Post on 06-Mar-2018
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)
Energy Systems Research Laboratory, FIU (c) Copyright by Prof. Osama A. Mohammed. All rights reserved. 4-2
•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
Energy Systems Research Laboratory, FIU (c) Copyright by Prof. Osama A. Mohammed. All rights reserved. 4-3
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
Energy Systems Research Laboratory, FIU (c) Copyright by Prof. Osama A. Mohammed. All rights reserved. 4-4
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
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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°
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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
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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
Energy Systems Research Laboratory, FIU (c) Copyright by Prof. Osama A. Mohammed. All rights reserved. 4-8
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
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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 = ?
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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
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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
Energy Systems Research Laboratory, FIU (c) Copyright by Prof. Osama A. Mohammed. All rights reserved. 4-12
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
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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
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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
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THREE PHASE CIRCUITS
A → B → C
A B
C
V AN
N
V ph = V L
B
A
C
WYE
DELTA
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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
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Y-connection
L I ph I
L V phase V
=
= 3
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∆-connection
3 L I
ph I
L V ph V
=
=
ph ph I V S * 3 =
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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
Energy Systems Research Laboratory, FIU (c) Copyright by Prof. Osama A. Mohammed. All rights reserved. 4-20
∆-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
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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
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j0.4 0.3 Load
Load ~
~
Source
Source
One Line Diagram
100 kVA 0.8 p.f. lagging
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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
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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