Excercises Electric Machinery Fundamentals

Chapter 2 Problem 2-6



In [1]:

    
%pylab inline
%precision 4









    



Populating the interactive namespace from numpy and matplotlib






    Out[1]:





'%.4f'

Description

A 1000-VA 230/115-V transformer has been tested to determine its equivalent circuit. The results of the tests are shown below:

Open-circuit test	Short-circuit test
(on secondary side)	(on primary side)
$V_{OC} = 115\,V$	$V_{SC} = 17.1\,V$
$I_{OC} = 0.11\,A$	$I_{SC} = 8.7\,A$ (should really be 4.35 A)
$P_{OC} = 3.9\,W$	$P_{SC} = 38.1\,W$

Note: As was correctly pointed out in the lecture, the current $I_{sc}$ is actually to large by a factor of 2. I assume it simply being an oversight an the correct value given should be $I_{SC} = 4.35\,A = I_n$ (i.e., equal to the nominal current which can be calculated from $I_n = \frac{1000\,VA}{230\,V}$).

The following solution is based on the incorrectly given current of $I_{SC} = 8.7\,A$ but you can simply set the value of Isc below to 4.35 and run "Cell → Run all" to recaculate all values based on the correct $I_{SC}$.

See also Example 2-2 in the book which is similar to this problem but where the correct values were used.



In [2]:

    
Voc = 115.0  # [V] 
Ioc =   0.11 # [A]
Poc =   3.9  # [W]
Vsc =  17.1  # [V] 
Isc =   8.7  # [A]   replace with 4.35 so see alternative(correct solutions)
Psc =  38.1  # [W]

(a)

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

(b)

Find the transformer’s voltage regulation at rated conditions and
1. for 0.8 PF lagging
2. for 1.0 PF
3. for 0.8 PF leading.

(c)

Determine the transformer’s efficiency at rated conditions and 0.8 PF lagging.

SOLUTION

(a)

Solution based on active and reactive power:

Open-circuit test results (ignore $R_{EQ}$ and $X_{EQ}$ and all values referred to the secondary side):



In [3]:

    
Rc = Voc**2/Poc
Rc









    Out[3]:





3391.0256



In [4]:

    
Soc = Voc*Ioc
Qoc = sqrt(Soc**2 - Poc**2)
Xm = Voc**2/Qoc
Xm









    Out[4]:





1098.9873

Short-circuit test results (ignore $R_C$ and $X_M$ and all values referred to the primary side):



In [5]:

    
Req = Psc/Isc**2
Req









    Out[5]:





0.5034



In [6]:

    
Ssc = Vsc*Isc
Qsc = sqrt(Ssc**2 - Psc**2)
Xeq = Qsc/Isc**2
Xeq









    Out[6]:





1.9000

Solution based on complex angle:



In [7]:

    
Yex_amp = Ioc/Voc
Yex_amp









    Out[7]:





0.0010



In [8]:

    
theta_ex = arccos(Poc/Soc)
Yex = Yex_amp * exp(-1j*theta_ex)
Yex









    Out[8]:





(0.00029489603024574667-0.00090992866136550029j)



In [9]:

    
Rc = 1/real(Yex)
Rc









    Out[9]:





3391.0256



In [10]:

    
Xm = -1/imag(Yex)
Xm









    Out[10]:





1098.9873



In [11]:

    
Zeq_amp = Vsc/Isc
Zeq_amp









    Out[11]:





1.9655



In [12]:

    
theta_eq = arccos(Psc/Ssc)
Zeq = Zeq_amp * exp(1j*theta_eq)
Zeq









    Out[12]:





(0.50336900515259597+1.8999678078354438j)



In [13]:

    
Req = 1*real(Zeq)
Req









    Out[13]:





0.5034



In [14]:

    
Xeq = 1*imag(Zeq)
Xeq









    Out[14]:





1.9000

The resulting equivalent circuit is:

To convert the equivalent circuit to the secondary side, divide each series impedance by the square of the turns ratio ( a = 230/115 = 2). Note that the excitation branch elements are already on the secondary side. The resulting equivalent circuit is shown below:



In [15]:

    
a = 230/115
a









    Out[15]:





2.0000



In [16]:

    
Rcs = Rc # measurements were already done on the secondary side
Rcs









    Out[16]:





3391.0256



In [17]:

    
Xms = Xm # measurements were already done on the secondary side
abs(Xms)









    Out[17]:





1098.9873



In [18]:

    
Reqs = Req/a**2
Reqs









    Out[18]:





0.1258



In [19]:

    
Xeqs = Xeq/a**2
Xeqs









    Out[19]:





0.4750

(b)

To find the required voltage regulation, we will use the equivalent circuit of the transformer referred to the secondary side. The rated secondary current is



In [20]:

    
Sn = 1000.0    # [VA] nominal apparent power
Vs = 115.0     # [VA] nominal secondary voltage
Is_amp = Sn/Vs # amplitude of the current
Is_amp









    Out[20]:





8.6957

We will now calculate the primary voltage referred to the secondary side and use the voltage regulation equation for each power factor. The calulations are taking the secondary voltage $V_S = 115\,V\angle 0^\circ$ as a reference. $$ \vec{V_P}' = \vec{V_S} + Z_{EQ} \vec{I_S}$$

1) 0.8 PF lagging



In [21]:

    
PF = 0.8
theta = -arccos(PF)
Is = Is_amp *exp(1j*theta)
print('Is = {:.1f} A ∠{:.1f}°'.format(
            abs(Is), theta/pi*180))









    



Is = 8.7 A ∠-36.9°



In [22]:

    
Vps_lagg = Vs + (Reqs + 1j*Xeqs)*Is
Vps_lagg_angle = arctan(imag(Vps_lagg)/real(Vps_lagg))
print('Vps = {:.1f} V ∠{:.2f}°'.format(
            abs(Vps_lagg), Vps_lagg_angle/pi*180))









    



Vps = 118.4 V ∠1.28°

And the voltage regulation with: $$ VR = \frac{V_P/a-V_S}{V_S} = \frac{V_P'-V_S}{V_S} $$ is therefore:



In [23]:

    
VR = (abs(Vps_lagg) - abs(Vs)) / abs(Vs) * 100  # in percent
print('VR = {:.2f} %'.format(VR))
print('===========')









    



VR = 2.94 %
===========

2) 1.0 PF



In [24]:

    
PF = 1.0
theta = arccos(PF)
Is = Is_amp *exp(1j*theta)
print('Is = {:.1f} A ∠{:.1f}°'.format(
            abs(Is), angle(Is, deg=True)))









    



Is = 8.7 A ∠0.0°



In [25]:

    
Vps = Vs + (Reqs + 1j*Xeqs)*Is
Vps_angle = arctan(imag(Vps)/real(Vps))
print('Vps = {:.1f} V ∠{:.2f}°'.format(
            abs(Vps), Vps_angle/pi*180))









    



Vps = 116.2 V ∠2.04°

And the voltage regulation with: $VR = \frac{V_P/a-V_S}{V_S} = \frac{V_P'-V_S}{V_S}$ is therefore:



In [26]:

    
VR = (abs(Vps) - abs(Vs)) / abs(Vs) * 100  # in percent
print('VR = {:.2f} %'.format(VR))
print('===========')









    



VR = 1.02 %
===========

3) 0.8 PF leading



In [27]:

    
PF = 0.8
theta = +arccos(0.8)
Is = Is_amp *exp(1j*theta)
print('Is = {:.1f} A ∠{:.1f}°'.format(
            abs(Is), theta/pi*180))









    



Is = 8.7 A ∠36.9°



In [28]:

    
Vps_lead = Vs + (Reqs + 1j*Xeqs)*Is
Vps_lead_angle = arctan(imag(Vps_lead)/real(Vps_lead))
print('Vps = {:.1f} V ∠{:.1f}°'.format(
            abs(Vps_lead), Vps_lead_angle/pi*180))









    



Vps = 113.5 V ∠2.0°

And the voltage regulation with: $VR = \frac{V_P/a-V_S}{V_S} = \frac{V_P'-V_S}{V_S}$ is therefore:



In [29]:

    
VR = (abs(Vps_lead) - abs(Vs)) / abs(Vs) * 100  # in percent
print('VR = {:.2f} %'.format(VR))
print('============')









    



VR = -1.33 %
============

(c)

At rated conditions and 0.8 PF lagging, the output power of this transformer is: $$ P_\text{OUT} = V_s I_s \cos(\theta) = V_s I_s \cdot PF $$



In [30]:

    
Pout = abs(Vs) * abs(Is) * PF
Pout









    Out[30]:





800.0000

watts. The copper and core losses of this transformer are $P_\text{CU} = I_s^2 \cdot R_{\text{EQ}_s}$:



In [31]:

    
Pcu = abs(Is)**2 * Reqs
Pcu









    Out[31]:





9.5155

watts. And the core losses are $P_\text{core} = V_P'^2/R_c$:



In [32]:

    
Pcore = abs(Vps_lagg)**2 / Rcs
Pcore









    Out[32]:





4.1328

watts. Therefore the efficiency of this transformer at these conditions is $$ \eta = \frac{P_\text{OUT}}{P_\text{OUT}+P_\text{CU}+P_\text{core}} \times 100\,\%$$



In [33]:

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









    



η = 98.3 %
==========

Note:

One might argue that $115\,V\angle 0^\circ$ really should be seen as the no-load voltage on the secondary side and not as the full-load voltage. I.e., $V_P/a = 115\,V\angle 0^\circ$. Think about what kind of effect this has on the voltage regulation and efficiency calculations in (b) and (c) above. What would be changed in the equations above in order to calculate this variant?