Requirements of a Balanced 3-Phase Set

Following are the requirements that must be satisified in order for a set of 3 sinusoidal variables (usually voltages or currents) to be a "balanced 3-phase set"

1. All 3 variables have the same amplitude
2. All 3 variables have the same frequency
3. All 3 variables are 120^o in phase

In terms of the time domain, a set of balance 3-phase voltages has the following general form.

v_a = V_m cos ( t + )

v_b = V_m cos ( t + - 120^o )

v_c = V_m cos ( t + - 240^o ) = V_m cos ( t + +120^o )

Notice that we have assumed (and will continue to assume) positive (abc) phase sequence, i.e., phase "b" follows 120^o behind "a" & phase "c" follows 120^o behind phase "b"

Figure 1 below illustrates the balanced 3-phase voltages in time domain.

Figure 1: Balanced 3-Phase Variables in Time Domain
In terms of phasors, we write the same balanced set as follows. Note that the phasors are in rms, as will be assumed throughout this course.

V_a = V_{m m}

V_b = V_m - 120^o

V_c = V_m - 240^o = V_m +120^o

Thus,

V_b = V_a (1 -120^o) , and V_c = V_a (1 +120^o)

Figure 2 below illustrates the balanced 3-phase phasors graphically.

Figure 2: Balanced 3-Phase Phasors

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Requirements of a Balanced 3-Phase Circuit

Following are the requirements that must be satisified in order for a 3-phase system or circuit to be balanced

1. All 3 sources are reprensented by a set of balanced 3-phase variables
2. All loads are 3-phase with equal impedances
3. Line impedances are equal in all 3 phases

Having a balanced circuit allows for simplified analysis of the 3-phase circuit. In fact, if the circuit is balanced, we can solve for the voltages, currents, and powers, etc. in one phase using circuit analysis. The values of the corresponding variables in the other two phases can be found using some basic equations. This type of solution is accomplished using a "one-line diagram", which will be discussed later. If the circuit is not balanced, all three phases should be analyzed in detail.

Figure 3 illustrates a balanced 3-phase circuit and some of the naming conventions to be used in this course

Figure 3: A Balanced 3-Phase Circuit

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Terms and Naming Conventions

Phase

describes or pertains to one element or device in a load, line, or source. It is simply a "branch" of the circuit and could look something like this .

Line

refers to the "transmission line" or wires that connect the source (supply) to the load. It may be modeled as a small impedance (actually 3 of them), or even by just a connecting line.

Neutral

the 4th wire in the 3-phase system. It's where the phases of a Y connection come together.

Phase Voltages & Phase Currents

the voltages and currents across and through a single branch (phase) of the circuit. Note this definition depends on whether the connection is Wye or Delta!

Line Currents

the currents flowing in each of the lines (I_a, I_b, and I_c). This definition does not change with connection type.

Line Voltages

the voltages between any two of the lines (V_ab, V_bc, and V_ca). These may also be referred to as the line-to-line voltages. This definition does not change with connection type.

Line to Neutral Voltages

the voltages between any lines and the neutral point (V_a, V_b, and V_c). This definition does not change with connection type, but they may not be physically measureable in a Delta circuit.

Line to Neutral Currents

same as the line currents (I_a, I_b, and I_c).

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Where Does that Come From?

Let us determine the relationships between the line and line to neutral voltages. By applying Kirchoff's Voltage Law (KVL) to the top "loop" of the source section in Figure 3, we get

V_ab = V_a - V_b = V_m - V_m - 120^o

Now, without loss of generality, let = 0^o

thus, V_a = V_m 0^o, and V_b = V_m -120^o, so

V_ab = V_m 0^o - V_m - 120^o = V_m (1 - 1 - 120^o ) = V_m (1 - (cos 120^o - j sin 120^o))

= V_m (1 - (-1/2) + j ( / 2 ) ) = V_m (3 / 2) + j ( / 2 ))

Converting to polar form,

V_ab = V_m Sqrt[ (3/2)² + ( / 2)² ] tan^-1 {( / 2) / (3/2) }

= V_m Sqrt[ 9/4 + 3/4 ] tan^-1 {1/ }

= V_m tan^-1 {(1 / 2) / ( /2) } = V_m tan^-1 {(sin 30^o) / (cos 30^o) }

= V_m tan^-1 {tan 30^o } = V_m 30^o

Thus we have the general equation (for abc sequence anyway)

V_ab = V_a 30^o
The relationships between the currents can be developed similarly. Summing currents at the "A" node in Figure 3 yields the starting equation,

I_a = I_AB - I_CA

This time choose I_a to be the phasor reference (at 0^o). The final result is:

I_a = I_AB -30^o These relationships can also be remembered graphically using Figures 4 and 5 below. Figure 4 illustrates the voltage relationship. By looking at the phasor equation as the sum of two vectors (V_a and -V_b ) we obtain the resulting V_ab shown in the figure.

Since V_ab is longer, we know . . . . |V_ab| = |V_a| ,

and since V_ab is ahead of V_a, we know that, . . . . (the angle of V_ab) = (the angle of V_a) + 30^o

Figure 4: Graphical Voltage Relationship
Figure 5 illustrates the current relationship. Now the phasor equation is the sum of two vectors (I_ab and -I_ca ) we obtain the resulting I_a shown in the figure.

Since I_a is longer, we know

|I_a| = |I_ab| ,

and since I_a is behind I_ab, we know that,

(the angle of I_a) = (the angle of I_ab) - 30^o

Figure 5: Graphical Current Relationship

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Wyes and Deltas

A summary of the characteristics of the two types of 3-phase circuit connections are given below.

The Wye = Y = "Star" connection	____	The Delta = connection
each phase is connected between a line and the neutral		each phase is connected between two lines
Figure 6: A Y Circuit		Figure 7: A Circuit
Phase voltages = Line to neutral voltages (Va, etc.) Phase currents = Line currents (Ia, etc.) Neutral connects the 3 phases		Phase voltages = Line voltages (Vab, etc.) Phase currents = currents from line to line (Iab, etc.) Neutral is not present

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Y to Conversions

In terms of power, currents & line voltages, the following sources are the same and may be used interchangably in most cases. Note, the Y connection should be used in a one-line diagram.

Wye connected source	____	Delta connected source
Figure 8: A Y Source		Figure 9: A Source
VA = Vab / ( 30o)		Vab = VA ( 30o)

Similarly, the two loads given below are the same in terms of the resulting power, line currents and line voltages and can usually be substituted as desired. Note that the Y connection is the one needed for the one-line diagram!

Wye connected load	____	Delta connected load
Figure 10: A Y Impedance Load		Figure 11: A Impedance Load

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The One-Line Diagram

If the circuit is balanced, all corresponding sets of 3-phase voltages and currents are balanced, and the neutral current will be zero.

I_N = I_a + I_b + I_c

Why must this be so? Because the sum of a balanced set of 3-phase variables is equal to zero. This can be verified mathematically using the definition, or visually by considering using vector addition to add the balanced set in Figure 5.

Because the neutral current is zero, this means that if the neutral in the load is connected to the neutral in the source, no current will flow. Thus, the voltage at each of the neutrals must be the same. This means they can be considered to be the same point.

Now consider the circuit of Figure 12. In general, any circuit with a source, load, and line configuration can be converted to a circuit of this type by replacing any Delta -connected sources or loads with the equivalent Wye connected sources or loads.

Figure 12: Completely Y-Connected Circuit Including Neutral
If the Neutral points in Figure 12 are actually the same point, Figure 12 can be redrawn as shown in Figure 13. Figure 13: ReDrawn All-Y Circuit
From this figure we see that each of the phase currents depends only on the source in the corresponding loop. In other words,

I_a = V_a / ( Z_line + Z_Y) , I_b = V_b / ( Z_line + Z_Y) , and I_c = V_c / ( Z_line + Z_Y)

Notice that these equations are VERY similar.

Recall that in balanced set of variables, once we know one variable, the other two can be found by simply adding and subtracting 120^o. Thus, we only need to consider and solve one loop of Figure 13 --- this is the one-line diagram!

Figure 14 shows the one-line diagram for the circuit of Figure 13. Usually the one line that is considered is the "a" phase. The "b" phase quantities are then found by subtracting 120^o, and the "c" phase quantities are found by adding 120^o.

Figure 14: The One-Line Diagram

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3-Phase Power

The 3-phase (3) power of a circuit is simply the sum of the power in the three individual phases. Thus for a Wye circuit, the equation is

S₃ = S_a + S_b + S_c

and for a Delta circuit, the equation is

S₃ = S_ab + S_bc + S_ca

Another adavantage of having a balanced circuit is that each phase has the same power. That is,

S = S_ab = S_bc = S_ca = S_a = S_b = S_c

so that,

S₃ = 3 S = 3 S_ab = 3 S_a

Just in case you didn't know, right now you should be thinking "This is very cool!"

The single phase power can be found using either

S = V_a I_a* or S = V_ab I_ab*

We can do some interesting rearrangements to get the power in terms of the line voltage (V_ab) and line current (I_a) only.

S = V_a I_a* = |V_a| | I_a| = {|V_ab| / }| I_a| = S

Thus, S₃ = 3 S = 3 {|V_ab| / }| I_a| = |V_ab| | I_a|

Note:

In balanced systems, all the S's and S₃ have the same power factor (pf) and thus the same power factor angle = impedance angle = .

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Collection of Important 3-Phase Equations

If X_a, X_b, and X_c are a balanced set,

X_b = X_a (1 -120^o) , and

X_c = X_a (1 +120^o)

In general,

S₃ = S_a + S_b + S_c

S₃ = S_ab + S_bc + S_ca

For a balanced system,

S₃ = 3 S

S = V_a I_a*

S = V_ab I_ab*

For the balanced and positive sequence case,

V_ab = V_a 30^o

I_a = I_AB -30^o

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What to Assume

If you are given a voltage, current, or power value and not told specifically which variable it is, you should assume that you have been given a "line" value. That is, assume the following:

Voltage => Line voltage = |V_ab|

Current => Line current = |I_a|

Power => Three Phase Power = S₃, P₃, or Q₃

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Unbalanced Circuits

When we have an unbalanced circuit, we CANNOT use the one-line diagram to solve for "a" phase values and then get the answers for the other phases by adding or subtracting 120^o.

In general, a unbalanced three-phase circuit requires that you draw the complete circuit including all 3-phase and single-phase loads and perform a circuit analysis of the whole thing. Normal methods such as "meshes" or "node voltages" may be used. If you have the simple case in which a balanced 3-phase load is connected directly to a source and a single phase load is connected in parallel to the same source, you may calculate the currents in the balanced load using a one-line method. The single phase current is calculated separately and then individual line currents can be found by summing the currents at certain nodes in the system.

Remember any circuit that does not have all loads with the same impedance in all three branches is an unbalanced circuit.

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Wattmeters

The schematic for a wattmeter is given in Figure 15 below. Note that in order to measure power, we need to measure a current and a voltage. The wattmeter doesn't care which current or which voltage you use. It will give you a reading regardless of whether or not it means anything. It is up to the user (you) to make sure the meter is sensing the correct voltage and current to give a meaningful measurement!

Figure 15: The Basic Wattmeter
The meter reading will be

W = I V cos (_V - _I )

Under balanced conditions and conditions in which there are only three wires in the system, we can measure the power in all three phases of a load (or source) by using only two meters. This is called the "Two Wattmeter Method."

This method is quite convenient when all you have access to are the three wires going into a three-phase motor (for example). You want to measure P₃, where do you connect your meter?

To measure the 3-phase power correctly using two meters:

connect the current coils in two of the phases connect the positive terminals of the voltage coils to the same two phases (where you're measuring the current) connect both of the negative voltage terminals to the third phase.

Figures 16 and 17 below show two possible connections with phases "b" and "c" respectively, used as the voltage reference. Note that the "plus-minus" symbol marks the positive voltage terminal & the negative terminal is generally unmarked.

Figure 16: Two Wattmeter Connection with "b" as Reference
Figure 17: Two Wattmeter Connection with "c" as Reference

The meter readings for Figure 16 are:

W₁ = |I_a| |V_ab| cos (_Vab - _Ia ) W₂ = |I_c| |V_cb| cos (_Vcb - _Ic )

The meter readings for Figure 17 are:

W₁ = |I_a| |V_ac| cos (_Vac - _Ia ) W₂ = |I_b| |V_bc| cos (_Vbc - _Ib )

The three phase real power is found using . . . P₃ = W₁ + W₂

similarly, for the balanced condition, the magnitude three phase reactive power can be found using . . . |Q₃| = |W₁ - W₂|

the sign of Q may vary depending on how the wattmeters are connected. So, it is generally safer to determine the sign using other means. 

http://eece.ksu.edu/~starret/581/3phase.html