Excercises Electric Machinery Fundamentals

Chapter 1 Problem 1-19



In [1]:

    
%pylab inline
%precision %.4g









    



Populating the interactive namespace from numpy and matplotlib






    Out[1]:





'%.4g'

Description

Figure P1-14 shows a simple single-phase ac power system with three loads. The voltage source is $\vec{V} = 240\,V\angle 0^\circ$, impedances of these three loads are:

$$\vec{Z}_1 = 10\,\Omega\angle 30^\circ \quad \vec{Z}_2 = 10\,\Omega\angle 45^\circ \quad \vec{Z}_3 = 10\,\Omega\angle -90^\circ $$



In [2]:

    
V  = 240  # [V]
Z1 = 10.0 * exp(1j* 30/180*pi)  
Z2 = 10.0 * exp(1j* 45/180*pi)  
Z3 = 10.0 * exp(1j*-90/180*pi)

Answer the following questions about this power system.

(a)

Assume that the switch shown in the figure is initially open, and calculate the current I , the power factor, and the real, reactive, and apparent power being supplied by the source.

(b)

How much real, reactive, and apparent power is being consumed by each load with the switch open?

(c)

Assume that the switch shown in the figure is now closed, and calculate the current I , the power factor, and the real, reactive, and apparent power being supplied by the source.

(d)

How much real, reactive, and apparent power is being consumed by each load with the switch closed?

(e)

What happened to the current flowing from the source when the switch closed? Why?

SOLUTION

(a)

With the switch open, only loads 1 and 2 are connected to the source. The current $\vec{I}_1$ in Load 1 and the current $\vec{I}_2$ in Load 2 are:



In [3]:

    
I1 = V/Z1
I2 = V/Z2
I1_angle = arctan(I1.imag/I1.real)
I2_angle = arctan(I2.imag/I2.real)
print('''I1 = {:.1f} A ∠{:.1f}°
I2 = {:.1f} A ∠{:.1f}°'''.format(
        abs(I1), I1_angle/pi*180,
        abs(I2), I2_angle/pi*180))









    



I1 = 24.0 A ∠-30.0°
I2 = 24.0 A ∠-45.0°

Therefore the total current from the source is $\vec{I} = \vec{I}_1 + \vec{I}_2$:



In [4]:

    
I = I1 + I2
I_angle = arctan(I.imag/I.real)
print('I = {:.1f} A ∠{:.1f}°'.format(
        abs(I), I_angle/pi*180))
print('==================')









    



I = 47.6 A ∠-37.5°
==================

The power factor supplied by the source is:



In [5]:

    
PF = cos(-I_angle)
PF









    Out[5]:





0.7934

lagging (because current laggs behind voltage).

Note that the angle $\theta$ used in the power factor and power calculations is the impedance angle, which is the negative of the current angle as long as voltage is at $0^\circ$.

The real, reactive, and apparent power supplied by the source are

$$S = VI^* \quad P = VI\cos\theta = real(S) \quad Q = VI\sin\theta = imag(S)$$



In [6]:

    
So = V*conj(I)    # I use index "o" for open switch
So









    Out[6]:





(9061.2413854328806+6952.935059634513j)

Let's pretty-print that:



In [7]:

    
print('''
So = {:>7.1f} VA
Po = {:>7.1f} W
Qo = {:>7.1f} var
================'''.format(abs(So), So.real, So.imag))









    



So = 11421.4 VA
Po =  9061.2 W
Qo =  6952.9 var
================

(b)

The real, reactive, and apparent power consumed by Load 1 and by Load 2 respectively are:



In [8]:

    
S1o = V*conj(I1)
S1o









    Out[8]:





(4988.3063257983667+2879.9999999999991j)



In [9]:

    
S2o = V*conj(I2)
S2o









    Out[9]:





(4072.9350596345143+4072.9350596345143j)

Let's pretty-print that:



In [10]:

    
print('''
S1o = {:>6.1f} VA
P1o = {:>6.1f} W
Q1o = {:>6.1f} var
----------------
S2o = {:>6.1f} VA
P2o = {:>6.1f} W
Q2o = {:>6.1f} var
================'''.format(abs(S1o), S1o.real, S1o.imag,
                           abs(S2o), S2o.real, S2o.imag))









    



S1o = 5760.0 VA
P1o = 4988.3 W
Q1o = 2880.0 var
----------------
S2o = 5760.0 VA
P2o = 4072.9 W
Q2o = 4072.9 var
================

As expected, the real and reactive power supplied by the source are equal to the sum of the real and reactive powers consumed by the loads.

(c)

With the switch closed, all three loads are connected to the source. The current in Loads 1 and 2 is the same as before. The current $\vec{I}_3$ in Load 3 is:



In [11]:

    
I3 = V/Z3
I3_angle = arctan(I3.imag/I3.real)
print('I3 = {:.1f} A ∠{:.1f}°'.format(abs(I3), I3_angle/pi*180))









    



I3 = 24.0 A ∠90.0°

Therefore the total current from the source is $\vec{I} = \vec{I}_1 + \vec{I}_2 + \vec{I}_3$:



In [12]:

    
I = I1 + I2 + I3
I_angle = arctan(I.imag/I.real)
print('I = {:.1f} A ∠{:.1f}°'.format(abs(I), I_angle/pi*180))
print('=================')









    



I = 38.1 A ∠-7.5°
=================

The power factor supplied by the source is:



In [13]:

    
PF = cos(-I_angle)
PF









    Out[13]:





0.9914

lagging (because current laggs behind voltage).

The real, reactive, and apparent power supplied by the source are

$$S = VI^* \quad P = VI\cos\theta = real(S) \quad Q = VI\sin\theta = imag(S)$$



In [14]:

    
Sc = V*conj(I)    # I use index "c" for closed switch
Sc









    Out[14]:





(9061.2413854328806+1192.9350596345134j)

Let's pretty-print that:



In [15]:

    
print('''
Sc = {:.1f} VA
Pc = {:.1f} W
Qc = {:.1f} var
==============='''.format(abs(Sc), Sc.real, Sc.imag))









    



Sc = 9139.4 VA
Pc = 9061.2 W
Qc = 1192.9 var
===============

(d)

The real, reactive, and apparent power consumed by Load 1, Load 2 and by Load 3 respectively are:



In [16]:

    
S1c = V*conj(I1)
S1c









    Out[16]:





(4988.3063257983667+2879.9999999999991j)



In [17]:

    
S2c = V*conj(I2)
S2c









    Out[17]:





(4072.9350596345143+4072.9350596345143j)



In [18]:

    
S3c = V*conj(I3)
S3c









    Out[18]:





(3.5269827815443778e-13-5760j)



In [19]:

    
print('''
S1c = {:>7.1f} VA
P1c = {:>7.1f} W
Q1c = {:>7.1f} var
-----------------
S2c = {:>7.1f} VA
P2c = {:>7.1f} W
Q2c = {:>7.1f} var
-----------------
S3c = {:>7.1f} VA
P3c = {:>7.1f} W
Q3c = {:>7.1f} var
================='''.format(abs(S1c), S1c.real, S1c.imag,
                            abs(S2c), S2c.real, S2c.imag,
                            abs(S3c), S3c.real, S3c.imag))









    



S1c =  5760.0 VA
P1c =  4988.3 W
Q1c =  2880.0 var
-----------------
S2c =  5760.0 VA
P2c =  4072.9 W
Q2c =  4072.9 var
-----------------
S3c =  5760.0 VA
P3c =     0.0 W
Q3c = -5760.0 var
=================

(e)

The current flowing decreased when the switch closed, because most of the reactive power being consumed by Loads 1 and 2 is being supplied by Load 3. Since less reactive power has to be supplied by the source, the total current flow decreases.