Current-and-Voltage-Transformers: VT-Tutorial Notes | EduRev

: Current-and-Voltage-Transformers: VT-Tutorial Notes | EduRev

 Page 1


Module 2 : Current and Voltage Transformers
Lecture 9 : VT Tutorial
   Objectives
   In this lecture we will solve tutorial problems to:
Design a CCVT.
Find out the value of tuning inductance.
Find out ratio error and phase angle error.
Performance analysis of VT.
Example 1 :
 
Design a CCVT for a 132kV transmission line using the following data. Resistive Burden  = 150VA,
frequency deviation to be subjected to , phase angle error  = 40 minutes. Consider four
choices of V
2
 as 33 kV, 11 kV, 6.6 kV and 3.3 kV. Transmission line voltage V = 132 kV. The standardized
VT secondary voltage is 110 volts (L - L).
 
 
Answer:
 
Let V
2
 (L - N) be the voltage to be produced by the capacitive potential divider with capacitance values C
1
and C
2
. Let L be the value of tuning inductor. Our first task is to come up with a value of L. Here the
specification for phase angle error  is 40 minutes. Variation in frequency can be upto 
approximately. Phase angle error for change in  from  by  in the above equivalent circuit can be
calculated as follows:
 
 
 
At tuning frequency,  
 
Page 2


Module 2 : Current and Voltage Transformers
Lecture 9 : VT Tutorial
   Objectives
   In this lecture we will solve tutorial problems to:
Design a CCVT.
Find out the value of tuning inductance.
Find out ratio error and phase angle error.
Performance analysis of VT.
Example 1 :
 
Design a CCVT for a 132kV transmission line using the following data. Resistive Burden  = 150VA,
frequency deviation to be subjected to , phase angle error  = 40 minutes. Consider four
choices of V
2
 as 33 kV, 11 kV, 6.6 kV and 3.3 kV. Transmission line voltage V = 132 kV. The standardized
VT secondary voltage is 110 volts (L - L).
 
 
Answer:
 
Let V
2
 (L - N) be the voltage to be produced by the capacitive potential divider with capacitance values C
1
and C
2
. Let L be the value of tuning inductor. Our first task is to come up with a value of L. Here the
specification for phase angle error  is 40 minutes. Variation in frequency can be upto 
approximately. Phase angle error for change in  from  by  in the above equivalent circuit can be
calculated as follows:
 
 
 
At tuning frequency,  
 
Substituting  
 
 Example 1: (contd..)
 
From figure 9.3, . For small enough , tan =  and hence from
its % phase angle error    x 100 --- (1)
Using this equation the value L for different values of V
2
 is found out.
(1) Let V
2
 be 33kV (L - N) Then; 
 
 
 
From eqn (1)
 
Example 1 : (contd..)
 
(2) 
 
 
(3) 
 
 
Page 3


Module 2 : Current and Voltage Transformers
Lecture 9 : VT Tutorial
   Objectives
   In this lecture we will solve tutorial problems to:
Design a CCVT.
Find out the value of tuning inductance.
Find out ratio error and phase angle error.
Performance analysis of VT.
Example 1 :
 
Design a CCVT for a 132kV transmission line using the following data. Resistive Burden  = 150VA,
frequency deviation to be subjected to , phase angle error  = 40 minutes. Consider four
choices of V
2
 as 33 kV, 11 kV, 6.6 kV and 3.3 kV. Transmission line voltage V = 132 kV. The standardized
VT secondary voltage is 110 volts (L - L).
 
 
Answer:
 
Let V
2
 (L - N) be the voltage to be produced by the capacitive potential divider with capacitance values C
1
and C
2
. Let L be the value of tuning inductor. Our first task is to come up with a value of L. Here the
specification for phase angle error  is 40 minutes. Variation in frequency can be upto 
approximately. Phase angle error for change in  from  by  in the above equivalent circuit can be
calculated as follows:
 
 
 
At tuning frequency,  
 
Substituting  
 
 Example 1: (contd..)
 
From figure 9.3, . For small enough , tan =  and hence from
its % phase angle error    x 100 --- (1)
Using this equation the value L for different values of V
2
 is found out.
(1) Let V
2
 be 33kV (L - N) Then; 
 
 
 
From eqn (1)
 
Example 1 : (contd..)
 
(2) 
 
 
(3) 
 
 
(4) 
 
 
 
 
 Example 1 : (contd..)
 
The values of L,  for different values of V
2
 are tabulated below.
 
V
2
L in H
 
33 kV 6722.2 0.00151
11 kV 747.2 0.0136
6.6 kV 269 0.0377
3.3 kV 67.25 0.151
 
From the above table it is clear that smaller the value of V
2
 , the smaller is the value of L and higher the
value of C
1
 and C
2
 for tuning condition. If we select too low value of V
2
 and L then capacitance values
will be beyond available limits, and if we select higher value of V
2
 and L, then CCVT's inductor will
become bulky. So a compromise solution is necessary and let us select V
2
 = 6.6 kV
For V
2
 = 6.6 kV
L = 269 H
Now, 
In this design, we have explained the basic concept for CCVT design and we assumed the transformer to
be ideal. However in real life design, the value of magnetizing impedance of transformer, resistance of
reactor etc have to be taken into account, as the ratio error  and the phase angle error  will also
get affected by these values. The next example brings out these issues.
Example 2:
The equivalent circuit of a CCVT
is shown in fig 9.4. The values of
C
1
 and C
2
 are 0.0018 and
0.0186 respectively. Tuning
Page 4


Module 2 : Current and Voltage Transformers
Lecture 9 : VT Tutorial
   Objectives
   In this lecture we will solve tutorial problems to:
Design a CCVT.
Find out the value of tuning inductance.
Find out ratio error and phase angle error.
Performance analysis of VT.
Example 1 :
 
Design a CCVT for a 132kV transmission line using the following data. Resistive Burden  = 150VA,
frequency deviation to be subjected to , phase angle error  = 40 minutes. Consider four
choices of V
2
 as 33 kV, 11 kV, 6.6 kV and 3.3 kV. Transmission line voltage V = 132 kV. The standardized
VT secondary voltage is 110 volts (L - L).
 
 
Answer:
 
Let V
2
 (L - N) be the voltage to be produced by the capacitive potential divider with capacitance values C
1
and C
2
. Let L be the value of tuning inductor. Our first task is to come up with a value of L. Here the
specification for phase angle error  is 40 minutes. Variation in frequency can be upto 
approximately. Phase angle error for change in  from  by  in the above equivalent circuit can be
calculated as follows:
 
 
 
At tuning frequency,  
 
Substituting  
 
 Example 1: (contd..)
 
From figure 9.3, . For small enough , tan =  and hence from
its % phase angle error    x 100 --- (1)
Using this equation the value L for different values of V
2
 is found out.
(1) Let V
2
 be 33kV (L - N) Then; 
 
 
 
From eqn (1)
 
Example 1 : (contd..)
 
(2) 
 
 
(3) 
 
 
(4) 
 
 
 
 
 Example 1 : (contd..)
 
The values of L,  for different values of V
2
 are tabulated below.
 
V
2
L in H
 
33 kV 6722.2 0.00151
11 kV 747.2 0.0136
6.6 kV 269 0.0377
3.3 kV 67.25 0.151
 
From the above table it is clear that smaller the value of V
2
 , the smaller is the value of L and higher the
value of C
1
 and C
2
 for tuning condition. If we select too low value of V
2
 and L then capacitance values
will be beyond available limits, and if we select higher value of V
2
 and L, then CCVT's inductor will
become bulky. So a compromise solution is necessary and let us select V
2
 = 6.6 kV
For V
2
 = 6.6 kV
L = 269 H
Now, 
In this design, we have explained the basic concept for CCVT design and we assumed the transformer to
be ideal. However in real life design, the value of magnetizing impedance of transformer, resistance of
reactor etc have to be taken into account, as the ratio error  and the phase angle error  will also
get affected by these values. The next example brings out these issues.
Example 2:
The equivalent circuit of a CCVT
is shown in fig 9.4. The values of
C
1
 and C
2
 are 0.0018 and
0.0186 respectively. Tuning
 
inductor has an inductance of
497H and resistance of 4620 .
X
m
 of the VT referred to 6.6 kV
side is 1M , core loss = 20
watts per phase, VA burden =
150VA per phase. Value of C
m
 for
compensating the current drawn
by X
m
 is equal to .
(a) Verify the appropriateness of choice of L and C
m
.
(b) Find out the nominal value of V/V
2
(c) If the frequency drops from 50Hz to 47Hz, what would be the values of ratio error and phase angle error?
 
Answer (a):
 
If and  then the value L of tuning inductor is given by
 
 
where  and  = nominal frequency. Thus, 
 
= 496.7 H which is equal to the given value of L. 
Now,  C
m
 has to be in parallel resonance with X
m
. Therefore, 
 
The value is also same as the selected value of C
m
. Hence, the selection of both L and C
m
 is appropriate.
 Answer (b):
 
 = 11.33 
 
V = 11.33 x 6.6  
Thus, this VT is connected to a 132 kV bus.
 Example 2: (contd..)
 Answer (c):
 Core loss = 20 W
 VA burden = 150VA (resistive)
Page 5


Module 2 : Current and Voltage Transformers
Lecture 9 : VT Tutorial
   Objectives
   In this lecture we will solve tutorial problems to:
Design a CCVT.
Find out the value of tuning inductance.
Find out ratio error and phase angle error.
Performance analysis of VT.
Example 1 :
 
Design a CCVT for a 132kV transmission line using the following data. Resistive Burden  = 150VA,
frequency deviation to be subjected to , phase angle error  = 40 minutes. Consider four
choices of V
2
 as 33 kV, 11 kV, 6.6 kV and 3.3 kV. Transmission line voltage V = 132 kV. The standardized
VT secondary voltage is 110 volts (L - L).
 
 
Answer:
 
Let V
2
 (L - N) be the voltage to be produced by the capacitive potential divider with capacitance values C
1
and C
2
. Let L be the value of tuning inductor. Our first task is to come up with a value of L. Here the
specification for phase angle error  is 40 minutes. Variation in frequency can be upto 
approximately. Phase angle error for change in  from  by  in the above equivalent circuit can be
calculated as follows:
 
 
 
At tuning frequency,  
 
Substituting  
 
 Example 1: (contd..)
 
From figure 9.3, . For small enough , tan =  and hence from
its % phase angle error    x 100 --- (1)
Using this equation the value L for different values of V
2
 is found out.
(1) Let V
2
 be 33kV (L - N) Then; 
 
 
 
From eqn (1)
 
Example 1 : (contd..)
 
(2) 
 
 
(3) 
 
 
(4) 
 
 
 
 
 Example 1 : (contd..)
 
The values of L,  for different values of V
2
 are tabulated below.
 
V
2
L in H
 
33 kV 6722.2 0.00151
11 kV 747.2 0.0136
6.6 kV 269 0.0377
3.3 kV 67.25 0.151
 
From the above table it is clear that smaller the value of V
2
 , the smaller is the value of L and higher the
value of C
1
 and C
2
 for tuning condition. If we select too low value of V
2
 and L then capacitance values
will be beyond available limits, and if we select higher value of V
2
 and L, then CCVT's inductor will
become bulky. So a compromise solution is necessary and let us select V
2
 = 6.6 kV
For V
2
 = 6.6 kV
L = 269 H
Now, 
In this design, we have explained the basic concept for CCVT design and we assumed the transformer to
be ideal. However in real life design, the value of magnetizing impedance of transformer, resistance of
reactor etc have to be taken into account, as the ratio error  and the phase angle error  will also
get affected by these values. The next example brings out these issues.
Example 2:
The equivalent circuit of a CCVT
is shown in fig 9.4. The values of
C
1
 and C
2
 are 0.0018 and
0.0186 respectively. Tuning
 
inductor has an inductance of
497H and resistance of 4620 .
X
m
 of the VT referred to 6.6 kV
side is 1M , core loss = 20
watts per phase, VA burden =
150VA per phase. Value of C
m
 for
compensating the current drawn
by X
m
 is equal to .
(a) Verify the appropriateness of choice of L and C
m
.
(b) Find out the nominal value of V/V
2
(c) If the frequency drops from 50Hz to 47Hz, what would be the values of ratio error and phase angle error?
 
Answer (a):
 
If and  then the value L of tuning inductor is given by
 
 
where  and  = nominal frequency. Thus, 
 
= 496.7 H which is equal to the given value of L. 
Now,  C
m
 has to be in parallel resonance with X
m
. Therefore, 
 
The value is also same as the selected value of C
m
. Hence, the selection of both L and C
m
 is appropriate.
 Answer (b):
 
 = 11.33 
 
V = 11.33 x 6.6  
Thus, this VT is connected to a 132 kV bus.
 Example 2: (contd..)
 Answer (c):
 Core loss = 20 W
 VA burden = 150VA (resistive)
 The equivalent circuit can be represented as shown below.
 
 
 at = 50 Hz
 
Example 2 : (contd..)
The frequency of interest is 47Hz. Hence values of X
m
 and other impedance can be calculated at 47Hz.
Figure 9.5 can be simplified as figure 9.6. 
Where  
 
 
 
 
 
 
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