AC Circult Theory - Basic Theory of Electromagnetically Induced Voltages

2.2.1 Case Study

H. If there is a car parked near the powerline, measure the AC voltage between the steel frame of the car and the ground pin

3.2 Basic Theory of Electromagnetically Induced Voltages

3.2.1 AC Circult Theory

3.2.1 AC Circult Theory

Before starting a detailed discussion of how AC voltages are electromagnetically induced on a pipeline, it is important to understand some of the basic principles of AC circuits.

Consider first the case of a transformer. The voltage induced in the secondary winding (VS) relative to the primary voltage (Vp) is equal to the ratio of the number of turns in the secondary winding to those in the primary winding (see Equation 3-4a).

p S p

N N V

V = [3-6a]

Alternatively, one may think of this relationship in the following terms: the volts/turns ratio for the primary winding is equal to the volts/turns ratio for the secondary winding, or windings, in the case of a transformer having multiple secondary windings.

S S p p

N V N

V = [3-6b]

For example, 100V is applied to the primary winding of the transformer in Figure 3-16. Because this winding has 100 turns, then the volts/turns ratio is 1V/turn.

This same volts/turns ratio also applies to the secondary winding having only five turns, so that it will develop a voltage of 5V.

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100 V

100 Turns

5 turns 5 turns 5 turns

1 V/turn 1 V/turn

V_S= V_P N_S N_P

⎛

⎝ ⎜ ⎞

⎠ ⎟ =100V 5 100

⎛ ⎝ ⎞

⎠ = 5V

Figure 3-16: Determination of Voltage on a Transformer Secondary

Now consider the case where several secondary windings of the transformer are connected together in series (Figure 3-17). Assuming that all of the secondary windings have been wound onto the transformer core in the same direction, then the voltages are additive.

100 V

100 Turns

5 turns 5 turns 5 turns

1 V/turn 1 V/turn

V1 V2 V3

V= V₁+ V₂+ V₃

= 5V + 5V + 5V

= 15V

Figure 3-17: Effect of Interconnecting the Secondary Windings

This is analogous to adding the voltages of a number of batteries that are connected together in series, as shown in Figure 3-18a. If, however, one of the batteries was reversed (Figure 3-18b), then its voltage would be added to the voltages of the other batteries as a negative value.

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1.5V 1.5 V 1.5 V

V= 1.5V +1.5V+1.5V = 4.5V

1.5V 1.5 V 1.5 V

(a) (b)

- - +

V = 1.5V -1.5V +1.5V =1.5V

Figure 3-18: Effect of Polarity on a Series Combination of DC Voltage Sources

Consider once again the case of the transformer in Figure 3-17. Although AC voltages and currents exhibit no polarity in the DC sense, reversing the connections to one of the secondary windings would be akin to reversing the polarity of one battery in Figure 3-18—as Figure 3-19 shows.

100 V

100 Turns

5 turns 5 turns 5 turns

1 V/turn 1 V/turn

V= 5V - 5V + 5V = 5V

Figure 3-19: Effect of “Polarity” on a Series Combination of AC Voltage Sources

Even though the polarity of an AC voltage continuously alternates from positive to negative, the voltages in the specific case shown can still be added and subtracted just as if they exhibited a constant polarity. When working with AC circuits, the term “polarity” is generally not used because it applies only to a particular instant in time. Instead, waveforms are compared in terms of their phase relationships.

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AC waveforms are said to be either in-phase or out-of-phase with one another.

“In-phase” means that two waveforms have the same polarity at every instant in time (Figure 3-20). In the case of the transformer in Figure 3-17, the voltages in all three secondary windings must be in-phase with one another because they are all generated by the same magnetic field. When the connections are reversed on the middle winding, its voltage waveform becomes exactly out-to-phase with the waveforms of the other two windings. In other words: the waveforms start at the same instant in time but, when one goes positive, the other goes negative.

When viewed on an oscilloscope, an AC waveform indicates the variation of the voltage or current versus time. When the frequency of the waveform is 60 Hz, the period of the waveform is 1/60 s or 16.67 mS. When solving interference problems related to AC power systems, the frequency of all waveforms will be the same (either 50 Hz or 60 Hz, depending upon the country). Therefore, frequency (and time) can be omitted from the analysis. Instead, it is easier to discuss waveforms in terms of phase angle (Figure 3-20): one full cycle of the sinusoidal waveform is comprised of 360 degrees or, alternatively, 2π radians.

-1 0 1

0 90 180 270 360

Voltage or Current

Angle (Degrees)

0 π/2 π 3π/2 2π

Angle (Radians) Time (ms) at 60 Hz

0 4.2 8.3 12.5 16.6

-1 0 1

0 90 180 270 360

Voltage or Current

Angle (Degrees)

0 π/2 π 3π/2 2π

Angle (Radians) Time (ms) at 60 Hz

0 4.2 8.3 12.5 16.6

Figure 3-20: In-Phase 60 Hz AC Waveforms

To better understand the use of phase angles in AC circuits, consider the case of a typical distribution transformer that would be used to supply a residential electrical service (Figure 3-21). The secondary of the distribution transformer consists of

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two windings, both of which supply approximately 120 V. Recognize that the secondary windings would exhibit no voltage with respect to ground in the configuration shown because all terminals of the secondary windings are floating.

In order for there to be any relationship between the voltages across the secondary windings and ground, a connection must be made between one of the secondary terminals and earth. In a residential electrical service, this connection is made at the connection between the two windings (Terminal B)—as shown in Figure 3-22.

4140 V

120 V 120 V 4140 V

Figure 3-21: Typical Electrical Distribution Transformer 4140 V

120 V 4140 V

120 V

-120 +120 V

A B C

Figure 3-22: Typical Residential Electrical Service

By connecting Terminal B to ground, the voltage at this point is forced to become 0 V with respect to earth. Furthermore, because a voltage of 120 V exists across each of the windings, there is now a voltage of 120 V with respect to earth at both terminals A and C. It is important to realize, however, that the voltages at these two terminals are completely out-of-phase with one another. At a moment when the voltage across the windings is rising with the polarity indicated by the arrows,

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Terminal A will have a positive polarity while Terminal B will have a negative polarity. The AC voltage waveforms with respect to earth would therefore appear as shown in Figure 3-23

-1 0 1

0 90 180 270 360

Voltage

Angle (Degrees)

0 π/2 π 3π/2 2π

Angle (Radians) Terminal A

Terminal C

Figure 3-23: AC Waveforms on a Residential Electrical Service

The two waveforms in Figure 3-23 are said to be 180 degrees out-of-phase with one another because any point on one waveform is found on the other waveform to be shifted horizontally by 180 degrees. When waveforms are in-phase with one another, such as those in Figure 3-20, they have a phase shift of 0 degrees between them.

In general, any sinusoidal AC voltage waveform v(t) can be described by the following equation:

) (

cos )

(t = V ωt+φ

v _m [3-7]

where Vm is the peak voltage, ω is the angular frequency, t is time, and φ is the phase angle of the waveform. Angular frequency simply expresses frequency in either degrees/second—or, more commonly, in radians/second—rather than in cycles per second and is therefore simply:

πf

ω 2 [3-8]

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The phase angle is the angle in degrees (or radians) that the subject waveform is shifted from a pure cosine wave. Equation 3-7 is plotted in Figure 3-24 for the cases where the phase shift is equal to 0 degrees and –45 degrees.

-1 0 1

0 90 180 270 360

-Vm

ωt (Degrees) φ=0° φ=-45°

Figure 3-24: Plot of General Equation for Sinusoidal AC Waveforms

When performing analysis on AC waveforms having the same frequency, the only quantities required to distinguish a waveform are its amplitude and its phase angle.

Therefore, instead of using Equation 3-7 to denote a waveform’s characteristics, phasor notation is used. A sinusoidal AC voltage waveform can be identified in phase notation as:

∠

= V

V [3-9]

where V is the voltage phasor, V is the magnitude of the voltage waveform, and φ is its phase angle with respect to a pure cosine wave. Therefore, recalling the example of a residential electrical service in Figure 3-22, the correct way to refer to the voltages at terminals A and C would be 120V /0° and 120V /180°, respectively, rather than +120V and –120V. Phasors can also be represented graphically using a phasor diagram. The graphical representation is essentially a vector, having its origin at (0,0), having a length equal to the voltage magnitude, and an angle with respect to the positive x-axis is equal to the phase angle. For example, two waveforms having the phasor notations 1V /0° and 1V /45° would be represented graphically as shown in Figure 3-25.

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Y-axis

X-axis 1V /45°

1V /0° +1 -1

-1 +1

Figure 3-25: Typical Phasor Diagram

Phasor diagrams are important because they help illustrate how AC voltages and currents must be added and subtracted. In the case of determining the total voltage of a series of batteries (Figure 3-18), the voltages are simply added or subtracted based on their polarities. The same is true when dealing with AC voltages that are either in-phase, or 180 degrees out-of-phase—as was the case with the transformer windings in figures 3-17 and 3-18. However, when determining a voltage difference between two points in an AC circuit—where the phase difference between the waveforms is something other than 0 degrees or 180 degrees—the rules of vector algebra must be applied.

Example Calculation:

Two AC voltage sources are connected together in series, as shown in Figure 3-26. Both have a voltage output of 1 V; however, the waveforms are 45°

out-of-phase with one another. Determine the total voltage of the system.

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1V /0° 1V /45°

A B

VAB

Figure 3-26: Series Combination of AC Voltage Sources

The voltages of the two AC sources can be added together using the phasor diagram in Figure 3-27 and the rules of vector algebra. In order to add the two phasors together, the vector for source B is shifted horizontally so that its tail is placed at the head of the vector for Source A—as shown in Figure 3-27. The total voltage for the two sources is then provided by the new vector running from the origin to the head of the vector for source B.

Y-axis

X-axis VB= 1V /45°

VA=

1V /0° +1 -1

-1 +1

VAB

φΑΒ

Figure 3-27: Phasor Diagram for Problem in Figure 3-26

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If this diagram was drawn to scale, then VAB could be determined using a ruler to measure its magnitude, ⏐VAB⏐, and a protractor to measure its phase angle, /φAB. An easier and more accurate way, however, is to use trigonometry.

The vectors for VA and VB are broken down into x- and y- components. The x- and y- components for VAB are then determined by adding the x-components together and then adding the y-x-components together, as follows:

X-component of VAB: 1⋅ cos(0º) + 1⋅ cos(45º) = 1+ 0.71 = 1.71 Y-component of VAB: 1⋅ sin(0º) + 1⋅ sin(45º) = 0 + 0.71 = 0.71

The magnitude and phase angle for the new vector is then determined as follows:

Note that most scientific calculators can now add and subtract vector quantities directly, without the need to break the vectors down into components and then reassemble them.

Phasor multiplication and division are much simpler procedures than addition and subtraction. To determine the product of two phasor quantities, their magnitudes are multiplied together while their angles are added together.

Division is done similarly, as shown below:

Α∠φ × Β∠θ = Α ⋅ Β∠( φ + θ) [3-10]

Α∠φ ÷ Β∠θ = Α ÷ Β∠( φ − θ) [3-11]

In addition to vector algebra, the analysis of AC circuits requires the use of complex mathematics involving the use of both real and imaginary numbers.

This is a difficult concept for some to understand, but it can be simplified as follows.

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In complex mathematics, the x- and y- axes in the phasor diagram of Figure 3-25 are renamed the real and imaginary axes, respectively. Any phasor A/φ can then be broken down into real and imaginary components as follows:

A = x + jy [3-12]

where the real part of A, Re(A) is determined as:

x = ⏐A⏐cosφ [3-13]

and where the imaginary part of A, Im(A) is determined as:

y = ⏐A⏐sinφ [3-14]

The significance of j is to denote the imaginary component of the phasor, and it is known as the complex operator. The mathematical equivalent of j is:

−1

j [3-15]

hence it is referred to as an imaginary number. Note that in mathematics, an imaginary number is typically represented by the symbol I; but, so as not to confuse it with the symbol for current, electrical engineers have adopted the symbol j.

Although the mathematical definition for j can be perplexing, its use in AC circuit calculations is not. In essence, j is simply a phasor having a magnitude of unity and a phase angle of 90 degrees; that is,

j = 1/90º [3-16]

Therefore, if a phasor is multiplied by j, the magnitude of the phasor is unaffected; however, the phase angle becomes increased by 90º. Similarly, if a phasor is divided by j, its magnitude is unaffected but its phase angle will be decreased by 90º. Multiplication by -j has the same effect as dividing by +j, as shown below:

A/φ × j = A/φ × 1/90º = A/φ + 90º [3-17]

A/φ × (-j) = A/φ × -1/90º = A/φ + 90º = A/φ – 90º [3-18]

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A/φ ÷ j = A/φ ÷ 1/90º = A/φ – 90º [3-19]

One of the uses for the complex operator is the calculation of impedances, along with determining what effect an impedance has on the phase angle of an AC waveform. Just as the two AC voltage sources could not be added together directly in Figure 3-26, a resistance and a capacitive reactance connected together in series cannot be simply added together to determine their combined impedance.

Consider again the case of a capacitor in an AC circuit (Figure 3-28), as previously discussed. The magnitude of the reactance offered by the capacitor is given by Equation 3-3; therefore, the magnitude of the current I from the voltage source V can already be determined. However, to determine the effect that the capacitor has on phase angle, the following formula must be used:

Figure 3-28: Determination of Current through a Capacitor

The current is then calculated using Ohm’s Law as:

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By studying Equation 3-21, the capacitor is found to shift the phase angle of the current by 90 degrees relative to the voltage waveform. This is illustrated graphically in Figure 3-29.

-1 0 1

0 90 180 270 360

-Vm

ωt (Degrees) Applied Voltage 90° Shift

Resulting Current

Figure 3-29: Voltage and Current Waveforms for a Purely Capacitive Circuit

Because the current waveform is starting 90 degrees before the voltage waveform, it is said that the current leads the voltage in a capacitor.

Now consider the case of an inductor, L (Figure 3-30). In the same way that reactance in a capacitor results from the accumulation of electrical charges, reactance in an inductor results from the generation of a magnetic field. An inductor is generally a coil of wire; however, any conductor (such as a pipeline) has an inductive component.

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Figure 3-30: Determination of Current through an Inductor

The reactance of an inductor is calculated as:

fL j

X_L = 2π [3-22]

Because inductive reactance is proportional to frequency, an inductor has the opposite behavior to that of a capacitor; in effect, the inductor appears as a short circuit to DC and as an open-circuit to very high-frequency currents.

Once again, Ohm’s Law can be used to determine the current through the inductor:

− π ∠ π =

= 90

2 2

V fL

j V X

I V

L [3-23]

The phase angle for the current in Equation 3-23 indicates that an inductor shifts the phase angle of the current by –90 degrees relative to the voltage waveform; it is therefore said that the current lags the voltage in an inductor. This is illustrated graphically in Figure 3-31.

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Figure 3-31: Voltage and Current Waveforms for a Purely Inductive Circuit

The effect of capacitors and inductors on the phase angle of the current can be remembered by the following simple pneumonic device: “ELI the ICE man.” This indicates that E leads I in an inductive (L) circuit, whereas I leads E in a capacitive (C) circuit.

As Section 3.1 notes, this discussion of AC interference applies primarily to 3φ AC power lines having voltage waveforms that are 120 degrees apart—as shown in Figure 2. The phasor representation of such a 3φ circuit is shown in Figure 3-32.

Figure 3-32: Phasor Representation of a Three-Phase Circuit

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When dealing with 3φ powerlines, the system voltage is always specified in terms of the line-to-line voltage: that is, from phase A to phase B, phase B to phase C, or phase C to phase A. As the phasor diagram shows, the phase-to-phase voltage (from the tip of one phasor to the tip of another) is greater than the phase-to-ground voltage (the length of any one phasor). To determine the relationship between the phase-to-phase and phase-to-ground voltages, apply the rules of vector algebra as follows:

Vφ–φ = VA – VB

= V_φ–G /0º – V_φ–G /120º

= (V_φ–G cos 0º + jV_φ–G sin 0º) – (V_φ–G cos 120º + jV_φ–G sin 120º)

= (1.5V – j0.866V) V_Φ₋_G

( )

V _G V _G

V_φ₋_φ = + − ⋅ _φ₋ = _φ₋

∴ 1.5² .866 ² 3 [3-24]

Example Calculation:

Determine the line-to-ground voltage of a 500-kV transmission line.

kV 289

In document NACE CP Interference January 2008 (Page 133-148)