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Old Jan 31, 2008, 01:15 PM
Texas Buzzard
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Efficiency of a ducted fan

Is "effeciency " of a ducted fan standardized?

How is this measured by American Mfg?

I found the following in an add by GWS. Does it make sense?
Motor Volts (V): Amps (A): Thrust (g/oz): Power (W): Efficiency (g/W - oz/W):
EM400 (Stock) 7.2 6.6 147 / 5.19 55.44 2.65 / .09
EM400 (Stock) 8.4 7.9 170 / 6.00 75.84 2.24 / .08
EM400 (Stock) 9.6 9.3 210 / 7.41 100.44 2.09 / .07
EM400 (Stock) 10.8 11.2 246 / 8.68 134.40 1.83 / .07
EM400 (Stock) 12 12.5 281 / 9.91 165.00 1.70 / .06

If the Watts input to motor is measured to be ( 11v x 20 amps) = 240 Watts. And the test stand gave a measured thrust of 20 ounces, then o can we say effeciency is 0.08/Watt or about about 0.08 or 8% 8% is not good, eh?

Is this semi-valid? Thrust/Watt sounds good at first approximation.




- - - - - -
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Old Jan 31, 2008, 03:26 PM
Life begins at transition
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Thrust per power is only usefull if you're comparing two fans of the same size, with the same outlet. A large fan running slowly will produce x thrust much easier than a small fan flat out.

Real efficiencies need to be non-dimensional, i.e. power out divided by power in. Only with non-dimensional numbers can you scale up and down without too many concerns
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Old Jan 31, 2008, 06:58 PM
Texas Buzzard
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Good stuff

Quote:
Originally Posted by Odysis
Thrust per power is only usefull if you're comparing two fans of the same size, with the same outlet. A large fan running slowly will produce x thrust much easier than a small fan flat out.

Real efficiencies need to be non-dimensional, i.e. power out divided by power in. Only with non-dimensional numbers can you scale up and down without too many concerns
.........>>><<<<<........

Yes I was comparing two fans, but using the Same Motor. They both had the same I.D. but one had the option of having 5 blades rather than 3. Both are purchased with 3 blades per fan.

Am I correct in saying the 5-blade fan turning at the same rpm as the 3-blade fan will draw more current, Thus more Watts??
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Old Jan 31, 2008, 08:59 PM
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Would depend on pitch of blades. They are likely different. The Alpha 3 and 5 blades are different animals and for the same fan housing. Honestly don't know if pitch is different as I only have the old 3 blade.
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Old Jan 31, 2008, 09:24 PM
jrb
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Efficiency is and always will be power out divided by power in -- same units above and below!

Power in is easy -- use a Whattmeter.

Power out isn't hard either; though two measurements are needed -- thrust & e-flux velocity -- speed of the exit flow!

Propulsion power is thrust times e-flux velocity.

Use a Kestral 1000 or similar.
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Old Feb 01, 2008, 10:03 AM
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TB, let me give it a shot, first, efficiency has no units, like JRB said same unit above and below and if you divide the same unit, result is non-dimensional, no unit, so it is represented by %.
ok, you try to compare two different fans with same motor and try to find out which one has better efficiency, say first motor with a input (watt), give you "A" ozs of thrust and on the second fan with same motor and SAME input watt give our "B" ozs of thrust, and if "A" is higher than "B", usually we say fan "A" has higher efficiency than fan "B", does this make sense?
You note I use the word usually, because there are lot other factor come in, like amp draw, rpm voltage drop, so higher effeciceny sometimes is not necessary the best setup. confusing, right?
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Old Feb 01, 2008, 12:04 PM
Texas Buzzard
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Not confusing at all

Quote:
Originally Posted by hokonghing
TB, let me give it a shot, first, efficiency has no units, like JRB said same unit above and below and if you divide the same unit, result is non-dimensional, no unit, so it is represented by %.
ok, you try to compare two different fans with same motor and try to find out which one has better efficiency, say first motor with a input (watt), give you "A" ozs of thrust and on the second fan with same motor and SAME input watt give our "B" ozs of thrust, and if "A" is higher than "B", usually we say fan "A" has higher efficiency than fan "B", does this make sense?
You note I use the word usually, because there are lot other factor come in, like amp draw, rpm voltage drop, so higher effeciceny sometimes is not necessary the best setup. confusing, right?
>>>>>>>>>>>>>><<<<<<<<<<<<<<<<<

No Kong, you were NOT confusing at all. You would make a good H.S. teacher. Why? You talk on a level that is understandable to the mentally handicaped like me - HA HA HA just joking. But your post on effeciency of two fans w/ the same motor and same watts is good for me!
.................................................. ........
When anyone solves a problem is is common courtesy to show your calculation for two reasons.
1. If a novice is reading then he might glean more by seeing the progression.
2. If a reader disagrees with what is written then that reader can be specific in his critique, therefore I wrote as if I was being informative & courteous.


Watts is easy, but to measure thrust isn't. Back in my glow days I had a dial type spring balance ( a thing to measue the mass of an object ) since it was calibrated in kg. Kg is a mass unit. 1.0kg = 1000 gms. Max. scaling was 10 kg or since 1.0 kg = 2.2 lbs it went up to 22.0 lbs.

1.So first I have to find it, hang it from ceiling and hang the fan/motor unit with electronics facing so the thrust is aimed downward - right? The difference in the reading before fan is turned on AND after it is turned on will give the actual measured thrust.
2. Since effeciency is nondeminsional (as you said) Watts IN / Watts out will give a fraction lessss tham unity ( 1.00 ) - right? So multiply the quotent by 100 to get percent effeciency....don't you agree?
3. But there is a small problem, how to convert thrust into Watts OR Watts into Force units since Thrust is measured in Force units - right?
4. We can say that the fan is pumping air out of the tailpipe. Essencially this fan is doing WORK on the air. The fan is applying a Force on the air causing the air to be moved through a distance. The rate (speed) at which the air is moving is it's velocity since we know the direction the air is moving. O.K.,(I am thinking out loud now!) Since Work is F=ma or force x distance and Power is the rate of doing the Work or in English Syst: pounds feet per sec (ft lb/sec). We are getting closer to the conversion now.
5. Volts x Amps = Watts - right? And Force x Distance = Work - O.K.?
6. We have to do something with the thrust we measured to change it to Watts so we can divide Watts IN by Watts OUT.

7. When we use a spring balance to measure the thrust we probably are using an English balance calibrated in ounces. Lets say the measured thrust was 10.0 ounces. Since one ounce is equavalent to ABOUT 29 gms of mass we can say that 10 oz x 29 gms/ounce = 290 grams - but that is Not a force unit! 290 gms x the acceleration of gravity (980 cm/sec^2) gives us the conversion of Ounces into Metric Units or 290 gms x 980 cm/sec^2. This gives us 284,200 dynes. The dyne is a unit of Force in the MKS system.. 1.0 gm resting on your table pushes down with a force of 980 dynes.
So our thrust being converted from ounces to dynes . That is a big number. Lets do better. 10 oz x 29 gms/oz = 290 gms thrust. Can we change it to JOULES which is more convenient? Yes I think so. A Joule is a unit of work JUST LIKE WATTS IS A UNIT OF WORK.
1 Joule = 1.0 newton meter ( The force of 1 newton moving thru' one meter. Like ft. pounds is 1 pound of force moved thru' 1 ft /sec- this is 1 ft lb/sec ... 550 ft lbs/sec = 1 H.P.
Going back to the 290 gms of thrust... it happens to be = to 0.290 Kg! When using Kgs we use the MKS system or 9.8 m/sec^2. So multiply acceleration of gravity 9.8m/sec^2 x 0.290 kg , it's = 2.842 kg m/ sec^2
This is a force unit. POWER EQUALS FORCE x DISTANCE/TIME OR THE RATE OF DOING WORK. So if you push a trunk across the floor with a force of 550 pounds and move it 1.0 foot in 1.0 sec then you exhibilted 1.0 H.P. In metric 1.0 Joule of Work done per second = 1.0 Watt. So think that a WATT IS EQUAVALENT TO H.P. or just Power.

1 joule = 1 newton meter .... a newton is a force unit having the unit kg m/sec^2. If some of you are engineers you may want to use the engineering system and stick to pounds, slugs and feet etc. I don't
want to do that.

** Work being done by the fan = 2.84 kg m/ sec^2

Since 1 newton meter = 1 Joule we are in business.

1 Joule/sec = 1 watt That is: work done per sec = power

So the 2.84 Joules/sec has to = 2.84 watts being done per sec on the air by the fan unit.

If watts IN was 100 watts and watts OUT was 2.84 watts

100 watts/2.84 watts = 35.2 or 35.2% effecient
.


.................................................. .........................

Everyone except two have said that watts IN/watts OUT is the way to go. It does simplify it for us and for all practical purposes the slight differences in Temp, Hunidity and air Pressure doesn't make a significant difference. The NASA engineer at Langley in the wind tunnel needs more than we can give him with our spring balance and watt meter. Nuff said.

PLEASE - ANYONE, if you find an error make me aware of it by referencing the line in which the error appeared. 35.2% seems to be a low effeciency but after all it was a hypothetical question.

8. Remember the thrust was 10 ounces. .[U] We just measured the thrust didn't we? The amt. of work done by the fan on the air was ~ 2.84 Joules ( force x distance OR newton meters) What about the time? It is taken care of since when we measured WATTS IN we have the Amt. of Work being done per second by the battery. Ha Ha , we just converted OUNCES TO WORK BEING DONE ON THE AIR! iT WAS DONE IN JOULES OF WORK ! A WATT IS A UNIT OF WORK/sec or what we call POWER. Looks like we are just about there![/U]

By definition the Joule is a unit of work done/sec .

1 H.P. = 746 Joules = 746 Joules /sec
again - - - -- 1 joule/sec = 1 watt.

9. Watts divided by Watts is unitless! You CAN COMPARE TWO DIFFERENT FAN UNITS WHEN YOU KNOW WATTS IN and THRUST produced .
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Old Feb 01, 2008, 12:31 PM
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Quote:
Originally Posted by Texas Buzzard
... Since Work is F=ma ...
And who again gave you a physics degree ???????


http://www.rcgroups.com/forums/showp...2&postcount=18
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Old Feb 01, 2008, 01:44 PM
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Quote:
Originally Posted by Texas Buzzard
POWER EQUALS FORCE x DISTANCE/TIME OR THE RATE OF DOING WORK. 1.0 HORSE POWER = 550 FOOT POUND OF WORK DONE /SEC. So if you push a trunk across the floor with a force of 550 pounds and move it 1.0 foot in 1.0 sec then you exhibilted 1.0 H.P. In metric 1.0 Joule of Work done per second = 1.0 Watt. So think that WATT IS EQUAVALENT TO H.P. or just Power.
Work is Ws (WattSecond) = Power times duration.
And 1 H.P. is ~735W.

And by the way. Use International SI units , not imperial units.
The benefit is that you never need to scale units in between and the units will exactly resolve into the unit you need for the result.
Link 1 (sorry, in german, but the tables for derived SI units is missing in the english wikipedia)

This is an incredible help to find out if your thoughts got scewed.
If the result is not showng the righ units you screwed up.

So work [J] = m^2 * kg * s^(-2) = N * m

aka work spend to move something with one Newton force for one meter.

Power [W] = m^2 ∙ kg ∙ s^(-3)

aka work_done divided by time_spend (ask you boss what he considers your work power)

thrust: [N] = m∙kg∙s^(-2)
e-flux: m/s = m ∙ s^(-1)

thrust * e-flux = m∙kg∙s^(-2) * m/s = m^2 * kg * s^(-3) = [W]att

QED (jrb told it already)



Get your thrust by converting the mass units on your scale into Newton

force = kg * g ("g" is earths gravity ~9,81 m∙s^(-2) ) and you get [N]ewton.
1kg = 9,81N

regards Peter

P.S.
Thats the reason I will never understand americans sticking to imperial units.
It's a mess in physics classes
Not even NASA is able to solve this in imperial units.
May be that why a German was needed to get an American on the moon


P.P.S.
My physics teacher in highschool would have thrown me in detention for so many "mistakes"
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Old Feb 01, 2008, 02:27 PM
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we spent far too much time on Maths, weekend is come up, too many model, too little time, let build and fly some planes,
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Old Feb 01, 2008, 03:21 PM
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Kick me if I'm wrong but the problem with all this is..... it's not in the plane flying... the only thing you'ill get this way is a compairative measurement at static. what its doing at speed is completely different. real world, you need to do testing in a wind tunnel like DS did.

sorry
joe
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Old Feb 01, 2008, 03:24 PM
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Quote:
Originally Posted by wireit
Kick me if I'm wrong but the problem with all this is..... it's not in the plane flying... the only thing you'ill get this way is a compairative measurement at static. real world, you need to do testing in a wind tunnel like DS did.

sorry
joe
Careful Joe, he's gonna want to know where your AND Daniel Scheubeler's Degree's were given/earned.
-Mike
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Old Feb 01, 2008, 04:27 PM
jrb
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Thrust Power Derivation

Thrust time velocity is all that's needed; doesn't even matter what units you use!

Since you're primarily making comparisons, and made via the same tools and techniques a simply "scaling" (multiplier) is good enough to get it into a resonalbe range.

Here's some info from one of the other posters here:

"For thrust measurement it's enough to hang the plane from the ceiling of your garage with two wires. The deflection from the vertical will give you the thrust, you don't need a scale (In any case upside down placement on a kitchen scale will only work if the there is no front intake).

For efflux speed measurements we use a $2 electric motor hooked up to an ordinary voltmeter. It needs to be calibrated, otherwise it will give you only relative speed measurements:

http://aeneas.ps.uci.edu/edf/fan/p8300058.jpg

It's accurate enough to measure small deviations from uniform speed throughout the fan diameter.

Thrust=weight * tan (angle). So 45 degrees deflection -> thrust = weight etc.

(rpm of sensor probe microfan) proportional to (efflux speed) ;
also (voltage out) proportional to (rpm of sensor probe).

Thus measured voltage is lclose to linear in efflux speed. Static thrust can also be measured locally, it is proportional to the rpm^2 or measured voltage^2 .


http://www.rcgroups.com/forums/showt...threadid=32426

"

And, I've posted this in the past:

"Thrust Power Derivation

Fundamentally, thrust is proportional to rpm squared times diameter squared >> T = rpm^2 * D^2. There’s a few constants that need to come along though: rho ()/density (temp & pressure effects), pi (), a prop constant, etc.

Most often the prop constant gets determined empirically, and is usually expressed in terms of Pitch and Diameter. The most common form for thrust though doesn’t use pitch, but rather diameter squared.

Using R as rpm from hence forth.

And then, most folks don’t talk about the fundamentals so what you’ll find written is >> T = k*D^4*R^2. A few folks write as suggested above T = k*P*D^3*R^2; not much difference when P and D are close in value, like a “square” prop 5x5.

Power is actually more important in understanding a prop’s performance in E-Flight, and is the parameter that the various Calc programs use to predict a configuration’s performance. Again, it is rarely shown in its fundamental form, being written typically as Watts = k*P*D^4*R^3.

By 1st principles power is velocity times thrust. Maybe you’ve noticed the real fundamental parameter by now – Velocity!

Thrust for any propulsion system (i.e. EDF, or rocket) written from 1st principles is k*rho*V^2*A^2 – velocity (e-flux) squared applied over an area squared.

A prop turns R into V >> V = k*P*R which leads to all of the above.

Pitch speed (velocity) is commonly calculated as P*R/1000 using pitch in inches; this is an approximation that neither applies a “k”, nor use the proper constants for unit conversion."
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Old Feb 01, 2008, 04:29 PM
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Did you see my "Open Jet" wind tunnel pictured above?
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Old Feb 01, 2008, 05:26 PM
jrb
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Missed this in my copy & paste; from one of my previous postings:

"Many folks remember F=MA; but that is not what Newton actually derived.

Newton actually stated that a force is equal to the time rate of change of momentum (mass times velocity) = d/dt mass * velocity – this difference is very important with respect to propulsion.

In fundamental units, force is mass * length divided by time squared = M*L/T/T.

Velocity = L/T

And, then Power = v * T = L/T * M*L/T/T = M*L*L/T/T/T!

Just as shown below:

So force divided by power leaves you with a parameter that has the fundamental units of T/L. = Velocity!


Efficiency, like Reynolds is a parameter w/o units; its makes no sense if units are when T/W!"
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