# Gotta figure this out or I'm duct.

October 16, 2011 6:03 PM Subscribe

How do I calculate the static pressure resulting from the transition of a 12" air duct to an 8" duct, given a target CFM of 2,000 CFM?

I have a 12" diameter high output duct fan putting out 2,020 CFM when static pressure is 1/8". I need to vent this through an 8" opening in the roof. How do I calculate the static pressure resulting from the transition from 12" to 8"?

I know that this is very high flow for a duct this small. I also know that the physical nature of the transition can influence entrance loss, so I'm also hoping for any guidance with regard to that.

This seems like it could be a common HVAC problem, so I hope that there's a table out there somewhere (that I can't find) that can help me determine this.

tl dr:

1. How do I calculate the static pressure resulting from the transition of a 12" air duct to an 8" duct, given a target CFM of 2,000 CFM?

2. Do you have any guidance about the design of such a transition?

I have a 12" diameter high output duct fan putting out 2,020 CFM when static pressure is 1/8". I need to vent this through an 8" opening in the roof. How do I calculate the static pressure resulting from the transition from 12" to 8"?

I know that this is very high flow for a duct this small. I also know that the physical nature of the transition can influence entrance loss, so I'm also hoping for any guidance with regard to that.

This seems like it could be a common HVAC problem, so I hope that there's a table out there somewhere (that I can't find) that can help me determine this.

tl dr:

1. How do I calculate the static pressure resulting from the transition of a 12" air duct to an 8" duct, given a target CFM of 2,000 CFM?

2. Do you have any guidance about the design of such a transition?

Response by poster: Thanks for the spreadsheet -- I think it could be helpful on another project.

But I need to clarify my question. Instead of "target" I should have said "input". IOW:

posted by buzzv at 6:45 PM on October 16, 2011

But I need to clarify my question. Instead of "target" I should have said "input". IOW:

**1. How do I calculate the static pressure resulting from the transition of a 12" air duct to an 8" duct, given an***input*flow of 2,000 CFM?posted by buzzv at 6:45 PM on October 16, 2011

"Input flow" doesn't make any sense here. The input flow is exactly the same as the output flow; the same volume comes out as goes in. Your fan can push 2020 CFM when it's not fighting any significant resistance, but when you connect the ductwork and the static pressure goes up, the flow will drop throughout the system.

At this point you don't have enough information to calculate the flow you'll get when you connect it to the ductwork. It depends on the horsepower of the motor and the design of the fan itself. Before you can figure out how many CFMs you'll get from this arrangement, you'll need a graph of how much air your fan can move against various static pressures. The manufacturer may be able to help you with that. Ask for a "fan curve."

posted by jon1270 at 7:11 PM on October 16, 2011

At this point you don't have enough information to calculate the flow you'll get when you connect it to the ductwork. It depends on the horsepower of the motor and the design of the fan itself. Before you can figure out how many CFMs you'll get from this arrangement, you'll need a graph of how much air your fan can move against various static pressures. The manufacturer may be able to help you with that. Ask for a "fan curve."

posted by jon1270 at 7:11 PM on October 16, 2011

Response by poster: I see your point, Jon. I guess I was better off with the original question, given that I'm looking to have input=output=2000CFM. And the next step is the fan curve--that's why I'd like to find out the static pressure drop.

posted by buzzv at 7:29 PM on October 16, 2011

posted by buzzv at 7:29 PM on October 16, 2011

This page suggests that the table you want is the Duct Fitting Database published by ASHRAEâ€” maybe you can find a copy of their

Presumably the engineering approach is to start at the exhaust and work backwards (adding losses from fittings, viscosity, etc) to figure out what pressure the fan will be working against, then find a fan that can put out 2000 CFM against that pressure (whatever it turns out to be). But it might be easier to simply build the duct, attach the fan you have, and measure its performance.

posted by hattifattener at 8:18 PM on October 16, 2011

*Handbook*in a library or the like? (Scroll down to "Dynamic losses".)Presumably the engineering approach is to start at the exhaust and work backwards (adding losses from fittings, viscosity, etc) to figure out what pressure the fan will be working against, then find a fan that can put out 2000 CFM against that pressure (whatever it turns out to be). But it might be easier to simply build the duct, attach the fan you have, and measure its performance.

posted by hattifattener at 8:18 PM on October 16, 2011

What is the angle of the offset? I will take a look when i get to work. I dont think its published in the handbook, but i have some tables from somewhere. There is also going to be a lot of static on the discharge of that when it goes from 8" to free. (1/8" isnt going to work. Do you have a fan curve to see what it can actually put out at higher static?)

posted by ihadapony at 11:07 PM on October 16, 2011

posted by ihadapony at 11:07 PM on October 16, 2011

It's gonna be noisy!

1/8" mercury, I presume?

Do you have any gear to make the measurement? Are you trying to predict performance before you buy the motor? Do you just want to know what CFM you are achieving with this setup, which sounds from your post like you have it in hand? I am a little confused.

(Normally, I am a lot confused.)

posted by FauxScot at 4:29 AM on October 17, 2011

1/8" mercury, I presume?

Do you have any gear to make the measurement? Are you trying to predict performance before you buy the motor? Do you just want to know what CFM you are achieving with this setup, which sounds from your post like you have it in hand? I am a little confused.

(Normally, I am a lot confused.)

posted by FauxScot at 4:29 AM on October 17, 2011

Response by poster: Yes, ultimately I am trying to figure out the expected performance of the selected fan if it is exhausted through an 8" hole. In order to estimate this, I am seeking an answer to the following question:

If there was a reduction in duct diameter from 12" to 8" using a reduction rate of 1:7, what would the resulting static pressure be (in water column inches), given an exit flow volume of 2000 CFM?

With this information, I can apply that amount of SP to the fan curve (which I now have) to estimate the resulting flow. That's my intent. And I appreciate all the ideas so far--thank you.

posted by buzzv at 12:25 PM on October 17, 2011

If there was a reduction in duct diameter from 12" to 8" using a reduction rate of 1:7, what would the resulting static pressure be (in water column inches), given an exit flow volume of 2000 CFM?

With this information, I can apply that amount of SP to the fan curve (which I now have) to estimate the resulting flow. That's my intent. And I appreciate all the ideas so far--thank you.

posted by buzzv at 12:25 PM on October 17, 2011

Best answer: Ok. So the constriction isn't that bad. Its the free discharge and the run at the smaller pipe size that is killer.

So a 12" hole is 133 sq in. And a 8" hole is 50 sq in. 2000 CFM through a 8" hole travels at a velocity of 5730 FPM. This has a velocity of pressure of 2.06. I looked at a 30 deg reduction. This has a factor of C=.06. Pressure loss in this fitting is Velocity pressure multiplied by C which equals .1236.

A abrupt end of the duct at the far side of the wall has a C=1. So you are looking at a drop of 2.06 CFM there.

Also for the 8" section of pipe you have a 6" pressure drop for each 100' of run.

All pressures above are in inches water column.

posted by ihadapony at 4:10 PM on October 17, 2011 [1 favorite]

So a 12" hole is 133 sq in. And a 8" hole is 50 sq in. 2000 CFM through a 8" hole travels at a velocity of 5730 FPM. This has a velocity of pressure of 2.06. I looked at a 30 deg reduction. This has a factor of C=.06. Pressure loss in this fitting is Velocity pressure multiplied by C which equals .1236.

A abrupt end of the duct at the far side of the wall has a C=1. So you are looking at a drop of 2.06 CFM there.

Also for the 8" section of pipe you have a 6" pressure drop for each 100' of run.

All pressures above are in inches water column.

posted by ihadapony at 4:10 PM on October 17, 2011 [1 favorite]

Response by poster: Thanks, ihadapony!

So, the interesting thing that I've learned is that the pressure loss at the fitting is based on

But final question -- do I just treat this pressure loss as another static pressure to add up for the entire run? I ask because all the fan curves are based on SP.

posted by buzzv at 4:50 PM on October 17, 2011

So, the interesting thing that I've learned is that the pressure loss at the fitting is based on

*dynamic*pressure, not static, which completely makes sense now.But final question -- do I just treat this pressure loss as another static pressure to add up for the entire run? I ask because all the fan curves are based on SP.

posted by buzzv at 4:50 PM on October 17, 2011

I really should proof read my answers before i post them. I think the meaning was mostly clear. hole should be duct. That should be 2.06 in wg, not cfm. 0.1236 is in in. wg.

The calculated value above is a total pressure loss, which includes static loss and gain from changes in the dynamic pressure. This plus the static loss from duct friction, measured from center line to centerline of the fitting, is what the fan has to produce in static pressure.

So, just add them. The fan curve is a CFM at a static pressure, so velocity pressure in there. If you change sizes up from your fan outlet you may see some static regain where the velocity pressure drops and static rises. But you didn't mention any fitting off of the fan so just add them.

Let me know if that clears it up or if any of that needs more explanation.

posted by ihadapony at 7:54 PM on October 17, 2011

The calculated value above is a total pressure loss, which includes static loss and gain from changes in the dynamic pressure. This plus the static loss from duct friction, measured from center line to centerline of the fitting, is what the fan has to produce in static pressure.

So, just add them. The fan curve is a CFM at a static pressure, so velocity pressure in there. If you change sizes up from your fan outlet you may see some static regain where the velocity pressure drops and static rises. But you didn't mention any fitting off of the fan so just add them.

Let me know if that clears it up or if any of that needs more explanation.

posted by ihadapony at 7:54 PM on October 17, 2011

Response by poster: Yes, you've made that completely clear. Thanks.

I know that I haven't mentioned anything downstream in the 8" duct--what will happen is it will exhaust about 5 feet from the constricting fixture. So, wrt the "abrupt end" of the of the duct that you mention, by that do you mean a complete closure at that point (as if the duct was capped and sealed), or did you mean a complete termination of the duct (i.e. an 8" opening), which is the case in my system. The former causing 2.06" SP makes sense to me, but the latter would not make sense to me.

By the way, you're being very helpful--thanks again!

posted by buzzv at 7:23 AM on October 18, 2011

I know that I haven't mentioned anything downstream in the 8" duct--what will happen is it will exhaust about 5 feet from the constricting fixture. So, wrt the "abrupt end" of the of the duct that you mention, by that do you mean a complete closure at that point (as if the duct was capped and sealed), or did you mean a complete termination of the duct (i.e. an 8" opening), which is the case in my system. The former causing 2.06" SP makes sense to me, but the latter would not make sense to me.

By the way, you're being very helpful--thanks again!

posted by buzzv at 7:23 AM on October 18, 2011

ihadapony is much more knowledgeable in this area than I, but I thought I'd offer this: His 2.06" SP assumes an 8" opening, not a sealed cap. A lot of turbulence is created as slightly compressed air moving smoothly along at 5730 feet per minute exits that tube and suddenly expands, runs smack-dab into air that's hardly moving at all, and stalls. What happens at that exit point has a much larger effect than you might imagine.

As you may already realize, you're going to need a different fan to do what you're contemplating.

posted by jon1270 at 8:52 AM on October 18, 2011

As you may already realize, you're going to need a different fan to do what you're contemplating.

posted by jon1270 at 8:52 AM on October 18, 2011

Jon has it. I don't know if i really believe that 2.06 number either. It is probably pretty conservative, but it is what the SMACNA table says. I wouldn't doubt its at least 1 to 1.5 in w.c.

posted by ihadapony at 5:18 PM on October 19, 2011

posted by ihadapony at 5:18 PM on October 19, 2011

This thread is closed to new comments.

1/8" static pressure is basically nothing. Your fan can probably push something close to 2000 CFM if the ductwork is a short, straight run with nice, smooth transitions. If the air has to pass through baffles, filters, elbows, etc. then I think you're going to fall short of your target.

Here's a spreadsheet for calculating static pressure. It's meant to help design dust collection systems, but might be useful anyhow.

posted by jon1270 at 6:36 PM on October 16, 2011