Lecture (5) Power Factor,three- phase circuits, and Per...

Energy Systems Research Laboratory, FIU (c) Copyright by Prof. Osama A. Mohammed. All rights reserved. 5-1

Lecture (5) Power Factor,three-

phase circuits, and Per Unit Calculations


Make this plant take power at 0.95 p.f. lagging

1000 volts @ 60 Htz

Motor Light

20 kVA 0.7 p.f. lagging

10kW unity p.f.

From source

P = ? Q = ? S = ?

Repeating the Example on Power Factor Correction (Given last Class)


MOTOR S = 20 ∠ cos -1 (0.7) kVA

= 20 ∠ 45.6 ° kVA P = 20 cos 45.6° kW = 14 kW

Q = 20 sin 45.7 ° = 14.28 kVAR 0 10 j S L + =

Example Solution

kVA

Light


TOTAL

P = 14 + 10 = 24 Kw Q = 14.28 KVAR

S = KVA 9 . 27 2 28 . 14 2 24 = +

° = = − 75 . 30 24 28 . 14 tan 1 θ

p.f. = cos 30.75° = 0.86 lagging

This is the total for this particular problem, before Power Factor Correction


Qnew

When Adding a Capacitor

Need a capacitor that produces 6.38 KVAR at 1000 volts

18.19 ° θ new = cos -1 .95 = 18.19 °

Qnew = P (tan 18.19°) = 24 x 103 x tan 18.19° = 7.9 kVAR

Qcap = Qold – Qnew = 14.28 - 7.9 = 6.38 kVAR


Find the value of capacitance c = ?

2

* * 1

Q V

Q * V V

V Q V

I V

j ω c -jxc = =

= = =

= 2 ω V

Q c = 2 V

Q ω c = 2 V

- jQ - j ω c

( ) = × =

60 2 2

1000

3 10 38 . 6 π

c 16.9 µF


THREE PHASE CIRCUITS

A → B → C

A B

C

V LN

N V ph = V L

Y-Connection Delta-Connection


THREE PHASE CIRCUITS

3 phase V =

V ph - ph = V L = V AB = phase to phase (or line voltage)

V phase = = phase to neutral

Y-Connection

S = 3 Vph Iph*

VL

VLN = VAN


Y-connection

phase V = 3

VAB=VBC=VCA=VL

IL=Iph

N VAB=VAN-VBN

Iph=IL

VL

A

B

C


∆-connection

3 ph I

ph V

=

=

ph ph I V S * 3 =

IA

IB

IC

IL

VL A

B

C


Y- Connection

* 3

* 3

3

* 3

IL VL

IL VL Iph Vph S

=

= =

Power in ∆- and Y Connection

If it is not specified in the problem, that the connection is“∆ or Y”, assume that The connection is a Y-connection

* 3

3 3

* 3

L L

h p h p

I V

IL VL

I V S

=

=

=

*

∆- Connection


Example

100 kVA 0.8 p.f. lagging 3 phase

Source

13.8 kV

0.4j

0.4j 0.4j

0.3 0.3 0.3


j0.4 0.3 Load

Load ~

~

Source

Source

One Line Diagram

100 kVA 0.8 p.f. lagging


Example Solution

A L I

A I

I

L

L

8 . 36 18 . 4

8 . 0 cos 3 10 8 . 13 3 10 100

3 10 8 . 13 3 10 100 *

1

° − ∠ =

− ∠ ×

× =

∠ ×

× =

−

θ 3

3

Energy Systems Research Laboratory, FIU (c) Copyright by Prof. Osama A. Mohammed. All rights reserved.

1991-1998 4-15

On the other hand, using Y-connection,

IL * = 100 x 103 ∠cos-1 0.8A = 4.18∠36.8° Α

3 13.8 x103 3

S = 3 Vp Ip* = 3 Vp IL

*

i.e. IL = 4.18 ∠−36.8° Α


Example Solution

ph I * ph V S 3 = * V S I

L L 3

= I S L L

* 3 =

° ∠ + ° ∠ = 33 . 16 09 . 2 0 43 . 7967 source V

( ) ( ) ° ∠ ° − ∠ + ° ∠ = 13 . 53 5 . 0 8 . 36 18 . 4 0 43 . 7967 source V

) 3

( ) ( + ° − ∠ + ° ∠ = 4 3 . 0 8 . 36 18 . 4 0 43 . 7967 source j V

( ) ( ) + ° − ∠ + ° ∠ × = 4 3 . 0 8 . 36 18 . 4 0 3

10 8 . 13 source j V

V


P.U. (per unit)

find Z eq = ? I = ? in p.u. on rated base value

100 kVA @ 0.8p.f. lagging

I

13.8

kV

Based values for “S” and “V”

PER UNIT SYSTEM


S base = 100 kVA V base = 13.8 kV

S p u S actual S base

. . . = = = 100 100 1 0

V p u V actual V base

. . . . . = = = 13 8

13 8 1 0

the load is operating at 1 p.u.

S V I = × ∗


I S V

S p u V p u

p u * . .

. . . . = = = = ∠−36.8 °

1 1 0

I actual

=

= ∠ ° ×

×

I p u I base

× . .

. . .

1 0 -36.8 100 10 3

13 8 10 3 x

= 7.246 ∠ -36.8 ° A


=

old S V

u p Z actual Z 2

. . base new

base old

Z p u new Z

actual Z . . =

Z p u new Z . . = = Z

actual V S

2 S base new

V base new 2

Z p u new Z p u old S

S base new

V base new . . . .

=

2 V2

base old


Calculate Z at a Different Base

=

new base

new base

base old

base old old u p new u p V

S

S

V Z Z

2

2

. . . .

× = new base

new base

base old

base old base old base new V

S

S

V Z Z

2

2

Z

Z Z pu

base

actual =


Z = 0.2 + j0.2 p.u. on 200 kVA at 34.5 kV (use as base)

Calculate Z at 100 kVA at 13.8 kV base. Z actual = Z P.U. old ( Z base old )

100 kVA @ 0.8 p.f. lagging. (use as base)

Source Feeder Load

~

Example


=

Z p u new j . . . . .

. = + ×

×

×

×

0 2 0 2 34 5 10 3

200 10 3

100 10 3

13 8 10 3

Z p u new . . + j . .

0 625 0 625

Z p u new . . = ∠ ° 0 .88 45 p.u.

2

2

Z p.u. new = (0.2+0.2j) 34.5 1 13.8

2

2


Example on Per Unit

Load p.f. 0.8 lagging

S = 100 MVA S b = 100 MVA V = 13.8 kV S b = 13.8 kV

Load

Ref T.L. R XL


θ ∠ = S S 0 ∠ = V V

θ ∠ = I I

lagging = “-” leading = “+”

θ ∠ = =

=

V I V

V I V Z

I V Z

2 2

Calculating Impedance Using S and V


S b = Base MVA

V b = Base Voltage

. . 1 3 10 8 . 13

3 10 8 . 13

. . 1 6 10 100

6 10 100

u p base V

actual V pu V

u p base S

actual S pu S

= × × = =

= × × = =


( )

. . 1

36. 87° 1.90

8 . 0 cos 10 100 10 8 . 13 1

6

2 3

2

u p Z Z

Z

Z

Z

V I V

Z

base

actual = =

∠ =

∠ × × =

∠ =

−

θ

Single Phase


P

Q

cos-1 0.8

1.

cos-1 0.8

2.

V

I

R

X

cos-1 0.8

3.

Power Factor Angle

Power Triangle

Impedance Triangle

Voltage and

Current


Base at generator S

b = 100 MVA V = 13.8 kV

XL =j0.3 p.u. on 240 kV base and 250 MVA base at the transmission Line

Ref

T.L.

13.8/240

Load

T.L. R XL

Another Example on Per Unit Calculations


Calculate Z(p.u.) on a new base

=

new base

new base

base old

base old old u p new u p V

S S V

Z Z 2

2

. . . .

× = new base

new base

base old

base old base old base new V

S S

V Z Z 2

2

Z

Z Z pu

base

actual =


29 pu . 36 90 . 1 12 . 69

100 8 . 13 12 . 69

2 . . = = = = base

actual u p Z

Z Z

( )( ) 4.230

1025010240

6

23

..@ =××

=LTbaseZ

Zactual = Zp.u. Zbase = 0.3 x 230.4 = 69.12 Ohms

Example Solution

Ohms

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