Excercises Electric Machinery Fundamentals

Chapter 5 Problem 5-1

Note: You should first click on "Cell → Run All" in order that the plots get generated.



In [1]:

    
%pylab notebook
%precision %.4g









    



Populating the interactive namespace from numpy and matplotlib






    Out[1]:





'%.4g'

Description

A 480-V, 60 Hz, 400-hp 0.8-PF-leading eight-pole $\Delta$-connected synchronous motor has a synchronous reactance of $0.6\,\Omega$ and negligible armature resistance. Ignore its friction, windage, and core losses for the purposes of this problem. Assume that $|\vec{E}_A|$ is directly proportional to the field current $I_F$ (in other words, assume that the motor operates in the linear part of the magnetization curve), and that $|\vec{E}_A| = 480\,V$ when $I_F = 4\,A$.



In [2]:

    
Vt   = 480         # [V]
PF   =   0.8
fse  =  60         # [Hz]
p    =   8
Pout = 400 * 746   # [W] using the official "electrical horsepower" conversion
Xs   =   0.6       # [Ohm]

(a)

What is the speed of this motor?

(b)

If this motor is initially supplying 400 hp at 0.8 PF lagging.

What are the magnitudes and angles of $\vec{E}_A$ and $\vec{I}_A$ ?

(c)

How much torque is this motor producing?
What is the torque angle $\delta$ ?
How near is this value to the maximum possible induced torque of the motor for this field current setting?

(d)

If $|\vec{E}_A|$ is increased by 30 percent.

What is the new magnitude of the armature current?
What is the motor’s new power factor?

(e)

Calculate and plot the motor’s V-curve for this load condition.

SOLUTION

(a)

The speed of this motor is given by:

$$n_m = \frac{120f_{se}}{P}$$



In [3]:

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









    



n_m = 900 r/min
===============

(b)

If losses are being ignored, the output power is equal to the input power. This situation is shown in the phasor diagram below:

The line current flow under these circumstances is:

$$I_L = \frac{P}{\sqrt{3}V_T PF}$$



In [4]:

    
Pin = Pout
il = Pin / (sqrt(3) * Vt * PF)
il # [A]









    Out[4]:





448.6

Because the motor is $\Delta$-connected, the corresponding phase current is:



In [5]:

    
ia = il / sqrt(3)
ia # [A]









    Out[5]:





259

The angle of the current is:



In [6]:

    
Ia_angle = -arccos(PF)
Ia_angle /pi *180 # [degrees]









    Out[6]:





-36.87



In [7]:

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









    



Ia = 259 A ∠-36.87°

The internal generated voltage $\vec{E}_A$ is: $$\vec{E}_A = \vec{V}_\phi - jX_S\vec{I}_A$$



In [8]:

    
EA = Vt - Xs * 1j * Ia
EA_angle = arctan(EA.imag/EA.real)
print('''
EA = {:.0f} V ∠{:.1f}°
=================='''.format(abs(EA), EA_angle/pi*180))









    



EA = 406 V ∠-17.8°
==================

(c)

The induced torque is:

$$\tau_\text{ind} = \frac{P_\text{out}}{\omega_m}$$



In [9]:

    
w_m = n_m * (1.0 / 60.0) * (2.0*pi/1.0)
tau_ind = Pout / w_m
print('''
tau_ind = {:.0f} Nm
================='''.format(tau_ind))









    



tau_ind = 3166 Nm
=================

The maximum possible induced torque for the motor at this field setting is the maximum possible power divided by $\omega_m$:

$$\tau_\text{ind,max} = \frac{3V_\phi E_A}{\omega_mX_S}$$



In [10]:

    
tau_ind_max = (3*Vt*abs(EA)) / (w_m * Xs)
print('''
tau_ind_max = {:.0f} Nm
======================'''.format(tau_ind_max))









    



tau_ind_max = 10345 Nm
======================

The current operating torque is about



In [11]:

    
tau_ind/tau_ind_max









    Out[11]:





0.3061

times the maximum possible torque.

(d)

If the magnitude of the internal generated voltage $E_A$ is increased by 30%, the new torque angle can be found from the fact that $E_A\sin{\delta} \propto P =$ constant.



In [12]:

    
Ea_2 = 1.30 * abs(EA)
Ea_2 # [V]









    Out[12]:





528.1

$$\delta_2 = \arcsin\left(\frac{E_{A1}}{E_{A2}}\sin{\delta_1}\right)$$



In [13]:

    
delta_1 = EA_angle
delta_2 = arcsin(abs(EA) / Ea_2 * sin(delta_1))
delta_2/pi *180 # [degrees]









    Out[13]:





-13.62

The new armature current is: $$\vec{I}_{A2} = \frac{\vec{V}_\phi - \vec{E}_{A2}}{jX_S}$$



In [14]:

    
EA2 = Ea_2 * (cos(delta_2)+sin(delta_2)*1j)
Ia_2 = (Vt - EA2) / (Xs*1j)
Ia_2_angle = arctan(Ia_2.imag/Ia_2.real)
PF2 = cos(Ia_2_angle)
print('''
Ia_2 = {:.0f} A ∠{:.2f}°
====================
PF2  = {:.3f}
============'''.format(abs(Ia_2), Ia_2_angle/pi*180, PF2))









    



Ia_2 = 215 A ∠14.98°
====================
PF2  = 0.966
============

(e)

A Python program to calculate and plot the motor’s V-curve is shown below:

Initialize values:



In [15]:

    
Ea_plot = linspace(0.90, 1.70, 81)  * EA

Calculate $\delta_2$



In [16]:

    
delta_plot = arcsin(abs(EA) / Ea_plot * sin(delta_1))

Calculate the phasor $E_A$



In [17]:

    
EA_plot = Ea_plot * (cos(delta_plot)+sin(delta_plot)*1j)

Calculate $I_A$



In [18]:

    
Ia_plot = (Vt - Ea_plot) / (Xs*1j)

Plot the v-curve



In [19]:

    
title(r'Armature current versus $E_A$')
xlabel(r'$E_A$ [kV]')
ylabel(r'$I_A$ [A]')
plot(abs(EA_plot)/1000,abs(Ia_plot),  linewidth = 2)
grid()