How do I compute thrust of a rotor disk and the necessary power to drive it??

Question

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.

Answer 1

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 of two rotors, each with: $$ A = 3.14 m^2, \rho = 1.225, T = 981 N $$ the ideal power is:

$$P_{ideal} = N_{rotors} \cdot \left(\frac{(T)^{3/2}}{\sqrt{2 \cdot \rho \cdot A}}\right) = 2 \cdot \left(\frac{(981)^{3/2}}{\sqrt{2 \cdot 1.225 \cdot 3.14}} \right) = 22.2 kW $$

From J. Gordon Leishman, Principles of Helicopter Aerodynamics:

$$C_T = \frac{T}{\rho \cdot A \cdot \Omega^2 \cdot R^2} \tag{CT}$$
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$^2$) = 0.0045. Corresponding Figure of Merit = 0.55. So the power to drive both rotors would be $$22.2 kW / 0.55 = 40.3 kW$$

The rotors 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 * 40,283 = 0.403 kW.

$$Torque = P / \Omega = 40,685 / 238 = 170.9 Nm$$

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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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EDIT2

Clarified that Ideal power and Thrust coefficient are being calculated for two rotors/disks with half the load each. original question asked for 100kg (981N) thrust each rotor, or 200kg total. So Ideal power increases from 7.8kW to 22.2kW. $C_T$ was already calculated for each disk at 981N.

Answer 2

Use this formula:

(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...