic foot. This is important because coil selection software calculates required coil capacities based upon pounds per hour (lb/hr) of air passing through the coil, not CFM.

The constant 4.5 is used to convert CFM to lbs/hr

4.5 = (60 min/hr)/ 13.33

or (60min x 0.075 lbs/cu ft)

Where 13.33 is the specific volume of standard air (cu.ft/lb) and 0.075 is the density (lbs/cu.ft)

If the air being measured is not standard air, the air density will vary with the barometric pressure and the absolute temperature. To recalculate the air density, measure the temperature and obtain the barometric pressure use the following formula:

Air Density (lb/cf) = 1.325 x Bp/Tabs

Where: 1.325 (Constant to keep consistent units)

Bp = Barometric Pressure

Tabs = Temperature (Absolute)

Example 1.325 x 29.92/(70°F + 460°F) = 0.0748 ~ 0.075 lb/cu ft

This is how standard air density is calculated

If you were measuring air coming out of a furnace, and the air was 154° F the air density would change as follows

1.325 x 29.92/(154°F + 460°F) = 0.0645

If heated air were used in this formula, the constant would be:

(60min x .0645 lbs/cu ft) = 3.87 instead of 4.5 used for standard air.

The constant used in the sensible heat formula 1.08 is used to convert CFM to lbs/hr and factor in .24 the specific heat of standard air (BTU/lb/°F)

Where: 1.08 = (0.24 x 60)/13.33 or 0.24 x 4.5

4.5 = (60 min/hr)/13.33 or (60min x 0.075 lbs/cu ft)

0.24 BTU = specific heat of standard air (BTU/lb/°F)

The constant 0.68 used in the latent heat formula is used to factor out the amount of heat contained in water vapor in BTU/LB

Where 0.68 = (60/13.33) x (1060/7000) or 4.5 x (1060/7000)

Where: 13.33 is the specific volume of standard air (cu.ft/lb)

1060 = average latent heat of water vapor. (Btu/LB)

7000 = grains per lb or water.

4.5 = (60 min/hr)/13.33

or (60min x 0.075 lbs/cu ft)

The specific heat of the air remains fairly constant for air from -58 to 104 degrees

Whew, hope that helps

Jim Bergmann

306too much to remember. When I was doing some air balancing I used to carry a handy slide rule type calculator put out by Dwyer. Working in the mountains you have to allow for changes in density on every job.

6On a Beckett Burner with a .65 nozzle, at 120 PSI, is according to the documentation on the burner, is supposed to burn just over 1/2 gallon or 80 ounces of fuel per hour. I have checked several of these burners and they are actually burning 1.25 gallon per hour. My question is- Do they figure their consumption on a 1/2 hour per time per hour or is the .65 nozzle supposed to be 1.25 gallon per hour? Thanks for all your help.

638the .65 rating is at 100PSI at 120 PSI it should be putting .71 Gallons per hour or 90.88 Oz of oil

6Do you know how pressure it is running at if it is using 1.25 gallons per hour?

2,398on that score, Gene. My hands-on regarding oil is at best imaginary. I would never ask myself an oil burner question; I go to the other guys.

-Ernie White, my Dad

638I can tell you that for a .65 nozzle to output 1.25 gph yyou would have to have in excess of 300 PSI and that cannot be done with a residential pump and if your unit was doing this consistantly then you would not bre able to fire that large of a fire in that small of a chamber for long, without sooting up the inside. Having said that I had one unit that would push in excess of 300PSI on startup, I know because I blew up two 300 PSI gauges trying to trouble shoot it. I am thinking that someting is off in the calculations somewhere.

4Saw this thread and thought that some of the more experienced personnel could answer this question. My wife and I are adding a "great Room' onto the back of our house, ~700 sqft. We have steam heating from an old Kohler boiler. Over the last couple of years I have kept my eyes open and was able to get my hands on some used cast iron radiator identicle to the ones in the rest of the house. Brand is AERO and the plates are about 1" X 8" X 24". Based on this can someone give me an idea of how many radiators containing how many plates I'm going to need to effectively heat the room. Some relative placement suggestions wouldn't hurt either. Thanks.

2,398You should post this as a new topic ("New Thread") of your very own and get a wider audience targeted to your needs.

Brad

-Ernie White, my Dad

592I'm gonna print this and re-read a few times before bed. Want to pass that test in the morning.

Serioulsy, will read and try to understand it.

513F2=F1 X (P2/P1)^.5

WHich doesn't translate too well to ascii typing.

F2= new flow rate

F1 = rated flow rate

P2 = new pressure

P1= rated pressure (100psi)

eg .65 nozzle @ 140 psi= .65 X (140/100)^.5 ( to the .5 power)

=.65 X 1.18 = .77 gph

(Did someone already say that? I believe they did ))