No Slide Title - Florida International...

Energy Systems Research Laboratory, FIU (c) Copyright by Prof. Osama A. Mohammed. All rights reserved. 4-1

Complex power, power factor and three-phase

circuits

Lecture (4)


•Definitions •Determination •Correction of Power Factor •Improvement of Load Operation •Voltage Drop •Examples

Overview of Today’s Lecture

SINGLE AND THREE PHASE CIRCUITS

• Power, Voltage, Current • Connections • Single Phase Equations • Three Phase Equations


Example

12 Sixty watt lights (60 x 12 watts) 16 Kw heating elements 5 hp Induction Motor at 0.72 p.f. lagging Efficiency 82% Supplied from 240 V, 60hz source


Example (Continued)

From Source

Motor P out = 5hp x 746 Watt= 3730 Watts

Light P out = 60 x 12 = 720 Watts

Load

Motor Heat Load

Lighting Load


Example (Continued)

P out = 16000 Watts

Power Factor

Heating Element p.f. = 1 Lights p.f. = 1

i.e. θ = 0 ° i.e. θ = 0 °

Heat load

Motor p.f. =0.72 IM lags VM in 43.94°


Example (Continued)

=

=

η

η

=

× =

4549 0.82

746 5

Watts

P P

P P

out in

in

out = =

4384 sin θ

VARS S Q

( ) ° =

− =

94 . 43 72 . 0 1 cos θ

=

=

=

=

6318 0.72 4549 cos

cos

θ

θ

VA

P S

S P

S θ

Motor


Example (Continued)

Light Load P = 12 x 60 = 720 watts Q = 0 VARS

Heating Load P = 16 Kw = 16000 watts Q = 0 VARS

Totals for “P” and “Q” P = 4549 + 720 + 16000 = 21269 Q = 0 + 0 + 4384 = 4384 VAR


Example (Continued)

= 21,716.32 VA 2 4384 2 21269

2 2

+ =

+ = Q P S

° = −1 = −1 = 6 . 11 21269 4384 tan tan

P Q θ

θ

P

Q S

p.f. = cos θ = cos 11.6 ° = 0.979 lagging


Another Example:

Power Factor Correction

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

1000 volts @ 60 Hz

Motor Light

20 kVA 0.7 p.f. lagging

10kW unity p.f.

From source

P = ? Q = ? S = ?


MOTOR S = 20 ∠ cos -1 (0.7) kVA

= 20 ∠ 45.6 ° kVA

P = 20 cos 45.6° kW = 14 kW

Q = 20 sin 45.6 ° = 14.28 kVAR 0 kVA 10 j S L + =

Example Solution

Light


TOTAL LOAD

P = 14 + 10 = 24.0 kW Q = 14.28 kVAR

S = kVA 9 . 27 2 28 . 14 . 242 = +

° = = − 75 . 30 24.0 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


Impedance and Power Triangle

θ θ θ θ ∠ = ∠ = ∠ = ∠ = I

V I V V

I V Z Z 2

V V

P = 24 Kw

30.75 θ

Q cap Q

14.28 S

Q new

18.9

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

1991-1998 4-13

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 °

Q cap = Q old - Q new

= 14.28 - 7.9 = 6.38 kVAR

Q new = P (tan 18.19°) = 24 x 10 3 tan 18.19° = 7.9 kVARS

P=24 kW


Find the value of capacitance c = ?

2

* * 1

Q V

Q V * V

V Q V

I V

j ω c x c = =

= = =

= 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

f=60Hz


THREE PHASE CIRCUITS

A → B → C

A B

C

V AN

N

V ph = V L

B

A

C

WYE

DELTA


THREE PHASE CIRCUITS

ph ph I V S 3 =

3 L

V phase

V =

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

V phase = V AN = phase to neutral

Y-Connection


Y-connection

L I ph I

L V phase V

=

= 3


∆-connection

3 L I

ph I

L V ph V

=

=

ph ph I V S * 3 =


Y Connection

* 3

* 3

3

* 3 h p h p

IL VL

IL VL I V S

=

= =

Y- Connection

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

The connection is a Y-connection


∆-Connection

* 3

* 3

* 3 3

3

* 3

L L

ph ph

L L

ph

h p h p

I V

I V

I V

IL V

I V S

=

=

=

=

=

Vph=VL


Another 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 4-24

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 V 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

_

= 7969.43+j 0.59 V

No Slide Title - Florida International...

Documents

Transcript of No Slide Title - Florida International...