Excercises Electric Machinery Fundamentals

Chapter 4 Problem 4-6



In [1]:

    
%pylab inline









    



Populating the interactive namespace from numpy and matplotlib

Description

The internal generated voltage $E_A$ of a 2-pole, $\Delta$-connected, 60 Hz, three phase synchronous generator is $14.4\,kV$, and the terminal voltage $V_T$ is $12.8\,kV$. The synchronous reactance of this machine is $4\,\Omega$, and the armature resistance can be ignored.



In [2]:

    
p       = 2
fm      = 60     # [Hz]
Ea      = 14.4e3 # [V]
V_phase = 12.8e3 # [V]
Xs      = 4.0    # [Ohm]

(a)

If the torque angle of the generator $\delta = 18^\circ$

How much power is being supplied by this generator at the current time?

(b)

What is the power factor of the generator at this time?

(c)

Sketch the phasor diagram under these circumstances.

(d)

Ignoring losses in this generator

What torque must be applied to its shaft by the prime mover at these conditions?

SOLUTION

(a)

If resistance is ignored, the output power from this generator is given by:

$$P = \frac{3V_\phi E_A}{X_S} \sin{\delta}$$



In [3]:

    
delta = 18 /180.0 * pi # [rad]
P = (3*V_phase*Ea) / Xs * sin(delta)
print('''
P = {:.1f} MW
==========='''.format(P/1e6))









    



P = 42.7 MW
===========

(b)

The phase current flowing in this generator can be calculated from:

$$\vec{E}_A = \vec{V}_\phi + jX_S\vec{I}_A$$$$\vec{I}_A = \frac{\vec{E}_A-\vec{V}_\phi}{jX_S}$$



In [4]:

    
EA = Ea * (cos(delta) + sin(delta) * 1j)
Ia = (EA - V_phase) / (Xs*1j)
Ia_angle = arctan(Ia.imag/Ia.real)
print('Ia = {:.0f} ∠{:.1f}° A'.format(abs(Ia), Ia_angle/pi*180))
print('''
Therefore the impedance angle θ = {:.1f}°, and the power factor is cos({:.1f}°) = {:.2} lagging.
=============================================================================================
'''.format(-Ia_angle/pi*180, Ia_angle/pi*180, cos(Ia_angle)))









    



Ia = 1135 ∠-11.4° A

Therefore the impedance angle θ = 11.4°, and the power factor is cos(-11.4°) = 0.98 lagging.
=============================================================================================

(c)

The phasor diagram is:

(d)

The induced torque is given by the equation:

$$P_\text{conv} = \tau_\text{ind} \omega_m$$

With no losses,

$$\tau_\text{app} = \tau_\text{ind} = \frac{P_\text{conv}}{\omega_m}$$



In [5]:

    
wm = 2*pi *fm
tau = P/ wm
print('''
τ_app = {:.0f} Nm
================='''.format(tau))









    



τ_app = 113314 Nm
=================