Integration help for helicopter rotor

[jmmccain]

Hi,

This is my first post here, but I've been enjoying the site for some time now. I've run into a snag with an equation I'm trying to develop. Its purpose is to help determine power requirements for radio controlled, model helicopter rotors:

P = ( π² Θ³ C_d ρ A / 2 ) ( R⁴ - r⁴ )

P = power (Watts)
π = 3.14159
Θ = rotor head speed (revolutions per second)
C_d = drag coefficient (unitless)
ρ = air density (kilograms per cubic meter)
A = rotor blade area (square meters; span times chord)
F_d = drag force (Newtons)
V = velocity (meters per second)
R = rotor blade outside (tip) radius (meters)
r = rotor blade inside radius (meters)
τ = rotor head torque (Newton meters)

I started with the relatively well known equation for fixed wing drag:

F_d = C_d ρ A V² / 2

Then, knowing that the velocity varies with radius:

V = 2 π r Θ

The following may be obtained:

F_d = C_d ρ A ( 2 π r Θ )² / 2

This may be rearranged as:

F_d = ( 2 π² Θ² C_d ρ A ) r²

Torque equals force times radius, so:

τ = F_d r = ( 2 π² Θ² C_d ρ A ) r³

It appears to me that this equation only works for a small area at a specific radius. For instance, it might provide a reasonable estimate for a model helicopter's flybar paddle. In order to make it work for large areas over a range of radii, it needs to be integrated with respect to radius, correct? This yields:

τ = ( π² Θ² C_d ρ A / 2 ) ( R⁴ - r⁴ )

R and r now represent the limits of integration. Since power equals torque times angular velocity:

P = τ Θ = ( π² Θ³ C_d ρ A / 2 ) ( R⁴ - r⁴ )

Since r⁴ << R⁴, r⁴ may be neglected (typically) and this becomes:

P = π² Θ³ C_d ρ A R⁴ / 2

Now, here's my problem. Unit analysis tells me I have an extra meter in there. I get Watt-meters instead of Watts. I know that it appeared when integrating. Am I integrating incorrectly? Do I need to integrate at all? It's readily apparent that I've missed something obvious, but I'm sure what or where. It's been about 20 years since I last took a calculus class, so go easy on me.

Yes, I know the equation is simplistic. Determing a drag coefficient for a rotor blade isn't easy. However, one could use such an equation to "measure" the effects of changes to head speed and blade dimensions. I.e., if the blade length is increased, how much must the head speed decrease in order to return to the original power level - with the stipulation that the drag coefficient doesn't change significantly.

On second thought, don't go easy on me. Gimme the third degree! I will appreciate any help that comes my way.

Thanks,
Joseph

 

[gamesguru]

Here's your error:

It appears to me that this equation only works for a small area at a specific radius. For instance, it might provide a reasonable estimate for a model helicopter's flybar paddle. In order to make it work for large areas over a range of radii, it needs to be integrated with respect to radius, correct?

I must admit, it was quite a creative way to approach the problem, perhaps a little more complex than the way I would have chosen.

Here's what I would have done:

Imagine there is just a single blade (1/4th of what there is in reality, we can quadruple it in the end). The blade is rotating around a pivot point, each differential (infinitesimal) segment of the blade which we will call $dr$ experiences a force,
$$dF_d=\frac{1}{2}C_d\rho (hdr) v^2$$,
where:
$dF_d$ is a small contribution to the total frictional force the blade encounters,
$h$ is the height of the blade (h dr gives the area of that small piece),
$v$ is the velocity around the pivot point at which the small section travels,
and the other variables are the same as you listed.
We must now recognize that $v=r\omega$ where $\omega$ is the angular velocity of the blade. We now see:
$$dF_d=\frac{1}{2}C_d\rho h \omega^2 r^2 dr$$.
Integrate this from from r=0 to r=r,
$$F_d=\frac{1}{6}C_d\rho h \omega^2 r^3$$.
Recalling that we have four blades,
$$F_{d_{total}}=\frac{2}{3}C_d\rho h \omega^2 r^3$$.
Check my integration and work, hope that helps.
Works dimensionally so it should be good.

 

[jmmccain]

Hi gamesguru,

That looks fine, but I would need an integral for torque, not drag. I think I know what happened though. Torque has the radius in it as well:

τ = F_d r = ( 2 π² Θ² C_d ρ A ) r³

In other words, I integrated with the variable, r, "appearing" on both sides of the equation. It seems like this should work, but it has been far too long for me to remember all of the rules. Apparently it doesn't...

It looks like I need to obtain the drag force, as you have done, and then determine a single radial point at which to consider the force to be applied. I'm pretty sure I can handle this approach. If not, in the words of the gov'nuh, "I'll be back"!

Thanks again,
Joseph

 

[gamesguru]

Hi gamesguru,

That looks fine, but I would need an integral for torque, not drag. I think I know what happened though. Torque has the radius in it as well:

τ = F_d r = ( 2 π² Θ² C_d ρ A ) r³

In other words, I integrated with the variable, r, "appearing" on both sides of the equation. It seems like this should work, but it has been far too long for me to remember all of the rules. Apparently it doesn't...

It looks like I need to obtain the drag force, as you have done, and then determine a single radial point at which to consider the force to be applied. I'm pretty sure I can handle this approach. If not, in the words of the gov'nuh, "I'll be back"!

Thanks again,
Joseph

Yea, to get the power of the blade, just use the formula $P=Fv$ and realize that in this case, $v=r\omega$ so,
$$P=F_dv=\frac{2}{3}C_d\rho h \omega^3 r^4$$.
Easy as that, no integration involved. Do not integrate unless both sides have a differential. In situations where you need to carve something up into little pieces, always start off with $\Delta$'s. For instance I would have started off by slicing the blade into little pieces with cross-section $A$ as:
$$\Delta F_d=\frac{1}{2}\rho \Delta A C_d (r\omega)^2$$.
Then just express $\Delta A=h\Delta r$, let everything go to 0 and you have a simple, one-variable integral to do.
Always ask if it makes physical sense, that's the biggest part.
Hope that helps.

 

[jmmccain]

Well, here are results. I've reworked the equation for a single blade, rpm instead of angular velocity, and chord instead of thickness:

P = π³ Θ³ C_d ρ C R⁴ / 162000

Results for one of the more popular models (TREX 450):

P = 3.14^3 3000^3 0.01 1.2 0.032 0.35^4 / 162000 = 29.8 Watts

This model has two rotors (neglecting the flybar and tail) for a calculated total of about 60 Watts. Being aware that the flybar and tail were neglected, this is about one half of a real world value for the 3000 rpm head speed. This may be attributed to the value of the drag coefficient. The value I used is about right for a fixed wing at moderate angles of attack. I fully expect helicopter rotors to have higher values.

I'm now ready to collect some data from fellow modelers in order to narrow the range of values for the drag coefficient.

Thank you very much for the help, gamesguru. My original attempt predicted about 1 Watt for the same rotor. Much improved! Pfft... Just multiply by velocity... /slaps self!

Regards,
Joseph

P.S. Why use chord instead of thickness or projected frontal area? Most drag coefficients for airfoils are provided with the expectation of wing area being used (or so I've read). Here, this would imply the use of chord.

 

[jmmccain]

My apologies for the double post. A thunderstorm interrupted my effort to edit the previous post.

"My original attempt predicted about 1 Watt" -meter!

I'm a little obsessive-compulsive when it comes to writing.

I am now much better prepared to help one or several thousand modelers. Wish me luck!