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Three-phase power AC power is typically generated and delivered via 3φ systems. There are two principal advantages to using 3φ power.
- The power delivered to a balanced 3φ load does not vary with time as do 1φ loads. This results in less vibration and lower stresses in 3φ motors and in systems driven by them.
- A 3φ system can deliver the same amount of power at lower current levels than can 1φ systems. This allows the use of smaller conductors.
A three-phase source is made from three independent sources. Each of these sources is a sinusoidal source of a certain frequency—all three sources have the same ω, and differing phases—thus the term “three-phase.”
Three-phase source This 3φ source is connected in the Y-configuration. Here the "Y" can be viewed sideways.
Another 3φ connection, the ∆-configuration, is also possible. For sources, the Y-configuration is most often used.
The source is "balanced" if Vpa = Vpb = Vpc = Vp and if θa, θb, and θc differ by 120°
If any of these conditions are not met, then the 3φ source is unbalanced.
To be a balanced 3φ source, the source amplitudes must be equal and their phases must differ from each other by 120°.
Three-Phase Loads Y-configuration. The Y-connected 3φ load is said to be "balanced" if all impedances are equal. That is, if Z a = Z b = Z c = Z Y.
If these impedances are not equal, the load is unbalanced.
∆-configuration. Likewise, the ∆ -connected 3φ load is said to be "balanced" only if all impedances are equal. That is, if Z ab = Z bc = Z ca = Z ∆.
If these impedances are not equal, the load is unbalanced.
A balanced 3φ, ∆-connected load can be converted to an equivalent Y-connected load by dividing Z ∆ by 3.
A three-phase system—the source and load—is said to be balanced only if both the source and load are balanced. If either is unbalanced, the system is also unbalanced.
- For balanced 3φ systems, we can simplify our task quite a bit if we are familiar with the characteristics of balanced 3φ system.
- On the other hand, the way to treat an unbalanced 3φ system is like any other phasor circuit.
KVL and KCL give the relations between phase quantities and line quantities.
Y: I L = I φ VL = √3Vφ the line voltage’s phase leads the phase of phase voltages by 30 °
KVL gives V ab + V bn - V an = 0
Re-arranging gives: V ab = V an - V bn
∆ : V L = V φ IL = √3 Iφ the line current/s phase lags the phase of phase voltages by 30 °
KCL
I a - I ab + I ca = 0
Re-arranging gives: I a = I ab - I ca
V an
V cn
-V bn
V ab
V bn
I ab
I b c
-I ca
I a
I ca
Example A 208V 3φ supply feeds a balanced ∆-connected load with Z = 120∠ 35 °^ Ω. Determine i) the line and phase voltages and currents ii) the real, reactive and apparent power supplied to the 3φ load.
Balanced Y-Y System (with neutral connection removed)
Analyzing balanced 3φ systems
1 φ equivalent from 3φ ß If loads are given as impedances, convert all ∆ s to Y s by dividing Z ∆ by 3. ß If loads are given as powers, the powers will be 3φ powers. Divide 3φ power by 3 to obtain 1φ power. ß Divide all line voltages by v3 to produce Y phase voltages. ß Analyze the 1φ equivalent system. ß Multiply resulting phase voltages by v3 to produce line voltage results. ß Multiply resulting powers by 3 to give 3φ power. 1 φ equivalent circuit
Balanced Y-∆ System
1 φ equivalent circuit
There are usually two main factors in the cost of electricity: i) Demand charge which is based on the maximum kVA demanded in the month. ii) Energy charge which is based on the total kWh consumed in the month.
Electric bills are greatly reduced if the pf is corrected to near unity. It is worth noting that power factors are not corrected to exactly unity because this would lead to the pf being over-correction when the load is small and produce a leading pf.
Example The following diagram shows two loads supplied from a common 3φ feeder. Load 1 is 10 MW @ 0.65 lag, while load 2 is 15 MVA @ 0.75 lag. Both loads are supplied at 34.5 kV.
Notice that the meter is placed before the feeder. This means that the feeder losses will be part of the electricity bill. Determine: i) Current in the feeder ii) Voltage regulation iii) Cost of electricity ($/yr) for constant loads, energy costs 2.5 ¢/kWh, and demand charge is $7.00/KVA/mth. iv) The capacitance (μF/ph) of a wye -connected capacitor bank needed to improve the combined load pf to 0.98 lag. v) Repeat parts (i) - (iii) with the new pf.
Meter
2 Ω j6 Ω
2 Ω (^) j6 Ω
2 Ω (^) j6 Ω
Three
Phase
Supply
Load
Load
General formulas, nomenclature, units, and conventions in power
Time domain v(t) = V (^) m cos (ωt + θv) = Vp cos (ωt + θv) i(t) = Im cos (ωt + θi) = Ip cos (ωt + θi)
Phasors with magnitude in peak value V p = Vp∠ θv I p = Ip∠ θi
Phasors in rms V = V∠θv = (Vp/√2)∠θv I = I∠θi = (Ip/√2)∠θi
If a phasor current or voltage is given, without other qualification, it’s assumed to be in rms.
Power S = S∠(θv-θi)?= S∠θs?= Pav + jQ
S ~ complex power, in VA S ~ apparent power, in VA Pav ~ average power, in W Q ~ reactive power, in VAR θs ??= θv - θI = θz
If power is referred to, without any other qualification, it’s assumed to be average power
S = VI * = V∠θv I∠−θi = VI∠(θv-θi) = VI cos?(θv-θi) + j VI sin(?θv-θi)
S = VI Pav = VI cos?(θv-θi) = VI cos θs Q = VI sin?(θv-θi) = VI sin θs
power factor = cos?(θv-θi) lagging for θv>θI (θ>0, Q>0, inductive loads) leading for θv<θI (θ<0, Q<0, capacitive loads)
3 φ power
ß Unless otherwise specified, voltages and currents are assumed to be line quantities. ß Power, unless otherwise specified, is assumed to be average 3φ power. ß Any power, unless otherwise specified, is assumed to be 3φ.
A 3φ system is balanced if
- Phase voltages are equal in magnitude and displaced 120° in phase from each other.
- Phase load impedances are equal.
for balanced Y-connection, VL = √3Vp, IL = Ip for balanced ∆-connection, VL = Vp, IL = √3Ip
θs = θz = θv – θi S = S∠θs = P + jQ =√ 3 VI ***** =√3 VI cos θs + j√3 VI sin θs
