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. WhereRead More

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