Silicon Productions

Motor Power, Torque and synchronous speed analysis

__Analysis by Warren Winovich__Text enclosed within [ ] braces has been added by Mike (including graphs)

[ In this documant we find an equation that relates Torque to motor speed using specifications of the controller and motor. From this equation we calculate :

Motor speed at peak power (S

_{pp}= 4683 rpm)

Torque at peak power (T

_{pp}= 45 pound feet, 61 Nm)

Starting Torque (T

_{s}= 54 pound feet, 73 Nm)

and Maximum expected speed of the vehicle in mph (54.42 mph, 87 km/h)

]

**Torque/ Speed Relation: AC31-01 Motor, Curtis 1238 Inverter**

__Appendix 2.__Modern inverter/controllers can maintain the ratio of applied voltage to frequency constant for motor speeds up to one-third of the synchronous speed. When the ratio cannot be maintained constant, motor torque decreases as the frequency increases, up to the operating point. At some motor speed, above one-third of the synchronous speed, the condition of peak power occurs. As motor speed exceeds the peak-power point, power decreases until the steady-speed operating point is established for the applied load. Motors are selected, for powering electric cars, to havce ratings considerably greater than that needed for steady-speed operation at the top speed of the vehicle. At the steady-speed operation, rotor slip is typically 2% to 4% for induction motors used to power electric cars.

The torque/speed relation that approximates that of the modern inverter/controller installed in today's electric cars can be represented by a simple equation.

T/T

_{s}= 1 - (S/N)

^{n}

The value of the exponent determines how rapidly the torque decreases with motor speed. As noted above, the torque is maintained constant for motor speeds up to one third of the synchronous speed. The value of the exponent, n, that maintains torque within 0.995 of the starting torque for a motor speed ratio of one third is n = 5. [This can be found graphically by plotting the equation above for various values of n

]

It follows that an approximation of the torque/speed relation for the motor/control system is:

T/T

_{s}= 1 - (S/N)

^{5}

[ T is the torque at a given motor speed (S) in rpm

T

_{s}is the starting torque of the motor

N is the synchronous speed of the motor in rpm

]

The rating of the AC31-01 induction motor is specified as :

9kW @ 6500 rpm, continuous; 9.75 poundfeet @ 6500 rpm

30kW @ peak power; motor speed unspecified

Motor speed at peak power can be found by using the torque/speed equation shown above. Power is given by the basic relation:

P = T x S/7040, kW

[ The 7040 conversion factor comes from :

Power (in kW) = Torque (in Nm) x 2pi x rpm / 60000

Torque : 1 foot pound = 1.356 Nm

Hence, if T is in foot pounds then

Power (in kW) = Torque (in ft pounds) x 1.356 x 2pi x rpm / 60000

Power (in kW) = Torque (in ft pounds) x rpm / 7040

]

Substitute torque given by the simple torque/speed equation and differentiate the resulting power expression and setting the result to zero gives the motor speed ratio for peak power. That is, for the condition of peak power:

(S

_{pp}/N) = (1/6)

^{1/5}= 0.699

[ steps to get to above result...

T/T

_{s}= 1 - (S/N)

^{5}

T = T

_{s}- T

_{s}x (S/N)

^{5}

P = T x S/7040

P = T

_{s}x S/7040 - T

_{s}x (S/N)

^{5}x S/7040

P = T

_{s}x S/7040 - T

_{s}/7040 x S

^{6}/N

^{5}

dP/dS = T

_{s}/7040 - T

_{s}/7040 x 6 x S

^{5}/N

^{5}

set dP/dS = 0, meaning max power at a given S called S

_{pp}I.E. S

_{peakpower}

T

_{s}/7040 = T

_{s}/7040 x 6 x S

_{pp}

^{5}/N

^{5}

1 = 6 x S

_{pp}

^{5}/N

^{5}

(S

_{pp}/N) = (1/6)

^{1/5}

]

Synchronous motor speed (i.e. a rotational speed of the magnetic field) is determined by taking the slip at the motor rated condition as 3%. This gives the synchronous speed as N = 6700 rpm. [this is 6500 + 3% of 6500 approx ]

It follows that the motor speed at peak power is:

S

_{pp}= 0.699 x 6700 = 4683 rpm

[ This means that peak power will occur at approx 47mph given the fixed gearing on the car]

[ Here is a plot of P = T

_{s}x S/7040 - T

_{s}/7040 x S

^{6}/N

^{5}where T

_{s}has been chosen to make the peek power occur at 30kW. T

_{s}= 54 pound feet

]

Torque at peak power follows directly from the motor equation:

T

_{pp}= 45 pound-feet

[ this comes from P = T x S/7040 where P = 30kW (see specs) and S = 4683 rpm

T = 30 x 7040 / 4683 = 45 ]

The starting torque, then, is found from the torque/speed relation:

T

_{pp}/T

_{s}= 1 - (0.699)

^{5}= 0.833;

and,

T

_{s}= 45/0.833 = 54 pound-feet

The appropriate value for starting torque of the AC31-01 induction motor in conjunction with the Curtis 1238 inverter/controller is 54 pound-feet.

Top speed for this system is found for 5% slip. Motor speed: 6365rpm. Top speed is found as 54.42 mph (= (6365/8)x(0.958)/(14.0056)).

[ the 6365 is (6700 - (5% of 6700), the 8 is the motor to wheel gear ratio, 0.958 is the radius of the wheel in feet. 14.0056 converts rpm (where the wheel radius is in feet) into mph (2*pi/60 x 3600/5280) where 5280 is number of feet in a mile, 3600 is seconds in an hour]

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