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).
Magnetic saturation leads to a nonlinear relationship between magnetic flux linkage $ \lambda $ and current $ i $. (An equivalent statement can be made for $ \vec{B} $ and $ \vec{H} $.) Despite being a magnetically nonlinear system, magnetic saturation does not correspond to any losses ($ W_{f\ell} = 0 $). Therefore, the coenergy approach to finding torque is valid since the coupling field is still conservative. (see pp. 8-10)
Problem 4
$ \boxed{\text{No}} $
Magnetic hysteresis causes magnetic flux linkage $ \lambda $ no longer to be a single-valued function of current $ i $. (An equivalent statement can be made for $ \vec{B} $ and $ \vec{H} $.) Traversing the $ \lambda-i $ loop corresponds to energy loss in the coupling field ($ W_{f\ell} > 0 $). Therefore, the coupling field is not conservative, and the coenergy approach to finding torque is invalid. (see pp. 13, 17)
Problem 5
$ \boxed{\text{Synchronous r.f.}} $
If and only 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. Since the QD0 variables in the synchronous r.f. are all constants with time and position, the time derivative will be zero in steady state. (see pp. 99-100)
Problem 6
$ \boxed{\text{Flux linkages must be a function of mechanical position}} $
In order to transfer power between the electrical and mechanical systems with a magnetic coupling field, there must be a voltage drop associated with the mechanical position of the machine through the coupling field and there must be an electromagnetic force or electromagnetic torque associated with the current or flux linkage in each electrical system through the coupling field. Physically, this means magnetic flux must pass through the mechanically moving object. The flux linkage is naturally a function of the electrical system currents, so it is known that $ \vec{\lambda}(\vec{i}, x) $ for a translational system or $ \vec{\lambda}(\vec{i}, \theta) $ for a rotational system. Thus, the coupling field energy $ W_f $ will contain terms depending on both $ \vec{i} $ and either $ x $ or $ \theta $. (pp. 17-22)
Problem 7
$ \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 8
- $ \boxed{\text{Stationary r.f.}} $
- $ \boxed{\text{Synchronous r.f.}} $
- $ \boxed{\text{Rotor r.f.}} $ (see pp. 97)
