Electrical – Can a BLDC generator power another identical brushless motor

Driving a motor blind is a bad idea for several reasons:

1. It is inefficient. The most efficient way to run the motor is for the magentic field to be 90° ahead of the rotor. Put another way, the torque on the rotor is the cross product of the driving magnetic field and the rotor's magnetic orientation.

   With position feedback, the magnetic field can be kept near the optimum angle, which means the current goes to actually pushing the motor instead of holding it in place. Put another way, the amplitude is just what it needs to be to keep the motor spinning at the desired speed in the maximum torque configuration. When you don't know where the rotor is, you end up over-driving the motor.

   Another way to look at this is that the driving field has a sine and cosine component. Let's say the cosine is the part 90° ahead of the rotor and the sine part is where the rotor currently is. Any phase angle can be thought of as just a different mix of the sine and cosine components. However, only the cosine component moves the motor. The sine component only causes heating and represents wasted power.

2. Once you loose lock, the game is over. With a fixed drive, the drive angle will only be a little ahead of the rotor at low torque. As the speed increases (and the effective drive voltage automatically decreases due to back EMF) or the load increases, the fixed drive will get up to 90° ahead of the rotor.

   However, at this point it's right on the edge and any change will cause less torque. If the load on the motor increases, the rotor will get more than 90° behind, which causes less torque, which causes it to get even more behind. Over the next 1/4 turn of slip, the forward torque will decrease to zero. Then for the next 1/2 turn after that, the driving torque actually pushes the rotor backwards.

   At this point you're totally screwed. Remember you got into this predicament in the first place because the driving torque couldn't keep up with load, and you've just experienced a net negative drive over the last 3/4 turn. If the load is suddenly removed and if you're very lucky, the rotor might be able to speed up to sync up with the drive in the next 1/4 cycle, but certainly not if whatever condition caused the problem in the first place is still present.

   Once the rotor gets out of sync, the net torque over any one rotation is 0. The product of two sine waves of different frequency always averages to 0 regardless of the phase angle between them.

The load angle

Your BLDC behaves like a synchronous machine, the only difference is that you give it block voltages instead of sinusoidal voltages as input. That means that the principles of the synchronous machine can be used on your machine.

The load angle is a measure (as the name says) of the loading of the machine. As you can see from the figure below it is the relative displacement of the rotor in regard to the field axis.

The second figure shows you how the torque(or power) of the machine depends on the load angle. You can see that there is a maximum at \$90^\circ\$. After this points the flux lines tear apart, and the torque reduces. That means if you pass this point your motor will loose torque and slow down.

The simplified equation describing this behavior is: $$ T=\frac{3UE}{X_s\omega}\sin\delta $$

For references and a short explanation look this link.

EDIT: In regard to the comment

The synchronicity is irrelevant, that is true. The thing that is worth thinking about is the stability. Let's do a little calculation, the induced voltage is divided between the short circuit resistance and the reactance of the machine(we will neglect \$R_{SC}\$ as needed):

\begin{equation} \underline{E}={{jX_s}\underline{I}+R_{SC}\underline{I}}\nonumber\\ E\angle\delta={{jX_s}I\angle\varphi+R_{SC}I\angle\varphi}\\ I\angle\varphi=\frac{E\angle\delta}{{jX_s}+R_{SC}}; R_{SC}<<X_s\\ I\angle\varphi=\frac{E\angle(\delta-90^\circ)}{{X_s}}\\ I\angle-\varphi=\frac{E\angle(\delta-90^\circ-2\varphi)}{{X_s}} \end{equation}

Let's multiply by the current to get the losses (multiply by\$\underline{I}^*\$): $$ IE\angle(\delta-\varphi)={{jX_s}I^2+R_{SC}I^2} $$

The losses are the last term \$P_{loss}=R_{SC}I^2\$, we can equalize them with the real part of the rest of the equation: $$ P_{loss}=\mathfrak{R}\{IE\angle(\delta-\varphi)-{jX_s}I^2\}\\ P_{loss}=\mathfrak{R}\{\frac{E\angle(\delta)E\angle(\delta-90^\circ-2\varphi)}{{X_s}}\}\\ P_{loss}=\frac{E^2\cos(2\delta-2\phi-90)}{{X_s}} $$

These are the losses, not really comparable to the previos torque equation. Let's look what happens to the load angle in the meantime: $$ 1\angle\delta=\frac{{jX_s}I\angle\varphi+R_{SC}I\angle\varphi}{E}; R_{SC}<<X_s\\ 1\angle\delta=\frac{{X_s}I\angle(\varphi+90)}{E}\\ \delta=\varphi+90 $$ For an ideal short, it will be unstable as you can see from the last equation. Put some resistors instead of a short, it should work better. Capacitors would help too, just calculate it first with the given equations. Watch out for three phases, this is the one phase calculation!