Chapter 12: Three-Phase Circuits 12.1 Introduction 12.2 Balanced ...
Chapter 12: Three-Phase Circuits 12.1 12.2 12.3 12.4 12.7
Introduction Balanced Three-Phase Voltages Balanced Wye-Wye connection Balanced Wye-Delta Connection Power in a Balanced System
12.1 INTRODUCTION A single-phase ac power system consists of a generator connected through a pair of wires (i.e., a transmission line) to a load as shown in figure below:
Circuits or systems in which the ac sources operate at the same frequency but different phases are known as polyphase. The following figures below shows a two-phase three-wire system and a three-phase four-wire system.
As distinct from a single-phase system, a two-phase system is produced by a generator consisting of two coils placed perpendicular to each other so that the voltage generated lags the other by 90º. Similarly, a three-phase system is produced by a generator consisting of three sources having a same amplitude and frequency but out of phase with each other by 120º. Since the three-phase system is by far the most prevalent and most economical polyphase system, hence all the discussion in this chapter will be mainly on three-phase systems.
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Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
Three-phase systems are important for several reasons: • Nearly all electric power is generated and distributed in three-phase at the operating frequency of 60Hz or ω=377rad/s (i.e., US) or 50Hz or ω=314rad/s (i.e., East Japan). • When single-phase or two-phase inputs are required, they can be taken from the threephase system rather than generated independently. If more phases like 48 phases are required for industry usage, three-phase supplied can be manipulated to provide 48 phases • The instantaneous power in a three-phase system can be constant (i.e., not pulsating).This results in uniform power transmission and less vibration of three-phase machines. • For the same amount of power, the three-phase system is more economical than the single-phase system. • The amount of wire required for a three-phase system is less than that required for an equivalent single-phase system.
12.2 BALANCED THREE_PHASE VOLTAGES This section will begin with a discussion of balanced three-phase voltages. Three-phase voltages are often produced with a three-phase ac generator (or alternator) whose cross-sectional view is shown below.
The generator basically consists of a rotating magnet called the rotor surrounded by a stationary winding called the stator. Three separate windings or coils with terminal a-a’, b-b’ and c-c’ are physically placed 120º apart around the stator. As the rotor rotates, its magnetic field ‘cuts’ the flux from the three coils and induces voltages in the coils. Since the coils are placed 120º apart, the induced voltages in the coils are equal in magnitude but out of phase by 120º shown below.
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Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
Since each coil can be regarded as a single-phase generator by itself, the three-phase generator can supply power to both single-phase and three-phase loads. A 3-phase voltage system is equivalent to three single-phase systems. The voltage sources (generator connection) can be connected in wye-connected shown in (a) or delta-connected as in (b)
The voltages Van, Vbn and Vcn are between lines a, b and c and the neutral line, n respectively. Van, Vbn and Vcm are called phase voltages. It is called balanced if they are same in magnitude and frequency and are out of phase from each other by 120º such that V an + V bn + V cn = 0 ….. (12.1) Van = Vbn = Vcn ….. (12.2) There are two possible combinations; positive sequence and negative sequence
Positive/ abc Sequence (Balanced) ° Let Van = V p ∠0 vector diagram for phase abc
Negative / acb Sequence (Balanced) ° Let Van = V p ∠0 vector diagram for phase acb
phase sequence gives
phase sequence gives
Assuming Van as the reference hence
V an = V p ∠0 and Vp = Vp ∠θ °
°
Where V p =amplitude of phase voltage in rms Thus;
V an = V p ∠0°
Where V p =amplitude of phase voltage in rms
V an = V p∠0°
= Vp ∠θ °
V bn = V an∠− 240°
= Vp ∠(θ ° −120° ) = Vp ∠(θ ° + 240° ) ….(12.3) V cn = V an∠ − 240°
= Vp ∠(θ ° − 240° ) = Vp ∠(θ ° + 120° ) …. (12.4) V cn = V an∠−120°
= Vp ∠(θ ° − 240° ) = Vp ∠(θ ° + 120° ) This sequence is produced when the rotor rotates in the counterclockwise direction.
V an = V p ∠0° and Vp = Vp ∠θ ° Thus;
= Vp ∠θ °
V bn = V an∠ −120°
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Assuming Van as the reference hence
= Vp ∠(θ ° −120° ) = Vp ∠(θ ° + 240° ) This sequence is produced when the rotor rotates in the clockwise direction.
Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
Mathematically, both phase sequences in Eq. (12.3) and Eq. (12.4) satisfy Eq. (12.1). For example, from Eq. (12.4)
V an +V bn +V cn = Vp ∠θ ° + Vp ∠(θ ° − 240° ) + Vp ∠(θ ° −120° ) = Vp ∠θ ° (1∠0° +1∠− 240° +1∠−120° )
= Vp ∠θ ° (1+ (−0.5 + j0.866) + (−0.5 − j0.866)) = 0 Like the generator connections, loads can also be connected in Y or ∆. Balanced load is where the phase impedances are equal in magnitude and are in phase
Y-connected Load (Balanced) Three-phase load configuration:
∆-connected Load (Balanced) Three-phase load configuration:
For balanced Y-connected load:
Z1 = Z 2 = Z 3 = Z Y
….. (12.5) Where ZY is the Total load impedance per phase
For balanced Y-connected load:
Za = Zb = Zc = ZΔ
….. (12.6) Where Z Δ is the Total load impedance per phase Recall from earlier chapter 9, Eq. (9.69) that
1 Z Δ …..(12.7) 3 Hence, Y-connected load can be transformed into a ∆-connected load or vice versa using Eq. (12.7)
Z Δ = 3 Z Y
or
ZY =
Since both sources and loads can be connected in either Y or ∆, there are 4 possible connections: • Y-Y connection (i.e., Y-connected source with Y-connected load) • Y-∆ connection (i.e., Y-connected source with ∆-connected load) • ∆-∆ connection (i.e., ∆-connected source with ∆-connected load) • ∆-Y connection (i.e., ∆-connected source with Y-connected load)
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Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
12.3 BALANCED WYE-WYE CONNECTION The balanced four-wire Y-Y system shown below has a Y-connected source on the left and Yconnected load on the right. Assume a balanced load (i.e., impedances are equal).
Z Y = Total load impedance per phase, Ω/phase Z s = Source impedance per phase, Ω/phase Z l = Line impedance per phase, Ω/phase Z L = Load impedance per phase, Ω/phase Z n = impedance of the neutral line per phase, Ω/phase Thus in general
ZY = Z s + Zl + Z L
….. (12.8)
Z s and Z l are often very small compared with Z L therefore, one can assume that Z Y = Z L if no source or line impedance is given. By lumping the impedances together, the Y-Y system shown above can be simplified as below
Voltages: Assuming the source rotates in positive sequence, the phase voltages (i.e., line-to-neutral voltages) are (from Eq. (12.3)) where V p = V p ∠θ °
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V an = V p∠0°
V an = Vp ∠θ ° ,
V an = V p∠−120° or
V bn = Vp ∠(θ ° −120° ) = Vp ∠(θ ° + 240° )
V an = V p∠+120°
V cn = Vp ∠(θ ° − 240° ) = Vp ∠(θ ° +120° )
Chapter 12: Three-Phase Circuits
….. (12.9)
Dr. C.S.Tan
From the phasor diagram, the line voltages (i.e., line to line voltages) V ab , V bc and V ca are related to the phase voltages. For example, ° V ab = V an + V nb = V an − V bn = Vp ∠θ ° − Vp ∠ (θ − 120)
(
)
= Vp ∠θ ° (1∠0° − 1∠ −120° ) = Vp ∠θ ° (1 − (−0.5 − j 0.866) )
(
)
(
= Vp ∠θ ° (1.5 + j 0.866) = Vp ∠θ ° = 3 Vp ∠ (θ + 30) =
(
3∠30°
V bc = V bn V ca = V cn
( ( (
) (V )
) ( 3∠30 ) = ( V 3∠30 ) = ( V
)(
3∠30°
)( + 120 ) ) (
p
∠(θ ° − 120° )
p
∠(θ °
°
°
)
( proved )
an
3∠30° = Vp ∠θ °
3∠30°
°
Similarly, we can get Vbc and Vca as follows:
V ab = V an
)(
°
Note: The magnitude of the line voltages VL is
)
= 3 Vp ∠(θ ° + 30° )
) 3∠30 )
3∠30°
= 3 Vp ∠(θ ° − 90° )
°
= 3 Vp ∠(θ ° + 150° )
….. (12.10)
3 times the magnitude of the phase voltages V p
and the line voltages lead their corresponding phase voltage by 30° (See phasor diagram below) VL = 3V p Where
Vp = V an = V bn = V cn and VL = V ab = V bc = V ca
From Phasor diagram (Positive Sequence)
V
ab
= V
an
+V
nb
V
bc
= V
bn
+V
nc
V
ca
= V
cn
+V
na
….. (12.11)
Phase Voltages:
V an = Vp ∠θ ° V bn = V an∠−120° = V an∠+ 240° = Vp ∠(θ ° −120° ) = Vp ∠(θ ° + 240° ) ….. (12.12) V cn = V an∠− 240° = V an∠+120° = Vp ∠(θ ° − 240° ) = Vp ∠(θ ° +120° )
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Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
Currents: Applying KVL to each phase, we obtain the line currents as:
Balanced Y-Y connection:
V an ZY V bn V an∠−120° = = = I a∠ −120° ….. (12.12) I b ZY ZY V cn V an∠+ 120° = = I a ∠+ 120° Ic = ZY ZY Ia =
And (See phasor diagram on the left)
Ia + Ib + Ic = 0 Line Current, I a = Phase Current, I AB . Line Current = Current in each line Phase Current= Current in each phase of the source or load
(
)
I n = − I a + I b + I c = 0 ….. (12.13)
or
V nN = Z n I n = 0
This implies that the voltage across the neutral wire is equal to zero. Therefore the neutral line can be removed without affecting the system (balanced Y-Y). Since it is a balanced Y-Y system, the system can be represented or analyzed using a single phase equivalent circuit with or without neutral line (see below)
. The single-phase analysis yields the line current, I a as I a
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Chapter 12: Three-Phase Circuits
=
V an ZY
….. (12.14)
Dr. C.S.Tan
12.4 BALANCED WYE-DELTA CONNECTION Balanced Y-∆ connection is a 3-phase system with balanced Y-connected sources supplying ∆connected load. From the figure above, there is no neutral connection from source to load in this type of connection. Assuming positive sequence, the phase voltages (source) are (i.e., Eq. (12.9)) Balanced Y-∆ connection: V an = Vp ∠θ ° , V bn = Vp ∠(θ −120 ) = Vp ∠(θ + 240 ) °
°
°
°
V cn = Vp ∠(θ ° − 240° ) = Vp ∠(θ ° +120° ) As shown and derive in Section 12.3, the line voltages across the source are (i.e., Eq. (12.10)) V ab = 3 Vp ∠(θ ° + 30° ) V ab = 3 Vp ∠(θ ° + 30° )
V bc = 3 Vp ∠(θ ° − 90° ) Phasor Diagram: Relationship between phase and line voltages
or
V ca = 3 Vp ∠(θ ° +150° )
V bc = V ab∠−120° V ca = V ab∠+120°
Looking at the balanced Y-∆ connection, it shows that the line voltages (at the source) are equal to the voltage across the load impedances. Assuming positive sequence, V ab = V AB = 3V an ∠30° V bc = V BC = 3V bn ∠30° = 3V an ∠ − 90° ….. (12.15) V ca = V CA = 3V cn ∠30° = 3V an ∠ + 150°
From these voltages (at the load), we can obtain the phase currents as V AB V BC V CA I AB = , I BC = , I CA = ….. (12.16) Z+ Z+ Z + Note: Using KVL, the phase currents can be determined. For example, applying KVL around loop aABbna to find phase currents
I AB
Thus, taking VAB as the reference
I BC I CA
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V AB ∠0° Z+ V BC V AB ∠ − 120° = = = I AB ∠ − 120° ….. (12.17) Z+ Z+ V CA V AB ∠ + 120° = = = I AB ∠ + 120° Z+ Z+
I AB =
Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
To obtain the line currents, I a , I b , and I c . Applying KCL at nodes A, B, and C we obtain
Balanced ∆ connection at the load
At node A : I a = I AB − I CA At node B : I b = I BC − I AB ….. (12.18)
At nodeC : I c = I CA − I BC Thus
I a = I AB − I CA = I AB (1 − 1∠ − 240° ) = I AB (1 + 0.5 − j 0.866)
= 3 I AB ∠ − 30
Note: I L = line current , I a , I b , I c
°
I p = phase current , I AB , I BC , I CA
Phasor Diagram: Relationship between phase and line currents
Similarly,
I b , and I c , gives
I b = 3 I BC ∠ − 30° I c = 3 I CA∠ − 30°
Generally,
I L = 3I p ….. (12.19)
IL = I a = I b = I c
and
I = I AB = I BC = I CA p
Note: The magnitude of the line current I L is 3 times the magnitude of the phase current I p and the line currents lag their corresponding phase currents by 30° (See phasor diagram beside) Another way to analyze the Y-∆ circuit is by transforming the ∆-connected load to an equivalent Y-connected load. Using the transformation formula in Eq. (12.7), we will have a Y-Y system. Hence, the three-phase Y-∆ system can be replaced by a single-phase equivalent circuit as shown below. where
Z
Y
=
Z + 3
….. (12.20)
This will allows us to calculate the line currents. The phase currents are then obtained using Eq. (12.19) with the fact that each of the phase currents leads the corresponding line current by 30º.
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Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
12.7 POWER IN A BALANCED SYSTEM The total instantaneous power in the load is the sum of instantaneous powers in the three phases: that is
p = pa + pb + pc = vAN ia + vBN ib + vCN ic =
(
2Vp cos ωt +
(
)(
) (
2I p cos(ωt − θ ) +
2Vp cos(ωt + 120° )
)(
2Vp cos(ωt − 120° )
2I p cos(ωt − θ + 120° )
)
)(
)
2I p cos(ωt − θ − 120° )
Applying trigonometric identity, gives (If you are interested to derive please referred to pp.520)
p = 3Vp I p cos θ
….. (12.21)
where Vp = phase voltage I p = phase current
θ = angle of the load impedance = angle between the phase voltage and the phase current From Eq. (12.21), one can notice that the total instantaneous power in a balanced three-phase system is constant. In other words, it does not change with time as one can see in the instantaneous power of each phase. This result is true for both Y and ∆-connected loads.
Since the total instantaneous power is independent of time, the average power per phase Pp either for ∆-connected load or Y-connected load is p / 3 , yields Pp = p / 3 = V p I p cos θ
….. (12.22)
Q p = V p I p sin θ
….. (12.23)
S p = Vp I p
….. (12.24)
And the average reactive power per phase is The average apparent power per phase is Total real power in a balanced 3-phase system is:
P = Pa + Pb + Pc = 3V p I p cos θ
….. (12.25)
Total reactive power in a balanced 3-phase system is: Q = Qa + Qb + Qc = 3V p I p sin θ ….. (12.26) Total apparent/complex power in a balanced 3-phase system is: ∗ p
S = S a + S b + S c = 3V p I = 3I Z p = 2 p
3V p2 ∗
….. (12.27)
Zp
where Z p = Z p ∠φ ° = load impedance per phase Alternatively, we may write Eq. (12.27) as
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S = P + jQ
Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
Power Balance Y-Y Connection
Power Balanced for Y-∆ Connection
Assuming abc sequence V L = 3V p∠30°
Assuming abc sequence V L = 3V p∠30°
I L = 3I p∠− 30°
I L = 3I p∠− 30°
power absorbed by the load is ∗
∗
∗
S1φ = V p I p = I p ZL I p = ZL I p
power absorbed by the load is ∗
S1φ = V p I p = I p ZL I p = ZL I p
2
2 (12.28) S 3φ = 3S 1φ = 3 I L Z L .....
Or the above 3-phase can be represented as per phase diagram since they are balanced
= ZL
2
2
1 2 = ZL IL 3 3
Ip
2 S 3φ = 3S 1φ = Z L I L ..... (12.29)
Or the above 3-phase can be represented as per phase diagram since they are balanced
Where
Where
V p =Van
From the single-phase one-line diagram above, power absorbed by the load is S 1φ = Z L I L
2
S 3φ = 3S 1φ S 3φ = 3Z L I L
Where
IL = Ia
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2
V p =Van
2
= 3Z L I a
From the single-phase one-line diagram above, power absorbed by the load is Z 2 S 1φ = L I L 3 S 3φ = 3S 1φ 2
2
S 3φ = I L Z L = I a Z L
Where
IL = Ia
Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
Introduction Balanced Three-Phase Voltages Balanced Wye-Wye connection Balanced Wye-Delta Connection Power in a Balanced System
12.1 INTRODUCTION A single-phase ac power system consists of a generator connected through a pair of wires (i.e., a transmission line) to a load as shown in figure below:
Circuits or systems in which the ac sources operate at the same frequency but different phases are known as polyphase. The following figures below shows a two-phase three-wire system and a three-phase four-wire system.
As distinct from a single-phase system, a two-phase system is produced by a generator consisting of two coils placed perpendicular to each other so that the voltage generated lags the other by 90º. Similarly, a three-phase system is produced by a generator consisting of three sources having a same amplitude and frequency but out of phase with each other by 120º. Since the three-phase system is by far the most prevalent and most economical polyphase system, hence all the discussion in this chapter will be mainly on three-phase systems.
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Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
Three-phase systems are important for several reasons: • Nearly all electric power is generated and distributed in three-phase at the operating frequency of 60Hz or ω=377rad/s (i.e., US) or 50Hz or ω=314rad/s (i.e., East Japan). • When single-phase or two-phase inputs are required, they can be taken from the threephase system rather than generated independently. If more phases like 48 phases are required for industry usage, three-phase supplied can be manipulated to provide 48 phases • The instantaneous power in a three-phase system can be constant (i.e., not pulsating).This results in uniform power transmission and less vibration of three-phase machines. • For the same amount of power, the three-phase system is more economical than the single-phase system. • The amount of wire required for a three-phase system is less than that required for an equivalent single-phase system.
12.2 BALANCED THREE_PHASE VOLTAGES This section will begin with a discussion of balanced three-phase voltages. Three-phase voltages are often produced with a three-phase ac generator (or alternator) whose cross-sectional view is shown below.
The generator basically consists of a rotating magnet called the rotor surrounded by a stationary winding called the stator. Three separate windings or coils with terminal a-a’, b-b’ and c-c’ are physically placed 120º apart around the stator. As the rotor rotates, its magnetic field ‘cuts’ the flux from the three coils and induces voltages in the coils. Since the coils are placed 120º apart, the induced voltages in the coils are equal in magnitude but out of phase by 120º shown below.
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Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
Since each coil can be regarded as a single-phase generator by itself, the three-phase generator can supply power to both single-phase and three-phase loads. A 3-phase voltage system is equivalent to three single-phase systems. The voltage sources (generator connection) can be connected in wye-connected shown in (a) or delta-connected as in (b)
The voltages Van, Vbn and Vcn are between lines a, b and c and the neutral line, n respectively. Van, Vbn and Vcm are called phase voltages. It is called balanced if they are same in magnitude and frequency and are out of phase from each other by 120º such that V an + V bn + V cn = 0 ….. (12.1) Van = Vbn = Vcn ….. (12.2) There are two possible combinations; positive sequence and negative sequence
Positive/ abc Sequence (Balanced) ° Let Van = V p ∠0 vector diagram for phase abc
Negative / acb Sequence (Balanced) ° Let Van = V p ∠0 vector diagram for phase acb
phase sequence gives
phase sequence gives
Assuming Van as the reference hence
V an = V p ∠0 and Vp = Vp ∠θ °
°
Where V p =amplitude of phase voltage in rms Thus;
V an = V p ∠0°
Where V p =amplitude of phase voltage in rms
V an = V p∠0°
= Vp ∠θ °
V bn = V an∠− 240°
= Vp ∠(θ ° −120° ) = Vp ∠(θ ° + 240° ) ….(12.3) V cn = V an∠ − 240°
= Vp ∠(θ ° − 240° ) = Vp ∠(θ ° + 120° ) …. (12.4) V cn = V an∠−120°
= Vp ∠(θ ° − 240° ) = Vp ∠(θ ° + 120° ) This sequence is produced when the rotor rotates in the counterclockwise direction.
V an = V p ∠0° and Vp = Vp ∠θ ° Thus;
= Vp ∠θ °
V bn = V an∠ −120°
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Assuming Van as the reference hence
= Vp ∠(θ ° −120° ) = Vp ∠(θ ° + 240° ) This sequence is produced when the rotor rotates in the clockwise direction.
Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
Mathematically, both phase sequences in Eq. (12.3) and Eq. (12.4) satisfy Eq. (12.1). For example, from Eq. (12.4)
V an +V bn +V cn = Vp ∠θ ° + Vp ∠(θ ° − 240° ) + Vp ∠(θ ° −120° ) = Vp ∠θ ° (1∠0° +1∠− 240° +1∠−120° )
= Vp ∠θ ° (1+ (−0.5 + j0.866) + (−0.5 − j0.866)) = 0 Like the generator connections, loads can also be connected in Y or ∆. Balanced load is where the phase impedances are equal in magnitude and are in phase
Y-connected Load (Balanced) Three-phase load configuration:
∆-connected Load (Balanced) Three-phase load configuration:
For balanced Y-connected load:
Z1 = Z 2 = Z 3 = Z Y
….. (12.5) Where ZY is the Total load impedance per phase
For balanced Y-connected load:
Za = Zb = Zc = ZΔ
….. (12.6) Where Z Δ is the Total load impedance per phase Recall from earlier chapter 9, Eq. (9.69) that
1 Z Δ …..(12.7) 3 Hence, Y-connected load can be transformed into a ∆-connected load or vice versa using Eq. (12.7)
Z Δ = 3 Z Y
or
ZY =
Since both sources and loads can be connected in either Y or ∆, there are 4 possible connections: • Y-Y connection (i.e., Y-connected source with Y-connected load) • Y-∆ connection (i.e., Y-connected source with ∆-connected load) • ∆-∆ connection (i.e., ∆-connected source with ∆-connected load) • ∆-Y connection (i.e., ∆-connected source with Y-connected load)
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Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
12.3 BALANCED WYE-WYE CONNECTION The balanced four-wire Y-Y system shown below has a Y-connected source on the left and Yconnected load on the right. Assume a balanced load (i.e., impedances are equal).
Z Y = Total load impedance per phase, Ω/phase Z s = Source impedance per phase, Ω/phase Z l = Line impedance per phase, Ω/phase Z L = Load impedance per phase, Ω/phase Z n = impedance of the neutral line per phase, Ω/phase Thus in general
ZY = Z s + Zl + Z L
….. (12.8)
Z s and Z l are often very small compared with Z L therefore, one can assume that Z Y = Z L if no source or line impedance is given. By lumping the impedances together, the Y-Y system shown above can be simplified as below
Voltages: Assuming the source rotates in positive sequence, the phase voltages (i.e., line-to-neutral voltages) are (from Eq. (12.3)) where V p = V p ∠θ °
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V an = V p∠0°
V an = Vp ∠θ ° ,
V an = V p∠−120° or
V bn = Vp ∠(θ ° −120° ) = Vp ∠(θ ° + 240° )
V an = V p∠+120°
V cn = Vp ∠(θ ° − 240° ) = Vp ∠(θ ° +120° )
Chapter 12: Three-Phase Circuits
….. (12.9)
Dr. C.S.Tan
From the phasor diagram, the line voltages (i.e., line to line voltages) V ab , V bc and V ca are related to the phase voltages. For example, ° V ab = V an + V nb = V an − V bn = Vp ∠θ ° − Vp ∠ (θ − 120)
(
)
= Vp ∠θ ° (1∠0° − 1∠ −120° ) = Vp ∠θ ° (1 − (−0.5 − j 0.866) )
(
)
(
= Vp ∠θ ° (1.5 + j 0.866) = Vp ∠θ ° = 3 Vp ∠ (θ + 30) =
(
3∠30°
V bc = V bn V ca = V cn
( ( (
) (V )
) ( 3∠30 ) = ( V 3∠30 ) = ( V
)(
3∠30°
)( + 120 ) ) (
p
∠(θ ° − 120° )
p
∠(θ °
°
°
)
( proved )
an
3∠30° = Vp ∠θ °
3∠30°
°
Similarly, we can get Vbc and Vca as follows:
V ab = V an
)(
°
Note: The magnitude of the line voltages VL is
)
= 3 Vp ∠(θ ° + 30° )
) 3∠30 )
3∠30°
= 3 Vp ∠(θ ° − 90° )
°
= 3 Vp ∠(θ ° + 150° )
….. (12.10)
3 times the magnitude of the phase voltages V p
and the line voltages lead their corresponding phase voltage by 30° (See phasor diagram below) VL = 3V p Where
Vp = V an = V bn = V cn and VL = V ab = V bc = V ca
From Phasor diagram (Positive Sequence)
V
ab
= V
an
+V
nb
V
bc
= V
bn
+V
nc
V
ca
= V
cn
+V
na
….. (12.11)
Phase Voltages:
V an = Vp ∠θ ° V bn = V an∠−120° = V an∠+ 240° = Vp ∠(θ ° −120° ) = Vp ∠(θ ° + 240° ) ….. (12.12) V cn = V an∠− 240° = V an∠+120° = Vp ∠(θ ° − 240° ) = Vp ∠(θ ° +120° )
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Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
Currents: Applying KVL to each phase, we obtain the line currents as:
Balanced Y-Y connection:
V an ZY V bn V an∠−120° = = = I a∠ −120° ….. (12.12) I b ZY ZY V cn V an∠+ 120° = = I a ∠+ 120° Ic = ZY ZY Ia =
And (See phasor diagram on the left)
Ia + Ib + Ic = 0 Line Current, I a = Phase Current, I AB . Line Current = Current in each line Phase Current= Current in each phase of the source or load
(
)
I n = − I a + I b + I c = 0 ….. (12.13)
or
V nN = Z n I n = 0
This implies that the voltage across the neutral wire is equal to zero. Therefore the neutral line can be removed without affecting the system (balanced Y-Y). Since it is a balanced Y-Y system, the system can be represented or analyzed using a single phase equivalent circuit with or without neutral line (see below)
. The single-phase analysis yields the line current, I a as I a
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Chapter 12: Three-Phase Circuits
=
V an ZY
….. (12.14)
Dr. C.S.Tan
12.4 BALANCED WYE-DELTA CONNECTION Balanced Y-∆ connection is a 3-phase system with balanced Y-connected sources supplying ∆connected load. From the figure above, there is no neutral connection from source to load in this type of connection. Assuming positive sequence, the phase voltages (source) are (i.e., Eq. (12.9)) Balanced Y-∆ connection: V an = Vp ∠θ ° , V bn = Vp ∠(θ −120 ) = Vp ∠(θ + 240 ) °
°
°
°
V cn = Vp ∠(θ ° − 240° ) = Vp ∠(θ ° +120° ) As shown and derive in Section 12.3, the line voltages across the source are (i.e., Eq. (12.10)) V ab = 3 Vp ∠(θ ° + 30° ) V ab = 3 Vp ∠(θ ° + 30° )
V bc = 3 Vp ∠(θ ° − 90° ) Phasor Diagram: Relationship between phase and line voltages
or
V ca = 3 Vp ∠(θ ° +150° )
V bc = V ab∠−120° V ca = V ab∠+120°
Looking at the balanced Y-∆ connection, it shows that the line voltages (at the source) are equal to the voltage across the load impedances. Assuming positive sequence, V ab = V AB = 3V an ∠30° V bc = V BC = 3V bn ∠30° = 3V an ∠ − 90° ….. (12.15) V ca = V CA = 3V cn ∠30° = 3V an ∠ + 150°
From these voltages (at the load), we can obtain the phase currents as V AB V BC V CA I AB = , I BC = , I CA = ….. (12.16) Z+ Z+ Z + Note: Using KVL, the phase currents can be determined. For example, applying KVL around loop aABbna to find phase currents
I AB
Thus, taking VAB as the reference
I BC I CA
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V AB ∠0° Z+ V BC V AB ∠ − 120° = = = I AB ∠ − 120° ….. (12.17) Z+ Z+ V CA V AB ∠ + 120° = = = I AB ∠ + 120° Z+ Z+
I AB =
Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
To obtain the line currents, I a , I b , and I c . Applying KCL at nodes A, B, and C we obtain
Balanced ∆ connection at the load
At node A : I a = I AB − I CA At node B : I b = I BC − I AB ….. (12.18)
At nodeC : I c = I CA − I BC Thus
I a = I AB − I CA = I AB (1 − 1∠ − 240° ) = I AB (1 + 0.5 − j 0.866)
= 3 I AB ∠ − 30
Note: I L = line current , I a , I b , I c
°
I p = phase current , I AB , I BC , I CA
Phasor Diagram: Relationship between phase and line currents
Similarly,
I b , and I c , gives
I b = 3 I BC ∠ − 30° I c = 3 I CA∠ − 30°
Generally,
I L = 3I p ….. (12.19)
IL = I a = I b = I c
and
I = I AB = I BC = I CA p
Note: The magnitude of the line current I L is 3 times the magnitude of the phase current I p and the line currents lag their corresponding phase currents by 30° (See phasor diagram beside) Another way to analyze the Y-∆ circuit is by transforming the ∆-connected load to an equivalent Y-connected load. Using the transformation formula in Eq. (12.7), we will have a Y-Y system. Hence, the three-phase Y-∆ system can be replaced by a single-phase equivalent circuit as shown below. where
Z
Y
=
Z + 3
….. (12.20)
This will allows us to calculate the line currents. The phase currents are then obtained using Eq. (12.19) with the fact that each of the phase currents leads the corresponding line current by 30º.
EEEB123
Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
12.7 POWER IN A BALANCED SYSTEM The total instantaneous power in the load is the sum of instantaneous powers in the three phases: that is
p = pa + pb + pc = vAN ia + vBN ib + vCN ic =
(
2Vp cos ωt +
(
)(
) (
2I p cos(ωt − θ ) +
2Vp cos(ωt + 120° )
)(
2Vp cos(ωt − 120° )
2I p cos(ωt − θ + 120° )
)
)(
)
2I p cos(ωt − θ − 120° )
Applying trigonometric identity, gives (If you are interested to derive please referred to pp.520)
p = 3Vp I p cos θ
….. (12.21)
where Vp = phase voltage I p = phase current
θ = angle of the load impedance = angle between the phase voltage and the phase current From Eq. (12.21), one can notice that the total instantaneous power in a balanced three-phase system is constant. In other words, it does not change with time as one can see in the instantaneous power of each phase. This result is true for both Y and ∆-connected loads.
Since the total instantaneous power is independent of time, the average power per phase Pp either for ∆-connected load or Y-connected load is p / 3 , yields Pp = p / 3 = V p I p cos θ
….. (12.22)
Q p = V p I p sin θ
….. (12.23)
S p = Vp I p
….. (12.24)
And the average reactive power per phase is The average apparent power per phase is Total real power in a balanced 3-phase system is:
P = Pa + Pb + Pc = 3V p I p cos θ
….. (12.25)
Total reactive power in a balanced 3-phase system is: Q = Qa + Qb + Qc = 3V p I p sin θ ….. (12.26) Total apparent/complex power in a balanced 3-phase system is: ∗ p
S = S a + S b + S c = 3V p I = 3I Z p = 2 p
3V p2 ∗
….. (12.27)
Zp
where Z p = Z p ∠φ ° = load impedance per phase Alternatively, we may write Eq. (12.27) as
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S = P + jQ
Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
Power Balance Y-Y Connection
Power Balanced for Y-∆ Connection
Assuming abc sequence V L = 3V p∠30°
Assuming abc sequence V L = 3V p∠30°
I L = 3I p∠− 30°
I L = 3I p∠− 30°
power absorbed by the load is ∗
∗
∗
S1φ = V p I p = I p ZL I p = ZL I p
power absorbed by the load is ∗
S1φ = V p I p = I p ZL I p = ZL I p
2
2 (12.28) S 3φ = 3S 1φ = 3 I L Z L .....
Or the above 3-phase can be represented as per phase diagram since they are balanced
= ZL
2
2
1 2 = ZL IL 3 3
Ip
2 S 3φ = 3S 1φ = Z L I L ..... (12.29)
Or the above 3-phase can be represented as per phase diagram since they are balanced
Where
Where
V p =Van
From the single-phase one-line diagram above, power absorbed by the load is S 1φ = Z L I L
2
S 3φ = 3S 1φ S 3φ = 3Z L I L
Where
IL = Ia
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2
V p =Van
2
= 3Z L I a
From the single-phase one-line diagram above, power absorbed by the load is Z 2 S 1φ = L I L 3 S 3φ = 3S 1φ 2
2
S 3φ = I L Z L = I a Z L
Where
IL = Ia
Chapter 12: Three-Phase Circuits
Dr. C.S.Tan
