Balanced & Unbalanced Polyphase Systems

Single Phase Power Equations

Apparent Power or Complex Power

P: real power (W) Q: reactive power (VAR) S: apparent power (VA)

• Power Factor

• Real Power

• Reactive Power

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For the following circuit, find the (a) real power, and (b) reactive power. (c) Draw the power triangle.

Example: Single-Phase Power

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For the following circuit, find the (a) real power, and (b) reactive power. (c) Draw the power triangle.

(a) The real power is:

(b) The reactive power is:

P = VR2

R= (120 V)2

10 Ω= 1440 W

Q = VL2

XL

= (120 V)2

4 Ω= 3600 VAR

Example: Single-Phase Power

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(c) The power triangle is:

In this example, the impedance of the inductor has a lagging current, so the current has a negative phase angle. The complex conjugate of the current has a positive phase angle, so the reactive power, Q, is positive and the power triangle is in the first quadrant. For a leading current (which has a positive phase angle compared to the voltage) the power triangle has a negative imaginary part and a negative power angle, so it is in the fourth quadrant.

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Current I creates drop (+) (-) as shown VAB defined as the rise B A

Polyphase Circuits E.E. element convention

ABC (+) sequence Find VAB Via KVL

(+) (-) V

A B I

←“VAB”

3031200

0

∠=−∠−∠=

−==+−+

PPP

BNANAB

BNANAB

VVV

VVVVVV

(+120 )

(150 )

(30 )

(0 )

(-120 )

- +

-

-

+

+

+ (-90 )

+

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Note: Voltages as given are line to line voltages unless otherwise specified

ACB (-) sequence Find VAC via KVL

In both examples VA is the reference phasor. If VB or VC are designated the reference phasor then the phasor relationships will differ. (Try CBA)

(+120 )

(150 )

(30 )

(0 )

CNV(-120 )

- +

-

- +

(-90 )

+ +

- 0

0 120

3 30

AC AN CN

AC AN CN

P P

AC

V V V

V V V

V V

V Vp

+ − + =

= −

= ∠ − ∠ −

= ∠

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Three-Phase Systems

Three-phase power systems have three wires to transmit power. Some systems also have a neutral wire. In a balanced three-phase system, the voltages on the three wires are all the same magnitude but are out of phase by 120 degrees. In a balanced system, the currents are also the same magnitude but are out of phase by 120 degrees. The vector sum of the voltages and currents in a balanced three-phase system are zero.

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Example: Three-Phase System

The voltage on phase A of a balanced, three-phase power system is: °∠37220

What are the voltages and angles of the other phases?

(Assume that the phase sequence is ABC)

( ) °−∠=°∠−°∠= 8322012037220BV

( ) °∠=°∠+°∠= 15722012037220CV

°−∠= 83220BV°∠= 37220AV

°∠= 157220BV

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Three-Phase Wye System A three-phase wye configuration has three phases and a neutral wire. The “phase” voltages are between the legs and are named VAB, VBC, and VCA. The line to neutral voltages are between the lines and the neutral and are designated VAN, VBN and VCN.

Generator Load

line-lineline-neutral 3

VV =

Line-to-line voltages and line-to-neutral voltages are 30 degrees out of phase. The phase voltage leads the line-neutral voltage by 30 degrees and the line-neutral voltage lags the phase voltage by 30 degrees.

°∠= 303 ANAB VV °−∠= 303

ABAN

VV

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Problem: Three-Phase Wye

1-11

How do we know it’s a wye system?

Problem: Three-Phase Power

Key Points to Remember •Answer (A) has the

correct amplitude but does not include the phase in the equation.

• In answers (C) and (D), the line-to-line voltage equations are used Answer is (B)

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Three-Phase Delta System

Three-phase delta configuration has three phases and no neutral wire. The only voltages to be considered are the line-to-line voltages. There are two currents to be considered: the line currents which flow in the lines, and the phase currents which flow in the resistors, shown as load in the diagram.

Generator Load

The line currents are out of phase with the line-line (phase) currents:

°−∠= 303 ABA II °∠= 303A

ABII

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Example: Delta System

PROBLEMS

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Example: Delta System

SOLUTIONS

PROBLEMS

1-15

Wye Delta Conversions

To convert balanced Wye to Delta:

For balanced loads where all the impedances are equal, the conversion from Delta to Wye and vice versa is simplified.

WyeDelta RR 3=

To convert balanced Delta to Wye: DeltaWye RR

31

=

In the figure to the right, in order for the two loads to be equivalent:

131 RRA = ARR 31 =

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Power in Three-Phase Systems For balanced systems, three-phase power can be computed using any of the following equations:

jQPS += LineLineLine IVS −= 3LineLineLine IVS *3 −=

In a balanced load where the load is the same in all three phases, the three-phase real power can be computed as three times the power in any single phase.

θCosIVP LineLineLineLine −= Cos θ is the power factor

3LineLine

NeutralLineVV −

− = LineTotal PP 3=θCosIVP LineLineLine

Line 3−=

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Example: Power Computation

PROBLEMS

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Example: Power Computation

SOLUTIONS

PROBLEMS

1-19

Balanced Polyphase Example

3 Ø motor takes 10 kVA at 0.6PF lagging from a 220 V source. The motor is in parallel with a balanced Δ load with an impedance of R = 16 xc = 12 in each phase. Find the total VA, power, power factor.

For the load I phase AjV

ZVI LL 87.3611

1216220

∠=−

==

IMOTOR

x

R

R

x MOTOR

x R

LOAD

ILOAD LINE

ILOAD PHASE

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VAAVIVS LINELLLOAD 87.362420)87.3611)(220(*1 −∠=−∠==φ

Note that the angle of reactive power is given by load

9.361612tan 1 == −θ

So )(9.367260 leadingSLOAD∠=

leading because capacitive

is given as 10KVA @ 0.6PF lagging

Angle of SMOTOR is a cos-1 0.6 = 53.13 (lagging)

15.17357,1287.36726013.531010 3 −∠=∠+−∠=+= xSSS LOADMOTORTOTAL

Note: This is a leading power factor diagram. It is not a phasor diagram.

12

16

20

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VAVASS LOADLOAD 87.36726087.36342033 13 ∠=∠•== φφ

Problem can also be solved by converting delta to wye and computing parallel impedance.

Impedance of motor Ω=== 84.424.26

3/220

C

LC

IVZ

Ω=== 904.2)6.0)(84.4(cosθzRMOTOR Ω== 872.3sin84.4 θMOTORx

Parallel impedance

AIT 13.1742.3213.1793.33/220

−∠=∠

=

VAIVS LLL 13.17357,12)13.1744.32)(220(33 * ≅∠==

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Alternate Solution Method

Unbalanced Polyphase Example Unbalanced Polyphase Load Delta

Find phase currents 100 0 100 0 10 53.1 A6 8 10 53.1

ABab

ab

VIZ j

∠ ∠= = = = ∠ −

+ ∠

100 120 5 120 A20

CAca

cc

VIZ

∠= = = ∠

IC

IB

A

B

C c

IA Iab Ica Ibc

b a

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13.832087.365

12010034120100

−∠=−∠

−∠=

−−∠

==jZ

VIbc

BCbc

Compute line currents via KCL at nodes

1-24

Unbalanced Wye Example Unbalanced Wye Connected Load, Four Wire A three phase four wire 150Y, CBA system has a Y connected load. VAN has an angle of -90 The load impedances are

Obtain all the line currents and neutral current

A

N

B

C IC

IB

IA

IN VAN

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16721.1010532.17043.149043.14

10532.174551506.86

043.14306

306.86

9043.1406

906.86

1506.861503

1501503

306.86303

150303

906.86903

150903

−∠=∠+∠+−∠=++=

∠=∠∠

==

∠=∠

∠==

−∠=∠

−∠==

∠=∠=∠=

∠=∠=∠=

−∠=−∠=−∠=

CBAN

C

CNC

B

BNB

A

ANA

LNCN

LNBN

LNAN

IIIIZVI

ZVI

ZVI

VV

VV

VV

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Unbalanced Three Wire, Wye Connection Unbalanced Three Wire Connection, Wye Load.. 150V CBA system, & connected load,

A

B

C IC

IB

IA

N (-120 )

(0 )

'BNV

NAV ′

(120 )

30

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What types of loads are these?

Find the all voltages and currents °∠= 16.2767.66:Given 'BNV

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120150

0150

120150

∠=

∠=

−∠=

CA

BC

AB

VVV

The line to line voltages may be determined by inspection:

Find VAN’ using KVL:

96.987.100

16.2767.66120150

0

−∠=

∠+−∠=

+==−−

′

′

′′

′′

NA

NA

NBABNA

NBABNA

VV

VVVVVV

Find VCN’ using KVL:

4.16159.95

16.2715092.9864.100

0

∠=

∠+−∠=

+==−−

′

′

′

′′

NC

NC

CAANNC

NACANC

VV

VVVVVV

92.9878.1606

96.987.100∠=

∠−∠

== ′

A

NAA Z

VI

84.213.11306

16.2767.66−∠=

∠∠

== ′

B

NBB Z

VI

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( ) 01 =++ ′′′

C

NC

B

NB

A

NA

ZV

ZV

ZV

Find the currents via KCL at central node:

4.11612.19455

4.16159.95∠=

∠∠

== ′

B

NBC Z

VI

How far is the neutral offset?

41.39197.2096.987.100906.86

96.987.1009053

150

∠=

−∠−∠=

−∠−∠=

′NNV

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What happens if you connect the neutral of the load to the neutral of the generator (supply)?

Power Afternoon Sample Questions A 3-phase, 4-wire, neutral-grounded wye-connected utility line has a phase-to-phase voltage of 13.2 kV. A complex load of (200 + ƒ100) kVA is connected between Phase A and neutral; identical load is connected between Phase B and neutral. The neutral current (amperes) is nearly: (A) 0 (B) 9.8 (C) 16.9 (D) 29.3

56.14634.29120

3102.13

10)100200(

56.2634.290

3102.13

10)100200(

3

3

3

3

∠=−∠

+==

∠=∠

+==

VxVAxj

VSI

VxVAxj

VSI

B

BB

A

AA

Neutral current is the sum of IA + IB + IC =IN

Answer (0)

Note that answer (C) is the result of incorrectly using VLL rather than VLN

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Power Afternoon Sample Questions

A 3-phase, 3-wire, ungrounded, 13.2kV (phase-to-phase) wye-connected course is connected to a balanced delta load that is grounded on Corner A. The voltage measured between Corner B and ground is most nearly: (A) Half the phase-to-phase voltage (B) 7.62 kV (C) 13.2 kV (D) cannot be determined

Use KVL:

(sign independent) Answer (C)

A A

C B

B

Grounding

C

VAB -

+

- + VAB

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Power Afternoon Sample Questions

The only load on a 3-phase, 4-wire system is placed between Phase B and Phase C. The phase-to-phase voltage is 13.2 kV. The load is 500kVA at 0.85 lagging power factor. The magnitude of the line current in Phase C amperes is most nearly: (A) 65.6 (B) 55.8 (C) 37.9 (D) 32.2

Power factor not relevant

VxVAx

VSI 3

3

102.1310500

==

500KVA 0.85 lag

A

B

C

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Voltage Regulation

Voltage regulation is a measure of the degree to which the voltage at a load is held constant as the amount of load varies from no load to full load.

100% ×−

=FL

FLNLR V

VVVVNL is the no load voltage

VFL is the full load voltage

The concept of regulation can be applied to a problem with a component designed to hold steady voltage (voltage regulator) or to a problem with a general source without a regulator supplying load.

Line regulation measures the ability to maintain a constant output voltage regardless of changes in the input voltage.

Load regulation measures the ability to maintain a constant output voltage regardless of changes in size of the load (current draw).

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Example: Voltage Regulation

A transmission line supplies 60 MVA of load at 0.8 power factor, lagging. The voltage at the receiving end of the line is 138 kV. The transmission line impedance is 10+j30. What is the % regulation? (Use Per-Phase Analysis)

20 MVA 0.8 Lag

10+j30 Ω

PROBLEM

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Example: Voltage Regulation

A transmission line supplies 60 MVA of load at 0.8 power factor, lagging. The voltage at the receiving end of the line is 138 kV. The transmission line impedance is 10+j30. What is the % regulation?

20 MVA 0.8 Lag

10+j30 Ω

Solve the problem considering only one phase of the system. The 60 mVA load is 20 mVA per phase.

kVVVLN 67.79310138 3

=×

= Load line to neutral voltage

*1 LLN IVS =φ A

VVA

VSI

LNL 87.36251

1067.7987.361020

3

61

*

−∠=×−∠×

== φ

°== − 87.368.0cos 1θ Finding phase angle

Sending end voltage

VjVS °∠×=°∠×+°−∠+= 31031.8601067.79)87.36251)(3010( 33

%33.810067.79

67.7931.86% =×−

=RV

SOLUTION

PROBLEM

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