8/16/2019 Y-Bus Exam 2 Fall 2010 Solution

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Solution Exam 2 ELCT 751 Fall 2010

This is a take-home exam that is due on Wednesday November 17, 2010. Work independently. You are

encouraged to use computer tools such as Matlab, especially for the matrix calculations. Please contact theinstructor if you lack access to an appropriate tool.

Three bus system: 230 kV lines with line impedances given below:

Branch From To x [Ω/mile] Length1 1 2 0.70 100 mile

2 1 3 0.65 110 mile

3 2 3 0.72 80 mile4 0 1 G+T

Transformer and generator at bus 1 represented as series impedance: Generator: x = 25% on 500 MVA base

(rating) and transformer: x = 10% on 500 MVA base (rating).

Put all impedances in per unit on a 100 MVA base. Assume the zero sequence line impedances are 2.5 times the

positive sequence impedances. The Y and Z matrices were constructed as part of the example used in the

November 08, 2010 lecture. Verify the per-unit matrices shown below are correct for this system.

-29.241 7.557 7.399

Y = j 7.557 -16.741 9.184

7.399 9.184 -16.583

-55.982 3.023 2.959

Y0  = j 3.023 -6.696 3.674

2.959 3.674 -6.633

A new generator and its step-up transformer are added at bus 2. The generator is rated at 780 MVA and 20 kV,

and the step-up transformer is rated at 750 MVA with voltage ratings of 20 kV (Delta) to 230 kV (Y-grounded).

The generator has a reactance of 25% on a base given by its rating, and the transformer has a reactance of 12%

on a base given by its rating. Convert both of these to per unit on a 100 MVA base, using the voltage base to be

21

3

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230 kV on the high-voltage side. Use the same approach as was done for the generator at 1 for the new

generator.

A new transmission line is added from bus 2 to bus 3 having the same impedance as the existing line.Recalculate the following nodal admittance matrices Y and Y0 for the 230 kV buses (there will still be only 3

buses). Calculate Z and Z0 by inverting the Y and Y0 matrices.

Perform a fault study for three-phase and single-phase-to-ground short circuits at each bus. Calculate the bus

voltages at each bus. Calculate the transmission line currents. Perform the study in per unit and then convert

the currents to A or kA.

Solution: One one-line diagram the node numbers shown with boxes, but the branch numbers with no box.

Branch From To X1 [pu] X0 [pu]

1 1 2 0.132 0.331

2 1 3 0.135 0.338

3 2 3 0.109 0.272

4 2 3 0.109 0.272

5 0 1 0.070 0.020

6 0 2 0.048 0.016

Three-phase short-circuit study:

Fault current for fault at each bus in turn (first in per unit, then in kA):

Y

29.241j−

7.557j

7.399j

7.557j

46.736j−

18.368j

7.399j

18.368j

25.767j−

 

 

 

 

=   Z

0.04500j

0.01716j

0.02516j

0.01716j

0.03627j

0.03078j

0.02516j

0.03078j

0.06798j

 

 

 

 

=

Y0

55.982j−3.023j

2.959j

3.023j

72.870j−

7.347j

2.959j

7.347j

10.307j−

 

 

 

 

=   Z0

0.01827j

0.00139j

0.00623j

0.00139j

0.01489j

0.01101j

0.00623j

0.01101j

0.10667j

 

 

 

 

=

2

1

3

1

24

3

5

6

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Branch currents for fault at bus f in per unit and in kA:

f = 1 f = 2 f = 3

Single-phase-to-ground short-circuit study: Fault current for fault at each bus in turn (zero sequence, then

phase A first in per unit, then in kA) on faulted phase (A):

I1f 

-22.223j

-27.571j

-14.711j

=   I3φf I b⋅

-5.578j

-6.921j

-3.693j

kA

=

 

I' br 

4.675j

3.263j

-1.631j

-1.631j

-14.286j

-7.937j

=   I' br  I b⋅

1.173j

0.819j

-0.409j

-0.409j

-3.586j

-1.992j

kA

=   I' br -3.981j

-2.779j

1.389j

1.389j

-6.760j

-20.811j

=   I' br  I b⋅

-0.999j

-0.697j

0.349j

0.349j

-1.697j

-5.224j

kA

=   I' br -0.626j

-4.661j

-5.025j

-5.025j

-5.286j

-9.424j

=   I' br  I b⋅

-0.157j

-1.170j

-1.261j

-1.261j

-1.327j

-2.366j

kA

=

21

3

1

24

3

56

6.92 kA

5.22 kA0.35 kA

1.00 kA

0.35 kA

21

3

1

24

35

6

5.58 kA

1.17 kA

0.82 kA

3.59 kA

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