Review of 3-Phase Circuits

Index

• Requirements of a Balanced 3-Phase Set

• Requirements of a Balanced 3-Phase Circuit

• Terms and Naming Conventions

• Where Does that Come From?

• Wyes and Deltas

• Y to Conversions

• The One-Line Diagram

• 3-Phase Power

• Collection of Important 3-Phase Equations

• What to Assume

• Unbalanced Circuits

• Wattmeters

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 )

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

Figure 1: Balanced 3-Phase Variables in Time Domain

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: Balanced 3-Phase Phasors

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

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).

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

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

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

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

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

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

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

Figure 13: ReDrawn All-Y Circuit

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

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 = .

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

Voltage => Line voltage = |V_ab|

Current => Line current = |I_a|

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

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.

Figure 15: The Basic Wattmeter

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 )

3

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.

Shelli Starrett

starret@eece.ksu.edu

September 18, 1997