Efficiency of a ducted fan - Page 2 - RC Groups
Feb 01, 2008, 08:41 PM
Texas Buzzard
Quote:
 Originally Posted by jrb 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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Thanks, All very true jrb. Can youi help with actual calculations in solving this little problem OR are you trying to say that the problem can't be solved using only watts / watts out?
Feb 01, 2008, 09:02 PM
Texas Buzzard

# Charging batteries

Looks like a beautiful day tomorrow. Maybe the people who frequent this thread are tired of a bit of the real-world approach.

Some are throwing in factors that we can't measure. How can we measure thrust while the plane is flying.

My friend from GERMANY should be specific since he says his H.S. teacher would punish him for what I wrote. BE SPECIFIC in critique please.

This started out as an honest question, no one would repond with calculations so I finally did as a courtesy to all interested parties.

Vampire67's post reminds me of the day I asked a question in a Seminar Class at the Natl Labratory at White Sands, NM. There were several PhD there in the class who I thought were competent. In answering my question the prof used the chalk board for a time, then turned aroud and said, "There are 9 physistists her and I think each of the 9 would have a slightly different comment".

I say that to emphasise that your training might be different fom others thus, your explanation would be different but still would arrive at a common answer.

The German commented on the use of Imperial or English Units and that the CGS and the MKS was superior. DID HE NOTICE that I converted to MKS as soon as I could? If he had attended major universities in the U.S. he would know that in Physics we haven't used the English System ( slugs,feet & seconds) since WW II. But, the engineers still use what's called the Engineering System which actually the English System. Of course MKS os more convenient to use.
 Feb 02, 2008, 12:44 AM Registered User T B - You started your first post of this two-part thread with a deception; you implied ignorance. You asked for the opinions of others and you got their opinions, cased closed? no. It is not a great feat to mathmatics to follow energy input from the battery, through the ESC, motor, and fan and finally as thrust out the tailpipe. But, it is fruitless to do so, when a bathroom scale is available. As a number of participants in this thread have accurately noted; (and I paraphrase) static thrust means diddily-squat, the proof of this pudding is in flight. So, was there a real question in there somewhere, or was it just a demonstration of mathmatical agility? Ron
Feb 02, 2008, 01:17 AM
The blade numbers go up to 11
Quote:

OHG! There's so much totally wrong in there! I don't have time to teach highschool physics - I'm sure there's a wiki page for it - but I'll fix up a few fundamental errors which may lead people astray.

1. work is NOT ma, force = ma or mass x acceleration
2. force is force, work is force x distance, don't confuse them
3. the Joule is a unit measure of energy or work, not a measure of force, you cannot change thrust into Joules!!!!!
4.Watts IS NOT A UNIT OF WORK!!!!!
5. Here's the biggy: IF POWER IN WAS 100WATTS AND POWER OUT WAS 2.84WATTS, EFFICIENCY IS 2.84%!!!!! ie 2.84 / 100 . Unfortunately, this is the efficiency of some people's brains.

Fan efficiency is nothing more than the ratio of actual thrust divided by the theoretical thrust available with a given amount of shaft power. Note that I said shaft power, motor efficiency doesn't come into it. Unfortunately the subject is far too involved to get into here. If you really need to know, get some propulsion textbooks and study real hard.

Stu.
Feb 02, 2008, 01:32 PM
Member
Bingo!

There are dozens of threads around that speak to the physics; note it works for either EDFs or Props!

Here’s a few that might be of interest:

Any unit of measure can converted from one system to another via a multiplication factor: feet to meters, pounds to grams, pounds to Newtons, etc.

So, in the end simply measure the thrust, and velocity to get Propulsion Output Power. Compare that to Watts In and you’ll have great measure of efficiency!

You can scale it via a multiplier for easy use for display, or if wanted to convert it to its actual magnitude.

Here’s the above cleaned up just a little:

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

Velocity = L/T

Power is M*L*L/T/T/T

Velocity times Thrust as fundamental units = L/T * M*L/T/T;

Which = M*L*L/T/T/T -- Power!

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