SECTION 7: THREE-PHASE CIRCUIT FUNDAMENTALSweb.engr.oregonstate.edu/~webbky/ENGR202_files/SECTION...
Transcript of SECTION 7: THREE-PHASE CIRCUIT FUNDAMENTALSweb.engr.oregonstate.edu/~webbky/ENGR202_files/SECTION...
ENGR 202 β Electrical Fundamentals II
SECTION 7: THREE-PHASE CIRCUIT FUNDAMENTALS
K. Webb ENGR 202
Balanced Three-Phase Networks2
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Balanced Three-Phase Networks
We are accustomed to single-phase power in our homes and offices A single line voltage referenced to a neutral
Electrical power is generated, transmitted, and largely consumed (by industrial customers) as three-phase power Three individual line voltages and (possibly) a neutral Line voltages all differ in phase by Β±120Β°
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Ξ- and Y-Connected Networks
Two possible three-phase configurations Applies to both sources and loads
Y-Connected Source π«π«-Connected Source
Y-connected network has a neutral node Ξ-connected network has no neutral
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Line-to-Neutral Voltages
A three-phase network is balanced if Sources are balanced The impedances connected to each phase are equal
In the Y network, voltages ππππππ, ππππππ, and ππππππ are line-to-neutral voltages
A three-phase source is balanced if Line-to-neutral voltages have equal
magnitudes Line-to-neutral voltage are each 120Β°
out of phase with one another
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Line-to-Neutral Voltages
The line-to-neutral voltages are
π½π½ππππ = πππΏπΏπΏπΏβ 0Β°
π½π½ππππ = πππΏπΏπΏπΏβ β 120Β°
π½π½ππππ = πππΏπΏπΏπΏβ β 240Β° = πππΏπΏπΏπΏβ + 120Β°
This is a positive-sequence or abc-sequencesource
π½π½ππππ leads π½π½ππππ, which leads π½π½ππππ
Can also have a negative- or acb-sequencesource
π½π½ππππ leads π½π½ππππ, which leads π½π½ππππ
Weβll always assume positive-sequence sources
Positive-Sequence Phasor Diagram:
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Line-to-Line Voltages The voltages between the three phases are line-to-
line voltages Apply KVL to relate line-to-line voltages to line-to-
neutral voltagesπ½π½ππππ β π½π½ππππ + π½π½ππππ = πππ½π½ππππ = π½π½ππππ β π½π½ππππ
We know thatπ½π½ππππ = πππΏπΏπΏπΏβ 0Β°
andπ½π½ππππ = πππΏπΏπΏπΏβ β 120Β°
so π½π½ππππ = πππΏπΏπΏπΏβ 0Β° β πππΏπΏπΏπΏβ β 120Β° = πππΏπΏπΏπΏ 1β 0Β° β 1β β 120Β°
π½π½ππππ = πππΏπΏπΏπΏ 1 β β12 β ππ
32 = πππΏπΏπΏπΏ
32 + ππ
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π½π½ππππ = 3πππΏπΏπΏπΏβ 30Β°
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Line-to-Line Voltages
Again applying KVL, we can find π½π½πππππ½π½ππππ = π½π½ππππ β π½π½πππππ½π½ππππ = πππΏπΏπΏπΏβ β 120Β° β πππΏπΏπΏπΏβ 120Β°
π½π½ππππ = πππΏπΏπΏπΏ β12β ππ
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β β12
+ ππ3
2
π½π½ππππ = πππΏπΏπΏπΏ βππ 3
π½π½ππππ = 3πππΏπΏπΏπΏβ β 90Β°
Similarly,
π½π½ππππ = 3πππΏπΏπΏπΏβ 150Β°
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Line-to-Line Voltages
The line-to-line voltages, with ππππππ as the reference:
π½π½ππππ = 3πππΏπΏπΏπΏβ 30Β°π½π½ππππ = 3πππΏπΏπΏπΏβ β 90Β°π½π½ππππ = 3πππΏπΏπΏπΏβ 150Β°
Line-to-line voltages are 3 times the line-to-neutral voltage
Can also express in terms of individual line-to-neutral voltages:
π½π½ππππ = 3π½π½ππππβ 30Β°
π½π½ππππ = 3π½π½ππππβ 30Β°
π½π½ππππ = 3π½π½ππππβ 30Β°
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Line Currents in Balanced 3ππ Networks
π°π°ππ =π½π½π¨π¨π¨π¨ππππ
=πππΏπΏπΏπΏβ 0Β°ππππ
π°π°ππ =π½π½π©π©π¨π¨ππππ
=πππΏπΏπΏπΏβ β 120Β°
ππππ
π°π°ππ =π½π½πͺπͺπ¨π¨ππππ
=πππΏπΏπΏπΏβ + 120Β°
ππππ
Line currents are balanced as long as the source and load are balanced
We can use the line-to-neutral voltages to determine the line currents Y-connected source and load Balanced load β all
impedances are equal: ππππ
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Neutral Current in Balanced 3ππ Networks
Apply KCL to determine the neutral current
π°π°ππ = π°π°ππ + π°π°ππ + π°π°ππ
π°π°ππ =πππΏπΏπΏπΏππππ
1β 0Β° + 1β β 120Β° + 1β 120Β°
π°π°ππ =πππΏπΏπΏπΏππππ
1 + β12 β ππ
32 + β
12 + ππ
32
π°π°ππ = 0
The neutral conductor carries no current in a balanced three-phase network
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Delta- & Wye-Connected Networks12
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Three-Phase Network Configurations
As for sources, three-phase loads can also be connected in two different configurations
Y-Connected Load π«π«-Connected Load
The Y load has a neutral connection, but the Ξ load does not Currents in a Y-connected load are the line currents we just
determined Next, weβll look at currents in a Ξ-connected load
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Balanced Ξ-Connected Loads
We can use line-to-line voltages to determine the currents in Ξ-connected loads
π°π°π¨π¨π©π© =π½π½π¨π¨π©π©ππΞ
=3π½π½π¨π¨π¨π¨β 30Β°
ππΞ=
3πππΏπΏπΏπΏβ 30Β°ππΞ
π°π°π©π©πͺπͺ =π½π½π©π©πͺπͺππΞ
=3π½π½π©π©π¨π¨β 30Β°
ππΞ=
3πππΏπΏπΏπΏβ β 90Β°ππΞ
π°π°πͺπͺπ¨π¨ =π½π½πͺπͺπ¨π¨ππΞ
=3π½π½πͺπͺπ¨π¨β 30Β°
ππΞ=
3πππΏπΏπΏπΏβ 150Β°ππΞ
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Balanced Ξ-Connected Loads
Applying KCL, we can determine the line currentsπ°π°ππ = π°π°π¨π¨π©π© β π°π°πͺπͺπ¨π¨
π°π°ππ =3πππΏπΏπΏπΏππΞ
1β 30Β° β 1β 150Β°
π°π°ππ =3πππΏπΏπΏπΏππΞ
32 + ππ
12 β β
32 + ππ
12 =
3πππΏπΏπΏπΏππΞ
3 =3πππΏπΏπΏπΏππΞ
The other line currents can be found similarly:
π°π°ππ =3πππΏπΏπΏπΏβ 0Β°
ππΞ= 3π°π°π¨π¨π©π©β β 30Β°
π°π°ππ =3πππΏπΏπΏπΏβ β 120Β°
ππΞ= 3π°π°π©π©πͺπͺβ β 30Β°
π°π°ππ =3πππΏπΏπΏπΏβ 120Β°
ππΞ= 3π°π°πͺπͺπ¨π¨β β 30Β°
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Ξ β ππ Conversion
Analysis is often simpler when dealing with ππ-connected loads Would like a way to convert Ξ loads to ππ loads (and vice
versa)
For a ππ load and a Ξ load to be equivalent, they must result in equal line currents
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Ξ β ππ Conversion
Line currents for a ππ-connected load:
π°π°ππ =πππΏπΏπΏπΏβ 0Β°ππππ
π°π°ππ =πππΏπΏπΏπΏβ β 120Β°
ππππ
π°π°ππ =πππΏπΏπΏπΏβ 120Β°
ππππ
For a Ξ-connected load:
π°π°ππ =3πππΏπΏπΏπΏβ 0Β°
ππΞ
π°π°ππ =3πππΏπΏπΏπΏβ β 120Β°
ππΞ
π°π°ππ =3πππΏπΏπΏπΏβ 120Β°
ππΞ
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Ξ β ππ Conversion
Equating any of the three line currents, we can determine the impedance relationship
πππΏπΏπΏπΏβ 0Β°ππππ
=3πππΏπΏπΏπΏβ 0Β°
ππΞ
ππππ = ππΞ3
and ππΞ = 3ππππ
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Line-to-Neutral Schematics
For balanced networks, we can simplify our analysis by considering only a single phase A per-phase analysis Other phases are simply shifted by Β±120Β°
For example, a balanced ππ-ππ circuit:
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One-Line Diagrams
Power systems are often depicted using one-line diagrams or single-line diagrams Not a schematic β not all wiring is shown
For example:
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Example Problems21
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Determine: ππππ ππππππ ππππππ
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Determine: ππππ ππππππ ππππππ ππππππ
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Power in Balanced 3ππ Networks27
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Instantaneous Power
Weβll first determine the instantaneous power supplied by the source Neglecting line impedance, this is also the power absorbed by the load
The phase ππ line-to-neutral voltage is
π£π£ππππ π‘π‘ = 2πππΏπΏπΏπΏ cos πππ‘π‘ + πΏπΏ The phase ππ current is
ππππ π‘π‘ = 2πΌπΌπΏπΏ cos πππ‘π‘ + π½π½where π½π½ depends on the load impedance
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Instantaneous Power
The instantaneous power delivered out of phase ππof the source is
ππππ π‘π‘ = π£π£ππππ π‘π‘ ππππ π‘π‘
ππππ π‘π‘ = 2πππΏπΏπΏπΏπΌπΌπΏπΏ cos πππ‘π‘ + πΏπΏ cos πππ‘π‘ + π½π½
ππππ π‘π‘ = πππΏπΏπΏπΏπΌπΌπΏπΏ cos πΏπΏ β π½π½ + πππΏπΏπΏπΏπΌπΌπΏπΏ cos 2πππ‘π‘ + πΏπΏ + π½π½
The ππ and ππ phases are shifted by Β±120Β° Power from each of these phases is
ππππ π‘π‘ = πππΏπΏπΏπΏπΌπΌπΏπΏ cos πΏπΏ β π½π½ + πππΏπΏπΏπΏπΌπΌπΏπΏ cos 2πππ‘π‘ + πΏπΏ + π½π½ β 240Β°
ππππ π‘π‘ = πππΏπΏπΏπΏπΌπΌπΏπΏ cos πΏπΏ β π½π½ + πππΏπΏπΏπΏπΌπΌπΏπΏ cos 2πππ‘π‘ + πΏπΏ + π½π½ + 240Β°
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Instantaneous Power
The total power delivered by the source is the sum of the power from each phase
ππ3ππ π‘π‘ = ππππ π‘π‘ + ππππ π‘π‘ + ππππ π‘π‘
ππ3ππ π‘π‘ = 3πππΏπΏπΏπΏπΌπΌπΏπΏ cos πΏπΏ β π½π½+πππΏπΏπΏπΏπΌπΌπΏπΏ[cos 2πππ‘π‘ + πΏπΏ + π½π½+ cos 2πππ‘π‘ + πΏπΏ + π½π½ β 240Β°+ cos 2πππ‘π‘ + πΏπΏ + π½π½ + 240Β° ]
Everything in the square brackets cancels, leaving
ππ3ππ π‘π‘ = 3πππΏπΏπΏπΏπΌπΌπΏπΏ cos πΏπΏ β π½π½ = ππ3ππ
Power in a balanced ππππ network is constant In terms of line-to-line voltages, the power is
ππ3ππ = 3πππΏπΏπΏπΏπΌπΌπΏπΏ cos πΏπΏ β π½π½
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Complex Power
The complex power delivered by phase ππ is
πΊπΊππ = π½π½πππππ°π°ππβ = πππΏπΏπΏπΏβ πΏπΏ πΌπΌπΏπΏβ π½π½ β
πΊπΊππ = πππΏπΏπΏπΏπΌπΌπΏπΏβ πΏπΏ β π½π½πΊπΊππ = πππΏπΏπΏπΏπΌπΌπΏπΏ cos πΏπΏ β π½π½ + πππππΏπΏπΏπΏπΌπΌπΏπΏ sin πΏπΏ β π½π½
For phase ππ, complex power is
πΊπΊππ = π½π½πππππ°π°ππβ = πππΏπΏπΏπΏβ πΏπΏ β 120Β° πΌπΌπΏπΏβ π½π½ β 120Β° β
πΊπΊππ = πππΏπΏπΏπΏπΌπΌπΏπΏβ πΏπΏ β π½π½πΊπΊππ = πππΏπΏπΏπΏπΌπΌπΏπΏ cos πΏπΏ β π½π½ + πππππΏπΏπΏπΏπΌπΌπΏπΏ sin πΏπΏ β π½π½
This is equal to πΊπΊππ and also to phase πΊπΊππ
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Complex Power
The total complex power isπΊπΊππππ = πΊπΊππ + πΊπΊππ + πΊπΊπππΊπΊππππ = 3πππΏπΏπΏπΏπΌπΌπΏπΏβ πΏπΏ β π½π½
πΊπΊππππ = 3πππΏπΏπΏπΏπΌπΌπΏπΏ cos πΏπΏ β π½π½ + ππ3πππΏπΏπΏπΏπΌπΌπΏπΏ sin πΏπΏ β π½π½
The apparent power is the magnitude of the complex power
ππ3ππ = 3πππΏπΏπΏπΏπΌπΌπΏπΏ
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Complex Power
Complex power can be expressed in terms of the real and reactive power
πΊπΊππππ = ππ3ππ + ππππ3ππ
The real power, as weβve already seen is
ππ3ππ = 3πππΏπΏπΏπΏπΌπΌπΏπΏ cos πΏπΏ β π½π½
The reactive power is
ππ3ππ = 3πππΏπΏπΏπΏπΌπΌπΏπΏ sin πΏπΏ β π½π½
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Advantages of Three-Phase Power
Advantages of three-phase power: For a given amount of power, half the amount of wire
required compared to single-phase No return current on neutral conductor
Constant real power Constant motor torque Less noise and vibration of machinery
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Three-Phase Power β Example
Determine Load voltage, π½π½π¨π¨π©π© Power triangle for the load Power factor at the load
Weβll do a per-phase analysis, so first convert the Ξ load to a ππ load
ππππ =ππΞ3
= 1 + ππ0.5 Ξ©
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Three-Phase Power β Example
The per-phase circuit:
The line current is
π°π°π³π³ =π½π½ππππ
πππΏπΏ + ππππ=120β 0Β° ππ1.1 + ππ1 Ξ©
=120β 0Β° ππ
1.45β 42.3Β° Ξ©
π°π°π³π³ = 80.7β β 42.3Β° π΄π΄
The line-to-neutral voltage at the load is
π½π½π¨π¨π¨π¨ = π°π°π³π³ππππ = 80.7β β 42.3Β° π΄π΄ 1 + ππ0.5 Ξ©
π½π½π¨π¨π¨π¨ = 80.7β β 42.3Β° π΄π΄ 1.12β 26.6Β° Ξ©
π½π½π¨π¨π¨π¨ = 90.25β β 15.71Β° V
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Three-Phase Power β Example
Calculate the line-to-line voltage from the line-to-neutral voltage
π½π½π¨π¨π©π© = 3π½π½π¨π¨π¨π¨β 30Β°
π½π½π¨π¨π©π© = 156β 14.3Β° ππ
Alternatively, we could calculate line-to-line voltage from the two line-to-neutral voltages. The line-to-neutral voltage at phase π΅π΅ is
π½π½π©π©π¨π¨ = 90.25β β 135.71Β° ππ
So the line-to-line voltage is given by
π½π½π¨π¨π©π© = π½π½π¨π¨π¨π¨ β π½π½π©π©π¨π¨ = 156β 14.3Β° ππ
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Three-Phase Power β Example
The complex power absorbed by the load is
πΊπΊππππ = 3πΊπΊπ¨π¨ = 3π½π½π¨π¨π¨π¨π°π°π³π³β
πΊπΊππππ = 3 90.25β β 15.71Β° ππ 80.7β β 42.3Β° π΄π΄ β
πΊπΊππππ = 21.85 β 26.6Β° πππππ΄π΄
πΊπΊππππ = 19.53 + πππ.78 πππππ΄π΄
The apparent power:ππ3ππ = 21.85 πππππ΄π΄
Real power:ππ = 19.53 ππππ
Reactive power:ππ = 9.78 πππ£π£ππππ
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Three-Phase Power β Example
The power triangle at the load:
ππ = 9.78 kvar
ππ = 19.53 kW
26.6Β°
The power factor at the load is
ππ.ππ. = cos 26.6Β° =ππππ
=19.53 ππππ21.85 πππππ΄π΄
ππ. ππ. = 0.89 ππππππππππππππ
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Example Problems40
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Determine complex power: From the source To the load To the line
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Line-to-line voltage at the load is maintained at 4.16 kV. What is the voltage at the source? How much complex power is delivered by the source?
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Line-to-line voltage at the load is maintained at 4.16 kV. Determine: Power factor at load Power triangle at load Loss in Lines
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Add power factor correction to improve p.f. to 0.98, lagging.Determine: Power triangle at load Loss in Lines
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