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

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http://www.dmack.net/mazda/index.html

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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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