lec16

EE2022 Electrical Energy Systems

Panida Jirutitijaroen Department of Electrical and Computer Engineering

3/15/2013 EE2022: Per Unit Analysis by P. Jirutitijaroen 1

Lecture 16: Per Unit Analysis – Single Phase 19-03-2013

Detailed Syllabus

3/15/2013 EE2022: Three-phase circuit by P. Jirutitijaroen 2

31/01/2013 Three-phase power systems: Introduction to three-phase circuit. Balanced three-phase

systems. Delta-Wye connection. Relationship between phase and line quantities

05/02/2013 Three-phase power systems: Per-phase analysis. Three-phase power calculation. Examples.

07/02/2013 Generation: Simple generator concept. Equivalent circuit of synchronous generators

14/02/2013 Generation: Operating consideration of synchronous generators, i.e. excitation voltage

control, real power control, and loading capability

21/02/2013 Generation: Principle of asynchronous generators. Examples.

Transmission: Overhead VS Underground cable.

05/03/2013 Transmission: Four basic parameters of transmission lines.

07/03/2013 Transmission: Long transmission line model, Medium-length transmission line model, Short

transmission line model. Operating consideration of transmission lines i.e. voltage regulation,

line loadability, efficiency. Examples.

12/03/2013 Distribution: Principle of transformer. Ideal transformer.

14/03/2013 Distribution: Reflected load. Impedance matching. Practical transformer. Examples

19/03/2013 Per unit analysis: Single-phase per unit analysis.

21/03/2013 Per unit analysis: Three-phase transformer. Three-phase per unit analysis. Examples.

02/04/2013 2nd Mid-term test

Generators, transmission lines, and transformers (Tutorials 4-6)

Review of Electrical Energy Systems

G1 G2

Load

Transmission Line Transmission Line

Step-up Transformer Step-up Transformer

Step-down Transformer

A Single-line diagram

3/15/2013 3 EE2022: Per Unit Analysis by P. Jirutitijaroen

IN THIS LECTURE

Learning outcomes

Outline

Reference


Learning Outcomes

• Apply the concepts of per-phase analysis and per-unit analysis to solve three-phase balanced circuit problems in power engineering.


References

• Glover, Sarma, and Overbye, “Power System Analysis and Design”.

– Chapter 3


SINGLE PHASE PER UNIT ANALYSIS

Per Unit Quantity

Base Value

Change of Base

Steps of Calculation


Motivations • Transformer introduces various voltage levels. • So far we can only reflect the load from one side of the

transformer to another. Still we need to use turns ratio to find voltage and current at each side of the transformer.

• It is difficult to calculate voltage and current of the system at various points.

• It is even more difficult for system operators to observe the current situation of the system.

EE2022: Per Unit Analysis by P. Jirutitijaroen 8 3/15/2013

30 kV – good or bad? 30 kV – good or bad?

15kV:150kV 150kV:30kV 30kV:300 V 300V:150 V Generation Transmission Load Distribution

Per Unit System

• Per unit system is when we normalize the voltage and current at each location.

• The normalization typically follows transformer ratings. • This usually makes the per unit value of both voltage

and current to be around 1.0 per unit. • Per unit system allows system operators to overlook

abnormalities in the system easily.


30 kV = 1.0 per unit 30 kV = 0.2 per unit

15kV:150kV 150kV:30kV 30kV:300 V 300V:150 V

Per Unit Quantity

• The per unit quantity of voltage, current, power and impedance is found from dividing the actual quantity by a base value of that quantity.

• Per unit value is denoted by ‘p.u.’.

• All base values are real numbers, denoted by subscript ‘B’.

• The base value is used only to normalize the quantity.


quantity of valuebase

quantity actualquantityunit per

Base Value for Voltage

• Transformers separate overall circuit to different zones with different voltage levels.

• We typically set the base value quantity for voltage following transformers’ voltage ratio.

• Note that the per unit values of the voltage at both sides of an ideal transformer are the same . Why?


Example 1

• Consider the following electrical energy system.

– How many zones (with different voltage level) does the system has?

– Find the base value of the voltage at each zone.

15kV:150kV 150kV:30kV 30kV:300 V 300V:150 V

Zone 1 15kV

Zone 2 150kV

Zone 3 30kV

Zone 4 300V

Zone 5 150V


Example 2

• Consider a single phase 480/120 V transformer. – Choose the base value of voltage on the primary side to be

480 and that of the secondary side to be 120.

– If the voltage at primary is measured to be 432 V, which is 0.9 per unit, the voltage at secondary side is 108 V.

– What is the per unit quantity on the secondary side?


480:120

9.0120

108quantityunit per

Example 3

• Consider the following electrical energy system.

– Find the base value of the voltage at each zone.

15kV:150kV 150kV:30kV 30kV:300 V 320V:150 V

Zone 1 15kV

Zone 2 150kV

Zone 3 30kV

Zone 4 300V

Zone 5

V 625.140

300320

150


Base Value for Complex Power

• First, choose voltage base values following transformer voltage ratings.

• Select only single base complex power in the system.

• The base value of power is used to normalize the quantity. Thus, the base values of real power, reactive power, and complex power are all the same real number.


111

BBB SQP

1

BS

Base Value for Current and Impedance

• Current base values are calculated from the base power and base voltage.

• Impedance base values (same value for impedance, resistance, or reactance) are calculated from voltage and current.

B

BB

V

SI

1

1

2

B

B

B

BB

S

V

I

VZ BBB ZXR


KVL, KCL, Complex Power Calculation

• We can still apply KVL, KCL, complex power calculation to the per unit value.

• The actual quantity is simply found from multiplying the per unit quantity (normalized quantity) with the base value.


BBB IZV

BBB IVS

p.u.p.u.p.u. IZV

*

p.u.p.u.p.u. IVS

Think of Base value as ‘Normalization’.

Example 4: Per Unit Value

• A single-phase 20kVA, 480/120 V, 60 Hz transformer has an equivalent leakage impedance referred to 120-volt winding of Ω. Using the transformer rating as base values, find per-unit leakage impedance.


13.780525.0eq2Z

V 4801 BVkVA 20BS

V 1202 BV

52.11

1

2

11

B

BB

S

VZ

72.0

1

2

22

B

BB

S

VZ

p.u. 13.780729.0

72.0

13.780525.0

2

2

p.u.

B

eq

Z

ZZ

eq2Z

Example 4: Per Unit Circuit


V 4801 BV

kVA 20BS

V 1202 BV

52.111BZ 72.02BZ p.u. 13.780729.0p.u. Z

There is no transformer in the per unit equivalent circuit.

Per unit equivalent circuit

eq2Z4:1 p.u.Z

P.U. Equivalent Circuit of a Transformer

• Ideal transformer model

• Practical transformer model


a:1

V₁(p.u.) V₂(p.u.)

i₁ (p.u.) i₂(p.u.)

+ +

- -

Zeq (p.u.)

Y (p.u.) V₁(p.u.) V₂(p.u.)

i₁ (p.u.) i₂(p.u.)

+ +

- -

a:1 Zeq

Y

Change of Base Value

• Manufacturers usually specify equipment impedances in per unit values together with voltage ratings (V) and apparent power rating (VA).

• The impedance base values can be found from the ratings of the equipment.

• Different equipment has different ratings.

• We may need to calculate per unit values on the new basis.


new

B

new

p.u.

old

B

old

p.u.actual ZZZZZ new

B

old

B

old

p.u.new

p.u.Z

ZZZ

You need to practice this in tutorial problems.

Steps of Per Unit Analysis

1. Choose for the system.

2. Select for different zones (usually follows transformer voltage ratings).

3. Calculate for different zones.

4. Express all quantities in p.u.

5. Draw impedance diagram and solve for p.u. quantities.

6. Convert back to actual quantities if needed.

EE2022: Per Unit Analysis by P. Jirutitijaroen 22

1

BS

BV

BZ

3/15/2013

Example 5: 1Ф, Per Unit Analysis

• Three zones of a single-phase circuit are shown below. Use base value of 30 kVA and 240 V in zone 1, draw per unit circuit and find per unit value of source voltage and all impedances.


V 0220 sVT1: 30 kVA 240/480 V

Xeq = 0.1 p.u.

T2: 20 kVA 460/115 V

Xeq = 0.1 p.u.

2jZline 2.09.0 jZload

Example 5: Base Values of Each Zone


V 2401 BV

kVA 30BS

V 480240240

4802

BV

48.0

2

33

B

BB

S

VZ

T1: 30 kVA 240/480 V

Xeq = 0.1 p.u.

T2: 20 kVA 460/115 V

Xeq = 0.1 p.u.

V 120480460

1153

BV

68.7

2

22

B

BB

S

VZ

p.u. 0.260422

2

.., jZ

jZ

B

upline

p.u. 4167.0875.1

2.09.0

2

..,

j

Z

jZ

B

upload

V 00.9167

240

0220..,

upsV

Example 5: P.U. Transformer Reactance


T1: 30 kVA 240/480 V

Xeq = 0.1 p.u.

T2: 20 kVA 460/115 V

Xeq = 0.1 p.u.

V 2401 BV

kVA 30BS

V 4802 BV V 1203 BV

p.u. 1378.048.0

20000

1151.0

2

new

B

old

B

old

p.u.new

p.u.T2,

X

XXX

68.72BZ 48.03BZ

For T1, S base and and V base are the same as of the circuit. In this case, we don’t need to change the base. The per unit value of reactance is the same = 0.1 p.u.

p.u. 1378.068.7

20000

4601.0

2

new

B

old

B

old

p.u.new

p.u.T2,

X

XXX

Example 5: Per Unit Circuit


~

V 2401 BV V 4802 BV V 1203 BV

p.u. 1378.0new

p.u.T2, jZ p.u. 1.0p.u.T1, jZ

p.u. 0.26042.., jZ upline p.u. 4167.0875.1.., jZ upload

V 00.9167.., upsV

A 125240

300001 BI A 62.5

480

300002 BI A 250

120

300002 BI

p.u.01.260.43954167.0875.11378.026042.01.0

00.9167..

jjjjI up

We can find current at any part of the circuit by simply multiplying the per unit value with the base value.

Advantages of Per Unit Analysis

• Simplify calculation by eliminating transformers.

• Helps to spot errors in the data – p.u. is more uniform

compare to actual impedance value of different sizes of equipments.

• Helps to detect abnormality in the system – Operator at control center

can spot over/under voltage/current rating easily.


Inside nationalgrid, UK


Source: http://www.youtube.com/watch?v=vX0G9F42puY&feature=related March 2012. “The National Grid is a high-voltage electric power transmission network, connecting power stations and 340+ substations ensuring supply and demand is in the balance. Nigel Williams speaks to Robert Llewellyn about how the National Grid works in Great Britain, and what challenges it faces with the rapidly changing world. Fully Charged is an online show hosted by Robert Llewellyn (Red Dwarf, Scrapheap Challenge, Carpool), sponsored by British Gas: Looking After Your World.”

http://www.youtube.com/watch?v=vX0G9F42puY&feature=related



Summary

• Per unit system helps to eliminate transformers in the circuit analysis.

• Per unit value is found by normalize the actual value by base value.

• We can divide circuits into zones according to transformer voltage ratings.

• Choose only single base power. Voltage, current, and impedance base value is calculated for each zone.

• Actual value is found from multiplying the per unit value to its corresponding base value in its zone.


Next Lecture

• Three-phase transformers

• Three-phase per unit analysis


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