CFM vs BTU output.

[nevada]

i know we have some engineers here, and some otherwise very smart experienced people.

I need some help with a project im working on.

I need to know how many CFM of air I need to displace 6300 BTU's of heat.

the goal is enough air flow, but not so much that it will end up blowing cool/cold air.

ive done quite a few searches online, but haven't found anything that helps, or helps in a way I can understand haha.

any ideas?

thanks guys.

 

[4Reak Show]

*warning* I do NOT have an engineering degree...

However, I feel like the 2 units are incomparable... It's like asking how many gallons of water does it take to generate 10 watts of power... There are tons of variables like generator size, gravitation flow rate, overall temperature (for my example).

The best way I can figure is to measure the temperature for static BTU output you desire and then experiment by moving the heated air at various CFM rates and see what the temperature drop is. I would imagine the ventilation length and overall dimensions would play a key role. I don't think it's as simple as A=B.

Again, zero fancy papers hanging so now I'm curious from a physics stand point.

Subscribed.

Edit: 2.4 seconds Google search gave me this:

http://www.oceanhvac.com/cfms/

Turns out TIME is the missing factor:i.e. BTU's PER HOUR vs cubic feet PER MINUTE.

 

[nevada]

*warning* I do NOT have an engineering degree...

However, I feel like the 2 units are incomparable... It's like asking how many gallons of water does it take to generate 10 watts of power... There are tons of variables like generator size, gravitation flow rate, overall temperature (for my example).

The best way I can figure is to measure the temperature for static BTU output you desire and then experiment by moving the heated air at various CFM rates and see what the temperature drop is. I would imagine the ventilation length and overall dimensions would play a key role. I don't think it's as simple as A=B.

Again, zero fancy papers hanging so now I'm curious from a physics stand point.

Subscribed.

Edit: 2.4 seconds Google search gave me this:

CFM calculator

Turns out TIME is the missing factor:i.e. BTU's PER HOUR vs cubic feet PER MINUTE.

yes, sorry I thought BTU's was generally accepted as per hour.

and yes, I realize they are incomparable figures. however, I imagine for HVAC use, its got to be a standard conversion type. heaters have x amount of BTU output, there has to be a way to configure how many CFM's are needed to properly move that heat.


edit: that's a great calculator. thanks. I will see what I can come up with. I will have to get the actual temps, as I do not have those. the only info I currently have is from the sticker, of 6300 BTU's. though, I imagine there has to be a way to convert that to a temperature also. hmm..

 

[4Reak Show]

yes, sorry I thought BTU's was generally accepted as per hour.

and yes, I realize they are incomparable figures. however, I imagine for HVAC use, its got to be a standard conversion type. heaters have x amount of BTU output, there has to be a way to configure how many CFM's are needed to properly move that heat.

Check out the link in my post. Has a fancy calculator deal. Or google "BTU to CFM conversion calculator". Brought up a bunch of fancy words and numbers haha

 

[theoutbackdream]

Dont have an engineering degree, but i've taken some physics classes that have dealt with this kind of stuff before.

You'll need to account for:
-Altitude
-Temperature change of what you want to heat up
-Humidity of the air
-Area of the tubing you plan to flow the heat through.

An online calculator will give you a pretty good ballpark though.

but use this equation....

CFM = q/(p*60(cp)(delta)T

q being the flow rate of the fan.
p will be 0.0075lbs for standard air
60 for 1 hour = 60minutes
cp, dry air is 0.24, while humid air is 0.244
(delta)T will be the temperature difference entering and exiting the fan.
(Exit temp - enter temp)

Alternatively, you can do

CFM = BTU / ((delta)T * cp*p * 60)

 

[1engineer]

i know we have some engineers here, and some otherwise very smart experienced people.

I need some help with a project im working on.

I need to know how many CFM of air I need to displace 6300 BTU's of heat.

the goal is enough air flow, but not so much that it will end up blowing cool/cold air.

ive done quite a few searches online, but haven't found anything that helps, or helps in a way I can understand haha.

any ideas?

thanks guys.

Several ways to work this problem. Charles' Law or the Ideal Gas Law could do it but there would be too many assumptions and it gets a bit complicated. Plus you have to work in moles and that's just a PITA.

Here's a quote from a contractor forum that might help:

"Heating airflow can be calculated several ways. The easiest method we have found is the cfm per 10,000 Btu of input method. Example: An induced draft furnace should move 130 cfm per 10,000 Btu at sea level. So, say we have a 100,000 Btu Input induced draft furnace. To find required airflow divide the rated furnace input by 10,000 Btu. 100,000 divided by 10,000 equals a factor of 10. 10 Times 130 cfm equals 1300 cfm.

This airflow rule is applied to determine furnace airflow for three types of furnaces. Condensing furnaces require 150 cfm per 10, 000 Btu input.
Induced draft furnaces require 130 cfm per 10,000 Btu input, and natural draft furnaces require 100 cfm per 10, 000 Btu input."

I don't know what type of heater you are using, but if we take the middle of the road draft furnace value of 130 cfm per 10,000 Btu input, we can use a little algebra to calculate a simple ratio for you:

Here it is in step form:
130/10000=X/6300
0.013=X/6300
0.013*6300=X
81.9=X
So you need 81.9 CFM required to displace 6300 BTU

Hope this helps. If you have a different type of heater we can work it differently. Also, this is at sea level. We can apply a correction factor depending on altitude.

 

[-JD-]

As a general rule about 400 CFM per ton (12,000 BTU/hour)

 

[nevada]

Several ways to work this problem. Charles' Law or the Ideal Gas Law could do it but there would be too many assumptions and it gets a bit complicated. Plus you have to work in moles and that's just a PITA.

Here's a quote from a contractor forum that might help:

"Heating airflow can be calculated several ways. The easiest method we have found is the cfm per 10,000 Btu of input method. Example: An induced draft furnace should move 130 cfm per 10,000 Btu at sea level. So, say we have a 100,000 Btu Input induced draft furnace. To find required airflow divide the rated furnace input by 10,000 Btu. 100,000 divided by 10,000 equals a factor of 10. 10 Times 130 cfm equals 1300 cfm.

This airflow rule is applied to determine furnace airflow for three types of furnaces. Condensing furnaces require 150 cfm per 10, 000 Btu input.
Induced draft furnaces require 130 cfm per 10,000 Btu input, and natural draft furnaces require 100 cfm per 10, 000 Btu input."

I don't know what type of heater you are using, but if we take the middle of the road draft furnace value of 130 cfm per 10,000 Btu input, we can use a little algebra to calculate a simple ratio for you:

Here it is in step form:
130/10000=X/6300
0.013=X/6300
0.013*6300=X
81.9=X
So you need 81.9 CFM required to displace 6300 BTU

Hope this helps. If you have a different type of heater we can work it differently. Also, this is at sea level. We can apply a correction factor depending on altitude.

this!

this is what I was looking for perfect. thanks.

I guess my heater would be considered natural draft. its a propane furnace in an old camp trailer. non forced air, just a burner, with a grill front that lets the heat escape.

I have placed small pancake fans on the face of the grill to help draw the heat out and into the room, instead of just straight up the wall as it was originally designed.

the system works well, and the trailer is MUCH warmer, i have no complaints really. I just thought maybe I was leaving some heat "on the table", by using fans that were rated too low. but, based on your numbers, I think im right in there, with the pair of 40cfm fans.

I would assume I can simply plug different BTU numbers into that equation to solve for different heat sources, without changing any other factors?

As a general rule about 400 CFM per ton (12,000 BTU/hour)

thats oddly simple. though, a bit different than what was posted above.

 

[1engineer]

this!

this is what I was looking for perfect. thanks.

I guess my heater would be considered natural draft. its a propane furnace in an old camp trailer. non forced air, just a burner, with a grill front that lets the heat escape.

I have placed small pancake fans on the face of the grill to help draw the heat out and into the room, instead of just straight up the wall as it was originally designed.

the system works well, and the trailer is MUCH warmer, i have no complaints really. I just thought maybe I was leaving some heat "on the table", by using fans that were rated too low. but, based on your numbers, I think im right in there, with the pair of 40cfm fans.

I would assume I can simply plug different BTU numbers into that equation to solve for different heat sources, without changing any other factors?

thats oddly simple. though, a bit different than what was posted above.

Yep, you can just plug different values into the formula to get correct results.
I think JD's rule of thumb is more for heavier commercial applications. Glad it's working!

 

[nevada]

Yep, you can just plug different values into the formula to get correct results.
I think JD's rule of thumb is more for heavier commercial applications. Glad it's working!

I just re-read your original comment, it says BTU INPUT. wouldn't the number needed be based on the output heat, not input? maybe for electric heat sources they are one and the same?

my propane furnace is old, its rated at 70%. my input is 9k, but the output is only 6300.

I have read the newer ones are 90%, which is amazing. but I think a lot of that extra efficiency is due to the fan system that is able to pull out more of the heat, instead of losing it out the exhaust. which, IMO, is the same thing im doing with mine, though on a much lesser scale.

 

[-JD-]

Number was given to me by my HVAC instructor. More detailed calculations would involve temperature differential, humidity, etc. I could look those up when I'm home later if you need them, but that's a good general guideline

 

[1engineer]

I just re-read your original comment, it says BTU INPUT. wouldn't the number needed be based on the output heat, not input? maybe for electric heat sources they are one and the same?

my propane furnace is old, its rated at 70%. my input is 9k, but the output is only 6300.

I have read the newer ones are 90%, which is amazing. but I think a lot of that extra efficiency is due to the fan system that is able to pull out more of the heat, instead of losing it out the exhaust. which, IMO, is the same thing im doing with mine, though on a much lesser scale.

Output of the heater, input-ed heat into the forced circulating air.

 

[nevada]

Output of the heater, input-ed heat into the forced circulating air.

aha!
yes, that makes sense.