DC shunt motor load loss

Question

I have three DC shunt-wound motors in series. Each motor has a coupling to a transmission. I am trying to work out (mathematically, as opposed to intuitively), what happens if the coupling was to drop in one motor (motor 2) resulting in no load on that motor.

Intuitively I know that that motor will speed up and eventually the centrifugal force will result in destruction of that motor if the supply isn't shut off in time.

The supply voltage is 600 V so there is 200 V across each DC motor.

1. When the coupling drops and there is no load, does the current result in excessive torque and increase the DC motor's RPM?
2. As the motor's RPM increases, does the back EMF increase, resulting in a reduction in the current in the motor armature?
3. What is happening to the voltage across the motor? Increasing or decreasing? Is it P=VI, so current decreases due to back EMF, so the voltage across that motor increases? Where does the back EMF come into play mathematically?

Answer 1

There are two major effects here you might be neglecting:

1. Shunt motors self-regulate (when driven by constant voltage). If the load drops, speed increases => back EMF increases => current lowers => torque lowers => speed no longer increases.
2. Current through each motor in the string is the same, so one affects the other.

Effect (2) means effect (1) is compromised a little - motor 2 will start to consume more and more of the string's voltage before it reaches self-regulation.

Thus, intuitively what will happen is motor 2 will start to speed up due to loss of load. Back EMF will increase which will reduce current available to the whole string. Reduction in string current will reduce back EMF in motors 1 and 2 slightly, increasing voltage available. String will find new stable point where voltage is shared and current is same for each motor.

So when each motor was loaded it might look like this:

V1 = 200V
V2 = 200V
V3 = 200V
I = 10A

Then when motor 2 loses its load, the distribution of voltages shifts to something like this:

V1 = 100V
V2 = 400V
V3 = 100V
I = 5A

So motors 1 and 3 have halved in torque and therefore halved in speed (same load), while motor 2 has doubled in speed with half the torque (load dropped by a quarter). Motor 2 is essentially robbing the string of its power by over-spinning.