Excercises Electric Machinery Fundamentals

Chapter 4 Problem 4-14



In [1]:

    
%pylab inline









    



Populating the interactive namespace from numpy and matplotlib

Description

During a short-circuit test, a Y-connected synchronous generator produces 100 A of short-circuit armature current per phase at a field current of 2.5 A. At the same field current, the open-circuit line voltage is measured to be 440 V.

(a)

Calculate the saturated synchronous reactance under these conditions.

(b)

If the armature resistance is $0.3\,\Omega$ per phase, and the generator supplies 60 A to a purely resistive Y-connected load ~~of 3 Ohms per phase at this~~ at a matching field current setting,

Determine the voltage regulation under these load conditions.

SOLUTION

(a)

The saturated synchronous reactance at a field current of 2.5 A can be found from the information supplied in the problem. The open circuit line voltage at $I_F = 2.5\,A$ is 440 V, and the short-circuit current is 100 A.



In [2]:

    
If   =   2.5 # [A]
Vocc = 440.0 # [V]
Isc  = 100.0 # [A]

Since this generator is Y-connected, the corresponding phase voltage is:



In [3]:

    
Vphi = Vocc / sqrt(3)
print('Vphi = {:.0f} V'.format(Vphi))









    



Vphi = 254 V

and the armature current is $I_A = I_{SC}$ . Therefore, the saturated synchronous reactance is:



In [4]:

    
Ia = Isc
Xs = Vphi / Ia
print('''
Xs = {:.2f} Ω
==========='''.format(Xs))









    



Xs = 2.54 Ω
===========

(b)

Assume that the desired line voltage is 440 V, which means that the phase voltage $\vec{V}_\phi = 254\,V\angle 0°$ (as calulated above). The armature current is $\vec{I}_A = 60\,A \angle 0°$, so the internal generated voltage is:

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



In [5]:

    
Vl_desired = 440   # [V]
Ra         =   0.3 # [Ohm]
Ia_b       =  60.0 # [A]
EA         = Vphi + Ra * Ia_b + Xs*1j * Ia_b
EA_angle   = arctan(EA.imag/EA.real)
print('EA = {:.0f} V ∠{:.1f}°'.format(abs(EA), EA_angle/pi*180))









    



EA = 312 V ∠29.3°

This is also the phase voltage at no load conditions. The corresponding line voltage at no load conditions would be:



In [6]:

    
Vl_nl = abs(EA) * sqrt(3)
print('Vl_nl = {:.0f} V'.format(Vl_nl))









    



Vl_nl = 540 V

The voltage regulation is:

$$VR = \frac{V_\text{T,nl} - V_\text{T,fl}}{V_\text{T,fl}} \cdot 100\%$$



In [7]:

    
VR = (Vl_nl - Vl_desired) / Vl_desired
print('''
VR = {:.1f} %
==========='''.format(VR*100))









    



VR = 22.7 %
===========