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Hi Ed,
I think there must be an unwarranted assumption. Namely that at max
speed (Thrust=Drag) the speed of the air coming out of the prop is
equal to the air going into the prop. It seems more likely that Dv is
the amount needed to produce the amount of thrust required to equal the
drag. The flow of air behind the prop should be higher than the
forward speed of the aircraft. There has to be some slippage for the
math to make sense.
I must admit that I haven't read any of the text books, but I don't see
how it can be any other way.
Bob W.
On Thu, 4 Mar 2010 07:45:32 -0500
"Ed Anderson" <eanderson@carolina.rr.com> wrote:
> I’ll give it a shot, George – even thought I believe I have previously
> admitted to not understanding this prop stuff.
>
>
>
> (1)
>
> You are correct V is the velocity of incoming air flow to the prop disc
> which if the aircraft is not moving is equal to zero.
>
> So if the air coming into the prop and the air going out of the prop disc
> are equal - then that means the prop did not provide any acceleration or
> additional velocity to the air mass or in other words Dv (or the change in
> air velocity) = 0.
>
>
>
> So if Dv = 0 and every term in the equation is multiplied by this factor, if
> it is zero then the entire equation = 0 and that leaves us with Thrust T =
> 0. So if no change in air velocity then no change in momentum (pDv)of the
> air mass and therefore again no thrust produced.
>
>
>
> Perhaps another way to look at it is the prop didn’t push against the air
> and therefore didn’t change (increase) its velocity, therefore no thrust –
> if it had pushed against the air then the air would have gained in velocity
> or Dv <> 0
>
>
>
> Strickly looking at the formula for Thrust above you can see that all terms
> of the equation or multiplied by the tail end part p*Dv. Well that part is
> actually the equation for momentum or rather since its Dv rather than V, it
> is the change in momentum imparted to the air mass by the spinning prop.
> Since the density p is constant, then the change in momentum is entirely due
> to the Dv factor. Since Dv = v2 – v1 change in velocity – that is also the
> definition of Acceleration = Dv/Dt – here the Dt (change in time is assumed
> to be the same unit time factor for all factors in the equation) and so is
> not shown).
>
>
>
> We know the area for a circle is P R2 or equivalently P (D/2)2 = P D2/4
> note that this is the first term in the above formula - the area of the
> prop disc. So if we now multiply this by p (air density) we get the air
> mass that the prop disc is trying to impart a change in momentum to or P
> D2/4 * p that leaves (v + Dv/2) *Dv – since the aircraft is not moving then
> V = 0 leaving Dv/2*Dv or Dv2/2 . if we move the density factor p back to
> this part of the equation we have p* Dv2/2 . This is the formula for the
> kinetic energy increase imparted to the air mass by the additional velocity
> imparted to the air by the spinning prop.
>
>
>
> So probably have not helped your understanding – but in summary, if there
> is no change in velocity Dv imparted to the air mass flowing through the
> prop disc, then there is no thrust T. Or at least that is what the equation
> appears to say to me.
>
>
>
> Best Regards
>
>
>
> Ed
>
>
>
> Ed Anderson
>
> Rv-6A N494BW Rotary Powered
>
> Matthews, NC
>
> eanderson@carolina.rr.com
>
> <http://www.andersonee.com> http://www.andersonee.com
>
> <http://www.dmack.net/mazda/index.html>
> http://www.dmack.net/mazda/index.html
>
> http://www.flyrotary.com/
>
> <http://members.cox.net/rogersda/rotary/configs.htm>
> http://members.cox.net/rogersda/rotary/configs.htm#N494BW
>
> http://www.rotaryaviation.com/Rotorhead%20Truth.htm
> <http://www.dmack.net/mazda/index.html>
>
> _____
>
> From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On
> Behalf Of George Lendich
> Sent: Thursday, March 04, 2010 3:49 AM
> To: Rotary motors in aircraft
> Subject: [FlyRotary] Re: Fixed Pitch Prop Thrust was RE: [FlyRotary] Re:
> single rotor
>
>
>
> Ed,
>
> Propellers are something else I don't understand and need to learn, so a
> couple of question if I may.
>
> 1. I assume 'Velocity of incoming flow ' means speed of the aircraft.
>
> 2. Additional velocity, acceleration by propeller - don't quite understand
> that one,
>
> Could you explain how one gets 2. value.
>
> TIA
>
> George ( down under)
>
> Arggg! Props!
>
>
>
> Ok, Bill – here is my stab at it. The following equation tells the tale –
> well some of it.
>
>
>
> (1)
>
>
> where:
>
>
> T
>
> thrust
>
> [N]
>
>
> D
>
> propeller diameter
>
> [m]
>
>
> v
>
> velocity of incoming flow
>
> [m/s]
>
>
>
>
> additional velocity, acceleration by propeller
>
> [m/s]
>
>
>
>
> density of fluid
>
> [kg/m³]
>
>
>
>
> (air: = 1.225 kg/m³, water: = 1000 kg/m³)
>
>
>
>
>
>
>
> Thrust is about increasing the momentum of the air mass passing through the
> prop disc. Static thrust occurs while sitting still - therefore v (the air
> velocity of air in front of the prop disc) = 0. So the addition momentum
> imparted to the air mass by our spinning prop is p*DV. Since the air mass p
> during our run up is essentially constant ), that leaves two variables - the
> diameter of the prop D and the amount the spinning prop accelerates the air
> (Dv) to affect the thrust (T) generated.
>
>
>
> The following are extracts from some of the better article (more
> understandable) material I have read about props and static thrust. As it
> concludes and Tracy points out Static Thrust does not really tell the whole
> story.
>
>
>
> For a typical, fixed pitch propeller, the largest induced velocity occurs
> under static conditions, where the efficiency is small. It decreases with
> increasing flight speed, until it reaches zero: no thrust is generated.
>
> For a given power P, it is always desirable to use the largest possible
> propeller diameter D, which may be limited by mechanical restrictions
> (landing gear height) or aerodynamic constraints (tip Mach number). That's
> why most man or solar powered airplanes use large, slowly turning props.
> These catch a large volume of air and accelerate it only slightly to achieve
> the maximum efficiency.
>
> As long as an aircraft does not move, its propeller operates under static
> conditions. There is no air moving towards the propeller due to the flight
> speed, the propeller creates its own inflow instead. A propeller, with its
> chord and twist distribution designed for the operating point under flight
> conditions, does not perform very well under static conditions.
>
> As opposed to a larger helicopter rotor, the flow around the relatively
> small propeller is heavily distorted and even may be partially separated.
> From the momentum theory of propellers we learn, that the efficiency at
> lower speeds is strongly dependent on the power loading (power per disk
> area), and this ratio for a propeller is much higher than that for a
> helicopter rotor. We are able to achieve about 80-90% of the thrust, as
> predicted by momentum theory for the design point, but we can reach only 50%
> or less of the predicted ideal thrust under static conditions.
>
> So much for theory. My personal experience when I went from the faster
> turning 68x72” prop to the slower turning (2.85 gear box) 76x88 prop – my
> take off acceleration increased significantly indicating (in my opinion)
> more thrust was being generated. With the 76x88 prop and my old 13B I
> would generate 5800 rpm static (for what its worth), with it cut down to
> 74x88 I picked up 200 rpm for a static of 6000 rpm. Plus I got another inch
> of ground clearance – needed on my nose geared Rv-6a.
>
>
>
> Interestingly enough the larger slower turning prop not only did not hurt my
> top speed it actually increase around 4 mph – perhaps due to the increased
> HP due to high rpm of the lighter loaded engine?
>
>
>
> Ok, Bill that’s my take and what I could pull out of references. Don’t know
> if it really tells us a whole lot – there are some good NACA studies on Prop
> – but the math makes my head hurt.
>
>
>
> Ed
>
> Ed Anderson
>
> Rv-6A N494BW Rotary Powered
>
> Matthews, NC
>
> eanderson@carolina.rr.com
>
> <http://www.andersonee.com> http://www.andersonee.com
>
> <http://www.dmack.net/mazda/index.html>
> http://www.dmack.net/mazda/index.html
>
> http://www.flyrotary.com/
>
> <http://members.cox.net/rogersda/rotary/configs.htm>
> http://members.cox.net/rogersda/rotary/configs.htm#N494BW
>
> http://www.rotaryaviation.com/Rotorhead%20Truth.htm
> <http://www.dmack.net/mazda/index.html>
>
>
> _____
>
>
> From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On
> Behalf Of Bill Bradburry
> Sent: Wednesday, March 03, 2010 5:01 PM
> To: Rotary motors in aircraft
> Subject: [FlyRotary] Re: single rotor
>
>
>
> I would like to get some educational (for me) discussions going on this.
>
> A prop of 76 X 88 is pretty common in our usage. Tracy, Ed, and I have a
> Performance Prop in this dimension. Dennis and maybe others have a Catto
> prop in this dimension. We all seem to be getting static rpm of about
> 52-5400 rpm (except for Dennis with his new DIE manifold). Tracy and Ed had
> their prop cut down to 74 X 88 and are getting increased static to around
> 6000 rpm. Higher rpm = higher HP for the rotary. We should get higher
> thrust with a slightly smaller diameter prop? This has something to do with
> the idea of sizing the prop to the engine. I wonder what is the proper
> size? What is the proper static rpm for best performance with the rotary?
> What did Tracy and Ed lose in prop performance and what did they gain in
> total performance when they cut the prop down?
>
>
>
> It seems to me that a prop sized for climb would allow around 7500 rpm at
> about Vx or Vy? Max speed would require 7500 rpm at WOT sea level? I
> wonder what rpm our props allow at these speeds? If you had a prop that
> would do the above, I wonder what the static rpm would be? Then since most
> of us have fixed pitch props, I wonder where we should try to be for the
> best of both worlds (a compromise)?
>
>
>
> We have some really good engineers in this group and they have made these
> selections. I know they know why they made the selection they did. How
> about sharing? :>)
>
> Don’t worry, you can not ramble on too much for me!
>
>
>
> Bill B
>
>
>
>
> _____
>
>
> From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On
> Behalf Of Tracy Crook
> Sent: Wednesday, March 03, 2010 2:32 PM
> To: Rotary motors in aircraft
> Subject: [FlyRotary] Re: single rotor
>
> Al is correct about it taking HP to make static thrust with a prop but the
> assumption about the relationship between HP and static thrust is subject to
> a lot of variables. There is no fixed relationship between static thrust
> and HP. If there were, you could not account for the ability of most
> helicopters to hover.
>
>
>
> You could easily increase static thrust by 1.18 by increasing the diameter
> of the prop and the reduction ratio of the redrive with NO increase in HP.
>
>
>
> But my real point was that static thrust is not a very useful measurement to
> us.
>
>
>
> Tracy
>
> On Wed, Mar 3, 2010 at 11:06 AM, Al Gietzen <ALVentures@cox.net> wrote:
>
> Looking at the two sizes of the engine, it takes 1.6 times as much
> horsepower to develop 1.18 times as much static thrust! Somehow this does
> not compute for me….I always doubt the performance figures in a sales
> presentation and when they don’t make sense to me…..???
>
>
>
> Bill B (hoping this generates an educational experience for me :>)
>
>
>
> We’re talking about the amount of force exerted by the prop with the plane
> (motor) standing still.
>
> So, it seems to make sense to me that the power needed to accelerate the air
> to generate the thrust would go as the cube root; and the cube root of 1.6
> is very close 1.18.
>
>
>
> To move the amount of air it takes to generate the thrust certainly does
> take horsepower. Very much the same as the power it takes to drive the pump
> (or generator) on a dyno. So I don’t know how Tracy was interpreting the
> question.
>
>
>
> Al
>
>
>
>
>
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--
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