Tony K

Posted by Mark Powell 2 on 23/08/2012 10:33:14: No Ps, Vs, or anything else needed. Put your hand six inches in front of a propellor and then six inches behind it. You will be left in no doubt that the speed does increase. So the UoS writer is not correct in stating that V2 = V3. That would mess up the momentum equation on the third page and how do you apply the continuity equation?

The person who never made a mistake never did anything.

Posted by Grasshopper on 23/08/2012 08:24:13: In fact I intend doing two thrust tests, one with the naked motor and one with my thrust tube fitted, to gain some knowledge of the effect of fitting said tube. Peter, please do that. I, for one, would be very interested in the results.

Posted by Mark Powell 2 on 23/08/2012 05:54:35: Froude (or Sidney University) says "as this air passes through the rotating blades its speed increases" A poster here questions that. Sometimes a little practical experience, as in (1) flying a 747 or (2) building a model with a propellor on it gives a clearer perspective. Yes, I question that, particularly as further down the page it is clearly stated that V2 = V3. As it is also stated that section 2 is "the front of the propellor disc" and section 3 is "just behind the disc", how can speed increase from V2 to V3? Also, if the velocity is the same at V2 and V3 how does the pressure change?

Posted by John Olsen 1 on 22/08/2012 21:03:54: Tony, you said " As thrust equals drag there must be a thrust being produced although there is no acceleration of the air before, through, or after the system." I'm sorry but this is a breach of Newtons laws of motion..."For every action there must be an equal and opposite reaction" For the aircraft to experience a thrust, there must be a thrust on something else, and presuming the ground is out of reach, that must be a thrust on the air, which being a fluid, must move in response to the thrust. Actually even if we did thrust against something solid, like the ground, that too must move, although the ratio of the masses tends to be such as to make the movement unmeasureable. John, I am not sure if I completely understand your point here. Isn't it the air moving which causes the thrust? I didn't say the air was not moving, I said it was not accelerating. Let me try to visualise it another way. Imagine the column of air which passes through the duct is a rope. With the model held stationary the rope is driven through the duct and its weight times the speed it moves creates a force. That is static thrust. Now take the two ends of the rope and tie them to fixed objects. The model will now move along the rope at the speed at which aerodynamic drag is equal to the force pushing it along. To the ouside spectator the rope is not moving but from the models point of view it is (but not accelerating) and, as before, the weight of the rope times the speed it moves gives you the force acting in the opposite direction. That is dynamic thrust.   Edited By Tony K on 23/08/2012 01:33:23 Edited By Tony K on 23/08/2012 01:34:23

Posted by Erfolg on 22/08/2012 15:28:18: The concept of V2 & 3 being essentially identical, is not unique, it the basis of integration. Yes, but are P2 and P3 the same? If not, why not?

Posted by Erfolg on 21/08/2012 12:28:12: Swiss Flyer A quick look at the equations ( in your link) seems to suggest that the air stream ratio of V1 to V4 should be in the region of 1 to 3, at max efficiency (for the fan duct system, although that ignores duct losses, as opposed to the free air equations used). Erfolg, how do you know the values of V1 and V4? UoS quote "The pressures at (1) and (4) are equal to the free stream value." That suggests to me that the velocities are also equal to the free stream value.   Edited By Tony K on 22/08/2012 14:47:18

I have had a look at the Sidney university paper on Froude's theory (Swissflyer's link). Quote, "As this air passes through the rotating blades...its speed increases" Does it? Further down the page is the statement, "V2 = V3". Quote, "Since the disc is thin and the area of cross section at (2) and (3) are equal... If V2 =V3 and the cross sections are equal how can P2 be different than P3? What is P anyway? I assume it is the dynamic pressure, 1/2 rho v^2. The density doesn't change at the speeds we are talking about and V2 =V3 so P2 must be equal to P3. In my opinion the written equation F = A(P3 - P2) = mdot(V4 -V1) is meaningless because A(P3 -P2) = 0. Perhaps one of you better educated chaps can tell me where I am going wrong.

John, I don't think there is any argument against static thrust being produced by acceleration of static air into the inlet, but remember that V-0 = V. Look at it this way. The EDF is a power converter. It converts electrical power into an ouput power. As you probably know, power is the rate of doing work. In the static case, what is having work done to it? The air of course, it is being forced to enter and flow through a tube. The work being done can be measured as static thrust. In the dynamic (flying) case, our power converter is still using the same amount of electrical power so we can assume that the output power is roughly the same. As the incoming air is already moving it does not need any work done to it, so where is the ouput power going? It is being used to overcome the drag of pushing the model through the air at a speed. The work is now being done on the model to push it against the drag. As thrust equals drag there must be a thrust being produced although there is no acceleration of the air before, through, or after the system. Therefore, thrust can only be derived from the velocity of the air through the system, hence thrust (static and dynamic) equals mass flow times mean velocity.     Edited By Tony K on 21/08/2012 12:13:55 Edited By Tony K on 21/08/2012 12:15:27

Posted by Erfolg on 20/08/2012 22:38:33: Like it or not, a EDF is just that, a fan, which produce a reactive force. It is quite simple. An air mover in a tube, just like a vacuum cleaner.

Posted by Mark Powell 2 on 20/08/2012 21:49:58: Nor does mass (flow, if you must), by itself. Thrust is mass x exhaust velocity. raise one and lower the other, the thrust will be the same. Does not matter if that mass is expanded, as by heat in a jet engine, or not, as in a EDF. Speed is a function of thrust and drag and nothing else. Sorry Mark, in the interests of accuracy:- mass x velocity = momentum (kg.m/s) mass flow x velocity = thrust (kg.m/s^2)

Posted by Simon Chaddock on 20/08/2012 19:28:02: I have been trying to find out what is the actual efflux velocity for a modern airliner engine. The only figures I have found are for a JT9D. 868 km/hr (539 mph) for the fan and 1120 (696) for the turbine. Given that the fan generates the majority of the thrust and the 747 with this engine has a maximum speed 594 mph it does rather confirm that planes can travel faster than their efflux velocity. Simon, I will bet you a pint that at flight level 350 the fan contributes very little, if anything, to the total thrust. Earlier on this thread Richard Sharman suggested that my mathematical modelling of EDF theory was nothing more than modelling vacuum cleaners. He was right, an EDF is an air pump in a tube, as is a vacuum cleaner. The thrust produced is the mass flow times the mean velocity (in the closed system). The only way an EDF model can fly faster than the efflux velocity is if the mean velocity is greater than the the free stream velocity (model speed). The only way that mean velocity can exceed free stream velocity is if the inlet area is so small that the system becomes a sucking device instead of a blowing device. It is possible, in that case, that the efflux velocity could be less than the free stream.     Edited By Tony K on 20/08/2012 22:15:30

Posted by Richard Sharman on 18/08/2012 12:45:27: The question is: why does it fly so well ? Considering it's only foam, has standard components, poor aerofoil section, and lot's of other technical deficiencies, (presumably to keep the cost down) it shouldn't be as good as it is. But it takes off quickly, flies smoothly in up to 20mph winds, and lands predictably. That is a good question Richard. Has there ever been an analysis of of exactly what makes one model fly better than another. The basic parameters must be understood; wing loading, aspect ratio, thrust line, tail moment, control surface size, etc. Are there other properties which should be considered, eg. polar moment of inertia, or is it just the inter-reaction of the basics?

I have half a tin of yacht varnish left over from a DIY project. I was wondering if anyone had used it for finishing balsa with tissue or glass fibre. I tried a small test piece with tissue. The finish was smooth and quite hard with no distortion but, before I use it on a whole fuselage, has anyone else used it? The main ingredient is 2-butanone oxime, whatever that is.

Ton, thanks for the pictures, I can work with those. As soon as I have finished my current project, which may take a couple of weeks, I will make a start on the Starlet. It must be very frustrating to get a design published and then for the published plan to be such a mess ( certainly not worth 10 GBP).

I have had a look at Schuebeler's website, those big units are very impressive. Your figures seem to fit the theory. By my calculations, the output power is in the region of 7,5 kW so the efficiency is nearer 80%. This you could expect to be reduced with duct losses. If a 14 cell lipo gives you about 52v, then 9,4 kW will draw about 180 amps. What size cables do you need to carry that current? BTW, I used to race 100cc karts, I couldn't imagine bolting one of those engines to a model aeroplane!

Posted by Mark Powell 2 on 15/08/2012 08:38:27: Power input 9.4 Kw Nine thousand four hundred Watts?

Posted by John Olsen 1 on 14/08/2012 01:33:08: So if you try to measure the air velocity at the entrance to the duct and at the exit, there is not going to be as much difference as you might think. For practical purposes though, we could measure the speed of the aircraft and regard that as the starting speed of the air. Then if we measure the speed of the air at the exit, there should be a difference. Given then that we know the diameter of the duct and the speed of the air leaving it, we should be able to come up with a reasonable figure for the mass flow, and then have some idea of how much thrust the system is producing. I made up a spreadsheet which with three inputs, inlet area, outlet area, and outlet velocity, will give a theoretical thrust and output power. On 26/11/2011, I posted (on this thread) some theoretically obtained figures which matched quite closely to test figures provided.

Posted by John Olsen 1 on 14/08/2012 01:33:08: Tony, the dv figure is not actually acceleration, it is change in velocity, so the equation you give is actually correct for the thrust from a mass flow, eg thrust is mass flow per second times change in velocity. The units come out correctly, eg your result is in kg.m/s^2 which is force in Newtons. A change in velocity is not in itself an acceleration, although it does imply that one must have taken place. John, you are right of course. I must try not to post after returning from the pub. Acceleration is dv/dt. What I was trying to say is that if the inlet and outlet are the same size, the inlet velocity and outlet velocity will be the same. Therefore no dv and no acceleration. The statement ,"thrust is mass flow times change in velocity",cannot be correct because you can obtain a thrust when the change in velocity is zero.

Posted by Alan B on 16/06/2008 10:57:00: Just for useless information - both retailers are owned by two German brothers. Years ago they both owned one of the supermarket chains together (cant remember which). They then had a major bust up and split. The other brother then started the other chain as competition to his brother. Just stumbled upon this thread and for the sake of accuracy, although this is probably of no interest to anyone, the above statement is wrong. ALDi was owned by two brothers named Albrecht and they did have a bust up over the policy of selling tobacco at a discount. The busines was divided into two groups, ALDI NORD and ALDI SUD ( north and south). UK shops are under the control of Aldi Sud. Lidl (rhymes with needle) was started by a chap named Schwarz but as the name Schwarz Markt (black market) did not seem to be appropriate for a discount retailer, Schwarz bought the rights to use the name of a business partner named Lidl.

Posted by John Olsen 1 on 11/08/2012 06:23:05: Tony K asked a question a few posts back about a constant section duct with therefore a constant mass flow before and after the fan, the question being how could it create thrust? John, the point I was trying to make is that the equation often quoted, ie. thrust equals mass flow x acceleration (dv) is incorrect. Thrust is mass flow x mean or average velocity (I am not sure which so I use the average) through the system. The equation is: T = kg/s x m/s = kg.m/s^2 The system is that between the inlet and the outlet. The average velocity is inlet velocity plus outlet velocity divided by two. The frame of reference being the model in flight.

Roger Daltry, one of the few old codgers who can still sing.

Ton, thanks for replying to my post. It would be interesting to see how many builders choose to use CNC parts. I must have too much free time because choosing the wood and creating the individual parts is, to me, very satisfying. Not only that but I do not like spending cash when I can do something myself. I have had a good look at the plan and apart from the wing structure it is a fairly conventional design. I can see a few areas where I think I can simplify the parts. Some of the parts, as you mentioned, do look a bit odd but that is not too much of a problem, except for the wing ribs. Are they supposed to be a flat bottom or semi symmetrical?

Interesting piece on the BBC news site about funding. Hope the link works. First attempt.

I bought the KS230 when the local DIY shop was offering 15% discount and I am very pleased with it. I have cut 6mm ply without a problem and with nice square edges. The only annoying thing is that with the guard in place I can only cut 11mm thick material.