ECE PhD QE PE1 2011 Problem2 - Rhea

Problem 2

Coenergy Calculation

Coenergy will be calculated in steps using a sequential ramping process. Before any steps can be completed, the contribution to coenergy of fixing the mechanical system should be documented (zero unless the mechanical system can store energy in its position).

$ \begin{equation} W_{c,0} = 0 \end{equation} $

The first step in the coenergy calculation will ramp dummy variable $ i_{as}' $ from $ 0 $ to its final value of $ i_{as} $ while dummy variable $ i_{bs}' = 0 $ is held at its initial value.

$ \begin{align} W_{c,1} &= \left.\int_{i_{as}' = 0}^{i_{as}} \lambda_{as}\left(i_{as}', i_{bs}', \theta_{rm}\right) \, di_{as}'\right|_{i_{bs}' = 0} + \left.\int_{i_{as}' = 0}^{i_{as}} \lambda_{bs}\left(i_{as}', i_{bs}', \theta_{rm}\right) \, di_{as}'\right|_{i_{bs}' = 0} \\ W_{c,1} &= \int_{i_{as}' = 0}^{i_{as}} \left[5i_{as}' - 3\cos(2\theta_{rm}) \left(i_{as}' + \cancelto{0}{i_{bs}'}\right)^\frac{1}{2} + 2\cos(\theta_{rm})\right] \, di_{as}' + \int_{i_{as}' = 0}^{i_{as}} \left[5\cancelto{0}{i_{bs}'} - 3\cos(2\theta_{rm}) \left(i_{as}' + \cancelto{0}{i_{bs}'}\right)^\frac{1}{2} + 2\sin(\theta_{rm})\right] \, di_{as}' \\ W_{c,1} &= \frac{5}{2} i_{as}^2 - 3 \left(\frac{3}{2}\right)^{-1} \cos(2\theta_{rm}) i_{as}^\frac{3}{2} + 2\cos(\theta_{rm}) i_{as} - 3 \left(\frac{3}{2}\right)^{-1} \cos(2\theta_{rm}) i_{as}^\frac{3}{2} + 2\sin(\theta_{rm}) i_{as} \\ W_{c,1} &= \frac{5}{2} i_{as}^2 - 4 \cos(2\theta_{rm}) i_{as}^\frac{3}{2} + 2\left[\cos(\theta_{rm}) + \sin(\theta_{rm})\right] i_{as} \end{align} $

The second step in the coenergy calculation will ramp dummy variable $ i_{bs}' $ from $ 0 $ to its final value of $ i_{bs} $ while dummy variable $ i_{as}' = i_{as} $ is held at its final value.

$ \begin{align} W_{c,2} &= \left.\int_{i_{bs}' = 0}^{i_{bs}} \lambda_{as}\left(i_{as}', i_{bs}', \theta_{rm}\right) \, di_{bs}'\right|_{i_{as}' = i_{as}} + \left.\int_{i_{bs}' = 0}^{i_{bs}} \lambda_{bs}\left(i_{as}', i_{bs}', \theta_{rm}\right) \, di_{bs}'\right|_{i_{as}' = i_{as}} \\ W_{c,2} &= \int_{i_{bs}' = 0}^{i_{bs}} \left[5i_{as} - 3\cos(2\theta_{rm}) \left(i_{as} + i_{bs}'\right)^\frac{1}{2} + 2\cos(\theta_{rm})\right] \, di_{bs}' + \int_{i_{bs}' = 0}^{i_{bs}} \left[5i_{bs}' - 3\cos(2\theta_{rm}) \left(i_{as} + i_{bs}'\right)^\frac{1}{2} + 2\sin(\theta_{rm})\right] \, di_{bs}' \\ W_{c,2} &= 5 i_{as} i_{bs} - 3 \left(\frac{3}{2}\right)^{-1} \cos(2\theta_{rm}) \left(i_{as} + i_{bs}\right)^\frac{3}{2} + 2 \cos(2\theta_rm) i_{as}^\frac{3}{2} + 2\cos(\theta_{rm}) i_{bs} + \frac{5}{2} i_{bs}^2 - 3 \left(\frac{3}{2}\right)^{-1} \cos(2\theta_{rm}) \left(i_{as} + i_{bs}\right)^\frac{3}{2} + 2 \cos(2\theta_rm) i_{as}^\frac{3}{2} + 2\sin(\theta_{rm}) i_{bs} \\ W_{c,2} &= 5 i_{as} i_{bs} + \frac{5}{2} i_{bs}^2 - 4 \cos(2\theta_{rm}) \left(i_{as} + i_{bs}\right)^\frac{3}{2} + 4\cos(2\theta_{rm}) i_{as}^\frac{3}{2} + 2\left[\cos(\theta_{rm}) + \sin(\theta_{rm})\right] i_{bs} \end{align} $

The evaluation of the previous integral has terms for $ i_{bs}' = 0 $ (lower bound evaluation) that must not be forgotten. The final step is the sum all the individual coenergy contributions $ W_c(i_{as}, i_{bs}, \theta_{rm}) = \sum_{k = 0}^{2} W_{c,k}(i_{as}, i_{bs}, \theta_{rm}) $. Recall that $ i_{as}, i_{bs} \geq 0 $ to remove any fears of producing a complex result.

$ \begin{align} W_c(i_{as}, i_{bs}, \theta_{rm}) &= W_{c,0} + W_{c,1} + W_{c,2} \\ W_c(i_{as}, i_{bs}, \theta_{rm}) &= \begin{split} &{}0 + \frac{5}{2} i_{as}^2 - 4 \cos(2\theta_{rm}) i_{as}^\frac{3}{2} + 2\left[\cos(\theta_{rm}) + \sin(\theta_{rm})\right] i_{as} \\ &{}+ 5 i_{as} i_{bs} + \frac{5}{2} i_{bs}^2 - 4 \cos(2\theta_{rm}) \left(i_{as} + i_{bs}\right)^\frac{3}{2} + 4\cos(2\theta_{rm}) i_{as}^\frac{3}{2} + 2\left[\cos(\theta_{rm}) + \sin(\theta_{rm})\right] i_{bs} \\ \end{split} \\ W_c(i_{as}, i_{bs}, \theta_{rm}) &= \frac{5}{2} i_{as}^2 + 5 i_{as} i_{bs} + \frac{5}{2} i_{bs}^2 - 4 \cos(2\theta_{rm}) \left(i_{as} + i_{bs}\right)^\frac{3}{2} + 2\left[\cos(\theta_{rm}) + \sin(\theta_{rm})\right] \left(i_{as} + i_{bs}\right) \end{align} $

Electromechanical Torque Calculation

Electromechanical torque is just the partial derivative of coenergy with respect to mechanical rotor position.

$ \begin{align} T_e &= \frac{\partial W_c(i_{as}, i_{bs}, \theta_{rm})}{\partial \theta_{rm}} \\ T_e &= 0 + 0 + 0 + 8\sin(2\theta_{rm}) \left(i_{as} + i_{bs}\right)^\frac{3}{2} + 2\left[\cos(\theta_{rm}) - \sin(\theta_{rm}\right] \left(i_{as} + i_{bs}\right) \end{align} $

A quick burst phasor addition will show that $ \cos(\theta_{rm}) - \sin(\theta_{rm}) = \sqrt{2} \cos\left(\theta_{rm} + \frac{\pi}{4}\right) $. Thus, the electromechanical torque equation is obtained for this device.

$ \begin{equation} \boxed{T_e = 8\sin(2\theta_{rm}) \left(i_{as} + i_{bs}\right)^\frac{3}{2} + 2\sqrt{2} \cos\left(\theta_{rm} + \frac{\pi}{4}\right) \left(i_{as} + i_{bs}\right)} \end{equation} $

Discussion

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