Excercises Electric Machinery Fundamentals

Chapter 4 Problem 4-13



In [1]:

    
%pylab inline
%precision 1









    



Populating the interactive namespace from numpy and matplotlib






    Out[1]:





'%.1f'

Description

A 25-MVA, 12.2-kV, 0.9-PF-lagging, three-phase, two-pole, Y-connected, 60-Hz synchronous generator was tested by the open-circuit test, and its air-gap voltage was extrapolated with the following results:

Open-citcuit test

Field current [A]	Line voltage [kV]	Extrapolated air-gap voltage [kV]
275	12.2	13.3
320	13.0	15.4
365	13.8	17.5
380	14.1	18.3
475	15.2	22.8
570	16.0	27.4

Short-circuit test

Field current [A]	Armature current [A]
275	890
320	1040
365	1190
380	1240
475	1550
570	1885

The armature resistance is $0.6\,\Omega$ per phase.

(a)

Find the unsaturated synchronous reactance of this generator in ohms per phase and in per-unit.

(b)

Find the approximate saturated synchronous reactance $X_S$ at a field current of 380 A. Express the answer both in ohms per phase and in per-unit.

(c)

Find the approximate saturated synchronous reactance at a field current of 475 A. Express the answer both in ohms per phase and in per-unit.

(d)

Find the short-circuit ratio for this generator.

(e)

What is the internal generated voltage of this generator at rated conditions?

(f)

What field current is required to achieve rated voltage at rated load?



In [2]:

    
Sbase = 25e6   # [VA]
Vbase = 12.2e3 # [V]
PF = 0.9
Ra = 0.6       # [Ohm]

SOLUTION

(a)

The unsaturated synchronous reactance of this generator is the same at any field current, so we will look at it at a field current of 380 A.



In [3]:

    
if_a = 380.0 # [A]

The extrapolated air-gap voltage at this point is 18.3 kV, and the short-circuit current is 1240 A



In [4]:

    
Vag_a = 18.3e3 # [V]
isc_a = 1240.0 # [A]

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



In [5]:

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









    



Vphi_a = 10566 V

and the armature current is:



In [6]:

    
Ia_a = isc_a
print('Ia_a = {:.0f} A'.format(Ia_a))









    



Ia_a = 1240 A

Therefore, the unsaturated synchronous impedance $Z_{s} = \sqrt{R_a^2 + X_s^2}$ is:



In [7]:

    
Zsu_a = Vphi_a / Ia_a
print('Zsu_a = {:.2f} Ω'.format(Zsu_a))









    



Zsu_a = 8.52 Ω

Which leads to the unsaturated syncronous reactance $X_{s} = \sqrt{Z_s^2 - R_a^2}$:



In [8]:

    
Xsu_a = sqrt(Zsu_a**2 - Ra**2)
print('''
Xsu_a = {:.2f} Ω
==============
'''.format(Xsu_a))









    



Xsu_a = 8.50 Ω
==============

As you can see the impact of the armature resistance is negligible small. This is also the reason why $R_a$ is often simply ignored in calculations of the synchronous reactance. Especially for larger machines.

The base impedance of this generator is:

$$Z_\text{base} = \frac{3V^2_{\phi,\text{base}}}{S_\text{base}}$$



In [9]:

    
Vphi_base = Vbase/sqrt(3)
Zbase = 3*Vphi_base**2 / Sbase
print('Zbase = {:.2f} Ω'.format(Zbase))









    



Zbase = 5.95 Ω

Therefore, the per-unit unsaturated synchronous reactance is:



In [10]:

    
xsu_a = Xsu_a / Zbase
print('''
xsu_a = {:.2f}
============
'''.format(xsu_a))









    



xsu_a = 1.43
============

(b)

The saturated synchronous reactance at a field current of 380 A can be found from the OCC and the SCC. The OCC voltage at $I_F = 380 A$ is 14.1 kV, and the short-circuit current is 1240 A.



In [11]:

    
If_b   = 380.0  # [A]
Vocc_b = 14.1e3 # [V]
isc_b  = 1240.0 # [A]

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



In [12]:

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









    



Vphi_b = 8141 V

and the armature current is:



In [13]:

    
Ia_b = isc_b
print('Ia_b = {:.0f} A'.format(Ia_b))









    



Ia_b = 1240 A

Therefore, the saturated synchronous reactance is:



In [14]:

    
Zs_b = Vphi_b / Ia_b
Xs_b = sqrt(Zs_b**2 - Ra**2)
print('''
Xs_b = {:.2f} Ω
=============
'''.format(Xs_b))









    



Xs_b = 6.54 Ω
=============

and the per-unit unsaturated synchronous reactance is:



In [15]:

    
xs_b = Xs_b / Zbase
print('''
xs_b = {:.2f}
===========
'''.format(xs_b))









    



xs_b = 1.10
===========

(c)

The saturated synchronous reactance at a field current of 475 A can be found from the OCC and the SCC. The OCC voltage at $I_F = 475 A$ is 15.2 kV, and the short-circuit current is 1550 A.



In [16]:

    
If_c   = 475.0  # [A]
Vocc_c = 15.2e3 # [V]
isc_c  = 1550.0 # [A]

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



In [17]:

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









    



Vphi_c = 8776 V

and the armature current is:



In [18]:

    
Ia_c = isc_c
print('Ia_c = {:.0f} A'.format(Ia_c))









    



Ia_c = 1550 A

Therefore, the saturated synchronous reactance is:



In [19]:

    
Zs_c = Vphi_c / Ia_c
Xs_c = sqrt(Zs_c**2 - Ra**2)
print('''
Xs_c = {:.2f} Ω
=============
'''.format(Xs_c))









    



Xs_c = 5.63 Ω
=============

and the per-unit unsaturated synchronous reactance is:



In [20]:

    
xs_c = Xs_c / Zbase
print('''
xs_c = {:.3f}
============
'''.format(xs_c))









    



xs_c = 0.946
============

(d)

The rated voltage of this generator is 12.2 kV, which requires a field current of 275 A.



In [21]:

    
If_d = 275.0 # [A]

The rated line and armature current of this generator is:



In [22]:

    
Il = Sbase / (sqrt(3) * Vbase)
print('Il = {:.0f} A'.format(Il))









    



Il = 1183 A

The field current required to produce such short-circuit current is about 365 A. Therefore, the short-circuit ratio of this generator is:



In [23]:

    
If_d_2 = 365.0 # [A]
SCR = If_d / If_d_2
print('''
SCR = {:.2f} 
==========
'''.format(SCR))









    



SCR = 0.75 
==========

(e)

The internal generated voltage of this generator at rated conditions would be calculated using the saturated synchronous reactance.



In [24]:

    
Xs_e = Xs_b
If_e = If_b
Ia_e = Il   # rated current as calculated in part d

Since the power factor is 0.9 lagging, the armature current is:



In [25]:

    
IA_e_angle = -arccos(PF)
IA_e = Ia_e * (cos(IA_e_angle) + sin(IA_e_angle)*1j)
print('IA_e = {:.0f} Ω ∠{:.2f}°'.format(*(abs(IA_e), IA_e_angle/ pi*180)))









    



IA_e = 1183 Ω ∠-25.84°

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



In [26]:

    
EA = Vphi_base + Ra*IA_e + Xs_e*IA_e*1j
EA_angle = arctan(EA.imag / EA.real)
print('''
EA = {:.0f} V ∠{:.1f}°
===================
'''.format(*(abs(EA), EA_angle/pi*180)))









    



EA = 12901 V ∠31.0°
===================

(f)

If the internal generated voltage $E_A$ is



In [27]:

    
abs(EA)









    Out[27]:





12901.0

Volts per phase, the corresponding line value would be:



In [28]:

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









    



Vline_f = 22345 V

This would require a field current of about (determined by usind the two-point form of $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)$):



In [29]:

    
If_f=(475-380)/(22.8e3-18.3e3)*(abs(EA)*sqrt(3)-18.3e3)+380
print('''
If_f = {:.0f} A
============
'''.format(If_f))









    



If_f = 465 A
============