Power Systemanalysis and Control (EET 408) Laboratory Module s1

Power SystemAnalysis and Control (EET 408) Laboratory Module

EXPERIMENT 6

ANALYSIS THE UNSYMMETRICAL FAULT

1. OBJECTIVE:

To analysis the Unsymmetrical fault by means of MiPower soft ware

2. EQUIPMENT:

MiPower software

3. INTRODUCTION:

For unsymmetrical fault, the solution be used the symmetrical components, by applying Thevenin’s Theorem, which allows us to find the current in the fault by replacing the entire system with a single generator and series impedance, and the bus impedance matrix is applied to the analysis of unsymmetrical faults. Actually in a symmetrical fault we used the positive-sequence network to solve for the fault current, and voltages and currents at various other busses during the fault. It is also possible to create bus impedance matrices for negative- and zero- sequence networks, and they can be strung together in different ways to determine voltages and currents for unsymmetrical faults in a power system. And we have seen that when the network is balanced, the symmetrical components impedances are diagonal, so, that is possible to calculate Zbus separately for zero-, positive-, and negative- sequence networks. And also, we have observed that for a fault at bus k, the diagonal element in the k axis of the bus impedance matrix Zbus is the Thevenin impedance to the point of fault. In order to obtain a solution for the unsymmetrical faults, the bus impedance matrix for each sequence network is obtained separately. The fault formulas for various unsymmetrical faults are summarized as below.

1. Assumed the fault occurs at phase a for single line-to ground, phase b and c for double line-to-ground and double line-to-ground fault.

2. The fault occurs at bus k through fault impedance Zf.

3. The diagonal impedance are Zkk,0, Zkk,1 and Zkk,2 for zero-, positive- and negative sequence networks.

KOLEJ UNIVERSITI KEJURUTERAAN UTARA MALAYSIA – Exp.1 (Revision 1) 1 Power SystemAnalysis and Control (EET 408) Laboratory Module

4. Voltage for prefault condition in bus k = Vf So, we can give for each unsymmetrical fault using [Z] as below

1. SINGLE LINE-TO-GROUND FAULT

Consider a fault between phase a and ground through an impedance Zf at bus k as shown in Fig1. The line to-ground fault requires that positive-, negative- and zero- sequence networks. Thus, in general for a fault at bus k, the symmetrical components of fault current is

V fk,a I k,a0  I k,a1  I k ,a2  Z kk ,a0  Z kk,a1  Z kk,a2  3Z f 1

The fault phase current is

I k,abc  CI k,012

Bus k of network

Fig 1 Single line-to-ground fault at bus k

2 DOUBLE LINE FAULT

For this fault, the configure sequence network as Fig. 2.

Bus k of network

Fig 2 Double line fault at bus k

The symmetrical components of the fault current as given below

I k,0  0

Vk, f I k,1  I k,2  Z kk,1  Z kk,2  Z f 3 And the fault phase current is

3. DOUBLE LINE-TO-GROUND FAULT USING [z]

The diagram sequence network for double line-to-ground fault through an impedance Zf as Fig 3. follow

Bus k of network

Fig 3. A double line-to-ground fault through an impedance Zf at bus k

The symmetrical components of the fault current as given

Vf k,a1 I ka1  Z kk,a2 (Z kk,a0  3Z f ) Z kk ,a1  Z kk,a2  Z kk,a0 5

I k,a1Z kk,a1 Vfk,a1 I k,a2  Z kk ,a2 6 and I Z V k,a1 kk,a1 fk, a1 I k,a0  Z kk,a0 7 The phase fault current at bus k

Using the sequence components of the fault current the symmetrical components of the ith bus voltages during fault are obtained

Vif ,a0  0  Z ik,a0 I k,a0

Vif ,a1  Vif ,a1  Z ik ,a1I fk ,a1

Vif ,a2  0  Z ik ,a2 I fk ,a2 9 where

Vif,a1 = phase voltage at phase a when prefault at bus i. The phase voltages during fault are

Vi,abc  CVi,012

The symmetrical components of fault current in line i to j is given by

Vif ,a0 V jf ,a0 I ij,a0  zij,a0

Vif ,a1 V jf ,a1 I ij,a1  zij,a1 11

Vif ,a2 V jf ,a2 I ij,a2  zij,a2 where zij,a0, zij,a1 and zij,a2 are the zero-, positive- and negative- sequence components of the actual line impedance between buses i and j. having obtained the symmetrical components of current, the phase fault current in line i to j is

Iij,abc  CIij,012

PROCEDURE

Figure 6.1 Line Bus +ve Seq +ve Seq. Zero Seq Zero Seq. No Structure Code Impedance Susceptance Impedance Susceptance Ref. No. p – q

1 1 1 – 4 0.035 +0.225 0.0065 0.175+j0.45 0.0039

2 2 1 - 6 0.04 + 0.215 0.0055 0.2+j0.430 0.0033

3 3 1 - 5 0.025+j0.105 0.0045 0.125 +j0.21 0.0027 4 4 4 – 6 0.028 +j0.125 0.0035 0.14 +j0.25 0.0021

5 5 5 – 6 0.026 +j0.175 0.030 0.13 +j0.35 0.018

* De-Rated MVA for all Transmission Line is 100 MVA

Table 6.1

Bus Data

Bus Number 1 2 3 4 5 6 Bus Name Bus 1 Bus 2 Bus 3 Bus 4 Bu 5 Bus 6 Nominal Voltage 11 3.3 3.3 11 11 11 Table 6.2 Transformer data Transformer Number 1 2 Transformer Name 2T1 2T2 From Bus-to Bus- 2 – 4 3 - 5 Control Bus 2 3 Manufacturer Ref No. 1 2 De-rated MVA 100 100 Nominal tap setting 5 5 +ve /zero Sequence Impedance 0.035 0.042 +ve/zero Sequence R to X Ratio 20 20 Table 6.3 Generator Data Name GEN-1 GEN 2 GEN 3 Bus Number 1 2 3 Manufacturer Ref. No. 20 21 22 De-rated MVA (MVA) 100 200 100 Specified Voltage (kV) 11 3.3 3.3 Schedule Power (MW) 0 150 100 Reactive Power Minimum (Mvar) 0 0 0 Reactive Power Maximum (Mvar) 100 140 90 Direct axis Transient Reactance 0.2 0.15 0.25 Table 6.4

Figure 6.1 shows a single line diagram of 6 bus power system. Find the fault currents and voltages for the following type of Faults at Bus 5. 1. Single Line to Ground Fault. 2. Line-to- line fault. 3. Double Line to Ground Fault.

1. Double click the MiPower icon on your pc screen. 2. Follow the procedure in first experiment (Introduction to MiPower) to draw and input all the data and elements. 3. For transmission line, load and generation database, use the data from Table 6.1 6.2 and 6.3. 4. To solve unsymmetrical fault, on network editor screen, click Solve  Short circuit Analysis, Short circuit analysis screen analysis will appear. 5. Click Study Info button. In the short circuit studies screen, specify a type of fault. For first case, select Single Line to Ground fault. Select bus to simulate on fault. 6. On short Circuit Output Options screen, select details data as shown below.

7. After click Execute. Draw a single line drawing together with the short circuit data. Print the result. 8. From the result obtain, fill in and complete Table6.4 in the answer sheet.

EXPERIMENT 5 ANALYSIS OF UNSYMMETRICAL FAULT Name: ______Date:______Matrix No ______Experiment No: ______

Fault at Bus 5

Single Line to Line to Line Double Line to Ground fault Fault Ground Fault Sequence +ve Fault -ve Current zero

Phase Phase A Fault Phase B Current Phase C

FAULT MVA

TOTAL FAULT CURRENT

QUESTIONS

1. Repeat the above procedure for Fault at Bus 4. Obtain line current for every transmission line and voltage magnitude for each bus. 2. A synchronous machine A generating I pu voltage is connected through a star- star transformer, reactance 0.12pu, to two lines in parallel. The other ends of the line are connected through a star-star transformer of reactance 0.1 pu to a

second machine B, also generating 1 pu voltage. for both transformer X1=X2=X0.

Machine A 0.3 0.2 0.05 machine B 0.25 0.15 0.03

For each line X1=X2=X0= 0.70

Figure 1

a) Draw an equivalent circuit for positive, negative sequence and zero sequence for above system b) Calculate the fault current.

DISCUSSION

CONCLUSION

KOLEJ UNIVERSITI KEJURUTERAAN UTARA MALAYSIA – Exp.1 (Revision 1) 11