I wanted to calculate the joule heating while the motor is in motion and putting out mechanical watts (before calculating iron loss). Are my calculations correct?
If I have a 90kv BLDC hub motor which is 0.1ohm lead to lead, a 50v battery, ESC with 20a motor amp limit, 20a battery amp limit, presently 2000rpm and full throttle
90kv 0.1ohm 50v battery 20a motor limit 20a battery limit 2000rpm
1/90kv = bemf v per rpm = 0.01111111 v
2000rpm * 0.01111111v = 22.2222222v bemf
(AB)/C=D
A=B+(C*D)
A = pwm effective v
B = bemf v
C = ohms winding resistance
D = motor current
A=B+(C*D)
A=22.2222222v bemf+(0.1ohm *20a motor amps)
A=24.2222222v pwm effective v
24.2222222v pwm effective v * 20a motor current = 484.444444w electrical watts
484.444444w electrical watts / 50v battery = 9.68888888a battery amps
(24.2222222v pwm effective / 50v battery) * 100 = 48.44 % duty cycle
battery amps = motor amps * duty cycle %
9.68888888a battery amps = 20a motor amps * (48.44 [duty cycle %] / 100)
Kt (torque per amp) = 60/(2*pi*KV)
KT = 0.10610339
20a motor amps * 0.10610339kt torque per amp = 2.122 newton meters torque
(2000rpm*2*pi)/60=209.43951 radians per second
2.122 newton meters torque * 209.43951 radians per second = 444.4306w mechanical
24.2222222v pwm effective v * 20a motor current = 484.444444w electrical watts
(444.4306w mechanical / 484.44444w electrical ) * 100 = 91.7402% electrical to mechanical conversion efficiency
I^2R=W
(20a motor amps * motor amps 20a)*0.1ohm = 40w joule heating
444.4306w mechanical + 40w joule heating = 484.4306w
9.68888a battery amps * 50v = 484.44w electrical
Simply I calculate that with 20a motor amps at 2000rpm with the 90kv 0.1ohm motor— of the 484.44 electrical watts drawn from the battery, 40w is converted to joule heating in the motor, while 444.4306w is converted into rotational mechanical watts.
Additionally I observe the battery draw is 9.6888a battery amps @ 50v, motor current is 20a, back emf is 22.22222v, pwm effective v is 24.22222v. Electrical to mechanical conversion efficiency is 91.74% which is numerically the same as the ratio of BEMF to PWM effective V —(22.22222 / 24.22222)*100=91.74 — interestingly before calculating core losses the ratios of mechanical to electrical watts and back emf v to effective v are always the same. correct?
Calculating motor heating?

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Re: Calculating motor heating?
Also, if I have an electric skateboard with a 50v battery, and a 90kv (90 max rpm per volt noload) hub motor which is 0.1 ohm leadtolead, and a reprogrammable VESC electronic speed controller....
& in the reprogrammable settings for the electronic speed controller, my friend chooses the following settings:
Battery Amp Limit: 20a
Motor Amp Limit: 20a
If I stall the motor and use full throttle, how many watts of heating will the motor experience (1000w or 40w)? How many amps will be drawn from the battery? What is the duty cycle? Torque at stall?
Watts heating:
I^2R=W
20a^2*0.1ohm=40w
Battery Amps:
40w/50v=0.8a battery amps
Duty Cycle:
40w / 20a motor amps = 2v effective pwm volts
(2v pwm effective / 50v battery) * 100 = 4% duty cycle
correct so far?
&
battery amps = motor amps * duty cycle %
0.8a [battery amps] = 20a [motor amps] * (4 [duty cycle %] / 100)
Torque at stall:
Kt (torque per amp) = 60/(2*pi*KV)
KT = 0.10610339
20a motor amps * 0.10610339kt torque per amp = 2.122 newton meters torque
correct?
& in the reprogrammable settings for the electronic speed controller, my friend chooses the following settings:
Battery Amp Limit: 20a
Motor Amp Limit: 20a
If I stall the motor and use full throttle, how many watts of heating will the motor experience (1000w or 40w)? How many amps will be drawn from the battery? What is the duty cycle? Torque at stall?
Watts heating:
I^2R=W
20a^2*0.1ohm=40w
Battery Amps:
40w/50v=0.8a battery amps
Duty Cycle:
40w / 20a motor amps = 2v effective pwm volts
(2v pwm effective / 50v battery) * 100 = 4% duty cycle
correct so far?
&
battery amps = motor amps * duty cycle %
0.8a [battery amps] = 20a [motor amps] * (4 [duty cycle %] / 100)
Torque at stall:
Kt (torque per amp) = 60/(2*pi*KV)
KT = 0.10610339
20a motor amps * 0.10610339kt torque per amp = 2.122 newton meters torque
correct?

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Re: Calculating motor heating?
In your calculation you don't speak about cooling.
1. Cooling by the metallic mass of the motor.
2. Cooling by the speed that aircool your hub.
3. Cooling if your motor use a fan (drone, skate,...) motors.
Cooling is very important and you have to take it into account, but often it's very difficult to evaluate.
Some motor manufacturers like Maxon give you thermal constants of their motors.
1. Thermal time constant winding
2. Thermal time constant motor
They also give you:
3. Thermal resistance housingambient
4. Thermal resistance windinghousing
5. Max. winding temperature
Hope it helps.
Thierry
1. Cooling by the metallic mass of the motor.
2. Cooling by the speed that aircool your hub.
3. Cooling if your motor use a fan (drone, skate,...) motors.
Cooling is very important and you have to take it into account, but often it's very difficult to evaluate.
Some motor manufacturers like Maxon give you thermal constants of their motors.
1. Thermal time constant winding
2. Thermal time constant motor
They also give you:
3. Thermal resistance housingambient
4. Thermal resistance windinghousing
5. Max. winding temperature
Hope it helps.
Thierry

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Re: Calculating motor heating?
thanks. any tips on calculating or estimating iron/core loss? switching loss? what are the factors I should use to guesstimate?
once i have an estimate for core loss and switching loss, is that subtracted from the mechanical watts or added to the electrical watts/battery amps, etc?
do my above calculations all seem correct before factoring cooling, iron/core loss and switching loss?
once i have an estimate for core loss and switching loss, is that subtracted from the mechanical watts or added to the electrical watts/battery amps, etc?
do my above calculations all seem correct before factoring cooling, iron/core loss and switching loss?

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Re: Calculating motor heating?
I think your calculation is correct: Rxi² = 0.1 * 20² = 40W joule losses.
As for the iron/friction losses, you can estimate them with a constant drag torque of 12% of the max motor torque, depending on motor quality obviously.
You could also measure your motor noload power consumption vs RPM, and assume these losses are the same when the motor drives a load.
As for the iron/friction losses, you can estimate them with a constant drag torque of 12% of the max motor torque, depending on motor quality obviously.
You could also measure your motor noload power consumption vs RPM, and assume these losses are the same when the motor drives a load.
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