Excercises Electric Machinery Fundamentals

Chapter 2 Problem 2-8

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



In [1]:

    
%pylab notebook









    



Populating the interactive namespace from numpy and matplotlib

Description

A 150-MVA 15/200-kV single-phase power transformer has a per-unit resistance of 1.2 percent and a perunit reactance of 5 percent (data taken from the transformer’s nameplate). The magnetizing impedance is $j80$ per unit.



In [2]:

    
V_base =  15e3   # [V] primary voltage
S_base = 150e6   # [VA] apparent power
Req_0  =   0.012 # The per-unit resistance of power transformer
Xeq_0  =   0.05  # The per_unit reactance of power transformer
Xm_0   =  80.0   # The magnetizing impedance (per unit)

(a)

Find the equivalent circuit referred to the low-voltage side of this transformer.

(b)

Calculate the voltage regulation of this transformer for a full-load current at power factor of 0.8 lagging.

(c)

Calculate the copper and core losses in transformer at the conditions in (b).

(d)

Assume that the primary voltage of this transformer is a constant 15 kV.

Plot the secondary voltage as a function of load current for currents from no-load to full-load.
Repeat this process for power factors of 0.8 lagging, 1.0, and 0.8 leading.

SOLUTION

(a)

The base impedance $Z_\text{base}$ of this transformer referred to the primary (low-voltage) side is $$Z_\text{base} = \frac{V_\text{base}^2}{S_\text{base}}$$



In [3]:

    
Z_base = V_base**2 / S_base  # [Ω]
print('Z_base = {:.1f} Ω'.format(Z_base))









    



Z_base = 1.5 Ω

So the resistance $R_\text{EQ}$, reactance $X_\text{EQ}$ and the magnetizing imoedance $X_M$ can be calculated:



In [4]:

    
Req = Req_0 * Z_base      # [Ω]
Xeq = Xeq_0 * Z_base * 1j # [Ω], Xeq is not supposed to be a real value here.
Xm  = Xm_0  * Z_base      # [Ω]

The equivalent circuit referred to the low-voltage side of the transformer is:



In [5]:

    
print('R_EQ,P =  {:>5.3f} Ω'.format(Req))
print('X_EQ,P = j{:>5.3f} Ω'.format(Xeq.imag))
print('X_M    =  {:5.0f} Ω'.format(Xm))
print('R_C    =  not specified')









    



R_EQ,P =  0.018 Ω
X_EQ,P = j0.075 Ω
X_M    =    120 Ω
R_C    =  not specified

(b)

If the load on the secondary side of the transformer is 150 MVA at 0.8 PF lagging, and the referred secondary voltage is 15 kV:



In [6]:

    
S_load = 150e6        # Load [VA]
VS = 15.0e3 + 0.0j    # [V]
PF_b = 0.8            # Lagging

Then the referred secondary current is $$I'_S = \frac{S_\text{LOAD}}{V_S} = \frac{P_\text{LOAD}}{V_S PF} $$



In [7]:

    
Is = S_load / abs(VS)                        # The absolute value of I_S [A]
Is_angle = - arccos(PF_b)                    # The angle of I_S [rad]
IS = Is*cos(Is_angle) + Is*sin(Is_angle)*1j
print('I_s = {:.0f} A ∠{:.2f}°'.format(
                abs(IS), Is_angle/pi*180))  # Turn the angle from radian into degree









    



I_s = 10000 A ∠-36.87°

The voltage on the primary side of the transformer is: $$V'_P = V'_S + I'_S Z_{\text{EQ}_P}$$



In [8]:

    
VP_b = VS + IS * (Req + Xeq)                 # [V]
VP_b_angle = arctan(imag(VP_b)/real(VP_b))
print('V_P = {:.0f} V ∠{:.2f}°'.format(
            abs(VP_b), VP_b_angle/pi*180)) # return the calculated angle in degrees









    



V_P = 15602 V ∠1.81°

Therefore the voltage regulation of the transformer is:



In [9]:

    
VR = (abs(VP_b) - abs(VS)) / (abs(VS)) * 100      # [%]
print('VR = {:.2f} %'.format(VR))
print('===========')









    



VR = 4.01 %
===========

(c)

The copper loss can be calculated with: $$P_{Cu} = (I_S)^2 R_{EQ}$$



In [10]:

    
P_Cu = abs(IS)**2 * Req
print('P_Cu = {:.1f} kW'.format(P_Cu/1000))
print('================')









    



P_Cu = 1800.0 kW
================

The core loss couldn't been found due to the unspecified parameter $R_C$

(d)

This problem is repetitive in nature, and is ideally suited for Python. A program to calculate the secondary voltage of the transformer as a function of load is shown below:

Define the values for the transformer:

(the equivalent R and X are defined above)



In [11]:

    
amps = linspace(0, 12500 ,101)  # Current values [A]
VP_d = 15000                    # Primary voltage [V]

Calculate the current calues for the three power factors.

The first array of I contains the lagging currents, the second array contains the unity currents, and the third array contains the leading currents.



In [12]:

    
I = amps * array ([[0.8 - 0.6j],   # Lagging
                   [1.0],          # Unity
                   [0.8 + 0.6j]])  # Leading

Calculate VS referred to the primary side for each current and power factor.



In [13]:

    
aVS = VP_d - (Req*I + Xeq*I)

Refer the secondary voltage back to the secondary side using the turns ratio.



In [14]:

    
VS = aVS * (200/15)

Plot the secondary voltage (in kV!) versus load:



In [15]:

    
title('Secondary voltage vs load current')
xlabel('Load current [A]')
ylabel('Secondary voltage (kV)')
plot(amps,abs(VS[0,]/1000), 'b-', linewidth = 2)
plot(amps,abs(VS[1,]/1000), 'k--',linewidth = 2)
plot(amps,abs(VS[2,]/1000), 'r-.',linewidth = 2)
legend(('0.8 PF lagging', '1.0 PF', '0.8 PF leading'), loc=3);
grid()