The electrical part of a PM machine is described as a symmetric three-phase Y-connected circuit with floating neutral point, $e=Ldi/dt+Ri+v$, where $e$, $i$ and $v$ are vectors of emf, current and terminal voltage of three phases, and $L$ is inductor and $R$ is resistor. The mechanical part is described as $J\dot{\omega}=T_m-T_e$, where $J$ is inertia, $\omega$ is the speed, $T_m$ is mechanical torque and $T_e$ is electromagnetic torque. Two parts are coupled by $T_e=<e,i>/\omega$ and $e$ having magnitude $\sqrt{2}k_e\omega$ and frequency $p\omega$ where $k_e$ is emf constant (rms value) and $p$ is machine's pole-pair number.

Now I've been doing a simulation of the machine, where the terminal voltage was set to $v=0$, and a constant mechanical torque $T_m$ was applied for a short period of time $t_1$. The speed $\omega$ will first go up and then go down, and after a long time $t_2$ become 0. The same does the current. From energy's point of view, the energy into the machine is $E_{in}=\int_0^{t_{1}}T_m\omega(t)dt$; after $t_2$, because $\omega=0$ and $i=0$, there is no energy stored in $J$ and $L$, and thus the energy consumed by $R$ should be equal to $E_{in}$. I calculate the energy on $R$ using $E_{out}=3\int_0^{t_{2}}i^2(t)Rdt$, but I found $E_{in}\neq E_{out}$. I'm wondering whether I was using correct formula to calculate $E_{out}$. Thanks for any help.

Ps. The above integrations were calculated numerically.

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    $\begingroup$ If you put up a circuit diagram I can probably answer this. $\endgroup$ – DanielSank Mar 18 '15 at 23:49
  • $\begingroup$ Problem solved. It's an issue of the numeric integration. $\endgroup$ – user3703018 Mar 19 '15 at 11:09

As DanielSank said, I cannot say anything quantitative about the question without a diagram. That being said, your formula, in general, is correct if the topology of the circuit is symmetric. Without any details about how mechanical energy is coupled to the circuit, I can only guess the general reason for the apparent non-conservation of energy.

This circuit having capacitors it will have a certain amount of reactance. Thus, the energy flow in the circuit will not be trivial. Part of the mechanical energy coming in the circuit will be stored in the reactive elements up to a stationnary state that will be reached when the dissipated power in the resistors equate the mechanical power coupled. The presence of reactive power will affect the mechanical part of the generator. Be sure to verify that the model you are using for your generator takes that into account.

  • $\begingroup$ Thanks. I've updated the problem description. Hopefully it's clear even without a diagram. $\endgroup$ – user3703018 Mar 19 '15 at 9:44

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