Excercises Electric Machinery Fundamentals

Chapter 2 Problem 2-3



In [1]:

    
%pylab inline
%precision 4









    



Populating the interactive namespace from numpy and matplotlib






    Out[1]:





'%.4f'

Description

Consider a simple power system consisting of an ideal voltage source, an ideal step-up transformer, a transmission line, an ideal step-down transformer, and a load. The voltage of the source is $\vec{V}_S = 480 \angle 0° V$ . The impedance of the transmission line is $Z_\text{line} = 3 + j4 \,\Omega$, and the impedance of the load is $Z_\text{load} = 30 + j40\,\Omega$ .



In [2]:

    
VS = 480.0 * exp(0j)  # [Ohm] using polar syntax
Zline = 3.0 + 4.0j    # [Ohm] using cartesian syntax
Zload = 30.0 + 40.0j  # [Ohm] using cartesian syntax

(a)

Assume that the transformers are not present in the circuit.

What is the load voltage and efficiency of the system?

(b)

Assume that transformer 1 is a 1:5 step-up transformer, and transformer 2 is a 5:1 step-down transformer.



In [3]:

    
a_b = 1.0/5.0
a_b









    Out[3]:





0.2000

What is the load voltage and efficiency of the system?

(c)

What transformer turns ratio would be required to reduce the transmission line losses to 1% of the total power produced by the generator?

SOLUTION

(a)

The equivalent circuit of this power system is shown below:

The load current in this system is: $$\vec{I}_\text{load} = \frac{\vec{V}_S}{Z_\text{line} + Z_\text{load}} $$



In [4]:

    
Iload_a = VS / (Zline+Zload)
Iload_a_angle = arctan(Iload_a.imag / Iload_a.real) # angle of Iload_a [rad]
print('Iload_a = {:.3f} A ∠{:.2f}°'.format(
            abs(Iload_a), degrees(Iload_a_angle)))









    



Iload_a = 8.727 A ∠-53.13°

The load voltage is:

$$\vec{V}_\text{load} = \vec{I}_\text{load}Z_\text{load}$$



In [5]:

    
Vload_a = Iload_a * Zload  
print('Vload_a = {:.1f} V ∠{:.0f}°'.format(
            abs(Vload_a), angle(Vload_a, deg=True)))









    



Vload_a = 436.4 V ∠0°

The power consumed by the load is: $$P_\text{load} = I_\text{load}^2R_\text{load}$$



In [6]:

    
Rload = Zload.real
Pload_a = abs(Iload_a)**2 * Rload
print('Pload_a = {:.1f} W'.format(Pload_a))









    



Pload_a = 2285.0 W

The power consumed by the transmission line is: $$P_\text{line} = I_\text{load}^2R_\text{line}$$



In [7]:

    
Rline = Zline.real
Pline_a = abs(Iload_a)**2 * Rline
print('Pline_a = {:.1f} W'.format(Pline_a))









    



Pline_a = 228.5 W

The efficiency of the power system is: $$\eta = \frac{P_\text{OUT}}{P_\text{IN}} \cdot 100\% = \frac{P_\text{load}}{P_\text{load}+P_\text{line}} \cdot 100\%$$



In [8]:

    
eta_a = Pload_a / (Pload_a + Pline_a) * 100  # [%]
print('η_a = {:.1f} %'.format(eta_a))









    



η_a = 90.9 %

(b)

The equivalent circuit of this power system is shown below:

The line impedance referred to primary side of $T_1$ is: $$Z'_\text{line} = a^2Z_\text{line}$$



In [9]:

    
Z_line_b = a_b**2 * Zline
print('Z_line_b = {:.2f} Ω'.format(Z_line_b))









    



Z_line_b = 0.12+0.16j Ω

The load impedance referred to primary side of $T_1$ is the same as the actual impedance, since the turns ratios of the step-up and step-down transformers undo each other’s changes.



In [10]:

    
Z_load_b = Zload

The resulting equivalent circuit referred to the primary side of $T_1$ is:

The load current in this system is: $$\vec{I}_\text{load} = \frac{\vec{V}_S}{Z'_\text{line}+Z'_\text{load}}$$



In [11]:

    
Iload_b = VS / (Z_line_b+Z_load_b)
Iload_b_angle = arctan(Iload_b.imag/Iload_b.real) # angle of Iload_b [rad]
print('Iload_b = {:.3f} A ∠{:.2f}°'.format(
          abs(Iload_b), degrees(Iload_b_angle)))









    



Iload_b = 9.562 A ∠-53.13°

The load voltage is:



In [12]:

    
Vload_b = Iload_b * Z_load_b
Vload_b_angle = arctan(Vload_b.imag/Vload_b.real) # angle of Vload_b [rad]
print('Vload_b = {:.0f} V ∠{:.1f}°'.format(
            abs(Vload_b), degrees(Vload_b_angle)))









    



Vload_b = 478 V ∠0.0°

The power consumed by the load is:



In [13]:

    
R_load_b = Z_load_b.real
Pload_b = abs(Iload_b)**2 * R_load_b
print('Pload_b = {:.1f} W'.format(Pload_b))









    



Pload_b = 2742.8 W

The power consumed by the transmission line is:



In [14]:

    
R_line_b = Z_line_b.real
Pline_b = abs(Iload_b)**2 * R_line_b
print('Pline_b = {:.1f} W'.format(Pline_b))









    



Pline_b = 11.0 W

The efficiency of the power system is:



In [15]:

    
eta_b = Pload_b / (Pload_b+Pline_b) * 100  # [%]
print('η_b = {:.1f} %'.format(eta_b))









    



η_b = 99.6 %

(c)

Since the power in a resistor is given by $P = I^2R$, the total power consumed in the line resistance will be directly proportional to the ratio of the line resistance to the total resistance in the circuit. The load resistance is $30\,\Omega$, and that must be $99\,\%$ of the total resistance in order for the efficient to be $1\,\%$.

Therefore, the referred line resistance must be:

$$\frac{R'_\text{line}}{R_\text{load}} = \frac{0.01}{0.99}$$



In [16]:

    
Rload = Zload.real
R_line_c = 0.01/0.99 * Rload
print('R_line_c = {:.3f} Ω'.format(R_line_c))









    



R_line_c = 0.303 Ω

Since the referred line resistance is



In [17]:

    
R_line_c









    Out[17]:





0.3030

and the actual line resistance is



In [18]:

    
Rload









    Out[18]:





30.0000

the turns ration must be: $$a^2 = \frac{R'_\text{line}}{R_\text{line}}$$



In [19]:

    
Rline = Zline.real
a_c = sqrt(R_line_c/ Rline)
print('a = {:.3f}'.format(a_c))

in order for 1% of the power to be consumed in the tranmission line.