Power Systems

Chapter 2
Per-Unit Systems and Power-System Component Models

Bachelors Program in Electrical Engineering
Power Systems   •   Bachelor

Per-Unit Systems and Power-System Component Models

Introduction

Power systems interconnect equipment having widely different voltage, current and power ratings. A generating unit may operate at 13.8 kV, transmit power through a 132 kV network and supply consumers at 400 V. Direct calculations in physical units are possible, but transformer ratios, large numerical ranges and unit-conversion errors make them cumbersome.

The per-unit system normalizes electrical quantities to selected base values. It simplifies transformer calculations, allows direct comparison of equipment impedances and provides the standard numerical framework for load-flow, fault and stability studies. Reliable modeling also requires consistent phasor, complex-power and current-direction conventions.

Learning Objectives

After studying this chapter, students should be able to:

  • calculate balanced three-phase complex power from phasors;
  • select and derive compatible per-unit bases;
  • convert per-unit impedances between bases;
  • develop steady-state models of generators and transformers;
  • represent overhead lines and cables using appropriate equivalent circuits;
  • calculate cable charging current and reactive-power generation;
  • model loads, shunts and network interconnections;
  • apply consistent ABCD current-direction conventions; and
  • assemble component models for practical power-system analysis.

1. Phasors and Complex Power

1.1 RMS phasors

Under sinusoidal steady-state conditions, a voltage can be represented by an RMS phasor V = |V|∠θV. Phasors convert differential relationships into algebraic ones. For example, an inductor of inductance L has impedance jωL, while a capacitor of capacitance C has impedance 1/(jωC), where ω = 2πf and f is frequency in hertz.

Unless stated otherwise, this chapter uses RMS phasors and the passive sign convention: current entering the positive-voltage terminal gives positive absorbed power.

1.2 Single-phase and three-phase complex power

For a single phase,

S = VI^* = P + jQ

where:

  • S is complex power in volt-amperes;
  • P is active power in watts;
  • Q is reactive power in vars;
  • V and I are RMS phasors; and
  • the superscript * denotes complex conjugation.

For a balanced three-phase system,

S3φ = 3VphIph^*

The corresponding magnitude relationships are

P = √3VLLIL cosφ
Q = √3VLLIL sinφ
|S| = √3VLLIL

Here, VLL is line-to-line RMS voltage, IL is line current and φ is the angle by which current lags voltage. For a lagging load, Q > 0; for a leading load, Q < 0.

Power factor is

pf = P/|S| = cosφ

Example: load power

A balanced load takes 20 MVA at 0.8 lagging power factor. Therefore,

P = 20 × 0.8 = 16 MW
Q = 20√(1-0.82) = 12 MVAr

Thus S = 16 + j12 MVA.

2. The Per-Unit System

2.1 Definition

The per-unit value of a quantity is its actual value divided by a selected base value of the same dimension:

xpu = xactual/xbase

A percentage value is 100xpu. Thus, an impedance of 0.12 pu is also 12% impedance.

For balanced three-phase studies, select:

  1. a three-phase apparent-power base SB; and
  2. a line-to-line voltage base VB at one bus.

The other bases follow from electrical relationships:

IB = SB/(√3VB)
ZB = VB2/SB
YB = 1/ZB = SB/VB2

When SB is in MVA and VB is in kV, ZB is directly obtained in ohms. Current is in kA.

2.2 Why the impedance-base formula works

Using per-phase quantities, the phase-voltage base is Vph,B=VB/√3 and the current base is IB=SB/(√3VB). Hence,

ZB = Vph,B/IB
= (VB/√3)/(SB/(√3VB))
= VB2/SB

The formula therefore applies consistently when SB is the total three-phase power base and VB is the line-to-line voltage base.

2.3 Base calculation example

For SB=25 MVA and VB=11 kV,

IB = 25×106/(√3×11×103) = 1.312 kA
ZB = 112/25 = 4.84 Ω

An actual impedance of 0.968 Ω is therefore 0.968/4.84=0.20 pu.

2.4 Change of base

If an impedance is known in per unit on old bases, its value on new bases is

Zpu,new = Zpu,old(SB,new/SB,old)(VB,old/VB,new)2

Both voltage bases must refer to the same side of the same component.

Example: change of base

A reactance is 0.08 pu on 30 MVA, 11 kV bases. On 50 MVA, 10.5 kV bases,

Xpu,new = 0.08(50/30)(11/10.5)2 = 0.1463 pu

2.5 Advantages and limitations

Per-unit calculation offers several advantages:

  • transformer turns ratios disappear when voltage bases follow rated ratios;
  • equipment impedances occupy relatively narrow numerical ranges;
  • data from differently rated equipment can be compared directly;
  • dimensional errors become easier to detect; and
  • numerical conditioning is improved in computer studies.

A per-unit number is meaningful only when its bases are known. Mixing single-phase and three-phase bases, using phase voltage where line voltage is intended, or changing power base without changing impedance values produces incorrect results.

3. Generator Models

A synchronous generator converts mechanical input into electrical output. Its positive-sequence steady-state model consists of an internal generated voltage behind stator impedance.

Generator steady-state equivalent circuit E Rₐ + jXₛ V I
Generator equivalent circuit with internal voltage behind impedance
Figure: Generator equivalent circuit with internal voltage behind impedance

Figure: Per-phase generator model, with current directed out of the generator terminal.

With current I directed out of the generator,

E = V + (Ra + jXs)I

where E is internal generated voltage, V is terminal voltage, Ra is armature resistance and Xs is synchronous reactance. For many steady-state studies, Ra is small compared with Xs and is neglected.

Different reactances serve different studies. Synchronous reactance is appropriate for balanced steady state; transient and subtransient reactances are used during corresponding post-disturbance periods. Substituting one for another without regard to time scale is physically incorrect.

A generator connected to a strong grid is commonly represented as a controlled-voltage bus: active power and voltage magnitude are specified, while reactive power is calculated subject to capability limits. If a reactive limit is reached, the bus is treated as a specified-power bus in load-flow analysis.

4. Transformer Models

4.1 Ideal transformer and impedance referral

For an ideal transformer with turns ratio a=N1/N2, voltage magnitudes satisfy V1/V2=a, while current magnitudes satisfy I1/I2=1/a. An impedance on side 2 referred to side 1 is

Z'2 = a2Z2

A practical transformer model adds winding resistances, leakage reactances and a shunt magnetizing branch. The shunt branch contains core-loss resistance Rc and magnetizing reactance Xm. In many transmission studies, no-load current is small, so the shunt branch is omitted and the transformer is represented by its series leakage impedance.

4.2 Transformer per-unit invariance

If the power base is common on both sides and voltage bases are selected in the transformer’s rated voltage ratio, the per-unit series impedance has the same value on either side. This property eliminates explicit ideal-transformer ratios from many network diagrams.

An impedance-test result of 10% means |Z|=0.10 pu on the transformer’s own MVA and rated-voltage bases. It does not necessarily mean that the reactance alone is 0.10 pu; resistance must be included when its value is material.

4.3 Taps and phase shifts

An off-nominal transformer ratio is represented by a complex tap

t = τejθ

where τ is the magnitude ratio and θ is phase shift. Tap-changing transformers regulate voltage, while phase-shifting transformers influence active-power flow. Tap position, reference side and sign convention must be documented in network data.

5. Overhead Transmission-Line Models

A transmission line has distributed series impedance and shunt admittance. Per unit length,

z = r + jωL
y = g + jωC

where r is resistance, L is inductance, g is leakage conductance and C is capacitance, all per unit length. At power frequency, overhead-line leakage conductance is usually negligible.

5.1 Short, medium and long lines

The model should depend on electrical length and required accuracy, rather than on a rigid distance boundary.

ModelRepresentationTypical use
Short lineSeries Z onlyShunt charging negligible
Nominal πSeries Z with Y/2 at each endMedium-length lines and load flow
DistributedHyperbolic-function modelLong lines or high-accuracy studies

5.2 ABCD convention

This chapter uses

[VS; IS] = [A B; C D][VR; IR]

Here IS enters the line at the sending end, while IR leaves the receiving-end port and flows toward the receiving load. Thus, for a series-only line,

VS = VR + ZIR
IS = IR

If both two-port currents are instead defined as entering the network, the receiving current must appear as -IR in the transmission equations. Mixing these conventions changes signs and invalidates results.

For the nominal-π model,

A = D = 1 + YZ/2
B = Z
C = Y(1 + YZ/4)

For a distributed line of length ℓ,

A = D = cosh(γℓ)
B = Zc sinh(γℓ)
C = (1/Zc)sinh(γℓ)

where γ=√(zy) is the propagation constant and Zc=√(z/y) is the characteristic impedance.

6. Cable Models and Charging Current

Power cables have the same fundamental distributed parameters as overhead lines but normally have much larger capacitance because their conductors are close to grounded metallic screens and separated by high-permittivity insulation. Charging current can therefore constrain usable cable length and converter or transformer ratings.

6.1 Cable equivalent circuit

For a cable of length ℓ,

Z = (r + jωL)ℓ
Y = (g + jωC)ℓ

A practical power-frequency model is a nominal-π circuit with total series impedance Z and shunt admittances Y/2 at both ends. For long HVAC cables, several cascaded π sections or the distributed-parameter model should be used.

Nominal pi cable equivalent circuit Z = R + jX Y/2 Y/2 Sending Receiving
Nominal pi equivalent circuit for a power cable
Figure: Nominal pi equivalent circuit for a power cable

Figure: Nominal-π cable model with half of the shunt admittance at each end.

6.2 Charging-current calculation

Neglecting dielectric conductance and series voltage drop, the charging-current magnitude per phase is approximately

Ic = ωCtotVph

where Ctot=C'ℓ is capacitance per phase for the complete cable length. Since Vph=VLL/√3, the three-phase reactive power generated by the cable is

Qc = 3ωCtotVph2 = ωCtotVLL2

Here Qc denotes positive capacitive injection. Under the passive load convention, the cable shunt absorbs -Qc. In a π model, each end shunt contributes approximately half the total charging current at its local voltage.

Worked example: 132 kV cable

A 40 km, 132 kV, 50 Hz cable has capacitance 0.20 μF/km per phase. Thus,

Ctot = 0.20×40 = 8.0 μF
Ic = 2π(50)(8×10-6)(132000/√3) = 191.6 A

The three-phase capacitive reactive-power injection is

Qc = 2π(50)(8×10-6)(132000)2 = 43.8 MVAr

This charging current exists even when the receiving-end load is zero. Shunt reactors may therefore be required to absorb excess reactive power and control voltage.

7. Load and Shunt Models

7.1 Constant-power loads

A balanced constant-power load satisfies

I = S^*/(3Vph^*)

or, in per unit on compatible three-phase bases,

Ipu = Spu^*/Vpu^*

A constant-power model draws approximately the specified P and Q as voltage varies. Its current therefore increases as voltage falls, which can contribute to voltage instability.

7.2 Constant-impedance and constant-current loads

For a balanced load consuming total three-phase power S at line voltage VLL, its equivalent per-phase impedance is

Z = VLL2/S^*

For constant impedance, power varies with the square of voltage magnitude. A constant-current load has power approximately proportional to voltage magnitude when its current angle relative to voltage remains fixed.

The ZIP model combines these behaviors:

P/P0 = aZv2 + aIv + aP
Q/Q0 = bZv2 + bIv + bP

where v=|V|/V0, and each coefficient group normally sums to one.

7.3 Shunts

A shunt admittance is

Ysh = G + jB

A capacitive shunt has B>0 and injects reactive power into the network. For negligible conductance, the three-phase per-unit reactive injection is approximately

Qinj,pu = Bpu|Vpu|2

A shunt reactor has negative susceptance and absorbs reactive power. Because shunt reactive power varies with voltage squared, a fixed capacitor cannot maintain constant MVAr output during depressed voltage.

8. Network Interconnection and the Bus-Admittance Model

After conversion to a common base, component models can be assembled into a network. The nodal equation is

Ibus = YbusVbus

For a branch admittance yik between buses i and k, the off-diagonal entries receive -yik, while the corresponding diagonal entries receive +yik. Shunt admittances are added only to the relevant diagonal entries.

For a branch with series admittance y and a complex tap t on the from side, its contributions are

Yff = y/|t|2, Yft = -y/t^*, Ytf = -y/t, Ytt = y

Any line-charging admittances must also be added at the appropriate terminals. Correct bus numbering, tap orientation and sign conventions are essential for reproducible software models.

9. Illustrative Case: Generator-to-Load Interconnection

Background and problem

This illustrative case considers a generator supplying a 132 kV load through a step-up transformer and transmission line. The purpose is to develop a common-base model and estimate internal generator voltage and line loss.

Use a 50 MVA system base. The data are:

  • generator: 40 MVA, 13.8 kV, Xs=0.15 pu on its own base;
  • transformer: 50 MVA, 13.8/132 kV, leakage reactance 0.10 pu;
  • line: Z=8+j40 Ω;
  • receiving load: 35+j15 MVA at VR=1∠0° pu.

Line charging and component resistances other than line resistance are neglected.

Analysis

The generator reactance on 50 MVA base is

XG,new = 0.15(50/40) = 0.1875 pu

The 132 kV impedance base is

ZB = 1322/50 = 348.48 Ω

Therefore, line impedance is

Zline,pu = (8+j40)/348.48 = 0.02296+j0.11478

The total series impedance from internal generator voltage to the load bus is

Ztotal = 0.02296 + j(0.1875+0.10+0.11478)
= 0.02296+j0.40228 pu

The load power is

SL = (35+j15)/50 = 0.7+j0.3 pu

At VR=1∠0° pu,

I = SL^*/VR^* = 0.7-j0.3 pu

The internal generated voltage is

E = VR + ZtotalI
= 1 + (0.02296+j0.40228)(0.7-j0.3)
= 1.13676+j0.27471 pu

Thus,

|E| = 1.1695 pu, ∠E = 13.6°

Only the line resistance contributes to real-power loss in this simplified model:

Ploss,pu = Rline|I|2
= 0.02296(0.72+0.32)
= 0.01332 pu
Ploss = 0.01332×50 = 0.666 MW

Outcome and learning points

The common-base model removes the ideal transformer ratio and permits all series impedances to be added directly. The calculation also shows that reactive impedance strongly influences required internal voltage, while real loss depends on resistance. A complete operational study would include line charging, generator reactive limits, transformer taps and voltage-dependent load behavior.

10. Modeling Practice, Limitations and Contemporary Developments

A model is an approximation selected for a purpose. Positive-sequence models are efficient for balanced load flow but cannot represent phase unbalance or zero-sequence grounding paths. Constant-power loads may be suitable near nominal voltage but unrealistic during severe disturbances. Temperature changes conductor resistance, magnetic saturation changes transformer parameters, and cable capacitance varies with construction and installation.

Professional best practice includes:

  • recording MVA, voltage and frequency bases with every per-unit dataset;
  • stating RMS, phase and current-direction conventions;
  • validating transformer tap side and phase shift;
  • distinguishing rated data from measured operating parameters;
  • selecting line or cable models according to accuracy requirements;
  • checking computed voltages, currents and powers against equipment ratings; and
  • maintaining traceable network-model revisions.

Utilities generally use IEC- or IEEE-aligned equipment definitions, testing methods and protection requirements. Model governance has safety and regulatory significance: incorrect ratings or topology can lead to unsafe switching decisions, overloaded assets or misleading planning results.

Digital substations, synchronized phasor measurements and online parameter estimation increasingly improve model validation. Machine-learning tools can identify suspicious data or estimate changing load composition, but they do not replace physical constraints, engineering review or auditable sign and unit conventions.

Chapter Summary

Complex power combines active and reactive power through phasor relationships. The per-unit system normalizes quantities on common power and voltage bases, simplifies transformer coupling and supports network-wide computation. Generators are modeled as internal voltages behind appropriate reactances, while transformers require leakage impedance, ratio and tap information. Overhead lines and cables may be represented by series, nominal-π or distributed models. Cable capacitance can produce substantial charging current and reactive-power injection. Loads may be modeled as constant power, current, impedance or ZIP combinations, while shunts provide voltage-dependent reactive compensation. Consistent ABCD current directions, base quantities and network-data conventions are indispensable for accurate engineering analysis.

Key Terms

ABCD parameters: Constants relating sending-end and receiving-end voltage and current.

Base quantity: Selected reference value used to normalize an electrical variable.

Charging current: Capacitive current drawn by a line or cable shunt capacitance.

Complex power: Quantity S=P+jQ combining active and reactive power.

Constant-power load: Load modeled as maintaining specified active and reactive power.

Generator internal voltage: Generated emf represented behind the machine impedance.

Line charging: Reactive effect caused by transmission-line or cable capacitance.

Nominal-π model: Equivalent circuit with series impedance and half the shunt admittance at each end.

Off-nominal tap: Transformer ratio differing from the nominal system voltage ratio.

Per unit: Actual quantity divided by its compatible base value.

Phasor: Complex representation of a sinusoidal RMS quantity and phase angle.

Power factor: Ratio of active power to apparent-power magnitude.

Reactive power: Imaginary component of complex power associated with energy storage.

Shunt admittance: Conductance and susceptance connected from a bus to reference.

Synchronous reactance: Steady-state generator reactance used in balanced analysis.

Transformer leakage impedance: Series impedance produced primarily by leakage flux and winding resistance.

Voltage base: Reference voltage selected at a system bus.

ZIP model: Load model combining constant-impedance, constant-current and constant-power behavior.

Review Questions

Short Answer Questions

  1. Define per-unit value and percentage value.
  2. State the passive sign convention for complex power.
  3. Why does transformer per-unit impedance remain unchanged across windings under compatible bases?
  4. What current direction is assigned to IR in the ABCD convention used in this chapter?
  5. Why is cable charging usually more significant than overhead-line charging?

Long Answer Questions

  1. Derive the balanced three-phase current and impedance bases from selected power and voltage bases.
  2. Explain the construction, assumptions and limitations of the synchronous-generator steady-state model.
  3. Compare short-line, nominal-π and distributed-parameter transmission models.
  4. Develop the practical transformer equivalent circuit and explain when its shunt branch may be omitted.
  5. Compare constant-power, constant-current, constant-impedance and ZIP load models.

Analytical / Application Questions

  1. A 100 MVA, 220 kV base is selected. Calculate current, impedance and admittance bases, and convert an actual impedance of 15+j60 Ω to per unit.
  2. A 66 kV, 50 Hz cable is 30 km long and has capacitance 0.18 μF/km per phase. Estimate total charging current and three-phase reactive-power injection.
  3. Construct a common-base reactance diagram for a generator, transformer and line of your choice. Calculate the generator internal voltage for a specified receiving-end load and voltage.

Multiple Choice Questions with Answers

Question 1

A balanced three-phase load consumes 30 MW and 15 MVAr. What is its complex power?

A. 30-j15 MVA

B. 15+j30 MVA

C. 30+j15 MVA

D. 45+j0 MVA

Correct Answer: C. 30+j15 MVA

Question 2

On a 100 MVA, 220 kV three-phase base, what is the impedance base?

A. 2.20 Ω

B. 48.4 Ω

C. 220 Ω

D. 484 Ω

Correct Answer: D. 484 Ω

Question 3

An impedance is 0.10 pu on a 20 MVA base. If the voltage base is unchanged, what is it on a 50 MVA base?

A. 0.04 pu

B. 0.10 pu

C. 0.25 pu

D. 0.40 pu

Correct Answer: C. 0.25 pu

Question 4

When voltage bases on the two sides of a transformer follow its rated ratio and the power base is common, which statement is correct?

A. The per-unit impedance is the same on both sides.

B. The per-unit impedance changes by the turns ratio.

C. The per-unit impedance changes by the square of the turns ratio.

D. The per-unit impedance becomes zero.

Correct Answer: A. The per-unit impedance is the same on both sides.

Question 5

A 20 MVA load operates at 0.8 lagging power factor. What reactive power does it consume?

A. 10 MVAr

B. 12 MVAr

C. 16 MVAr

D. 25 MVAr

Correct Answer: B. 12 MVAr

Question 6

For [VS;IS]=[A B;C D][VR;IR] and a series line satisfying VS=VR+ZIR, how is IR directed?

A. Into the receiving port from the load

B. Out of the receiving port toward the load

C. From ground into the sending bus

D. Opposite to both load and line current

Correct Answer: B. Out of the receiving port toward the load

Question 7

A 33 kV, 50 Hz cable is 10 km long and has capacitance 0.25 μF/km per phase. Neglecting series voltage drop, what is its approximate charging current per phase?

A. 4.99 A

B. 8.64 A

C. 14.96 A

D. 25.91 A

Correct Answer: C. 14.96 A

Question 8

A constant-impedance load consumes 10 MW at 1.0 pu voltage. Approximately how much active power does it consume at 0.9 pu voltage?

A. 8.1 MW

B. 9.0 MW

C. 10.0 MW

D. 11.1 MW

Correct Answer: A. 8.1 MW

Question 9

Under the convention that positive reactive injection supports the network, a capacitor with susceptance B>0 injects approximately what per-unit reactive power at voltage magnitude |V|?

A. -B|V|

B. B/|V|2

C. -B|V|2

D. B|V|2

Correct Answer: D. B|V|2

Question 10

Which nominal-π ABCD parameter set is correct for total series impedance Z and total shunt admittance Y?

A. [[MATH:A=D=1,

B=Z,

C=Y]]

B. [[MATH:A=D=1+YZ/2,

B=Z,

C=Y(1+YZ/4)]]

C. [[MATH:A=D=YZ,

B=Y,

C=Z]]

D. [[MATH:A=D=1-YZ/2,

B=1/Z,

C=1/Y]]

**Correct Answer: B. [[MATH:A=D=1+YZ/2,

B=Z,

C=Y(1+YZ/4)]]**

Power Systems
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