(Created page with start of coenergy calculation)
 
(Completed coenergy and electromechanical torque calculations)
 
Line 26: Line 26:
 
<math>\begin{align}
 
<math>\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} &= \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}</math>
 
\end{align}</math>
−
<!--%W_{c,1} &= \int_{i_1' = 0}^{i_1} \left[5i_1' + 10\left(1 - \frac{1}{1 + i_1' + \cancelto{0}{2i_2'}}\right) \cos(\theta_r)\right] \, di_1' + \int_{i_1' = 0}^{i_1} \left[\cancelto{0}{i_2'} + 20\left(1 - \frac{1}{1 + i_1' + \cancelto{0}{2i_2'}}\right) \cos(\theta_r)\right] \, di_1' \\
+
 
−
%W_{c,1} &= \frac{5}{2} i_1^2 + 10\left(i_1 - \ln|1 + i_1| + \cancelto{0}{\ln|1 + 0|}\right) \cos(\theta_r) + 20\left(i_1 - \ln|1 + i_1| + \cancelto{0}{\ln|1 + 0|}\right) \cos(\theta_r) \\
+
The second step in the coenergy calculation will ramp dummy variable <math>i_{bs}'</math> from <math>0</math> to its final value of <math>i_{bs}</math> while dummy variable <math>i_{as}' = i_{as}</math> is held at its final value.
−
%W_{c,1} &= \frac{5}{2} i_1^2 + 10\left(i_1 - \ln|1 + i_1|\right) \cos(\theta_r) + 20\left(i_1 - \ln|1 + i_1|\right) \cos(\theta_r)
+
−
The second step in the coenergy calculation will ramp dummy variable <math>i_2'</math> from <math>0</math> to its final value of <math>i_2</math> while dummy variable <math>i_1' = i_1</math> is held at its final value. Substitute <math>i_m = i_1 + 2i_2</math> as needed.
+
  
 
<math>\begin{align}
 
<math>\begin{align}
−
W_{c,2} &= \left.\int_{i_2' = 0}^{i_2} \lambda_1\left(i_1', i_2', \theta_r\right) \, di_2'\right|_{i_1' = i_1} + \left.\int_{i_2' = 0}^{i_2} \lambda_2\left(i_1', i_2', \theta_r\right) \, di_2'\right|_{i_1' = i_1} \\
+
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_2' = 0}^{i_2} \left[5i_1 + 10\left(1 - \frac{1}{1 + i_1 + 2i_2'}\right) \cos(\theta_r)\right] \, di_2' + \int_{i_2' = 0}^{i_2} \left[i_2' + 20\left(1 - \frac{1}{1 + i_1 + 2i_2'}\right) \cos(\theta_r)\right] \, di_2' \\
+
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_1 i_2 + 10\left(i_2 - \ln|1 + i_1 + 2i_2| + \ln|1 + i_1|\right) \cos(\theta_r) + \frac{1}{2} i_2^2 + 20\left(i_2 - \ln|1 + i_1 + 2i_2| + \ln|1 + i_1|\right) \cos(\theta_r)
+
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}</math>
 
\end{align}</math>
  
−
The final step is the sum all the individual coenergy contributions <math>W_c(i_1, i_2, \theta_r) = \sum_{k = 0}^{2} W_{c,k}(i_1, i_2, \theta_r)</math>. Recall that <math>i_1, i_2 \geq 0</math> to simplify the absolute value signs.
+
The evaluation of the previous integral has terms for <math>i_{bs}' = 0</math> (lower bound evaluation) that must not be forgotten. The final step is the sum all the individual coenergy contributions <math>W_c(i_{as}, i_{bs}, \theta_{rm}) = \sum_{k = 0}^{2} W_{c,k}(i_{as}, i_{bs}, \theta_{rm})</math>. Recall that <math>i_{as}, i_{bs} \geq 0</math> to remove any fears of producing a complex result.
  
 
<math>\begin{align}
 
<math>\begin{align}
−
W_c(i_1, i_2, \theta_r) &= W_{c,0} + W_{c,1} + W_{c,2} \\
+
W_c(i_{as}, i_{bs}, \theta_{rm}) &= W_{c,0} + W_{c,1} + W_{c,2} \\
−
W_c(i_1, i_2, \theta_r) &=
+
W_c(i_{as}, i_{bs}, \theta_{rm}) &=
 
\begin{split}
 
\begin{split}
−
   &{}0 + \frac{5}{2} i_1^2 + 10\left(i_1 - \cancel{\ln|1 + i_1|}\right) \cos(\theta_r) + 20\left(i_1 - \cancel{\ln|1 + i_1|}\right) \cos(\theta_r) \\
+
   &{}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_1 i_2 + 10\left(i_2 - \ln|1 + i_1 + 2i_2| + \cancel{\ln|1 + i_1|}\right) \cos(\theta_r) \\
+
   &{}+ 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} \\
−
  &{}+ \frac{1}{2} i_2^2 + 20\left(i_2 - \ln|1 + i_1 + 2i_2| + \cancel{\ln|1 + i_1|}\right) \cos(\theta_r)
+
 
\end{split} \\
 
\end{split} \\
−
W_c(i_1, i_2, \theta_r) &= \frac{5}{2} i_1^2 + 5 i_1 i_2 + \frac{1}{2} i_2^2 + 10\left[i_1 + i_2 - \ln(1 + i_1 + 2i_2)\right] \cos(\theta_r) + 20\left[i_1 + i_2 - \ln(1 + i_1 + 2i_2)\right] \cos(\theta_r)
+
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}</math>
 
\end{align}</math>
−
 
+
<!--
−
The last two terms have the same variable dependence, so the coefficients may be added.
+
This calculation is consistent with the known result of a magnetically linear system having a coenergy of <math>W_c(i_{as}, i_{bs}, \theta_{rm}) = \frac{1}{2} L_{asas}(\theta_{rm}) i_{as}^2 + L_{asbs}(\theta_{rm}) i_{as} i_{bs} + \frac{1}{2} L_{bsbs}(\theta_{rm}) i_{bs}^2 = \frac{1}{2} \sum_{j = 1}^2 \sum_{k = 1}^2 \lambda_j(i, \theta_{rm}) i_k</math>.
−
 
+
-->
−
<math>\begin{equation}
+
−
\boxed{W_c(i_1, i_2, \theta_r) = \frac{5}{2} i_1^2 + 5 i_1 i_2 + \frac{1}{2} i_2^2 + 30\cos(\theta_r)\left[i_1 + i_2 - \ln(1 + i_1 + 2i_2)\right]}
+
−
\end{equation}</math>
+
−
 
+
 
===<small>Electromechanical Torque Calculation</small>===
 
===<small>Electromechanical Torque Calculation</small>===
  
Line 63: Line 59:
  
 
<math>\begin{align}
 
<math>\begin{align}
−
T_e &= \frac{\partial W_c(i_1, i_2, \theta_r)}{\partial \cancelto{\theta_r}{\theta_{rm}}} \\
+
T_e &= \frac{\partial W_c(i_{as}, i_{bs}, \theta_{rm})}{\partial \theta_{rm}} \\
−
T_e &= 0 + 0 + 0 - 30\sin(\theta_r)\left[i_1 + i_2 - \ln(1 + i_1 + 2i_2)\right]
+
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}</math>
 
\end{align}</math>
  
−
Thus, the electromechanical torque equation is obtained for this device.
+
A quick burst phasor addition will show that <math>\cos(\theta_{rm}) - \sin(\theta_{rm}) = \sqrt{2} \cos\left(\theta_{rm} + \frac{\pi}{4}\right)</math>. Thus, the electromechanical torque equation is obtained for this device.
  
 
<math>\begin{equation}
 
<math>\begin{equation}
−
\boxed{T_e = -30\sin(\theta_r)\left[i_1 + i_2 - \ln(1 + i_1 + 2i_2)\right]}
+
\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}</math>
 
\end{equation}</math>
−
-->
+
 
 
----
 
----
 
==Discussion==
 
==Discussion==
Line 78: Line 74:
 
----
 
----
 
----
 
----
−
[[ECE_PhD_QE_ES1_2011|Back to PE-1, August 2011]]
+
[[ECE_PhD_QE_PE1_2011|Back to PE-1, August 2011]]

Latest revision as of 18:28, 26 January 2018


Answers and Discussions for

ECE Ph.D. Qualifying Exam PE-1 August 2011

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

Back to PE-1, August 2011

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