6.1 Polyphase System 6.2 Notations 6.3 Single-phase Three-wire Systems

6.1 Polyphase System6.1 Polyphase System

6.2 Notations 6.2 Notations

6.3 Single-phase Three-wire Systems6.3 Single-phase Three-wire Systems

6.4 Three Phase connection 6.4 Three Phase connection

6.5 The Delta ( ) Connection6.5 The Delta ( ) Connection

6.6 Power measurement6.6 Power measurement

YY

Engineering Circuit AnalysisEngineering Circuit Analysis

CH6 Polyphase CircuitsCH6 Polyphase Circuits


Polyphase system : system with polyphase sourcesSingle source (Vs)

Notice the instantaneous voltage maybe zero The instantaneous power will be zero

T2T T3

V

t

V

tT3T2T

321 ,, VsVsVsPoly sources ( )

They all have 120o phase differences The instantaneous power will never be zero.

Ch6 Polyphsae Circuits

321 VsVsVsV

• The incident with a zero instantaneous power has been exempted.

• The source power can be delivered more stably.

• The polyphase systems can provide multiple output voltage levels.

• Polyphase systems in practice certain sources which maybe approximated by ideal voltage sources, or ideal voltage sources in series with small internal impedances.



6.2 Notations6.2 Notations


A8

a b

c d e f

g h i j

k l

A5

A2

A4

A10

A2deI?cdIefI

ijIA6A3fjIA3 , A8A5 cdcd II

For note c :

For note f :

A7 ,A 3A4 efef II

For note j :

A7 ,A 10A4A3 ijij II

6.2 Notations6.2 Notations

VVan00100

VVbn0120100

VVcn0240100

V

VV

VV

VVV

bnan

nbanab

0

00

302.173

1201000100

Test with graphical analysis ? (Using the phasor diagram)

c

a

b

n

The voltage of point a with

respect to point ba +; b -;

Similarly, Iab denotes the current from point a to b.



nban VV

nbanab VVV 22

Function: allowing household electronics operating at two levels of voltages to be applied.

1-phase3-wire Source

anb

Voltage characteristics

1V

2V

a

n

b

V110 V220

nban VV bnan VV

0 bnan VV

Household electronics may either operate with

or with

Phase characteristics


aAbBNn III

pbB Z

VI 1

pAa Z

VI 1

0NnI

Current characteristics

This is no current in the neutral wire.

1V

1V

A

N

a

B

n

b

pZ

pZ

How if the two pZ are NOT equal, and all the wires have impedances ?

This is a more practical scenario.



,50 100 1020 j

1 3 1

Example 9.1 (P242)① Determine the power delivered to the

and the

② Determine the power lost in the three lines represented by respectively.

and

③ Determine the transmission efficiency?

Loads.

Hints: observe a structure with regular meshes and know the impedances, we can determine the currents I1, I2 and I3 in order to find out the power being lost and delivered!

1

350

100

20

10 10jV0115 0

V0115 01I

2I

3I

η = total power absorbed by the loadstotal power generated by the sources



rms

rms

03501V0115 312110 IIIII

0501001020 12322 IIIIIj

011003V0115 323130 IIIII

Apply KVL for the three meshes.

Rearranging them in a matrix form as

0

0

3

2

1

0115

0

0115

1041003

1001017050

35054

I

I

I

j



A80.2137.10

A47.24389.9

A83.1924.11

03

02

01

I

I

I

W176320

W117100

W20650

2

21020

2

23100

2

2150

IP

IIP

IIP

j

W2086

It can be calculated:

Hence, the average power delivered to each of the loads are:

Total loaded power

A27.2∠02.221oII A12.2∠08.123

oII



A3.2∠947.013oII

rms

rms

rms

rms

rms

rms

W333

W1081

W1261

2

13

2

2

3

2

1

IIIP

IP

IP

nNnN

bB

aA

W237

%100generatedpowertotal

loadthetodeliveredPower

W2323W1107W1216

80.21cos37.1011583.19cos24.11115 00

sourcesP

%8.89%100W2323

W2086

Power lost in three wires are:

Total lost power

Transmission efficiencyη

Total power generated by the two voltage sources is:

Transmission efficiency



6.4 Three Phase connection6.4 Three Phase connection YY AB

N

C

anVbnV

cnV

ab

n


0

cnbnan

cnbnan

VVV

VVV

Balanced three-phase sources(phasor voltages)


Positive phase sequence (abc)

0

0

0

240

120

0

pcn

pbn

pan

VV

VV

VV

Negative phase sequence (cba)

0

0

0

240

120

0

pcn

pbn

pan

VV

VV

VV

anV

bnV

cnV

pV02400120

0240

0120anV

cnV

bnV

6.4 Three Phase connection6.4 Three Phase connection YY


(Anti-clockwise rotation)

(clockwise rotation)

0

00

3032

3

2

3

60sin60cos600

ppp

pppppnbanab

VVjV

jVVVVVVVV

0

00

9032

3

2

1

2

3

2

1

60120

ppppp

ppncbnbc

VVjVVjV

VVVVV

0

00

210302

3

2

1

180240

pppp

ppnacnca

VVVjV

VVVVV

0 cabcab VVV

Line-to-line voltages (take the abc sequence as an example)

Hence verifies KVL.

anV

bnV

cnV

naV

nbV

ncV

abV

bcV

caV naV

ncV

nbV



Voltage types

Phase voltages ( )

LV

pV

Line-to-line voltages ( )

0303 pab VV 0903 pbc VV 02103 pca VV

magnitude Phasor difference

pV

pV30120

0120




A B

N

C

aAI

bBI

cCI

a b

n

pZ

PZ

c

pZ



pZConsider three impedances are connected between each line and

the neutral line.

p

anaA Z

VI 0

0

120120

aAp

an

p

bnbB I

Z

V

Z

VI

00

240240

aAp

p

p

cncC I

Z

V

Z

VI

0=++ cCbBaA IIIHence

When balanced impedances are applied to each of the three lines andthe neutral line carries no current.



Vrms240200,Vrms120200,Vrms0200 000 cnbnan VVV

Example 9.2 (P247)

line-to-line voltage:

Phase voltages:

Arms60260100

0200 00

0

p

anaA Z

VI

Arms1802 0bBI

Line currents:

Power absorbed by the three loads

Vrms2103200,Vrms903200,Vrms303200 000 cabcab VVV

Arms3002 0cCI

W60060cos22003 oP



The total instantaneous power is NEVER ZERO.

Example 9.2 (P247)

How about the instantaneous power?

Note: Van = 200V rms



Similarly , the instantaneous total power absorbed by the loads are :

W600

W1802cos4003002cos400602cos400600

ttt

tPtPtPtP CBA

W602cos400200

A60cos22Vcos2200

A60cos22

Vcos2200

0

0

0

t

tttitvtP

tti

ttv

aAanaA

aA

an

• Example 9.3 (P249)

A balanced three-phase system with a line voltage of 300Vrms is supplying a balanced Y-connected load with 1200W at a leading power factor (PF) of 0.8. Determine line cuurent IL and per-phase load impedance Zp.

The phase voltage is: Vp = 300/ Vrms.

The per-phase power is: 1200W/3 = 400W.

Therefore , and IL = 2.89Arms

The phase impedance is:

A leading PF of 0.8 implies the current leads the voltage, and the impedance angle is: -argcos(0.8) = -36.9o

and Zp = 60 -36.9o Ω

3

8.0×)(3

300=400 LI

Ω60=89.2

3300==||

L

PP I

VZ

∠

Note: the apparent power of a Y-Y connected load is P = Van × IAN (phase voltage × line current)



pZVrms3300pV

~

LI

6.5 The Delta ( ) Connection6.5 The Delta ( ) Connection The neural line dose not exist. Balanced impedances are connectedbetween each pair of lines.

A B

C

a b

n

pZ

PZ

c

pZ


03033 pabpL VVVV

cnbnanp VVVV


Phase voltages

Line voltages cabcabL VVVV

﹠

CABCABp IIII


Phase currents

Line currents pcCbBaAL IIIII 3



connections

Phase voltages

Line voltages

pIPhase currents

Line currents pL II 3

Y connections

pV pV

pL VV 3 pL VV 3

pI

pL II



√

√

√

√

• Example 9.5 (p251)

Determine the amplitude of line current in a three-phase system with a line voltage of 300Vrms that supplies 1200W to a Δ-connected load at a lagging PF of 0.8.

The per-phase average power is: 1200W/3 = 400W

Therefore, 400W = VL ∙ IP ∙ 0.8 = 300V ∙ IP ∙ 0.8, and IP = 1.667Arms

The line current is: IL = IP = 1.667A = 2.89Arms

Moreover, a lagging PF implies the voltage leads the current by argcos(0.8) = 36.9o

The impedance is:

Note: the apparent power of a Δ connected load is P = Vab × IAB (line voltage × phase current)

3 3

oo

P

PP I

VZ 9.36∠1809.36

667.1

300



VIP 6.6 Power measurement6.6 Power measurement

Wattmeter

measured by current coil

E.g.

W10004.1530cos18.11100

angangcos

Vrms0100

Arms4.15318.11

IVIVP

V

I

measured by potential coil

I

V

current coil

potential coil

Passive Network


A

B

C

c

b

a

PZ

PZ PZ1

2



aAI

cCI

bBI

ABI

BCI

CAI

Validate the power meter reads the actual power absorbed/delivered by the three impedances.

0

02

0

01

30cos

12090cosangangcos

30cos

30cosangangcos

LL

LLcCCBcCCB

LL

LLaAABaAAB

IV

IVIVIVP

IV

IVIVIVP



tg

tg

tg

tg

P

P

3

3

21

23

21

23

sin30sincos30cos

sin30sincos30cos

30cos

30cos00

00

0

0

2

1

12

123PP

PPtg

12

123PP

PParctg

inductive , 2

0 , 21

PP

reactive (PF=0) capacitive / inductive (0<PF<1) resistive (PF=1)

tg , 2

0 , 0 tg

21 PP 21 PP

tgtg , ,

2

2capacitive ,0

2 , 21 π

PP


Vrms0230 abV



with positive phase sequence.

(1) Find the reading of each wattmeter.

(2) The total power absorbed by the loads.With positive phase sequence , we know :

Vrms0230 abV

Vrms120230 caVVrms120230 bcV

Wattmeter 1 reads and :acVaAI

Vrms60230 caac VV

A1.105554.8154

303

230

154

jj

VI anaA

.4 15jA

c

b

a

1

2

.

.

B

C

.N




W13891.10560cos554.8230

angangcos1

aAacaAac IVIVP Wattmeter 1 reads :

A9.134554.8154

1503

230

154

jj

VI bnaB

Wattmeter 2 reads and :bcVbBI

W5.5129.134120cos554.8230

angangcos2

bBbcbBbc IVIVP

W5.87621 PPPHence ,

.4 15jA

c

b

a

1

2

.

.

B

C

.N

Q: Please try to prove the two wattmeters read the power associated with the three impedances.

6.1 Polyphase System 6.2 Notations 6.3 Single-phase Three-wire Systems

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Transcript of 6.1 Polyphase System 6.2 Notations 6.3 Single-phase Three-wire Systems