The dominant form of power generation, transmission, and distribution in the United States is via three-phase. Three-phase power uses three separate, energized conductors, 120° apart in phase to transmit power more efficiently than single-phase (line and neutral) power. The cost-effectiveness of three-phase power makes it worth its while, but the added complexity over single-phase power adds some challenges.
Why Three-Phase Power? - The benefits of three-phase power can be summarized succinctly: Three-phase power is better than single phase because it lets us transmit three times the power of single phase with one more wire. In other words, we get 200% added benefit for 50% added cost. That's a pretty great deal!
How does this work? Well, it comes back to the phase difference between voltages. Thanks to the fact that the voltages are all shifted by 120°, they algebraically sum to zero. This means that the current from all three phases will also algebraically sum to zero by Ohm's Law. In other words, there doesn't need to be a dedicated "return" conductor, like on a single-phase system. The phasor diagram below explains this in more detail.
Phasor Diagram of Three Voltages or Currents in a 3-Phase System
Another benefit of three-phase power is the steadiness of power supplied. While single phase power is volatile, alternating rapidly between its peak value and 0, three-phase power is much more uniform. Thanks to the fact that three-phase sources have different phases, the power received by the load looks much closer to constant. This is a big benefit for things like motors.
This might make you think, "If three-phase is better than single phase, can I just keep increasing the phases to make a better power system?"
Unfortunately, not really. Higher phase count systems will be able to transmit more power, but they rely on the same physics as the three phase system. This means we end up needing 9 wires to transport 9x the power of single phase. Our per-unit cost of transporting power is still the same as three-phase but we've now increased the complexity. That's no good.
Wye and Delta Connections - When we talk about single-phase power, its easy to understand how things wire together: every source and load has a positive and a negative. Make sure that everything connects with the right polarity and you're good to go.
With three-phase power, connections look different. A three-phase component (whether a source or a load) can be visualized by three single-phase elements wired together. There are two options for this connection method: Wye and Delta. The images below show both wiring options. The Wye connection, also known as the T-connection, has a center point common to all elements. This center point is known as the neutral point and provides a common reference for the A, B, and C phases. The delta connection is configured differently. Notice that there is no common neutral point for all three phases on the delta system.
Wye Connected Elements with Neutral Point N
Delta Connected Elements
Some terms that are used when referring to three-phase (wye and delta) connections are as follows:
The line-to-line voltage VL is the voltage between two phases (e.g. A to B). The line-to-line voltages on a balanced three-phase system will all have the same magnitude but be shifted in phase by 120°.
The line-to-neutral voltage VN is the voltage between one phase and the neutral point of the source or load (if applicable). Like the line-to-line voltage, the line-to-neutral voltages in the system will all have the same magnitude but be shifted in phase by 120°.
The magnitude of the line-to-neutral voltage | VN | is related to the magnitude of the line-to-line voltage | VL | by:
| VL | = √(3) | VL |
The line current IL is the current flowing on the conductors connecting sources to loads. Line currents will all have the same magnitude but be shifted in phase by 120°.
The phase current IP is the current that flows through the internal elements of a source or load. Phase currents will all have the same magnitude but be shifted in phase by 120°.
On a Wye-connected system, the line current is equal to the phase current by Kirchoff's Current Law. On a delta-connected system, the line current magnitude is related to the phase current magnitude by:
| IL | = √(3) | IP |
Summary of Variables for Three-Phase System
Complex Power with Three-Phase - The apparent power flowing in a three-phase system can be computed as follows, regardless of wye or delta connections:
S = √(3) VL IL
The real power and reactive power absorbed by a load can then be determined using the power factor of that load:
P = S cos(θ) and Q = S sin(θ)
All of these equations assume a typical, balanced three-phase system. By balanced, we mean that the loads have an equal impedance on all three branches and the sources have an equal supply voltage.
Per-Phase Analysis - With a balanced system, each phase behaves with the same magnitudes of voltages, currents, and impedances. The only difference between each phase is a shift of 120°. Instead of worrying about evaluating circuits on a three-phase basis, we can simplify things back down to a "per-phase" analysis by recognizing this symmetry. Once we solve for all variables on one phase, we know that it is descriptive of the entire system.
In order to complete a per-phase analysis, we must ensure that all loads and sources are in their wye-connection equivalents. This means that all voltages must be considered as line-to-neutral and all loads that are in a delta form must be converted to a wye form. In theory, this is a laborious task, but in practice we don't usually need to get into this level of detail. Most loads are specified in terms of their rated load currents instead of their impedances.
Example: A wye-connected load is rated for 480V line-to-line and has a load current of 10A. What is the line-to-neutral voltage across the load? What is the apparent power drawn by the load?
Solution: The line-to-neutral voltage is found by taking the line-to-line voltage and dividing by √(3).
| VN | = 480 / √(3) = 277 V
The apparent power of the three-phase load can also be directly calculated:
S = √(3) VL IL = √(3) * 480 * 10 = 8.31 kVA
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