Difference between revisions of "ECE PhD QE PE1 2014 Problem1b" - Rhea

Latest revision as of 16:41, 4 August 2018

Answers and Discussions for

ECE Ph.D. Qualifying Exam PE-1 August 2014

Problem 1, part b

Recycled Information from part a

The voltage equation and flux linkage equation for these five identical, magnetically coupled inductors in matrix form is reused(resistance is neglected and linear magnetics assumed).

$\begin{align} \vec{v}_{12345s} &= p \vec{\lambda}_{12345s} \\ \vec{\lambda}_{12345s} &= \mathbf{L}_{ss} \vec{i}_{12345s} \\ \begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \lambda_4 \\ \lambda_5 \end{bmatrix} &= \begin{bmatrix} L_s & M_{ns} & M_{fs} & M_{fs} & M_{ns} \\ M_{ns} & L_s & M_{ns} & M_{fs} & M_{fs} \\ M_{fs} & M_{ns} & L_s & M_{ns} & M_{fs} \\ M_{fs} & M_{fs} & M_{ns} & L_s & M_{ns} \\ M_{ns} & M_{fs} & M_{fs} & M_{ns} & L_s \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \\ i_3 \\ i_4 \\ i_5 \end{bmatrix} \end{align}$

Because the inductors are identical to each other, only three unique, constant entries exist in the inductance matrix $\mathbf{L}_{ss}$ : the self-inductance $$ L_s > 0 $$ , the mutual inductance between inductors $\frac{2\pi}{5} \, \text{rad}$ apart $M_{ns} > 0$ , and the mutual inductance between inductors $\frac{4\pi}{5} \, \text{rad}$ apart $M_{fs} < 0$ .

The results from the previous current excitation must also be retained. Based on the symmetry of the situation under this operating condition alone, it may be deduced from the flux linkage equation that $$ v_5(t) = v_2(t) $$ and $$ v_3(t) = v_4(t) $$ .

$\begin{align} \begin{bmatrix} i_1(t) \\ i_2(t) \\ i_3(t) \\ i_4(t) \\ i_5(t) \end{bmatrix} &= \begin{bmatrix} 10 \cos\left(100 \, \frac{\text{rad}}{\text{s}} t\right) \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \, \text{A} \\ \begin{bmatrix} v_1(t) \\ v_2(t) \\ v_3(t) \\ v_4(t) \\ v_5(t) \end{bmatrix} &= \begin{bmatrix} 10^3 L_s \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \\ 10^3 M_{ns} \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \\ 10^3 |M_{fs}| \cos\left(100 \, \frac{\text{rad}}{\text{s}} t + \frac{\pi}{2} \, \text{rad}\right) \\ 10^3 |M_{fs}| \cos\left(100 \, \frac{\text{rad}}{\text{s}} t + \frac{\pi}{2} \, \text{rad}\right) \\ 10^3 M_{ns} \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \end{bmatrix} \, \text{V} = \begin{bmatrix} 40 \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \\ 10 \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \\ 1 \cos\left(100 \, \frac{\text{rad}}{\text{s}} t + \frac{\pi}{2} \, \text{rad}\right) \\ 1 \cos\left(100 \, \frac{\text{rad}}{\text{s}} t + \frac{\pi}{2} \, \text{rad}\right) \\ 10 \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \end{bmatrix} \, \text{V} \end{align}$

Field Energy Calculation

The inductance matrix entries are solved based on the previous results: $L_s = +0.04 \, \text{H}$ , $M_{ns} = +0.01 \, \text{H}$ , and $M_{fs} = -0.001 \, \text{H}$ . The new current excitation is written in matrix form.

$\begin{equation} \begin{bmatrix} i_1(t) \\ i_2(t) \\ i_3(t) \\ i_4(t) \\ i_5(t) \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 10 \\ 0 \\ 2 \end{bmatrix} \, \text{A} \, \forall t \end{equation}$

Coenergy will first 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). This coenergy contribution is trivially zero since the coupled inductor circuit lacks a mechanical system.

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

The first step in the coenergy calculation will ramp dummy variable $i_{3}'$ from $$ 0 $$ to its final value of $i_{3}$ while dummy variable $i_{5}' = 0$ is held at its initial value. Many terms in the expanded equation have zero value and will be omitted.

$\begin{align} W_{c,1} &= \left. \sum_{j = 1}^5 \int_{i_{3}' = 0}^{i_{3}'} \lambda_{j}\left(\vec{i}_{12345s}'\right) \, di_{3}'\right|_{i_{5}' = 0} \\ W_{c,1} &= \int_{i_{3}' = 0}^{i_{3}} \left[M_{fs} i_{3}'\right] \, di_{3}' + \int_{i_{3}' = 0}^{i_{3}} \left[M_{ns} i_{3}'\right] \, di_{3}' + \int_{i_{3}' = 0}^{i_{3}} \left[L_s i_{3}'\right] \, di_{3}' + \int_{i_{3}' = 0}^{i_{3}} \left[M_{ns} i_{3}'\right] \, di_{3}' + \int_{i_{3}' = 0}^{i_{3}} \left[M_{fs} i_{3}'\right] \, di_{3}' \\ W_{c,1} &= \frac{1}{2} M_{fs} i_{3}^2 + \frac{1}{2} M_{ns} i_{3}^2 + \frac{1}{2} L_s i_{3}^2 + \frac{1}{2} M_{ns} i_{3}^2 + \frac{1}{2} M_{fs} i_{3}^2 \\ W_{c,1} &= \left(\frac{1}{2} L_s + M_{ns} + M_{fs}\right) i_{3}^2 \end{align}$

The second step in the coenergy calculation will ramp dummy variable $i_{5}'$ from $$ 0 $$ to its final value of $i_{5}$ while dummy variable $i_{3}' = i_{3}$ is held at its final value.

$\begin{align} W_{c,2} &= \left. \sum_{j = 1}^5 \int_{i_{5}' = 0}^{i_{5}'} \lambda_{j}\left(\vec{i}_{12345s}'\right) \, di_{5}'\right|_{i_{3}' = i_{3}} \\ W_{c,2} &= \int_{i_{5}' = 0}^{i_{5}} \left[M_{fs} i_{3} + M_{ns} i_{5}'\right] \, di_{5}' + \int_{i_{5}' = 0}^{i_{5}} \left[M_{ns} i_{3} + M_{fs} i_{5}'\right] \, di_{5}' + \int_{i_{5}' = 0}^{i_{5}} \left[L_s i_{3} + M_{fs} i_{5}'\right] \, di_{5}' + \int_{i_{5}' = 0}^{i_{5}} \left[M_{ns} i_{3} + M_{ns} i_{5}'\right] \, di_{5}' + \int_{i_{5}' = 0}^{i_{5}} \left[M_{fs} i_{3} + L_s i_{5}'\right] \, di_{5}' \\ W_{c,2} &= M_{fs} i_{3} i_{5} + \frac{1}{2} M_{ns} i_{5}^2 + M_{ns} i_{3} i_{5} + \frac{1}{2} M_{fs} i_{5}^2 + L_s i_{3} i_{5} + \frac{1}{2} M_{fs} i_{5}^2 + M_{ns} i_{3} i_{5} + \frac{1}{2} M_{ns} i_{5}^2 + M_{fs} i_{3} i_{5} + \frac{1}{2} L_s i_{5}^2 \\ W_{c,2} &= \left(L_s + 2M_{ns} + 2M_{fs}\right) i_{3} i_{5} + \left(\frac{1}{2} L_s + M_{ns} + M_{fs}\right) i_{5}^2 \end{align}$

Because the other currents remain at zero, no additional coenergy contributions need to be found. The final step is the sum all the individual coenergy contributions $W_c(\vec{i}_{12345s}) = \sum_{k = 0}^{2} W_{c,k}(\vec{i}_{12345s})$ .

$\begin{align} W_c(\vec{i}_{12345s}) &= W_{c,0} + W_{c,1} + W_{c,2} \\ W_c(\vec{i}_{12345s}) &= \begin{split} &{}0 + \left(\frac{1}{2} L_s + M_{ns} + M_{fs}\right) i_{3}^2 \\ &{}+ \left(L_s + 2M_{ns} + 2M_{fs}\right) i_{3} i_{5} + \left(\frac{1}{2} L_s + M_{ns} + M_{fs}\right) i_{5}^2 \\ \end{split} \\ W_c(\vec{i}_{12345s}) &= \left(\frac{1}{2} L_s + M_{ns} + M_{fs}\right) i_{3}^2 + \left(L_s + 2M_{ns} + 2M_{fs}\right) i_{3} i_{5} + \left(\frac{1}{2} L_s + M_{ns} + M_{fs}\right) i_{5}^2 \\ W_c(\vec{i}_{12345s}) &= (L_s + 2 M_{ns} + 2 M_{fs}) \left(\frac{1}{2} i_{3}^2 + \frac{1}{2} i_{5}^2 + i_{3} i_{5}\right) \end{align}$

This calculation is consistent with the known result of a magnetically linear system having a coenergy of $W_c(\vec{i}_{12345s}, \theta_{rm}) = \frac{1}{2} \vec{i}_{12345s}^\mathrm{T} \mathbf{L}_{ss}(\theta_{rm}) \vec{i}_{12345s}$ .

Because this is a magnetically linear system, the field energy is equal to the coenergy. The value of $L_s + 2M_{ns} + 2M_{fs} = +0.04 \, \text{H} + 2\left(+0.01 \, \text{H}\right) + 2\left(-0.001 \, \text{H}\right) = +0.058 \, \text{H}$ needs to be substituted in along with $i_3 = 10 \, \text{A}$ and $i_5 = 2 \, \text{A}$ .

$\begin{equation} W_f(\vec{i}_{12345s}) = (+0.058 \, \text{H}) \left(50 \, \text{A}^2 + 2 \, \text{A}^2 + 20 \, \text{A}^2\right) \end{equation}$

$\begin{equation} \boxed{W_f(\vec{i}_{12345s}) = 4.176 \, \text{J}} \end{equation}$

Discussion

Back to PE-1, August 2014

@@ Line 58: / Line 58: @@
 <math>\begin{align}
-W_{c,1} &= \left. \sum_{j = 1}^5 \int_{i_{3}' = 0}^{i_{3}'} \lambda_{j}\left(\vec{i}_j\right) \, di_{3}'\right|_{i_{5}' = 0} \\
+W_{c,1} &= \left. \sum_{j = 1}^5 \int_{i_{3}' = 0}^{i_{3}'} \lambda_{j}\left(\vec{i}_{12345s}'\right) \, di_{3}'\right|_{i_{5}' = 0} \\
 W_{c,1} &= \int_{i_{3}' = 0}^{i_{3}} \left[M_{fs} i_{3}'\right] \, di_{3}' + \int_{i_{3}' = 0}^{i_{3}} \left[M_{ns} i_{3}'\right] \, di_{3}' + \int_{i_{3}' = 0}^{i_{3}} \left[L_s i_{3}'\right] \, di_{3}' + \int_{i_{3}' = 0}^{i_{3}} \left[M_{ns} i_{3}'\right] \, di_{3}' + \int_{i_{3}' = 0}^{i_{3}} \left[M_{fs} i_{3}'\right] \, di_{3}' \\
-W_{c,1} &= \frac{1}{2} M_{fs} i_{3}^2 + \frac{1}{2} M_{ns} i_{3}^2 + \frac{1}{2} L_s i_{3}^2 + \frac{1}{2} M_{ns} i_{3}^2 + \frac{1}{2} M_{fs} i_{3}^2
+W_{c,1} &= \frac{1}{2} M_{fs} i_{3}^2 + \frac{1}{2} M_{ns} i_{3}^2 + \frac{1}{2} L_s i_{3}^2 + \frac{1}{2} M_{ns} i_{3}^2 + \frac{1}{2} M_{fs} i_{3}^2 \\
+W_{c,1} &= \left(\frac{1}{2} L_s + M_{ns} + M_{fs}\right) i_{3}^2
 \end{align}</math>
-<!--The evaluation of the previous integral has terms for <math>i_{1}' = 0</math> (lower bound evaluation) that must not be forgotten. 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.
+The second step in the coenergy calculation will ramp dummy variable <math>i_{5}'</math> from <math>0</math> to its final value of <math>i_{5}</math> while dummy variable <math>i_{3}' = i_{3}</math> is held at its final value.
 <math>\begin{align}
-W_{c,2} &= \left.\int_{i_{2}' = 0}^{i_{2}} \lambda_{1}\left(i_{1}', i_{2}', \theta_{rm}\right) \, di_{2}'\right|_{i_{1}' = i_{1}} + \left.\int_{i_{2}' = 0}^{i_{2}} \lambda_{2}\left(i_{1}', i_{2}', \theta_{rm}\right) \, di_{2}'\right|_{i_{1}' = i_{1}} \\
+W_{c,2} &= \left. \sum_{j = 1}^5 \int_{i_{5}' = 0}^{i_{5}'} \lambda_{j}\left(\vec{i}_{12345s}'\right) \, di_{5}'\right|_{i_{3}' = i_{3}} \\
-W_{c,2} &= \int_{i_{2}' = 0}^{i_{2}} \left[10i_{1} + 2 \left(5 + 2\sin(2\theta_{rm})\right) \frac{2i_{1} + i_{2}'}{\sqrt{1 + (2i_{1} + i_{2}')^2}}\right] \, di_{2}' + \int_{i_{2}' = 0}^{i_{2}} \left[5i_{2}' + \left(5 + 2\sin(2\theta_{rm})\right) \frac{2i_{1} + i_{2}'}{\sqrt{1 + (2i_{1} + i_{2}')^2}}\right] \, di_{2}' \\
+W_{c,2} &= \int_{i_{5}' = 0}^{i_{5}} \left[M_{fs} i_{3} + M_{ns} i_{5}'\right] \, di_{5}' + \int_{i_{5}' = 0}^{i_{5}} \left[M_{ns} i_{3} + M_{fs} i_{5}'\right] \, di_{5}' + \int_{i_{5}' = 0}^{i_{5}} \left[L_s i_{3} + M_{fs} i_{5}'\right] \, di_{5}' + \int_{i_{5}' = 0}^{i_{5}} \left[M_{ns} i_{3} + M_{ns} i_{5}'\right] \, di_{5}' + \int_{i_{5}' = 0}^{i_{5}} \left[M_{fs} i_{3} + L_s i_{5}'\right] \, di_{5}' \\
-W_{c,2} &= 10 i_{1} i_{2} + 2 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1} + i_{2})^2} - 2 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1})^2} + \frac{5}{2} i_{2}^2 + \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1} + i_{2})^2} - \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1})^2} \\
+W_{c,2} &= M_{fs} i_{3} i_{5} + \frac{1}{2} M_{ns} i_{5}^2 + M_{ns} i_{3} i_{5} + \frac{1}{2} M_{fs} i_{5}^2 + L_s i_{3} i_{5} + \frac{1}{2} M_{fs} i_{5}^2 + M_{ns} i_{3} i_{5} + \frac{1}{2} M_{ns} i_{5}^2 + M_{fs} i_{3} i_{5} + \frac{1}{2} L_s i_{5}^2 \\
-W_{c,2} &= 10 i_{1} i_{2} + \frac{5}{2} i_{2}^2 + 3 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1} + i_{2})^2} - 3 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1})^2}
+W_{c,2} &= \left(L_s + 2M_{ns} + 2M_{fs}\right) i_{3} i_{5} + \left(\frac{1}{2} L_s + M_{ns} + M_{fs}\right) i_{5}^2
 \end{align}</math>
-The evaluation of the previous integral has terms for <math>i_{2}' = 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_{1}, i_{2}, \theta_{rm}) = \sum_{k = 0}^{2} W_{c,k}(i_{1}, i_{2}, \theta_{rm})</math>.
+Because the other currents remain at zero, no additional coenergy contributions need to be found. The final step is the sum all the individual coenergy contributions <math>W_c(\vec{i}_{12345s}) = \sum_{k = 0}^{2} W_{c,k}(\vec{i}_{12345s})</math>.
 <math>\begin{align}
-W_c(i_{1}, i_{2}, \theta_{rm}) &= W_{c,0} + W_{c,1} + W_{c,2} \\
+W_c(\vec{i}_{12345s}) &= W_{c,0} + W_{c,1} + W_{c,2} \\
-W_c(i_{1}, i_{2}, \theta_{rm}) &=
+W_c(\vec{i}_{12345s}) &=
 \begin{split}
-    &{}0 + 5 i_{1}^2 + 3 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1})^2} - 3 \left(5 + 2\sin(2\theta_{rm})\right) \\
+    &{}0 + \left(\frac{1}{2} L_s + M_{ns} + M_{fs}\right) i_{3}^2 \\
-    &{}+ 10 i_{1} i_{2} + \frac{5}{2} i_{2}^2 + 3 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1} + i_{2})^2} - 3 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1})^2} \\
+    &{}+ \left(L_s + 2M_{ns} + 2M_{fs}\right) i_{3} i_{5} + \left(\frac{1}{2} L_s + M_{ns} + M_{fs}\right) i_{5}^2 \\
 \end{split} \\
-W_c(i_{1}, i_{2}, \theta_{rm}) &= 5 i_{1}^2 + 10 i_{1} i_{2} + \frac{5}{2} i_{2}^2 + 3 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1} + i_{2})^2} - 3 \left(5 + 2\sin(2\theta_{rm})\right)
+W_c(\vec{i}_{12345s}) &= \left(\frac{1}{2} L_s + M_{ns} + M_{fs}\right) i_{3}^2 + \left(L_s + 2M_{ns} + 2M_{fs}\right) i_{3} i_{5} + \left(\frac{1}{2} L_s + M_{ns} + M_{fs}\right) i_{5}^2 \\
+W_c(\vec{i}_{12345s}) &= (L_s + 2 M_{ns} + 2 M_{fs}) \left(\frac{1}{2} i_{3}^2 + \frac{1}{2} i_{5}^2 + i_{3} i_{5}\right)
 \end{align}</math>
-<!--
-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>.
---><!--
-===<small>Electromechanical Torque Calculation</small>===
-Electromechanical torque is just the partial derivative of coenergy with respect to mechanical rotor position.
+This calculation is consistent with the known result of a magnetically linear system having a coenergy of <math>W_c(\vec{i}_{12345s}, \theta_{rm}) = \frac{1}{2} \vec{i}_{12345s}^\mathrm{T} \mathbf{L}_{ss}(\theta_{rm}) \vec{i}_{12345s}</math>.
-<math>\begin{align}
+Because this is a magnetically linear system, the field energy is equal to the coenergy. The value of <math>L_s + 2M_{ns} + 2M_{fs} = +0.04 \, \text{H} + 2\left(+0.01 \, \text{H}\right) + 2\left(-0.001 \, \text{H}\right) = +0.058 \, \text{H}</math> needs to be substituted in along with <math>i_3 = 10 \, \text{A}</math> and <math>i_5 = 2 \, \text{A}</math>.
-T_e &= \frac{\partial W_c(i_{1}, i_{2}, \theta_{rm})}{\partial \theta_{rm}} \\
-T_e &= 0 + 0 + 0 + 12\cos(2\theta_{rm}) \sqrt{1 + (2i_{1} + i_{2})^2} - 12\cos(2\theta_{rm})
-\end{align}</math>
-Thus, the electromechanical torque equation is obtained for this device.
+<math>\begin{equation}
+W_f(\vec{i}_{12345s}) = (+0.058 \, \text{H}) \left(50 \, \text{A}^2 + 2 \, \text{A}^2 + 20 \, \text{A}^2\right)
+\end{equation}</math>
 <math>\begin{equation}
-\boxed{T_e = 12\cos(2\theta_{rm}) \left[\sqrt{1 + (2i_{1} + i_{2})^2} - 1\right]}
+\boxed{W_f(\vec{i}_{12345s}) = 4.176 \, \text{J}}
-\end{equation}</math>-->
+\end{equation}</math>
 ----