fundamentals grady notes june 2007 print

7/22/2019 Fundamentals Grady Notes June 2007 Print

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Table of Contents

1. Fundamentals p. 2

2. Three Phase Circuits p. 40

3. Transformers p. 56

4. Per Unit System and Sequence Networks p. 86

5. Transmission Lines p. 115

6. Symmetrical Components p. 165

7. More Sequence Networks p. 1718. System Matrices p. 175

9. Programming Considerations p. 213

10. Short Circuits and Voltage Sags p. 228

11. Loadflow p. 270

12. Stability p. 314

13. Lightning, Grounding, and Shock Energy p. 338

14. Harmonic Filters p. 386

Fundamentals of Electric Power Systems

Prof. Mack Grady

Dept. of Electrical & Computer Engineering

University of Texas at [email protected], www.ece.utexas.edu/~grady

June 2007

Grady 2007, p. 1



M a g =

1 0

1 2

8

A n g =

0

- 9 0

4 5

- 3 6 0

1 0

3 . 6

8 E - 1

5

5 . 6

5 6 8 5 4

1 5

. 6 5 6 8 5

- 3 5 8

9 . 9

9 3 9 0 8

0 . 4

1 8 7 9 4

5 . 4

5 5 9 8 7

1 5

. 8 6 8 6 9

- 3 5 6

9 . 9

7 5 6 4 1

0 . 8

3 7 0 7 8

5 . 2

4 8 4 7 2

1 6

. 0 6 1 1 9

- 3 5 4

9 . 9

4 5 2 1 9

1 . 2

5 4 3 4 2

5 . 0

3 4 5 6 3

1 6

. 2 3 4 1 2

- 3 5 2

9 . 9

0 2 6 8 1

1 . 6

7 0 0 7 7

4 . 8

1 4 5 2

1 6

. 3 8 7 2 8

- 3 5 0

9 . 8

4 8 0 7 8

2 . 0

8 3 7 7 8

4 . 5

8 8 6 1 1

1 6

. 5 2 0 4 7

- 3 4 8

9 . 7

8 1 4 7 6

2 . 4

9 4 9 4

4 . 3

5 7 1 1 2

1 6

. 6 3 3 5 3

- 3 4 6

9 . 7

0 2 9 5 7

2 . 9

0 3 0 6 3

4 . 1

2 0 3 0 5

1 6

. 7 2 6 3 2

- 3 4 4

9 . 6

1 2 6 1 7

3 . 3

0 7 6 4 8

3 . 8

7 8 4 7 7

1 6

. 7 9 8 7 4

- 3 4 2

9 . 5

1 0 5 6 5

3 . 7

0 8 2 0 4

3 . 6

3 1 9 2 4

1 6

. 8 5 0 6 9

- 3 4 0

9 . 3

9 6 9 2 6

4 . 1

0 4 2 4 2

3 . 3

8 0 9 4 6

1 6

. 8 8 2 1 1

- 3 3 8

9 . 2

7 1 8 3 9

4 . 4

9 5 2 7 9

3 . 1

2 5 8 4 9

1 6

. 8 9 2 9 7

- 3 3 6

9 . 1

3 5 4 5 5

4 . 8

8 0 8 4

2 . 8

6 6 9 4 4

1 6

. 8 8 3 2 4

- 3 3 4

8 . 9

8 7 9 4

5 . 2

6 0 4 5 4

2 . 6

0 4 5 4 5

1 6

. 8 5 2 9 4

- 3 3 2

8 . 8

2 9 4 7 6

5 . 6

3 3 6 5 9

2 . 3

3 8 9 7 4

1 6

. 8 0 2 1 1

- 3 3 0

8 . 6

6 0 2 5 4

6

2 . 0

7 0 5 5 2

1 6

. 7 3 0 8 1

- 3 2 8

8 . 4

8 0 4 8 1

6 . 3

5 9 0 3 1

1 . 7

9 9 6 0 8

1 6

. 6 3 9 1 2

- 3 2 6

8 . 2

9 0 3 7 6

6 . 7

1 0 3 1 5

1 . 5

2 6 4 7 2

1 6

. 5 2 7 1 6

- 3 2 4

8 . 0

9 0 1 7

7 . 0

5 3 4 2 3

1 . 2

5 1 4 7 6

1 6

. 3 9 5 0 7

- 3 2 2

7 . 8

8 0 1 0 8

7 . 3

8 7 9 3 8

0 . 9

7 4 9 5 5

1 6

. 2 4 3

- 3 2 0

7 . 6

6 0 4 4 4

7 . 7

1 3 4 5 1

0 . 6

9 7 2 4 6

1 6

. 0 7 1 1 4

- 3 1 8

7 . 4

3 1 4 4 8

8 . 0

2 9 5 6 7

0 . 4

1 8 6 8 8

1 5

. 8 7 9 7

- 3 1 6

7 . 1

9 3 3 9 8

8 . 3

3 5 9

0 . 1

3 9 6 1 9

1 5

. 6 6 8 9 2

- 3 1 4

6 . 9

4 6 5 8 4

8 . 6

3 2 0 7 8

- 0 . 1

3 9 6 2

1 5

. 4 3 9 0 4

- 3 1 2

6 . 6

9 1 3 0 6

8 . 9

1 7 7 3 8

- 0 . 4

1 8 6 9

1 5

. 1 9 0 3 6

- 3 1 0

6 . 4

2 7 8 7 6

9 . 1

9 2 5 3 3

- 0 . 6

9 7 2 5

1 4

. 9 2 3 1 6

- 3 0 8

6 . 1

5 6 6 1 5

9 . 4

5 6 1 2 9

- 0 . 9

7 4 9 5

1 4

. 6 3 7 7 9

- 3 0 6

5 . 8

7 7 8 5 3

9 . 7

0 8 2 0 4

- 1 . 2

5 1 4 8

1 4

. 3 3 4 5 8

- 3 0 4

5 . 5

9 1 9 2 9

9 . 9

4 8 4 5 1

- 1 . 5

2 6 4 7

1 4

. 0 1 3 9 1

- 3 0 2

5 . 2

9 9 1 9 3

1 0

. 1 7 6 5 8

- 1 . 7

9 9 6 1

1 3

. 6 7 6 1 6

- 3 0 0

5

1 0

. 3 9 2 3

- 2 . 0

7 0 5 5

1 3

. 3 2 1 7 5

- 2 9 8

4 . 6

9 4 7 1 6

1 0

. 5 9 5 3 7

- 2 . 3

3 8 9 7

1 2

. 9 5 1 1 1

- 2 9 6

4 . 3

8 3 7 1 1

1 0

. 7 8 5 5 3

- 2 . 6

0 4 5 5

1 2

. 5 6 4 6 9

- 2 9 4

4 . 0

6 7 3 6 6

1 0

. 9 6 2 5 5

- 2 . 8

6 6 9 4

1 2

. 1 6 2 9 7

- 2 9 2

3 . 7

4 6 0 6 6

1 1

. 1 2 6 2 1

- 3 . 1

2 5 8 5

1 1

. 7 4 6 4 2

- 2 9 0

3 . 4

2 0 2 0 1

1 1

. 2 7 6 3 1

- 3 . 3

8 0 9 5

1 1

. 3 1 5 5 7

- 2 8 8

3 . 0

9 0 1 7

1 1

. 4 1 2 6 8

- 3 . 6

3 1 9 2

1 0

. 8 7 0 9 2

- 2 8 6

2 . 7

5 6 3 7 4

1 1

. 5 3 5 1 4

- 3 . 8

7 8 4 8

1 0

. 4 1 3 0 4

- 2 8 4

2 . 4

1 9 2 1 9

1 1

. 6 4 3 5 5

- 4 . 1

2 0 3

9 . 9

4 2 4 6 3

- 2 8 2

2 . 0

7 9 1 1 7

1 1

. 7 3 7 7 7

- 4 . 3

5 7 1 1

9 . 4

5 9 7 7 6

- 2 8 0

1 . 7

3 6 4 8 2

1 1

. 8 1 7 6 9

- 4 . 5

8 8 6 1

8 . 9

6 5 5 6 3

- 2 7 8

1 . 3

9 1 7 3 1

1 1

. 8 8 3 2 2

- 4 . 8

1 4 5 2

8 . 4

6 0 4 2 8

- 2 7 6

1 . 0

4 5 2 8 5

1 1

. 9 3 4 2 6

- 5 . 0

3 4 5 6

7 . 9

4 4 9 8 4

- 2 7 4

0 . 6

9 7 5 6 5

1 1

. 9 7 0 7 7

- 5 . 2

4 8 4 7

7 . 4

1 9 8 6 1

- 3 0

- 2 0

- 1 0 0

1 0

2 0

3 0

- 3 6 0

- 2 7 0

- 1 8 0

- 9 0

0

9 0

1 8 0

2 7 0

3 6 0

- 3 0

- 2 0

- 1 0 0

1 0

2 0

3 0

- 3 6 0

- 2 7 0

- 1 8 0

- 9 0

0

9 0

1 8 0

2 7 0

3 6 0

Grady 2007, p. 2



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Grady, Definitions, June 2007, Page 1

Definitions

Definitions related to power. RMS values, active and reactive power, power factor, power flow

in transmission lines. Review of per unit system and the advantages that it offers.

1. Single-Phase Definitions

The root-mean-squared (RMS) value of a periodic voltage (or current) waveform is

T ot

ot

RMS dt t vT

V )(1 2

, where T is the period of v(t).

If )sin()( t V t v , where V is the peak value, then using2

)2cos(1)(sin 2 A

A

, RMS V

becomes

2

V V RMS .

Instantaneous power flowing to a load, using the sign convention shown in Figure 1, is defined

as

)()()( t it vt p .

v(t)

i(t)

+

-

p(t) = v(t) i(t)

--->

---> p(t)

Figure 1. Instantaneous Power Flowing Into a Load

Average power flowing to a load is defined as

T ot

ot

dt t p

T

P )(1

, where T is the period of )(t p .

If )sin()( t V t v , )sin()( t I t i , then the instantaneous power becomes

)2cos()cos(2

)()()( t VI

t it vt p .

Grady 2007, p. 12




Note that )(t p has double-frequency and time-invariant components. The average value of )(t p

is the time-invariant component, or

)cos(2

VI

P .

Since2

V V RMS and

2

I I RMS , then

)cos( RMS RMS I V P ,

where )cos( is known as the displacement power factor ( DPF ).

Reactive power Q is defined as

)sin( RMS RMS I V Q .

Since 1)(cos)(sin 22 x x , then

2222 RMS RMS I V Q P .

Complex power S is defined as

jQ P S ,

so that

S P alRe , S Q Imag .

The magnitude of S is

22|| Q P S ,

which is identical to

RMS RMS I V S || .

Using voltage and current phasors ||~

RMS V V and ||~

RMS I I , the product *~~ I V is

)sin(||||)cos(||||)(|||| RMS RMS RMS RMS RMS RMS I V j I V I V .

Therefore,

Grady 2007, p. 13




*~~ I V jQ P S .

When I ~

lags V ~

, Q is positive, and the power factor is lagging. When I ~

leads V ~

, Q is

negative, and the power factor is leading. Thus, an inductive load has a lagging power factor and

absorbs Q, while a capacitive load has a leading power factor and produces Q.

The total power factor DPF is defined as

S

P DPF .

For sinusoidal systems, total power factor is identical to displacement power factor defined

previously as the cosine of the relative phase angle between voltage and current.

If P and DPF are given, Q can be calculated using

11

2

22

2

2222

DPF

P P

DPF

P P S Q , or

11

2

DPF

P Q .

The relationships among P , Q, S , and pf are shown in the power factor triangle given in Figure 2.

P

QS

pf = ------ P

S

Figure 2. Power Factor Triangle

The impact of power factor on the Q/P ratio is given below in the Table 1.

Grady 2007, p. 14




DPF Q / P

1.0 0.00

0.9 0.48

0.8 0.75

0.707 1.00

0.6 1.33

Table 1. Impact of Power Factor on Reactive Power Q

Now, consider the power flow through a purely inductive circuit element, such as a lossless

transmission line or transformer shown in Figure 3.

j X

+

-

+

-

P + j Q -->

Vs Vr r s

Figure 3. Power Flow Through a Purely Inductive Circuit Element

The active and reactive power flows, measured at the sending end S, can be shown to be

)sin( RS RS

X

V V P , )cos( RS RS

S V V X

V Q .

Usually, )( RS is small, so that

)( RS P , RS S V V

X

V Q .

Therefore, in inductive circuit elements, P tends to be proportional to voltage angle difference,

and Q tends to be proportional to voltage magnitude difference.

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A load consists of a 47 resistor and 10mH inductor in series. The load is energized by a 120V,

60Hz voltage source. The phase angle of the voltage source is zero.a. Determine the phasor current

b. Determine the load P, pf, Q, and S.

c. Find an expression for instantaneous p(t)

Grady 2007, p. 22



Grady 2007, p. 23



Calculate the P and Q flows (in per unit) for the loadflow situation shown below.

0.05 + j0.15

pu ohms

j0.20 pumhos

j0.20 pumhos

P + jQ

V = 1.020 /0° V = 1.010 /-10°

Grady 2007, p. 24



Grady 2007, p. 25



Notes on Nodal Analysis

Definitions

Node: A point or set of points at the same potential that have at least two branches

connected to them.

Branch: A circuit element that connects nodes.

Major Node: A node with three or more branches connected to it.

Super Node: Two major nodes with an ideal voltage source between them.

Reference Node: The node to which all other node potentials are referenced. The relativevoltage of the reference node is zero.

Solution Procedure

1. Draw a neat circuit diagram and try to eliminate as many branch crossings as possible.

2. Choose a reference node. All other node voltages will be referenced to it. Ideally, it

should be the node with the most branches connected to it, so that the number of terms in

the admittance matrix is minimal.

3. If the circuit contains voltage sources, do either of the following:

Convert them to current sources (if they have series impedances)

Create super nodes by encircling the corresponding end nodes of each voltage source.

4. Assign a number to every major node (except the reference node) that is not part of asuper node (N1 of these).

5. Assign a number to either end (but not both ends) of every super node that does not touch

the reference node (N2 of these)

6. Apply KCL to every numbered node from Step 4 (N1 equations)

7. Apply KCL to every numbered super node from Step 5 (N2 equations)

8. The dimension of the problem is now N1 + N2. Solve the set of linear equations for the

node voltages. At this point, the circuit has been “solved.”

9. Using your results, check KCL for at least one node to make sure that your currents sum

to zero.

9. Use Ohm’s Law, KCL, and the voltage divider principle to find other node voltages,

branch currents, and powers as needed.

Grady 2007, p. 26



Notes on Mesh Analysis

Definitions

Branch: A circuit element that connects nodes

Planar Network: A network whose circuit diagram can be drawn on a plane in such a

manner that no branches pass over or under other branches

Loop: A closed path

Mesh: A loop that is the only loop passing through at least one branch

Solution Procedure

1. Draw a neat circuit diagram and make sure that the circuit is planar (if not

planar, then the circuit is not a candidate for mesh analysis)

2. If the circuit contains current sources, do either of the following:

A. Convert them to voltage sources (if they have internal impedances), or

B. Create super meshes by making sure in Step 3 that two (and not more than

two) meshes pass through each current source. SM super meshes.

3. Draw clockwise mesh currents, where each one passes through at least one

new branch. M meshes.

4. Apply KVL for every mesh that is not part of a super mesh (M – 2SM

equations)

5. For meshes that form super meshes, apply KVL to the portion of the loop

formed by the two meshes that does not pass through the current source (SMequations)

6. For each super mesh, write an equation that relates the corresponding meshcurrents to the current source (SM equations)

7. The dimension of the problem is now M. Solve the set of M linear equations

for the mesh currents. At this point, the network has been “solved.”

8. Using your results, check KVL around at least one mesh to make sure that the

net voltage drop is zero.

9. Use Ohm’s law, loop currents, KVL, KCL and the voltage divider principle to

find node voltages, branch currents, and powers as needed.

Grady 2007, p. 27



N o t e s o n T h e v e n i n E q u i v a l e n t s

T h e t h r e e c a s e s t o c o n s i d e r a r e

C a s e 1 . A l l s o u r c e s a r e i n d e p e n d e n t

C a s e 2 . T h e c i r c u i t h a s d e p e n d e n t s o u r c e s a n d i n d e p e n d e n t

s o u r c e s

C a s e 3 . T h e c i r c u i t h a s o n l y d e p e n d e n t s o u r c e s

D e p e n d i n g o n t h e c a s e , o n e o r m o r e o f t h e f o l l o w i n g m e t h o d s c a n b e

u s e d t o f i n d t h e T h e v e n i n

e q u i v a l e n t :

D i r e c t R t h . ( A p p l i e s o n l y t o C a s e 1 ) .

T u r n o f f a l l i n d e p

e n d e n t s o u r c e s ( i . e . , s e t V = 0 f o r v o l t a g e

s o u r c e s , a n d I = 0

f o r c u r r e n t s o u r c e s ) . N o t e - t h i s i s t h e

s a m e t h i n g a s r e p

l a c i n g v o l t a g e s o u r c e s w i t h s h o r t c i r c u i t s ,

a n d c u r r e n t s o u r c e s w i t h o p e n c i r c u i t s .

C o n n e c t a f i c t i t i o

u s o h m m e t e r a c r o s s t e r m i n a l s a - b ,

a n d

“ m e a s u r e ” R t h d i

r e c t l y .

F i n d I s c ( o r , a l t e r n a t i v e l y , f i n d V o c = V t h ) .

C o m p u t e V t h = V

o c = I s c • R t h ( o r , a l t e r n a t i v e l y , I s c = V o c /

R t h ) .

I f t i m e p e r m i t s , f i n d V t h ( o r , a l t e r n a t i v e l y , I s c ) d i r e c t l y f r o m

t h e c i r c u i t , a n d t h

e n d o u b l e - c h e c k w i t h t h e a b o v e .

V o c , I s c ( A p p l i e s t o C a s e s 1 a n d 2 ) .

F i n d V o c = V t h .

F i n d I s c .

C o m p u t e R t h = V

t h / I s c .

F i c t i t i o u s S o u r c e ( A p p l i e s t o a l l C a s e s )

A t t a c h a f i c t i t i o u s s o u r c e V a b a c r o s s t e r m i n

a l s a - b . F i n d a

l i n e a r e q u a t i o n w i t h t h e f o l l o w i n g f o r m :

a b

a

b

B I

A

V

.

B y d e f i n i t i o n t h e l i n e a r e q u a t i o n m u s t m a t c

h T h e v e n i n

e q u a t i o n

a b

t h

t h

a b

I

R

V

V

, t e r m b y t e r m

. T h u s ,

m a t c h i n g

t h e t e r m s y i e l d s

B

R A

V

t h

t h

,

.

C a s e

D i r e c t R t h .

V o c , I s c

F i c t i t i o u s

S o u r c e

C a s e 1

O K

O K

O K

C a s e 2

O K

O K

C a s e 3

O K

+

a b

t h

t h

a b

I

R

V

V

–

R t h

V t h

I a b

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Grady 2007, p. 39



Three Properties of Balanced Three-Phase Systems

Page 1 of 3

First, some trigonometry identities:

cos(A – B) = cos(A)cos(B) + sin(A)sin(B)

cos(A + B) = cos(A)cos(B) – sin(A)sin(B)

sin(A – B) = sin(A)cos(B) – cos(A)sin(B)

sin(A + B) = sin(A)cos(B) + cos(A)sin(B)

1. Because they form a balanced set, the a-b-c currents sum to zero. Thus, there is no

return current through the neutral or ground, which reduces wiring losses.

)cos()( t I t ia , )120cos()( t I t ib , )120cos()( t I t ic ,

)120cos()120cos()cos()()()()( t t t I t it it it i cba N ,

)sin()sin()cos()cos(2

)( t t I

t i N

)120sin()sin()120cos()cos( t t

)120sin()sin()120cos()cos( t t ,

)sin()sin()cos()cos(2

)( t t I

t i N

)120sin()cos()120cos()sin()sin()120sin()sin()120cos()cos()cos( t t

)120sin()cos()120cos()sin()sin()120sin()sin()120cos()cos()cos( t t ,

)120sin()120sin()sin()120cos()120cos(1)cos()cos(2

)( t I

t i N

)120sin()120sin()cos()120cos()120cos(1)sin()sin(2

t I

,

2

3

2

3)sin(

2

1

2

11)cos()cos(

2)( t

I t i N

2

3

2

3)cos(2

1

2

11)sin()sin(2 t

I .

Thus 0)( t i N .

Grady 2007, p. 40




Page 2 of 3

2. A N-wire system needs (N – 1) meters. A three-phase, four-wire system needs three

meters. A three-phase, three-wire system needs only two meters.

)()()()( t pt pt pt p cbatot ,

)()()()()()()()( ,,,, t it vt it vt it vt it v nref ncref cbref baref a ,

)()()()( t it it it icban

,

)()()()()()()( ,,, t it vt it vt it vt p cref cbref baref atot

)()()()(, t it it it v cbaref n ,

)()()()()()()()()()( ,,,,,, t it vt vt it vt vt it vt vt p cref nref cbref nref baref nref atot .

Thus, for three-phase, four-wire, three wattmeters are needed to compute

)()()()()()()( t it vt it vt it vt p ccnbbnaantot .

For three-phase, three-wire, the neutral wire is not present. Thus,

)()()( t it it i bac ,

and letting phase c become the reference, then

)()()()()()()()( t it it vt it vt it vt p bacnbbnaantot ,

)()()()()()()( t it vt vt it vt vt p bcnbnacnantot .

Thus, for three-phase, three-wire, two wattmeters are needed to compute

)()()()()( t it vt it vt p bbcaactot .

Three-phase,four wire system

a

bc

n

Reference

Grady 2007, p. 41




Page 3 of 3

More trigonometry identities:

)cos()cos(2

1)cos()cos( B A B A B A

)cos()cos(

2

1)sin()sin( B A B A B A

3. The instantaneous power is constant

)cos()( t V t va , )120cos()( t V t vb , )120cos()( t V t vc ,

)()()()( t pt pt pt p cbatot ,

)120cos()120cos()cos()cos( t t t t VI

)120cos()120cos( t t ,

)2402cos()cos()2cos()cos(2

t t VI

)2402cos()cos( t ,

)1202cos()1202cos()2cos(2

)cos(2

3 t t t

VI VI .

The terms in side the bracket form a balanced set, and thus equal zero. Thus, )(t ptot

is a

constant and equals its average value, which is

)cos(3)cos(2

3)()()()( rmsrmscbatot I V

VI t pt pt pt p .

Grady 2007, p. 42



2 b_

B a l a n c e d_ T h

r e e_ P h a s e_ P

h a s o r s . d o c

P a g e 1 o f 6

T h e p h a s o r s a r e r o t a t i n g c o u n t e r - c l o c k w i s e .

T h e m a g n i t u d e o f l i n e - t o - l i n e v o l t a g e p h a

s o r s i s

3 t i m e s t h e m a g n i t u d e

o f l i n e - t o - n e u t r a l v o l t a g e p h a s o

r s .

V b n

V a b

= V a n – V b n

V b c =

V b n – V c n

V a n

V c n

3

0 °

1 2 0 °

I m a g i n a r y

R e a l

V c a = V c n – V a n

Grady 2007, p. 43



2 b_


r e e_ P h a s e_ P

h a s o r s . d o c

P a g e 2 o f 6

C

o n s e r v a t i o n o f p o w e r r e q u i r e s t h a t t h e m a g n i t u d e s o f d e l t a c u r r e n t s I a b , I c a , a n d I b c a r e

3 1

t i m e s t h e

m a g n i t u d e o f l i n e c u r r e n t s I a , I b , I c .

V a

n

V b n

V c n

R e a l

I m a g i n a r y

V a b = V a n – V b n

V b c =

V b n – V c n

3 0 °

V c a = V c n – V a n

I a

I b

I c

I a b

I b c

I c a

I b I c

I a b

I c a

I b c

I a

a

c

b –

V a b

+

B a l a n c e d S e t s A d d

t o Z e r o i n B o t h

T i m e a n d P h a s o r D o m a i n s

I a + I b + I c = 0

V a n + V b n +

V c n = 0

V a b + V b c +

V c a = 0

L i n e c u r r e n t s I a , I b , a n d I c

D e l t a c u r r e n

t s I a b , I b c , a n d I c a

Grady 2007, p. 44



2 b_


r e e_ P h a s e_ P

h a s o r s . d o c

P a g e 3 o f 6

T h e T w o A b o v e L o a d s a r e E q u i v a l e n t i n B a l a n c e d S y s t e m s

( i . e . , s a m

e l i n e c u r r e n t s I a , I b , I c a n d p h a s e - t o - p h a s e v o l t a g e s V

a b , V b c , V c a i n b o t h c a s e s )

3 Z

3 Z

3 Z

a

c

b

–

V a b

+

I a I b I c

Z

Z

Z

a

c

b

–

V a b

+

I a I b I c

n

Grady 2007, p. 45



2 b_


r e e_ P h a s e_ P

h a s o r s . d o c

P a g e 4 o f 6

T h e T w o A b o v e S o

u r c e s a r e E q u i v a l e n t i n B a l a n c e d S y s t e m s

( i . e . , s a m e l i n e c u r r e n t s I a , I b , I c a

n d p h a s e - t o - p h a s e v o l t a g e s

V a b , V b c , V c a i n b o t h c a s e s )

a

c

b

–

V a b

+

I a I b I c

V a n

a

c

b

–

V a b

+

I a I b I c

n

+

–

Grady 2007, p. 46



2 b_


r e e_ P h a s e_ P

h a s o r s . d o c

P a g e 5 o f 6

Z

Z

Z

a

b

–

V a b

+

I a I b I c

c n

I n

K C L : I n =

I a + I b + I c

B u t f o r a b a l a n c e d s e t ,

I a + I b + I c

= 0 , s o I n = 0

G r o u n d

( i . e . , V = 0 )

T h e E x p e r i m e n t :

O p e n i n g a n d c l o s i n g t h e s w

i t c h h a s n o e f f e c t b e c a u s e

I n i s a l r e a d y z e r o f o r a t h r e e - p h a s e

b a l a n c e d s e t . S i n

c e n o c u r r e n t f l o w s , e v e n i f

t h e r e i s a r e s i s t a n c e i n t h e g r o u n d i n g p a t h , w e m u s t c o

n c l u d e t h a t

V n = 0 a t t h e n e u t r a l p o i n t ( o r e q u i v a l e n t n e u

t r a l p o i n t ) o f a n y b a l a n c e d

t h r e e p h a s e l o a d o r s o u r c e i n a b a l a n c e d

s y s t e m . T h i s a l l o

w s u s t o d r a w a “ o n e - l i n e ”

d i a g r a m ( t y p i c a l l y f o r p h a s e a ) a n d s o l v e a s i n g l e - p h a s

e p r o b l e m .

S o l u t i o n s f o r p h a s e s b a n d c f o l l o w f r o m t h e

p h a s e s h i f t s t h a t m u s t e x i s t .

Grady 2007, p. 47



2 b_


r e e_ P h a s e_ P

h a s o r s . d o c

P a g e 6 o f 6

B a l a n c e d t h r e e - p h a s e s y s t e m s , n o m a t t e r i f t h e y a r e d e l t a

c o n n e c t e d , w y

e c o n n e c t e d , o r a m i x , a r e e a s y t o s o l v e i f y o u

f o l l o w t h e s e s t e p s :

1 . C o n v e r t t h

e e n t i r e c i r c u i t t o a n e q u i v a l e n t w y e w i t h a

g r o u n d e d n e u t r a l .

2 . D r a w t h e o

n e - l i n e d i a g r a m f o r p h a s e a , r e c

o g n i z i n g t h a t

p h a s e a h a s o n e t h i r d o f t h e P a n d Q .

3 . S o l v e t h e o n e - l i n e d i a g r a m f o r l i n e - t o - n e u t r a l v o l t a g e s a n d

l i n e c u r r e n

t s .

4 . I f n e e d e d ,

c o m p u t e l i n e - t o - n e u t r a l v o l t a g e s a n d l i n e c u r r e n t s

f o r p h a s e s

b a n d c u s i n g t h e ± 1 2 0 ° r e l a t i o n

s h i p s .

5 . I f n e e d e d ,

c o m p u t e l i n e - t o - l i n e v o l t a g e s a n

d d e l t a c u r r e n t s

u s i n g t h e

3 a n d ± 3 0 ° r e l a t i o n s h i p s .

a n

a n

Z l o a d

+ V a n –

Z l i n e

I a a

c

b

–

V a b

+

3 Z l o a d

a

c

b

I b I a I c

Z l i n e

Z l i n e

Z l i n e

3 Z l o a d

3 Z l o a d

T

h e “ O n e - L i n e ”

D i a g r a m

Grady 2007, p. 48



Grady 2007, p. 49



Grady 2007, p. 50



Grady 2007, p. 51



A three-phase, 460V motor draws 5kW with a power factor of 0.80 lagging. Assuming that anV ~

has phase angle zero,

a. Find phasor currents a I ~

and ab I ~

(note – ab I ~

is inside the motor delta windings)

b. Find the three phase motor Q and S

c. How much capacitive kVAr (three-phase) should be connected in parallel with the motor toimprove the net power factor to 0.95?

d. Assuming no change in supply voltage, what will be the new a I ~

after the kVArs are added?

Grady 2007, p. 52



Grady 2007, p. 53



The 60Hz system shown below is balanced. The line-to-line voltage of the source is 460V.

Resistors R are each 5.

Part a. If each Z is (90 + j45), determine the three-phase complex power delivered by the

source, and the three-phase complex power absorbed by the delta-connected Z loads.

Part b. If anV ~

at the source has phase angle zero, find ''~

baV at the load.

Z

ZZ

Grady 2007, p. 54



Grady 2007, p. 55



Grady 2007, p. 56



Grady 2007, p. 57



Grady 2007, p. 58



Grady 2007, p. 59



Grady 2007, p. 60



Grady 2007, p. 61



Grady 2007, p. 62



Grady 2007, p. 63



Grady 2007, p. 64



Grady 2007, p. 65



Grady 2007, p. 66



Grady 2007, p. 67



Grady 2007, p. 68



Notes_on_ANSI_Transformer_Connections.doc

Page 1 of 1

N1

N1

N1

a

b

c

N2

N2 N2n

a’

b’c'

LV HV

N1

N1 N1n

N2

N2

N2

a

bc

b’

a’

c’

LV HV

Grady 2007, p. 69



Grady 2007, p. 70



The following data were taken for a 25kVA, 7200/240V transformer:

Short Circuit Test: 240V side short-circuited, 7200V side energized at reduced voltage.The measurements (on the 7200V side) are Irms = 3.47A (i.e., rated current), Vrms =

216V, P = 335W, Q = 669VAr.

Open Circuit Test: 7200V side open-circuited, 240V side energized. The measurements

(on the 240V side) are Vrms = 240V (i.e., rated voltage), Irms = 2.08A, P = 171W, Q =469VAr.

Draw the transformer equivalent circuit with all four circuit parameters (i.e., R, Ll, Lm, R m)

shown on the 7200V side.

Grady 2007, p. 71



Grady 2007, p. 72

3.47



Grady 2007, p. 73



EE369/394J, Test #4, Sept. 24, 2004. Name _______________________________________One sheet of notes permitted. Show all steps.

The following data were taken for a 75kVA, 7200/480V transformer:

Short Circuit Test: 480V side short-circuited, 7200V side energized at reduced voltage. The

measurements (on the 7200V side) are Irms = 10.0A, Vrms = 144V, P = 750W.

Open Circuit Test: 7200V side open-circuited, 480V side energized. The measurements (on the

480V side) are Vrms = 480V, Irms = 1.50A, P = 500W.

Draw the transformer equivalent circuit with all four circuit parameters (i.e., R, Ll, Lm, R m) shown on

the 7200V side. Hint – remember that rmsrms I V S ,222 Q P S .

Grady 2007, p. 74



Grady 2007, p. 75



Open circuit and short circuit tests are performed on a single-phase, 7200/240V, 25kVA, 60Hz

distribution transformer. The results are:

Short circuit test (short circuit the low-voltage side, energize the high-voltage side so thatrated current flows, and measure Psc and Qsc). Measured Psc = 400W, Qsc = 200VAr.

Open circuit test (open circuit the high-voltage side, apply rated voltage to the low-voltage

side, and measure Poc and Qoc). Measured Poc = 100W, Qoc = 250VAr.

Determine the four impedance values (in ohms) for the transformer model shown.

Rs jXs

IdealTransformer

7200/240VRm jXm

7200V 240V

Grady 2007, p. 76



Grady 2007, p. 77



Grady, Transformers, June 2007, Page 1

Transformers

Transformers. Transformer phase shift. Wye-delta connections and impact on zero sequence.

Inductance and capacitance calculations for transmission lines. GMR, GMD, L, and C matrices,

effect of ground conductivity. Underground cables.

Equivalent Circuits

The standard transformer equivalent circuit used in power system simulation is shown below,

where the R and X terms represent the series resistance and leakage reactance, and N1 and N2

represent the transformer turns. Note that the shunt terms are usually ignored in the model..

R jX

N1 N2

Figure 1. Power System Model for Transformer

Three-phase transformers can consist of either three separate single-phase transformers, or three

windings on a three-legged, four-legged, or five-legged core. The high-voltage and low-voltage

sides can be connected independently in either wye or delta.

A B C

High-Voltage Side

Low-Voltage Side

Figure 2. A Three-Phase Ground-Wye Grounded-Wye Transformer

Grady 2007, p. 78




A B C

High-Voltage Side

Low-Voltage Side

Figure 3. A Three-Phase Delta-Delta Transformer

The transformer impedances consist of winding resistances and leakage reactances. There are no

mutual resistances, and the mutual leakage reactances between the separate phase a-b-c coils are

negligible. Hence, in symmetrical components, S = R + jX , and M = 0, so that S + 2M = S - M

= R + jX , so therefore the positive and negative sequence impedances of a transformer are

jX R Z Z 21 .

One must remember that no zero sequence currents can flow into a three-wire connection.

Therefore, the zero sequence impedance of a transformer depends on the winding connections.

In the case where one side of a transformer is connected grounded-wye, and the other side is

delta, circulating zero sequence currents can be induced in the delta winding. In that case, the

zero sequence impedance "looking into" the transformer is different on the two sides.

The zero sequence equivalent circuits for three-phase transformers is given in Figure 4.

Grady 2007, p. 79




Grounded Wye - Grounded Wye

Grounded Wye - DeltaR + jX

Grounded Wye - Ungrounded WyeR + jX

R + jXUngrounded Wye - Delta

R + jXDelta - Delta

R + jX

Figure 4. Zero Sequence Impedance Equivalent Circuits for Three-Phase Transformers

A wye-delta transformer connection introduces a 30o phase shift in positive/negative sequence

voltages and currents because of the relative shift between line-to-neutral and line-to-ground

voltages. Transformers are labeled so that

1. High side positive sequence voltages and currents lead those on the low side by 30o.

2. High side negative sequence voltages and currents lag those on the low side by 30o.

3. There is no phase shift for zero sequence.

Transformer tap magnitudes can be adjusted to control voltage, and transformer phase shifts can

be adjusted to control active power flow. The effect of these "off-nominal" adjustments can be

incorporated into a pi-equivalent circuit model for a transformer.

Grady 2007, p. 80




y

Ii ---> Ik --->Bus i

Bus k t / :1

Bus k'

Figure 5. Off-Nominal Transformer Circuit Model

Assume that the transformer in Figure 5 has complex "off-nominal" tap t t and series

admittance y. The relationship between the voltage on opposite sides of the transformer tap is

t

ik

t

V V

~~

' , and since the power on both sides of the ideal transformer must be the same, then

*''

* ~~~~k k ii I V I V , implying that t ik t I I

~~' . Now, suppose that the transformer can be

modeled by the following pi-equivalent circuit of Figure 6:

yik

Ii --->

Bus i

yii

Bus k

<--- -Ik

ykk

Figure 6. Pi-Equivalent Model of Transformer

Admittances yii, yik , and ykk can be found so that the above circuit is equivalent to Figure 5. This

can be accomplished by forcing the terminal behavior to be the same. For the above circuit, the

appropriate equations are

iiiik k ii yV yV V I ~~~~

, and kk k ik ik k yV yV V I ~~~~

,

or in matrix form

k

i

ik kk ik

ik ik ii

k

i

V

V

y y y

y y y

I

I ~

~

~

~

.

For Figure 5, the terminal equations are

yV t

V yV V I k

t

ik k k

~~

~~~'

,

and since

Grady 2007, p. 81




t

k i

t

I I

~~

,

then

yt

V

t t

V I

t

k

t t

ii

~~~ .

In matrix form,

k

i

t

t t t

k

i

V

V

yt

y

t

y

t t

y

I

I ~

~

~

~

.

Comparing the two sets of terminal equations shows that equality can be reached if the shunt

branch in the equivalent circuit, yik , can have two values:

t ik

t

y y

from the perspective of Kirchhoff's current law at bus i,

t ik

t

y y

from the perspective of Kirchhoff's current law at bus k.

Note that if the tap does not include an off-nominal phase shift, thent

y yik from either

direction.

Next, solving for yii and ykk yields

1

1

t t t t t ii

t t

y

t

y

t t

y y

,

t t kk

t y

t

y y y

11 .

Neutral Grounding Impedance

If the wye-side of a transformer or wye-connected load is grounded through a grounding

impedance Z g , the grounding impedance is "invisible" to the positive and negative sequence

currents since their corresponding voltages at the wye-point is always zero due to symmetry.

However, since the neutral current is three-times the zero sequence current, the voltage drop on

the grounding impedance is 3 I ao. For that reason, the zero sequence equivalent circuit for a

grounding impedance must contain 3 Z g .

Grady 2007, p. 82




Z Z Z

Zg

Zao = Z + 3Zg+

-

Vao = 3 Iao Zg

Iao Iao Iao

Za1 = Za2 = Z

Figure 7. Effect of Grounding Impedance on Sequence Impedances

Grady 2007, p. 83



3l_ANSI_Transformer_Wye_Delta_Labeling.doc

Page 1 of 1

N1

N1

N1

a

b

c

N2

N2 N2n

a’

b’c'

LV HV

N1

N1 N1n

N2

N2

N2

a

bc

b’

a’

c’

LV HV

Grady 2007, p. 84



Grady 2007, p. 85



Grady, Per Unit, June 2007, Page 1

1. Single-Phase Per Unit System

Advantages of the per unit system:

1. Transformers can be replaced by their equivalent series impedances.

2. Equipment impedances can be easily estimated since their per unit impedances lie within a

relatively narrow range.

Define four quantities in per unit of their respective base values:

volts

volts

V

V V

base pu ,

amps

amps

I

I I

base pu ,

voltamps

voltamps

S

S S

base

pu ,ohms

ohms

Z

Z Z

base

pu .

The relationships among the bases are

basebasebase I V S , andbase

basebase

I

V Z .

Once two base variables are specified, the other two base variables may be calculated.

A convenient relation, derived from the two above equations, is

base

base

base

basebasebasebasebase

Z V

Z V V I V S

2

.

Consider a transformer with an 21 : N N turns ratio and series impedance, reflected on side 1,

equal 1 L Z . 1 L Z can be reflected to side 2 using

1

2

1

22 L L Z

N

N Z

.

Let side 1 and side 2 have base values designated by subscripts S1 and S2. Then

1

21

1base

basebase

S

V Z ,

2

22

2base

basebase

S

V Z ,

Expressing the transformer impedance on the two respective bases yields

Grady 2007, p. 86




21

111

base

base L PU L

V

S Z Z ,

22

222

base

base L PU L

V

S Z Z .

If 21 B B S S , the two above equations may be combined so that

PU Lbase

base PU L Z

V

V

N

N Z 1

2

2

12

1

22

.

Substituting the relation between 1 L Z and 2 L Z yields

PU Lbase

base PU L Z

V

V

N

N Z 1

2

2

12

1

22

.

Therefore, if12

12

N

N

V

V

basebase , then PU L PU L Z Z 12 .

Hence, if a common voltampere base is chosen on both sides of the transformer, and if the

voltage bases are chosen so that they vary according to the transformer turns ratio, then the per

unit series impedance of the transformer is the same value on both sides.

When analyzing a circuit with many transformers, a common voltampere base should be chosen

throughout the circuit, and a voltage base should be chosen at one location. The voltage base

must vary across the circuit according to the transformer turns ratios.

When analyzing a circuit in per unit, if the bases are chosen according to the above rules,

transformers can be replaced by their equivalent per unit series impedances, and their turns can

be ignored.

A manufacturer usually provides the impedance of a transformer on the transformer's rated

voltage and power bases. However, when solving a power network circuit, the power and

voltage bases must vary according to the above rules, and they may not equal the manufacturer-

specified bases. Per unit impedances, specified on one base, may be converted to a new base as

follows:

Given

old base

old PU Z

Z Z

,

, , on bases old baseV , and old baseS , , new PU Z , on new bases

newbaseV , and newbaseS , is

2,

,

,

2,

,,

,,

,,

newbase

newbase

old base

old baseold PU

newbase

old baseold PU

newbase

ohmsnew PU

V

S

S

V Z

Z

Z Z

Z

Z Z .

Grady 2007, p. 87




2. Three-Phase Per Unit System

The same advantages apply to a three-phase system if the following rules are obeyed:

1. A common three-phase voltampere base is used throughout the system, where

1,3, 3 basebase S S .

2. Once selected at a point in the network, the three-phase voltage base must vary according to

the line-to-line transformer turns ratios.

Convenient formulas relating single-phase to three-phase bases are given below.

baseneutral linebasebase I V S ,1, ,

1,3, 3 basebase S S ,

3,

2,

3,

,

2

1,

2,

3/

3/

base

linelinebase

base

linelinebase

base

neutral linebasebase

S

V

S

V

S

V Z

.

Grady 2007, p. 88



Grady 2007, p. 89



Grady 2007, p. 90



Grady 2007, p. 91



Grady 2007, p. 92

4 of 5



Grady 2007, p. 93

5 of 5



Find the magnitude of the line-to-line voltages on the 12.47kV side of the transformer.

Z = 0.05pu

15kVA

12.47kV/480V,

X/R = 1

Transformer

(GY-GY)

V(line-to-line) = 460V,

three-phase P = 10kW,

pf = 0.90 lagging

Supply network

Grady 2007, p. 94



Grady 2007, p. 95



Information for a small power system is shown below. Per unit values are given on the equipment bases.

Using a 138kV, 100MVA base in the transmission lines, draw the per unit diagram. Assume that no

current is flowing in the network, so that all generator and motor voltages are 1.0pu in your final diagram.

Trans1

Gen1

X” = 0.15

40MVA

20kV

Trans1

X = 0.16

60MVA

18kV/138kV

TLine1

Line1

R = 10

X = 60

Trans2Trans2

X=0.14

50MVA

20kV/138kV

TLine2

Line2

R = 10

X = 60

Trans3

Trans3X = 0.10

40MVA

13.8kV/138kV

Motor

X” = 0.10

25MVA

13.2kV

Gen2

X” = 0.15

35MVA

22kV

Grady 2007, p. 96



Grady 2007, p. 97



4g_Grainger_Stevenson_Problems.doc

Should be 50 MVA

Bus1 Bus2

Bus3

Bus5Bus6Bus4

Grady 2007, p. 98




Bus1 Bus2

Bus3

Bus4 Bus5

Bus6 Bus7Bus8 Bus9

Grady 2007, p. 99




Short Circuit Calculation

(Based upon Grainger/Stevenson Problem 6.15/11.11)

SC Problem 1. A three-phase balanced fault, with ZF = 0, occurs at Bus 4. Determine

a. F a I 4 (in per unit and in amps)

b. Phasor abc line-to-neutral voltages at the terminals of Gen 1

c. Phasor abc currents flowing out of Gen 1 (in per unit and in amps)

SC Problem 2. Repeat Problem 1, again with ZF = 0, but with a single-phase-to-ground

fault at Bus 4, phase a.

In both Problems 1 and 2, make sure that you add in the transformer phase shifts,

assuming that pre-fault Van in the transmission line has reference phase angle 0.

Grady 2007, p. 100



Grady 2007, p. 101



Grady 2007, p. 102



Grady 2007, p. 103



Grady 2007, p. 104



Grady 2007, p. 105



Grady 2007, p. 106



Grady 2007, p. 107



Grady 2007, p. 108



Grady 2007, p. 109



Grady 2007, p. 110



Grady 2007, p. 111



Grady 2007, p. 112



Grady 2007, p. 113



Grady, Transmission Lines, June 2007, Page 1

Transmission Lines

Inductance and capacitance calculations for transmission lines. GMR, GMD, L, and C matrices,

effect of ground conductivity. Underground cables.

1. Equivalent Circuit for Transmission Lines (Including Overhead and Underground)

The power system model for transmission lines is developed from the conventional distributed

parameter model, shown in Figure 1.

+

-

v

R/2 L/2

G C

R/2 L/2

i --->

<--- i

+

-

v + dv

i + di --->

<--- i + di

R, L, G, C per unit length

< >dz

Figure 1. Distributed Parameter Model for Transmission Line

Once the values for distributed parameters resistance R, inductance L, conductance G, and

capacitance are known (units given in per unit length), then either "long line" or "short line"

models can be used, depending on the electrical length of the line.

Assuming for the moment that R, L, G, and C are known, the relationship between voltage andcurrent on the line may be determined by writing Kirchhoff's voltage law (KVL) around the

outer loop in Figure 1, and by writing Kirchhoff's current law (KCL) at the right-hand node.

KVL yields

02222

t

i Ldz i

Rdz dvv

t

i Ldz i

Rdz v

.

This yields the change in voltage per unit length, or

t i L Ri

z v

,

which in phasor form is

I L j R z

V ~~

.

Grady 2007, p. 115




KCL at the right-hand node yields

0

t

dvvCdz dvvGdz diii

.

If dv is small, then the above formula can be approximated as

t

vCdz vGdz di

, or

t

vC Gv

z

i

, which in phasor form is

V C jG z

I ~~

.

Taking the partial derivative of the voltage phasor equation with respect to z yields

z

I

L j R z

V

~~

2

2

.

Combining the two above equations yields

V V C jG L j R z

V ~~~

2

2

2

, where jC jG L j R , and

where , , and are the propagation, attenuation, and phase constants, respectively.

The solution for V ~

is

z z Be Ae z V )(~ .

A similar procedure for solving I ~

yields

o

z z

Z

Be Ae z I

)(

~ ,

where the characteristic or "surge" impedance o Z is defined as

C jG

L j R Z o

.

Constants A and B must be found from the boundary conditions of the problem. This is usually

accomplished by considering the terminal conditions of a transmission line segment that is d

meters long, as shown in Figure 2.

Grady 2007, p. 116




d< >

+

-

+

-

Vs Vr

Is ---> Ir --->

<--- Is <--- Ir

Sending End Receiving End

Transmission

Line Segment

z = 0z = -d

Figure 2. Transmission Line Segment

In order to solve for constants A and B, the voltage and current on the receiving end is assumed

to be known so that a relation between the voltages and currents on both sending and receiving

ends may be developed.

Substituting z = 0 into the equations for the voltage and current (at the receiving end) yields

o R R

Z

B A I B AV

~

,~

.

Solving for A and B yields

2

~

,2

~ Ro R Ro R I Z V

B I Z V

A

.

Substituting into the )(~

z V and )(~

z I equations yields

d I Z d V V R RS sinh~

cosh~~

0 ,

d I d Z

V I R

o

RS cosh

~sinh

~~

.

A pi equivalent model for the transmission line segment can now be found, in a similar manner

as it was for the off-nominal transformer. The results are given in Figure 3.

Grady 2007, p. 117




d< >

+

-

+

-

Vs Vr

Is ---> Ir --->

<--- Is <--- Ir

Sending End Receiving End

z = 0z = -d

Ysr

Ys Yr

o RS

Z

d

Y Y

2

tanh

, d Z

Y o

SR sinh

1 ,

C jG

L j R Z o

, C jG L j R

R, L, G, C per unit length

Figure 3. Pi Equivalent Circuit Model for Distributed Parameter Transmission Line

Shunt conductance G is usually neglected in overhead lines, but it is not negligible in

underground cables.

For electrically "short" overhead transmission lines, the hyperbolic pi equivalent model

simplifies to a familiar form. Electrically short implies that d < 0.05 , where wavelength

Hz f

sm

r

/103 8

= 5000 km @ 60 Hz, or 6000 km @ 50 Hz. Therefore, electrically short

overhead lines have d < 250 km @ 60 Hz, and d < 300 km @ 50 Hz. For underground cables,

the corresponding distances are less since cables have somewhat higher relative permittivities(i.e. 5.2r ).

Substituting small values of d into the hyperbolic equations, and assuming that the line losses

are negligible so that G = R = 0, yields

2

Cd jY Y RS

, and

Ld jY SR

1 .

Then, including the series resistance yields the conventional "short" line model shown in Figure

4, where half of the capacitance of the line is lumped on each end.

Grady 2007, p. 118




< >d

Cd

2

Cd

2

Rd Ld

R, L, C per unit length

Figure 4. Standard Short Line Pi Equivalent Model for a Transmission Line

2. Capacitance of Overhead Transmission Lines

Overhead transmission lines consist of wires that are parallel to the surface of the earth. To

determine the capacitance of a transmission line, first consider the capacitance of a single wire

over the earth. Wires over the earth are typically modeled as line charges l Coulombs permeter of length, and the relationship between the applied voltage and the line charge is the

capacitance.

A line charge in space has a radially outward electric field described as

r o

l ar

q E ˆ

2 Volts per meter .

This electric field causes a voltage drop between two points at distances r = a and r = b away

from the line charge. The voltage is found by integrating electric field, or

a

bqar E V

o

l r

br

ar

ab ln2

ˆ

V.

If the wire is above the earth, it is customary to treat the earth's surface as a perfect conducting

plane, which can be modeled as an equivalent image line charge l q lying at an equal distance

below the surface, as shown in Figure 5.

Grady 2007, p. 119




Surface of Earth

h

a

b

ai bi

A

Bh

Conductor with radius r, modeled electricallyas a line charge ql at the center

Image conductor, at an equal distance below

the Earth, and with negative line charge -ql

Figure 5. Line Charge l q at Center of Conductor Located h Meters Above the Earth

From superposition, the voltage difference between points A and B is

bia

aibq

ai

bi

a

bqa E a E V

o

l

o

l r

bir

air

ir

br

ar

ab ln2

lnln2

ˆˆ

.

If point B lies on the earth's surface, then from symmetry, b = bi, and the voltage of point A with

respect to ground becomes

a

aiqV

o

l ag ln

2 .

The voltage at the surface of the wire determines the wire's capacitance. This voltage is found

by moving point A to the wire's surface, corresponding to setting a = r , so that

r

hqV

o

l rg

2ln

2 for h >> r .

The exact expression, which accounts for the fact that the equivalent line charge drops slightly

below the center of the wire, but still remains within the wire, is

r

r hhqV

o

l rg

22

ln2

.

Grady 2007, p. 120




The capacitance of the wire is defined asrg

l

V C

which, using the approximate voltage

formula above, becomes

r

hC o

2ln

2

Farads per meter of length.

When several conductors are present, then the capacitance of the configuration must be given in

matrix form. Consider phase a-b-c wires above the earth, as shown in Figure 6.

a

ai

b

bi

c

ci

Daai

Dabi

Daci

Dab

Dac

Surface of Earth

Three Conductors Represented by Their Equivalent Line Charges

Images

Conductor radii ra, rb, rc

Figure 6. Three Conductors Above the Earth

Superposing the contributions from all three line charges and their images, the voltage at the

surface of conductor a is given by

ac

acic

ab

abib

a

aaia

oag

D

Dq

D

Dq

r

DqV lnlnln

2

1

.

The voltages for all three conductors can be written in generalized matrix form as

c

b

a

cccbca

bcbbba

acabaa

ocg

bg

ag

q

qq

p p p

p p p p p p

V

V V

2

1 , or abcabc

oabc Q P V

2

1 ,

where

Grady 2007, p. 121




a

aaiaa

r

D p ln ,

ab

abiab

D

D p ln , etc., and

ar is the radius of conductor a,

aai D is the distance from conductor a to its own image (i.e. twice the height of

conductor a above ground),

ab D is the distance from conductor a to conductor b,

baiabi D D is the distance between conductor a and the image of conductor b (which

is the same as the distance between conductor b and the image of

conductor a), etc.

A Matrix Approach for Finding C

From the definition of capacitance, CV Q , then the capacitance matrix can be obtained via

inversion, or

12

abcoabc P C .

If ground wires are present, the dimension of the problem increases proportionally. For example,

in a three-phase system with two ground wires, the dimension of the P matrix is 5 x 5. However,

given the fact that the line-to-ground voltage of the ground wires is zero, equivalent 3 x 3 P and

C matrices can be found by using matrix partitioning and a process known as Kron reduction.

First, write the V = PQ equation as follows:

w

v

c

b

a

vwabcvw

vwabcabc

o

wg

vg

cg

bg

ag

q

q

q

q

q

P P

P P

V

V

V

V

V

)2x2(|)3x2(

)2x3(|)3x3(

2

1

0

0 ,

,

,

or

vw

abc

vwabcvw

vwabcabc

ovw

abc

Q

Q

P P

P P

V

V

,

,

2

1

,

where subscripts v and w refer to ground wires w and v, and where the individual P matrices are

formed as before. Since the ground wires have zero potential, then

Grady 2007, p. 122




vwvwabcabcvwo

Q P Q P

,

2

1

0

0

,

so that

abcabcvwvwvw Q P P Q ,1 .

Substituting into the abcV equation above, and combining terms, yields

abcabcvwvwvwabcabco

abcabcvwvwvwabcabcabco

abc Q P P P P Q P P P Q P V ,1

,,1

,2

1

2

1

,

or

abcabcoabc

Q P V '

2

1

, so that

abcabcabc V C Q ' , where 1'' 2

abcoabc P C .

Therefore, the effect of the ground wires can be included into a 3 x 3 equivalent capacitance

matrix.

An alternative way to find the equivalent 3 x 3 capacitance matrix 'abcC is to

obtain the 5 x 5 C matrix by inverting the 5 x 5 P , and then

Kron reduce the 5 x 5 C directly.

Computing 012 Capacitances from Matrices

Once the 3 x 3 'abcC matrix is found by either of the above two methods, 012 capacitances can

be determined by averaging the diagonal terms, and averaging the off-diagonal terms of, 'abcC to

produce

S M M

S S M

M M S

avg abc

C C C

C C C

C C C

C .

avg abc

C has the special symmetric form for diagonalization into 012 components, which yields

Grady 2007, p. 123




M S

M S

M S avg

C C

C C

C C

C

00

00

002

012 .

The Approximate Formulas for 012 Capacitances

Asymmetries in transmission lines prevent the P and C matrices from having the special form

that allows their diagonalization into decoupled positive, negative, and zero sequence

impedances. Transposition of conductors can be used to nearly achieve the special symmetric

form and, hence, improve the level of decoupling. Conductors are transposed so that each one

occupies each phase position for one-third of the lines total distance. An example is given below

in Figure 7, where the radii of all three phases are assumed to be identical.

a b c a cthen then

then then

bthen

b a c

b c a c a b c b a

where each configuration occupies one-sixth of the total distance

Figure 7. Transposition of A-B-C Phase Conductors

For this mode of construction, the average P matrix (or Kron reduced P matrix if ground wires

are present) has the following form:

cc

acaa

bcabbb

bb

bccc

abacaa

cc

bcbb

acabaaavg abc

p

p p

p p p

p

p p

p p p

p

p p

p p p

P 6

1

6

1

6

1

aa

abbb

acbccc

aa

accc

abbcbb

bb

abaa

bcaccc

p

p p

p p p

p

p p

p p p

p

p p

p p p

6

1

6

1 ,

where the individual p terms are described previously. Note that these individual P matrices aresymmetric, since baabbaab p p D D , , etc. Since the sum of natural logarithms is the same

as the logarithm of the product, P becomes

S M M

M S M

M M S avg

abc

p p p

p p p

p p p

P ,

Grady 2007, p. 124




where

3

3

ln3 cba

ccibbiaaiccbbaa s

r r r

D D D P P P p

,

and

3

3

ln3 bcacab

bciaciabibcacabM

D D D

D D D P P P p

.

Since avg

abc P has the special property for diagonalization in symmetrical components, then

transforming it yields

M S

M S

M S avg

p p p p

p p

p p

p

P 00

00

002

0000

00

2

1

0

012 ,

where

33

33

3

3

3

3

lnlnlnbciaciabicba

bcacabccibbiaai

bcacab

bciaciabi

cba

ccibbiaaiM s

D D Dr r r

D D D D D D

D D D

D D D

r r r

D D D p p .

Inverting avg

P 012

and multiplying by o 2 yields the corresponding 012 capacitance matrix

M S

M S

M S

ooavg

p p

p p

p p

p

p

p

C

C

C

C

100

01

0

002

1

2

100

01

0

001

2

00

00

00

2

1

0

2

1

0

012 .

When the a-b-c conductors are closer to each other than they are to the ground, then

bciaciabiccibbiaai D D D D D D ,

yielding the conventional approximation

2,1

2,1

3

3

21 lnlnGMR

GMD

r r r

D D D p p p p

cba

bcacabM S ,

Grady 2007, p. 125




where 2,1GMD and 2,1GMR are the geometric mean distance (between conductors) and

geometric mean radius, respectively, for both positive and negative sequences. Therefore, the

positive and negative sequence capacitances become

2,1

2,121ln

22

GMR

GMD p pC C

o

M S

o

Farads per meter.

For the zero sequence term,

3

3

3

3

0 ln2ln2bcacab

bciaciabi

cba

ccibbiaaiM s

D D D

D D D

r r r

D D D p p p

3

2

2

ln

bcacabba

bciaciabiccibbiaai

D D Dr r r

D D D D D D .

Expanding yields

9

2

2

0 ln3

bcacabba

bciaciabiccibbiaai

D D Dr r r

D D D D D D p

9ln3cbcababcacabba

cbicaibaibciaciabiccibbiaai

D D D D D Dr r r

D D D D D D D D D ,

or

0

00 ln3

GMR

GMD p ,

where

90 cbicaibaibciaciabiccibbiaai D D D D D D D D DGMD ,

9

0 cbcababcacabcba D D D D D Dr r r GMR .

The zero sequence capacitance then becomes

o

o

M S

o

GMR

GMD p pC

00

ln

2

3

1

2

2

Farads per meter,

Grady 2007, p. 126




which is one-third that of the entire a-b-c bundle by because it represents the average

contribution of only one phase.

Bundled Phase Conductors

If each phase consists of a symmetric bundle of N identical individual conductors, an equivalent

radius can be computed by assuming that the total line charge on the phase divides equally

among the N individual conductors. The equivalent radius is

N N eq NrAr

11 ,

where r is the radius of the individual conductors, and A is the bundle radius of the symmetric set

of conductors. Three common examples are shown below in Figure 8.

Double Bundle, Each Conductor Has Radius r

A

rAr eq 2

Triple Bundle, Each Conductor Has Radius r

A

3 23rAr eq

Quadruple Bundle, Each Conductor Has Radius r

A

4 34rAr eq

Figure 8. Equivalent Radius for Three Common Types of Bundled Phase Conductors

Grady 2007, p. 127




3. Inductance

The magnetic field intensity produced by a long, straight current carrying conductor is given by

Ampere's Circuital Law to be

r

I H

2 Amperes per meter,

where the direction of H is given by the right-hand rule.

Magnetic flux density is related to magnetic field intensity by permeability as follows:

H B Webers per square meter,

and the amount of magnetic flux passing through a surface is

sd B Webers,

where the permeability of free space is 7104 o .

Two Parallel Wires in Space

Now, consider a two-wire circuit that carries current I, as shown in Figure 9.

I I

Two current-carying wires with radii r

D< >

Figure 9. A Circuit Formed by Two Long Parallel Conductors

The amount of flux linking the circuit (i.e. passes between the two wires) is found to be

r

r D I dx

x

I dx

x

I or D

r

or D

r

o ln22

Henrys per meter length.

From the definition of inductance,

I

N L

,

Grady 2007, p. 128




where in this case N = 1, and where N >> r , the inductance of the two-wire pair becomes

r

D L o ln


A round wire also has an internal inductance, which is separate from the external inductanceshown above. The internal inductance is shown in electromagnetics texts to be

8

int int L Henrys per meter length.

For most current-carrying conductors, oint so that int L = 0.05µH/m. Therefore, the total

inductance of the two-wire circuit is the external inductance plus twice the internal inductance of

each wire (i.e. current travels down and back), so that

4

14

1

lnlnln41ln

82ln

re

Der

Dr

Dr

D L oooootot

.

It is customary to define an effective radius

r rer eff 7788.04

1

,

and to write the total inductance in terms of it as

eff

otot

r D L ln


Wire Parallel to Earth’s Surface

For a single wire of radius r , located at height h above the earth, the effect of the earth can be

described by an image conductor, as it was for capacitance calculations. For a perfectly

conducting earth, the image conductor is located h meters below the surface, as shown in Figure

10.

Grady 2007, p. 129




Surface of Earth

h

h

Conductor of radius r, carrying current I

Note, the image

flux exists only

above the Earth

Image conductor, at an equal distance below the Earth

Figure 10. Current-Carrying Conductor Above Earth

The total flux linking the circuit is that which passes between the conductor and the surface of

the earth. Summing the contribution of the conductor and its image yields

r

r h I

rh

r hh I

x

dx

x

dx I ooh

r

r h

h

o 2ln

2

2ln

22

2

.

For 2h r , a good approximation is

r

h I o 2ln

2

Webers per meter length,

so that the external inductance per meter length of the circuit becomes

r

h L o

ext 2

ln2


The total inductance is then the external inductance plus the internal inductance of one wire, or

4

1

2ln

24

12ln

28

2ln

2

re

h

r

h

r

h L oooo

tot

,

or, using the effective radius definition from before,

eff

otot

r

h L

2ln

2


Grady 2007, p. 130




Bundled Conductors

The bundled conductor equivalent radii presented earlier apply for inductance as well as for

capacitance. The question now is “what is the internal inductance of a bundle?” For N bundled

conductors, the net internal inductance of a phase per meter must decrease as

N

1 because the

internal inductances are in parallel. Considering a bundle over the Earth, then

N

eq

o

eq

o

eq

oo

eq

otot

er

he

N r

h

N r

h

N r

h L

4

14

12

ln2

ln12

ln24

12ln

28

2ln

2

.

Factoring in the expression for the equivalent bundle radius eqr yields

N N eff

N N N N N N

eq A Nr A Nree NrAer

11

1

14

1

4

1114

1

Thus, eff r remains 4

1

re , no matter how many conductors are in the bundle.

The Three-Phase Case

For situations with multiples wires above the Earth, a matrix approach is needed. Consider thecapacitance example given in Figure 6, except this time compute the external inductances, rather

than capacitances. As far as the voltage (with respect to ground) of one of the a-b-c phases is

concerned, the important flux is that which passes between the conductor and the Earth's surface.

For example, the flux "linking" phase a will be produced by six currents: phase a current and its

image, phase b current and its image, and phase c current and its image, and so on. Figure 11 is

useful in visualizing the contribution of flux “linking” phase a that is caused by the current in

phase b (and its image).

Grady 2007, p. 131




a

ai

b

bi

Dab

D bg

D bgDabi

g

Figure 11. Flux Linking Phase a Due to Current in Phase b and Phase b Image

Grady 2007, p. 132




S M M

S S M

M M S avg abc

L L L

L L L

L L L

L ,

so that

M S

M S

M S avg

L L

L L

L L

L

00

00

002

012 .

The Approximate Formulas for 012 Inductancess

Because of the similarity to the capacitance problem, the same rules for eliminating ground

wires, for transposition, and for bundling conductors apply. Likewise, approximate formulas for

the positive, negative, and zero sequence inductances can be developed, and these formulas are

2,1

2,121 ln

2 GMR

GMD L L o

,

and

0

00 ln

23

GMR

GMD L o

.

It is important to note that the GMD and GMR terms for inductance differ from those of

capacitance in two ways:

1. GMR calculations for inductance calculations should be made with 4

1

rer eff .

2. GMD distances for inductance calculations should include the equivalent complex depth for

modeling finite conductivity earth (explained in the next section). This effect is ignored in

capacitance calculations because the surface of the Earth is nominally at zero potential.

The Complex Depth Method for Modeling Imperfect Earth

The effect of the Earth's non-infinite conductivity should be included when computinginductances, especially zero sequence inductances. Because the Earth is not a perfect conductor,

the image current does not actually flow on the surface of the Earth, but rather through a cross-

section. The higher the conductivity, the narrower the cross-section.

The simplest method to account for finite conductivity is via a complex depth cd , where the

equivalent earth surface is assumed to be an additional cd meters below the actual earth surface.

Grady 2007, p. 134




Using complex depth, if a conductor is d meters above earth, its image is cd d 2 meters

below the conductor.

The complex depth cd is related to resistive skin depth by

jd c

1

= o45

2

=

22

j ,

where

f

1 .

At f = 60 Hz andm

1

01.0 (typical for limestone), the skin depth is 650 meters, so that

2

650

2

650 jd c = 325 – j325 m.

Skin depth in typical soils for 50 - 60 Hz varies from 500 to 2000 meters. The effect on

positive/negative sequence inductances is not great, but the effect on zero sequence inductances

is quite significant.

Since cd is a frequency-dependent complex number, the self- and image-distance terms are

complex, their natural logarithms are complex, and the L matrix contains complex numbers. The

questions now are “how to take the natural log of a complex number and how to interpret the

results?”

Begin with the natural log of a complex number je z z , which is

Earth’s Surface

d c

Equivalent Earth’s Surface

d c

d

d

Grady 2007, p. 135




j z e z z j ln)ln(ln)ln( , in radians.

Thus, for example, 4

325ln)325325ln(

j j = 5.78 – j0.785.

To be consistent with the perfect Earth case (i.e., 0, ), the real part of the natural

logarithm must produce the inductive term. Then, the imaginary term yields the resistance of the

Earth’s current path, which must be added to the series resistances of the overhead conductors.

In matrix form, the calculations are made accoding to

c

cci

bc

bci

b

bbi

ac

aci

ab

abi

a

aai

oabcabc

r Dcomplex

D

Dcomplex

r

Dcomplex

D

Dcomplex

D

Dcomplex

r

Dcomplex

j L j R

ln

lnln

lnlnln

2)()(

,

where complex depth cd is included in the D terms. Because of the j multiplier, the real

terms become positive, frequency-dependent resistances that account for the losses in the Earth.

Off-diagonal resistances account for the fact that currents in neighboring phases contribute to the

total voltage drop along the Earth, per meter, as seen by any given phase. The imaginary terms

are inductive reactances.

4. Electric Field at Surface of Overhead Conductors

Ignoring all other charges, the electric field at a conductor’s surface can be approximated by

r

q E

or

2 ,

where r is the radius. For overhead conductors, this is a reasonable approximation because the

neighboring line charges are relatively far away. It is always important to keep the peak electric

field at a conductor’s surface below 30kV/cm to avoid excessive corono losses.

Going beyond the above approximation, the Markt-Mengele method provides a detailed

procedure for calculating the maximum peak subconductor surface electric field intensity for

three-phase lines with identical phase bundles. Each bundle has N symmetric subconductors ofradius r . The bundle radius is A. The procedure is

1. Treat each phase bundle as a single conductor with equivalent radius

N N eq NrAr /11 .

Grady 2007, p. 136




2. Find the C(N x N) matrix, including ground wires, using average conductor heights above

ground. Kron reduce C(N x N) to C(3 x 3). Select the phase bundle that will have the

greatest peak line charge value ( lpeak q ) during a 60Hz cycle by successively placing

maximum line-to-ground voltage V max on one phase, and – V max/2 on each of the other

two phases. Usually, the phase with the largest diagonal term in C(3 by 3) will have the

greatest lpeak q .

3. Assuming equal charge division on the phase bundle identified in Step 2, ignore

equivalent line charge displacement, and calculate the average peak subconductor surface

electric field intensity using

r N

q E

o

lpeak peak avg

2

1,

4. Take into account equivalent line charge displacement, and calculate the maximum peak

subconductor surface electric field intensity using

A

r N E E peak avg peak )1(1,max, .

5. Resistance and Conductance

The resistance of conductors is frequency dependent because of the resistive skin effect.

Usually, however, this phenomenon is small for 50 - 60 Hz. Conductor resistances are readily

obtained from tables, in the proper units of Ohms per meter length, and these values, added to

the equivalent-earth resistances from the previous section, to yield the R used in the transmission

line model.

Conductance G is very small for overhead transmission lines and can be ignored.

6. Underground Cables

Underground cables are transmission lines, and the model previously presented applies.

Capacitance C tends to be much larger than for overhead lines, and conductance G should not be

ignored.

For single-phase and three-phase cables, the capacitances and inductances per phase per meter

length are

a

bC r o

ln

2 Farads per meter length,

and

Grady 2007, p. 137




a

b L o ln

2

Henrys per meter length,

where b and a are the outer and inner radii of the coaxial cylinders. In power cables,a

b is

typically e (i.e., 2.7183) so that the voltage rating is maximized for a given diameter.

For most dielectrics, relative permittivity 5.20.2 r . For three-phase situations, it is

common to assume that the positive, negative, and zero sequence inductances and capacitances

equal the above expressions. If the conductivity of the dielectric is known, conductance G can

be calculated using

C G Mhos per meter length.

Grady 2007, p. 138



Grady 2007, p. 139



Grady 2007, p. 140



Grady 2007, p. 141



Grady 2007, p. 142



Grady 2007, p. 143



Grady 2007, p. 144



Grady 2007, p. 145



Grady 2007, p. 146



Grady 2007, p. 147



Procedure for Computing Positive/Negative/Zero Sequence Line Constants for

Transmission Lines in Air

Page 1 of 6

POSITIVE/NEGATIVE SEQUENCE CALCULATIONS

Assumptions

Balanced, far from ground, ground wires ignored. Valid for identical single conductors per

phase, or for identical symmetric phase bundles with N conductors per phase and bundle radiusA.

Computation of positive/negative sequence capacitance

/

//

ln

2

C

o

GMR

GMDC

farads per meter,

where

3/ bcacab D D DGMD meters,

where bcacab D D D ,, are

distances between phase conductors if the line has one conductor per phase, or

distances between phase bundle centers if the line has symmetric phase bundles,

and where

/C GMR is the actual conductor radius r (in meters) if the line has one conductor per phase, or

N N C Ar N GMR 1

/

if the line has symmetric phase bundles.

Computation of positive/negative sequence inductance

/

// ln

2 L

o

GMR

GMD L

henrys per meter,

where /GMD is the same as for capacitance, and

for the single conductor case, / LGMR is the conductor gmr r (in meters), which takes

into account both stranding and the 4/1e adjustment for internal inductance. If gmr r is

not given, then assume 4/1 rer gmr , and

for bundled conductors, N N gmr L Ar N GMR 1

/

if the line has symmetric phase

bundles.

Grady 2007, p. 148





Page 2 of 6

Computation of positive/negative sequence resistance

R is the 60Hz resistance of one conductor if the line has one conductor per phase. If the line has

symmetric phase bundles, then divide the one-conductor resistance by N.

Some commonly-used symmetric phase bundle configurations

ZERO SEQUENCE CALCULATIONS

Assumptions

Ground wires are ignored. The a-b-c phases are treated as one bundle. If individual phaseconductors are bundled, they are treated as single conductors using the bundle radius method.

For capacitance, the Earth is treated as a perfect conductor. For inductance and resistance, the

Earth is assumed to have uniform resistivity . Conductor sag is taken into consideration, and a

good assumption for doing this is to use an average conductor height equal to (1/3 the conductor

height above ground at the tower, plus 2/3 the conductor height above ground at the maximum

sag point).

The zero sequence excitation mode is shown below, along with an illustration of the relationship

between bundle C and L and zero sequence C and L. Since the bundle current is actually 3Io, the

zero sequence resistance and inductance are three times that of the bundle, and the zero sequence

capacitance is one-third that of the bundle.

A A A

N = 2 N = 3 N = 4

Grady 2007, p. 149





Page 3 of 6

Computation of zero sequence capacitance

0

00

ln

2

3

1

C

C

o

GMR

GMDC

farads per meter,

where 0C GMD is the average height (with sag factored in) of the a-b-c bundle above perfect

Earth. 0C GMD is computed using

9222

0 ibc

iac

iab

iccibbiaaC D D D D D DGMD meters,

where iaa D is the distance from a to a-image, iab

D is the distance from a to b-image, and so

forth. The Earth is assumed to be a perfect conductor, so that the images are the same distance

below the Earth as are the conductors above the Earth. Also,

92223

/0 bcacabC C D D DGMRGMR meters,

where /C GMR , ab D , ac D , and bc D were described previously.

Computation of zero sequence inductance

00 ln

23

L

eo

GMR

D L

Henrys per meter,

+V

o –

Io

Io

Io 3Io

C bundle

+V

o –

Io

Io

Io 3Io

3Io

L bundle

+Vo

–

Io

Io

Io 3Io

Co Co Co

Io

+Vo

–

Io

Io 3Io

3Io Lo

Lo

Lo

Grady 2007, p. 150





Page 4 of 6

where f

De

4.658 meters (see Bergen), is the resistivity of the Earth (a good assumption

is 100-m), and f is the frequency (Hz). e D takes into account the fact that the Earth is

resistive and that zero sequence currents flow deep in the Earth. In most cases, e D is so large

that the actual height of the conductors makes no difference in the calculations. For example, for

60Hz and = 100-m, e D = 850m.

The geometric mean bundle radius is computed using

92223

/0 bcacab L L D D DGMRGMR meters,

where / LGMR , ab D , ac D , and bc D were shown previously.

Computation of zero sequence resistance

There are two components of zero sequence line resistance. First, the equivalent conductorresistance is the 60Hz resistance of one conductor if the line has one conductor per phase. If the

line has symmetric phase bundles with N conductors per bundle, then divide the one-conductor

resistance by N.

Second, the effect of resistive Earth is included by adding the following term to the conductorresistance:

f 7

10869.93

ohms per meter (see Bergen),

where the multiplier of three is needed to take into account the fact that all three zero sequence

currents flow through the Earth.

Electric Field at Surface of Overhead Conductors

Ignoring all other charges, the electric field at a conductor’s surface can be approximated by

r

q E

o

r

2

,

where r is the radius. For overhead conductors, this is a reasonable approximation because the

neighboring line charges are relatively far away. It is always important to keep the peak electric

field at a conductor’s surface below 30kV/cm to avoid excessive corono losses.

Going beyond the above approximation, the Markt-Mengele method provides a detailed

procedure for calculating the maximum peak subconductor surface electric field intensity forthree-phase lines with identical phase bundles. Each bundle has N symmetric subconductors of

radius r . The bundle radius is A. The procedure is

Grady 2007, p. 151





Page 5 of 6

1. Treat each phase bundle as a single conductor with equivalent radius

N N eq NrAr /11 .

2. Find theC(N x N)

matrix, including ground wires, using average conductor heights above

ground. Kron reduce C(N x N) to C(3 x 3). Select the phase bundle that will have the

greatest peak line charge value ( lpeak q ) during a 60Hz cycle by successively placing

maximum line-to-ground voltage V max on one phase, and – V max/2 on each of the other

two phases. Usually, the phase with the largest diagonal term in C(3 by 3) will have the

greatest lpeak q .

3. Assuming equal charge division on the phase bundle identified in Step 2, ignore

equivalent line charge displacement, and calculate the average peak subconductor surfaceelectric field intensity using

r N

q E

o

lpeak peak avg

2

1,

4. Take into account equivalent line charge displacement, and calculate the maximum peaksubconductor surface electric field intensity using

A

r N E E peak avg peak )1(1,max, .

Grady 2007, p. 152





Page 6 of 6

Practice Problem – a 345kV double-circuit configuration that is commonly used in Texas

Grady 2007, p. 153



Consider the 69kV transmission line shown below. The phases have single conductors.

a. Find the positive sequence R, L, and C per meter for the transmission line shown.

b. Convert the above values to series (R + jX), and shunt MVAr, per km, using a base of 69kV,

100MVA.

a

b

c

12m

3m

3m

= 100-m

Conductors have outer radius 1cm, and resistance 0.06 per km

Grady 2007, p. 154



Each phase of the 345kV transmission configuration shown has two bundled 795ACSR

conductors, separated by 45.7cm. The characteristics of a 795ACSR conductor are

r = 1.407cm, r gmr = 1.137cm,

R 60Hz = 0.0728/km.

The conversion from feet to meters is 3.281 feet per meter.

a. Find the positive/negative sequence C, L, and R permeter of length

b. Draw the corresponding pi-equivalent circuit, in ohms

@ 60Hz, for a 100km long line.

ft.

Grady 2007, p. 156



Grady 2007, p. 157



The a-b-c phase bundles of a three-phase, 345kV transmission line have the spacings and

average heights above Earth as shown. Each conductor has outer radius 1cm, and the twoconductors per phase are separated by 45cm. Ignoring ground wires, and making the “far from

Earth” assumption, compute the positive/negative sequence inductance and capacitance per

meter length (express your answer in µH/m and pF/m). Important - include the internal

inductance in the µH/m calculation.

20m

6m 6m

Grady 2007, p. 158



Grady 2007, p. 159



Grady 2007, p. 160



A 345kV, 300km, 60Hz three-phase overhead transmission line has positive/negative sequence L

= 1µH/m and C = 11.5pF/m.

The receiving-end voltage and load are 345kV, 500MW, pf = 0.90 lagging. Use the ABCD

matrix to determine the sending end

voltage, current,

P,

and Q.

R

R

o

o

S

S

I V

x Z

x j

x jZ x

I V

~

~

cossin

sincos

~

~

Grady 2007, p. 161



Grady 2007, p. 162



Grady 2007, p. 163



Grady 2007, p. 164



Grady, Symmetrical Components, June 2007, Page 1

Symmetrical Components

Transformation matrices and the decoupling that occurs in balanced three-phase systems.

Physical significance of zero sequence.

1. Transformation Matrix

Fortescue's Theorem: An unbalanced set of N related phasors can be resolved into N systems of

phasors called the symmetrical components of the original phasors. For a three-phase system

(i.e. N = 3), the three sets are:

1. Positive Sequence - three phasors, equal in magnitude, 120o apart, with the same sequence

(a-b-c) as the original phasors.

2. Negative Sequence - three phasors, equal in magnitude, 120o apart, with the opposite

sequence (a-c-b) of the original phasors.

3. Zero Sequence - three identical phasors (i.e. equal in magnitude, with no relative phase

displacement).

The original set of phasors is written in terms of the symmetrical components as follows:

210~~~~aaaa V V V V ,

210~~~~bbbb V V V V ,

210

~~~~

cccc

V V V V ,

where 0 indicates zero sequence, 1 indicates positive sequence, and 2 indicates negative

sequence.

The relationship among the sequence components for a-b-c are

Positive Sequence Negative Sequence Zero Sequence

o11 1201

~~ ab V V o

22 1201~~

ab V V 000~~~cba V V V

o11 1201

~~

ac V V o

22 1201~~

ac V V

The symmetrical components of all a-b-c voltages are usually written in terms of the

symmetrical components of phase a by defining

o1201 a , so that oo2 12010421 a , and oo3 010631 a .

Grady 2007, p. 165




Substituting into the previous equations for cba V V V ~

,~

,~

yields

210~~~~aaaa V V V V ,

21

2

0

~~~~

aaab V aV aV V ,

22

10~~~~aaac V aV aV V .

In matrix form, the above equations become

2

1

0

2

2

~

~

~

1

1

111

~

~

~

a

a

a

c

b

a

V

V

V

aa

aa

V

V

V

,

or in matrix form

012~~V T V abc , and abcV T V

~~ 1012 ,

where transformation matrix T is

2

2

1

1

111

aa

aaT , and

aa

aaT 2

21

1

1

111

3

1 .

If abcV ~ represents a balanced set (i.e. aab V aV V ~1201~~ 2o , ~ ~ ~V V aV c a a 1 120o ),

then substituting into abcV T V ~~ 1

012 yields

0

~0

~

~

~

1

1

111

3

1

~

~

~

2

2

2

2

1

0

a

a

a

a

a

a

a

V

V a

V a

V

aa

aa

V

V

V

.

Hence, balanced voltages or currents have only positive sequence components, and the positive

sequence components equal the corresponding phase a voltages or currents.

If abcV ~

is an identical set (i.e. cba V V V ~~~

), substituting into abcV T V ~~ 1

012 yields

0

0

~

~

~

~

1

1

111

3

1

~

~

~

2

2

2

1

0 a

a

a

a

a

a

a V

V

V

V

aa

aa

V

V

V

,

Grady 2007, p. 166




which means that cba V V V ,, have only zero sequence components, and that these components

are identical and equal to aV .

Notice from the top row of abcV T V ~~ 1

012 that 0V is one-third of the sum of the three phase

voltages or currents. Therefore, since the sum of three line-to-line voltages is identically zerodue to Kirchhoff's voltage law, line-to-line voltages can have no zero sequence components.

1. Relationship Between Zero Sequence Currents and Neutral Current

Consider the relationship between zero sequence current and neutral current The zero sequence

current is

cbaa I I I I ~~~

3

1~0 ,

and, from Kirchhoff's current law, the neutral current is

cban I I I I ~~~~

.

Because the positive and negative sequence components of the a-b-c currents sum to zero, while

the zero sequence components are additive, then 0~

3~

an I I . Therefore, in a four-wire, three-

phase system, the neutral current is three-times the zero sequence current. In a three-wire, three-

phase system, there is no zero sequence current.

Ia

Ib

Ic

In = 3Io

a

b

c

n

3 Phase,

4 Wire

In = 3Io = Ia + Ib + IcSystem

Figure 1. Relationship Between Zero Sequence Currents and Neutral Current

(note – the neutral current is shown above as flowing out)

3. Decoupling in Systems with Balanced Impedances

In a three-phase system with balanced impedances, the relationship among voltage, current, and

impedance has the form

c

b

a

c

b

a

I

I

I

S M M

M S M

M M S

V

V

V

~

~

~

~

~

~

, or abcabcabc I Z V ~~

,

Grady 2007, p. 167




where S represents the self impedances of the phases, and M represents the mutual impedances.

This equation can be expressed in terms of sequence components by substituting 012~~V T V abc

and 012~~

I T I abc , yielding

012012

~~

I T Z V T abc .

Premultiplying by 1T yields

0120120121

012~~~ I Z I T Z T V abc ,

where T Z T Z abc 1012 . The symmetric form of abc Z given above yields

M S

M S

M S

Z

00

00

002

012 ,

which means that when working in sequence components, a circuit with symmetric impedances

is decoupled into three separate impedance networks with M S Z 20 , and M S Z Z 21

. Furthermore, if the voltages and currents are balanced, then only the positive sequence circuit

must be studied.

In summary, symmetrical components are useful when studying either of the following two

situations:

1. Symmetric networks with balanced voltages and currents. In that case, only the positivesequence network must be studied, and that network is the "one-line" network.

2. Symmetric networks with unbalanced voltages and currents. In that case, decoupling

applies, and three separate networks must be studied (i.e. positive, negative, and zero

sequences). The sequence components of the voltages and currents can be transformed back

to a-b-c by using the T transformation matrix.

4. Power

For a three-phase circuit, with voltages referenced to neutral,

**** ~~~~~~~~abc

T abcccnbbnaanabc I V I V I V I V P .

Substituting in 012~~V T V abc and 012

~~ I T I abc yields

*012

*012

~~ I T T V P

T T abc .

Grady 2007, p. 168




Since

2

2

1

1

111

aa

aaT , then T T T . Also,

aa

aaT 2

2*

1

1

111

. Therefore,

300

030003*T T T , so that *

012012~

300

030003~

I V P T abc

, or

*22

*11

*00

~~~~~~3 aaaaaaabc I V I V I V P .

Note the factor of three. If desired, the following power invariant transformation can be used to

avoid the factor of three:

2

2

1

1

111

31

aa

aaT ,

aa

aaT 2

21

1

1

111

31 ,

so that *22

*11

*00

~~~~~~aaaaaaabc I V I V I V P .

Grady 2007, p. 169



G r a d y , S y m m e t r i c a l C o m p o n e n t s , J u

n e 2 0 0 7 , P a g e 6

Grady 2007, p. 170



Problem 1. Draw the positive, negative, and zero-sequence one-line diagrams for the system

shown. Use 69kV, 100MVA as the base in the transmission line. Generator, transformer, and

motor impedances are given in per unit on their own bases.

Generator

50MVA, 20kV

Subtransient reactances

X1 = X2 = 0.15pu

X0 = 0.12pu

Generator is connected

GY through a j0.3grounding reactor

X = 0.05pu

60MVA

18kV/69kV

Transformer

(Delta-GY)

X = 0.05pu

5MVA

69kV/4.16kV

Transformer

(GY-GY)

Motor

4MVA, 4.16kV


X1 = X2 = 0.10pu

X0 = 0.08pu

delta connection

Transmission lineX1 = X2 = 10

X0 = 30

Grady 2007, p. 171



Grady 2007, p. 172



All impedances are given in per unit on system base. The generators have impedances

j0.10pu, and the transformers have impedances j0.08pu. The grounding reactor hasimpedance j0.05pu. Draw the zero-sequence network, showing impedance values. Identify

all nodes (i.e., P,Q,R,S,T,U,V,W, and ground).

Grady 2007, p. 173



Grady 2007, p. 174



Grady, System Matrices, June 2007, Page 1

Power System Matrices and Matrix Operations

Nodal equations using Kirchhoff's current law. Admittance matrix and building algorithm.

Gaussian elimination. Kron reduction. LU decomposition. Formation of impedance matrix by

inversion, Gaussian elimination, and direct building algorithm.

1. Admittance Matrix

Most power system networks are analyzed by first forming the admittance matrix. The

admittance matrix is based upon Kirchhoff's current law (KCL), and it is easily formed and very

sparse.

Consider the three-bus network shown in Figure that has five branch impedances and one current

source.

1 2 3

ZE

ZA

ZB

ZC

ZDI3

Figure 1. Three-Bus Network

Applying KCL at the three independent nodes yields the following equations for the bus voltages

(with respect to ground):

At bus 1, 0211

A E Z

V V

Z

V ,

At bus 2, 032122

C A B Z

V V

Z

V V

Z

V ,

At bus 3, 3233 I

Z

V V

Z

V

C D

.

Collecting terms and writing the equations in matrix form yields

Grady 2007, p. 175




33

2

1

0

0

1110

11111

0111

I V

V

V

Z Z Z

Z Z Z Z Z

Z Z Z

DC C

C C B A A

A A E

,

or in matrix form,

I YV ,

where Y is the admittance matrix, V is a vector of bus voltages (with respect to ground), and I is a

vector of current injections.

Voltage sources, if present, can be converted to current sources using the usual network rules. If

a bus has a zero-impedance voltage source attached to it, then the bus voltage is already known,

and the dimension of the problem is reduced by one.

A simple observation of the structure of the above admittance matrix leads to the following rule

for building Y :

1. The diagonal terms of Y contain the sum of all branch admittances connected directly to the

corresponding bus.

2. The off-diagonal elements of Y contain the negative sum of all branch admittances connected

directly between the corresponding busses.

These rules make Y very simple to build using a computer program. For example, assume thatthe impedance data for the above network has the following form, one data input line per branch:

From To Branch Impedance (Entered

Bus Bus as Complex Numbers)

1 0 ZE

1 2 ZA

2 0 ZB

2 3 ZC

3 0 ZD

The following FORTRAN instructions would automatically build Y , without the need of

manually writing the KCL equations beforehand:

COMPLEX Y(3,3),ZB,YB

DATA Y/9 * 0.0/

1 READ(1,*,END=2) NF,NT,ZB

Grady 2007, p. 176




YB = 1.0 / ZB

C MODIFY THE DIAGONAL TERMS

IF(NF .NE. 0) Y(NF,NF) = Y(NF,NF) + YB

IF(NT .NE. 0) Y(NT,NT) = Y(NT,NT) + YB

IF(NF .NE. 0 .AND. NT .NE. 0) THEN

C MODIFY THE OFF-DIAGONAL TERMS

Y(NF,NT) = Y(NF,NT) - YB

Y(NT,NF) = Y(NT,NF) - YB

ENDIF

GO TO 1

2 STOP

END

Of course, error checking is needed in an actual computer program to detect data errors and

dimension overruns. Also, if bus numbers are not compressed (i.e. bus 1 through bus N ), then

additional logic is needed to internally compress the busses, maintaining separate internal andexternal (i.e. user) bus numbers.

Note that the Y matrix is symmetric unless there are branches whose admittance is direction-

dependent. In AC power system applications, only phase-shifting transformers have this

asymmetric property. The normal 30o phase shift in wye-delta transformers creates asymmetry.

2. Gaussian Elimination and Backward Substitution

Gaussian elimination is the most common method for solving bus voltages in a circuit for which

KCL equations have been written in the form

YV I .

Of course, direct inversion can be used, where

I Y V 1 ,

but direct inversion for large matrices is computationally prohibitive or, at best, inefficient.

The objective of Gaussian elimination is to reduce the Y matrix to upper-right-triangular-plus-

diagonal form (URT+D), then solve for V via backward substitution. A series of row operations

(i.e. subtractions and additions) are used to change equation

N N N N N N

N

N

N

N V

V

V

V

y y y y

y y y y

y y y y

y y y y

I

I

I

I

3

2

1

,3,2,1,

,33,32,31,3

,23,22,21,2

,13,12,11,1

3

2

1

Grady 2007, p. 177




into

N N N

N

N

N

N V

V

V

V

y

y y

y y y

y y y y

I

I

I

I

3

2

1

,'

,3'

3,3'

,2'

3,2'

2,2'

,13,12,11,1

'

'3

'2

1

000

00

0

,

in which the transformed Y matrix has zeros under the diagonal.

For illustrative purposes, consider the two equations represented by Rows 1 and 2, which are

N N

N N

V yV yV yV y I

V yV yV yV y I

,233,222,211,22

,133,122,111,11

.

Subtracting (1,1

1,2

y

y Row 1) from Row 2 yields

N N N

N N

V y y

y yV y

y

y yV y

y

y yV y

y

y y I

y

y I

V yV yV yV y I

,11,1

1,2,233,1

1,1

1,23,222,1

1,1

1,22,211,1

1,1

1,21,21

1,1

1,22

,133,122,111,11

.

The coefficient of 1V in Row 2 is forced to zero, leaving Row 2 with the desired "reduced" form

of

N N V yV yV y I ' ,23' 3,22' 2,2'2 0 .

Continuing, Row 1 is then used to "zero" the 1V coefficients in Rows 3 through N, one row at a

time. Next, Row 2 is used to zero the 2V coefficients in Rows 3 through N, and so forth.

After the Gaussian elimination is completed, and the Y matrix is reduced to (URT+D) form, the

bus voltages are solved by backward substitution as follows:

For Row N ,

N N N N V y I ' ,' , so '',

1 N

N N N I

yV .

Next, for Row N-1,

N N N N N N N V yV y I ',11

'1,1

'1 , so N N N N

N N

N V y I y

V ',1

'1'

1,1

11

.

Grady 2007, p. 178




Continuing for Row j, where 2,,3,2 N N j ,

N N j j j j j j j j V yV yV y I ',1

'1,

',

' , so

N N j j j j j j j

j V yV y I y

V ',1

'1,

'

',

1

,

which, in general form, is described by

k k j

N

jk

j

j j

j V y I y

V ',

1

'

',

1 .

A simple FORTRAN computer program for solving V in an N-dimension problem using

Gaussian elimination and backward substitution is given below.

COMPLEX Y(N,N),V(N),I(N),YMM

C GAUSSIAN ELIMINATE Y AND I

NM1 = N - 1

C PIVOT ON ROW M, M = 1,2,3, ... ,N-1

DO 1 M = 1,NM1

MP1 = M + 1

YMM = 1.0 / Y(M,M)

C OPERATE ON THE ROWS BELOW THE PIVOT ROW

DO 1 J = MP1,N

C THE JTH ROW OF II(J) = I(J) - Y(J,M) *YMM * I(M)

C THE JTH ROW OF Y, BELOW AND TO THE RIGHT OF THE PIVOT

C DIAGONAL

DO 1 K = M,N

Y(J,K) = Y(J,K) - Y(J,M) * YMM * Y(M,K)

1 CONTINUE

C BACKWARD SUBSTITUTE TO SOLVE FOR V

V(N) = I(N) / Y(N,N)

DO 2 M = 1,NM1

J = N - M

C BACKWARD SUBSTITUTE TO SOLVE FOR V, FOR

C ROW J = N-1,N-2,N-3, ... ,1

V(J) = I(J)

JP1 = J + 1

DO 3 K = JP1,N

V(J) = V(J) - Y(J,K) * V(K)

Grady 2007, p. 179




3 CONTINUE

V(J) = V(J) / Y(J,J)

2 CONTINUE

STOP

END

One disadvantage of Gaussian elimination is that if I changes, even though Y is fixed, the entire

problem must be re-solved since the elimination of Y determines the row operations that must be

repeated on I . Inversion and LU decomposition to not have this disadvantage.

3. Kron Reduction

Gaussian elimination can be made more computationally efficient by simply not performing

operations whose results are already known. For example, instead of arithmetically forcing

elements below the diagonal to zero, simply set them to zero at the appropriate times. Similarly,

instead of dividing all elements below and to the right of a diagonal element by the diagonal

element, divide only the elements in the diagonal row by the diagonal element, make the

diagonal element unity, and the same effect will be achieved. This technique, which is actually a

form of Gaussian elimination, is known as Kron reduction.

Kron reduction "pivots" on each diagonal element ',mm y , beginning with 1,1 y , and continuing

through 1,1 N N y . Starting with Row m = 1, and continuing through Row m = N - 1, the

algorithm for Kron reducing YV I is

1. Divide the elements in Row m, that are to the right of the diagonal, by the diagonal

element ',mm y . (Note - the elements to the left of the diagonal are already zero).

2. Replace element 'm I with

',

'

mm

m

y

I .

3. Replace diagonal element ',mm y with unity.

4. Modify the 'Y elements in rows greater than m and columns greater than m (i.e. below

and to the right of the diagonal element) using

',

',

',

', k mm jk jk j y y y y , for j > m, k > m.

5. Modify the ' I elements below the mth row according to

'',

''mm j j j I y I I , for j > m.

6. Zero the elements in Column m of 'Y that are below the diagonal element.

Grady 2007, p. 180




A FORTRAN code for Kron reduction is given below.

COMPLEX Y(N,N),V(N),I(N),YMM

C KRON REDUCE Y, WHILE ALSO PERFORMING ROW OPERATIONS ON I

NM1 = N - 1

C PIVOT ON ROW M, M = 1,2,3, ... ,N-1DO 1 M = 1,NM1

MP1 = M + 1

YMM = 1.0 / YMM

C DIVIDE THE PIVOT ROW BY THE PIVOT

DO 2 K = MP1,N

Y(M,K) = Y(M,K) * YMM

2 CONTINUE

C OPERATE ON THE I VECTOR

I(M) = I(M) * YMM

C SET THE PIVOT TO UNITY

Y(M,M) = 1.0

C REDUCE THOSE ELEMENTS BELOW AND TO THE RIGHT OF THE PIVOT

DO 3 J = MP1,N

DO 4 K = MP1,N

Y(J,K) = Y(J,K) - Y(J,M) * Y(M,K)

4 CONTINUE

C OPERATE ON THE I VECTOR

I(J) = I(J) - Y(J,M) * I(M)

C SET THE Y ELEMENTS DIRECTLY BELOW THE PIVOT TO ZERO

Y(J,M) = 0.0

3 CONTINUE

1 CONTINUE

4. LU Decomposition

An efficient method for solving V in matrix equation YV = I is to decompose Y into the product

of a lower-left-triangle-plus-diagonal (LLT+D) matrix L, and an (URT+D) matrix U , so that YV

= I can be written as

LUV = I .

The benefits of decomposing Y will become obvious after observing the systematic procedure for

finding V .

It is customary to set the diagonal terms of U equal to unity, so that there are a total of 2 N

unknown terms in L and U . LU = Y in expanded form is then

Grady 2007, p. 181




N N N N N

N

N

N

N

N

N

N N N N N y y y y

y y y y

y y y y

y y y y

u

uu

uuu

l l l l

l l l

l l

l

,3,2,1,

,33,32,31,3

,23,22,21,2

,13,12,11,1

,3

,23,2

,13,12,1

,3,2,1,

3,32,31,3

2,21,2

1,1

1000

100

10

1

0

00

000

.

Individual l and u terms are easily found by calculating them in the following order:

1. Starting from the top, work down Column 1 of L, finding 1,1l , then 1,2l , then 1,3l , ,

l N ,1 . For the special case of Column 1, these elements are 1,1, j j yl , j = 1,2,3, ,N .

2. Starting from the left, work across Row 1 of U , finding 2,1u , then 3,1u , then 4,1u , ,

N u ,1 . For the special case of Row 1, these elements are1,1

,1,1

l

yu

k k , k = 2,3,4, ,N .

3. Work down Column (k = 2) of L, finding 2,2l , then 2,3l , then 2,4l , , 2, N l , using

N k k k jul yl k m

k

mm jk jk j ,,2,1,,,

1

1

,,,

, Column N k k 2, .

4. Work across Row (k = 2) of U , finding 3,2u , then 4,2u , then 5,2u , , N u ,2 , using

N k k jl

ul y

uk k

k

m jmmk jk

jk ,,2+,1+=,,

1

1 ,,,

,

, Row )1(2, N k k .

5. Repeat Steps 3 and 4, first for Column k of L, then for Row k of U . Continue for all k =

3,4,5, ,( N 1) for L and U , then for k = N for L.

The procedure given above in Steps 1 - 5 is often referred to as Crout's method. Note that

elements of L and U never look "backward" for previously used elements of Y . Therefore, in

order to conserve computer memory, L and U elements can be stored as they are calculated in the

same locations at the corresponding Y elements. Thus, Crout's method is a memory-efficient "in

situ" procedure.

An intermediate column vector is needed to find V . The intermediate vector D is defined as

D = UV ,

so that

Grady 2007, p. 182




N

N

N

N

N V

V

V

V

u

uu

uuu

d

d

d

d

3

2

1

,3

,23,2

,13,12,1

3

2

1

1000

100

10

1

.

Since LUV = I , then LD = I . Vector D is found using forward-substitution from

N N N N N N N I

I

I

I

d

d

d

d

l l l l

l l l

l l

l

3

2

1

3

2

1

,3,2,1,

3,32,31,3

2,21,2

1,1

0

00

000

,

which proceeds as follows:

From Row 1, 11,1

1111,11

, I l

d I d l ,

From Row 2, 11,222,2

2222,211,21

, d l I l

d I d l d l ,

From Row k ,

j jk

k

jk k k k k k k k k k

d l I l

d I d l d l d l ,

1

1,,22,11,

1, .

Now, since D = UV , or

N

N

N

N

N V

V

V

V

u

uu

uuu

d

d

d

d

3

2

1

,3

,23,2

,13,12,1

3

2

1

1000

100

10

1

,

where D and U are known, then V is found using backward substitution.

An important advantage of LU decomposition over Gaussian elimination or Kron reduction is

that the I vector is not modified during decomposition. Therefore, once Y has been decomposed

into L and U , I can be modified, and V recalculated, with minimal work, using the forward and

backward substitution steps shown above.

Grady 2007, p. 183




A special form of L is helpful when Y is symmetric. In that case, let both L and U have unity

diagonal terms, and define a diagonal matrix D so that

Y = LDU ,

or

1000

100

10

1

000

000

000

000

1

0

01

001

0001

,3

,23,2

,13,12,1

,

3,3

2,2

1,1

3,2,1,

2,31,3

1,2

N

N

N

N N N N N

u

uu

uuu

d

d

d

d

l l l

l l

l

Y .

Since Y is symmetric, then T Y Y , and T T T T T T DLU L DU LDU LDU . Therefore,

an acceptable solution is to allow

T

U L . Incorporating this into the above equation yields

N N

N

N

N

N N N d

l d d

l d l d d

l d l d l d d

l l l

l l

l

Y

,

3,3,33,3

2,2,22,32,22,2

1,1,11,31,11,21,11,1

3,2,1,

2,31,3

1,2

000

00

0

1

0

01

001

0001

,

which can be solved by working from top-to-bottom, beginning with Column 1 of Y , as follows:

Working down Column 1 of Y ,

1,11,1 d y ,

1,11,21,21,11,21,2 /so, d yl d l y ,

1,11,1,1,11,1, /so, d yl d l y j j j j .

Working down Column 2 of Y ,

1,21,11,22,22,22,21,21,11,22,2so, l d l yd d l d l y ,

N jd l d l yl d l l d l y j j j j j j 2,/so, 2,21,21,11,2,2,2,22,1,21,11,2, .

Working down Column k of Y ,

Grady 2007, p. 184




mk mm

k

m

mk k k k k k k mk mm

k

m

mk k k l d l yd d l d l y ,,

1

1,,,,,,

1

1,, so,

,

N jk d l d l yl l d l yk k mk mm

k

m m jk jk jmk mm

k

m m jk j

,/so,

,,,

1

1 ,,,,,

1

1 ,,

.

This simplification reduces memory and computation requirements for LU decomposition by

approximately one-half.

5. Bifactorization

Bifactorization recognizes the simple pattern that occurs when doing "in situ" LU

decomposition. Consider the first four rows and columns of a matrix that has been LU

decomposed according to Crout's method:

4,33,44,22,44,11,44,44,43,22,43,11,43,43,42,11,42,42,41,41,4

3,34,22,34,11,34,34,33,22,33,11,33,33,32,11,32,32,31,31,3

2,24,11,24,24,22,23,11,23,23,22,11,22,22,21,21,2

1,14,14,11,13,13,11,12,12,11,11,1

/

/////

ul ul ul yl ul ul yl ul yl yl

l ul ul yuul ul yl ul yl yl

l ul yul ul yuul yl yl l yul yul yu yl

.

The pattern developed is very similar to Kron reduction, and it can be expressed in the following

steps:

1. Beginning with Row 1, divide the elements to the right of the pivot element, and in the

pivot row, by 1,1l , so that

1,1

,1',1

y

y y

k k , for k = 2,3,4, , N .

2. Operate on the elements below and to the right of the pivot element using

',11,,

', k jk jk j y y y y , for j = 2,3,4, , N , k = 2,3,4, , N.

3. Continue for pivots m = 2,3,4, , ( N - 1) using

N mmk y

y y

mm

k mk m ,,2,1,

',

'

,', ,

followed by

N mmk N mm j y y y y k mm jk jk j ,,2,1;,,2,1,',

',

',

', ,

Grady 2007, p. 185




for each pivot m.

When completed, matrix Y' has been replaced by matrices L and U as follows:

N N N N N l l l l

ul l l uul l

uuul

Y

,3,2,1,

5,33,32,31,3

5,23,22,21,2

5,13,12,11,1

'

,

and where the diagonal u elements are unity (i.e. 1,3,32,21,1 N N uuuu ).

The corresponding FORTRAN code for bifactorization is

COMPLEX Y(N,N),YMM

C DO FOR EACH PIVOT M = 1,2,3, ... ,N - 1

NM1 = N - 1

DO 1 M = 1,NM1

C FOR THE PIVOT ROW

MP1 = M + 1

YMM = 1.0 / Y(M,M)

DO 2 K = MP1,N

Y(M,K) = Y(M,K) * YMM

2 CONTINUE

C BELOW AND TO THE RIGHT OF THE PIVOT ELEMENT

DO 3 J = MP1,N

DO 3 K = MP1,N

Y(J,K) = Y(J,K) - Y(J,M) * Y(M,K)

3 CONTINUE

1 CONTINUE

STOP

END

6. Shipley-Coleman Inversion

For relatively small matrices, it is possible to obtain the inverse directly. The Shipley-Coleman

inversion method for inversion is popular because it is easily programmed. The algorithm is

1. For each pivot (i.e. diagonal term) m, m = 1,2,3, … , N , perform the following Steps 2 - 4.

2. Kron reduce all elements in Y , above and below, except those in Column m and Row musing

Grady 2007, p. 186




mk N k m j N j y

y y y y

mm

k mm jk jk j ,,,3,2,1;,,,3,2,1,

,

',

','

,', .

3. Replace pivot element ',mm y with its negative inverse, i.e.

' ,

1

mm y

.

4. Multiply all elements in Row m and Column m, except the pivot, by ',mm y .

The result of this procedure is actually the negative inverse, so that when completed, all terms

must be negated. A FORTRAN code for Shipley-Coleman is shown below.

COMPLEX Y(N,N),YPIV

C DO FOR EACH PIVOT M = 1,2,3, ... ,N

DO 1 M = 1,N

YPIV = 1.0 / Y(M,M)C KRON REDUCE ALL ROWS AND COLUMNS, EXCEPT THE PIVOT ROW

C AND PIVOT COLUMN

DO 2 J = 1,N

IF(J .EQ. M) GO TO 2

DO 2 K = 1,N

IF(K. NE. M) Y(J,K) = Y(J,K) - Y(J,M) * Y(M,K) * YPIV

2 CONTINUE

C INVERT THE PIVOT ELEMENT AND NEGATE IT

YPIV = -YPIV

Y(M,M) = YPIV

C WORK ACROSS THE PIVOT ROW AND DOWN THE PIVOT COLUMN,

C MULTIPLYING BY THE NEW PIVOT VALUE

DO 3 K = 1,N

IF(K .EQ. M) GO TO 3

Y(M,K) = Y(M,K) * YPIV

Y(K,M) = Y(K,M) * YPIV

3 CONTINUE

1 CONTINUE

C NEGATE THE RESULTS

DO 4 J = 1,N

DO 4 K = 1,N

Y(J,K) = -Y(J,K)

4 CONTINUE

STOP

END

Grady 2007, p. 187




The order of the number of calculations for Shipley-Coleman is 3 N .

7. Impedance Matrix

The impedance matrix is the inverse of the admittance matrix, or

1 Y Z ,

so that

ZI V .

The reference bus both Y and Z is ground. Although the impedance matrix can be found via

inversion, complete inversion is not common for matrices with more than a few hundred rows

and columns because of the matrix storage requirements. In those instances, Z elements are

usually found via Gaussian elimination, Kron reduction, or, less commonly, by a direct building

algorithm. If only a few of the Z elements are needed, then Gaussian elimination or Kronreduction are best. Both methods are described in following sections.

8. Physical Significance of Admittance and Impedance Matrices

The physical significance of the admittance and impedance matrices can be seen by examining

the basic matrix equations YV I and ZI I Y V 1 . Expanding the jth row of YV I yields

N

k

k k j j V y I

1, . Therefore,

k m N mmV k

jk j

V I y

,,,2,1,0

,

,

where, as shown in Figure 2, all busses except k are grounded, k V is a voltage source attached to

bus k , and j I is the resulting current injection at (i.e. flowing into) bus j. Since all busses that

neighbor bus k are grounded, the currents induced by k V will not get past these neighbors, and

the only non-zero injection currents j I will occur at the neighboring busses. In a large power

system, most busses do not neighbor any arbitrary bus k . Therefore, Y consists mainly of zeros

(i.e. is sparse) in most power systems.

Grady 2007, p. 188




+

-

Vk

Bus jApplied Voltage

Power System

Ij

All Other Busses

Grounded

at Bus k

Induced Current at

A

Figure 2. Measurement of Admittance Matrix Term k j y ,

Concerning Z , the kth row of ZI I Y V 1 yields

N

j

j jk k I z V

1

, . Hence,

k m N mm I k

jk j

I

V z

,,,2,1,0

,

,

where, as shown in Figure 3, I k is a current source attached to bus k , V j is the resulting voltage at

bus j, and all busses except k are open-circuited. Unless the network is disjoint, then current

injection at one bus, when all other busses open-circuited, will raise the potential everywhere in

the network. For that reason, Z tends to be full.

Power System

All Other Busses

Open Circuited

Applied Current atInduced Voltage

+

-

V

Bus k at Bus j

Ik

Vj

Figure 3. Measurement of Impedance Matrix Term k j z ,

9. Formation of the Impedance Matrix via Gaussian Elimination, Kron Reduction,and LU Decomposition

Gaussian Elimination

An efficient method of fully or partially inverting a matrix is to formulate the problem using

Gaussian elimination. For example, given

Grady 2007, p. 189




I YZ ,

where Y is a matrix of numbers, Z is a matrix of unknowns, and I is the identity matrix, the

objective is to Gaussian eliminate Y , while performing the same row operations on I , to obtain

the form

I Z Y ,

where Y' is in (URT+D) form. Then, individual columns of Z can then be solved using backward

substitution, one at a time. This procedure is also known as the augmentation method, and it is

illustrated as follows:

Y is first Gaussian eliminated, as shown in a previous section, while at the same time performing

identical row operations on I . Afterward, the form of I Z Y is

N N N N N

N N

N

N N

N N

N

z z z z

z z z z

z z z z

z z z z

y

y y y y y

y y y y

,3,2,1,

,33,32,31,3,23,22,21,2

,13,12,11,1

,'

,3'

3,3' ,2

'

3,2

'

2,2

',13,12,11,1

000

000

'

,

'

3,

'

2,

'

1,

'3,3

'2,3

'1,3

'2,2

'1,2

0

00

0001

N N N N N I I I I

I I I

I I

where Rows 1 of Y' and I' are the same as in Y and I . The above equation can be written in

abbreviated column form as

''3

'2

'1321

,'

,3'

3,3'

,2'

3,2'

2,2'

,13,12,11,1

000

00

0

N N

N N

N

N

N

I I I I Z Z Z Z

y

y y

y y y

y y y y

,

where the individual column vectors Z i and I i have dimension N x 1. The above equation can be

solved as N separate subproblems, where each subproblem computes a column of Z . For

example, the kth column of Z can be computed by applying backward substitution to

Grady 2007, p. 190




',

',3

',2

',1

,

,3

,2

,1

,'

,3'

3,3'

,2'

3,2'

2,2'

,13,12,11,1

000

00

0

k N

k

k

k

k N

k

k

k

N N

N

N

N

I

I

I

I

Z

Z

Z

Z

y

y y

y y y

y y y y

.

Each column of Z is solved independently from the others.

Kron Reduction

If Kron reduction, the problem is essentially the same, except that Row 1 of the above equation

is divided by '1,1 y , yielding

',

',3

',2

'

,1

,

,3

,2,1

,'

,3'

3,3'

,2'

3,2'

2,2' ,13,12,11,1

000

00

0

k N

k

k k

k N

k

k k

N N

N

N N

I

I

I

I

Z

Z

Z

Z

y

y y

y y y

y y y y

.

Because backward substitution can stop when the last desired z element is computed, the process

is most efficient if the busses are ordered so that the highest bus numbers (i.e. N , N-1, N-2, etc.)

correspond to the desired z elements. Busses can be ordered accordingly when forming Y to take

advantage of this efficiency.

LU Decomposition

Concerning LU decomposition, once the Y matrix has been decomposed into L and U , then we

have

I LUZ YZ ,

where I is the identity matrix. Expanding the above equation as I UZ L yields

N N N N N

N

N

N

N N N N N uz uz uz uz

uz uz uz uz

uz uz uz uz

uz uz uz uz

l l l l

l l l

l l

l

,3,2,1,

,33,32,31,3

,23,22,21,2

,13,12,11,1

,3,2,1,

3,33,21,3

2,21,2

1,1

0

00

000

.

Grady 2007, p. 191




1000

0100

0010

0001

The special structure of the above equation shows that, in general, UZ must be (LLT+D) in form.

UZ can be found by using forward substitution on the above equation, and then Z can be found,

one column at a time, by using backward substitution on Z U UZ , which in expanded form

is

N N N N N

N

N

N

N

N

N

z z z z

z z z z

z z z z

z z z z

u

uu

uuu

,3,2,1,

,33,32,31,3

,23,22,21,2

,13,12,11,1

,3

,23,2

,13,12,1

1000

100

10

1

N N N N N uz uz uz uz

uz uz uz

uz uz

uz

,3,2,1,

3,32,31,3

2,21,2

1,1

0

00

000

.

10. Direct Impedance Matrix Building and Modification Algorithm

The impedance matrix can be built directly without need of the admittance matrix, if the physical

properties of Z are exploited. Furthermore, once Z has been built, no matter which method was

used, it can be easily modified when network changes occur. This after-the-fact modification

capability is the most important feature of the direct impedance matrix building and modification

algorithm.

The four cases to be considered in the algorithm are

Case 1. Add a branch impedance between a new bus and the reference bus.

Case 2. Add a branch impedance between a new bus and an existing non-reference bus.

Case 3. Add a branch impedance between an existing bus and the reference bus.

Case 4. Add a branch impedance between two existing non-reference busses.

Direct formation of an impedance matrix, plus all network modifications, can be achieved

through these four cases. Each of the four is now considered separately.

Grady 2007, p. 192




Case 1, Add a Branch Impedance Between a New Bus and the Reference Bus

Case 1 is applicable in two situations. First, it is the starting point when building a new

impedance matrix. Second, it provides (as does Case 2) a method for adding new busses to an

existing impedance matrix.

Both situations applicable to Case 1 are shown in Figure 4. The starting-point situation is on the

left, and the new-bus situation is on the right.

Power System N Busses

Zadd

Bus 1

ReferenceBus

Zadd

Bus (N+1)

Situation 1 Situation 2

ReferenceBus

Figure 4. Case 1, Add a Branch Impedance Between a New Bus and the Reference Bus

The impedance matrices for the two situations are

Situation 1 Zadd Z 1,1 ,

Situation 2:

111

1

,1,1

000

0

0

0

x xN

Nx

NxN N N N N

Zadd

Z Z

.

The effect of situation 2 is simply to augment the existing N N Z , by a column of zeros, a row of

zeros, and a new diagonal element Zadd . New bus ( N+1) is isolated from the rest of the system.

Case 2, Add a Branch Impedance Between a New Bus and an Existing Non-Reference Bus

Consider Case 2, shown in Figure 5, where a branch impedance Zadd is added from existing

non-reference bus j to new bus ( N+1). Before the addition, the power system has N busses, plusa reference bus (which is normally ground), and an impedance matrix N N Z , .

Grady 2007, p. 193




+

-

Vk

Bus j

Power System

at Bus k

A

Induced Voltage

Injected Current at

I(N+1)

Zadd New Bus (N+1)

Figure 5. Case 2, Add a Branch Impedance Between New Bus ( N+1) and Existing Non-

Reference Bus j

Since all of the current 1 N I injected at new bus ( N+1) flows through Zadd and into Bus j, the

original N power system busses cannot distinguish between current injections j I and 1 N I .

Therefore,

)1(,1

N k I

V

I

V

j

k

N

k

,

meaning that impedance matrix elements jk z , and 1, N k z are identical for busses k N ( )1 ,

and that the effect on the impedance matrix is to augment it with an additional Row ( N+1) that is

identical to Row j.

Likewise, since bus ( N+1) is an open-circuited radial bus stemming from bus j, a current injected

at another bus k creates identical voltage changes on busses j and ( N+1). This means that

)1(,1 N k I

V

I

V

k

j

k

N

,

so that the effect on the impedance matrix is to augment it with an additional Column ( N+1) that

is identical to Column j.

Next, due to the fact that all of injection current 1,1 N N I passes through new branch impedance

Zadd , the relationship between jV and 1 N V is

Zadd I

V

I

V

N

j

N

N

11

1

,

so that the ( N+1) diagonal term of the impedance matrix becomes

Zadd z z j j N N ,1,1 ,

Grady 2007, p. 194




where j j z , is the jth diagonal term of NxN Z .

Summarizing, the total effect of Case 2 is to increase the dimension of the impedance matrix by

one row and column according to

11,1,

1,,1,1 of Row

of Column

x j j xN N N

Nx N N NxN N N N N Zadd z Z j

Z j Z Z .

Case 3, Add a Branch Impedance Between an Existing Bus and the Reference Bus

Now, consider Case 3, where a new impedance-to-reference tie Zadd is added to existing Bus j.

The case is handled as a two-step extension of Case 2, as shown in Figure 4.6. First, extend

Zadd from Bus j to fictitious Bus ( N+1), as was done in Case 2. Then, tie fictitious Bus ( N+1)

to the reference bus, and reduce the size of the augmented impedance matrix back to ( N x N ).

Bus j

Power System

New Bus (N+1)

Step 1 Step 2

Bus j

Power System

Zadd

Zadd

Figure 6. Case 3. Two-Step Procedure for Adding Branch Impedance Zadd from Existing Bus jto the Reference Bus

Step 1 creates the augmented 1,1 N N Z matrix shown in Case 2. The form of equation

N N N N I Z V 1,11 is

REF

Nx

x j j xN N N

Nx N N NxN N N

REF

Nx

I

I

Zadd z Z j

Z j Z

V

V 1

11,1,

1,,1

of Row

of Column

0 ,

where REF j j REF I Zadd z V and,, , are scalars. Defining the row and column vectors as j R

and jC , respectively, yields

REF

Nx

x j j xN j

Nx NxN N N

REF

Nx

I

I

Zadd z R

Z

V

V 1

11,1

1 j,1C

0 .

At this point, scalar REF I can be eliminated by expanding the bottom row to obtain

Grady 2007, p. 195




Zadd z

I R I

j j

x Nx j

REF

,

111 .

Substituting into the top N rows yields

1,

1,

1111

Nx j j j j

NxN Nx j j j j

Nx NxN Nx I RC Zadd z

Z I RC Zadd z

I Z V

,

or

1'

1 Nx NxN Nx I Z V .

Expanding j j RC gives

N j j N j j N j j N j j N

N j j j j j j j j

N j j j j j j j j

N j j j j j j j j

N j j j j

j N

j

j

j

z z z z z z z z

z z z z z z z z

z z z z z z z z z z z z z z z z

z z z z

z

z

z z

,,3,,2,,1,,

,,33,,32,,31,,3

,,23,,22,,21,,2

,,13,,12,,11,,1

,3,2,1,

,

,3

,2

,1

,

so that the individual elements of the modified impedance matrix '1 Nx Z can be written as

N k N i Zadd z

z z

z z j j

k j ji

k ik i ,,3,2,1;,,3,2,1,,

,,

,

'

,

.

Note that the above equation corresponds to Kron reducing the augmented impedance matrix,

using element 1,1 N N z as the pivot.

Case 4, Add a Branch Impedance Between Two Existing Non-Reference Busses

The final case to be considered is that of adding a branch between existing busses j and k in an

N -bus power system that has impedance matrix N N Z , . The system and branch are shown in

Figure 4.7.

Grady 2007, p. 196




Power System

Zadd

Reference Bus

Bus j

Ij

Ib

Ij - Ib

Bus k

Ik Ik + Ib

Ib

with N Busses

Figure 7. Case 4, Add Branch Impedance Zadd between Existing Busses j and k

As seen in the figure, the actual current injected into the system via Bus j is b j I I , and the

actual injection via Bus k is bk I I , where

Zadd

V V I

k jb

, or bk j I Zadd V V .

The effect of the branch addition on the voltage at arbitrary bus m can be found by substituting

the true injection currents at busses j and k into I Z V N N , , yielding

b j jmmmmm I I z I z I z I z V ,33,22,11,

N N mbk k m I z I I z ,, ,

or

k k m j jmmmmm I z I z I z I z I z V ,,33,22,11,

bk m jm N N m I z z I z ,,, .

For busses j and k specifically,

b j jk k k k

b j j j j j j

k

j

I I z I z I z I z

I I z I z I z I z

V

V

,33,22,11,

,33,22,11,

N N k bk k k

N N jbk k j

I z I I z

I z I I z

,,

,,

.

Then, k j V V gives

Grady 2007, p. 197




,,33,3,22,2,11,1, b j jk j jk jk jk jk j I I z z I z z I z z I z z V V

N N k N jbk k k k j I z z I I z z ,,,, .

Substituting in bk j

I Zadd V V and combining terms then yields

0 ,,33,3,22,2,11,1, j jk j jk jk jk j I z z I z z I z z I z z

b jk k jk k j j N N k N jk k k k j I Zadd z z z z I z z I z z ,,,,,,,, .

All of the above effects can be included as an additional row and column in equation

I Z V N N , as follows:

1,,,,1,,

1,,,

1

1

f R f Row

f Colf Col

0 Nx jk k jk k j j xN N N N N

Nx N N N N NxN N N

Nx

Nx N

Zadd z z z z Z ok ow Z o j

Z ok Z o j Z V .

11

1

xb

Nx N

I

I .

As was done in Case 3, the effect of the augmented row and column of the above equation can be

incorporated into a modified impedance matrix by using Kron reduction, where element z N N 1 1,

is the pivot, yielding

Zadd z z z z

z z z z z z

jk k jk k j j

mk m jk i ji

mimi

,,,,

,,,,

,

'

, .

Application Notes

Once an impedance matrix is built, no matter which method is used, the modification algorithm

can very easily adjust Z for network changes. For example, a branch outage can be effectively

achieved by placing an impedance, of the same but negative value, in parallel with the actual

impedance, so that the impedance of the parallel combination is infinite.

11. Example Code YZBUILD for Building Admittance Matrix and Impedance Matrixc

c program yzbuild. 10/10/04.c implicit none

complex ybus,yline,ycap,ctap,yii,ykk,yik,ykicomplex lmat,umat,unity,czero,ytest,ysave,diff,uz,1 uzelem,zbus,elem,alphainteger nb,nbus,nfrom,nto,j,k,ipiv,jdum,kdum,jm1,kp1,linteger nlu,ipiv1,nm1,ny,nbext,nbint,iout,mxext,mxintinteger jp1,ipp1,ngy,ngureal eps4,eps6,eps9,pi,dr,qshunt,r,x,charge,fill,ymag,umag

Grady 2007, p. 198




dimension ybus(150,150),nbext(150),nbint(9999)dimension lmat(150,150),umat(150,150),ysave(150,150),1 uz(150,150),zbus(150,150),unity(150,150)data mxint/150/,mxext/9999/,nbus/0/, nbint/9999 * 0/, iout/6/,1 nbext/150 * 0/,czero /(0.0,0.0)/,eps4/1.0e-04/,2 eps6/1.0e-06/,eps9/1.0e-09/

data ybus /22500 * (0.0,0.0)/data zbus /22500 * (0.0,0.0)/data uz /22500 * (0.0,0.0)/data unity /22500 * (0.0,0.0)/

c pi = 4.0 * atan(1.0)

dr = 180.0 / pi

open(unit=1,file='uz.txt')write(1,*) 'errors'close(unit=1,status='keep')open(unit=1,file='zbus_from_lu.txt')write(1,*) 'errors'close(unit=1,status='keep')

open(unit=1,file='ybus_gaussian_eliminated.txt')write(1,*) 'errors'close(unit=1,status='keep')open(unit=1,file='unity_mat.txt')write(1,*) 'errors'close(unit=1,status='keep')open(unit=1,file='zbus_from_gaussian_elim.txt')write(1,*) 'errors'close(unit=1,status='keep')

open(unit=6,file='yzbuild_screen_output.txt')open(unit=2,file='ybus.txt')open(unit=1,file='bdat.csv')write(6,*) 'program yzbuild, 150 bus version, 10/10/04'

write(2,*) 'program yzbuild, 150 bus version, 10/10/04'write(6,*) 'read bus data from file bdat.csv'write(6,*) 'number b'write(2,*) 'read bus data from file bdat.csv'write(2,*) 'number b'

cc read the bdat.csv filec 1 read(1,*,end=5,err=30) nb,qshunt

if(nb .eq. 0) go to 5write(6,1002) nb,qshunt

write(2,1002) nb,qshunt1002 format(i7,5x,f10.4)

if(nb .le. 0 .or. nb .gt. mxext) thenwrite(6,*) 'illegal bus number - stop'write(2,*) 'illegal bus number - stop'write(*,*) 'illegal bus number - stop'stop

endifnbus = nbus + 1if(nbus .gt. mxint) thenwrite(6,*) 'too many busses - stop'write(2,*) 'too many busses - stop'write(*,*) 'too many busses - stop'stop

Grady 2007, p. 199




endifnbext(nbus) = nbnbint(nb) = nbusycap = cmplx(0.0,qshunt)ybus(nbus,nbus) = ybus(nbus,nbus) + ycapgo to 1

5 close(unit=1,status='keep')

c open(unit=1,file='ldat.csv')

write(6,*)write(6,*) 'read line/transformer data from file ldat.csv'write(6,*) ' from to r x',1 ' b'write(2,*) 'read line/transformer data from file ldat.csv'write(2,*) ' from to r x',1 ' b'

cc read the ldat.csv filec 10 read(1,*,end=15,err=31) nfrom,nto,r,x,charge

if(nfrom .eq. 0 .and. nto .eq. 0) go to 15

write(6,1004) nfrom,nto,r,x,chargewrite(2,1004) nfrom,nto,r,x,charge1004 format(1x,i5,2x,i5,3(2x,f10.4)) if(nfrom .lt. 0 .or. nfrom .gt. mxext .or. nto .lt. 0

1 .or. nto .gt. mxext) thenwrite(6,*) 'illegal bus number - stop'write(2,*) 'illegal bus number - stop'write(*,*) 'illegal bus number - stop'stop

endifif(nfrom .eq. nto) thenwrite(6,*) 'same bus number given on both ends - stop'write(2,*) 'same bus number given on both ends - stop'write(*,*) 'same bus number given on both ends - stop'

stopendifif(r .lt. -eps6) thenwrite(6,*) 'illegal resistance - stop'write(2,*) 'illegal resistance - stop'write(*,*) 'illegal resistance - stop'stop

endif if(charge .lt. -eps6) then

write(6,*) 'line charging should be positive'write(2,*) 'line charging should be positive'write(*,*) 'line charging should be positive'stop

endifif(nfrom .ne. 0) thennfrom = nbint(nfrom)if(nfrom .eq. 0) thenwrite(6,*) 'bus not included in file bdat.csv - stop'write(2,*) 'bus not included in file bdat.csv - stop'write(*,*) 'bus not included in file bdat.csv - stop'stop

endif endif

if(nto .ne. 0) then nto = nbint(nto)

Grady 2007, p. 200




if(nto .eq. 0) thenwrite(6,*) 'bus not included in file bdat.csv - stop'

write(2,*) 'bus not included in file bdat.csv - stop'write(*,*) 'bus not included in file bdat.csv - stop'stop

endif endif

yline = cmplx(r,x)if(cabs(yline) .lt. eps6) go to 10yline = 1.0 / ylineycap = cmplx(0.0,charge / 2.0)

cc the line charging termsc if(nfrom .ne. 0) ybus(nfrom,nfrom) = ybus(nfrom,nfrom) + ycap

if(nto .ne. 0) ybus(nto ,nto ) = ybus(nto ,nto ) + ycapcc shunt elementsc if(nfrom .ne. 0 .and. nto .eq. 0) ybus(nfrom,nfrom) =

1 ybus(nfrom,nfrom) + ylineif(nfrom .eq. 0 .and. nto .ne. 0) ybus(nto ,nto ) =1 ybus(nto ,nto ) + yline

cc transmission linesc if(nfrom .ne. 0 .and. nto .ne. 0) then

ybus(nfrom,nto ) = ybus(nfrom,nto ) - ylineybus(nto ,nfrom) = ybus(nto ,nfrom) - ylineybus(nfrom,nfrom) = ybus(nfrom,nfrom) + ylineybus(nto ,nto ) = ybus(nto ,nto ) + yline

endifgo to 10

c

c write the nonzero ybus elements to file ybusc 15 close(unit=1,status='keep')

write(6,*)write(6,*) 'nonzero elements of ybus (in rectangular form)'write(2,*) 'nonzero elements of ybus (in rectangular form)'write(6,*) '-internal- -external-'write(2,*) '-internal- -external-'ny = 0

do j = 1,nbusdo k = 1,nbusysave(j,k) = ybus(j,k)if(cabs(ybus(j,k)) .ge. eps9) thenny = ny + 1write(6,1005) j,k,nbext(j),nbext(k),ybus(j,k)write(2,1005) j,k,nbext(j),nbext(k),ybus(j,k)

1005 format(2i5,3x,2i5,2e20.8)endif

end doend do

close(unit=2,status='keep') fill = ny / float(nbus * nbus)

write(iout,525) ny,fill

Grady 2007, p. 201




525 format(/1x,'number of nonzero elements in ybus = ',i5/1x,1'percent fill = ',2pf8.2/)

cc bifactorization - replace original ybus with luc nm1 = nbus - 1

do ipiv = 1,nm1write(iout,530) ipiv

530 format(1x,'lu pivot element = ',i5)ipiv1 = ipiv + 1alpha = 1.0 / ybus(ipiv,ipiv)do k = ipiv1,nbusybus(ipiv,k) = alpha * ybus(ipiv,k)

end dodo j = ipiv1,nbusalpha = ybus(j,ipiv)do k = ipiv1,nbusybus(j,k) = ybus(j,k) - alpha * ybus(ipiv,k)

end doend do

end do

write(iout,530) nbuswrite(iout,532)

532 format(/1x,'nonzero lu follows'/)open(unit=4,file='lu.txt')nlu = 0

do j = 1,nbusdo k = 1,nbusymag = cabs(ybus(j,k))if(ymag .gt. eps9) thennlu = nlu + 1write(iout,1005) j,k,nbext(j),nbext(k),ybus(j,k)

write(4,1005) j,k,nbext(j),nbext(k),ybus(j,k)endifend do

end do

fill = nlu / float(nbus * nbus)write(iout,535) nlu,fill

535 format(/1x,'number of nonzero elements in lu = ',i5/1x,1'percent fill = ',2pf8.2/)close(unit=4,status='keep')

cc check l times uc write(iout,560)560 format(1x,'lu .ne. ybus follows'/)

do j = 1,nbusdo k = 1,nbuslmat(j,k) = czeroumat(j,k) = czeroif(j .ge. k) lmat(j,k) = ybus(j,k)if(j .lt. k) umat(j,k) = ybus(j,k)

end doumat(j,j) = 1.0

end do

Grady 2007, p. 202




do j = 1,nbusdo k = 1,nbusytest = czerodo l = 1,kytest = ytest + lmat(j,l) * umat(l,k)

end do

diff = ysave(j,k) - ytestif(cabs(diff) .gt. eps4) write(iout,1005) j,k,nbext(j),

1 nbext(k),diffend do

end docc form uz (urt + diag)c write(iout,536) nbus536 format(1x,'uz pivot = ',i5)

uz(nbus,nbus) = 1.0 / ybus(nbus,nbus)nm1 = nbus - 1

do kdum = 1,nm1

k = nbus - kdumwrite(iout,536) kuz(k,k) = 1.0 / ybus(k,k)kp1 = k + 1do j = kp1,nbusuzelem = czerojm1 = j - 1alpha = 1.0 / ybus(j,j)do l = k,jm1uzelem = uzelem - ybus(j,l) * uz(l,k)

end douz(j,k) = uzelem * alpha

end doend do

write(iout,537)537 format(/1x,'nonzero uz follows'/)

open(unit=8,file='uz.txt')nlu = 0

do j = 1,nbusdo k = 1,nbusymag = cabs(uz(j,k))if(ymag .gt. eps9) thennlu = nlu + 1write(iout,1005) j,k,nbext(j),nbext(k),uz(j,k)write(8,1005) j,k,nbext(j),nbext(k),uz(j,k)

endifend do

end do

fill = nlu / float(nbus * nbus)write(iout,545) nlu,fill

545 format(/1x,'number of nonzero elements in uz = ',i5/1x,1'percent fill = ',2pf8.2/)close(unit=8,status='keep')

cc form zc

Grady 2007, p. 203




open(unit=10,file='zbus_from_lu.txt')

do kdum = 1,nbusk = nbus - kdum + 1write(iout,550) k

550 format(1x,'zbus column = ',i5)do jdum = 1,nbus

j = nbus - jdum + 1zbus(j,k) = czeroif(j .ge. k) zbus(j,k) = uz(j,k)if(j .ne. nbus) thenjp1 = j + 1do l = jp1,nbuszbus(j,k) = zbus(j,k) - ybus(j,l) * zbus(l,k)

end doend if

end doend do

write(iout,565)565 format(/1x,'writing zbus_from_lu.txt file')

do j = 1,nbusdo k = 1,nbuswrite(10,1005) j,k,nbext(j),nbext(k),zbus(j,k)write( 6,1005) j,k,nbext(j),nbext(k),zbus(j,k)

end doend do

close(unit=10,status='keep')cc check ybus * zbusc write(iout,5010)5010 format(/1x,'nonzero ybus * zbus follows')

do j = 1,nbusdo k = 1,nbuselem = czerodo l = 1,nbuselem = elem + ysave(j,l) * zbus(l,k)

end doymag = cabs(elem)if(ymag .gt. eps4) write(iout,1005) j,k,nbext(j),nbext(k),elem

end doend do

cc copy ysave back into ybus, and zero zbusc do j = 1,nbus

unity(j,j) = 1.0do k = 1,nbusybus(j,k) = ysave(j,k)zbus(j,k) = czero

end doend do

cc gaussian eliminate ybus, while performing the same operationsc on the unity matrixc

Grady 2007, p. 204




nm1 = nbus - 1do ipiv = 1,nm1write(iout,561) ipiv

561 format(1x,'gaussian elimination ybus pivot = ',i5)alpha = 1.0 / ybus(ipiv,ipiv)

ipp1 = ipiv + 1c

c pivot row operations for ybus and unityc do k = 1,nbus

if(k .gt. ipiv) thenybus(ipiv,k) = ybus(ipiv,k) * alphaelseunity(ipiv,k) = unity(ipiv,k) * alpha

endifend do

cc pivot elementc ybus(ipiv,ipiv) = 1.0c

c kron reduction of ybus and unity below the pivot row and toc the right of the pivot columnc do j = ipp1,nbus

alpha = ybus(j,ipiv)do k = 1,nbusif(k .gt. ipiv)

1 ybus(j,k) = ybus(j,k) - alpha * ybus(ipiv,k)if(k .lt. ipiv)

1 unity(j,k) = unity(j,k) - alpha * unity(ipiv,k)end do

end docc elements directly below the pivot element

c do j = ipp1,nbusalpha = ybus(j,ipiv)ybus(j,ipiv) = ybus(j,ipiv) - alpha * ybus(ipiv,ipiv)unity(j,ipiv) = unity(j,ipiv) - alpha * unity(ipiv,ipiv)

end do

end docc last rowc write(iout,561) nbus

alpha = 1.0 / ybus(nbus,nbus)

do k = 1,nbus unity(nbus,k) = unity(nbus,k) * alpha

end do

ybus(nbus,nbus) = 1.0ngy = 0ngu = 0write(iout,562)

562 format(/1x,'writing gaussian eliminated ybus and unity to files')open(unit=12,file='ybus_gaussian_eliminated.txt')open(unit=13,file='unity_mat.txt')

Grady 2007, p. 205




do j = 1,nbusdo k = 1,nbusymag = cabs(ybus(j,k))if(ymag .ge. eps9) thenwrite(12,1005) j,k,nbext(j),nbext(k),ybus(j,k)ngy = ngy + 1

endifumag = cabs(unity(j,k))if(umag .ge. eps9) thenwrite(13,1005) j,k,nbext(j),nbext(k),unity(j,k)ngu = ngu + 1

endifend do

end do

close(unit=12,status='keep')close(unit=13,status='keep')fill = ngy / float(nbus * nbus)write(iout,555) ngy,fill

555 format(/1x,'number of nonzero elements in gaussian eliminated',

1' ybus = ',i5/1x,1'percent fill = ',2pf8.2/)fill = ngu / float(nbus * nbus)write(iout,556) ngu,fill

556 format(/1x,'number of nonzero elements in gaussian eliminated',1' unity matrix = ',i5/1x,1'percent fill = ',2pf8.2/)

cc back substitute to find zc do k = 1,nbus

zbus(nbus,k) = unity(nbus,k)end do

do jdum = 2,nbusj = nbus - jdum + 1write(iout,550) jdo kdum = 1,nbusk = nbus - kdum + 1zbus(j,k) = unity(j,k)jp1 = j + 1do l = jp1,nbuszbus(j,k) = zbus(j,k) - ybus(j,l) * zbus(l,k)

end doend do

end do

write(iout,566)566 format(/1x,'writing zbus_from_gaussian_elim.txt file')

open(unit=11,file='zbus_from_gaussian_elim.txt')

do j = 1,nbusdo k = 1,nbuswrite(11,1005) j,k,nbext(j),nbext(k),zbus(j,k)write( 6,1005) j,k,nbext(j),nbext(k),zbus(j,k)

end doend do

close(unit=11,status='keep')

Grady 2007, p. 206




cc check ybus * zbusc write(iout,5010)

do j = 1,nbusdo k = 1,nbus

elem = czerodo l = 1,nbuselem = elem + ysave(j,l) * zbus(l,k)

end doymag = cabs(elem)if(ymag .gt. eps4) write(iout,1005) j,k,nbext(j),nbext(k),elem

end doend do

stop30 write(iout,*) 'error in reading bdat.csv file - stop'

write(*,*) 'error in reading bdat.csv file - stop'stop

31 write(iout,*) 'error in reading ldat.csv file - stop'

write(*,*) 'error in reading ldat.csv file - stop'stop

end

Grady 2007, p. 207



Fortran program yzbuild.f (and executable yzbuild.exe) reads bus datafile bdat.csv, and line/transformer data file ldat.csv, and then producesthe corresponding impedance matrix using LU decomposition, and again usingGaussian elimination. The program is dimensioned for systems as large as150 busses.

Comma-separated file bdat.csv contains the bus data, using the format:bus number , shunt q (in pu, where capacitance positive, and inductanceis negative)

Comma-separated file ldat.csv contains the branch data, using the format:from bus number , to bus number , r(pu) , x(pu) , line charging (in pu,for the total length of line)

It is OK to have ties to ground in ldat.csv. Ground is bus = 0.

Two sample systems are included. One has three busses, and the other haseighty-three busses. To try out yzbuild, simply copy and rename the datafiles for either system as bdat.csv and ldat.csv, and then click on yzbuild.exe.

Grady 2007, p. 208



The data below are given in per unit. R and X are the series impedances of the transmission lines

(in per unit ohms), and B is the line charging of the transmission lines (in per unit mhos).

Build the Y matrix. Show the complex terms individually. Do not perform any complex

arithmetic such as inversions, and do not combine the terms.

From Bus To Bus R – pu ohms jX – pu ohms jB – pu mhos

1 2 0.004 j0.053 j0

2 3 0.02 j0.25 j0.22

3 4 0.02 j0.25 j0.22

2 4 0.01 j0.15 j0.11

Grady 2007, p. 209



Grady 2007, p. 210



Use the definition of Z-matrix elements to find matrix elements z11, z12, z13, z14.

j21 2 3 4

j8

j6

j5

j4

j4

Grady 2007, p. 211



Grady 2007, p. 212



Grady, Programming Consideration, June 2007, Page 1

Programming Considerations

Methods for efficient programming. Sparsity techniques. Optimal bus ordering.

1. Sparsity Programming

Many power system matrices tend to be very large and very sparse. A typical example is the

admittance matrix Y . Each diagonal element of Y is non-zero, and each branch connected

between busses occurs twice in Y , so that a branch between busses j and k appears in y j k , and

yk j, . Additional parallel branches and shunt elements do not create additional y elements.

Consider a power system with 250 busses and 750 branches. Assume that none of the branches

are parallel branches. The dimension of Y is 250 x 250, or 62,500 elements. However, the

number of non-zero terms in Y equals 250 + 2(750), or 1750. Therefore, only 2.8% of the Y

elements are non-zero. The savings are even more impressive for larger systems, since Y grows

by N 2

, but the number of non-zero terms grows by N . For example, a 2500 bus, 7500 branch Yhas only 0.28% non-zero terms.

In most cases, the coding of large computer programs involves trade-offs between execution

time and memory requirements. Sparsity storage techniques, however, reduce both execution

time and memory requirements. The only drawback of sparsity programming is that it is more

difficult to code than full-matrix programming.

There are several techniques for sparsity programming. The method presented here is known as

the sequential trace method, and although the example is given for Y , it applies equally well to

the loadflow Jacobian matrix.

Four vectors are used for storing and accessing the sparse Y :

1. Instead of storing Y as a square matrix of dimension N x N , Y is stored as a column

vector, whose length M is estimated (at the time of program compilation) using typical

sparsity ratios of 1% - 5% of N x N . The location from the top of Y is known as the stack

location.

2. Vector NFIRST , of length N , points to the stack location in Y of the first non-zero element

of a row (in column order). For example, the stack position of the first non-zero element

in Row k is found in NFIRST(k).

3. Vector NCOL, of length M , gives the Y column number of the associated Y stack position.

4. Vector NEXT , of length M , gives the stack location in Y of the next non-zero element in

the corresponding row. A zero value indicates the end of the row.

The term "sequential trace" is derived from the fact that all elements of a Y row can be found, in

a column-order sequence, by starting with the lowest occupied column number of the row, and

continuing through the highest occupied column number of the row.

Grady 2007, p. 213




As an example, let Row 23 of Y have three elements: y235, in stack location 987, y23 23, in stack

location 560, and y23138, in stack location 1230. These elements can be found as follows:

1. The stack location of the first non-zero element (in column order) of Row 23 is found by

looking in NFIRST (23). In this example, NFIRST (23) equals 987.

2. The corresponding column number is found in NCOL(987), which equals 5. Thus, the

equivalent Y (23,5) element is found in stack position Y (987).

3. The stack position of the next non-zero Y element in Row 23 is found by looking at

NEXT (987). The value of NEXT (987) is 560.

4. The corresponding column number is found in NCOL(560), which equals 23. Thus, the

equivalent Y (23,23) element is found in stack position Y (560).

5. The stack position of the next non-zero Y element in Row 23 is found by looking at

NEXT (560). The value of NEXT (560) is 1230.

6. The corresponding column number is found in NCOL(1230), which equals 138. Thus,

the equivalent Y (23,138) element is found in stack position Y (1230).

7. A zero in NEXT (1230) indicates that the end Row 23 has been reached.

Table 1 gives the corresponding numbers for this example.

Table 1: Example of Sequential Trace Method for Sparsity Storage, where the First Non-Zero

Column for Row 23 is found in NFIRST (23) = 987

valuey01381230

valuey5605987

valuey123023560

NEXT NCOLLocationStack

23,138

23,5

23,23

Y

When Y is first built, the elements for each row are contiguous. However, as Y is Gaussianeliminated or LU decomposed, "fill-ins" are created, and they are normally added to the end of

the Y stack. The four vectors must be modified appropriately, and the elements of a row are no

longer contiguous.

Note - if Y is to be Gaussian eliminated, this can be performed simultaneously while Y is being

formed in row order, thereby never saving the LLT part of the eliminated Y matrix. For

example, Row 1 is formed and saved. Next, Row 2 is formed, then Gaussian eliminated by Row

Grady 2007, p. 214




1, and then saved. Next, Row 3 is formed, eliminated by Row 1, then by Row 2, and then saved,

etc.

2. Optimal Bus Ordering

Gaussian elimination and LU decomposition reduce matrices through a series of row operations.

Once the decision has been made to use sparsity programming, then it is very important in power

system applications to order the busses efficiently so that the advantages of sparsity

programming are not nullified by the creation of too many "fill-ins" (i.e. zero elements that

become non-zeros).

Consider the simple five-bus system given figuratively in Figure 1.

12

3

4

5

Figure 1. Simple Five-Bus System

The form of Y , for the given bus order, is

x x

x x x

x x x

x x

x x x x x

,

and the final form of the Gaussian eliminated Y is a full (URT+D) matrix, or

x x

x x

x

x x x x x

,

where the terms are "fill-ins."

However, if the bus numbers for busses 1 and 5 are interchanged, then the initial Y has the form

Grady 2007, p. 215




x x x x x

x x x

x x x

x x

x x

,

and the final form of the Gaussian eliminated Y is

x

x x

x x x

x x

x x

,

which has no fill-ins.

While this is a simple example, the importance of bus ordering cannot be overstated. Of course,

an efficient computer code does not require the user to order the busses ahead of time. Rather,

the user simply inputs his/her desired bus numbers, in no particular order, and the program

develops an "internal" (to the program) bus order. The following three methods are often used:

Method 1: Before Gaussian elimination begins, order the busses according to the number of

connected branches, ignoring parallel branches and shunt elements, so that the busses

with the fewest branches are first, and the busses with the most branches are last.

Method 2: Order the busses so that after each step of Gaussian elimination, the next bus in theorder is the one that has the fewest branches, including intermediate fill-in branches.

Method 3: Order the busses so that after each step of Gaussian elimination, the next bus in the

order is the one that creates the fewest fill-ins. This requires a simulated row-

reduction for all candidate busses.

Method 1 is the simplest and provides the highest incremental benefit. Method 2 usually

provides the best overall speed and memory benefits. Method 3 is more time-consuming to

compute, and the time and storage savings gained by Method 3 usually do not offset its

additional ordering computation time.

The impact of optimal ordering on the Gaussian elimination of Y for six sample systems is

shown in Table 2.

Grady 2007, p. 216




Table 2: Impact of Optimal Bus Ordering Methods on Gaussian Elimination of Y

Elements in ------------ Elements in Gaussian-Eliminated Y ------------

System in Y User Bus Order Method 1 Method 2 Method 3

18 Bus 52 66 41 35 35

83 Bus 303 592 292 264 267

125 Bus 421 1,357 614 419 422

264 Bus 976 5,736 1,204 1,005 1,004

515 Bus 1,761 14,142 1,992 1,763 1,751

2245 Bus 8,257 94,468 25,775 9,473 9,218*

* Estimated using similar results for loadflow Jacobian matrix

The impact of optimal bus ordering on the Jacobian matrix, before and after Gaussian

elimination, is shown in Figures 2a, 2b, and 2c, which correspond to optimal ordering methods 1,2, and 3, respectively.

Grady 2007, p. 217




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Column

Figure 2a. Jacobian Matrix for 125 Bus System, Before and After Gaussian Elimination, Using

Optimal Ordering Method 1

Grady 2007, p. 218




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Figure 2b. Jacobian Matrix for 125 Bus System, Before and After Gaussian Elimination, Using


Grady 2007, p. 219




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Figure 2c. Jacobian Matrix for 125 Bus System, Before and After Gaussian Elimination, Using


Grady 2007, p. 220




3. Example Code BORDER for Performing Optimal Ordering Methods 1, 2, and 3

cc Program BORDER optimally orders busses according to methodsc 1, 2, and 3. Reads formatted BDAT.BOR and LDAT.BOR files inc PCFLO fixed column format. Results placed in files border.1,

c border.2, and border.3.cc Mack Grady. See date below in variable versio.c character*6 versio/'981025'/

logical ybus,taken,ltrue,lfalse,ysavedimension ybus(100,100),nbext(100),nbint(9999),nbran(100),1 taken(100),ysave(100,100),jbus(100)

data mxint/100/,mxext/9999/,nbus/0/, nbint/9999 * 0/,1 nbext/100 * 0/,eps6/1.0e-06/data ybus /10000 * .false./,ltrue/.true./,lfalse/.false./data iout/15/pi = 4.0 * atan(1.0)dr = 180.0 / piopen(unit=1,file='bdat.bor')open(unit=iout,file='exlog.bor')write(6,*) 'Program BORDER, Version ',versiowrite(6,*) 'Read bus data from PCFLO fixed column format ',1 'file BDAT.BOR'write(6,*) 'number q shunt'write(iout,*) 'Program BORDER, Version ',versiowrite(iout,*) 'Read bus data from PCFLO fixed column format ',1 'file BDAT.BOR'write(iout,*) 'number q shunt'

cc read the bdat filec 1 read(1,1001,end=5) nb,qshunt1001 format(i4,t58,f8.0)

if(nb .eq. 0) go to 5write(6,1002) nb,qshuntwrite(iout,1002) nb,qshunt

1002 format(i7,5x,f10.2)if(nb .le. 0 .or. nb .gt. mxext) thenwrite(6,*) 'Illegal bus number - stop'write(iout,*) 'Illegal bus number - stop'stop

endifnbus = nbus + 1if(nbus .gt. mxint) thenwrite(6,*) 'Too many busses - stop'write(iout,*) 'Too many busses - stop'stop

endifnbext(nbus) = nbnbint(nb) = nbusybus(nbus,nbus) = ltruego to 1

5 close(unit=1,status='keep')c open(unit=1,file='ldat.bor')

write(6,*) 'Read line/transformer data from PCFLO fixed column ',1 'format file LDAT.BOR'

Grady 2007, p. 221




write(6,*) ' from to r x',1 ' charge tap phase shift'write(iout,*) 'Read line/transformer data from PCFLO fixed ',1 'column format file LDAT.BOR'write(iout,*) ' from to r x',1 ' charge tap phase shift'

c

c read the ldat filec 10 read(1,1003,end=15) nfrom,nto,r,x,charge,tap,phase1003 format(2i4,2x,3f12.0,t76,f7.0,f8.0)

if(nfrom .eq. 0 .and. nto .eq. 0) go to 15write(6,1004) nfrom,nto,r,x,charge,tap,phasewrite(iout,1004) nfrom,nto,r,x,charge,tap,phase

1004 format(1x,i5,2x,i5,4(2x,f10.4),3x,f10.2) if(nfrom .lt. 0 .or. nfrom .gt. mxext .or. nto .lt. 0

1 .or. nto .gt. mxext) thenwrite(6,*) 'illegal bus number - stop'write(iout,*) 'illegal bus number - stop'stop

endif

if(nfrom .eq. nto) thenwrite(6,*) 'same bus number given on both ends - stop'write(iout,*) 'same bus number given on both ends - stop'stop

endifif(r .lt. -eps6 .or. tap .lt. -eps6) thenwrite(6,*) 'illegal resistance or tap - stop'write(iout,*) 'illegal resistance or tap - stop'stop

endif if(charge .lt. -eps6) then

write(6,*) 'line charging should be positive'write(iout,*) 'line charging should be positive'stop

endifif(nfrom .ne. 0) thennfrom = nbint(nfrom)if(nfrom .eq. 0) thenwrite(6,*) 'bus not included in file bdat - stop'write(iout,*) 'bus not included in file bdat - stop'stop

endif endif

if(nto .ne. 0) then nto = nbint(nto)

if(nto .eq. 0) thenwrite(6,*) 'bus not included in file bdat - stop'write(iout,*) 'bus not included in file bdat - stop'stop

endif endif

phase = phase / drif(abs(phase) .gt. eps6 .and. abs(tap) .lt. eps6) tap = 1.0if((nfrom .eq. 0 .or. nto .eq. 0) .and. tap .gt. eps6) thenwrite(6,*) 'shunt elements should not have taps - stop'write(iout,*) 'shunt elements should not have taps - stop'stop

endif if(tap .gt. eps6 .and. charge .gt. eps6) then

Grady 2007, p. 222




write(6,*) 'transformers should not have line charging'write(iout,*) 'transformers should not have line charging'stop

endifcc the line charging termsc

if(nfrom .ne. 0) ybus(nfrom,nfrom) = ltrueif(nto .ne. 0) ybus(nto ,nto ) = ltrue

cc shunt elementsc if(nfrom .ne. 0 .and. nto .eq. 0) ybus(nfrom,nfrom) = ltrue

if(nfrom .eq. 0 .and. nto .ne. 0) ybus(nto ,nto ) = ltruecc transmission linesc if(nfrom .ne. 0 .and. nto .ne. 0 .and. tap .lt. eps6) then

ybus(nfrom,nto ) = ltrueybus(nto ,nfrom) = ltrueybus(nfrom,nfrom) = ltrue

ybus(nto ,nto ) = ltrueendifcc transformersc if(nfrom .ne. 0 .and. nto .ne. 0 .and. tap .ge. eps6) then

ybus(nfrom,nfrom) = ltrueybus(nto ,nto ) = ltrueybus(nfrom,nto ) = ltrueybus(nto ,nfrom) = ltrue

endifgo to 10

15 close(unit=1,status='keep')c

c count the number of nonzero terms in each rowc nterms = 0

do 20 j = 1,nbusnbran(j) = 0taken(j) = lfalsedo 20 k = 1,nbusysave(j,k) = ybus(j,k)if(ybus(j,k)) thennbran(j) = nbran(j) + 1nterms = nterms + 1

endif 20 continue write(6,1111) nterms

write(iout,1111) ntermscc simulate the gaussian elimination using file=bdat orderc ntfill = nterms

do 96 j = 1,nbustaken(j) = ltruedo 97 k = 1,nbusif(taken(k) .or. .not.ybus(k,j)) go to 97do 98 l = 1,nbusif(ybus(j,l) .and. .not.ybus(k,l)) then

Grady 2007, p. 223




ybus(k,l) = ltruentfill = ntfill + 1

endif98 continue

ybus(k,j) = lfalsentfill = ntfill - 1

97 continue

96 continue write(6,1113) ntfill

write(iout,1113) ntfill1113 format(1x,'Total reduced YBUS terms for BDAT.BOR bus order = ',

1 i5/)c do 99 j = 1,nbus

taken(j) = lfalsedo 99 k = 1,nbusybus(j,k) = ysave(j,k)

99 continuecc method 1c

open(unit=1,file='border.1')norder = 1nfound = 0nties = 2write(6,1111) ntermswrite(iout,1111) nterms

1111 format(/1x,'YBUS terms before reduction = ',i8/)1105 format(1x,'Method = ',i5,', nties = ',i5,

1 ', found = ',i5)30 do 25 j = 1,nbus

if(taken(j) .or. nbran(j) .ne. nties) go to 25taken(j) = ltruenfound = nfound + 1jbus(nfound) = j

write(1,1100) norder,nfound,nbext(j)write(6,1100) norder,nfound,nbext(j)write(iout,1100) norder,nfound,nbext(j)

1100 format(1x,'Method = ',i5,', int. bus = ',i5,', ext. bus = ',i5) 25 continue

write(6,1105) norder,nties,nfoundwrite(iout,1105) norder,nties,nfoundnties = nties + 1if(nfound .ne. nbus) go to 30close(unit=1,status='keep')

cc simulate the gaussian elimination for method 1 orderc do 35 j = 1,nbus

taken(j) = lfalse35 continue

ntfill = ntermsdo 36 jj = 1,nbusj = jbus(jj)taken(j) = ltruedo 37 k = 1,nbusif(taken(k) .or. .not.ybus(k,j)) go to 37do 38 l = 1,nbusif(ybus(j,l) .and. .not.ybus(k,l)) then

ybus(k,l) = ltrue

Grady 2007, p. 224




ntfill = ntfill + 1endif

38 continueybus(k,j) = lfalsentfill = ntfill - 1

37 continue36 continue

write(6,1122) norder,ntfillwrite(iout,1122) norder,ntfill

1122 format(/1x,'Total reduced YBUS terms for method = ',i5,1' equals ',i5/)

cc method 2c open(unit=1,file='border.2')

norder = 2do 40 j = 1,nbusnbran(j) = 0taken(j) = lfalsedo 40 k = 1,nbusybus(j,k) = ysave(j,k)

if(ybus(j,k)) nbran(j) = nbran(j) + 140 continuenfound = 0ntfill = nterms

nties = 2write(6,1111) ntermswrite(iout,1111) nterms

1108 format(1x,'Method = ',i5,', nties = ',i5,1 ', found = ',i5,', tot. terms = ',i5)

50 do 45 j = 1,nbusif(taken(j) .or. nbran(j) .ne. nties) go to 45taken(j) = ltruenfound = nfound + 1write(1,1100) norder,nfound,nbext(j)

write(6,1100) norder,nfound,nbext(j)write(iout,1100) norder,nfound,nbext(j)cc gaussian eliminate all the remaining rows with row jc do 60 k = 1,nbus

if(k .eq. j .or. taken(k)) go to 60if(ybus(k,j)) thendo 65 l = 1,nbusif(l .eq. j) go to 65if(ybus(j,l) .and. .not.ybus(k,l)) thenybus(k,l) = ltruenbran(k) = nbran(k) + 1ntfill = ntfill + 1

endif65 continue

cc for the term under the pivotc ybus(k,j) = lfalse

nbran(k) = nbran(k) - 1ntfill = ntfill - 1

endif 60 continue

write(6,1108) norder,nties,nfound,ntfill

Grady 2007, p. 225




write(iout,1108) norder,nties,nfound,ntfillif(nfound .eq. nbus) go to 59nties = 2if(nfound .eq. (nbus - 1)) nties = 1go to 50

45 continuenties = nties + 1

go to 5059 close(unit=1,status='keep')

write(6,1122) norder,ntfillwrite(iout,1122) norder,ntfill

cc method 3c open(unit=1,file='border.3')

norder = 3do 160 j = 1,nbustaken(j) = lfalsedo 160 k = 1,nbusybus(j,k) = ysave(j,k)

160 continue

nfound = 0nfills = 0ntfill = ntermswrite(6,1111) ntermswrite(iout,1111) nterms

170 write(6,1107) norder,nfills,nfound,ntfillwrite(iout,1107) norder,nfills,nfound,ntfill

1107 format(1x,'Method = ',i5,', nfills = ',i5,1 ', found = ',i5,', tot. terms = ',i5)

cc simulated gaussian elimination using the remaining pivot candidatesc mfirst = 0

nxtbus = 0

do 165 m = 1,nbusif(taken(m)) go to 165lstbus = mnfills = 0do 180 k = 1,nbusif(k .eq. m .or. taken(k)) go to 180if(ybus(k,m)) thendo 185 l = 1,nbusif(l .eq. m) go to 185if(ybus(m,l) .and. .not.ybus(k,l)) nfills = nfills + 1

185 continuecc for the term under the pivotc nfills = nfills - 1

endif 180 continue

if(mfirst .eq. 0) thenmfirst = 1nxtbus = mminfil = nfills

endifif(nfills .lt. minfil) thennxtbus = mminfil = nfills

Grady 2007, p. 226




endif 165 continue

if(nxtbus .eq. 0) thenif(nfound .ne. (nbus - 1)) thenwrite(6,1120)write(iout,1120)

1120 format(/1x,'Error in method 3 ordering - stop')

stopendifnxtbus = lstbusminfil = 0

endif j = nxtbus

nfills = minfiltaken(j) = ltruenfound = nfound + 1write(1,1100) norder,nfound,nbext(j)write(6,1100) norder,nfound,nbext(j)write(iout,1100) norder,nfound,nbext(j)if(nfound .eq. nbus) thenwrite(6,1107) norder,nfills,nfound,ntfill

write(iout,1107) norder,nfills,nfound,ntfillgo to 175endif

cc gaussian eliminate all the remaining rows with row jc do 290 k = 1,nbus

if(k .eq. j .or. taken(k)) go to 290if(ybus(k,j)) thendo 295 l = 1,nbusif(l .eq. j) go to 295if(ybus(j,l) .and. .not.ybus(k,l)) thenybus(k,l) = ltruenbran(k) = nbran(k) + 1

ntfill = ntfill + 1endif295 continue

cc for the term under the pivotc ybus(k,j) = lfalse

nbran(k) = nbran(k) - 1ntfill = ntfill - 1

endif 290 continue

go to 170175 close(unit=1,status='keep')

write(6,1122) norder,ntfillwrite(iout,1122) norder,ntfillstopend

Grady 2007, p. 227



Grady, Motor Starting, June 2007, Page 1

Motor Starting

Steady-state analysis of the motor starting problem.

1. Characteristics of Motors During Start-Up

During start-up, motors behave like fixed impedances that, at rated voltage, draw approximately

five-times their full-load running current, and at a very low power factor. If the motor terminal

voltage drops to less than 80% during start-up, most motors will not start.

2. Determination of Motor Start-Up Voltage

The equivalent circuit for calculating motor starting voltages is shown in Figure 1.

Zm

Zth

Vm+

-

Vth+

-

Figure 1. Motor Starting Equivalent Circuit

This Thevenin equivalent circuit has open-circuit voltage V TH = 1.0 pu, and Thevenin

impedance Z TH , which is the "short circuit," positive-sequence impedance of the system as

"seen" at the motor terminals. Z TH

is the motor-bus diagonal element of the positive-sequence

impedance matrix, which is computed with all other large machines represented by their

respective subtransient impedances.

The motor is represented at start-up as fixed impedance Z R jX M M M , where R X M M and

are calculated to give approximately five-times full-load running current @ 1.0 pu voltage, and

at a startup-up power factor of approximately 0.20.

Given the circuit parameters, the motor start-up voltage is then computed from the voltage

divider equation, or

V V Z

Z Z M TH M

TH M .

V M should not be less than 0.80 pu. if the motor is start properly.

This problem can be worked "in reverse," where the objective is to determine how large a motor

will start at a bus. A FORTRAN code for this purpose is shown below.

c

Grady 2007, p. 228




c program motors.f reads zbus1 and vsoln files, and then computesc the kw motor that can be started at each bus without dragging thec voltage below vmin (in percent). the results are written to filec mstart.c complex zzz,zmotor

character*12 anames,an

dimension anames(9999),hpmin(9999)data anames /9999 * ' '/, hpmin /9999 * -1.0/

cc abase = base current,c vbase = base voltage - volts (three-phase),c pbase = base power - va (three-phase),c arun = running amps per kw,c spf = motor power factor when starting,c rpf = motor power factor when running,c start = multiplier of run current when starting,c vmin = minimum acceptable starting voltage - pu.c data pbase/100000.0/,vbase/450.0/,spf/0.2/,start/5.0/

data vmin/0.8/,rpf/0.85/eps/1.0e-03/

c abase = pbase / 3.0 / vbase * sqrt(3.0)arun = 1000.0 / 3.0 / rpf / vbase * sqrt(3.0)write(6,5007) abase,arun

5007 format(1x,'base current = ',f10.2/1x,'running current ',1'for 1 kw = ',f10.2)

cc zmotor = motor starting impedance for a 1 kw motorc zmotor = abase / arun / start *

1 cmplx(spf,sqrt(1.0 - spf * spf))write(6,5006) zmotor

5006 format(1x,'z start for a 1 kw motor = ',2e15.6)rm = real(zmotor)

xm = aimag(zmotor)open(unit=1,file='vsoln')open(unit=2,file='zbus1')open(unit=3,file='mstart')

1 read(1,5000,end=2) kbus,an5000 format(1x,i4,2x,a12)

if(kbus .le. 0 .or. kbus .gt. 9999) go to 500anames(kbus) = ango to 1

2 read(2,5002,end=3) jbus,kbus,zzz5002 format(1x,2i5,2e15.6)

if(jbus .ne. kbus) go to 2rs = real(zzz)xs = aimag(zzz)a = (vmin * vmin - 1) * (rm * rm + xm * xm)b = 2.0 * vmin * vmin * (rs * rm + xs * xm)c = vmin * vmin * (rs * rs + xs * xs)rad = sqrt(b * b - 4.0 * a * c)h1 = 1.0 / ((-b + rad) / 2.0 / a)h2 = 1.0 / ((-b - rad) / 2.0 / a)h = -1.0if(h1 .lt. 0.0 .and. h2 .gt. 0.0) h = h2if(h2 .lt. 0.0 .and. h1 .gt. 0.0) h = h1if(h1 .gt. 0.0 .and. h2 .gt. 0.0) h = amin1(h1,h2)if(h .le. 0.0) go to 501

Grady 2007, p. 229




if(hpmin(jbus) .lt. 0.0) hpmin(jbus) = hif(h .lt. hpmin(jbus)) hpmin(jbus) = hwrite(6,5003) jbus,anames(jbus),zzz,h

5003 format(1x,'num = ',i4,', name = ',a12,', zsys = ',2e12.4,1 ', kw = ',f7.1)vcheck = cabs(zmotor / h / (zzz + zmotor / h))if(vcheck .lt. (vmin - eps) .or. vcheck .gt. (vmin + eps))

1 go to 502go to 2

3 do 5 jbus = 1,9999if(hpmin(jbus) .lt. 0.0) go to 5write(3,5011) jbus,anames(jbus),hpmin(jbus)

5011 format(1x,'bus number = ',i5,', name = ',a12,1', kw starting cap. = ',f7.1)5 continuestop

500 write(6,5001)5001 format(1x,'error in vsoln file')

stop501 write(6,5005)

5005 format(1x,'error in kw calculation')

stop502 write(6,5010)5010 format(1x,'error in vcheck')

stopend

Grady 2007, p. 230



Grady 2007, p. 231



Grady 2007, p. 232



B e r g e n , C h a p t e r 1 2 , p . 4 6 8

Grady 2007, p. 233



Grady, Short Circuits, June 2007, Page 1

-

Short Circuits

1. Introduction

Voltage sags are due mostly to faults on either transmission systems or distribution feeders.

Transmission faults affect customers over a wide area, possibly dozens of miles, but distributionfaults usually affect only the customers on the faulted feeder or on adjacent feeders served by the

same substation transformer.

Single-phase faults (i.e., line-to-ground) are the most common type of faults, followed by line-to-

line, and three-phase. Since single-phase and line-to-line faults are unbalanced, their resulting

sag voltages are computed using symmetrical components. Transformer connections affect the propagation of positive, negative, and zero sequence components differently. Thus, the

characteristics of a voltage sag changes as it propagates through a network.

Typically, a transmission voltage sag passes through two levels of transformers before reaching a

480V load (e.g., 138kV:12.47kV at the entrance to the facility, and 12.47kV:480V at the load).120V loads likely have a third transformer (e.g., 480V:120V). It is not intuitively obvious howthe sag changes, but the changes can be computed using symmetrical components and are

illustrated in this report.

2. Symmetrical Components

An unbalanced set of N related phasors can be resolved into N systems of phasors called the

symmetrical components of the original phasors. For a three-phase system (i.e. N = 3), the threesets are:

1. Positive Sequence - three phasors, equal in magnitude, 120o apart, with the same sequence

(a-b-c) as the original phasors.

2. Negative Sequence - three phasors, equal in magnitude, 120o apart, with the opposite

sequence (a-c-b) of the original phasors.

3. Zero Sequence - three identical phasors (i.e. equal in magnitude, with no relative phasedisplacement).

The original set of phasors is written in terms of the symmetrical components as follows:

210~~~~aaaa V V V V ,

210~~~~bbbb V V V V ,

210~~~~cccc V V V V ,

where 0 indicates zero sequence, 1 indicates positive sequence, and 2 indicates negative

sequence.

The relationships among the sequence components for a-b-c are

Grady 2007, p. 234




-

Positive Sequence Negative Sequence Zero Sequence~ ~V V b a1 1 1 120o ~ ~

V V b a2 2 1 120o000

~~~cba V V V

~ ~V V c a1 1 1 120o ~ ~

V V c a2 2 1 120o

The symmetrical components of all a-b-c voltages are usually written in terms of the symmetricalcomponents of phase a by defining

a 1 120o , so that a2 1 1 240 120o o , and a3 1 1 360 0o o .

Substituting into the previous equations for~

,~

,~

V V V a b c yields

~ ~ ~ ~V V V V a a a a 0 1 2 ,

212

0~~~~aaab V aV aV V ,

~ ~ ~ ~V V aV a V

c a a a

0 1

2

2 .

In matrix form, the above equations become

2

1

0

2

2

~

~

~

1

1

111

~

~

~

a

a

a

c

b

a

V

V

V

aa

aa

V

V

V

,

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

~

~

~

1

1

111

3

1

~

~

~

2

2

2

1

0

(1)

or in matrix form

~ ~V T V abc 012 , and ~ ~V T V abc012 1 , (2)

where transformation matrix T is

T a a

a a

1 1 1

1

1

2

2

, and T a a

a a

1 2

2

1

3

1 1 1

1

1

. (3)

If ~V abc represents a balanced set (i.e.

~ ~ ~V V a V b a a 1 2120o ,

~ ~ ~V V aV c a a 1 120o ), then substituting into ~ ~V T V abc012 1 yields

~

~

~

~

~

~

~V

V

V

a a

a a

V

a V

aV

V

a

a

a

a

a

a

a

0

1

2

2

2

21

3

1 1 1

1

1

0

0

.

Grady 2007, p. 235




-

Hence, balanced voltages or currents have only positive sequence components, and the positive

sequence components equal the corresponding phase a voltages or currents.

However, balanced voltages are rare during voltage sags. Most often, one phase is affected

significantly, and the other two less significantly. Thus, all three sequence voltages

210

~,

~,

~aaa V V V exist during most sags, and these sequence voltages are shifted differently by

transformers when propagating through a system. When recombined to yield phase voltages

cba V V V ~

,~

,~

, it is clear that the form of phase voltages must also change as transformers are

encountered.

3. Transformer Phase Shift

The conventional positive-sequence and negative-sequence model for a three-phase transformer

is shown below. Admittance y is a series equivalent for resistance and leakage reactance, tap t is

the tap (in per unit), and angle is the phase shift.

y

Ii ---> Ik --->Bus i

Bus k t / :1

Bus k'

Figure 1. Positive- and Negative-Sequence Model of Three-Phase Transformer

For grounded-wye:grounded-wye and delta:delta transformers, is +0˚, and thus positive- and

negative-sequence voltages and currents pass through unaltered (in per unit). However, for wye-delta and delta-wye transformers, is sequence-dependent and is defined as follows:

For positive sequence, is +30˚ if bus i is the high-voltage side, or –30˚ if bus i is thelow-voltage side

and oppositely

For negative sequence, is –30˚ if bus i is the high-voltage side, or +30˚ if bus i is the

low-voltage side

In other words, positive sequence voltages and currents on the high-voltage side lead those onthe low-voltage side by 30˚. Negative sequence voltages and currents on the high-voltage side

lag those on the low-voltage side by 30˚.

For zero-sequence voltages and currents, transformers do not introduce a phase shift, but they

may block zero-sequence propagation as shown in Figure 2.

Grady 2007, p. 236




-

Grounded Wye - Grounded Wye

Grounded Wye - DeltaR + jX

Grounded Wye - Ungrounded WyeR + jX

R + jXUngrounded Wye - Delta

R + jXDelta - Delta

R + jX

Figure 2. Zero-Sequence Model of Three-Phase Transformer

It can be seen in the above figure that only the grounded-wye:grounded-wye transformer

connection permits the flow of zero-sequence from one side of a transformer to the other.

Thus, due to phase shift and the possible blocking of zero-sequence, transformers obviously playan important role in unbalanced voltage sag propagation.

4. System Impedance Matrices

Fault currents and voltage sags computations require elements of the impedance matrix Z for thestudy system. While each of the three sequences has its own impedance matrix, positive- andnegative-sequence matrices are usually identical. Impedance elements are usually found by

Grady 2007, p. 237




-

building the system admittance matrix Y, and then inverting it to obtain the entire Z,

or by

using Gaussian elimination and backward substitution to obtain selected columns of Z.

The admittance matrix Y is easily built according to the following rules:

The diagonal terms of Y contain the sum of all branch admittances connected directly tothe corresponding bus.

The off-diagonal elements of Y contain the negative sum of all branch admittances

connected directly between the corresponding busses.

The procedure is illustrated by the three-bus example in Figure 3.

1 2 3

ZE

ZA

ZB

ZC

ZDI3

Figure 3. Three-Bus Admittance Matrix Example

Applying KCL at the three independent nodes yields the following equations for the bus voltages(with respect to ground):

At bus 1, 0211

A E Z

V V

Z

V ,

At bus 2, 032122

C A B Z

V V

Z

V V

Z

V ,

At bus 3, 3233 I

Z

V V

Z

V

C D

.

Collecting terms and writing the equations in matrix form yields

Grady 2007, p. 238




-

33

2

1

0

0

1110

11111

0111

I V

V

V

Z Z Z

Z Z Z Z Z

Z Z Z

DC C

C C B A A

A A E

,

or in matrix form,

I YV ,

Besides being the key for fault calculations, the impedance matrix, 1 Y Z , is also physically

significant. Consider Figure 4.

Power System

All Other Busses

Open Circuited

Applied Current at

Induced Voltage

+

-

V

Bus k at Bus j

Ik

Vj

Figure 4. Physical Significance of the Impedance Matrix

Impedance matrix element k j z , is defined as

k m N mm I k

jk j

I

V z

,,,2,1,0

,

, (4)

where k I is a current source attached to bus k , jV is the resulting voltage at bus j, and all busses

except k are open-circuited. The depth of a voltage sag at bus k is determined directly bymultiplying the phase sequence components of the fault current at bus k by the matrix elements

k j z , for the corresponding phase sequences.

5. Short Circuit Calculations

Short circuit calculations require positive, negative, and zero sequence impedance information,

depending on whether or the fault is balanced or not. For example, the commonly-studied, but

relatively rare, three-phase fault is balanced. Therefore, only positive sequence impedances arerequired for its study.

Grady 2007, p. 239




-

Consider the three-phase fault represented by the one-line diagram in Figure 5, where TH V and

TH Z are the Thevenin equivalent circuit parameters for bus k.

+

-

Vth

Zth

ZFIF

Bus k

Figure 5. Three-Phase Fault at Bus k

The fault current and voltage are clearly

F TH

TH F k Z Z

V I

, and

F TH

F TH

F k TH TH

F k Z Z

Z V I Z V V .

In a large power system, the Thevenin equivalent impedance for a bus is the corresponding

diagonal impedance matrix element, and the Thevenin equivalent voltage is usually assumed to

be 1.0 /0 pu.

The type of machine models used when building impedance matrices affects the Thevenin

equivalent impedances and fault calculations. Rotating machines actually have time-varyingimpedances when subjected to disturbances. However, for simplification purposes, their

impedances are usually divided into three zones - subtransient (first few cycles), transient (5

cycles - 60 cycles), and steady-state (longer than 60 cycles). When performing fault studies, the

time period of interest is usually a few cycles, so that machines are represented by theirsubtransient impedances when forming the impedance matrices.

Developing the equations for fault studies requires adept use of both a-b-c and 0-1-2 forms of thecircuit equations. The use of sequence components implies that the system impedances (but not

the system voltages and currents) are symmetric. In general, there are six equations and sixunknowns to be solved, regardless of the type of fault studied.

It is common in fault studies to assume that the power system is initially unloaded and that allvoltages are 1.0 per unit. When there are multiple sources, this assumption requires that there

are no shunt elements connected, such as loads, capacitors, etc., except for rotating machines

(whose Thevenin equivalent voltages are 1.0 pu.).

Since wye-delta transformers shift positive, negative, and zero sequence components differently,

it is important to model transformers according to the rules given earlier. This means that the pre-fault voltages all have magnitude 1.0 pu., but that the pre-fault voltage angles can be

ooo 30or ,30,0 , depending upon the net transformer phase shift between them and the chosen

reference bus.

Grady 2007, p. 240




-

Balanced Three-Phase Fault

Consider the three-phase fault at bus k , as shown in Figure 6.

a

b

c

ZF ZF ZF

IkaF

IkcF

IkbF

Bus k

Figure 6: Three-Phase Fault at Bus k

The Thevenin equivalent circuit equation, assuming no other current injections in the system, is

F kc

F kb

F ka

kckckbkckakc

kckbkbkbkakb

kckakbkakaka

ekc

ekb

eka

F kc

F kb

F ka

I

I

I

z z z

z z z

z z z

V

V

V

V

V

V

,,,

,,,

,,,

Pr

Pr

Pr

,

or in sequence form,

F k

F k

F k

k k

k k

k k

ek

ek

ek

F k

F k

F k

I

I

I

z

z

z

V

V

V

V

V

V

2

1

0

2,2

1,1

0,0

Pr 2

Pr 1

Pr 0

2

1

0

00

00

00

.

In abbreviated form, the above equations are

F kabcabck k

ekabc

F kabc

I Z V V ,Pr

, and F k k k

ek

F k

I Z V V 012012,

Pr 012012 , (5)

where F k

V consists of the voltages at bus k during the fault, ek

V Pr consists of the pre-fault

voltages, F k

I gives the fault currents, and k k Z contains the individual impedance elements

extracted from the impedance matrix.

Grady 2007, p. 241




-

The above matrix equations represents three equations (repeated in abc and 012 form), but there

are six unknowns represented by F k

V and F k

I , so that three additional equations are required.

The additional equations are found by observing that

F

kabc F

F

kabc I Z V , or

F

k F

F

k I Z V 012012 .

Substituting into the Thevenin equation, and recognizing that all zero- and negative-sequence

voltages and currents are zero for a balanced fault yields

0

0

00

00

00

0

0

0

0

1

2,2

1,1

0,0Pr 11

F k

k k

k k

k k e

k F k F I

z

z

z

V I Z ,

so that the positive sequence fault current is found to be

0,0,20

1,1

Pr 1

1

F

k F k

F k k

ek F

k I I

Z z

V I . (6)

Substituting into Thevenin equation

F k k k

ek

F k

I Z V V 012012,

Pr 012012 (7)

yields the fault voltage at Bus k . Similarly, because the impedance matrix relates the voltages at

network busses to current injections at network busses, the voltage at any other bus j is foundusing

F k k j

e j

F j I Z V V

012012,Pr 012012 . (8)

Note that the minus sign is needed because the fault current has been drawn as positive outward.

Once the fault voltages are known at neighboring busses, the contribution currents through theconnected branches can be easily found.

Single-Phase to Ground Fault

Consider the single-phase fault at bus k , as shown in Figure 7.

Grady 2007, p. 242




-

a

b

c

ZF

IkaF

Bus k

Ikb

F = 0

IkcF = 0

Figure 7: Single-Phase Fault at Bus k , Phase a

As before, the Thevenin equivalent circuit equations, assuming no other current injections in the

system, is

F k k k

ek

F k

I Z V V 012012,

Pr 012012 .

Examining F k

I 012

shows that in this case

F ka

F ka

F ka

F kc

F kb

F ka

F kabc

F k

I

I

I

I

I

I

aa

aa I T I 31

0

0

1

1

111

31

2

21012 . (9)

Substituting into the Thevenin equation yields

3/

3/

3/

00

00

00

0

0

2,2

1,1

0,0

Pr 2

Pr Pr 1

Pr 0

2

1

0

F ka

F ka

F ka

k k

k k

k k

ek

eka

ek

ek

F k

F k

F k

I

I

I

z

z

z

V

V V

V

V

V

V

Add the three rows yields

2,21,10,0Pr

210 3

1k k k k k k

F ka

eka

F ka

F k

F k

F k

z z z I V V V V V .

From the circuit it is obvious that

Grady 2007, p. 243




-

F F ka

F ka

Z I V ,

so that

2,21,10,0Pr

3

1k k k k k k

F ka

eka F

F ka z z z I V Z I .

Solving for F ka

I yields

F k k k k k k

eka F

ka Z z z z

V I

3

3

2,21,10,0

Pr

. (10)

Now, using

3210

F

ka F k

F k

F k I I I I ,

all network voltages can be found from

F k k j

e j

F j I Z V V

012012,Pr 012012 .

Note that if 2,10,0 k k k k z z , a single-phase fault will have a higher value than does a three-

phase fault.

Line-to-Line Fault

Consider the line-to-line fault at bus k , as shown in Figure 8.

a

b

c

IkcF

Bus k

IkaF = 0

ZF

IkbF

Figure 8. Line-to-Line Fault Between Phases b and c at Bus k

Examining F k

I 012

shows that in this case

Grady 2007, p. 244




-

aa I

aa I

I I

I

I

aa

aa I T I

F kb

F kb

F kb

F kc

F kb

F ka

F kabc

F k

2

2

2

21012

0

3

10

1

1

111

3

1 . (11)

Note that there is no zero sequence fault current.

Substituting into the Thevenin equation yields

3/

3/

0

00

00

00

0

0

2

2

2,2

1,1

0,0Pr

2

1

0

aa I

aa I

z

z

z

V

V

V

V

F kb

F kb

k k

k k

k k e

ka F k

F k

F k

.

Subtracting the last two rows of the Thevenin equation yields

33

2

2,2

2

1,1Pr

21

aa I z

aa I z V V V

F kb

k k

F kb

k k e

ka F k

F k

,

or

eka

F k

F k

k k k k F kb

V V V aa z aa z

I Pr 21

21,1

22,2

33

.

From the circuit, we see that

F

F kc

F kb F

kb Z

V V I

.

Using 012TV V abc , we find that

22

21

aaV aaV V V F k

F k

F kc

F kb

,

so that

F

F k

F k F

kb F

F k

F k F

kb Z

aaV V I

Z

aaV aaV I

221

22

21 or , .

Combining equations yields

Grady 2007, p. 245




-

eka

F F kbk k k k F

kb V

aa

Z I aa z aa z I Pr

2

21,1

22,2

33

.

Collecting terms yields

eka

F k k k k F kb

V aa

Z aa z aa z I Pr

2

21,1

22,2

33

,

or

ekak k k k

F F kb

V z z aa

aa

Z I Pr

2,21,1

2

2 3

.

Simplifying yields

and where 0, F ka

F kb

F kc

I I I . All network voltages can now be found from

F k k k k

eka F

kb Z z z

V j I

2,21,1

Pr 3 , (12)

F k k j

e j

F j I Z V V 012012,

Pr 012012 .

Grady 2007, p. 246



Grady, Voltage Sag Propagation, June 2007, Page 1

-

Voltage Sag Propagation Along Feeders

1. Introduction

Short circuit equations provide the theoretical framework for determining the voltage sag at a bus

due to a fault anywhere in the system. However, the short circuit equations by themselves provide little insight. We now proceed with examples to provide this insight by showing how a

sag propagates for various transformer situations.

2. Impact of Transmission System Faults on Customers

Consider the typical situation shown in Figure 1. A fault occurs at bus k of the transmissionsystem, causing a voltage sag that affects a substation (bus j) and the customers connected to its

feeders. There can be as many as three transformers between the customer’s load and the

transmission fault point, and each of these transformers can have a 30o phase shift. Typically,all three of the transformers shown (i.e., T1, T2, and T3) are delta connected on the high side,

and grounded-wye connected on the low side.

Figure 1. Example System for Analyzing the Propagation of Transmission Voltage Sags into

Customer Low-Voltage Busses

The standard assumption for fault calculations is that

the circuit is initially unloaded, or at least that the voltages are all close to 1.0 per unit.

Using this assumption, and further assuming that there are no significant contributors of faultcurrent on the feeders, then the actual location of the customer is not important because all

points on the three 12.47kV feeders shown (including the substation 12.47kV bus) willexperience the same sag. Furthermore, the sag experienced on the substation 12.47kV bus will

Transmission

System

T2: 12.47kV

-----

480V

T3: 480V

-----120/208V

T1: 138kV-----

12.47kV

SubstationFeeders

Customer Level 1.480V loads

Customer Level 2.

120/208V loads

12.47kV

connectionFault

Substation

transmission bus j

Faulted bus k

Grady 2007, p. 247




-

be the same as on substation 138kV bus j, except for possible zero-sequence component blocking

and positive/negative-phase shifts.

The significance of the above paragraph is that

for transmission faults, one monitor at either the substation 138kV bus or at the substation12.47kV bus is adequate to predict voltage sag levels anywhere on the substation’s feeders, provided there are no significant contributors of fault current on the feeders.

If the transmission fault is electrically far away, then the sag experienced at the substation and at

the customer site will be small. Alternatively, if the fault is immediately at substation 138kV bus

j, then the sag will be the most severe possible. Thus, it is reasonable to assume that an electrical“proximity” factor exists, where a proximity factor of zero (i.e., 0%) indicates that the fault is at

substation 138kV bus j, and a proximity factor of unity (i.e., 100%) indicates that the

transmission fault bus k is very far away. From knowledge of the physical significance of theimpedance matrix, and from examining Thevenin equations

F k k k

ek

F k

I Z V V 012012,

Pr 012012 ,

F k k j

e j

F j I Z V V

012012,Pr 012012 ,

this proximity factor P is approximated using the ratio of positive-sequence impedances

1,1

1,11

k k

k j

z

z P . (1)

By coding the short circuit equations into a Visual Basic program, and employing (1), voltage

sag propagation for the situation described in Figure 1 can now be illustrated. Assuming that thetransmission fault is relatively close to the substation (i.e., proximity factor = 25%), and that T1,

T2, and T3 are all delta:grounded-wye transformers, the line-to-neutral voltages for single-phase,

phase-to-phase, and three-phase transmission faults are shown in Figures 2 – 4, respectively.

Both phasor plots and magnitude bar charts are given.

It is important to note that if a transformer is connected grounded-wye:grounded-wye or

delta:delta, then the voltage sag on the low-voltage side of the transformer is the same as on thehigh-voltage side, as illustrated in Figure 5 for the single-phase fault.

Grady 2007, p. 248




-

Figure 2. Propagation of Close-In Single Phase Fault on the Transmission System (all three

transformers have delta:grounded-wye connections)

Note the voltage swell on phases b and c at the substation 138kV bus. Note also that two phases

are affected after the first transformation, then one phase is affected after two transformations,and again two phases are affected after three transformations.

Grady 2007, p. 249




-

Figure 3. Propagation of Close-In Phase-to-Phase Fault on the Transmission System (all three


Grady 2007, p. 250




-

Figure 4. Propagation of Close-In Three-Phase Fault on the Transmission System (all three


Grady 2007, p. 251




-

Figure 5. Situation in Figure 2 Repeated, but with all Three Transformers Having Grounded-

Wye:Grounded-Wye Connections

Grady 2007, p. 252




-

3. Impact of Distribution System Faults on Adjacent Feeders

Now, consider the situation in Figure 6 where a fault occurs on an adjacent feeder, and a monitor

records the voltage waveform at the substation 12.47kV bus.

Figure 6. Substation Monitor Records Voltages when a Sag Occurs on an Adjacent Feeder.

As in Section 2, unless the customer’s feeder has significant contributors to the fault current, the

voltage sag at the substation 12.47kV bus will appear everywhere along the customer’s 12.47kVfeeder. However, to predict the voltage sag at Customer Levels 1 and 2, the a-b-c line-to-neutral

voltages at the substation 12.47kV bus must be

converted to positive/negative/zero-sequence components,

shifted with the appropriate transformer phase shifts,

converted back to a-b-c.

Ideally, the monitor should report phasor voltage magnitudes and angles during the sag

condition, rather than time-sampled data points for which phase angles must be estimated.

The conversion/shifting/re-conversion procedure has been coded into a Visual Basic program

and is illustrated in Figure 7 to illustrate the propagation of a sag due to a single-phase faultthrough two levels of delta:grounded-wye transformers inside a customer’s facility. Thesubstation 12.47kV voltage entered is approximately the same as predicted in Figure 2.

T2: 12.47kV

-----

480V

T3: 480V

-----120/208V

T1: 138kV

-----

12.47kV

Substation

Customer Level 1.

480V loads

Customer Level 2.

120/208V loads

12.47kV

connection

138kV

Fault on adjacent

feeder

Monitor observes

voltage waveform

Grady 2007, p. 253




-

Figure 7. Propagation of Voltage Sag Due to Single-Phase Fault on Adjacent Feeder

Each vertical column in Figure 7 indicates the following voltages: 1. on the feeder, 2. after one

level of delta:grounded-wye transformer, and 3. after a second level of delta:grounded-wye

transformer. The four vertical columns correspond to various locations down the feeder, provided that sequence impedances of the feeder are entered. For Figure 7, all four columns are

identical since no feeder currents and feeder impedances were entered.

4. Impact of Distribution System Faults on the Customer’s Feeder

To analyze voltage sags on a faulted feeder, feeder currents must be known. Consider thesituation in Figure 8. The Thevenin-equivalent impedance at Customer 1 is Z1, followed by

incremental impedances dZ between the other customers that are distributed down the feeder.

Grady 2007, p. 254




-

Figure 8. Model of Faulted Feeder with Four Distributed Customers

If the fault is to the right of the last customer (as shown), and if the customer load currents aresmall in relation to the measured fault I at the substation transformer, then the voltage drops

along the feeder can be easily calculated using current times impedance, as illustrated in Figure

9. More study is needed, supplemented with field tests, to determine when and with what

certainty the latter assumption can be made.

Figure 9. Propagation of Voltage Sag Due to Single-Phase Fault on Customer’s Feeder

T2: 12.47kV

-----

480V

T3: 480V

-----

120/208V

T1: 138kV

-----

12.47kV

Substation

138kV Fault

T2: 12.47kV

-----

480V

T3: 480V

-----

120/208V

T2: 12.47kV

-----

480V

T3: 480V

-----

120/208V

T2: 12.47kV

-----

480V

T3: 480V

-----

120/208V

Z dZ dZdZ

I

Voltage

measurement

Grady 2007, p. 255



The positive-sequence one-line diagram for a network is shown below. The ground ties

at busses 1 and 4 represent the subtransient impedances of machines. Prefault voltagesare all 1.0pu.

a. Use the definition

k mm I k

j

k

j jk I

V I V z

,0

to fill in column 1 of the Z matrix.

Now, a solidly-grounded three-phase fault occurs at bus 1.

b. Compute the fault currentc. Use the fault current and Z matrix terms to compute the voltages at busses 2 and 3.

d. Find the magnitude of the current flowing in the line connecting busses 2 and 3.

1 2 3

1

2

3

j0.2

j0.3

j0.1

j0.05

Bus 1

Bus 2

Bus 3

j0.1

+

1/0

–

+1/0

–

Grady 2007, p. 257



Grady 2007, p. 258



A 30MVA, 12kV generator is connected to a delta - grounded wye transformer. The generator and

transformer are isolated and not connected to a “power grid.” Impedances are given on equipment bases.

A single-phase to ground fault, with zero impedance, suddenly appears on phase a of the 69kV

transformer terminal. Find the resulting a-b-c generator currents (magnitude in amperes and phase).

Regarding reference angle, assume that the pre-fault phase a voltage on the transformer’s 69kV bus has

angle = 0.

Gen

30MVA, 12kV


X1 = X2 = 0.18puX0 = 0.12pu

Generator is connected

GY through a j0.5 ohm

grounding reactor

X = 0.05pu

30MVA

12kV/69kV

Single phase to

ground fault

occurs on phase aTransformer(Delta-GY)

Grady 2007, p. 259



Grady 2007, p. 260



Grady 2007, p. 261



Grady 2007, p. 262



A one-line diagram for a two-machine system is shown below.

The transmission line between busses 2 and 3 has X1 = X2 = 0.12pu, X0 = 0.40pu on a

100MVA, 345kV base.

Using a base of 100MVA, 345kV in the transmission line, draw one line diagrams in per unit for positive, negative, and zero-sequences.

Using the results from above,

a. Compute the phase a fault current (in pu) for a three-phase bolted fault at bus 2.

b. Compute the phase a fault current (in pu) for a line-to-ground fault at bus 2, phase a.

All values in pu

on equipment

base

Grady 2007, p. 263



Grady 2007, p. 264



Stevenson Six-Bus Short Circuit Example. Fault at Bus 4.

Grady 2007, p. 265



u . t . a u s t i n p o w e r s

y s t e m e n g i n e e r i n g

,

f a u l t s v e r s i o

n =

6 . 0

c a p a b i l i t i e s =

5 0

0 0

b u s s e s ,

1 2 5 0 0

l i n e s a n

d t r a n s f o r m e r s

" - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - -

t h r e e p h a s e

f a u l t

a t

b u s =

4 , n a m e =

b u

s # 4

( p e r u n i t

i m p e d a n c

e s

i n r e c t a n g u l a r

f o r m )

( p e r u n i t v o l t a g e s

a n d c u r r e n t s

i n p o l a r

f o

r m )

0 1 2 s y s t e m

i m p e d a n

c e

( p u ) =

0 . 0 0 0 0 0 E + 0 0

0

. 1 5 2 4 8 E + 0 0

- 0 . 5 1 9 3 8 E - 0 8

0

. 2 2 4 1 1 E + 0 0

0 . 5 1 9 3 8 E - 0 8

0

. 2 2 4 1 1 E + 0 0

f a u l t

i m p e d a n c e

( p

u ) =

0 . 0 0 0 0 0 E + 0 0

0 . 0

0 0 0 0 E + 0 0

0 1 2 v o l t a g e =

0 . 0 0 0 0 0

0 . 0

0

. 0 0 0 0 0

0 . 0

0 . 0 0 0 0

0

0 . 0

a b c v o l t a g e =

0 . 0 0 0 0 0

0 . 0

0

. 0 0 0 0 0

0 . 0

0 . 0 0 0 0

0

0 . 0

0 1 2 c u r r e n t =

0 . 0 0 0 0 0

0 . 0

4

. 4 6 2 0 0

- 9 0 . 0

0 . 0 0 0 0

0

0 . 0

a b c c u r r e n t =

4 . 4 6 2 0 0

- 9 0 . 0

4

. 4 6 2 0 0

1 5 0 . 0

4 . 4 6 2 0

0

3 0 . 0

f r o m s u b t r a n s i e n t

i m p e d a n c e

0 1 2 c u r r e n t =

0 . 0 0 0 0 0

0 . 0

0

. 0 0 0 0 0

0 . 0

0 . 0 0 0 0

0

0 . 0


0 . 0 0 0 0 0

0 . 0

0

. 0 0 0 0 0

0 . 0

0 . 0 0 0 0

0

0 . 0

a t n e i g h b o r i n g

b u s

=

1 , n a m e = g e n # 1

0 1 2 v o l t a g e =

0 . 0 0 0 0 0

0 . 0

0

. 2 8 5 7 1

- 3 0 . 0

0 . 0 0 0 0

0

0 . 0


0 . 2 8 5 7 1

- 3 0 . 0

0

. 2 8 5 7 1 - 1 5 0 . 0

0 . 2 8 5 7

1

9 0 . 0

f a u l t c o n t r i b u t i o n

f r o m c i r c u i t =

1 a t

b u s

=

4

0 1 2 c u r r e n t =

0 . 0 0 0 0 0

0 . 0

1

. 4 2 8 5 7

- 9 0 . 0

0 . 0 0 0 0

0

0 . 0


1 . 4 2 8 5 7

- 9 0 . 0

1

. 4 2 8 5 7

1 5 0 . 0

1 . 4 2 8 5

7

3 0 . 0

v - i

i m p e d a n c e r a t i

o

f o r c i r c u i t =

1 a t

b u s

=

1

0 1 2

i m p e d a n c e r

a t i o =

a b c

i m p

e d a n c e r a t i o =

0 . 0 0 0 0 0 E + 0 0

0

. 0 0 0 0 0 E + 0 0

0 . 1 7 3 2 1

E + 0 0

0 . 1 0 0 0 0 E + 0 0

0 . 1 7 3 2 1 E + 0 0

0

. 1 0 0 0 0 E + 0 0

0 . 1 7 3 2 1

E + 0 0

0 . 1 0 0 0 0 E + 0 0

0 . 0 0 0 0 0 E + 0 0

0

. 0 0 0 0 0 E + 0 0

0 . 1 7 3 2 1

E + 0 0

0 . 1 0 0 0 0 E + 0 0

Grady 2007, p. 266




b u s

=

5 , n a m e =

b u s # 5

0 1 2 v o l t a g e =

0 . 0 0 0 0 0

0 . 0

0

. 2 5 0 5 6

0 . 0

0 . 0 0 0 0

0

0 . 0


0 . 2 5 0 5 6

0 . 0

0

. 2 5 0 5 6 - 1 2 0 . 0

0 . 2 5 0 5

6

1 2 0 . 0



1 a t

b u s

=

4

0 1 2 c u r r e n t =

0 . 0 0 0 0 0

0 . 0

3

. 0 3 3 4 3

- 9 0 . 0

0 . 0 0 0 0

0

0 . 0


3 . 0 3 3 4 3

- 9 0 . 0

3

. 0 3 3 4 3

1 5 0 . 0

3 . 0 3 3 4

3

3 0 . 0

v - i


o


1 a t

b u s

=

5

0 1 2

i m p e d a n c e r

a t i o =

a b c

i m p


0 . 0 0 0 0 0 E + 0 0

0

. 0 0 0 0 0 E + 0 0

- 0 . 2 1 3 8 6

E - 1 5

0 . 8 2 6 0 0 E - 0 1

0 . 5 1 2 4 1 E - 0 8

0

. 8 2 6 0 0 E - 0 1

0 . 6 5 0 2 1

E - 0 8

0 . 8 2 6 0 0 E - 0 1

0 . 0 0 0 0 0 E + 0 0

0

. 0 0 0 0 0 E + 0 0

0 . 8 8 7 0 2

E - 0 8

0 . 8 2 6 0 0 E - 0 1

e n d o f t h r e e p h a s e

f a u l t r e p o r t

- - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - -

l i n e - t o - l i n e

f a u l t

a t

b u s =

4 , n a m e =

b

u s # 4

( p e r u n i t

i m p e d a n c

e s


f o r m )



i n p o l a r

f o

r m )

0 1 2 s y s t e m

i m p e d a n

c e

( p u ) =

0 . 0 0 0 0 0 E + 0 0

0

. 1 5 2 4 8 E + 0 0

- 0 . 5 1 9 3 8 E - 0 8

0

. 2 2 4 1 1 E + 0 0

0 . 5 1 9 3 8 E - 0 8

0

. 2 2 4 1 1 E + 0 0

f a u l t

i m p e d a n c e

( p

u ) =

0 . 0 0 0 0 0 E + 0 0

0 . 0

0 0 0 0 E + 0 0

0 1 2 v o l t a g e =

0 . 0 0 0 0 0

0 . 0

0

. 5 0 0 0 0

0 . 0

0 . 5 0 0 0

0

0 . 0


1 . 0 0 0 0 0

0 . 0

0

. 5 0 0 0 0 - 1 8 0 . 0

0 . 5 0 0 0

0

1 8 0 . 0

0 1 2 c u r r e n t =

0 . 0 0 0 0 0

0 . 0

2

. 2 3 1 0 0

- 9 0 . 0

2 . 2 3 1 0

0

9 0 . 0


0 . 0 0 0 0 0

0 . 0

3

. 8 6 4 2 1

1 8 0 . 0

3 . 8 6 4 2

1

0 . 0


i m p e d a n c e

0 1 2 c u r r e n t =

0 . 0 0 0 0 0

0 . 0

0

. 0 0 0 0 0

0 . 0

0 . 0 0 0 0

0

0 . 0


0 . 0 0 0 0 0

0 . 0

0

. 0 0 0 0 0

0 . 0

0 . 0 0 0 0

0

0 . 0


b u s

=

1 , n a m e = g e n # 1

0 1 2 v o l t a g e =

0 . 0 0 0 0 0

0 . 0

0

. 6 4 2 8 6

- 3 0 . 0

0 . 3 5 7 1

4

3 0 . 0

Grady 2007, p. 267




0 . 8 7 7 7 3

- 9 . 4

0

. 8 7 7 7 3 - 1 7 0 . 6

0 . 2 8 5 7

1

9 0 . 0



1 a t

b u s

=

4

0 1 2 c u r r e n t =

0 . 0 0 0 0 0

0 . 0

0

. 7 1 4 2 9

- 9 0 . 0

0 . 7 1 4 2

9

9 0 . 0


0 . 0 0 0 0 0

0 . 0

1

. 2 3 7 1 8 - 1 8 0 . 0

1 . 2 3 7 1

8

0 . 0

v - i


o


1 a t

b u s

=

1

0 1 2

i m p e d a n c e r

a t i o =

a b c

i m p


0 . 0 0 0 0 0 E + 0 0

0

. 0 0 0 0 0 E + 0 0

0 . 1 2 1 2 4

E + 0 1

- 0 . 2 0 0 0 0 E + 0 0

0 . 7 7 9 4 2 E + 0 0

0

. 4 5 0 0 0 E + 0 0

0 . 6 0 6 2 2

E + 0 0

0 . 1 0 0 0 0 E + 0 0

0 . 4 3 3 0 1 E + 0 0

- 0

. 2 5 0 0 0 E + 0 0

0 . 8 4 9 0 5

E - 0 7

0 . 4 0 0 0 0 E + 0 0


b u s

=

5 , n a m e =

b u s # 5

0 1 2 v o l t a g e =

0 . 0 0 0 0 0

0 . 0

0

. 6 2 5 2 8

0 . 0

0 . 3 7 4 7

2

0 . 0


1 . 0 0 0 0 0

0 . 0

0

. 5 4 5 0 6 - 1 5 6 . 5

0 . 5 4 5 0

6

1 5 6 . 5



1 a t

b u s

=

4

0 1 2 c u r r e n t =

0 . 0 0 0 0 0

0 . 0

1

. 5 1 6 7 2

- 9 0 . 0

1 . 5 1 6 7

2

9 0 . 0


0 . 0 0 0 0 0

0 . 0

2

. 6 2 7 0 3 - 1 8 0 . 0

2 . 6 2 7 0

3

0 . 0

v - i


o


1 a t

b u s

=

5

0 1 2

i m p e d a n c e r

a t i o =

a b c

i m p


0 . 0 0 0 0 0 E + 0 0

0

. 0 0 0 0 0 E + 0 0

0 . 0 0 0 0 0

E + 0 0

0 . 0 0 0 0 0 E + 0 0

0 . 2 2 5 5 2 E - 0 7

0

. 4 1 2 2 6 E + 0 0

0 . 1 9 0 3 3

E + 0 0

0 . 8 2 6 0 0 E - 0 1

- 0 . 2 9 7 5 4 E - 0 8

- 0

. 2 4 7 0 6 E + 0 0

- 0 . 1 9 0 3 3

E + 0 0

0 . 8 2 6 0 0 E - 0 1

e n d o f

l i n e - t o - l i n

e


- - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - -

l i n e - t o - g r o u n d

f a u

l t a t

b u s =

4 , n a m e =

b u s # 4

( p e r u n i t

i m p e d a n c

e s


f o r m )



i n p o l a r

f o

r m )

0 1 2 s y s t e m

i m p e d a n

c e

( p u ) =

0 . 0 0 0 0 0 E + 0 0

0

. 1 5 2 4 8 E + 0 0

- 0 . 5 1 9 3 8 E - 0 8

0

. 2 2 4 1 1 E + 0 0

0 . 5 1 9 3 8 E - 0 8

0

. 2 2 4 1 1 E + 0 0

f a u l t

i m p e d a n c e

( p

u ) =

0 . 0 0 0 0 0 E + 0 0

0 . 0

0 0 0 0 E + 0 0

0 1 2 v o l t a g e =

0 . 2 5 3 8 3

1 8 0 . 0

0

. 6 2 6 9 1

0 . 0

0 . 3 7 3 0

9

1 8 0 . 0


0 . 0 0 0 0 0

0 . 0

0

. 9 4 6 0 3 - 1 1 3 . 7

0 . 9 4 6 0

3

1 1 3 . 7

Grady 2007, p. 268



0 1 2 c u r r e n t =

1 . 6 6 4 7 1

- 9 0 . 0

1

. 6 6 4 7 1

- 9 0 . 0

1 . 6 6 4 7

1

- 9 0 . 0


4 . 9 9 4 1 2

- 9 0 . 0

0

. 0 0 0 0 0

0 . 0

0 . 0 0 0 0

0

0 . 0


i m p e d a n c e

0 1 2 c u r r e n t =

0 . 0 0 0 0 0

0 . 0

0

. 0 0 0 0 0

0 . 0

0 . 0 0 0 0

0

0 . 0


0 . 0 0 0 0 0

0 . 0

0

. 0 0 0 0 0

0 . 0

0 . 0 0 0 0

0

0 . 0


b u s

=

1 , n a m e = g e n # 1

0 1 2 v o l t a g e =

0 . 0 0 0 0 0

0 . 0

0

. 7 3 3 5 1

- 3 0 . 0

0 . 2 6 6 4

9 - 1 5 0 . 0


0 . 6 4 3 1 0

- 5 1 . 0

0

. 6 4 3 1 0 - 1 2 9 . 0

1 . 0 0 0 0

0

9 0 . 0



1 a t

b u s

=

4

0 1 2 c u r r e n t =

1 . 2 6 9 1 5

- 9 0 . 0

0

. 5 3 2 9 8

- 9 0 . 0

0 . 5 3 2 9

8

- 9 0 . 0


2 . 3 3 5 1 1

- 9 0 . 0

0

. 7 3 6 1 7

- 9 0 . 0

0 . 7 3 6 1

7

- 9 0 . 0

v - i


o


1 a t

b u s

=

1

0 1 2

i m p e d a n c e r

a t i o =

a b c

i m p


0 . 0 0 0 0 0 E + 0 0

0

. 0 0 0 0 0 E + 0 0

0 . 5 4 1 6 3

E + 0 0

0 . 4 3 8 1 2 E + 0 0

0 . 1 1 9 1 9 E + 0 1

0

. 6 8 8 1 2 E + 0 0

0 . 0 0 0 0 0

E + 0 0

0 . 0 0 0 0 0 E + 0 0

0 . 4 3 3 0 1 E + 0 0

- 0

. 2 5 0 0 0 E + 0 0

0 . 1 0 8 3 3

E + 0 1

- 0 . 3 9 6 2 0 E - 0 7


b u s

=

5 , n a m e =

b u s # 5

0 1 2 v o l t a g e =

0 . 1 6 7 9 9

1 8 0 . 0

0

. 7 2 0 4 0

0 . 0

0 . 2 7 9 6

0

1 8 0 . 0


0 . 2 7 2 8 0

0 . 0

0

. 9 4 9 1 3 - 1 1 4 . 2

0 . 9 4 9 1

3

1 1 4 . 2



1 a t

b u s

=

4

0 1 2 c u r r e n t =

0 . 3 9 5 5 6

- 9 0 . 0

1

. 1 3 1 7 3

- 9 0 . 0

1 . 1 3 1 7

3

- 9 0 . 0


2 . 6 5 9 0 2

- 9 0 . 0

0

. 7 3 6 1 7

9 0 . 0

0 . 7 3 6 1

7

9 0 . 0

v - i


o


1 a t

b u s

=

5

0 1 2

i m p e d a n c e r

a t i o =

a b c

i m p


- 0 . 4 7 3 2 0 E - 2 4

- 0

. 4 2 4 7 0 E + 0 0

0 . 1 7 0 1 8

E - 2 4

0 . 1 0 2 5 9 E + 0 0

0 . 1 5 4 7 7 E - 0 7

0

. 6 3 6 5 4 E + 0 0

- 0 . 1 1 7 6 4

E + 0 1

0 . 5 2 7 5 8 E + 0 0

- 0 . 2 9 7 5 4 E - 0 8

- 0

. 2 4 7 0 6 E + 0 0

0 . 1 1 7 6 4

E + 0 1

0 . 5 2 7 5 8 E + 0 0

e n d o f

l i n e - t o - g r o

u n d


Grady 2007, p. 269



Grady, Loadflow, June 2007, Page 1

Loadflow

Formulation of the loadflow problem. Gauss-Seidel, Newton-Raphson, and Stott's algorithm.

Calculation of line flows, system losses, and area interchange.

1 Formulation of the Problem

The loadflow problem is one of the classic power system engineering problems. During the

early days of digital computers, many advances in techniques for solving large sets of equations

were brought about specifically to help solve the loadflow problem.

In most electrical circuit analyses, the network consists of known impedances, voltage sources,

and current sources. However, in the loadflow problem, active and reactive powers, rather than

shunt impedances, are specified at most network busses, because most loads behave, on average,

as constant power loads (active and reactive power), as long as their applied voltage remains

within reasonable ranges. Consider, for example, the air conditioning load of a building. A

certain amount of energy is required to maintain T between inside and outside temperatures.Even though the air conditioner cycles on-and-off, and the voltage may change slightly, the air

conditioning load appears, on the average, as a fixed power load, rather than as a fixed

impedance load.

Power system loads are closely monitored at substations, at large customers, and for total electric

utility companies. Loads tend to have predictable daily, weekly, and seasonal patterns. Annual

peak demands and energies for electric utilities are forecasted for generation and planning

purposes.

The purpose of the loadflow program is to compute bus voltages and line/transformer/cable

power flows once network topology, impedances, loads, and generators have been specified.Ideally, the computed bus voltages for the study system should remain within acceptable ranges,

and line/transformer/cable power flows should be below their rated values, for a reasonable set

of outage contingencies.

From a loadflow perspective, there are four parameters at every bus - voltage magnitude V ,

voltage angle , active power P , and reactive power Q. Two may be specified, and the other

two calculated. For most busses, P and Q are specified, and V and are calculated. Obviously,

P and Q cannot be specified at all busses because that would imply that system losses are known

a priori. Therefore, the loadflow problem must include one "swing bus" at which the P can

assume any value so that it "makes up" system losses. The swing bus is usually a centrally-

located large generator whose voltage magnitude and phase angle (usually = zero) arespecified.

Although any two of the four parameters can be specified, the usual way in which power system

busses are classified is given in Table 1.

Grady 2007, p. 270




Table 1. Loadflow Classification of Power System Busses

Classification Knowns Unknowns

PQ (Load Bus) P, Q V,

PV (Generator Bus) P, V Q,

V (Swing Bus) V, P, Q

The loadflow program solves for the set of unknowns that produces power balance at all busses,

or as illustrated for bus i in Figure 1,

calci

calci

speci

speci jQ P jQ P ,

where

*ii

calci

calci I V jQ P .

In other words, the power specified at each bus must equal the power flowing into the system.

Note in Figure 1 that specified power is drawn as positive generation, to be consistent with KCL

equation YV=I .

Total Current Flowing From Bus i into the System is

P + jQi i

spec spec

IB1

IB2

IB3

Bus iBranch

Currents

Into

System

I = IB1 + IB2 + IB3i

= V Ii i*

Vi i| |

Figure 1. Power Balance for Bus i

Since there are two unknowns at every bus, the size of the loadflow problem is 2 N , where N is

the number of busses. Obviously, to solve the problem, there must be two equations for every bus. These come from KCL, which for any bus i have the form

*

1

,*

N

j j jiiii

calci

calci

speci

speci V yV I V jQ P jQ P .

Separating into real and imaginary components yields two equations for bus i,

Grady 2007, p. 271




ji ji j ji

N

j

i speci V yV P ,,

1

cos

,

ji ji j ji

N

j i

spec

i V yV Q

,,1

sin

,

where ji ji jiiiiiii y yV V V V ,,,,, .

The problem now is now to find the set of bus voltages that satisfies the above 2 N equations.

2. Gauss-Seidel Method

Gauss-Seidel is an early formulation of the loadflow problem that requires little memory and it is

easily programmed. However, it is usually slower than other methods. It is based upon the idea

of expanding the complex form of the power balance equation as follows:

*

,1,,

*

1,

*

N

i j j j jiiiii

N

j j jiiii

speci

speci V yV yV V yV I V jQ P ,

or

N

i j j

j jiiiiii speci

speci V yV V yV jQ P

,1

,*

,* ,

so that

N

i j j

j ji

i

speci

speci

iii V y

V

jQ P

yV

,1

,*,

1 .

The solution procedure is to:

1. Initialize the bus voltages. For load busses, use V = 1 + j0. For generator busses

(including the swing bus), use 0 jV V spec .

2. One-by-one, update the individual bus voltages using

N

i j j

j ji

i

speci

speci

iii V y

V

jQ P

yV

,1

,*,

1.

For PV busses, update the voltage angle, while holding the voltage magnitude constant at

the specified value. Do not update the swing bus.

Grady 2007, p. 272




3. Check the mismatch P and Q at each bus. If all are within tolerance (typical tolerance is

0.00001 pu), a solution has been found. Otherwise, return to Step 2.

Convergence is usually faster if an acceleration factor is used. For example, assume that the

voltage at bus i at iteration m is miV , and that the updating equation in Step 2 computes new

jV .

Instead of using new jV directly, accelerate the update with

mi

newi

mi

mi V V V V 1 ,

where acceleration factor is in the range of 1.2 to 1.6.

3 Newton-Raphson Method

The Newton-Raphson method is a very powerful loadflow solution technique that incorporates

first-derivative information when computing voltage updates. Normally, only 3 to 5 iterations

are required to solve the loadflow problem, regardless of system size. Newton-Raphson is themost commonly used loadflow solution technique.

An easy way to illustrate the Newton-Raphson technique is to solve a simple equation whose

answers are already known. For example, consider

0991 x x ,

which when expanded becomes

0991002 x x .

The objective is to find x so that

099100)( 2 x x x f .

Of course, in this case, the two solutions are known a priori as x = 1, and x = 99.

The Newton-Raphson procedure is based on Taylor's expansion, truncated past the first

derivative, which gives

x x

x f

x f x x f x

)(

)()( .

Clearly, the above equation gives a straight-line approximation for )( x x f .

The objective is to find x so that )( x x f is the desired value (which in this example is

zero). Solving for x yields

Grady 2007, p. 273




x x

x f

x f x x f x

)(

)()( ,

which for this example is

x x x

x f

x f

x

x f

x f x

)(

)(

)(

)(0

.

The update equation for iteration (m + 1) is then

)(

)()()()1(

)(

)(

m x

mmmm

x

x f

x f x x x x

,

where in this example

1002)(

x x

x f

.

If a starting point of x = 2 is chosen, then the solution proceeds as follows:

Iteration - m x f(x) x

x f

)(

0 2 -97 -96

1 0.9896 1.0193 -98.02

2 0.9999 0.0098 -98.00

Additional iterations can be performed if tighter solution tolerance is needed. Note that if a

starting point of x = 50 had been chosen, the partial derivative would have been zero, and the

method would have failed.

If x = 80 is the starting point, then the process yields the following:

Iteration - m x f(x) x x f

)(

0 80 -1501 60

1 105.02 626.2 110.04

2 99.33 32.45 98.66

3 99.00 0 98.00

Grady 2007, p. 274




Therefore, the starting point greatly affects the ability of a Newton-Raphson method to converge,

and the answer to which it converges. Fortunately, in the loadflow problem, most voltages are

near 1.0 pu in magnitude and 0.0 degrees, so that we are able to accurately estimate starting

values.

For the loadflow problem, the Newton-Raphson method is expanded in matrix form. Forexample, consider a set of N nonlinear equations and N unknowns,

N N N

N

N

y x x x f

y x x x f

y x x x f

,,,

,,,

,,,

21

2212

1211

.

The task is to find the set of unknown N x x x ,,, 21 , given the known set N y y y ,,, 21 , and

given a starting point)0()0(

2)0(

1 ,,, N x x x .

Applying Taylor's theorem as before, truncated after the first derivative, yields for Row i

N N ii x x x x x x f y)0(

2)0(

21)0(

1 ,,, )0()0(2

)0(1

,,, N i x x x f

+

)0()0(22

)0(11

x N

i N

x

i

x

i

x

f x

x

f x

x

f x

,

where x(0) represents the starting estimate set)0()0(

2)0(

1,,, N x x x .

N similar equations in matrix form are

)0()0(2

)0(1

)0()0(2

)0(12

)0()0(2

)0(11

2

1

21

2

2

2

1

2

1

2

1

1

1

2

1

,,,

,,,

,,,

N N

N

N

N

N

N N N

N

N

N x x x f

x x x f

x x x f

x

x

x

x

f

x

f

x

f

x

f

x

f

x

f

x

f

x

f

x

f

y

y

y

,

or

Grady 2007, p. 275




N N N N

N

N

x

x

x

J

x x x f y

x x x f y

x x x f y

2

1

)0()0(2

)0(1

)0()0(2

)0(122

)0()0(2

)0(111

,,,

,,,

,,,

,

where J is an N x N matrix of partial derivatives, known as the Jacobian matrix. Therefore, in an

iterative procedure, the above equation is used to update the X vector according to

)(1)()()()()1( m spec

mmmmmY Y J X X X X

,

where

)()(2

)(1

)()(2

)(12

)()(2

)(11

)(2

1

,,,

,,,

,,,

,

m N

mm N

m N

mm

m N

mm

m

N

spec

x x x f

x x x f

x x x f

Y

y

y

y

Y

.

In the loadflow problem, the matrix update equation is symbolically written in mixed

rectangular-polar form as

m N

m N

mm

mm

m N

m N

mm

mm

m Ncalc Nspec

mcalc spec

mcalc spec

m Ncalc Nspec

mcalc spec

mcalc spec

V V

V V

V V V

Q J

Q J

V

P J

P J

QQ

QQ

QQ

P P

P P

P P

1

21

2

11

1

1

21

2

11

1

43

21

)(

)(22

)(11

)(

)(22

)(11

or, in abbreviated form,

V J J

J J

Q

P

43

21 .

Grady 2007, p. 276




The dimension of the above problem is actually 2 bussesPVof Number 2 N since V

updates at PV busses are not required, and since V and updates at the swing bus are not

required.

In highly inductive power systems, P is closely related to voltage angles, and Q is closely related

to voltage magnitudes. Therefore, in the above mixed rectangular-polar formulation, the terms

in 41 and J J tend to have larger magnitudes than those in 32 and J J . This feature makes the

Jacobian matrix more diagonally dominant, which improves robustness when Gaussian

eliminating or LU decomposing J .

The above formulation of the Jacobian matrix is often modified to take advantage of symmetry

in the partial derivatives. This modification is

V

V

V QQ

V

V

P P

V

Q

P

1

1

.

The partial derivatives are derived from

ji ji j ji

N

j

icalc

i V yV P ,,

1

cos

,

ji ji j ji

N

j

icalci V yV Q ,,

1

sin

,

and have the following form

For 1 J ,

.,sin

sin

,,

,1

,,

ik V yV P

V yV P

k ik ik k ii

k

i

N

i j j

ji ji j jiii

i

For 2 J ,

Grady 2007, p. 277




.,cos

cos2cos

,,

,,1

,,,

ik yV V

P

yV V yV

P

k ik ik iik

i

ii

N

i j j

iii ji ji j jii

i

For 3 J ,

.,cos

cos

,,

,1,,

ik V yV Q

V yV Q

k ik ik k iik

i

N

i j j

ji ji j jiii

i

For 4 J ,

.,sin

sin2sin

,,

,,1

,,,

ik yV V

Q

yV V yV

Q

k ik ik iik

i

ii

N

i j j

iii ji ji j jii

i

Note the symmetry in the J terms. If V is used as an updating parameter rather than , then

the expressions for 1 J are

.,sin

sin

,,

,1,,

ik yV V

P

V yV

P

k ik ik iik k

i

N

i j j

ji ji j jiii

i

and for 3 J ,

.,cos

cos

,,

,1,,

ik V yV

Q

V yV

Q

k ik ik k ik i

i

N

i j j

ji ji j jiii

i

and there is even more symmetry in J .

The diagonal dominance of 1 J and 4 J can be observed by the examining the partial derivatives

as follows: the differences between voltage angles at adjacent busses is usually small, so that the

Grady 2007, p. 278




ji terms are small. The angles found in the admittance matrix are usually large (i.e. near

o90 ) because most power systems are reactive. Therefore, the sine terms in the matrix update

equation tend to be near unity, while the cosine terms tend to be near zero. Decoupled loadflow

programs use only 1 J and 4 J , treating the P and Q problems separately.

So that the benefits of optimal bus ordering can be fully exploited, non-decoupled loadflow

Jacobian matrices are usually formulated in the following alternating-row form, rather than that

described symbolically above:

N

N N

N

N

V

V

V

V

V

V

J

Q

P

Q

P

Q

P

2

22

1

11

2

2

1

1

.

The solution procedure for the Newton-Raphson loadflow proceeds with:

1. Initialize the bus voltages. For load busses, use V = 1 + j0. For generator busses

(including the swing bus), use 0 jV V spec .

2. Form the Jacobian matrix, and update all bus voltage magnitudes and phase angles,

except for those at the swing bus, and except for the voltage magnitudes at PV busses.

3. Check the mismatch P and Q at each bus. If all are within tolerance (typical tolerance is

0.00001 pu), a solution has been found. Otherwise, return to Step 2.

4 Stott's Algorithm

Stott's algorithm takes advantage of the strong decoupling of P and Q that occurs in most high-

voltage power systems. His formulation is based upon the assumption that P is primarily related

to , and that Q is primarily related to V .

He begins by writing the expression for power at bus i, which is

*

1,

N

j

j jiicalci

calci V yV jQ P .

He then defines

Grady 2007, p. 279




ji ji ji jBG y ,,, .

Expanding the power expression yields

N

j ji ji jiii

calci

calci jBGV V jQ P

1,, ,

or, expanding in real and imaginary form

N

j

ji ji ji ji jicalc

i BGV V P

1

,,,, sincos

N

j

ji ji ji ji jicalci BGV V Q

1

,,,, cossin ,

where ji ji , .

From the above equations, the necessary loadflow partial derivatives are

ji ji ji ji ji

i j j

i BGV V P

,,,, cossin

,

and

ji ji ji ji ji

N

i j ji

i BGV V P

,,,,

,1

cossin

,

which, by comparing to the Q equation simplifies to

ii jiiiiiii

i

i BGV Q P

,,,,2

cossin

= iiii BV Q ,

2 .

Similarly, for Q,

i j

j

i

j ji ji ji jii

i j j

i P

V BGV

V

Q

1cossin ,,,, ,

and

Grady 2007, p. 280




iiiiiiiii

N

i j j

ji ji ji ji ji

i BGV BGV V

Q,,,,

,1

,,,, cossin2cossin

.

After simplifying, the above equation becomes

iiii

i

i

i BV V

Q

V

Q,

.

The form of the Jacobian update equation is now

V

V L

H

Q

P 1

0

0 ,

where

ji BGV V L H ji ji ji ji ji ji ji ,cossin ,,,,,, ,

iiiiii BV Q H ,2

, ,

iiiiii BV Q L ,2

, .

Now, since the angular differences are small, then ji ji ,, sincos . In reactive power

systems, ji ji

G B,,

, so that ji ji ji ji

G B,,,,

sincos . Also, in most cases,

iii Q B , , so that iiii Q BV ,2 . Substituting these approximations into the above

expressions for H and L yields

ji ji ji ji BV V L H ,,, ,

iiiiiii BV L H ,2

,, .

These lead to the following simple form of the update equation:

N N N N N N

N

N

N V

V

V

B B B

B B B

B B B

V

V

V

P

2

1

2

1

,2,1,

,22,21,2

,12,11,1

2

1

00

00

00

00

00

00

Grady 2007, p. 281




V VB ' ,

N

N N N N N N

N

N

N

V

V

V

V

V

V

V

V

V

B B B

B B B

B B B

V

V

V

Q

2

2

1

1

2

1

,2,1,

,22,21,2

,12,11,1

2

1

00

00

00

00

00

00

V

V V VB '' .

Stott proposes the following additional simplifications:

1. Omit from B' those network elements which primarily affect MVAr flows (i.e. shunt

reactances, off-nominal transformer taps, etc.)

2. Omit from B'' the angle-shifting effects of phase-shifting transformers, which primarily

affect MW flows.

3. Since the voltage magnitudes are close to unity, write

'' BV BV

P (first one-half iteration),

and

V BV

V V B

V

Q

'''' (second one-half iteration).

4. Neglect series resistances when calculating B' .

At this point, B' and B'' remain constant throughout the solution. Therefore, B' and B'' should be

LU decomposed once, and re-used for each half iteration. The solutions steps are

1. LU decompose B' and B'' .

2. Calculate

V

P .

3. Update using

' BV

P .

Grady 2007, p. 282




4. Check convergence, and continue if not converged.

5. Calculate

V

Q .

6. Update V using V BV

Q

''

7. Check convergence, and return to Step 2 if not converged.

5. Other Considerations

Current and Power Flow in Transmission Lines and Cables

Once the bus voltages throughout the system have been calculated, then the loadflow program

must calculate power flows through lines/transformers/cables. The standard pi-equivalentmodels are used for this purpose as follows:

R jX

Icapj Icapk

Ijk IkjVj/ j Vk/k

Q(pu)

2

Q(pu)

2 ,

so from side j, 0902

j jk k j j jk QV

jX R

V V I

,

and from side k, 0902

k k

j jk k kj

QV

jX R

V V I

.

Note that the current on the two opposite ends of the line are not exactly the same due to the fact

that the capacitor currents are not, in general, equal to each other.

The corresponding power flows for the two ends of the line are ** , kjkjkj jk jk jk I V S I V S .

Current and Power Flow in Transformers

The transformer equivalent circuit taken from Section 3 is

Grady 2007, p. 283




y

Ijk IkjBus j

Bus k t / :1

Bus k'

Vj/ jVk/ k

,

so for side k, yt

V V I

j jk k kj

,

and for side j,

t

I I

kj jk .

The corresponding power flows for the two ends of the transformer are

** , kjkjkj jk jk jk I V S I V S .

Area Interchange

Large-scale power systems usually consist of several, or perhaps many, individual electric utility

companies. In these cases, each area may have a desired net input or output power to satisfy

sales and purchase agreements.

The area interchange feature in a loadflow program sums the tie-line flows into each area for a

net area power input. If these nets are not the desired values, to within a few megawatts, then an

area-control generator within the area is adjusted by the error amount. The sum of all input

powers for all areas must, of course, be zero.

Convergence Criteria

Most loadflow data cases are developed using a 100 MVA base, so that a 1% load corresponds to

1 MW. A typical convergence criteria is that the highest individual bus P and Q mismatches are

within 0.001 - 0.01%, or 0.00001 - 0.0001 pu.

Grady 2007, p. 284



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,

P r o f . M a c k G r a d y , w w w . e c e . u t e x a s . e d u / ~ g r a d y , M a y 3 0 , 2 0 0 7

P a g e 1 o f 2 4

1 . S u m m a r y o f P C F L O F i l e s

I n p u t ( n o t e – a n

y l i n e s i n i n p u t f i l e s t h a t b e g i n w i t h a c o l o n i n c o l u m n 1 a r e

t r e a t e d a s c o m m e n t s a n d a r e

s k i p p e d )

A D A T . C S V : L o a d f l o w a r e a i n p u t d a t a . ( P C F L

O a l s o c r e a t e s t e m p o r a r y f i l e A D A T . T M P )

B D A T . C S V : B u s d a t a .

L D A T . C S V : L i n e a n d t r a n s f o r m e r d a t a .

O P T I O N S . C S V

: S o l u t i o n o p t i o n s .

S P E C T R A . C S V

: U s e r - s p e c i f i e d h a r m o n i c c u r r e n t i n j e c t i o n s p e c t r a . ( P C F L O a

l s o c r e a t e s t e m p o r a r y f i l e S P E C

T R A . T M P )

O u t p u t

A S O L N . C S V : S o l v e d a r e a d a t a f o r l o a d f l o w s .

B O R D E R . C S V : F i l e b u i l t b y P C F L O t h a t l i s t s

t h e b u s s e s i n o p t i m a l o r d e r .

E X L O G . C S V : E c h o p r i n t o f s c r e e n m e s s a g e s .

F R E P . T X T : O u

t p u t o f p r o g r a m F A U L T S .

H P A_

L A S T_ C

A S E . C S V , H P A_ S

U M M A R Y . C

S V . O u t p u t o f h a r m o n i c p o w e r a n a l y z e r p r o g r a m H P A .

I S O L N . C S V , V S O L N . C S V : S o l v e d b r a n c h c u r r e n t s a n d b u s v o l t a g e s f o r l o a d f l o w a n d h a r m o n i c s s t u d i e s . ( S i n e s e r i e s f o r m a t

f o r F o u r i e r s e r i e

s )

O U T 1 . C S V , O U

T 2 . C S V , O U T 3 . C S V : E c h o p r

i n t o f i n p u t d a t a f o r l o a d f l o w , s h o r t c i r c u i t , a n d h a r m o n i c s , a l o n

g w i t h p e r t i n e n t

m e s s a g e s a n d e r r o r s .

O U T 4 . C S V : F u

l l l o a d f l o w o u t p u t d a t a ( i f r e q u e s t e d ) .

O U T 5 . C S V : L o a d f l o w s u m m a r y o u t p u t u s e d f o

r a n a l y z i n g t h e i m p a c t o f p o w e r t r a n s a c t i o n s a c r o s s a p o w e r g r i d .

T H D V . C S V , T H

D I . C S V : S o l v e d t o t a l h a r m o n i c v o l t a g e a n d c u r r e n t d i s t o r t i o n s f o r h a r m o n i c s s t u d i e s .

Z B U S 0 . C S V , Z

B U S 1 . C S V , Z B U S 2 . C S V : S o l v e d z e r o / p o s i t i v e / n e g a t i v e i m

p e d a n c e m a t r i x e l e m e n t s f o r s h o r t c i r c u i t ( i n

r e c t a n g u l a r f o r m

) o r h a r m o n i c s t u d i e s ( i n p o l a r f o r m ) , i n p e r u n i t .

Grady 2007, p. 285



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 2 o f 2 4

T e m p o r a r y F i l e s C r e a

t e d D u r i n g E x e c u t i o n

B D A T . T M P : U

n f o r m a t t e d b u s d a t a f i l e b u i l t b y P C F L O a n d r e a d b y F A U L T S .

F F R E P 1 . T X T ,

F F R E P 2 . T X T , F F R E P 3 . T X T .

T e m p o r a r l y f i l e s u s e d d u r i n g s h o r t c i r c u i t s t u d i e s ( p r o d u c

e d b y p r o g r a m

F A U L T S )

L D A T . T M P : U

n f o r m a t t e d l i n e a n d t r a n s f o r m e r d a t a f i l e b u i l t b y P C F L O a n d r e a d b y F A U L T S .

S P E C T R A . T M P : T e m p o r a r y f i l e b u i l t b y P C F L

O d u r i n g h a r m o n i c s t u d i e s .

Grady 2007, p. 286



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 3 o f 2 4

2 . P r

e p a r e d C a s e s , R e a d y t o R u n

L o a d f l o w

5 B u s S t e v e n s o n * L o a d f l o w E x a m p l e , p p . 2 0 0 - 2 0 5 . F i l e E x t e n s i o n_ S

5 .

4 8 6 0 B u s L o a d f

l o w C a s e . F i l e E x t e n s i o n_

S C R

E W B E A N . U s e d f o r p o w e r g r i d s t u d i e s .

S h o r t C i r c u i t

6 B u s G r a i n g e r - S t e v e n s o n * * S h o r t C i r c u i t E x a m p l e ( P r o b . 3 . 1 2 , p . 1 3 9 , a n

d c o n t i n u e d w i t h P r o b . 1 1 . 1 7 , p . 4 6 9 ) . F i l e

E x t e n s i o n_ S

6

9 B u s G r a i n g e r - S t e v e n s o n * * S h o r t C i r c u i t E x a

m p l e ( P r o b . 3 . 1 3 , p p . 1 3 9 - 1 4 0 ,

a n d c o n t i n u e d w i t h P r o b . 1 1 . 1 8 , p . 4 6 9 ) . F i l e

E x t e n s i o n_ S

9 .

H a r m o n i c s

5 B u s T u t o r i a l .

F i l e E x t e n s i o n s_

F I V E a n d_ F I

V E_ F

I L T E R .

3 3 B u s A S D E x

a m p l e . F i l e E x t e n s i o n_ D

F W .

1 8 B u s O i l P i p e l i n e A S D E x a m p l e . F i l e E x t e n s

i o n_ P

I P E L I N E .

1 1 1 B u s 3 r d H a r m o n i c R e s o n a n c e E x a m p l e . F i l e E x t e n s i o n_ T

V .

1 7 B u s S m a l l S k i A r e a E x a m p l e . F i l e E x t e n s i o n s_ S

K I A ,_ S

K I B ,_ S

K I C .

4 5 4 B u s L a r g e S k i A r e a E x a m p l e . F i l e E x t e n s i o n s_ O

L Y M P I C S_ A

( u n f i l t e r e d

) ,_ O

L Y M P I C S_

D ( f i l t e r e d ) .

8 7 B u s D i s t r i b u t i o n E x a m p l e w i t h 3 r d H a r m o n i c

P r o b l e m D u e t o S i n g l e - P h a s e L

o a d s . F i l e E x t e n s i o n_ P

C .

*

W i l l i a m D . S t e v e n s o n , J r . ,

E l e m e n t s o f P o w e r S y s

t e m A n a l y s i s , F o u r t h E d i t i o n , M

c G r a w - H i l l , N e w Y o r k , 1 9 8 2 .

* *

J o h n J . G r a i n g e r , W i l l i a m D . S t e v e n s o n , J r . ,

P o w e

r S y s t e m A n a l y s i s , M c G r a w - H i l l , N e w Y o r k , 1 9 9 4 .

Grady 2007, p. 287



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 4 o f 2 4

3 . C

o n t e n t o f I n p u t D a t a F i l e s

B U S D A T A

( F i l e = B D A T . C

S V , o n e r e c o r d p e r b u s . C S V

f o r m a t )

V a r i a b l e

C o m m

e n t s

N u m b e r

I n t e g e r

N a m e

U p t o 1 2 c h a r a c t e r s

T y p e

1 = S w

i n g B u s

2 = P V

B u s

3 = P Q

B u s

L i n e a r P G e n e r a t i o n

P e r c e n

t

L i n e a r Q G e n e r a t i o n

P e r c e n

t

L i n e a r P L o a d

P e r c e n

t

L i n e a r Q L o a d

P e r c e n

t

D e s i r e d V o l t a g e

P e r u n i t

S h u n t R e a c t i v e Q L o a d

P e r c e n

t @ v o l t a g e = 1 p e r u n i t

M a x i m u m Q G e n e r a t i o n

P e r c e n

t

M i n i m u m Q G e n e r a t i o n

P e r c e n

t

C o n t r o l A r e a

I n t e g e r

Grady 2007, p. 288



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 5 o f 2 4

R e m o t e - C o n t r o l l e d B u s N u m b e r

U s e d f

o r c o n t r o l l i n g v o l t a g e a t a r e m o t e b u s . F o r t h e s e

c a s e s , t h e d e s i r e d v o l t a g e s p e c i f i e d a b o v e a p p l i e s t o t h e

r e m o t e

b u s .

C o n n e c t i o n T y p e f o r S h u n t R e a c t i v e Q L o a d

0 o r 1 = G r o u n d e d W y e .

O t h e r w

i s e , u n g r o u n d e d w y e o r d e l t a ( i . e . n o z e r o

s e q u e n

c e p a t h )

S u b t r a n s i e n t R , X ( P o s . S e q u e n c e )

S e r i e s

i m p e d a n c e o f m o t o r o r g e n e r a t o

r , i n p e r u n i t

S u b t r a n s i e n t R , X ( N e g

. S e q u e n c e )

S e r i e s



S u b t r a n s i e n t R , X ( Z e r o

S e q u e n c e )

S e r i e s



( i g n o r

i n g c o n n e c t i o n t y p e a n d g r o u n d

i n g i m p e d a n c e s )

( d o n o

t m u l t i p l y b y 3 )

C o n n e c t i o n T y p e f o r S u b t r a n s i e n t I m p e d a n c e s

0 o r 1 = G r o u n d e d W y e .

O t h e r w

i s e , u n g r o u n d e d w y e o r d e l t a ( i . e . n o z e r o

s e q u e n

c e p a t h )

G r o u n d i n g I m p e d a n c e R , X f o r S u b -

S e r i e s

i m p e d a n c e f r o m w y e p o i n t t o g r o u n d , i n p e r u n i t .

T r a n s i e n t I m p e d a n c e s

( d o n o t m u l t i p l y b y 3 ) .

N o n l i n e a r D e v i c e P G e n e r a t i o n

P G e n e r a t i o n t h a t c o m e s f r o m t h e n o n l i n e a r d e v i c e ,

i n p e r c

e n t .

N o n l i n e a r D e v i c e P L o a d

P L o a d

o f t h e n o n l i n e a r d e v i c e ,

i n p e r c

e n t .

N o n l i n e a r D e v i c e D i s p l a c e m e n t P o w e r F a c t o r

P e r U n

i t ( p o s i t i v e f o r l a g g i n g i s p o s i t i v e , n e g a t i v e f o r

l e a d i n g ) .

Grady 2007, p. 289



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 6 o f 2 4

N o n l i n e a r D e v i c e T y p e

1

=

2 - p u l s e c u r r e n t s o u r c e

2

=

2 - p u l s e v o l t a g e s o u r c e

3

=

6 - p u l s e c u r r e n t s o u r c e

4

=

6 - p u l s e v o l t a g e s o u r c e

5

=

1 2 - p u l s e c u r r e n t s o u r c e

6

=

1 2 - p u l s e v o l t a g e s o u r c e

7

=

1 8 - p u l s e c u r r e n t s o u r c e

8

=

1 8 - p u l s e v o l t a g e s o u r c e

9

=

D i v e r s i f i e d 6 - p u l s e c u r r e n t s o u

r c e ( l i k e

t y p e 3 , b u t w i t h t h e 1 1 & 1 3 t h h a r m o n i c s

m u l t i p l i e d b y 0 . 7 5 , t h e 1 7 & 1 9 t h h a r m o n i c s

b y 0 . 5 0 , a n d a l l h i g h e r h a r m o n i c s b y 0 . 2 5 )

1 0 =

S i n g l e - p h a s e e l e c t r o n i c G Y - G Y

1 1 =

S i n g l e - p h a s e e l e c t r o n i c D e l t a - G Y

1 2 =

M a g n e t i c f l u o r e s c e n t G Y - G Y

1 3 =

M a g n e t i c f l u o r e s c e n t D e l t a - G Y

1 4 =

U s e r - s p e c i f i e d t y p e 1 4

. . .

. . .

. . .

3 3 =

U s e r - s p e c i f i e d t y p e 3 3

N o n l i n e a r D e v i c e P h a s e S h i f t

D e g r e e s f o r p o s i t i v e s e q u e n c e , o n s y s t e m s i d e w i t h

r e s p e c t t o d e v i c e s i d e . T h i s i s t h e a d d i t i o n a l p h a s e s h i f t

b y w h i c h t h e c u r r e n t i n j e c t i o n p h a s e a n g l e s w i l l b e

a d v a n c e d f o r p o s i t i v e s e q u e n c e , a n d d e l a y e d f o r

n e g a t i v e s e q u e n c e .

C o n n e c t i o n T y p e f o r H a r m o n i c s M o d e l

0 o r 1 = G r o u n d e d W y e . O t h e r w i s e , u

n g r o u n d e d w y e

o f L i n e a r L o a d

o r d e l t a ( i . e . n o z e r o s e q u e n c e p a t h )

Grady 2007, p. 290



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 7 o f 2 4

C S V H e a d e r a n d S t r u

c t u r e f o r I n p u t F i l e B D A T . C S

V ( u s i n g s a m p l e f i l e B D A T_ H

E A D E R . c s v )

: B u s

D a

t a

: :

L i n e a r

L i n e a r

L i n e a r

L i n e a r

S h u n

t

M a x

i m u m

:

P

Q

P

Q

D e s

i r e d

R e a c

t i v e

Q

: B u s

B u s

B u s

G e n e r a

t i o n

G e n e r a

t i o n

L o a

d

L o a

d

V o

l t a g e

Q L o a

d

G e n e r a

t i o n

: N u m

b e r

N a m e

T y p e

( % )

( % )

( % )

( % )

( p u

)

( % )

( % )

: ( I )

( A )

( I )

( F )

( F )

( F )

( F )

( F )

( F )

( F )

c o n t i n u i n g a c r o s s ,

C o n n e c

t i o n

P o s

i t i v e

P o s

i t i v e

N e g a

t i v e

N e g a

t i v e

Z e r o

Z e r o

M i n i m u m

T y p e

f o r

S e q u e n c e

S e q u e n c e

S e q u e n c e

S e q u e n c e

S e q u e n c e

S e q u e n c e

Q

B u s

R

e m o

t e -

S h u n

t

S u

b t r a n s i e n

t S u

b t r a n s

i e n

t S u

b t r a n s

i e n

t S u b t r a n s

i e n

t S u

b t r a n s

i e n

t S u

b t r a n s i e n

t

G e n e r a

t i o n

C o n

t r o l

C

o n

t r o l l e d R e a c

t i v e

R

X

R

X

R

X

( % )

A r e a

B

u s

N o .

Q L o a

d

( p u

)

( p u

)

( p u

)

( p u

)

( p u

)

( p u

)

( F )

( I )

( I )

( I )

( F )

( F )

( F )

( F )

( F )

( F )


G r o u n

d i n g

G r o u n

d i n g

I m p e

d a n c e

I m p e

d a n c e

N o n

l i n e a r

N o n l i n

e a r

N o n

l i n e a r

C o n n e c

t i o n

R f o r

X f o r

D e v

i c e

D e v i c e

D e v

i c e

N o n

l i n e a r

L i n e a r

T y p e

f o r

S u

b t r a n s i e

n t S u

b t r a n s

i e n

t P

P

D i s p

l a c e m e n

t

N o n

l i n e a r

D e v

i c e

L o a

d

S u

b t r a n s

i e n

t I m p e

d a n c e

s

I m p e

d a n c e s

G e n

L o a

d

P o w e r

F a c

t o r

D e v

i c e

P h a s e

S h i f t C o n n e c

t i o n

R a n

d X

( p u

)

( p u

)

( % )

( % )

( p u

)

T y p e

( D

e g r e e s

)

T y p e

( I )

( F )

( F )

( F )

( F )

( F )

( I )

( F

)

( I )

Grady 2007, p. 291



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 8 o f 2 4

L I N E A

N D T R A N S F O R M E R D A T A

( F i l e = L D A T . C S

V , o n e r e c o r d p e r b r a n c h . C S

V f o r m a t )

V a r i a b l e

C o m

m e n t s

F R O M B U S N u m b e r

I n t e g

e r ( o r b l a n k i f n e u t r a l )

T O B U S N u m b e r

I n t e g

e r ( o r b l a n k , i f n e u t r a l )

C i r c u i t N u m b e r

I n t e g

e r ( o r b l a n k )

R , X ( P o s i t i v e / N e g a t i v e S e q u e n c e )

S e r i e

s i m p e d a n c e , i n p e r u n i t

C h a r g i n g ( P o s i t i v e / N e g

a t i v e S e q u e n c e )

P e r c e n t , f o r e n t i r e l e n g t h o f l i n e

R a t i n g

P e r c e n t

M i n i m u m T a p , o r M i n i m u m P h a s e S h i f t A n g l e

P e r u

n i t t a p , o r d e g r e e s , F R O M B U S

s i d e

w r t . T O B U S s i d e

M a x i m u m T a p , o r M a x i m u m P h a s e S h i f t A n g l e

P e r u

n i t t a p , o r d e g r e e s , o n F R O M B

U S s i d e

w r t . T O B U S s i d e

T a p S t e p S i z e , o r P h a s e

S h i f t S t e p S i z e

P e r u

n i t , o r d e g r e e s

T a p

P e r u

n i t , o n F R O M B U S s i d e

P h a s e S h i f t

D e g r e e s , F R O M B U S s i d e w r t . T O B

U S s i d e

V o l t a g e - C o n t r o l l e d B u s N u m b e r

U s e d

f o r c o n t r o l l i n g v o l t a g e a t a r e m

o t e b u s . F o r t h e s e

c a s e s , t h e d e s i r e d v o l t a g e s p e c i f i e d a

p p l i e s t o t h e

r e m o

t e b u s .

Grady 2007, p. 292



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 9 o f 2 4

V o l t a g e - C o n t r o l l e d B u s S i d e

W h e

n c o n t r o l l i n g t h e v o l t a g e a t a r e m o t e b u s , e n t e r 1

w h e n t h e r e m o t e b u s i s o n t h e F R O M

B U S s i d e o f t h e

t r a n s

f o r m e r . E n t e r 2 w h e n t h e r e m o t e b u s i s o n t h e

T O B

U S s i d e o f t h e t r a n s f o r m e r .

D e s i r e d V o l t a g e f o r V o

l t a g e - C o n t r o l l e d B u s ,

P e r u

n i t V o l t a g e , o r P e r c e n t A c t i v e P

o w e r ( F R O M B U S

o r D e s i r e d A c t i v e P o w e r F l o w f o r P h a s e

t o w a

r d T O B U S )

S h i f t i n g T r a n s f o r m e r

R , X ( Z e r o S e q u e n c e )

S e r i e

s i m p e d a n c e , i n p e r u n i t ( i g n o r i

n g c o n n e c t i o n t y p e

a n d g r o u n d i n g i m p e d a n c e s ) , ( d o n o t

m u l t i p l y b y 3 )

C h a r g i n g ( Z e r o S e q u e n

c e )

P e r c e n t , f o r e n t i r e l e n g t h o f l i n e

C o n n e c t i o n T y p e f o r T r a n s f o r m e r s a n d

F o r t r a n s f o r m e r s :

S h u n t E l e m e n t s

T y p e

F R O M B U S

T O B U S

0 o r 1

G Y

G Y

2

G Y

Y

3

Y

G Y

4

Y

Y

5

6

G Y

7

Y

8

G Y

9

Y

F o r s h u n t e l e m e n t s :

0 o r 1 = G r o u n d e d W y e .

O t h e

r w i s e , u n g r o u n d e d w y e o r d e l t a

( i . e . n o z e r o

s e q u e n c e p a t h )

G r o u n d i n g I m p e d a n c e R , X

S e r i e s i m p e d a n c e f r o m w y e p o i n t t o

g r o u n d , i n p e r u n i t

A p p l i e s t o w y e - c o n n e c t e d t r a n s f o r m e r s a n d s h u n t

Grady 2007, p. 293



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 1 0 o f 2 4

e l e m

e n t s . ( d o n o t m u l t i p l y b y 3 )

R e s i s t i v e S k i n E f f e c t F a c t o r f o r

H a r m

o n i c h ( h > 2 , f r a c t i o n a l v a l u e s

O K ) a t w h i c h

P o s i t i v e / N e g a t i v e S e q u e n c e

t h e c

o n d u c t o r r e s i s t a n c e i s d o u b l e t h

e f u n d a m e n t a l

f r e q u

e n c y r e s i s t a n c e .

R e s i s t i v e S k i n E f f e c t F a c t o r f o r

H a r m

o n i c h ( h > 2 , f r a c t i o n a l v a l u e s

O K ) a t w h i c h

Z e r o S e q u e n c e

t h e c

o n d u c t o r r e s i s t a n c e i s d o u b l e t h e f u n d a m e n t a l

f r e q u

e n c y r e s i s t a n c e . T h i s v a l u e a p p

l i e s t o t h e

c o m b i n e d c o n d u c t o r a n d g r o u n d i n g r e s i s t a n c e .

Grady 2007, p. 294



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 1 1 o f 2 4

H e a d e r a n d S t r u c t u r e

f o r I n p u t F i l e L D A T . C S V ( u s i n g s a m p l e f i l e L D A T_ H

E A D

E R . c s v )

: L i n e a n

d T r a n s

f o r m e r D a

t a

: : : :

P o s

i t i v e

P o s

i t i v e

P o s

/ N e g

:

S e q u e n c e

S e q u e n c e

S e q u e n c e

M i n i m u m

M

a x

i m u m

T a p

F i x e

d

: F r o m

T o

C

i r c u

i t

R

X

C

h a r g

i n g

R a

t i n g

T a p

T a p

S t e p

S i z e

T a p

: B u s

B u s

N

u m

b e r

( p u

)

( p u

)

( %

)

( % )

( p u

)

( p

u )

( p u

)

( p u

)

: ( I )

( I )

( I )

( F )

( F )

( F

)

( F )

( F )

( F

)

( F )

( F )


D e s

i r e d V o

l t a g e

a t V o

l t a g e

C o n

t . B u s

( p u

)

C o n n e c

t .

V o

l t a g e -

V o

l t a g e -

o r

D e s

i r e d P

Z e r o

Z e r o

Z e r o

T y p e

S e r i e s

S e r i e s

P h a s e

C o n

t .

C o n

t .

f o r

S e q u e n c e

S e q u e n c e

S e q u e n c e

f o r

T r a n s .

G r o u n

d i n g

G r o u n

d i n g

S h i f t

B u s

B u s

P h a s e

S h i f t e r

R

X

C h a r g

i n g

a n

d S h

u n

t R

X

( D e g r e e s

) N u m

b e r

S i d e

( % )

( p u

)

( p u

)

( % )

E l e m e n

t s

( p u

)

( p u

)

( F )

( I )

( I )

( F )

( F )

( F )

( F )

( I )

( F )

( F )

Grady 2007, p. 295



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 1 2 o f 2 4

A R E

A I N T E R C H A N G E D A T A

( F i l e = A D A T . C S V , o n e r e c o r d p e r l o a d f l o w a r e a .

C S V f o r m a t )

V a r i a b l e

C o m m

e n t s

N u m b e r

I n t e g e r

T i e - L i n e L o s s A s s i g n m

e n t

I f n o n - z e r o , t h e n p o w e r l o s s e s o n t i e l i n e s a r e a s s i g n e d

e q u a l l y b e t w e e n t h e t w o a r e a s . I f z e r o

, t h e T O B U S

a r e a f o

r e a c h t i e l i n e i s a s s i g n e d t h e l o s s ( i . e . , m e t e r a t

t h e F R

O M B U S ) .

C o n t r o l B u s N u m b e r

I n t e g e r

D e s i r e d E x p o r t P o w e r

P e r c e n

t

S o l u t i o n T o l e r a n c e f o r E x p o r t

P e r c e n

t

N a m e

U p t o 2 0 c h a r a c t e r s

Grady 2007, p. 296



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 1 3 o f 2 4


f o r I n p u t F i l e A D A T . C S V ( u s i n g s a m p l e f i l e A D A T_ H

E A D

E R . c s v )

: A r e a

I n t e r c

h a n g e

D a

t a

:

T i e L i n e

:

L o s s

:

A s s

i g n m .

:

( 1 f o r

:

S p

l i t t i n g ,

D e s

i r e d

S o

l u t i o n

:

0 f o r

T o

C

o n

t r o l

E x p o r t

T o

l e r a n c e

: A r e a

B u s

B u s

P o w e r

f o r

E x p o r t

A r e a

: N u m

b e r

O w n e r s

h i p N

u m

b e r

( % )

( % )

N

a m e

: ( I )

( I )

( I )

( F )

( F )

( A

)

Grady 2007, p. 297



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 1 4 o f 2 4

U S E R - S P E C I F I E D H

A R M O N I C C U R R E N T S P E C

T R A L D A T A

( F i l e = S P E C T R A . C S V , o n e r e c o

r d p e r h a r m o n i c p e r n o n l i n e a

r l o a d t y p e . C S V f o r m a t )

V a r i a b l e

C o

m m e n t s

T y p e o f S e r i e s

M u s t b e s i n f o r a s i n e s e r i e s , c o s f o r a c o s i n e s e r i e s .

A l l e n t r i e s i n t h i s f i l e m u s t b e e i t h e r s i n , o r c o s , a n d

c a n n o t b e m i x e d .

N o n l i n e a r L o a d T y p e

M u s t b e 1 4 , 1 5 , 1 6 , . . . , 3 3 .

H a r m o n i c O r d e r

1 , 2 , 3 , e t c .

C u r r e n t H a r m o n i c M a g n i t u d e

P e r u n i t . I f t h e f u n d a m e n t a l i s g i v e n , i t s m a g n i t u d e

m u

s t b e 1 . 0 , a n d t h e o t h e r h a r m o n i c m a g n i t u d e s f o r

t h e

s a m e n o n l i n e a r l o a d t y p e a r e a s s u m e d t o b e r e l a t i v e

t o 1 . 0 . T h e a c t u a l i n j e c t i o n c u r r e n t s w i l l b e s c a l e d

a c c o r d i n g t o b u s l o a d / g e n e r a t i o n .

I m

p o r t a n t : I f t h e f u n d a m e n t a l i s n o t g i v e n f o r a n o n l i n e a r l o a d

t y p

e , t h e n t h e h a r m o n i c m a g n i t u d e

s a r e a s s u m e d t o b e

g i v

e n o n t h e s y s t e m b a s e , r a t h e r t h

a n a s a f r a c t i o n o f t h e P f o r t h a t

b u s .

C u r r e n t H a r m o n i c P h a s

e A n g l e

D e

g r e e s , u s i n g l o a d c u r r e n t c o n v e n t i o n . I f t h e

f u n

d a m e n t a l a n g l e i s g i v e n , i t m u s t b e 0 . 0 . T h e a c t u a l

p h a s e a n g l e s w i l l b e a d j u s t e d i n t e r n a l l y a c c o r d i n g t o

b u s p o w e r f a c t o r a n d f u n d a m e n t a l

v o l t a g e a n g l e .

I f t h e f u n d a m e n t a l i s n o t g i v e n , t h e

p h a s e a n g l e s a r e

a s s

u m e d t o b e g i v e n w i t h r e s p e c t t o t h e b u s

f u n

d a m e n t a l f r e q u e n c y v o l t a g e p h a s e a n g l e .

Grady 2007, p. 298



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 1 5 o f 2 4


f o r I n p u t F i l e S P E C T R A . C S V ( u s i n g s a m p l e f i l e S P E C T R A

_ H E A D E R . c s v )

: H a r m o n

i c C u r r e n

t S p e c

t r a l D a

t a

:

C u r r e n

t

: T y p e o

f

C u r r e n

t

H a r m o n

i c

: S e r i e s

N o n

l i n e a r

H

a r m o n

i c

H a r m o n

i c

P h a s e

: ( S I N o r

L o a

d

O

r d e r

M a g .

A n g

l e

: C O S )

T y p e

( I n

t e g e r )

( p u

)

( D e g r e e s

)

: ( A )

( I )

( I )

( F )

( F )

Grady 2007, p. 299



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 1 6 o f 2 4

S O L U T I O N O P T I O N S

( F i l e = O P T I O N S . C S V . C S V f o r m a t . )

F o r L o a d f l o w ( u s i n g s

a m p l e f i l e O P T I O N S_ H

E A D E

R_ L

O A D F L O W . c s v )

L o a

d f l o w

S t u d y

C a s e - U

s e r

T i t l e G o e s o n

T h i s L i n e

: L o a

d f l o w

S o

l u t i o n

O p

t i o n s

:

V o

l t a g e

D i s a

b l e

: O p

t i m a

l

P

& Q

A c c e

l .

U p

d a

t e

D i s a

b l e

T r a n s

f .

: B u s

M

i s m a

t c h

F a c

t o r

C a p

P & Q

R e m o

t e

D i s a

b l e

I g n

o r e

T a p

: O r d e r i n g

G a u s s -

f o

r G a u s s -

f o r

G a u s s -

f o r

G a u s s -

M i s m a

t c h

V o

l t . R e g .

A r e a

Q L i m i t s

A d j u s

t . f o r

: M e

t h o

d

S e

i d e

l

S e

i d e

l

S e

i d e

l

S e

i d e

l

S o

l u t i o n

b y

P V

I n t r c h n g e

o n

P V

V o

l t a g e

: ( I n t e g e r )

S t a r t ?

S t a r t

S t a r t

S t a r t

T o

l e r a n c e

B u s s e s

?

P A d j u s

t ?

B u

s s e s

?

C o n

t r o l ?

: ( 1 - 2 - 3

)

( T o r

F )

( 0

. 5 p u

)

( 1 . 2

p u

)

( 0 . 0

0 5 p u

) ( 5 E - 0

6 p u

) ( T o r

F )

( T o r

F )

( T

o r

F )

( T o r

F )

: ( I )

( L )

( F

)

( F )

( F )

( F )

( L )

( L )

( L )

( L )

2 T

0 . 5

1 . 2

0 . 0

0 5 5

. 0 0 E - 0

6 F

F

F

F


L i m i t

O u

t p u

t

a n

d

a n

d

a n

d

a n

d

t o T h i s

T h i s

T h i s

T h i s

T h i s

C o n

t r o l

C o n

t r o l

C

o n

t r o l

C o n

t r o l

C o n

t r o l

A r e a

?

A r e a

?

A r e a

?

A r e a

?

A r e a

?

( I n t e g e r )

( I n t e g e r )

( I n

t e g e r )

( I n t e g e r )

( I n t e g e r )

( I )

( I )

( I )

( I )

( I )

0

0

0

0

0

Grady 2007, p. 300



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 1 7 o f 2 4

F o r S h o r t C i r c u i t ( u s i n g s a m p l e f i l e O P T I O N S_

H E A D E R_ S

H O R T_ C

I R C U I T . c s v )

S h o r t C i r c u i t S t u d y C a s e

- U s e r T i t l e G o e s o n T h i s L i n e

: S h o r t C i r c u i t S o l u t i o n O p t i o n s

: O p t i m a l

: B u s

: O r d e r i n g

: M e t h o d

E n t e r T f o r D i a g o n a l a n d N e i g h b o r Z B U S E l e m e

n t s O n l y ( r e c o m m e n d e d )

: ( I n t e g e r )

E n t e r F f o r A l l Z B U S E l e m e n t s ( n o t r e c o m m e n d

e d a n d n o t t o b e f o l l o w e d b y F A U L

T S )

: ( 1 - 2 - 3 )

( T o r F )

: ( I )

( L )

2 T

F o r F u l l H a r m o n i c S o

l u t i o n ( u s i n g s a m p l e f i l e O P T I O N S_

H E A D E R_ F

U L L_ H

A R

M O N I C_ S

O L U T I O N . c s v )

F u

l l H a r m o n

i c S o

l u t i o n S

t u d y

C a s e -

U s e r


T h i s L i n e

: F u

l l H a r m o n

i c S o

l u t i o n O

p t i o n s

: :

H a r m o n

i c L o a

d M o

d e

l

: O p

t i m a

l

P

& Q

A c c e

l .

U p

d a

t e

H i g h e s

t

f o r

P Q L i n e a r

L o a

d s

G l o b a l

: B u s

M

i s m a

t c h

F a c

t o r

C a p

P & Q

H a r m o n

i c

0 o r

1 :

R e s i s t

i v e - o n

l y ( r e c o m m e n

d e

d )

L i n e a r

G l o b a

l

: O r d e r i n g

G a u s s -

f o

r G a u s s -

f o r

G a u s s -

f o r

G a u s s -

M i s m a

t c h

o f

2 :

P a r a

l l e l R & L M o

d e

l

M o

t o r

R e s

i s t a n c e

: M e

t h o

d

S e

i d e

l

S e

i d e

l

S e

i d e

l

S e

i d e

l

S o

l u t i o n

I n t e r e s

t

3 :

S e r i e s

R &

L M o

d e

l

L o a

d

D o u

b l i n g

: ( I n t e g e r )

S t a r t ?

S t a r t

S t a r t

S t a r t

T o

l e r a n c e

( I n t e g e r )

4 :

I g n o r e

P Q

L o a

d s

( i . e .

N o

M o

d e

l )

M o

d e l i n g

H a r m o n

i c

: ( 1 - 2 - 3

)

( T o r

F )

( 0

. 5 p u

)

( 1 . 2

p u

)

( 0 . 0

0 5 p u

) ( 5 E - 0

6 p u

) ( 1 -

4 9 )

( 0 -

4 )

F r a c

t i o n

( I )

: ( I )

( L )

( F

)

( F )

( F )

( F )

( I )

( I )

( F )

( I )

2 T

0 . 5

1 . 2

0 . 0

0 5 5

. 0 0 E - 0

6

2 5

1

0

1 5

Grady 2007, p. 301



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 1 8 o f 2 4

F o r H a r m o n i c I m p e d a n c e S c a n ( u s i n g s a m p l e f i l e O

P T I O N S_

H E A D E R_ H

A R M O

N I C_ I

M P E D A N C E_

S C A N . c s v )

H a r m o n

i c I m p e

d a n c e

S c

a n

S t u d y

C a s e -

U s e r


T h i s L i n e

: H a r m o n

i c I m p e

d a n c e

S c a n

: : O p

t i m a

l

P

& Q

A c c e

l .

U p

d a

t e

L o w e s

t

H i g h e s

t

N u

m b e r

L i m i t t h e

: B u s

M

i s m a

t c h

F a c

t o r

C a p

P & Q

H a r m o n

i c

H a r m o n

i c

o f

S t e p s

O u

t p u

t t o

: O r d e r i n g

G a u s s -

f o

r G a u s s -

f o r

G a u s s -

f o r

G a u s s -

M i s m a

t c h

o f

o f

p e r

D i a g o n a

l

: M e

t h o

d

S e

i d e

l

S e

i d e

l

S e

i d e

l

S e

i d e

l

S o

l u t i o n

I n t e r e s

t

I n t e r e s

t

H a

r m o n

i c

E l e m e n

t s

: ( I n t e g e r )

S t a r t ?

S t a r t

S t a r t

S t a r t

T o

l e r a n c e

( I n t e g e r )

( I n t e g e r )

( I n

t e g e r )

O n

l y ?

: ( 1 - 2 - 3

)

( T o r

F )

( 0

. 5 p u

)

( 1 . 2

p u

)

( 0 . 0

0 5 p u

) ( 5 E - 0

6 p u

) ( 1 -

4 9 )

( 1 -

4 9 )

( 1

- 1 0 0 )

( T o r

F )

: ( I )

( L )

( F

)

( F )

( F )

( F )

( I )

( I )

( I )

( L )


H a r m o n i c L o a d M o d e l

f o r P Q L

i n e a r L o a d s

G l o b a l

0 o r 1 : R e s i s t i v e - o n l y ( r e c o m m e n d e d )

L i n e a r

G l o b a l

S c a n

a n d

a n d

a n d

a n d

2 :

P a r a l l e l R

& L M o d e l

M o t o r

R e s i s t a n c e

T h i s

T h i s

T h i s

T h i s

T h i s

3 :

S e r i e s R

& L M o d e l

L o a d

D o u b l i n g

B u s

B u s

B u s

B u s

B u s

4 :

I g n o r e P Q L o a d s ( i . e . N o M o d e l )

M o d e l i n g

H a r m o n i c

( I n t e g e r )

( I n t e g e r )

( I n t e g e r )

( I n t e g e r )

( I n t e g e r )

( 0

- 4 )

F r a c t i o n

( I )

( I )

( I )

( I )

( I )

( I )

( I )

( F )

( I )

Grady 2007, p. 302



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 1 9 o f 2 4

4 . H a r m o n i c - R e l a t

e d a n d S h o r t - C i r c u i t R e l a t e d O u t p u t F i l e s

( N o t e - f o r l o a d f l o w s t u d i e s , t h e f o r m a t s o f I S O L N a n

d V S O L N a r e d i f f e r e n t f r o m b e l o w b u t a r e s e l f - e x p l a n a t o r y w h e n v i e w i n g t h e

f i l e s . F o r s h o r t c i r c u i t s t u d i e s , t h e Z B U S f i l e s a r e s i m i l a r t o b e l o w , b u t w r i t t e n i n r e c t a n g u l a r f o r m )

C o m m a s s e p a r a t e t h e f i

e l d s s h o w n b e l o w t o f a c i l i t a t e t h e i r u s e w i t h M i c r o s o f t E x c e l .

I S O L N . C S V ( f o r h a r m

o n i c s )

D a t a F i e l d

( s t a r t i n g f r o m

t h e l e f t )

D e s c r i p t i o n

1

H a r m o n i c n u m b e r

2

F r o m b u s n u m b e r

3

T o b u s n u m b e r

4

C i r c u i t n u m b e r

5

C u r r e n t m a g n i t u d e - p

e r u n i t

6

C u r r e n t p h a s e a n g l e ( s i n e r e f e r e n c e ) - d e g r e e s

7

F r o m b u s n a m e ( a t t h e f i r s t o p p o r t u n i t y o n l y )

8

T o b u s n a m e ( a t t h e f i r s t o p p o r t u n i t y o n l y )

9

L o a d i n g l e v e l - p e r c e n t o f l i n e r a t i n g ( f o r

f u n d a m e n t a l f r e q u e n c

y o n l y )

V S O L N . C S V ( f o r h a r m o n i c s )

D a t a F i e l d


t h e l e f t )


1


2

B u s n u m b e r

3

V o l t a g e m a g n i t u d e - p e r u n i t

Grady 2007, p. 303



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 2 0 o f 2 4

4

V o l t a g e p h a s e a n g l e ( s i n e r e f e r e n c e ) - d e g r e e s

5

N o n l i n e a r d e v i c e l o a d

c u r r e n t m a g n i t u d e -

p e r u n i t

6

N o n l i n e a r d e v i c e l o a d

c u r r e n t p h a s e a n g l e

( s i n e r e f e r e n c e ) - d e g r e e s

7

B u s n a m e ( a t t h e f i r s t

o p p o r t u n i t y o n l y )

Z B U S 0 . C S V , Z B U S 1 . C S V , Z B U S 2 . C S V ( f o r h a r m o n i c s a n d s h o r t c i r c u i t )

D a t a F i e l d


t h e l e f t )


1


2

F r o m b u s n u m b e r

3

T o b u s n u m b e r

4

I m p e d a n c e m a g n i t u d e - p e r u n i t

5

I m p e d a n c e p h a s e a n g l e - d e g r e e s

6

F r o m b u s n a m e ( a t t h e f i r s t o p p o r t u n i t y o n l y )

7

T o b u s n a m e ( a t t h e f i r s t o p p o r t u n i t y o n l y )

T H D V . C S V ( f o r h a r m

o n i c s )

C o n t a i n s a l i s t o

f b u s n u m b e r s w i t h t h e i r c o r r e s p o n d i n g n a m e s a n d v o l t a g e d i s t o r t i o n s .

Grady 2007, p. 304



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 2 1 o f 2 4

E x a m p

l e S c r e e n s f o r t h e_ D

F W C a s e

Grady 2007, p. 305



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 2 2 o f 2 4

Grady 2007, p. 306



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 2 3 o f 2 4

Grady 2007, p. 307



P C F L O

_ V 6 ( 5 0 0 0 B u s ) U s e r M a n u a l ,


P a g e 2 4 o f 2 4

E x a m p l e I m p e d a n c e S c a n f o r B u s 3 o f t h e_ P

C c a s e

Grady 2007, p. 308



Five Bus Stevenson Loadflow Example

Grady 2007, p. 309



Five Bus Stevenson Loadflow Example

Grady 2007, p. 310



: F i v e

B u s

S t e v e n s o n

L o a

d f l o w

C a s e

: B u s

D a

t a

: :

L i n e

a r

L i n e a r

L i n e a r

L i n e a r

S h u n

t

M a x

i m u m

M i n i m u m

:

P

Q

P

Q

D e s

i r e d

R e a c

t i v e

Q

Q

B u s

R e m o

t e -

: B u s

B u s

B u s

G e n

e r a

t i o n

G e n e r a

t i o n

L o a

d

L o a

d

V o

l t a g e

Q L o a

d

G e n e r a

t i o n

G e n e r a

t i o n

C o n

t r o l

C o n

t r o l l e d

: N u m

b e r

N a m e

T y p e

( % )

( % )

( % )

( % )

( p u

)

( % )

( % )

( % )

A r e a

B u s

N o .

: ( I )

( A )

( I )

( F )

( F )

( F )

( F )

( F )

( F )

( F )

( F )

( I )

( I )

1 b i r c h

1

0

0

6 5

3 0

1 . 0

4

0

0

0

1

0

2 e

l m

3

0

0

1 1 5

6 0

0

0

0

0

1

0

3 m a p

l e

2

1 8 0

0

7 0

4 0

1 . 0

2

0

0

0

1

0

4 o a

k

3

0

0

7 0

3 0

0

0

0

0

1

0

5 p

i n e

3

0

0

8 5

4 0

0

0

0

0

1

0

: F i v e

B u s

S t e v e n s o n

L o a

d f l o w

C a s e

: L i n e a n

d T r a n s

f o r m e r

D a

t a

D e s

i r e d V o

l t a g e

:

a t V o

l t a g e

:

C o n

t . B u s

:

( p u

)

:

P o s

i t i v e

P o s

i t i v e

P o s

/ N e g

V o

l t a g e -

V o

l t a g e -

o r

D e s

i r e d P

:

S e q

u e n c e

S e q u e n c e

S e q u e n c e

M i n i m u m

M a x

i m u m

T a p

F i x e d

P h a s e

C o n

t .

C o n

t .

f o r

: F r o m

T o

C i r c u

i t

R

X

C h a r g

i n g

R a

t i n g

T a p

T a p

S t e p

S i z e

T a p

S h i f t

B u s

B u s

P h a s e

S h i f t e r

: B u s

B u s

N u m

b e r

( p u )

( p u

)

( % )

( % )

( p u

)

( p u

)

( p u

)

( p u )

( D e g r e e s

) N u m

b e r

S i d e

( % )

: ( I )

( I )

( I )

( F )

( F )

( F )

( F )

( F )

( F )

( F )

( F )

( F )

( I )

( I )

( F )

1

2

1

0 . 0

4 2

0 . 1

6 8

4 . 1

0

0

0

0

0

0

0

0

0

1

5

1

0 . 0

3 1

0 . 1

2 6

3 . 1

0

0

0

0

0

0

0

0

0

2

3

1

0 . 0

3 1

0 . 1

2 6

3 . 1

0

0

0

0

0

0

0

0

0

3

4

1

0 . 0

8 4

0 . 3

3 6

8 . 2

0

0

0

0

0

0

0

0

0

3

5

1

0 . 0

5 3

0 . 2

1

5 . 1

0

0

0

0

0

0

0

0

0

4

5

1

0 . 0

6 3

0 . 2

5 2

6 . 1

0

0

0

0

0

0

0

0

0

Grady 2007, p. 311



Screwbean Wind Farm Study

Page 1 of 2

The Screwbean 138kV substation is located in west Texas, halfway between Midland/Odessa

and El Paso, near Guadalupe Mountains National Park. It is about 400 miles from Austin. Thisis prime wind country, and several wind farms are already located in the area.

Your job is to examine the feasibility of transporting 50MW of power from a new wind farm

near Screwbean to the U.T. Austin campus. In particular, you are to determine the impact of thistransaction on the losses in individual control areas, and also determine if any high or low

voltages, or line overloads, are created by your transaction.

To perform the analysis, you will use a 5000 bus version of PCFLO, together with a summer

peak loadflow case (in which most bus names have been disguised). You should prepare a ½ to

1 page summary report of your study, as if you were going to submit it to your client. Tablesshould be attached as an appendix.

Explain to your client how many MW must be generated at Screwbean to deliver 50MW to U.T.Austin. Quantify the MW needed by each negatively-impacted control area to pay back for their

increased losses.

Here are the steps:

1. Go to www.ece.utexas.edu/~grady, and click the PCFLO_V6 link. Follow the download and

unzip instructions on the PCFLO page.

2. The _SCREWBEAN case is your “base case.” Solve it using PCFLO_V6_Interface.exe.

Using Excel, examine the output files produced, notably

exlog_SCREWBEAN.csv

asoln_SCREWBEAN.csv

vsoln_SCREWBEAN.csv isoln_SCREWBEAN.csv and

out5_SCREWBEAN.csv.

3. Print out asoln_SCREWBEAN.csv, using the landscape option. To verify your loadflow

result, check your power loss in asoln_SCREWBEAN.csv. It should be about 1215 MW.

4. Find the Screwbean 138kV substation (SCRWBEAN 138, bus 1095) and the U.T. Austin

Harris 69kV substation (HARRIS 69, bus 9204) in the out5_SCREWBEAN.csv file. Notetheir voltage magnitudes and phase angles, and the P and Q flows in lines/transformers

attached to these busses.

5. Copy files bdat_SCREWBEAN.csv to bdat_mod, ldat_SCREWBEAN.csv to ldat_mod, and

adat_SCREWBEAN.csv to adat_mod.

6. Add new PV bus SB WIND as bus 2 to bdat_mod.csv, using 20 for its control area. Put50MW (i.e., 50% on 100MVA base) of generation on this new bus, with a Max Q Gen of

25MVAr, and a Min Q Gen of negative 12.5MVAr. For the desired voltage, put a value that

is 0.005pu higher than the base case voltage at SCRWBEAN 138.

Grady 2007, p. 312



Screwbean Wind Farm Study

Page 2 of 2

7. Add new PQ bus UT CAMPUS as bus 3 to bdat_mod.csv, using 21 for its control area. Put50MW, 25MVAr of load on this new bus.

8. Connect new bus SB WIND to SCRWBEAN 138 through a line with impedance R =

0.001pu, X = 0.01pu, B = 0%.

9. Connect new bus UT CAMPUS to HARRIS 69 through a line with impedance R = 0.001pu,

X = 0.01pu, B = 0%.

10. Add control area SB as area 20 to adat_mod.csv, with a desired export of 50MW (i.e., 50%).

The area control bus number will be that of SB WIND. Use an export solution tolerance of0.1%.

11. Add control area UT as area 21 to adat_mod.csv, with a desired import of 50MW (i.e.,negative 50MW export). The area control bus number will be that of UT CAMPUS. Use an

export solution tolerance of 0.1%.

12. Re-run PCFLO_V6_Interface.exe, using _mod as the input case. Print out the new asoln.csv

file (using the landscape option), and using the new asoln file, tabulate area by area the

increase/decrease in each control area’s losses compared to the base case. The areas with

increased losses may reasonably expect MW payment from the wind power company. Thiscan be accomplished by putting in generator larger than 50MW, and exporting some power

to the control areas that are negatively impacted.

13. Use the loss increases from Step 12 to estimate how much actual generation would be needed

at SB WIND to deliver 50MW to UT CAMPUS and payback the extra losses to thenegatively-impacted control areas.

14. Check for any line overloads and high/low voltages created in the vicinity of SCRWBEAN138 and HARRIS 69. (In an actual study, there would have to be described and remedies

proposed. However, do not investigate remedies in your study.)

15. Repeat the above process, but this time reverse the transaction by putting a wind generator at

UT CAMPUS, and a 50MW load at SB WIND.

16. Describe the impacts of both transactions in your report.

Grady 2007, p. 313



Grady 2007, p. 314



Grady 2007, p. 315



Grady 2007, p. 316



Grady 2007, p. 317



Grady 2007, p. 319



Grady 2007, p. 320



Grady 2007, p. 321



Grady 2007, p. 322



Grady 2007, p. 323



Grady 2007, p. 324



Grady 2007, p. 325



Grady 2007, p. 326



Steady State Model for Round Rotor Machine

Va and Ia are the terminal voltage and current. Xs is the synchronous reactance. Resistance r is

the stator resistance. Ea is the Thevenin equivalent voltage behind synchronous reactance.

Steady-State Model for Salient Pole Machine

(No equivalent circuit. Phasor diagram only)

Steps when Va and Ia are known:

With Va, Ia, r, and Xq, find phasor a’ to locate the q-axis.

Then, compute Ia projections Iaq and Iad where

Iaq has magnitude aa I I cos and has phase angle ,

Iad has magnitude aa I I sin and has phase angle 90 ,

and where a I is the angle of Ia.

Then, find Ea.

+

Ea

–

+

Va

–

jXs r

Ia

Ea

jXsIa

rIa

Va

Ia

q-axis

d-axis

reference

Ea

jXqIa

rIa

Va

Ia

q-axis

d-axis

reference

Iaq

Iad jXdIad

jXqIaq

a’

Grady 2007, p. 327



Transient Stability Machine Model 1

Constant Voltage Magnitude E’ Behind Transient Reactance Xd’

This model is like the round rotor synchronous model, except that the transient reactance is used

instead of the synchronous reactance.

Transient Stability Machine Model 2

Salient Pole Rotor

Steps when Va and Ia are known: Begin with same steps used in the salient pole synchronous

machine model. Then, use Xd’ and Iad to find Ea’.

The magnitude of Ea’ varies according to a fd

do

a E E

T dt

E d

'

'1

, where Tdo’ is the direct-

axis transient open circuit time constant, and Efd is the field voltage (as seen from the stator). As

a first approximation, Efd is treated as a constant whose initial value is the same as the initial

value of a E .

+

E’

–

+

Va

–

jXd’ r

Ia

E’

j Xd’Ia

rIa

Va

Ia

q-axis

d-axis

reference

Ea

jXqIa

rIa

Va

Ia

q-axis

d-axis

reference

Iaq

Iad jXdIad

jXqIaq

a’

jXd’ Iad

Ea’

Grady 2007, p. 328



A synchronous generator is operating in steady-state with the following conditions (in per unit):

Va = 1.0/0°, Ia = 0.8/-30°, Xd = 1.0, Xq = 0.6, r = 0. Find S (at the machine terminals), and

phasors Iad, Iaq, Ea.

Grady 2007, p. 329



Grady 2007, p. 330



Bergen Problem 6.2. A generator with reactances Xd = 1.6 and Xq = 0.9 delivers SG = 1.0 ⁄45º

to a bus with voltage Va = 1.0 ⁄0º. Find Ia, Iad, Iaq, Ea, and the rotor angle o (i.e., the rotor

angle at t = 0).

Bergen Problem 6.4. Given a generator with Va = 1.0 ⁄0º, Ia = 1.0 ⁄60º, Xd = 1.0, Xq = 0.6, and r

= 0.1, find SG, Iad, Iaq, and Ea.

Grady 2007, p. 331



Grady 2007, p. 332



Grady 2007, p. 333



A new 32MVA, 60Hz steam generator was recently installed at the UT power plant. The

generator is connected through a large reactor to ERCOT at the Harris 12kV bus.

The electrical diagram and internal voltage (for the transient condition) that exists when the

generator is producing (25MW + j12.4MVAr) is shown below. Suddenly, at t = 0, a three-phase

fault with zero impedance occurs at the Harris 12kV bus. Assuming that E’, Vth, and Pmech are

constant, determine the critical clearing time (in cycles). Use the attached graph to show

important P and points and lines on the power curve, and show clearly which areas are equal.

Remember that during the fault, the electrical power output is zero, and thus the rotor accelerates

according to omech s t

H

P t

2

4)( , where )0( t o . Also, note that s =

377rad/sec, is in radians, and mech P is in pu. The formula for critical clearing angle is

)cos()sin()2(cos 1ooocrit .

E’/°

jXd’ jXr jXth

Vth/0°

Harris 12kV bus

Machine E’ = 1.100pu

Machine = 16.5°Machine P = 25MW (i.e., 0.781pu)

Machine jXd’ = j0.22pu

Machine H = 1.4seconds

Reactor jXr = j0.17pu

Thevenin jXth = j0.01pu

Thevenin Vth = (1.0 + j0)pu

Values given on the generator base

Grady 2007, p. 334



0

0.5

1

1.5

2

2.5

3

3.5

4

0 30 60 90 120 150 180

Grady 2007, p. 335



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Grady 2007, p. 382



Shock_Energy.xls

sigma Ipeak - kA Duration - microsec Rshoe Rbody (R2 - R1) - m Rfactor

0.01 40.00 22.00 1000 1000 1 0.000111111

Body Peak Body

R1 - m Voltage Joules

1 106103.30 82.56

1.1 91864.33 61.89

1.2 80381.28 47.381.3 70972.10 36.94

1.4 63156.72 29.25

1.5 56588.42 23.48

1.6 51011.20 19.08

1.7 46232.37 15.67

1.8 42104.48 13.00

1.9 38512.99 10.88

2 35367.77 9.17

2.1 32597.02 7.79

2.2 30142.98 6.66

2.3 27958.71 5.73

2.4 26005.71 4.962.5 24252.18 4.31

2.6 22671.64 3.77

2.7 21241.90 3.31

2.8 19944.23 2.92

2.9 18762.74 2.58

3 17683.88 2.29

Grady 2007, p. 383



Shock_Energy.xls

sigma Ipeak - kA Duration - microsec Rshoe Rbody (R2 - R1) - m Rfactor

0.001 40.00 22.00 1000 1000 1 0.000111111

Body Peak Body

R1 - m Voltage Joules

1 1061032.95 8255.80

2 353677.65 917.31

3 176838.83 229.334 106103.30 82.56

5 70735.53 36.69

6 50525.38 18.72

7 37894.03 10.53

8 29473.14 6.37

9 23578.51 4.08

10 19291.51 2.73

11 16076.26 1.90

12 13602.99 1.36

13 11659.70 1.00

14 10105.08 0.75

15 8841.94 0.5716 7801.71 0.45

17 6934.86 0.35

18 6204.87 0.28

19 5584.38 0.23

20 5052.54 0.19

21 4593.22 0.15

Grady 2007, p. 384



Grady 2007, p. 385



Grady 2007, p. 386



Grady 2007, p. 387

fundamentals grady notes june 2007 print

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