Excercises Electric Machinery Fundamentals

Chapter 4 Problem 4-8



In [1]:

    
%pylab inline
%precision 1









    



Populating the interactive namespace from numpy and matplotlib






    Out[1]:





'%.1f'

Description

A 200-MVA, 12-kV, 0.85-PF-lagging, 50-Hz, 20-pole, Y-connected water turbine generator has a per-unit synchronous reactance of 0.9 and a per-unit armature resistance of 0.1. This generator is operating in parallel with a large power system (infinite bus).



In [2]:

    
S   = 200e6 # [VA]
Vl  =  12e3 # [V]
PF  = 0.85  # lagging
fse = 50 # [Hz]
p   = 20
ra = 0.1 # [pu]
xs = 0.9 # [pu]

(a)

What is the speed of rotation of this generator’s shaft?

(b)

What is the magnitude of the internal generated voltage $E_A$ at rated conditions?

(c)

What is the torque angle of the generator at rated conditions?

(d)

What are the values of the generator’s synchronous reactance and armature resistance in ohms?

(e)

If the field current is held constant,

What is the maximum power possible out of this generator?
How much reserve power or torque does this generator have at full load?

(f)

At the absolute maximum power possible,

How much reactive power will this generator be supplying or consuming?
Sketch the corresponding phasor diagram. (Assume $I_F$ is still unchanged.)

SOLUTION

(a)

The speed of rotation of this generator’s shaft is:

$$n_\text{sync} = \frac{120f_{se}}{P}$$



In [3]:

    
n_sync = 120*fse / p
print('''
n_sync = {:.0f} r/min
=================='''.format(n_sync))









    



n_sync = 300 r/min
==================

(b)

The per-unit phase voltage at rated conditions is $\vec{v}_\phi = 1.0 \angle 0^\circ$



In [4]:

    
vphase = 1.0

and the per-unit phase current $\vec{i}_A$ at rated conditions is



In [5]:

    
ia_angle = -arccos(PF)  # negative because of lagging PF
ia = 1.0 * (cos(ia_angle) + sin(ia_angle)*1j)
print('ia = {:.1f} pu ∠{:.1f}°'.format(abs(ia), ia_angle/pi*180))









    



ia = 1.0 pu ∠-31.8°

, so the per-unit internal generated voltage is:

$$\vec{e_A} = \vec{v_\phi} + r_A\vec{i_A} + jx_S\vec{I_A}$$



In [6]:

    
ea = vphase + ra*ia + xs*1j*ia
ea_angle = arctan(ea.imag/ea.real)
print('ea = {:.2f} pu ∠{:.1f}°'.format(abs(ea), ea_angle/pi*180))









    



ea = 1.71 pu ∠24.6°

The base phase voltage is:



In [7]:

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









    



Vphase_base = 6928V

so the internal generated voltage is:



In [8]:

    
EA = ea * Vphase_base
print('''
EA = {:.0f} V ∠{:.1f}°
==================='''.format(abs(EA), ea_angle/pi*180))









    



EA = 11876 V ∠24.6°
===================

(c)

The torque angle of the generator is:



In [9]:

    
delta = ea_angle
print('''
δ = {:.1f}°
========='''.format(delta/pi *180))









    



δ = 24.6°
=========

(d)

The base impedance of the generator is:

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



In [10]:

    
Zbase = (3*Vphase_base**2) / S
print('Zbase = {:.2f} Ω'.format(Zbase))









    



Zbase = 0.72 Ω

Therefore the synchronous reactance is:



In [11]:

    
Xs = xs * Zbase
print('''
Xs = {:.3f} Ω
============'''.format(Xs))









    



Xs = 0.648 Ω
============

and the armature resistance is:



In [12]:

    
Ra = ra * Zbase
print('''
Ra = {:.3f} Ω
============'''.format(Ra))









    



Ra = 0.072 Ω
============

(e)

If the field current is held constant (and the armature resistance is ignored), the power out of this generator is given by:

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

The max power is given by:

$$P_\text{max} = \frac{3V_\phi E_A}{X_S}\sin{90°}$$



In [13]:

    
Pmax = (3*Vphase_base*abs(EA)) / Xs * sin(pi/2)
print('''
Pmax = {:.0f} MW
============='''.format(Pmax/1e6))









    



Pmax = 381 MW
=============

Since the full load power is:



In [14]:

    
P = S*PF
P/1e6









    Out[14]:





170.0

MW, this generator is supplying



In [15]:

    
P/Pmax*100









    Out[15]:





44.6

percent of the maximum possible power at full load conditions.

(f)

At the maximum power possible, the torque angle $\delta = 90^\circ$, so the phasor $\vec{E}_A$ will be at an angle of 90°, and the current flowing will be:

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



In [16]:

    
EA_angle_f = pi / 2 # [rad]
EA_f = abs(EA) * (cos(EA_angle_f) + sin(EA_angle_f)*1j)
Ia_f = (EA_f - Vphase_base) / (Ra + Xs *1j)
Ia_angle_f = arctan(Ia_f.imag/Ia_f.real)
print('Ia_f = {:.0f} A ∠{:.1f}°'.format(abs(Ia_f), Ia_angle_f/pi*180))









    



Ia_f = 21088 A ∠36.6°

The impedance angle $\theta$ is:



In [17]:

    
Ia_angle_f/pi*180









    Out[17]:





36.6

degrees, and the reactive power supplied by the generator is:

$$Q = 3V_\phi I_A\sin{\theta}$$



In [18]:

    
Q = 3 * Vphase_base * abs(Ia_f) * sin(Ia_angle_f)
print('''
Q = {:.0f} var
==========='''.format(Q/1e6))









    



Q = 261 var
===========