Chapter 12Chapter 12Three-Phase CircuitsThree-Phase Circuits

電路學電路學 (( 二二 ))

12.1 Introduction (1)12.1 Introduction (1)Single-Phase Systems

two-wire type

Three-Phase Four Wire Systems

three-wire type

12.1 Introduction (2)12.1 Introduction (2)• Nearly all electric power is generated and distributed in

3-phase.

• Instantaneous power in a 3- system can be constant.

• The 3- system is more economical than single-phase system.

12.2 Balanced Three-Phase 12.2 Balanced Three-Phase Voltages (1)Voltages (1)

A three-phase generate

Van Vbn Vcn

12.2 Balanced Three-Phase 12.2 Balanced Three-Phase Voltages (2)Voltages (2)

Y-connected source -connected source

12.2 Balanced Three-Phase 12.2 Balanced Three-Phase Voltages (2)Voltages (2)

0

120

240 120

an p

bn p

cn p p

V

V

V V

V

V

V

0an bn cn V V V

Balanced phase voltages are equal in magnitude and are out of phase with each other by 120

positive sequence正相序 negative sequence負向序

A Balanced Load is one in which the phase impedance are equal in magnitude and in phase.

a Y-connected load. a -connected load.

12.2 Balanced Three-Phase 12.2 Balanced Three-Phase Voltages (3)Voltages (3)

For a balanced Y-connected load Z1 = Z2 = Z3 = ZY

For a balanced -connected load Za = Zb = Zc = Z

Y- transformation 13 or 3Y Y Z Z Z Z

There are four possible connections in three-phase systems:• Y-Y connection• Y- connection- connection-Y connection

12.2 Balanced Three-Phase 12.2 Balanced Three-Phase Voltages (4)Voltages (4)

12.2 Balanced Three-Phase 12.2 Balanced Three-Phase Voltages (5)Voltages (5)

Example 1Determine the phase sequence of the set of voltages

200cos( 10 )200cos( 230 )200cos( 110 )

an

bn

cn

v tv tv t

12.3 Balanced Y-Y Connection (1)12.3 Balanced Y-Y Connection (1)

ZY = Zs + Zl + ZL

12.3 Balanced Y-Y Connection (2)12.3 Balanced Y-Y Connection (2)Assuming the positive sequence,

0 , 120

120an p bn p

cn p

V V

V

V V

V

The line-to-line voltages (line voltage)

0 120

3 30

ab an bn

p p

p

V V

V

V V V

3 90

3 210

bc bn cn p

ca cn an p

V

V

V V V

V V V

Similarly,

3L pV V

12.3 Balanced Y-Y Connection (3)12.3 Balanced Y-Y Connection (3)

120, 120

120120

an bn ana b a

Y Y Y

cn anc a

Y Y

V V VI I IZ Z ZV VI IZ Z

0( ) 0

0

a b c

n a b c

nN n n

I I II I I IV Z I

12.3 Balanced Y-Y Connection (4)12.3 Balanced Y-Y Connection (4)Example 2Calculate the line currents in the three-wire Y-Y system.

12.4 Balanced Y-12.4 Balanced Y- Connection (1) Connection (1)Assuming the positive sequence,

0 , 120

120an p bn p

cn p

V V

V

V V

V

3 30

3 90

3 150

ab p AB

bc p BC

ca p CA

V

V

V

V V

V V

V V

, ,BC CAABAB BC CA

V VVI I I

Z Z Z3 30

3 30

3 30

a AB CA AB

b BC AB BC

c CA BC CA

I I I I

I I I I

I I I I

3L pI I

p AB BC CAI I I I

L a b cI I I I

12.4 Balanced Y-12.4 Balanced Y- Connection (2) Connection (2)Example 3A balanced abc-sequence Y-connected source with Van = 10010 V is connected to a -connected balanced load (8 + j4) per phase. Calculate the phase and line currents.

12.5 Balanced 12.5 Balanced -- Connection (1) Connection (1)Assuming the positive sequence,

0 , 120

120ab p bc p

ca p

V V

V

V V

V

,ab AB bc BC

ca CA

V V V VV V

, ,ab BC bc CA caABAB BC CA

V V V V VVI I I

Z Z Z Z Z Z

3L pI I

3 30

3 30

3 30

a AB CA AB

b BC AB BC

c CA BC CA

I I I I

I I I I

I I I I

12.5 Balanced 12.5 Balanced -- Connection (2) Connection (2)Example 4 A balanced -connected load having an impedance 20 – j15 is connected to a -connected, positive-sequence generator having Vab = 3300 V. Calculate the phase currents of the load and the line currents.

12.6 Balanced 12.6 Balanced -Y Connection (1)-Y Connection (1)• Using KVL.

• Replacing the -connected source with its equivalent Y-connected source.

• Transforming the Y-connected load to an equivalent Y-connected load.

12.6 Balanced 12.6 Balanced -Y Connection (2)-Y Connection (2)

12.6 Balanced 12.6 Balanced -Y Connection (3)-Y Connection (3)Example 5 A balanced Y-connected load with a phase impedance 40 + j25 is supplied by a balanced, positive-sequence -connected source with a line voltage of 210 V. Calculate the phase currents. Use Vab as reference.

12.7 Power in a Balanced System 12.7 Power in a Balanced System (1)(1)

For a Y-connected load, the phase voltages are

2 cos , 2 cos( 120 )

2 cos( 120 )

AN p BN p

CN p

v V t v V t

v V t

The advantage of 3-phase systems for power distribution

• The total instantaneous power in a balanced 3-phas system is constant.

• The 3-phase system uses a lesser amount of wire than the single-phase system for the same line voltage VL and the same absorbed power PL.

If ZY = Z, the phase currents

2 cos( ), 2 cos( 120 )

2 cos( 120 )

a p b p

c p

i I t i I t

i I t

2 [cos cos( ) cos( 120 )cos( 120 )

cos( 120 )cos( 120 )]

a b c AN a BN b CN c

p p

p p p p v i v i v iV I t t t t

t t

[3cos cos cos cos 240 sin sin 240

cos cos 240 sin sin 240 ]3 cos

p p

P P

p V I

V I

1cos cos [cos( ) cos( )]2

A B A B A B Appling

12.7 Power in a Balanced System 12.7 Power in a Balanced System (2)(2)

12.7 Power in a Balanced System 12.7 Power in a Balanced System (3)(3)

2* 2

*

33 3 3 3p

p p p p p L LP

VI V I S S V I Z

Z

*p p p p pP jQ S V I

The complex per phase

The total complex power

where Vp, Ip, VL, and IL are all in rms values and is the angle of the load impedance.

for Y-connected loads, ,3L

p p LVV I I

for -connected loads, ,3L

p L pIV V I

12.7 Power in a Balanced System 12.7 Power in a Balanced System (4)(4)

22

22 2 Lloss L

L

PP I R RV

2 2' ' 2 ' ' '

2 23( ) 33L L

loss LL L

P PP I R R RV V

' '

2loss

loss

P RP R

2 '2

'2 2

2 / 2/r rr r

2 2

'2 '2

Material for 1- 2( ) 2Material for 3- 3( ) 3

2 (2) 1.3333

r rr r

12.7 Power in a Balanced System 12.7 Power in a Balanced System (5)(5)

Example 6Determine the total average power, reactive power, and complex power at the source and at the load.

12.7 Power in a Balanced System 12.7 Power in a Balanced System (6)(6)

Example 7A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW when the line voltage is 220 V and the line current is 18.2 A. Determine the power factor of the motor.

12.7 Power in a Balanced System 12.7 Power in a Balanced System (7)(7)Example 8

Two balanced loads are connected to a 240-kV rms 60-Hz line, as shown in the figure (a). Load 1 draws 30 kW at a power factor of 0.6 lagging, while load 2 draws 45 kVAR at a power factor 0.8 lagging. Assuming the abc sequence, determine: (a) the complex, real and reactive powers absorbed by the combined load, (b) the line currents, and(c) the kVAR rating of the three capacitors -connected in parallel with the load that will raise the power factor to 0.9 lagging and the capacitance of each capacitor.

12.10 Applications (1)12.10 Applications (1)• Three-Phase Power Measurement.

12.10 Applications (2)12.10 Applications (2)Consider the balanced Y-connected load

*1 Re[ ] cos( 30 )

cos( 30 )ab a ab a

L L

P V IV I

V I

*2 Re[ ] cos( 30 )

cos( 30 )cb c cb c

L L

P V IV I

V I

1 2 cos( 30 ) cos( 30 )

3 coscb c cb c

L L

P P V I V I

V I

1 2 cos( 30 ) cos( 30 )sincb c cb c

L L

P P V I V IV I

1 2

2 13( )T

T

P P P

Q P P

12.10 Applications (3)12.10 Applications (3)2 2

2 1

2 1

tan 3

T T T

T

T

S P QQ P PP P P

• If P2 = P1, the load is resistive.• If P2 > P1, the load is inductive.• If P2 < P1, the load is capacitive.

12.10 Applications (4)12.10 Applications (4)Example 9The two-wattmeter method produces wattmeter readings P1 = 1560 W and P2 = 2100 W when connected to a -connected load. If the line voltage is 220 V, calculate: (a) the per-phase average power, (b) the per-phase reactive power, (c) the power factor, and (d) the phase impedance.

12.10 Applications (5)12.10 Applications (5)Example 10The three-phase balanced load in the figure has impedance per phase of ZY = 8 + j6 . If the load is connected to 208-V lines, predict the readings of W1 and W2. Find PT and QT.