Answers and Discussions for
Contents
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(\vec{r}, t) $. A conservative vector field has the following three properties as a consequence:
- A line integral of the conservative vector field is path independent, thus a circulation (line integral over a closed path) is zero.
- The conservative vector field is irrotational: $ \vec{\nabla} \times \vec{F}(\vec{r}, t) = \vec{0} $.
- 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_c(\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, but phasor analysis requires time-varying quantities with sinusoidal steady state behavior. Therefore, an asynchronous reference frame such as the stationary r.f. can bring equations into phasor form using synchronous r.f. relationships. (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 easiest way to produce the third and sixth lines are to perform phasor analysis by converting the complex expressions from rectangular form to polar form. 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
- $ \boxed{\text{Stationary r.f.}} $
- $ \boxed{\text{Synchronous r.f.}} $
- $ \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)
