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as it seems, this question has been asked a few times here so please forgive me for asking again.

If I have 50 BHP shaft power available at 5000 RPM and 106 N.m torque, then can I convert it to 200 kgf of thrust?

I intend to use two sets of horizontal rotors, dividing 50 BHP between them and each generating 100 kgf thrust. Is it doable? What will be the diameter of propellers and their RPM to achieve this goal? What about pitch?

Any help, even if in part or help in the form of guidance to correct resource will be immensely helpful.

Thanks in advance :)

Edit: 1. I intend to provide vertical thrust and not horizontal as we do in aircraft. 2. I want to specifically know following things:-

Suppose I have a generic 2000 mm diameter rotor blade, (I have no idea what typical pitch value is). I want to know what RPM it needs to be turned at to produce 100 Kgf force and what will be the power or torque requirement.

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  • $\begingroup$ You are going to lose some (or a lot) of power in the mechanism you use to divide the power between the props. $\endgroup$ – Ron Beyer Nov 15 '17 at 18:54
  • $\begingroup$ What RPM will you be turning the blades at? If you're thinking 5000RPM, you'll need a very small diameter to keep the tips subsonic. If you're thinking less, then you'll need a gearbox mechanism (with losses) $\endgroup$ – Dan Pichelman Nov 15 '17 at 18:56
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    $\begingroup$ 5000 rpm are 523,6 rad/sec and 523,6 rad/sec x 106 N.m is 55,5 kW, around 75 hp, not 50... $\endgroup$ – xxavier Nov 15 '17 at 19:42
  • $\begingroup$ @xxavier yes, max torque is usually at a lower than max rpm. $\endgroup$ – Koyovis Nov 15 '17 at 22:28
  • $\begingroup$ What is the design speed of the craft? Horizontal rotors or vertical propellers? $\endgroup$ – Koyovis Nov 16 '17 at 1:56
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I'm answering the question of after the Edit. Simple momentum theory:

$$ T = C_T \cdot \rho \cdot A \cdot \Omega^2 \cdot R^2 \tag{Thrust}$$ $$ P = C_P \cdot \rho \cdot A \cdot \Omega^3 \cdot R^3 \tag{Power}$$ $$ C_P = \frac{{C_T}^{3/2}}{\sqrt2} \tag{Ideal Power}$$ $$ FM = \frac{Ideal Power}{InducedPower+ProfilePower} \tag{FigureOfMerit}$$

The Figure Of Merit is a dimensional unit and provides an efficiency measure. It always gives a better result for higher disk loadings, but does provide a correction for real life effects on the ideal power. So for your case, with A = 3.14 m$^2$, $\rho$ = 1.225, T = 981N, the ideal power is:

$$P_{ideal} = 2 \cdot \left(\frac{(T/2)^{3/2}}{\sqrt{2 \cdot \rho \cdot A}}\right) = 2 \cdot \left(\frac{(981/2)^{3/2}}{\sqrt{2 \cdot 1.225 \cdot 3.14}} \right) = 7.8 kW $$

From J. Gordon Leishman, Principles of Helicopter Aerodynamics:

enter image description here

$$C_T = \frac{T}{\rho \cdot A \cdot \Omega^2 \cdot R^2} \tag{CT}$$
Tip speed $V_{tip} = \Omega \cdot R$ should not exceed critical Mach = 0.7 * 340 m/s = 238 m/s. At this tip speed, $C_T$ = 981/(1.225 * 3.14 * 238$^2$) = 0.009 0.0045. Corresponding Figure of Merit = 0.75 0.55. So the power to drive the rotor would be $$7.8 / 0.55 = 14.2 kW$$

The rotor would turn at 238 rad/s = 2,270 rpm, hopefully that is around the rpm for max. torque for your engine. Otherwise you need to gear up/down as required, you have gearboxes anyway.

You would need to add transmission losses for driving the two rotors from your engine. Prouty gives a power loss per stage of 0.0025[max.power + actual power]. Each disk has 2 stages, so a power loss of about 0.005 * 2 * 14.2 = 0.14 kW per disk.

$$Torque = P / \Omega = 14,200 / 238 = 60 Nm$$

EDIT

A numerical error in calculating $C_T$ = 0.0045 not 0.009, so the FM is lower and torque may not be enough. In that case the rotor diameter should be reduced and the rpm increased.

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  • $\begingroup$ This is unbelievable, I can't really thank you enough kind stranger from the internet. Now I am going to study this momentum theory and see for myself how's and why's of its application in this case. Thank you very much again :) $\endgroup$ – Tony Stark Nov 16 '17 at 10:16
  • $\begingroup$ No worries mate. The momentum theory is only the beginning, it gets head spinningly complex later in the books. But this is kind of a quick Order Of Magnitude computation. $\endgroup$ – Koyovis Nov 16 '17 at 10:36
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Use this formula:

enter image description here

(From Wikipedia entry https://en.wikipedia.org/wiki/Disk_loading). Where P is the power in watts, T the thrust in Newtons, A the disk area of the rotor in sq. meters, and rho the air density in Kg/m3, around 1,22...

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    $\begingroup$ I would upvote this answer if it used the equation for a worked out solution. $\endgroup$ – Koyovis Nov 15 '17 at 22:30
  • $\begingroup$ The OP needs only to solve for A in order to know the disk area he needs, assuming 100% prop and transmission efficiency. Then he can split the area in two parts, one for each prop, and get the prop diameter. Very elementary, 12-year old algebra... $\endgroup$ – xxavier Nov 15 '17 at 22:37
  • $\begingroup$ Where is profile drag and induced drag in this equation? The huger the rotor disk, the lower the power required to drive it at a given thrust? $\endgroup$ – Koyovis Nov 15 '17 at 22:41
  • $\begingroup$ It's a simplified approach based on momentum theory. Setting a convenient value for prop efficiency would take things closer to reality. Yes, in general, the larger the rotor disk, the lower the power needed for a given thrust. That's the reason helicopters have big rotors... $\endgroup$ – xxavier Nov 15 '17 at 22:54
  • $\begingroup$ Well if I work this out with T= 981 N, area = pi sq. mtrs, density = 1.225 kg/m^3. Then it works out to Power = 981*((981/3.14)*(1/(2*1.225)))^(1/2) = 11077 Watts, and I have 25 KW available per shaft, this looks quite promising. Assuming props are 80% efficient then it's 13.843 KW, assuming 5% mechanical losses then it's 14.572 KW. This looks magical $\endgroup$ – Tony Stark Nov 16 '17 at 3:25

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