Excercises Electric Machinery Fundamentals

Chapter 4 Problem 4-7



In [1]:

    
%pylab inline
%precision 1









    



Populating the interactive namespace from numpy and matplotlib






    Out[1]:





'%.1f'

Description

A 100-MVA, 14.4-kV, 0.8-PF-lagging, 50-Hz, two-pole, Y-connected synchronous generator has a per-unit synchronous reactance of 1.1 and a per-unit armature resistance of 0.011.



In [2]:

    
Vl  = 14.4e3 # [V]
S   = 100e6  # [VA]
ra  = 0.011  # [pu]
xs  = 1.1    # [pu]
PF  = 0.8
p   = 2
fse = 50 # [Hz]

(a)

What are its synchronous reactance and armature resistance in ohms?

(b)

What is the magnitude of the internal generated voltage $E_A$ at the rated conditions?
What is its torque angle $\delta$ at these conditions?

(c)

Ignoring losses in this generator

What torque must be applied to its shaft by the prime mover at full load?

SOLUTION

The base phase voltage of this generator is:



In [3]:

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









    



Vphase_base = 8314 V

Therefore, the base impedance of the generator is:

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



In [4]:

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









    



Zbase = 2.074 Ω

(b)

The generator impedance in ohms are:



In [5]:

    
Ra = ra * Zbase
Xs = xs * Zbase
print('''
Ra = {:.4f} Ω      Xs = {:.3f} Ω
==============================='''.format(Ra, Xs))









    



Ra = 0.0228 Ω      Xs = 2.281 Ω
===============================

(b)

The rated armature current is:

$$I_A = I_L = \frac{S}{\sqrt{3}V_T}$$



In [6]:

    
Ia_amp = S / (sqrt(3) * Vl)
print('Ia_amp = {:.0f} A'.format(Ia_amp))









    



Ia_amp = 4009 A

The power factor is 0.8 lagging, so:



In [7]:

    
Ia_angle = -arccos(PF)
Ia = Ia_amp * (cos(Ia_angle) + sin(Ia_angle)*1j)
print('Ia = {:.0f} ∠{:.2f}° A'.format(abs(Ia), Ia_angle / pi *180))









    



Ia = 4009 ∠-36.87° A

It is very often the case that especially in larger machines the armature resistance $R_A$ is simply neclected and one calulates the armature voltage simply as:

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

But since in this case we were given the armature resistance explicitly we should also use it.

Therefore, the internal generated voltage is

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



In [8]:

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









    



EA = 15659.5 V ∠27.6°

Therefore, the magnitude of the internal generated voltage $E_A$ is:



In [9]:

    
abs(EA)









    Out[9]:





15659.5

V, and the torque angle $\delta$ is:



In [10]:

    
EA_angle/pi*180









    Out[10]:





27.6

degrees.

(c)

Ignoring losses, the input power would equal the output power. Since



In [11]:

    
Pout = PF * S
print('Pout = {:.1F} MW'.format(Pout/1e6))









    



Pout = 80.0 MW

and,

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



In [12]:

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









    



n_sync = 3000 r/min

the applied torque would be:

$$\tau_\text{app} = \tau_\text{ind} = \frac{P_\text{out}}{\omega_\text{sync}}$$



In [13]:

    
w_sync = n_sync * (2*pi/60.0)
tau_app = Pout / w_sync
print('''
τ_app = {:.0f} Nm
================='''.format(tau_app))









    



τ_app = 254648 Nm
=================