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Old Apr 29, 2008, 07:43 PM
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Exhaust ducts - Static thrust loss and efflux velocity

Hi all,

I'm going to be building a ducted fan model in about a month and was trying to figure out the optimum exhaust duct sizing. I've heard some rules of thumb but am trying to figure out where they come from. Using Bernoulli's equation and the fluid momentum equation, I came up with relations for static thrust loss and efflux velocity as functions of exit area.

I won't go through the derivations, but I will post some graphs that I created based on the equations I came up with.

What I found is that minimum static thrust loss (0) occurs when the exit area is exactly the same as the area of the ducted fan's exit. If you increase or decrease the exit area from that number, you will lose some thrust. Meanwhile, efflux velocity increases as exit area tends to 0.

My question is, what parameters should I use to decide the optimum exit area? It's obvious that I want to keep it less than or equal to the ducted fan area, because if it is greater I will lose thrust and efflux velocity. If it is less I will lose thrust quickly but also gain efflux velocity quickly.

Should I consider power output of the entire ducted fan system to determine the optimum area?

Thanks, and please let me know if my relations are incorrect.

Dan
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Old Apr 29, 2008, 09:25 PM
EDF rules... :)
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85% of fan swept area is optimum. Try using a constant diameter tube as a thrust tube and compare it to the 85% outlet size and you might find something interesting.

Eric B.
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Old Apr 29, 2008, 10:42 PM
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Thanks Eric. That's the number I've heard, but I'm wondering why it's the optimum. What's the criterion there? Also, do you mind explaining what you mean by "you might find something interesting"?

Perhaps my flow model is too simple and I'm missing something because of that (or, quite possibly, it's completely wrong ). It would certainly be interesting to set up a thrust stand and find out.

I plan on using the 85% rule for my plane, but I'm the type of person that likes to know why I should do something a certain way.

EDIT: I tried calculating the power output of the fan as a function of exit area, but it seems to depend not only on that but on fan performance parameters as well (static thrust and efflux velocity in the absence of a thrust tube, etc.), so there's no way to say that a certain exit area is best in general, at least when considering power. Perhaps the 85% number is just one that tends to work well in most cases?

I really should try to come by some testing equipment to figure these things out for myself. I wonder what my school has that I could use...

Dan
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Old Apr 30, 2008, 12:52 AM
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And read Eric's response closely! lol there is volumes of info and advice there! haha


Barry
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Old Apr 30, 2008, 09:18 AM
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Well it seems like something's quite wrong here. I plugged in some numbers for the Wemotec Mini Fan that someone reported and got something like an 800 lb thrust loss.

Regardless, my question still stands: why the 85% FSA at the nozzle? I assume it has to do with power and efficiency, but if anyone has some more insight that would be great...

@Barry-- I read Eric's response multiple times and don't see anything there besides the 85% FSA suggestion. Maybe I'm really dense and missing something or you guys are just messing with me.
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Old Apr 30, 2008, 09:34 AM
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85% is considered optimum because it's the best compromise between static thrust and efflux.

You won't see this on paper, however. You have to go fly with the numbers you're plugging into your calculator so what difference they make on the flight performance.

For example, I flew a Li'l MiG-29 and tried difference nozzle sizes. I originally flew with something like 90% fan swept area (FSA). I went to 100% FSA to see if the increase in static thrust improved the vertical climb and thereby improved aerobatic capabilities. It didn't. The lower efflux made the model slower overall, which yielded less performance.

Dan
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Old Apr 30, 2008, 12:54 PM
EDF rules... :)
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Montag DP,

Since you were seeking answers and using textbook physics to get an understanding of what your looking for in your outlet tube, I thought you might like to see graphically what Dan expalined.
Using a constant diameter tube as an outlet reduces the mass velocity from the fan as the air(fluid) moves into a larger area, this will result in less thrust force.
Air wont flow thru the motor holder and we must use the area where the air will flow(FSA=area of inner shroud less the area of the motor holder) to run any meaningful calculations.
The outlet duct will work from 100% FSA down to 85%, less than 85% will take a fan designed to much higher efficiency than the average fan available and like Dan stated there is a difference in performance with the change in velocity if the overall mass stays the same.
One more thing, The airframes drag is a deciding factor on how much outlet area/velocity is needed to make the aircraft perform up to its best.

Keep up the quest for knowledge...

Eric B.
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Old Apr 30, 2008, 01:07 PM
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promise not messing with you at all! Your following the proper path as far as I'm concerned and that is your using a logical approach and are trying to understand what your seeing. Were saying take that obeservative approach and sinter it with what you actually start to see as you put it together and run in whichever model it ends up in. So in that case think of Erics statement as pretty much covering it all. Yeah starting figures and usual practice will start you around %85 percent area, and from there test for yourself to see what really becomes optimum for you. So no worries! your not just slapping it all together and looking to be spoon fed, your actually thinking, your way ahead of the curve in my book!

Barry
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Old Apr 30, 2008, 02:48 PM
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Hey thanks a lot, guys. I appreciate it.

Eric, based on your response I noticed that I hadn't taken that area blocked by the motor mount into account when I did my initial work. Perhaps that is the reason the numbers didn't work out? That also explains why people use a motor fairing--that way the exhaust air isn't subjected to a sudden change of area, which no doubt wastes energy.

So when you say the outlet is about 85% FSA, that means the area blocked by the motor mount doesn't count. So if I have a Micro Fan with 50 mm diameter fan sweep circle, and the motor mount diameter is 27 mm, FSA is pi/4(50^2 - 27^2) = 1390.9 mm^2. 85% of that is 1182.3 mm^2, so that puts the exit diameter at 38.8 mm.

Is that correct? It seems lower than what most people have been doing.

I do intend to do some testing once I get a chance to start this ducted fan project.

Dan
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Old Apr 30, 2008, 07:38 PM
EDF rules... :)
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Well I suppose if there is enough cooling air arround the motor then a motor fairing would be apropriate but in most cases this is not. With enough motor mass to help metabolize the waste heat and short runs(>4min) they work. Think of the fan as a pump, once the rotor has moved the air there is minimal additional drag from expansion unlike in the inlet where expansion and contraction require much more expended energy, and to put this in perspective it adds less than can be meausured. More important to get the outlet area correct than streamlining in the outlet side of the duct.
Microfan outlet of 38mm is about right, I will have to take a look at my spreadsheet. Remember what Dan said about sizing on his Mig 29, the overal drag of the airframe will be the deciding factor on maximum speeds. The 85% (38mm) might not work with your airframe where as 90% might be a better alternative, in fact some experimentation must be performed to optimize your installation.

Eric B.
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Old Apr 30, 2008, 11:02 PM
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In case anyone is interested, I'm posting the analysis I used to derive the equations from the first post.

The point originally was to find out how static thrust and efflux velocity vary with changes in nozzle area. So V1 is the speed of the air leaving the ducted fan and entering the exit nozzle, and A1 is the FSA of the ducted fan.
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Old May 01, 2008, 02:59 AM
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That'd be why you're getting some weird results...

Bernoulli's can't be used through this control volume - it can only be used on flows with no work being done. Hopefully the fan is doing work!

Try using conservation of energy and mass. The rest should fall out.
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Old May 01, 2008, 06:27 AM
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what's happening in a duct.

All kind of information goes around and much of it is too theoretical for me.
I have made same drawings showing the change of airspeed in a duct before the fan moves.
all of the with an Ram airspeed of 1.0 for the catch area.
It gives you some idea of what to do when you make a duct for your next plane.
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Old May 01, 2008, 09:08 AM
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Quote:
Originally Posted by Odysis
That'd be why you're getting some weird results...

Bernoulli's can't be used through this control volume - it can only be used on flows with no work being done. Hopefully the fan is doing work!

Try using conservation of energy and mass. The rest should fall out.
Odysis,

The control volume chosen is just the exhaust duct after the fan. Essentially I'm letting the fan do its thing and trying to look at what happens afterwards.

I could try applying conservation of energy and the same thing would happen. Or, like you suggested, I could apply conservation of energy to a larger control volume including the ducted fan, which may be interesting but would be beyond the scope of what I was originally trying to analyze. I may do it though to see what kind of results I get.

Clausxpf, thanks for your advice.
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Old May 01, 2008, 10:00 AM
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Great advice! In my opinion, if a guy was to take Clauses drawings along with Odysises basic laws, and maybe add an explanation about the incompressability of air, you would end up with a pretty complete understanding of what happens in a duct.

Ron
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