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

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 Kmotor, keep it like this for a time and then open it and close Kloop and keep it closed for another interval of time then open it and close Kmotor and repeat the process. In PSIM (software used for simulating electrical circuits) I get zero current in the loop no matter that Kloop 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:

Ra (armature): Armature winding resistance, in Ohm

La (armature): Armature winding inductance, in H

Rf (field): Field winding resistance, in Ohm

Lf (field): Field winding inductance, in H

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

Vt (rated): Rated terminal voltage, in V

Ia (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:

Vt : terminal voltage

Ea: back emf

Ia: armature current

If: field current

Laf: mutual inductance between the field and the armature windings

The term Laf*If is often defined as kφ 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 Laf is calculated as follows based on the specified rated operating condition:

• You wrote "if I connect the required load to the output of the generator the power drawn from the DC source is too big ". Not my business, but why do you not feed that "required load" from the DC source, maybe through an electronic converter, if DC isn't good as is for the "required load"? Or (more preferably) directly from the primary source which feeds your DC source? In that way you wouldn't waste a single watt to friction nor accelerating masses in the motor and generator
– user33233
Commented Jul 24, 2022 at 18:17
• @user287001 , What I have to test is the generator not the load or the motor. If the endurance tests succeed the generator will be later rotated by a turbine, not an electric motor like now, and produce electricity that will be delivered to the required load. Commented Jul 24, 2022 at 19:47
• Oooooo, wow. Seems like you are trying to get something for nothing, Answer: No! Commented Jul 24, 2022 at 20:51
• @StainlessSteelRat , Have you read this: "I do not expect to run a loop with 100% efficiency but even a 75% efficiency will save me a lot of money" at the end of my question? Or, you imply that I have to pay some money in order to have an useful answer? Commented Jul 24, 2022 at 21:22
• @StainlessSteelRat, you have missed the point of the question. I took part in a similar exercise while at university and described it in the linked answer below. In my case we used a pair of synchronous machines. One was the motor and the other the generator. The generator generated back into the building three-phase supply that was powering the motor. The generator applies a real load to the motor but the power is not wasted as it is fed back to the motor to offset the demand from the grid. As the OP knows, this will allow full power testing of the equipment at a fraction of the energy cost. Commented Jul 25, 2022 at 8:20

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 Kloop falls to zero. At that point the generator voltage matches the motor voltage.
• Close Kloop.
• 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.

• In PSIM I can try your proposed steps. However, the real generator has a stator with permanent magnets so there is no way to modify the excitation of the stator. Commented Jul 24, 2022 at 19:58
• I think you need to add that information into the question along with other motor and generator details. Commented Jul 24, 2022 at 21:31

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

Eg =N*Kg

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

Respectively for the internal induced voltage in the motor we can write Em=N*Km

The current when the loop switch conducts and the battery is disconnected is (Eg -Em)/R where R is the sum of the generator, motor and wiring resistances. It can also be written as I = N(Kg - Km)/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(Kg -Km))2/R

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

or as well T = N*((Kg-Km)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. Km can be = Kg if the motor or the generator has adjustable magnetization.

If it happens that you succeed to set Kg=Km 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 Kg=Km 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 3rd 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.

• I will try in PSIM your schematic. However, is there a way to eliminate that battery from the circuit? Commented Jul 27, 2022 at 2:07
• A big enough capacitor will do the trick. Requires careful calculations based on the same math which is used in designing switch mode power supplies. Nothing trivial here! Start the design iterations by having an ideal constant voltage source in the place of the battery to find how S1 and S2 should be controlled so that a capacitor could do the job and to decide what inductors you need.
– user33233
Commented Jul 27, 2022 at 7:56