Can an electric motor act as a load for the electric generator rotated by that motor?

Question

I have an experimental generator and an off-the-shelf motor that I intend to arrange like in Fig. 1. The problem is that if I connect the required load to the output of the generator the power drawn from the DC source is too big and the electricity bill too high. I cannot afford it.

I intend to connect the motor as a load for the generator to run this generator much longer and pay just the losses in the motor and generator not the power dissipated in the load.

I tried to close K_motor, keep it like this for a time and then open it and close K_loop and keep it closed for another interval of time then open it and close K_motor and repeat the process. In PSIM (software used for simulating electrical circuits) I get zero current in the loop no matter that K_loop is closed or opened (some large ideal spikes appears but they disappear soon). Despite this zero current in the loop, both motor and generator quickly reach the nominal speed and voltage and stay like this.

I also added a flywheel with a big moment of inertia on the shaft of the generator starting from the idea that when the motor is not powered at all the only source of energy for the generator is the mechanical energy stored in the spinning rotors of the motor and generator and the flywheel. I got no positive results. The current in the loop continues to be zero.

I do not expect to run a loop with 100% efficiency but even a 75% efficiency will save me a lot of money.

Fig. 1. Generator powering the motor that rotates the generator.

UPDATE:

After asking the question and getting some answers I decided to use as motor and generator a predefined DC machine available in PSIM that has the following characteristics (unfortunately I get the same zero current in the loop):

and this symbol:

where:

R_a (armature): Armature winding resistance, in Ohm

L_a (armature): Armature winding inductance, in H

R_f (field): Field winding resistance, in Ohm

L_f (field): Field winding inductance, in H

Moment of Inertia: Moment of inertia J, in kg x m²

V_t (rated): Rated terminal voltage, in V

I_a (rated): Rated armature current, in A

N (rated): Rated mechanical speed, in RPM

If (rated): Rated field current, in A

This DC machine is described by the following equations:

where:

V_t : terminal voltage

E_a: back emf

I_a: armature current

I_f: field current

W_m: mechanical speed, in rad./sec.

L_af: mutual inductance between the field and the armature windings

The term L_af*I_f is often defined as k_φ in many text books. Note that the relationship between the flux phi and the field current If is assumed to be linear. Magnetic saturation is not considered. The mutual inductance L_af is calculated as follows based on the specified rated operating condition:

Answer 1

I think the procedure you want is:

• Start the motor. The only load will be the losses in the motor and the unloaded resistance of the generator.
• Adjust the generator excitation so that the voltage measured across K_loop falls to zero. At that point the generator voltage matches the motor voltage.
• Close K_loop.
• You can now increase the generator excitation a little and current will flow from it around the loop. The more you do that, the higher the load on the motor but, as you hoped, the supply will only be making up for the losses.

All this assumes a simple DC system. AC requires phase synchronisation. I've written about that in https://electronics.stackexchange.com/a/576412/7315.

Answer 2

You have planned an electric braking system. If we assume a simple linear DC machine model to be valid for the generator and for the motor we can write for the generator

E_g =N*K_g

where N is the rotation speed (=angular velocity), K_g is the induction constant (depends on magnetization) and E_g = induced voltage.

Respectively for the internal induced voltage in the motor we can write E_m=N*K_m

The current when the loop switch conducts and the battery is disconnected is (E_g -Em)/R where R is the sum of the generator, motor and wiring resistances. It can also be written as I = N(K_g - K_m)/R. The energy is dissipated by friction and the loss in R. If we ignore the friction, the dissipation power is P = (I²)R = (N(K_g -K_m))²/R

The braking torque T = P/N (assuming SI-units used, for ex. N is radians per second)

or as well T = N*((K_g-K_m)²)/R

This is actually the differential equation of exponential decaying. The rotation speed will decay exponentially towards zero except in case the induction factors happen to be equal. In that case only mechanical friction dissipates the rotation energy, so in real system the speed decays anyway. K_m can be = K_g if the motor or the generator has adjustable magnetization.

If it happens that you succeed to set K_g=K_m there's no current. Then you also do not need any loop wires nor the switch. Remove the loop wires and you do not have resistive braking, no matter is K_g=K_m or not. In that way the rotational energy dissipation rate is the slowest possible.

If we forget your not so useful circuit diagram and try to think how to reuse in the test the energy which is in the test produced by the generator we see that

• the generator should see the wanted test load
• the motor should run the generator with the nominal rotation speed
• an electronic switch mode converter with properly designed controller is needed to behave for the generator as the wanted load and to convert the generator output with low losses to a charge of the battery which runs the test. The battery is maintained by another switch mode charger branch and there's the 3^rd switch mode branch which feeds the motor from the battery so that the rotational speed is the wanted.

There should be no direct way how the generator output affects immediately to the motor input voltage like in your loop. Instead of it you set what's inputted to the motor and what load the generator sees.

If you happen to have some incredible luck all voltages (battery, induced ones in motor and generator, the available charge maintaining voltage) fit together without other inductive parts than the ones in the machines at the wanted rotation speed and generator load. In that case the auxiliary switching circuit could be simple. I do not believe anyone is so lucky, you'll need probably something like this:

Only the high frequency switching circuit is drawn but the actual controller for switches S1,S2 and S3 is left out. The controlling goals for the switches:

S1: the wanted rotation speed

S2: the wanted load current for the generator

S3: the battery charge is maintained

Designing the switching controller is a way more tricky than creating a single output switch mode power supply. I guess the design is still manageable even for an advanced hobbyist because the battery suppresses substantially the interactions of the different switching processes.

A practical battery with long wires may have too high inductance. That causes voltage ripple, so a capacitor must be added in parallel with it. In theory the whole battery could be replaced by a big enough capacitor.