ECE PhD QE PE1 2010 Problem3 11 - Rhea

General

Let r.f. be the abbreviation for "reference frame". All references are to Analysis of Electric Machinery and Drive Systems (Third Edition).

Problem 3

$ \boxed{\text{Yes}} $

A conservative vector field has the property that the field equation may be expressed as a gradient of a scalar potential field: $ \vec{F}(\vec{r}, t) = -\vec{\nabla} \phi(t) $. A conservative vector field has the following three properties as a consequence:

1. A line integral of the conservative vector field is path independent, thus a circulation (line integral over a closed path) is zero.
2. The conservative vector field is irrotational: $ \vec{\nabla} \times \vec{F}(\vec{r}, t) = \vec{0} $.
3. A conservative vector field has a corresponding conservative force such that no net work is done for a circulation (no losses to overcome).

Because hysteresis leads to energy loss such that $ W_{f\ell} > 0 $, magnetic hysteresis causes the coupling field to become non-conservative. (see pp. 13, 17)

Problem 4

$ \boxed{\text{No}} $

Magnetic saturation causes magnetic flux linkage $ \lambda $ to become a nonlinear function of current $ i $. (An equivalent statement can be made for $ \vec{B} $ and $ \vec{H} $.) Despite creating a magnetically nonlinear system, magnetic saturation does not correspond to any losses. Therefore, the coupling field remains conservative. (see pp. 8-10)

Problem 5

$ \boxed{\text{Field energy}} $

Field energy is expressed as $ W_f(\vec{\lambda}, x) = \sum_{j = 1}^J \int i(\vec{\lambda}, x) \, d\lambda $. Coenergy is expressed as $ W_f(\vec{i}, x) = \sum_{j = 1}^J \int \lambda(\vec{i}, x) \, di $. If $ i(\vec{\lambda}) $ is available, field energy is more direct to compute. (see pp. 16-22)

Problem 6

$ \boxed{\text{Stationary r.f.}} $

The analysis of an unsymmetric (or imbalanced) machine yields position-invariant circuit parameters only if the arbitrary reference frame is fixed in the r.f. where the imbalance exists. Namely, only the stationary r.f. with $ \omega = 0 $ is conducive for unsymmetric 3-phase machine analysis. (see pp. 91)

Problem 7

$ \boxed{\text{QD0 variables in the stationary r.f.}} $

If the arbitrary reference frame moves at synchronous speed $ \omega = \omega_e $, then the time-varying position-dependence of the circuit variables disappears for balanced, steady-state conditions. The only information affecting the three-phase, balanced set of circuit variables is the RMS value and initial phase angle. These are the two pieces of information retained in phasor analysis. (see pp. 99-100)

Problem 8

$ \boxed{\text{Yes}} $

The arbitary reference frame transformation is always reversible, meaning that the transformation matrix is always invertible.

$ \begin{align} \mathbf{K}_s &= \frac{2}{3} \begin{bmatrix} \cos\left(\theta\right) & \cos\left(\theta - \frac{2\pi}{3}\right) & \cos\left(\theta + \frac{2\pi}{3}\right) \\ \sin\left(\theta\right) & \sin\left(\theta - \frac{2\pi}{3}\right) & \sin\left(\theta + \frac{2\pi}{3}\right) \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{bmatrix} \\ \left(\mathbf{K}_s\right)^{-1} &= \begin{bmatrix} \cos\left(\theta\right) & \sin\left(\theta\right) & 1 \\ \cos\left(\theta - \frac{2\pi}{3}\right) & \sin\left(\theta - \frac{2\pi}{3}\right) & 1 \\ \cos\left(\theta + \frac{2\pi}{3}\right) & \sin\left(\theta + \frac{2\pi}{3}\right) & 1 \end{bmatrix} \end{align} $

To prove $ \mathbf{K}_s $ is always invertible, it must be shown $ \mathbf{K}_s \vec{v}_{abcs} = \vec{0} \, \implies \vec{v}_{abcs} = \vec{0} $.

The system of equations is as follows.

$ \begin{align} 0 &= \frac{2}{3} \left[\cos\left(\theta\right) v_{as} + \cos\left(\theta - \frac{2\pi}{3}\right) v_{bs} + \cos\left(\theta + \frac{2\pi}{3}\right) v_{cs}\right] \\ 0 &= \frac{2}{3} \left[\sin\left(\theta\right) v_{as} + \sin\left(\theta - \frac{2\pi}{3}\right) v_{bs} + \sin\left(\theta + \frac{2\pi}{3}\right) v_{cs}\right] \\ 0 &= \frac{1}{3} v_{as} + \frac{1}{3} v_{bs} + \frac{1}{3} v_{cs} \end{align} $

The last equation implies $ v_{cs} = -\left(v_{as} + v_{bs}\right) $. The top two equations in the system of equations may be simplified.

$ \begin{align} 0 &= \cos\left(\theta\right) v_{as} + \cos\left(\theta - \frac{2\pi}{3}\right) v_{bs} - \cos\left(\theta + \frac{2\pi}{3}\right) \left(v_{as} + v_{bs}\right) \\ 0 &= \left[\cos\left(\theta\right) - \cos\left(\theta + \frac{2\pi}{3}\right)\right] v_{as} + \left[\cos\left(\theta - \frac{2\pi}{3}\right) - \cos\left(\theta + \frac{2\pi}{3}\right)\right] v_{bs} \\ 0 &= \sqrt{3} \cos\left(\theta + \frac{\pi}{6}\right) v_{as} + \sqrt{3} \cos\left(\theta - \frac{\pi}{2}\right) v_{bs} \\ 0 &= \sin\left(\theta\right) v_{as} + \sin\left(\theta - \frac{2\pi}{3}\right) v_{bs} - \sin\left(\theta + \frac{2\pi}{3}\right) \left(v_{as} + v_{bs}\right) \\ 0 &= \left[\sin\left(\theta\right) - \sin\left(\theta + \frac{2\pi}{3}\right)\right] v_{as} + \left[\sin\left(\theta - \frac{2\pi}{3}\right) - \sin\left(\theta + \frac{2\pi}{3}\right)\right] v_{bs} \\ 0 &= \sqrt{3} \sin\left(\theta + \frac{\pi}{6}\right) v_{as} + \sqrt{3} \sin\left(\theta - \frac{\pi}{2}\right) v_{bs} \end{align} $

The system of equations must be valid for any two choices of $ \theta $.

• At $ \theta = 0 $, $ 0 = \sqrt{3} \frac{\sqrt{3}}{2} v_{as} = \frac{3}{2} v_{as} $ and $ 0 = \sqrt{3} \frac{1}{2} v_{as} - \sqrt{3} v_{bs} = \frac{\sqrt{3}}{2} v_{as} - \sqrt{3} v_{bs} $. The first equation is only solved if $ v_{as} = 0 $, which sets $ v_{bs} = 0 $ by the second equation.
• At $ \theta = \frac{\pi}{2} \, \textrm{rad} $, $ 0 = -\sqrt{3} \frac{1}{2} v_{as} + \sqrt{3} v_{bs} = -\frac{\sqrt{3}}{2} v_{as} + \sqrt{3} v_{bs} $ and $ 0 = \sqrt{3} \frac{\sqrt{3}}{2} v_{as} = \frac{3}{2} v_{as} $. The second equation is only solved if $ v_{as} = 0 $, which sets $ v_{bs} = 0 $ by the first equation.
• The pattern holds for all $ \theta \, \in \, \mathbb{R} $.

The relationships between $ v_{as} $, $ v_{bs} $, and $ v_{cs} $ hold for all values of $ \theta $, so $ \vec{v}_{abcs} = \vec{0} $. Therefore, $ \mathbf{K}_s $ is always invertible. (see pp. 89) $ \blacksquare $

Problem 9

1. $ \boxed{\text{Stationary r.f.}} $
2. $ \boxed{\text{Synchronous r.f.}} $
3. $ \boxed{\text{Rotor r.f.}} $ (see pp. 97)

Problem 10

$ \boxed{\text{No}} $

The line-to-line, zero-sequence voltage for a delta-connected component is $ v_{\ell\ell, 0s} = \frac{1}{3} \left(v_{ab} + v_{bc} + v_{ca}\right) = \frac{1}{3} \sum_{k \, \in \, \left\{ab, bc, ca\right\}} v_{\ell\ell, k} $. By Kirchhoff's Voltage Law (KVL) around the delta, $ \sum_{k \, \in \, \left\{ab, bc, ca\right\}} v_{\ell\ell, k} = 0 $. Therefore, no line-to-line, zero-sequence voltage can develop. (see pp. 111-113)

Problem 11

$ \boxed{\text{Yes}} $

Consider an example in which $ i_{ab} = 2 \, \textrm{A} $, $ i_{ba} = -1 \, \textrm{A} $, and $ i_{ab} = 1 \, \textrm{A} $. It is clear that $ \sum_{k \, \in \, \left\{ab, bc, ca\right\}} i_{\Delta, k} = 2 \, \textrm{A} \neq 0 $. With simple Kirchhoff's Current Law (KCL) equations, it is found that $ i_{a} = i_{ab} - i_{ca} = 1 \, \textrm{A} $, $ i_{b} = i_{bc} - i_{ab} = -3 \, \textrm{A} $, and $ i_{c} = i_{ca} - i_{bc} = 2 \, \textrm{A} $. The line currents are balanced at this moment, but the zero-sequence delta-branch current is $ i_{\Delta, k} = \frac{1}{3} \sum_{k \, \in \, \left\{ab, bc, ca\right\}} i_{\Delta, k} \neq 0 $. Therefore, there is a condition that can lead to zero-sequence current. (see pp. 111-113)

Discussion

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