# Excercises Electric Machinery Fundamentals

## 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
============