Power Protecction course notes

8/11/2019 Power Protecction course notes

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Examples

Problem 12.3

The problem requires us to find the three

currents Ia, Ib, and Ic, and also Vng.

The network is symmetric. Therefore, the

sequence circuits will be decoupled and we

can analyze them one at a time.

The zero sequence current must be zero

since the source is ungrounded and thereforeIa+Ib+Ic=.

!


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"ut the gi#en #oltages are unbalanced. $o

let%s hit them with an &'!to get the sequence

#oltages as follows(

=

=

+

)*.

),!-.

,)))).

!

!

!

!

!

!!!

)

!

*

*

jE

E

E

ecall that for loads and transmission lines,

/+=/'. $o we find the positi#e and negati#e

sequence currents using the below circuits.

0e obtain(=

=

+ !*+11+.*

),!-.

jI

=

= 2!**.

*

)*++.

jI

0e can then obtain the abc currents as

=

=

))).

.!3,!-.

.!31*-!.

2!**.

!*11.

!

!

!!!

*

*

c

b

a

I

I

I

4ow what about Vng5

*


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4otice from the positi#e and negati#e

sequence circuits that =+ngV and =ngV .

"ut consider ngV . The zero sequence circuit is

6rom the zero'sequence circuit, a 7V8equation results in

==

=+

,)))).,)))).

,)))).

ng

ng

V

V

4ow we ha#e all of the symmetrical

components of Vng, and we can therefore

obtain Vngas their sum(=++=++=

+,)))).,)))).

ngngngng VVVV

9larification( 0hen getting abc quantities

from +' quantities, we hit the +' quantities

with & :which contains terms in ; for the b'

and c'phase quantities


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quantities. &nd if we were to ha#e the c'

phase +' quantities, then the c'phase

quantity would ust be the sum of those c'phase +' quantities.

roblem 12.17

ecall that a 88> fault requires a parallel

combination of all three sequence networks.$o we must find all ) sequence networks.

?seful comments about the below networks.

0e get positi#e sequence phase shift of )@

going from low to highA negati#e sequence

of ')@ going from low to high.

The problem requires pre'fault, phase a

#oltage at the fault point be !.BC@.

- The positi#e sequence network is the

per'phase network for phase a. Therefore

this !.BC@ lies in the positi#e sequence

network.- Dre'fault implies no fault current, and

since we are ignoring load currents, there

is no current flow in the network.


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- In the teEt, FE.!*.1, because the fault

was on the line :in the middle of the

diagram! and >*,

the source #oltages had to be !BC')@.

- Gere, because the fault is at the

generator terminals, and therefore on the

same side of the phase shift as >!, >!

must be !BC@. "ecause there are equal andopposite phase shifts between the fault

point and >*, >* must also be !BC@.

The zero sequence network is

- to ground at each gen since both gens are

solidly grounded.

- open between each gen and the network

because of low side delta connection in the

Efmrs.

1


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The fault point, at the terminals of >! :on

the left


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sequence, and it will be the pre'fault #oltage

at the fault point, !BC@, for the positi#e

sequence network.

btaining the zero'sequence The#enin

network is also easy because of the open

circuit to the right of the fault pointA it is

shown below.

btaining positi#e, negati#e sequence The#.

impedances requires analyzing networks

with phase shifts. Gowe#er, because thephase shifts occur to #oltages and currents,

The#enin impedances may be computed

without consideration of phase shifts.

This is true for what your teEt calls normal

systems:see pp !1!'!1-


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$o the positi#e and negati#e sequence

networks to use in obtaining the The#enin

impedances are below. 4ote that the positi#e

sequence sources were idled.

These are identical networks. Therefore the

positi#e and negati#e sequence The#eninreatances are the same(

!*,.


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4ow we may draw the three The#enin

equi#alent networks.

&s mentioned at the beginning of thisproblem, a 88> fault requires a parallel

combination of all three sequence networks.

The connected circuits for analysis of a 88>

at the >! terminals is shown below.


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is easily done from the The#enin circuits

and knowledge of the fault currents, as seen

below.

&s a check on this work, let%s take a look at

the abc #oltages at the fault point.

!!


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=

=

=

23+2.

**3*.

**3*.

**3*.

!

!

!!!

*

*

c

b

a

V

V

V

Then we would use the sequence #oltages at

the fault point, in each sequence network, tocompute the currents in the #arious

elements. To do this, we would need to go

back to circuits with the phase shifts :see

page 2

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Power Protecction course notes