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 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 m 2 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:

$\begingroup$ 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 $\endgroup$

$\begingroup$ @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. $\endgroup$

$\begingroup$ Oooooo, wow. Seems like you are trying to get something for nothing, Answer: No! $\endgroup$

$\begingroup$ @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? $\endgroup$

$\begingroup$ @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. $\endgroup$