l11 polyphase circuits ch12

8/19/2019 L11 Polyphase Circuits Ch12

1/78

ENE 104

Electric Circuit Theory

Lecture 11:

Polyphase Circuits

http://webstaff.kmutt.ac.th/~dejwoot.kha/

(ENE) Mon, 26 Mar 2012 / (EIE) Wed, 4 Apr 2012

Week 12 : Dejwoot KHAWPARISUTH

http://webstaff.kmutt.ac.th/~dejwoot.kha/http://webstaff.kmutt.ac.th/~dejwoot.kha/http://webstaff.kmutt.ac.th/~dejwoot.kha/


2/78

Objectives : Ch12

Week #12

Page 2

ENE 104 Pol hase Circuits

• single-phase and polyphase systems

• Y- and Δ- connected three-phase system

• per-phase analysis of three-phase systems


3/78

Introduction:

Week #12

Page 3


An example set of three voltages, each of which is 120o out of

phase with the other two. As can be seen, only one of the voltages

will be zero at a particular instant.


4/78

Double-Subscript Notation:

Week #12

Page 4


(a) The definition of the

voltage Vab.

(b) Vad = Vab + Vbc+ Vcd

= Vab + Vcd.


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Week #12

Page 5


.V0100 anV

.V120100 bnV

.V240100 cnV


6/78


Week #12

Page 6


.V302.173)6.8650(100

.V1201000100

j

bnannbanab VVVVV


7/78

Practice: 12.1

Week #12

Page 7


Let

= 100∠0° V,

= 40∠80° V, and

= 70∠200° V. Find (a) ; (b) ; (c)


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Practice: 12.1

Week #12

Page 8



9/78

Practice: 12.2

Week #12

Page 9


Refer to the circuit of figure below and let

= 3 A, = 2 A, and ℎ = −6 A. Find (a),

; (b) ; (c)


10/78

Practice: 12.2

Week #12

Page 10



11/78

Single-Phase Three-Wire System:

Week #12

Page 11



12/78

Single-Phase Three-Wire System:

Week #12

Page 12


Since Van = Vnb

Then

And InN = IBb + IAa = IBb - IaA = 0

p

nb Bb

p

anaA

Z

VIZ

VI

Consider:


13/78

Example:

Week #12

Page 13


Effect of Finite Wire Impedance:

Analyze the system below and determine the power

delivered to each of the three loads as well as the power

lost in the neutral wire and each of the two lines.


14/78

Example:

Week #12

Page 14


0)(100)(30115

0)(50)(100)1020(0)(3)(500115

32313

12322

31211

IIIII

IIIII

IIIII

j


15/78

Example:

Week #12

Page 15


.A83.1924.111

I

.A47.24389.92 I

.A80.2137.103 I

.W206502

2150 II P

.W117100223100 II P

The average power drawn by each load:

.W1763202

21020 I j P


16/78

Example:

Week #12

Page 16


.W12612

1 IaA P

.A83.1924.111

I

.W108123 IbB P

.A47.24389.92 I

.A80.2137.103 I

The loss in each of the wires:

.W312

13 IInN P

P 17


17/78

Example:

Week #12

Page 17


The transmission efficiency, :

%8.892372086

2086

generated power total

load todelivered power total

P 18


18/78

Practice: 12.3

Week #12

Page 18


Modified Figure below by adding a 1.5 Ω

resistance to each of the two outer lines, and a2.5 Ω resistance to the neutral wire. Find the

average power delivered to each of the three

loads.

P 19


19/78

Practice: 12.3

Week #12

Page 19


Page 20


20/78

Practice: 12.3

Week #12

Page 20


h h CPage 21


21/78

Three-Phase Y-Y Connection:

Week #12

Page 21


A Y-connected three-phase four-wire source.

A balanced three-phase system:

may be defined as having

and

cnbnan VVV

0 cnbnan VVV

Th Ph Y Y C iPage 22


22/78


Week #12

Page 22


Positive, or abc,phase sequence.

Negative, or cba,phase sequence.

Th Ph Y Y C tiPage 23


23/78


Week #12

Page 23


Line-to-Line Voltage:

303

1200

pab

p p

bnanab

V

V V

V

VVV



24/78


Week #12

Page 24


Line-to-Line Voltage:

2103 pca V V

903 pbc V V

303 pab V V



25/78

A balanced three-phase system, connected

Y-Y and including a neutral.


Week #12

Page 25


p

anaA

Z

V

I

120

120

aA

p

an

p

bnbB

I

Z

V

Z

VI

240aAcC IITherefore:

0 cC bBaA Nn IIII

E lPage 26


26/78

Example:

Week #12

Page 26


Find the currents, voltages and the total power

dissipated in the load

E lPage 27


27/78

Example:

Week #12

Page 27


“Per -phase”:

.V0200 anV

.V120200 bnV

.V240200 cnV

2103 pca V V

903 pbc V V

303 pab V V 602 p

anaA

Z

VI

3002cC I

1802bBI

E lPage 28


28/78

Example:

Week #12

Page 28


“Per -phase”:

)0120cos(2200 t v AN

)60120cos(22 t i AN

E lPage 29


29/78

Example:

Week #12

Page 29


even though phase voltages and currents have

zero value at specific instants in time, theinstantaneous power delivered to the total load is

never zero.

)]60240cos(400200

)]60240cos()60[cos(400)60120cos()120cos(800)(

t

t t t ivt p AN AN A

)]300240cos(400200)( t t p B

)]180240cos(400200)( t t pC

E lPage 30


30/78

Example:

Week #12

g


the instantaneous power absorbed by the total

load is therefore

W t pt pt pt p C B A 600)()()()(

Practice: 12 4Page 31


31/78

Practice: 12.4

Week #12

g


A balanced three-phase three-wire system has aY-connected load. Each phase contains three

loads in parallel: −j100 Ω, 100 Ω, and 50 + j50 Ω.

Assume positive phase sequence with =

400∠0° V. Find (a) ; (b) ; (c) the total power

drawn by the load



32/78

Practice: 12.4

Week #12

g


Example:Page 33


33/78

A balanced three-phase system with a line

voltage of 300V is supplying a balanced Y-connected load with 1200W at a leading PF of

0.8. Find the line current and the per-phase load

impedance.

The phase voltage is

The per-phase power is

Example:

Week #12ENE 104 Pol hase Circuits

.3

300 V V p

.4003

1200 W

Example:Page 34


34/78

Example:


The line current is

The phase impedance is

.3

300 V V p .40031200 W P p

)cos()cos(

2

1 eff eff mm I V I V P

89.2

3

300

L

p

p I

V Z

.89.28.0

400

)cos(3

300 A

V

P I

p

p

L


35/78



36/78

Practice: 12.5


A balanced three-phase three-wire system has a

line voltage of 500 V. Two balanced Y-connected

loads are present. One is a capacitive load with

7 − j2 Ω per phase, and the other is an inductive

load with 4 + j2 Ω per phase. Find (a) the phasevoltage; (b) the line current; (c) the total power

drawn by the load; (d) the power factor at which

the source is operating.



37/78

Practice: 12.5


Example:Page 38


38/78

A balanced 600-W lighting load is added (in

parallel) to the previous example. Determine thenew line current.

per-phase circuit

Example:


9.3660 pZ

89.22 I

Example:Page 39


39/78

The amplitude of the lighting current is

determined by

So assume

And the line current is

Example:


155.11 I .A0155.11 I

0cos200 13300

I

)cos( eff eff I V P

.A9.3689.22 I

.A6.2687.321 III L



40/78

Practice: 12.6


Three balanced Y-connected loads are installed on a

balanced three-phase four-wire system. Load 1draws a total power of 6kW at unity PF, load 2

requires 10kVA at PF=0.96 lagging, and load 3

needs 7kW at 0.85 lagging. If the phase voltage at

the load is 135 V, if each line has a resistance of

0.1 Ω, and if the neutral has a resistance of 1 Ω, find

(a) the total power drawn by the load; (b) the

combined PF of the load; (c) the total power lost inthe four lines; (d) the phase voltage at the source;

(e) the power factor at which the source is operating.



41/78

Practice: 12.6




42/78

Practice: 12.6


The Delta (∆) Connection:Page 43


43/78

Where and

The Delta (∆) Connection:


A balanced -connected load is present on a three-wire

three-phase system. The source happens to be Y-

connected.

cn

bn

an pV

V

V

V

ca

bc

ab LV

V

V

V

p L V V 3 303 pab V V



44/78

The phase current

the line current:

and



CA ABaA III p

caCA

p

bc BC

p

ab AB

Z

VI

Z

V

I

Z

VI

CA BC AB p I III cC bBaA L I III

p L I I 3



45/78

An example with an inductive load



CA ABaA III

CA BC AB p I III

cC bBaA L I III

p L I I 3

∆-Connected:Page 46


46/78

∆ Connected:


total power,

cos

3

coscos

L L

p L

p p p

I V

I V I V P

cos3

3

L L

p

I V

P P

Y-connected:Page 47


47/78

Y connected:


total power,

cos

3

coscos

L L

L p

p p p

I V

I V I V P

cos3

3

L L

p

I V

P P



48/78

Practice: 12.7


Each phase of a balanced three-phase Δ-

connected load consists of a 0.2 H inductor in

series with the parallel combination of a 5 F

capacitor and a 200 Ω resistance. Assume zero

line resistance and a phase voltage of 200 V at

= 400 rad/s. Find (a) the phase current; (b)

the line current; (c) the total power absorbed by

the load



49/78

Practice: 12.7


Example 12 5:Page 50


50/78

Example 12.5:


Determine the amplitude of the line current in a

three-phase system with a line voltage of 300 Vthat supplies 1200 W to a ∆-connected load at a

lagging PF of 0.8

cos3

cos

cos

L L

p L

p p p

I V

I V

I V P

9.36180.,89.2 p L A I Z

Example 12 6:Page 51


51/78

Example 12.6:


Determine the amplitude of the line current in a

three-phase system with a line voltage of 300 Vthat supplies 1200 W to a Y-connected load at a

lagging PF of 0.8

cos3

cos

cos

L L

L p

p p p

I V

I V

I V P

9.3660.,89.2 p L A I Z3

D Z

Z Y



52/78

Practice: 12.8


A balanced three-phase three-wire system is

terminated with two Δ-connected load in

parallel. Load 1 draws 40kVA at lagging PF of

0.8, while load 2 absorbs 24 kW at a leading PF

of 0.9. Assume no line resistance, and let

= 440∠30° V. Find (a) the total power drawn

by the load; (b) the phase current for the

lagging load; (c) ; (d) .

Practice: 12.8Page 53


53/78

Practice: 12.8


Power Measurement:Page 54


54/78

Power Measurement:


wattmeter



55/78

Power Measurement:


IVIV ang ang P 22 cos



56/78

Practice: 12.9


Determine the wattmeter reading in Figure

below, state weather or not the potential coil hadto be reversed in order to obtain an upscale

reading, and identify the device or devices

absorbing or generating this power. The (+)

terminal of the wattmeter is connected to: (a) x;(b) y; (c) z.



57/78

Practice: 12.9




58/78

Practice: 12.9




59/78

Power Measurement:


The wattmeter in a Three-Phase System:


60/78



61/78


The wattmeter in a Three-Phase System:

dt ivivivT

dt iiivT

dt ivivivT

dt ivivivT

P P P P

cC CN bB BN aA

T

AN

T

cC bBaA NxcC CN bB BN aA

T

AN

cC CxbB BxaA

T

AxC B A

)(1

)(1)(1

)(1

0

00

0

NxCN Cx

Nx BN Bx

Nx AN Ax

vvv

vvv

vvv



62/78


The Two-Wattmeter Method:

aA ABaA AB ang ang P IVIV cos1 cC CBcC CB ang ang P IVIV cos2



63/78



in the case of a

balance load:

we can find PF

30cos

cos1

L L

aA ABaA AB

I V

ang ang P IVIV

30cos

cos2

L L

cC CBcC CB

I V

ang ang P IVIV



64/78



in the case of a

balance load:

we can find PF

30cos

30cos

2

1

P

P

12

123tan P P

P P

Example 12.7:Page 65


65/78

p


The balanced load is fed by a balanced three-

phase system having and positive phase sequence. Find the reading of each

wattmeter and the total power drawn by theload.

120230caV

120230bc

V

0230abV

0230abV

Example:Page 66


66/78

p


60230ac

V

W13891.10560cos554.8230

cos1

aAacaAac ang ang P IVIV

0230abV

1.105554.8

154

30

154

3230

j j

anaA

VI

120230caV

Example:Page 67


67/78

p


W5.5129.134120cos554.8230

cos2

bBbcbBbc ang ang P IVIV

1.105554.8aAI

120230bcV

9.134554.8bBI



68/78


For the circuit of Figure below, let the loads be

= 25∠60°Ω, = 50∠ − 60°Ω, = 50∠60°Ω, = 600∠0° V rms with (+) phase sequence, and

locate point x at C. Find (a) ; (b) ; (c)



69/78




70/78


Example: Final 2/47Page 71


71/78

ENE 104 AC Circuit Power Anal sis

เม อ แหล งจ ายในวงจร เปนแบบ balanced และ positive phase

sequenceจงหา

, , และ

the total complex powersupplied by the source

+ _

+ _

+ _

a

b

c

480 0

o

Vrms

A

C

B

10 ohm

-j10 ohm

j5 ohm

aAI bBI cC I

Week #12



72/78


+ _

+ _

+ _

a

b

c

480 0o Vrms

A

C

B

10 ohm-j10 ohm

j5 ohm

301.8310

ab AB

VI

3.1665

9038.831

j BC I

1201.83

10

15038.831

j

CAI

.V0480 anV

V120480 bnV

.V240480 cnV

21038.831caV

904803bcV

303 pab V V

Week #12



73/78


+ _

+ _

+ _

a

b

c

480 0o Vrms

A

C

B

10 ohm-j10 ohm

j5 ohm

3.166 BC I

301.83 ABI

1201.83CAI

CA ABaA III

AB BC bB III

BC CAcC III

***

CACA BC BC AB ABtotal IVIVIVS

Week #12



74/78


เม อ Wattmeters ในวงจร อ านค าได W.

และW.

เม อค าthe magnitude

ของthe

line voltage มค า 4160 V. จงหา

54.297,371 W

31.196,1392 W Z

12

123tan P P

P P

Week #12

Page 75

Ex:


75/78

ENE 104

+ _

+ _

+ _

a

b

c

480 0o Vrms

120o

480

480 -120o

Wm1

+-+-

Wm2

+-+-

Z A

ZB

ZC

A

B

C

ใหหา AB, BC, AB, BC, aA, cC และคาท อานไดท wattmeter ทั งสอง

Z A= 60 -30˚

ZB= 24 30˚

ZC= 80 0˚

AC Circuit Power Anal sis Week #12

Ex: Page 76


76/78

ENE 104

+ _

+ _

+ _

a

b

c

480 0o Vrms

120o

480

480 -120o

Wm1

+- +-

Wm2

+-+-

Z A

ZB

ZC

A

B

C

AC Circuit Power Anal sis Week #12

Hw: Page 77


77/78

ENE 104 AC Circuit Power Anal sis Week #12

Reference:Page 78

l11 polyphase circuits ch12

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