Excercises Electric Machinery Fundamentals

Chapter 2 Problem 2-9



In [1]:

    
%pylab inline









    



Populating the interactive namespace from numpy and matplotlib

Description

A 5000-kVA 230/13.8-kV single-phase power transformer has a per-unit resistance of 1 percent and a per-unit reactance of 5 percent (data taken from the transformer’s nameplate).



In [2]:

    
Sbase = 5000e3  # [VA]
Vp0 = 230e3     # [V]
Vs0 = 13.8e3    # [V]
Req_pu = 0.01
Xeq_pu = 0.05

The open-circuit test performed on the low-voltage side of the transformer yielded the following data:

$$V_{OC} = 13.8kV \qquad I_{OC} = 21.1A \qquad P_{OC} = 90.8kW$$



In [3]:

    
Voc = 13.8e3  # [V]
Ioc = 21.1    # [A]
Poc = 90.8e3  # [W]

(a)

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

(b)

If the voltage on the secondary side is 13.8 kV and the power supplied is 4000 kW at 0.8 PF lagging.

Find the voltage regulation of the transformer.
Find its efficiency.

SOLUTION

(a)

The open-circuit test was performed on the low-voltage side of the transformer, so it can be used to directly find the components of the excitation branch relative to the low-voltage side.

$$|Y_{EX}| = |G_{C} - jB_{M}| = \frac{I_{OC}}{V_{OC}}$$



In [4]:

    
yex = Ioc / Voc
print('yex = {:.6f}'.format(yex))









    



yex = 0.001529

$\theta = \arccos\frac{P_{OC}}{V_{OC}I_{OC}}$



In [5]:

    
theta = arccos(Poc/(Voc*Ioc))
print('theta = {:.2f}°'.format(theta/pi*180))  # print the angle in degrees









    



theta = 71.83°



In [6]:

    
Yex = yex * (cos(theta) - sin(theta)*1j)
print('Yex = {:.7f} S'.format(Yex))









    



Yex = 0.0004768-0.0014527j S

$$Y_{EX} = G_C - jB_M$$



In [7]:

    
Gc = Yex.real
Bm = -Yex.imag
Rc = 1/Gc
Xm = 1/Bm
print('Rc = {:4.0f} Ω'.format(Rc))
print('Xm = {:4.0f} Ω'.format(Xm))









    



Rc = 2097 Ω
Xm =  688 Ω

The base impedance of this transformer referred to the secondary side is:

$$Z_\text{base} = \frac{V^2_\text{base}}{S_\text{base}}$$



In [8]:

    
Vbase = Vs0
Zbase = Vbase**2 / Sbase
print('Zbase = {:.2f} Ω'.format(Zbase))









    



Zbase = 38.09 Ω

Thus:



In [9]:

    
Req = Req_pu * Zbase
Xeq = Xeq_pu * Zbase
print('Req = {:4.2f} Ω'.format(Req))
print('Xeq = {:4.2f} Ω'.format(Xeq))









    



Req = 0.38 Ω
Xeq = 1.90 Ω

The resulting equivalent circuit is shown below:



In [10]:

    
print('''
Req,s = {:.2f} Ω   Xeq,s = {:.1f} Ω 
Rc,s  = {:.0f} Ω    Xm,s = {:.0f} Ω 
==============================
'''.format(Req, Xeq, Rc, Xm))









    



Req,s = 0.38 Ω   Xeq,s = 1.9 Ω 
Rc,s  = 2097 Ω    Xm,s = 688 Ω 
==============================

(b)

If the load on the secondary side of the transformer is 4000 kW at 0.8 PF lagging and the secondary voltage is 13.8 kV, the secondary current is: $$I_S = \frac{P_{LOAD}}{V_SPF}$$



In [11]:

    
Pload_b = 4000e3 # [W]
VS_b    = 13.8e3 # [V]
PF      = 0.8

Is_b = Pload_b / (abs(VS_b) * PF)
print('Is_b = {:.1f} A'.format(Is_b))









    



Is_b = 362.3 A



In [12]:

    
Is_b_angle = -arccos(PF)    # [rad]
IS_b = Is_b * (cos(Is_b_angle) + sin(Is_b_angle)*1j)
print('IS_b = {:.1f} A ∠{:.2f}°'.format(
            abs(IS_b), Is_b_angle/pi*180))









    



IS_b = 362.3 A ∠-36.87°

The voltage on the primary side of the transformer (referred to the secondary side) is: $$\vec{V}'_P = \vec{V}_S + \vec{I}_SZ_{EQ}$$



In [13]:

    
Zeq = Req + Xeq*1j
V_P = VS_b + IS_b*Zeq
V_P_angle = arctan(V_P.imag/V_P.real)
print('V_P = {:.0f} V ∠{:.1f}°'.format(
            abs(V_P), V_P_angle/pi*180))









    



V_P = 14332 V ∠1.9°

There is a voltage drop under these load conditions.

Therefore the voltage regulation of the transformer is:

$$VR = \frac{V'_P-V_S}{V_S} \cdot 100\%$$



In [14]:

    
VR = (abs(V_P)-abs(VS_b)) / abs(VS_b) *100
print('VR = {:.2f} %'.format(VR))
print('===========')









    



VR = 3.86 %
===========

The transformer copper losses and core losses are:

$$P_{CU} = I^2_S R_{EQ,S}$$$$P_{core} = \frac{(V'_P)^2}{R_C}$$



In [15]:

    
Pcu = abs(IS_b)**2 * Req
Pcore = abs(V_P)**2 / Rc
print('Pcu   = {:.1f} kW'.format(Pcu/1000))
print('Pcore = {:.1f} kW'.format(Pcore/1000))









    



Pcu   = 50.0 kW
Pcore = 97.9 kW

Therefore the efficiency of this transformer at these conditions is:

$$\eta = \frac{P_{OUT}}{P_{OUT}+P_{CU}+P_{core}} \cdot 100\%$$



In [16]:

    
Pout = Pload_b
eta = Pout / (Pout + Pcu +Pcore) * 100
print('η = {:.1f} %'.format(eta))
print('==========')









    



η = 96.4 %
==========