2.1 INTRODUCTION

In modern AC electric power systems (Figure 2.1), power is generated, transmitted, and distributed as balanced, three-phase AC. The three-phase system was independently invented by Galileo Ferraris, Mikhail Dolivo-Dobrovolsky, Jonas Wenström, and Nikola Tesla in the 1880s and is the most widely used means of transferring power through power grids. The three-phase system has the advantage of economy over the single-phase system because more power can be transmitted with significant cost savings in conductors per unit line length. The three-phase system may be configured as three-wire star, four-wire star, or three-wire delta system.

Because of advances in electronics, the future electric power system is headed in the direction of microgrids and smart grids. In anticipation of demands, researchers and students need to be equipped with relevant knowledge on the emerging trends in this area. This chapter introduces the fundamentals of electric power systems and the basic computational tools needed for the design and analysis of the future-generation power system.

2.2 COMPLEX POWER CONCEPTS

In electrical power systems, we are mainly concerned with flow in the electrical circuit, such as Voltage (V), Frequency (f), Current (I), and Power (P). To treat sinusoidal, steady-state behavior of an electric current, some further definitions are necessary.

Let

(2.1)

and

(2.2)

and

(2.3)

where

(2.4)

In polar form,

(2.5)

or

(2.6)

It is important to recall the following trigonometric identities:

(2.7)

(2.8)

For a voltage signal represented in terms of root mean square (rms) value:

(2.9)

(2.10)

(2.11)

(2.12)

For impedance, if the circuit is purely resistive, inductive, or capacitive, there will be difference in current and angle phase shift.

2.2.1 Purely Resistive Circuit

In a purely resistive circuit, Figure 2.2, the current is in phase with voltage:

(2.13)

2.2.2 Purely Inductive Circuit

Current lags behind V by 90° as seen in Figure 2.3:

(2.14)

2.2.3 Purely Capacitive Circuit

In a purely capacitive circuit Figure 2.4, current leads voltage by 90°:

(2.15)

2.2.4 Instantaneous Power

The instantaneous power is given by:

(2.16)

(2.17)

(2.18)

(2.19)

where Cosϕ is the power factor.

2.2.5 Power Factor

The power factor of a system is defined as:

(2.20)

where P = VICosϕ and Q = VISinϕ.

Q is the reactive power measured in kilo-var (kvar).

2.2.6 Complex Power

The apparent power S is then:

(2.21)

(2.22)

and

(2.23)

(2.24)

If the load impedance is Z, then,

(2.25)

In terms of load admittance,

(2.26)

(2.27)

2.3 REVIEW OF AC-CIRCUIT ANALYSIS USING PHASOR DIAGRAMS

Consider the AC circuit of Figure 2.5. The load, operating at a voltage V_L, draws a current I_L from the source whose voltage is V_s, through resistance R and inductive reactance jX.

By Kirchhoff's voltage rule:

(2.28)

(2.29)

where

(2.30)

(2.31)

If the load is operated at power factor Cosϕ and voltage V_L then

(2.32)

(2.33)

Then Equation 2.29 may be rewritten as:

(2.34)

(2.35)

A phasor diagram Figure 2.6 may be constructed based on Equation 2.35 from which the source voltage V_s may be determined.

2.4 POLYPHASE SYSTEMS

One of the methods of transmitting and distributing AC electric power is by means of polyphase systems. This is a system with three or more energized AC currents carrying conductors with a phase deviation between them. For a balanced n-phase system, the phase difference is given by:

(2.36)

For two-, three- … six-phase systems, voltages will be out of phase by 90°, 120°, …60°, respectively.

Polyphase systems are particularly useful for transmitting power as more power can be transmitted than when a single phase is used.

2.4.1 Three-Phase Circuits

Power generation, transmission, and distribution are usually connected in a type of polyphase system for heavy utilization of AC electric power. These types of connections provide economic advantages as well as system stability and capacity control. Both voltage and currents are sinusoidal waveforms equal in magnitude, but are displaced from one another by 120° in time phase. Stator windings are connected in three-phase through a ground wire, leading to four wires. The center of the four wires (Figure 2.7) leads to a Y-connected system, where each is referred to as a phase and the fourth conductor is called the neutral wire, which has four-wire balanced connection.

2.4.2 Balanced Y-Connected Three-Phase Source

In vector (phasor) form

(2.37)

(2.38)

(2.39)

(2.40)

(2.41)

(2.42)

Similarly, for Δ-connected system

(2.43)

2.4.3 Phase and Line Voltages: Delta Connected

The line-neutral and line–line voltages with proper relations are shown in Figure 2.8.

(2.44)

Line-Neutral, V_BA leads V_NA by 30°, V_CB leads V_NB by 30°.

2.4.4 Equivalent Y-Connected Voltage Phasor Representation

Taking V_NA as a reference from Figure 2.9,

(2.45)

(2.46)

(2.47)

2.4.5 Mesh or Delta Connection

(2.48)

(2.49)

(2.50)

These connections are not properly balanced to maintain balanced voltage across each load. A delta-connected generator Figure 2.10 is possible but not desirable for two main reasons:

1. Grounding is not possible with a delta-connected generator. For safety, a generator-neutral point typical of Y is the logical point of connection to ground.
2. A delta connection of the coils of the generator provides a short-circuited path in which current can flow. Third harmonics in the coil voltages cause a disturbance, which produces power loss and lowers the efficiency of the generator.

2.5 THREE-PHASE IMPEDENCE LOADS

A delta-connected load or Y-connected load (Figure 2.11) uses the same configuration as discussed in Section 2.8.

(2.51)

(2.52)

(2.53)

From which the individual phase currents may be computed as

(2.54)

(2.55)

(2.56)

The line currents are

(2.57)

(2.58)

(2.59)

Also, under a balanced condition,

(2.60)

In general, the sum of the three line currents in a three-phase, three-wire system is zero. This implies that power is also balanced.

Total power

(2.61)

These are also given as real power for each phase, i.e.,

(2.62)

(2.63)

(2.64)

Accordingly, if

(2.65)

then |P_p| are equal and out of phase 120° apart and the line currents are . Similarly, voltages are equal and out of phase by 120°.

Therefore, we write: V_CB = V_Ph = V_Lfor delta connection

Total power

Therefore,

(2.66)

Cosθ is referred to as the power factor.

The same applies to a Y-connected system (Figure 2.12)

(2.67)

A Y-connected load can have four wires, as shown:

(2.68)

(2.69)

(2.70)

The phase and line currents have the same value

(2.71)

(2.72)

(2.73)

This leads to:

So that total power

(2.74)

(2.75)

(2.76)

Total power

(2.77)

Example 1

Find the positive-sequence, line-to-line voltages in the balanced Y-connected generator if the line-to-neutral voltages have a magnitude of 12 kV and phase a is the reference.

Hence,

Example 2

Given a Y-connected load of impedance Z_Y = 10∠30° Ω and a balanced positive sequence source with V_ab = 14.4∠30° kV, find the load currents.

Solution:

Example 3

Three noninductive resistances, each of 100 ohms, are connected in star to a three-phase 440 V supply. Three inductive coils, each of reactance 100 ohms connected in delta are also connected to the supply. Calculate the (a) line current and (b) power factor of the system.

Solution:

Voltage across each phase resistance = 

Current flowing through each resistor

For three inductive coils connected in delta, V_φ = V_L = 440V

Total line current drawn from the source = (2.53 − j7.62) = 8.032 A

2.6 TRANSFORMATION OF Y TO DELTA AND DELTA TO Y NETWORKS

In power-network computations, it is often required to convert a Y network to a delta (Δ) and vice versa. Consider the Y connection of impedances shown in Figure 2.13.

Given Z_A, Z_B, and Z_C,

(2.78)

(2.79)

(2.80)

For balance, if

then

(2.81)

For conversion of delta to Y, the diagram and process is as shown in Figure 2.14:

If

then

2.7 SUMMARY OF PHASE AND LINE VOLTAGES/CURRENTS FOR BALANCED THREE-PHASE SYSTEMS

Table 2.1 shows a summary of phase and line voltages and currents for balanced three phase systems.

Connection		Phase Voltages/Currents	Line
Y–Y	Voltage	VYAN = Vp∠0°		1.1
		VYBN = Vp∠ − 120°
		VYCN = Vp∠120°
	Current	Phase currents same as line currents	where + θ is for leading pf and − θ is for lagging pf
			IB = IA∠ − 120° ± θ
			IC = IA∠120° ± θ
Y − Δ	Voltage	VYAN = Vp∠0°		1.2
		VYBN = Vp∠ − 120°
		VYCN = Vp∠120°
	Current		IΔ1 = IA = IA∠ ± θ	1.3
			IΔ2 = IB = IA∠ − 120° ± θ
			IΔ3 = IA∠120° ± θ
Δ − Δ	Voltage		Same as phase voltage
	Current		IΔ1 = IA∠ ± θ	1.4
			IΔ2 = IA∠ − 120° ± θ
			IΔ3 = IA∠120° ± θ
	Voltage	VYAN = Vp∠0°		1.5
		VYBN = Vp∠ − 120°
		VYCN = Vp∠120°
	Current		IΔ1 = IA = IA∠ ± θ
			IΔ2 = IB = IA∠ − 120° ± θ
			IΔ3 = IA∠120° ± θ

Example 1

A balanced, delta-connected load having an impedance 20 − j15Ω is connected to a delta-connected, positive-sequence generator having V_ab = 330∠0°V. Calculate the phase currents of the load and the line currents.

Solution:

The load impedance per phase is: Z_Δ = 20 − j15 = 25∠ − 36.87°Ω.

Since V_AB = V_ab, the phase currents are:

For delta load, the line current always lags the corresponding phase current by 30° and has a magnitude times that of the phase current. Hence, the line currents are:

Example 2

A balanced abc sequence, Y-connected source with V_an = 100∠10° is connected to a Δ-connected, balanced load 8 + j4Ω per phase. Calculate the phase and line currents.

Option 1

If the phase voltage V_an = 100∠10°, then the line voltage is

or

The phase currents are

The line currents are

Option 2

Use a single-phase analysis approach.

Solution:

Convert the delta-connected load to the star using the delta–star conversion procedure where

hence

Recall that line current and phase current are the same in a star-connected system.

Hence, using the abc sequence, line currents are obtained as:

Also

Example 3

A balanced, delta-connected load with an impedance 20 − j5Ω is connected to a delta-positive sequence generator with V_ab = 330∠0°. Calculate the phase currents of the load and the line currents.

Solution:

The load impedance per phase is

For a delta-connected system, phase voltage is equal to line voltage, i.e., V_AB = V_ab, hence the phase currents are

For a delta load, the line current always lags the corresponding phase current by 30° and has a magnitude of times that of the phase current. Hence, the line currents are

Example 4

A balanced, Y-connected load with a phase impedance of 40 + j25Ω is supplied by a balanced, positive-sequence delta-connected source with a line voltage of 210 V. Calculate the phase currents. Use I_ab as a reference.

Solution:

The load impedance is

and the source voltage is V_ab = 210∠0°.

When the delta-connected source is transformed into a Y-connected source,

The line currents are

which are the same as the phase currents.

2.8 PER-UNIT SYSTEMS

The power-system quantities such as voltage, current, and power are normalized for ease of computation. In power-system analysis, per unit or percent of specified base values typically are expressed as measurements. The advantages of using per unit measurements are:

1. ease of computation and ease of comparison of results for various power systems that may have different base quantities,
2. early detection of calculation errors, especially when device parameters fall in small ranges, and
3. elimination of ideal transformers as windings, such that voltages, currents, and external impedances and admittances, expressed in per unit, do not change when they are referred from one side of the transformer to another. This leads to computational savings in a power system with hundreds of transformers and helps to eliminate errors in calculation.

The per unit definition allows an actual quantity to be normalized to unity and facilitates the comparison of all other measured values to that base-unit value. By definition, the per-unit value of a quantity X is given by

(2.82)

The actual quantity is the value of the quantity in actual units such as watts, vars, or hertz. The base values are determined by the mass quantity allowed for a device in terms of V, I, P, or f by the operation.

The base value has the same unit as the actual quantity, hence per unit is dimensionless. For electrical laws to be valid in the per unit system, the following relations must be used for other bases:

given that

where

(2.83)

then

(2.84)

such that

(2.85)

(2.86)

(2.87)

(2.88)

(2.89)

(2.90)

Note that:

1. The value of P_φ = Q_φ = S_φ applies to the entire system and is the same for the entire circuit.
2. The ratio of voltage bases on either side of a transformer is selected to be the same as the ratio of the transformer voltage ratings. With these two rules, it is evident that per unit impedance is the same for transformers when referred from one side to the other.

Example 1

Consider a 100 V sinusoidal source in series with a 3Ω resistor, an 8Ω inductor, and a 4Ω capacitor. Taking base voltage as 100 V and base VA as 500VA and working in per unit:

1. Draw the circuit diagram representing each component by its per unit value.
2. Compute the series impedance and the current flowing in the circuit.

Solution:

or

Converting the circuit component values to per unit:

The circuit is:

1.  

   

2. Z_p.u. = 0.15 + j(0.4 − 0.2)
      = 0.15 + j0.2
      0.25∠53.1°
3. 

   4.0∠ − 53.1°

Illustrative Problems and Examples

1. Three impedances of 4 + j3Ω are Δ connected and tied to a three-phase 208 V power line. Find:
   1. I_ph, I_L, P, S, Q
   2. power factor for this load

Solution:

For Δ connection,

2. If v is 141.4 cos(ωt + 60°) V and i is 14.14 sin(ωt + 120°) A, find both V and I:
   1. maximum value
   2. rms value
   3. phasor expression using voltage as the reference
   4. phasor expression using current as the reference
   5. whether circuit in which the voltage and current exist is capacitive or inductive

Solution:

1. 
2. 
3. 

   The voltage leads the current by 30°.

4. The current lags the voltage by 30°.
5. Hence the circuit is inductive.

3. Express each of these currents as a phasor
   1. 800 cos(ωt + 130°)A
   2. 480cos ωt − 640sin ωt A
   3. 550 sin(ωt + 40°) + 420 sin(ωt + 110°)A

Solution:

1. 
2. 
3. 

2.9 CHAPTER SUMMARY

In this chapter we presented the fundamentals of electric power network analysis concepts and principles, which are essential in analysis and computations in any electric power system. The nature and behavior of electric currents and voltage signals, and the effects of their interactions with discrete circuit components, are important base knowledge for students, as is awareness of the three-phase system as a better, more reliable carrier of electric power. Circuit analysis at this level is equally beneficial to dealing with more advanced electric power-system issues. Overall, fundamental computational skills needed for any analysis of the power system was presented here. This information is particularly applicable for competitive work in the micogrid.