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I've been attempting to calculate how much current a BLDC will draw given a known propeller size, thrust, internal resistance, Kv rating, Supply Voltage(Vin) and idle current. My propeller parameters are as follows;

Diameter=34;                               %Propeller diameter(In)
Pitch=11;                                  %Propeller pitch(In)
Pitch_m=Pitch*0.0254;                      %Propeller pitch(M)
Radius_m=0.0254*(Diameter/2);              %radius in meters
Area=pi*(Radius_m^2);                      %Disc Area

The information I have about the motors is it's idle current, internal winding resistance and its Kv rating.

%Motor Characteristics (Defined)
Io=1.30;                                       %No load current(I)
Rm=0.21;                                      %Internal Resistance(R)
Kv=100;                                        %Revolutions per Volt

I'm calculating the fluid speed using some basic thrust equations I grabbed off of a forum.

%Thrust equations
rho=1.225;
Pconst=1.15;
Thrust=130.9;                           %Reference thrust from experimental results
V=sqrt(Thrust/0.5/rho/Area);            %speed of fluid moved by propeller
P_aero=Thrust*V;                        %aerodynamic power
Pmec=P_aero*Pconst;                     %mechanical power
RPM=V/(Pitch_m/60);                     %revolutions per second

Finally, I am calculating the motor's current as follows (not everything below is used)

%Motor Equations
Ke=1/Kv;
Kt=Ke;                                  %Torque constant equals back emf constant
Vin=40;                    %Input Voltage(V)
a=-Rm/Pmec
b=(Vin+(Rm*Io))/Pmec
c=-1*((Vin*Io)/Pmec)
p= [a b c];
Iin= roots(p);
Vloss=Iin*Rm;

To explain the a, b , c and p I came to that by expanding out the following and finding the roots.

Pmec = (Vin - Rm * Iin) * (Iin - Io)

When I run this in MATLAB, the values I get are;

Mechanical power:     2874.5 Watts
Aeronautiical power:  25003 Watts
Vin:                  41V
Iin:                  174A

I get 174A for Iin, which is a colossal current value. On the manufacturer's website, the current draw for this motor when it's producing 13392 grams of thrust is 52 amps at 2092 Watts of power given a 40 Volt Supply.

Motor is a Tmotor U12KV100. The equations are from the forum linked below

https://www.rcgroups.com/forums/showthread.php?587549-Motor-formulas

edit: There was an order of magnitude error in the script. I have since corrected the error however the current consumption is still unrealistic and well off the experimental data provided by the manufacturer.

edit2: clarified some confusion on Vin. I was calculating how much voltage required even though I knew the supply voltage 40.

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  • \$\begingroup\$ Where did your formula(s) originate from? \$\endgroup\$
    – winny
    Commented Aug 21, 2017 at 11:45
  • \$\begingroup\$ From this forum post. rcgroups.com/forums/showthread.php?587549-Motor-formulas \$\endgroup\$
    – Ozymandias
    Commented Aug 21, 2017 at 11:45
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    \$\begingroup\$ Post that information in your question. \$\endgroup\$
    – winny
    Commented Aug 21, 2017 at 11:48
  • \$\begingroup\$ Hmm 41V/0.21 = 195 A stall/start current not 1950 A \$\endgroup\$ Commented Aug 21, 2017 at 14:31
  • \$\begingroup\$ I've made the correction. I was using 0.021 by accident. \$\endgroup\$
    – Ozymandias
    Commented Aug 21, 2017 at 14:34

1 Answer 1

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I think you have to go back to the physics instead of treating Matlab as some magic black box that spits out the answers.

For a given (say 40V) supply you can work out how the current varies as RPM, ((via the speed constant to give the back EMF, and the resistance to give the current) and hence both electrical power and mechanical power (= electric power minus loss in that resistance).

Your specific mistake is to calculate Iin as Vin/R. This is the current at speed=0, or the stall current. The motor will take this instantaneously on starting if the full voltage is applied. However as the propellor spins up, the motor generates back EMF (1V for every 100rpm) reducing the voltage across the winding resistance, and therefore reducing the current draw.

Your job is to find the point where the speed is high enough that there is only enough current to supply enough power to push the air, overcome mechanical losses (propellor efficiency) and generate waste heat in the winding resistance.

Now you also need to work from the mechanical side, to find how mechanical power required, thrust, and air velocity vary with propellor speed, given diameter (hence swept area) and pitch (hence nominal airspeed at a given RPM.

This is easiest if you convert everything into SI, so RPM into radians/second etc.

Airspeed * density * swept area gives you mass flow rate (mass/second),

0.5 * mass flow * airspeed^2 gives you power (via the usual formula for kinetic energy) - that's power (Pout) imparted to the air.

Then there's a fudge factor for propellor efficiency (Pconst=1.15) = Pmech is Pconst * Pout)

Put these two curves together, and you should find one speed where both expressions yield the same mechanical power. That will be the working speed of the motor.

Now you know the current at that speed from the first curve.

Also work out the thrust at that speed, from Pout/airspeed. Hopefully it'll match the assumption you originally made, of 130N.

I'm not spelling it out in full detail because I think you'll learn better if you have to think about the physics involved.

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  • \$\begingroup\$ That is what I'm doing at the moment. Pmec is the mechanical power that is produced at 130 newtons of thrust. I'm then using the mechanical power formula to find out how much current is needed to produce this much mechanical power and the only way I can think of at the moment is to expand the equation and find it's roots. \$\endgroup\$
    – Ozymandias
    Commented Aug 21, 2017 at 13:11
  • \$\begingroup\$ So, at what speed do you get that thrust? \$\endgroup\$
    – user16324
    Commented Aug 21, 2017 at 13:54
  • \$\begingroup\$ If you mean RPM, I get it at 4109RPM. I get this by re-arranging the following formula, V = n x pitch x (0.0254 / 60). About your previous post, are you suggesting I use Iin=(Vs-Ve)/Rm. Because this gives me the answer of 174 which is one of the solutions to the polynomial I solve above. To clarify, I'm only looking to solve the current for this particular thrust. \$\endgroup\$
    – Ozymandias
    Commented Aug 21, 2017 at 14:02
  • \$\begingroup\$ You'll have to explain further, there is no Ve above ... what is it and how did you compute it? Vin=40V, 4109/Kv=41.09V so this seems to be a windmill, generating power at the moment? \$\endgroup\$
    – user16324
    Commented Aug 21, 2017 at 14:46
  • \$\begingroup\$ Just to clarify, I know what Vin is going to be (40V). I was using Ve before and I wasn't getting a good number. I've made some further edits to the question to clarify any confusion. \$\endgroup\$
    – Ozymandias
    Commented Aug 21, 2017 at 15:00

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