Generator with Prime Mover

Electricity and Magnetism Level 4

A single-phase AC electric generator has a moving rotor with a magnet, which induces a voltage $V_t$ at its stator terminals. A resistor $R$ is connected across the stator terminals.

The machine model is as follows:

$\lambda = \frac{V_m}{\omega_0} \cos \theta \\ V_t = \dot{\lambda} \\ \frac{d}{dt} \Big(\frac{1}{2} I \dot{\theta}^2 \Big) = P_M-P_R$

In the model, $\lambda$ is the stator magnetic flux linkage as a function of rotor angular position $\theta$ . The terminal voltage $V_t$ is the time derivative of the flux linkage. The third equation states that the rotor is gaining kinetic energy at a rate equal to the difference between the mechanical input power and resistor power dissipation (a statement of energy conservation). The parameter $I$ is the machine inertia constant.

At time $t = 0$ , $\theta = 0$ and $\dot{\theta} = \omega_0$ . The mechanical input power $P_M$ is such that the machine maintains nominal speed $\omega_0$ (more or less). See the "details" section.

As the machine rotates, what is the magnitude of the maximum deviation of $\dot{\theta}$ from $\omega_0$ ? For practical purposes, restrict your simulation interval to $(0 \leq t \leq 10)$ , to limit the effects of any accumulated numerical errors.

Note: In this problem, we see that single-phase machines have an inherent wobble even during steady state operation. In three-phase machines, these wobbles cancel out, resulting in smoother operation.

Details and Assumptions (assume standard SI units):
1) $V_m = 120 \sqrt{2}$
2) $\omega_0 = 120 \pi$
3) $R = 15$
4) $I = 0.01$
5) $P_M = \frac{120^2}{R}$

The answer is 0.338.

1 solution

Karan Chatrath
Sep 24, 2019

The governing equation of the system is:

$\ddot{\theta} = \frac{P_m}{I\dot{\theta}} - A \sin^2(\theta)\dot{\theta}$

$A = \frac{V_m^2}{\omega_o^2RI}$

This can be re-written as a system of first-order ODE's by taking $x_1 = \theta$ and $x_2 = \dot{\theta}$ :

$\dot{x} = f(x) \implies\left[\begin{matrix} \dot{x}_1\\\dot{x}_2\end{matrix}\right] = \left[\begin{matrix} x_2\\\frac{P_m}{Ix_2} - A \sin^2(x_1)x_2 \end{matrix}\right]$

The initial conditions are: $x(0) = \left[\begin{matrix} x_1(0)\\x_2(0)\end{matrix}\right] = \left[\begin{matrix} 0\\\omega_o\end{matrix}\right]$

From here, numerical integration is used. Simulation code as follows:

import math
import numpy as np

#constants
wo  = 120*math.pi
Vm  = 120*math.sqrt(2)
R   = 15
I   = 0.01
A   = (Vm**2)/((wo**2)*R*I)
Pm  = (120**2)/R

# Initialize t and initial conditions

t  = 0
dt = 10.0**(-5.0)
xs = np.array([[0.0],[wo]])
max = 0

# Numerical Integration - Euler step:
while t <= 10.0:
    xs = xs +  np.array([xs[1],(Pm/(I*xs[1]))-(A*(math.sin(xs[0]))**2 *xs[1])])* dt

    # Computation of magnitude of deviation:
    Dev = abs(wo - xs[1])

    # Condition to check for maximum:
    if Dev > max:
        max = Dev

    t = t + dt



print(max)
# Result: 0.33840585

Generator with Prime Mover

The answer is 0.338.

1 solution

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