Alternating current three-phase circuits

(1)

11

Alternating current

three-phase

circuits

Unit 3. AC THREE-PHASE CIRCUITS

22

Unit 3. AC THREE-PHASE CIRCUITS

 Three-phase systems characteristics  Generation of three-phase voltages  Three-phase loads

 -Y and Y- transformation  Instantaneous power

 Three-phase power: S, P and Q

 Power measurement. Aaron connection  Power factor improvement

 Electrical measurements  Exercises

CONTENTS:

33

Unit 3. AC THREE-PHASE CIRCUITS

THREE-PHASE SYSTEM CHARACTERISTICS

44

THREE-PHASE SYSTEM CHARACTERISTICS

The electricity grid is made up of four main components:

 GENERATION: production of electricity from energy sources

such as coal, natural gas, hydropower, wind and solar.

 TRANSMISSION: the transmission system carries the electric

power from power plants over long distances to a distribution

system.

 DISTRIBUTION: the distribution system brings the power to the

customers.

 COSTUMERS: these are the consumers of electric power

(industry, service sector and residential uses).

(2)

55

THREE-PHASE SYSTEM CHARACTERISTICS



Instantaneous electric power has a sinusoidal shape with double

frequency than voltage or current.



SINGLE-PHASE AC CIRCUITS: instantaneous electric power is

negative twice a period (power flows from the load to the

generator) and positive twice a period, falling to zero.



BALANCED THREE-PHASE AC CIRCUITS: instantaneous

electric power is constant. Three-phase power never falls to

zero.



Three-phase electric motors perform better than single-phase AC

motors.



Three-phase power systems allow two voltage levels (L-L, L-N).



When electric power is transmitted, three-phase AC systems

require 25% less Cu/Al than single-phase AC systems.

66

GENERATION OF THREE-PHASE VOLTAGES



Three-phase generators contain three sinusoidal voltage sources

with voltages of the same frequency but a 120º-phase shift with

respect to each other.



This is achieved by positioning three coils separated by 120º

angles. There is only one rotor.



Amplitudes of the three phases are also equal.



The generator is then balanced.

77

INTRODUCTION

88

INTRODUCTION

R S T N

• N: neutral point

• R S T (or A B C) direct sequence or sequence RST

• V

_RS

, V

_ST

, V

_TR

: line voltages or line-to-line voltages

• V

_RN

, V

_SN

, V

_TN

: line-to-neutral voltages

• V

_line

=

₃

V

_{line-to-neutral} R S T N V V V V V V 120º TN SN RS TR ST RN vRN(t) = V0·cos(·t + 90º) V vSN(t) = V0·cos(·t - 30º) V v_TN(t) = V₀·cos(·t +210º) V

(3)

99

INTRODUCTION

50 Hz Usual system

V

_phase

= V

_{line-to-neutral} 230 volt

V

_line 400 volt Frequency 50 Hz    phase N to line line

3 .V

V



_ _ R S T N V V V V V V 120º TN SN RS TR ST RN 10 10

THREE-PHASE LOAD CLASSIFICATION

 WYE(two voltages) Balanced 3-wires 4-wires Unbalanced 3-wires 4-wires  DELTA(one voltage)

Balanced Unbalanced Z Z Z R N S T O=N IR IN IT IS Z Z Z R S T O=N R S T Z Z Z RS TR S R T ST I I I I I I R S T R S T Z Z Z RS TR S R T ST I I I I I I R S T RS ST TR 11 11

BALANCED WYE-CONNECTED LOAD

 The wye or star connection is made by connecting one end of each of the three-phase loads together.

 The voltage measured across a single load or phase is known as the phase voltage.

 The voltage measured between the lines is known as the line-to-line voltage or the line voltage.

 In a wye-connected system, the line voltage is higher than the load phase voltage by a factor of the square root of 3.

 In a wye-connected system, the phase current and line current are the same.

12 12

BALANCED DELTA-CONNECTED LOAD

 This connection received its name from the fact that a schematic diagram of it resembles the Greek letter delta ().

 In the delta connection, the line voltage and phase voltage in the load are the same.

 The line current of a delta connection is higher than the phase current by a factor of the square root of 3.

(4)

13 13

-Y TRANSFORMATION

R S T Z Z Z 1 2 3 13 23 12 Z Z Z 1 3 ₂ 23 13 12 13 12 1

· Z

Z





23 13 12 23 12 2

· Z

Z





23 13 12 23 13 3

· Z

Z





Balanced loads

: Z

_Y

= Z

_

/3

) ( ) ·( ) 1 ( 23 13 12 23 13 12 2 1 Z Z Z Z Z Z Z Z      Z between nodes 1 and 2:

) ( ) ·( ) 2 ( 23 13 12 23 12 13 3 1 Z Z Z Z Z Z Z Z      ) ( ) ·( ) 3 ( 23 13 12 13 12 23 3 2 Z Z Z Z Z Z Z Z     

From expressions (1), (2) and (3) the result is:

)

(

)

·(

23 13 12 23 13 12 2 , 1

Z





 Y:

Z

Y1,2



Z

1



Z

2  Z nodes 1-2: Z nodes 1-3: Z nodes 2-3: (1) + (2) - (3) (1) - (2) + (3) -(1) + (2) + (3) 14 14

BALANCED THREE/FOUR-WIRE WYE-CONNECTED LOAD

Z Z Z R N S T O=N I_R I_N IT I_S Z Z Z R S T O=N

0 )

I

(

I

Z

V

I

Z

V

I

Z

V

I

T S R N TN T SN S RN R











 



 





 



 



Q línia línia P línia línia º línia línia RN fase total

.sin

.I

.V

3 .cos

.I

.V

3 .I

.V

3 I

.

V

3. S

* R



j





 The three currents are balanced.  Thus the sum of them is always zero.  Since the neutral current of a balanced,

Y-connected three-phase load is always zero, the neutral conductor may be removed with no change in the results.

15 15

BALANCED THREE/FOUR-WIRE WYE-CONNECTED LOAD

Z Z Z R S T O=N

Example

A three-phase, RST system (400 V, 50 Hz), has a

three-wire Y-connected load for which Z = 10

30º

_{· Obtain the}

line currents and the complex power consumption·

A

3

40

10

3 400/

Z

V

I

60º 30º 90º RN R



A 3 40 10 3 400/ Z V I 60º 30º 30º SN S      A 3 40 10 3 400/ Z V I 180º 30º 210º TN T    (VAr) j8000 (watt) .41 3856 1 VA 000 6 1 ) 3 )·(40/ 3 3·(400/ I · V 3· S 3· S 30º * 60º 90º RN phase total *R        1watt 4 . 13856 º 30 ·cos 3 40 ·400· 3 ·cos ·I ·V 3 P_total _l _l



  VAr 8000 º 30 ·sin 3 40 ·400· 3 ·sin ·I ·V 3 Qtotal l l



  VA 16000 3 40 ·400· 3 ·I ·V 3 S_total _l _l  16 16

UNBALANCED FOUR-WIRE WYE-CONNECTED LOAD

Z Z Z R N S T O=N R S _T I_R I_N IS IT

0 )

I

(

I

Z

V

I

Z

V

I

Z

V

I

T S R N T TN T S SN S R RN R









* T * S * R

V

· I

V

· I

I

· V

S

total



RN



SN



TN

(5)

17 17

UNBALANCED THREE-WIRE WYE-CONNECTED LOAD

Z Z Z R S T O=N R S T IR IS I_T

0 I

I

R



S



T



T S R T TN S SN R RN T S R T TN S SN R RN ON

Y

· V

Y

· V

Y

· V

Z

1 Z

V

Z

V

Z

V









0 Z

V

Z

V

Z

V

T TO S SO R RO





0 Z

V

Z

V

Z

V

T ON TN S ON SN R ON RN













18 18

UNBALANCED THREE-WIRE WYE-CONNECTED LOAD

Z Z Z R S T O=N R S T IR IS I_T T ON TN T TO T S ON SN S SO S R ON RN R RO R

Z

V

Z

V

I

Z

V

Z

V

I

Z

V

Z

V

I

2)













* T * S * R

V

· I

V

· I

I

· V

S

3)

total



RO



SO



TO T S R T TN S SN R RN ON

Z

1 Z

V

Z

V

Z

V

)

1 





19 19

UNBALANCED THREE-WIRE WYE-CONNECTED LOAD

Example· A three-phase, RST system (400 V, 50 Hz), has a three-wire unbalanced Y-connected load for which Z_R= 100º_{, Z}

S= 100º and ZT= 1030º

· Obtain the line currents and the total complex power consumption.

O  N R S T IR IS IT ZR ZS ZT V 40.93 1 . 0 1 . 0 1 . 0 ·0.1 230 ·0.1 230 ·0.1 230 Y Y Y Y · V Y · V Y · V V 114.89º º 30 º 0 º 0 30º -210º 0º 30º -0º 90º T S R T TN S SN R RN ON             V_RO= V_RN–V_ON= 23090º_{- 40.93}114.89º_{= 193.64}84.90º_V V_SO= V_SN–V_ON= 230-30º_{- 40.93}114.89º_{= 264,54}-35.10º_V V_TO= V_TN–V_ON= 230210º_{- 40.93}114.89º_{= 237,18}-140.10º_V I_R= V_RO/Z_R= 193.6484.90º_/100º_{= 19,36}84.90º_A I_S= V_SO/Z_S= 264.54-35.10º_/100º_{= 26,45}-35.10º_A I_T= V_TO/Z_T= 237.18-140.10º_/1030º_{= 23.72}-170.10º_A

S_tot= V_RO·I_R* + V_SO·I_S* + V_TO·I_T* = 15619.56 W + j2812.72 VAr

1)

2)

3)

20 20

BALANCED DELTA-CONNECTED LOAD

R S T Z Z Z RS TR S R T ST I I I I I I R S T

Z

V

I

Z

V

I

Z

V

I

TR TR ST ST RS RS





 



 





 



 



Q línia línia P línia línia º línia línia RS fase total

·sin

·I

·V

3 ·cos

·I

·V

3 ·I

·V

3 I

· V

3·

S

3·

S

* RS



j





ST TR T RS ST S TR RS R

I













(6)

21 21

UNBALANCED DELTA-CONNECTED LOAD

R S T Z Z Z RS TR S R T ST I I I I I I R S T RS ST TR TR TR TR ST ST ST RS RS RS

Z

V

I

Z

V

I

Z

V

I



ST TR T RS ST S TR RS R

I













* TR * ST * RS

V

· I

V

· I

I

· V

S

total



RS



ST



TR 22 22

UNBALANCED THREE-WIRE

-CONNECTED LOAD

Example·A three-phase, RST system (400 V, 50 Hz), has an unbalanced -connected load for which Z_RS = 100º_{, Z}

ST = 1030º i ZTR= 10-30º · Obtain

the line currents and the complex power consumption·

R S T Z Z Z RS TR S R T ST I I I I I I R S T RS ST TR

A

40

10

400 Z

V

I

A

40

10

400 Z

V

I

A

40

10

400 Z

V

I

90º 30º 120º TR TR TR 30º 30º 0º ST ST ST 120º 0º 120º RS RS RS    



A

40.00 I

I

A

77.29 I

I

A

77.29 I

I

150º ST TR T 45º RS ST S 105º TR RS R  















43712.81VA

(VAr)

j0

)

43712.81(W

I

· V

I

· V

I

· V

S

* TR * ST * RS ST TR RS total











R S T N V V V V V V 120º TN SN RS TR ST RN 23 23

POWER MEASURMENT. Four-wire load

Unbalanced wye-connected, four-wire load

Z Z Z R S T O=N N W R S T W W R S T W_R= V_RN·I_R·cos( V_RN- I_R) WS= VSN·IS·cos( VSN - IS) WT= VTN·IT·cos( VTN - IT) Ptotal= WR+ WS+ WT Z Z Z R S T O=N N W

Balanced wye-connected, four-wire load

Ptotal= 3W W = VRN·IR·cos( VRN - IR)

24 24

POWER MEASURMENT. ARON CONNECTION

General 3-wire load. Two-wattmeter method (ARON connection)

R S T N V V V V V V 120º TN SN RS TR ST RN VRT VST VSR

)

º

30 cos(

V·I·

)

cos(

· ·I

V

W

)

º

30 cos(

V·I·

)

cos(

· ·I

V

W

S ST R RT I V S ST 2 I V R RT 1























)

cos(

30 º

)]

3 ·V·I·

cos

º

30 cos(

V·I·[

W

P

_TOTAL



₁



₂









, LOAD



)

cos(

º

)]

·V·I·

in

º

cos(

·V·I·[

]

W

·[W

Q

_TOTAL



3

₁



₂



3 

30 



30 



3 s

Demonstration done for a balanced 3-wire load

(7)

25 25

POWER MEASURMENT. BALANCED LOAD

, LOAD

)

W

3 arctg(

2 1 2 1











)

W

·(W

3 Q

W

P

2 1 total 2 1 total









Balanced load, general (Y/D, 3/4 wires). Two-wattmeter method (ARON connection)

,

LOAD

Unbalanced wye/delta-connected, three-wire load

P_total= W₁+ W₂ BALANCED

UNBALANCED

26 26

Aron cyclic permutations

POWER MEASURMENT: THE TWO-WATTMETER METHOD

W

3 Q

_TOT



W1 W2 V I V I RT R ST S SR S TR T TS T RS R W V I ST R TR S RS T

Q measurement: cyclic permutations

3 in · ·I V ) os(90º-· ·I V ) cos( · ·I V W TOTAL line line line line R ST Q s c IR VST         27 27

INSTANTANEOUS THREE-PHASE POWER

Single-phase load:

cosA·cosB = 0·5·[cos(A+B) + cos(A-B)] p(t) = v(t)·i(t) = V₀·cos(w·t + _V)·I₀·cos(wt + _) p(t) =1/2·V₀·I₀·cos(_V-_I) + 1/2·V₀·I₀·cos(2wt + _V+ _I) watt

Constant Oscillates at twice the mains frequency!

Three-phase wye balanced load:

p(t)= v_RN(t)·i_R(t) + v_SN(t)·i_S(t) + v_TN(t)·i_T(t) = =2V_p·cos(wt +_V)·2I_p·cos(wt +_)

+2V_p·cos(wt -120º+V)·2Ip·cos(wt -120º+) +2V_p·cos(wt +120º+V)·2Ip·cos(wt +120º+) =

V_p·I_p·cos(V-I) + Vp·Ip·cos(2wt +V+I) + V_p·I_p·cos(V-I) + Vp·Ip·cos(2wt -240º+V+I) + V_p·I_p·cos(_V-_I) + V_p·I_p·cos(2wt +240º+_V+_I)

= 3/2·V_p·I_p·cos(_V-_I) =3/2·V_p·I_p·cos = constant! ₂₈₂₈

INSTANTANEOUS POWER: SINGLE-PHASE LOAD

v(t)

i(t)

p(t)

(8)

29 29

INSTANTANEOUS THREE-PHASE POWER

v(t) i(t) p(t) pTOTAL= pR(t) + pS(t) + pT(t) 30 30

POWER LOSSES: THREE-PHASE/SINGLE PHASE

Single-phase line R1 R N LOAD R1 Three-phase line

Supposing same losses

1p 3p 3p 1p 2 1

S

2

1 S

S

2 R

2R







l





l





Single-phase line: 2 conductors of length l and section S1p

Three-phase line: 3 conductors of length l and section S_3p= 1/2S_1p As a result:

weight

_3p-cables

= 3/4weight

_1p-cables



2 2 2 load 1 2 1 losses ·cos V P · 2·R ·I 2·R P   V V



2 2 2 load 2 2 2 2 2 load 2 2 2 losses ·cos V P · R ·cos .V ) 3 ( P · 3·R ·I 3·R P     cos V P I load    cos ·V 3 P I load   31 31

Example 1

A Balanced load 1 capacitive C K C C U U1 ZL ZL ZL R S T _A 1 W1 Balanced load 2 inductive A2 W2

Three-phase balanced RST system for which A1= 1.633 A, A2= 5.773 A, W1= 6928.2 W, W2= 12000 W, U = 6000 V and Zline=4+j3· a) Obtain the complex power in the loads, as well as the ammeter A and the voltmeter U₁readings. b) Obtain the value of Cto improve the load’s PF to 1, assuming U = 6000 V.

48000VAr sin UI 3 Q º 12 . 53 cos UI 3 W 36000 W 3 P W 12000 cos UI 3 P º 45 sin UI 3 VAr 12000 Q VAr 12000 W 3 Q 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1                        VA 60000 6000 3 j 48000 ) Q j(Q ) P (P S 36.87º 2 1 2 1 2 1        A 774 . 5 773 . 5 633 . 1 I I I 1 2 90º 45º 90º 53.12º 90º 36.85º    _ _    TOTAL 6050V U 3 U V 97 . 3492 3 6000 ) 3 4 ( 774 . 5 U Z . I U 1 1,phase º 90 º 90 º 85 . 36 º 90 phase L , 1          j phase F 1.06 C ) .50.C 1/(2 (6000) -r -36000/3VA Q/3 Q 2 , 1    _     C R S T 32 32

Example 2

Three-phase 50 Hz system for which V = 400 V, W1= -8569.24 W, W2= -5286.36 W, AS = 21.56 A. Obtain a) the value of R. b) the reading of AR.c) the value of the inductance L 2 2 1 1 W ' W and W ' W            resultingin: R 23.095 R 400 2 W 13855.6 ' W ' W P a) 2 2 1 otal t A 32 . 17 095 . 23 400 R V I A 32 . 17 095 . 23 400 R V I b) 60º º 60 RT RT º 60 º 60 SR SR           A 30 I -I I : results It 90º SR RT R  L 90º L 180º º 60 L TS º 60 TS SR S X 400 j j15 66 . 8 X 400 32 . 17 jX V 32 . 17 I I I c)            H 0.2684 L : is result The 50L 2 84.308 X ) X 400 15 ( 8.66 A 21.56 I 2 _L L 2 S            AR R S T W1 R L W2 R AS AT V R S T R S T W₁ W₂ V I V I RT R ST S SR S TR T TS T RS R

(9)

33 33

Example 3

Varley phase-sequence indicator. Calculate the voltage in each element and deduce the practical consequences· Three-phase 400 V/50 Hz system·

V 171.55 5290 / 1 5290 / 1 3183 / 1 /5290 230 /5290 230 /3183 230 Y Y Y Y · V Y · V Y · V V 171.31º º 0 º 0 º 90 0º 210º 0º 30º -90º -90º T S R T TN S SN R RN ON             V_RO= V_RN–V_ON= 23090º_{- 171.55}171.31º₌_265.3450.28º_V V_SO= V_SN–V_ON= 230-30º_{- 171.55}171.31º₌_394.78-20.91º_V V_TO= V_TN–V_ON= 230210º_{- 171.55}171.31º₌_144.00-101.86ª_V

Conductor R is situated where the capacitor is placed, conductor S is situated where the brighter bulb is placed and T is the remaining conductor. C = 1F  X_C= 3183  R2bulbs= 2·V2/P = 2·2302/20 = 5290  R S T 0 _V 0N VR0 VT0 VS0 34 34

Example 4

A 400 V and 50 Hz three-phase line feeds two balanced loads through a line which has an internal impedance of ZL=0.5 + j·1· The -connected load has phase impedances whose values are 45+j·30 Ω, whereas the Y-connected load has phase impedances of 15–j·30 Ω· Determine: a) the reading of the ammeter A, b) the reading of the voltmeter V and c) the readings of watt-meters W1and W2.

R S T A 30 . 13 365 . 17 3 / 400 Z V I a) 82.875º º 125 . 7 º 90 TOT RN R    386.33V .7706) ·(13.30·16 3 ) ·(I·Z 3 V b)  //   ·1.154 3·13.30 ) W ·(W 3 Q ·16.731 3·13.30 W W P c) 2 2 1 LOAD 2 2 1 LOAD     

 The Aron connection results in:

W1= 4616,1 W, W2= 4262,5 W          Z (Z // Z ) Z (0.5 j·1) (16.731 j·1.154) 17.365 Z 7.125º // L Y L TOT Y Z W1 W2 V I V I RT R ST S SR S TR T TS T RS R W1 W2 ZL ZL ZL A V R S T 35 35

Example 5

Three-phase 400 V-50 Hz line. When switch K2is closed , WA= 4000 W. When K1and K3are closed, WA= 28352.6 W and WB= -11647.4 W. Determine: a) R2,b) R1and c) AT.

R S T j·34.64A 60 10 400 10 400 R V R V I I I º 60 º 180 1 RT 1 TS RT TS T1        R S T N WA AT WB K1 C1 R1 R1 R2 K2 K3 AS

a) K2closed: WA= VST·IS·cos(VST -IS)

4000 = 400·IS·cos(0º+30º)  IS= 11.55 A R2= VSN/IS= (400/3)/11.55 = 20  b) K₁and K₃closed: PTOT= W1+ W2= -WB+ WA= 40000 W = 2·4002/R1+ 4002/R2 R1= 10  c) K₁and K₃closed: j·0A 20 20 400 R V I º 180 2 TS T2    A 87.18 j·34.64 -80 j·0) (-20 j·34.64) 60 ( I I I -156.6º T2 T1 Ttotal          This results in AT= 87.18 A W1 W2 V I V I RT R ST S SR S TR T TS T RS R 36

Question 1

An electrical lineman is connecting three single-phase transformers in a Y(primary)-Y(secondary) configuration, for a business’s power service. Draw the connecting wires necessary between the transformer windings, and those required between the transformer terminals and the lines. Note: fuses have been omitted from this illustration for simplicity.

(10)

37

Question 2

Identify the primary-secondary connection configuration of these pole-mounted power transformers (i.e. Y-Y, Y-Delta, Delta-Y, etc.).

HV

LV

R S

T

These transformers are connected in a Yy configuration.

38

Question 3

These transformers are connected in a Yd configuration.

39

Question 3

These transformers are connected in open-delta configuration.

 Three single-phase transformers are not normally used because this is more expensive than using one three-phase transformer.

 However, there is an advantageous method called the open-Delta or

V-connection

 It functions as follows: a defective single-phase transformer in a Dd three-phase bank can be removed for repair. Partial service can be restored using the open-Delta configuration until a replacement transformer is obtained.

 Three-phase is still obtained with two transformers, but at 57.7% of the original power.

 This is a very practical transformer application for emergency conditions.

40

Question 4

One of the conductors connecting the secondary of a three-phase power distribution transformer to a large office building fails when open. Upon inspection, the source of the failure is obvious: the wire overheated at a point of contact with a terminal block, until it physically separated from the terminal. What is strange is that the overheated wire is the neutral conductor, not any one of the ”line” conductors. Based on this observation, what do you think caused the failure?

After repairing the wire, what would you do to verify the cause of the failure?

Here’s a hint (“pista”): if you were to repair the neutral wire and take current measurements with a digital instrument (using a clamp-on current probe, for safety), you would find that the predominant frequency of the current is 150 Hz, rather than 50 Hz. This scenario is all too common in modern power systems, as non-linear loads such as

switching power supplies and electronic power controls have become more prevalent. Special instruments exist to measure harmonics in power systems, but a simple DMM (digital multimeter) may be used as well to make crude assessments.

(11)

41 41

POWER MEASURMENT. ARON CONNECTION

General 3-wire load. Two-wattmeter method (ARON connection)

p(t) = vRN(t)·iR(t) + vSN(t)·iS(t) + vTN(t)·iT(t) p(t) = v_RN(t)·i_R(t) + v_SN(t)·i_S(t) + v_TN(t)·[-i_R(t) - i_S(t)]

p(t) = i_R(t)·[v_RN(t) - v_TN(t)] + i_S(t)·[v_SN(t) - v_TN(t)] = v_RT(t)·i_R(t) + v_ST(t)·i_S(t) Mean value

P_total= W₁+ W₂= V_RT·I_R·cos(V_RT-I_R) + V_ST·I_S·cos(V_ST-I_S)

,