Excercises Electric Machinery Fundamentals

Chapter 2 Problem 2-4



In [1]:

    
%pylab inline









    



Populating the interactive namespace from numpy and matplotlib

Description

The secondary winding of a real transformer has a terminal voltage of

$v_S(t) = \sqrt{2}\cdot 200 \sin(2\pi\cdot 60\,Hz\cdot t)\,V = 282.8 \sin(377t)\,V$.

The turns ratio of the transformer is

$100:200 \rightarrow a = 0.5$ .

If the secondary current of the transformer is

$i_S(t) = \sqrt{2}\cdot 5 \sin(377t - 36.87°)\, A$.

What is the primary current of this transformer?
What are its voltage regulation and efficiency?

The impedances of this transformer referred to the primary side are: $$R_{eq} = 0.20\,\Omega \qquad R_{C} = 300\,\Omega$$ $$X_{eq} = 0.80\,\Omega \qquad X_{M} = 100\,\Omega$$



In [2]:

    
VS_m     = sqrt(2) * 200 # [V]
VS_angle = 0             # [rad]|[deg]
IS_m     = sqrt(2)*5     # [A]
IS_deg   = -36.87        # angle of IS in [deg]
IS_rad   = IS_deg/180*pi # angle of IS in [rad]
a        =   0.5
Req      =   0.20        # [Ohm]
Rc       = 300.0         # [Ohm]
Xeq      =   0.80        # [Ohm]
Xm       = 100.0         # [Ohm]

SOLUTION

The equivalent circuit of this transformer is shown below. (Since no particular equivalent circuit was specified, we are using the approximate equivalent circuit referred to the primary side.)

The secondary voltage and current are:



In [3]:

    
VS = VS_m / sqrt(2) * (cos(VS_angle) + sin(VS_angle)*1j) # [V]
IS = IS_m / sqrt(2) * (cos(IS_rad) + sin(IS_rad)*1j)     # [A]
print('VS = {:>5.1f} V ∠{:>3.0f}°'.format(
            abs(VS), VS_angle))
print('IS = {:>5.1f} A ∠{:>3.2f}°'.format(
            abs(IS), (IS_deg)))









    



VS = 200.0 V ∠  0°
IS =   5.0 A ∠-36.87°

The secondary voltage referred to the primary side is: $$\vec{V}'_S = a\vec{V}_S$$



In [4]:

    
VSp = a * VS    
print('VSp = {:.1f} V ∠{:.0f}°'.format(
            abs(VSp), VS_angle)) # the turns ratio has no effect on the angle









    



VSp = 100.0 V ∠0°

The secondary current referred to the primary side is: $$\vec{I}'_S = \frac{\vec{I}_S}{a}$$



In [5]:

    
ISp = IS / a
print('ISp = {:.1f} A ∠{:.2f}°'.format(
            abs(ISp), IS_deg))  # the turns ratio has no effect on the angle









    



ISp = 10.0 A ∠-36.87°

The primary circuit voltage is given by:

$$\vec{V}_P = \vec{V}'_S + \vec{I}'_S(R_{eq} + jX_{eq})$$



In [6]:

    
VP = VSp + ISp * (Req + Xeq*1j)
VP_angle = arctan(VP.imag/VP.real)
print('VP = {:.1f} V ∠{:.1f}°'.format(
            abs(VP), degrees(VP_angle)))









    



VP = 106.5 V ∠2.8°

The excitation current of this transformer is: $$\vec{I}_{EX} = \vec{I}_C + \vec{I}_M$$



In [7]:

    
IC = VP / Rc
IM = VP / (Xm*1j)
Iex = IC + IM
Iex_angle = arctan(Iex.imag/Iex.real)
print('Iex = {:.2f} A ∠{:.1f}°'.format(
            abs(Iex), Iex_angle/pi*180))  # angle in [deg]









    



Iex = 1.12 A ∠-68.8°

Therefore, the total primary current of this transformer is:

$$\vec{I}_P = \vec{I}'_S + \vec{I}_{EX}$$



In [8]:

    
IP = ISp + Iex
IP_angle = arctan(IP.imag/IP.real)
print('VS = {:.1f} V ∠{:.1f}°'.format(
            abs(IP), IP_angle/pi*180)) # angle in [deg]
print('===================')









    



VS = 11.0 V ∠-40.0°
===================

The voltage regulation of the transformer at this load is:

$$VR = \frac{V_P - aV_S}{aV_S} \cdot 100\%$$



In [9]:

    
VR = (abs(VP) - a*abs(VS))/(a*abs(VS)) * 100
print('VR = {:.1f}%'.format(VR))
print('=========')









    



VR = 6.5%
=========

The input power to this transformer is: $$P_{IN} = V_PI_P\cos\theta$$



In [10]:

    
Pin = abs(VP) * abs(IP) * cos(VP_angle - IP_angle)
print('Pin = {:.0f} W'.format(Pin))









    



Pin = 858 W

The output power from this transformer is: $$P_{OUT} = V_SI_S\cos\theta$$



In [11]:

    
Pout = abs(VS) * abs(IS) * cos(VS_angle - IS_rad)
print('Pout = {:.0f} W'.format(Pout))









    



Pout = 800 W

Therefore, the transformer's efficiency is:

$$\eta = \frac{P_{OUT}}{P_{IN}} \cdot 100\,\%$$



In [12]:

    
eta = Pout / Pin * 100
print('η = {:.1f} %'.format(eta))
print('==========')









    



η = 93.3 %
==========