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Stored energy Induction motor

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I'm not sure the following calculations and assumptions are sound, please comment if you see a mistake or if you think it's a close enough approximation?

1/2HP single phase induction motor:
Inom =6A  (measured)
Vnom = 120V
LRA = 30A
R = 120/30 = 4 Ohms
Z = 120/6 = 20 Ohms
Xl = 16 Ohms
L = 16/377 = 0.04 H, round up to 0.05
Energy = L * I^2 = 0.05 * 30^2 = 45 J

Terry Bites:
The average electrical energy stored in the magnetic circuit(s) is zero.
Rotational kinetic energy is stored in the rotor. If you know the speed and rotor dimensions you can calculate it. labman.phys.utk.edu/phys135core/modules/m8/energy.html
It will be released as the motor decelerates.

If you cut the power to a small induction motor it will typically coast to a complete stop within a second or two, there's not much energy stored there.

The mechanical load i.e. compressor can kick back and, combined with rotor (decay) remanence does generate a huge impulse. It also depends if the motor has a run cap.
I've had small motors dish out 4kV impulses and rip apart contactors and arc across connectors. So a mathematical approach is interesting but just measure it on the bench, when tuning a snubber.

Your locked rotor amps are approximately half inductance and half resistance, but not always.

So you need to calculate 4 ohms of impedance at locked rotor conditions from the dc resistance of the windings, lets presume its 3 ohms. this leaves about 2.6 ohms for the inductance, yielding only 7mH, which makes only 0.12 joules at 6 amps, or 3.1 joules at locked rotor conditions.

the inductance and dc resistance add at 90 degree angles, so 3 squared plus 2.6 squared =16, = 4 ohms impedance.

Now lets presume you actually measured 16 ohms of inductive impedance at small signal voltages. that would be .9 joules at 6 amps. then factor in the motor is running at saturated conditions at nominal voltage and I bet you will find the energy stored in the inductance during operation to be less than half of .9 joules.


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