Gas and Oil Home Heating Furnaces - Duct sizing and CFM ratings

If CFM is cubic feet per minute and relates to the amount of air that a blower can move, what does CFM mean when representing a duct? For example I have seen 7" round ducts rated at 150 CFM, 10" round ducts at 400 CFM, but what does that really mean? Doesn't the CFM rating require that you know how big the blower behind the duct work is? Does the CFM rating of a duct refer to the maximum amount of CFM that a duct can support no matter how large the blower? Are the numbers that I listed above for 7" and 10" round ducts accurate?

One reason I ask, I just saw a rating for a 8x16 trunk line of 700 CFM. This was in the same table where I got the 7" and 10" round CFM numbers. I find that number extremely low for a 8x16" duct. Especially considering that I have a 1200 CFM blower that comes from the factory with a 8x16" duct attachment directly off of the blower. If the 700 CFM of a 8x16" duct is all a 1200 CFM blower could pump out, then the engineers at the fan mfg would have to have made a big mistake.

I am trying to determine how much duct work I should supply for input and output to the 1200 CFM blower.

Any help would be appreciated,

Mike.

Once you have somehow selected the furnace and air conditioner needed you can size the ductwork. Figure 400 cu. ft. per min for the AC per ton of cooling. (Example: a 2.5 to AC will need to move 1000 cfm for best results with a wet coil.) Divide up the air proportionately to the square footage served.

(Example: for a 1000 sq. ft. home each sq. ft will get 1 cfm...and a 140 sq ft bedroom will get 140 cfm.) Now size the ducts and branches so that the velocity is 600 to 900 ft. per minute for sheet metal and 600 ft per min. Maximum for flex duct.

(Example: a 7" dia. round flex duct has an area of 38.5 sq. in. {or .267 sq.ft.} and 140 cfm will pass through it with a velocity of 539 ft. per min.) This is rule of thumb information and experienced people would increase the airflow to key rooms (+10 to 20% for master BR and Family room and +30% to the kitchen). Then decrease the air to closets and bathrooms 10 to 30%. Finally if the proportion of windows is more in any room, that room will need more air to compensate. Trial and error learning can make this method work pretty good on smaller and uncomplicated homes.

Design like the Pros: you can buy Manuals C, J, S, D and T for residential design. Go to http://www.acca.org/redesign/catalog/Category.asp?cid=3 to order these reference manuals. They are well written for layman or pro to use.

(Sadly most AC contactors use the rule of thumb methods because the public does not know how complex good design can be and they are unwilling to pay for a time-consuming real design. Thus AC and heating discomfort is very common, especially in bigger and more complex homes. A quality contactor will guarantee comfort satisfaction in writing or Money back...IF you can find one.)

My situation is a little different. It is a simple input/output question. There will be no branches in this case, just a main in and main out with a 1240 cfm blower.

This is a secondary heat system that supplies about 65% of the heat to my house with a wood burning fireplace.

I have seen the number of 150 cfm used for 7" round pipe. If I supply 3 7" round air returns to the 1240 cfm blower, can I move enough air through those 3 7" round ducts to meet the 1240 cfm return air the blower wants? I think based on additional research that I have done it is possible but it depends on the velocity of the air.

On the output side I am running 1 7" round and 1 10" round. Again I think it is possible to move the 1200 cfm in this space, but I am not sure if the blower velocity is high enough to move that much air.

Using the numbers I have seen this equates to 450cfm ( 3 7" ducts ) on the input side and 550cfm on the output side ( 1 10" and 1 7"). I just want to make sure that my assumption of an actual higher cfm flow ( than the 450 and 550 mentioned ) based on velocity is valid. If it is valid, how to do I determine the actual velocity of the air? Is that a specification of the blower? As I mentioned the duct work out of the blower is 8x16, this would indicate to me that the air would be a relatively high velocity to get 1200 cfm in a 700 cfm space of the 8x16 duct ( according to the same chart where I got the 150cfm for 7" round and 400cfm for 10" round ).

I do not want to supply 1/2 of the required air in and air out to the blower. I think that would damage the blower over time. I can buy a variable speed switch to slow down the blower if I can not supply enough air to it. I just do not want to do that unless I need to.

I hope this explains the information I need a little better. I think it would be interesting to get those manuals mentioned, but I think it might be overkill for my situation.

Thanks,

Mike.

Cfm is short for Cubic Feet per minute. The rated cfm refers to the VOLUME of air that will pass through 100 feet of straight, smooth metal ductwork of a size listed on the reference table with a small friction loss. (Surface roughness will increase the friction losses and reduce airflow)

Friction loss is measured in inches of water column. The pressure needed to lift a column of water one-inch in a soda straw is one inch of water column. This is a very small pressure and errors in duct design are seldom economically fixable after the ductwork is covered up and inaccessible.

The design friction loss for the entire duct system (return and supply) should normally be less than 2/10ths of an inch of water column. Every turn and change in shape of the duct causes a friction loss "equivalent" to a length of straight duct of the size of the duct being used. Thus a common 7-inch round metal elbow has an equivalent length of 30 feet (of 7 inch round metal duct). Some other common fittings have equivalent lengths of 50, 70 or 100+ feet.

These friction losses add up and most duct systems end up with a total equivalent length of 200 to 400 feet. The mouth of the blower is inches long and is no guide for sizing a duct, which has a MUCH longer equivalent length. (Also remember that the sizing guide is based on only 100 feet and duct sizes must increase as the equivalent length increases or the airflow will drop. This is the reason why the longest and crookedest ducts often do not deliver enough air to properly meet the needs of the room being served.)

The common air handler and furnace fan can not deliver it rated airflow at pressures greater than ½ inch of external static pressure created by friction and turbulence in the ducts, fittings, grille and cooling coil. If the standard filter is in an internal location, the loss does not need to be considered when the actual installed location is external to the air handler or furnace.)

1) The blower is probably bigger than needed. Reducing the airflow drastically to 500 cfm WILL NOT HURT the blower. (The motor load will actually decrease because the mass of air being moved will be less. A smaller blower would use less electricity to do the same job but the wasted electricity used by the oversized blower ends up as heat anyway.)

2) The air velocity depends on the friction of the attached duct system. If the system is short and simple, build it in the garage first and test it out before assembling it inside of the house. The air velocities may be high and create more noise than you want. The duct system is a sound trap around the blower. The system will be quiet if the delivered velocities are about 600 feet per minute.

3) The blower mouth (8"x16") is .89 square feet and the design exit air velocity is 1240 cfm/.89 sq. ft. or about 1400 fpm AVERAGE. The associated peak velocity may be over 2000 fpm.

4) I assume that the unheated room air will pass through the blower.

5) The capacity of the stove is another factor, because the air can be heated too much which could be a safety issue…or too little, which would be a comfort issue.

Lynn,

Thanks for your help. I am slowly getting this through my thick skull. ;-)

I found my blower and the static pressure. According to this page:

http://www.grainger.com/Grainger/catalogpageview.jsp?xi=xi&CatPage=3728

It runs 1240 cfm at .4"SP using the 1/4 HP using motor 9F736, This is the performance numbers that I have seen listed. I thought the mouth was 8x16, after looking at it again, it is smaller at 9.5"x10.5". The mouth is ducted to a 8x16 duct with a sheet metal adapter. I do not think this dramatically affects the 1400fpm number you calculated. The 8x16 is .89 sq ft and the 9.5x10.5 is .7 sq ft. I guess that the actual fpm might be a little higher. Does that make this a high velocity blower?

Does that mean the velocity of the air is .4"SP? If I read your post correctly, does this mean it is 4x the velocity of a standard furnace blower at .1"SP? If so, then could you get 2800cfm ( 4x700cfm at .1"SP) in a 8x16 duct at .4"SP? I suspect the conversion is not that simple. How does the .4"SP affect the 7" round and 10" round numbers we discussed? Is it 4 times the airflow?

Regarding friction. There is 0 on the inlet side, 3 straight 7" round pipes are the supply. There is a bunch of friction on the outbound 1 - 10" and 1- 7" round. I guess this makes the inlet and outlet sides about even, given the amount of friction on the outbound side.

You are correct that the unheated room air passes through the blower.

The wood stove says 140,000 BTU. I have no reason to doubt that.

Again, thanks for you patients and help,

Mike.