VAA 2002 – Electrical Power Systems 1

Dr Horace KING

College of Engineering and ScienceOffice: 216

Phone: 99194696

Email:

LECTURE 5

Laboratory Times and The Test



The Per-Unit System

Power Transmission lines are operated at Voltage levels most commonly expressed in Kilo Volts (KV)

Terms used to express quantities in a power system are: KW,MW, Kilovolt Ampere (KVA), Megavolt Ampere (MVA)

Most important quantities in Power systems

Voltages

Currents

Impedances

These quantities are also very often expressed as aPercent (%) or Per-unit (p.u.) of a chosen base (reference) value

The per unit (p.u.) value representation of electrical variables in power system problems is favored in electric power systems

The numerical p.u. value of any quantity is its ratio to a chosen base quantity

p.u. values are expressed as decimals

Per-unit quantities are said to be normalised quantities with respect to the chosen base value

Worked Example 1

P.U. cont..

Knowing that the numerical per unit value of any actual quantity is the ration to a chosen base quantity i.e. actual/base

What is the main advantage of expressing quantities in p.u.?

Easier to handle!

When dealing with quantity products or ratios p.u. is easier to apply!

In D.C. given 0.7 p.u. (of current) and 0.8 p.u. (of voltage), what is the power?

0.7 x 0.8 = 0.56 p.u.(of power)

Some Calculations

Quantities (Voltages, Currents, Kilovolt Ampere, Impedance) are so related that if we specify the base values for any two, the other base values can be calculated

e.g. the base voltage is 100 volts and the base current is 5 amps What’s the base impedance and base power ? Ohm’s law anyone????

Base −impedance = R = VI = 1005 = 20Ω

Base −power = P = IV =100×5=500W

The above is a simple example for DC quantities!!! Ohm’s law anyone????

However, the per-unit system is most commonly used when expressing quantities in AC networks

When working with the AC quantities, extra attention is required because we have

Single-phase (Φ)

Three-phase(3Φ) quantities

Live voltages

Line currents

The Per-Unit System in AC

In AC, the following two base values are usually selected to specify the base

S = Complex Power (kVA) ( Remember the letter S is used for complex power)

V =Voltage (kV)

The rated values are usually taken as the base values ( Sbase = Srated)

If two base values are specified, the other quantities can be calculated

Sbase Vbase = Vbase2 Ibase = VZbase =

baseIbaseSbase

Example:

Given the base voltage (Vbase ) is 240 Volts and base complex power (S base ) is 12

kVA. What are the base Impedance and base Current?

Zbase = VSbasebase2 = (12000240)2 = 4.8 ΩIbase = VSbasebase = 12240kVAV = 50 Amps

The Per-Unit System in AC

Another Golden Rule: When taking base values for AC, always use the per-phasevalues

The voltage must be the Phase voltage (VΦ)

Remember “Phase” voltage is also called the “line to neutral” voltage (LN). So ,Vφ= VLN

The current must be the line current

The complex power (S) must be the complex power per phase (SΦ) Can be given 3Φ values in a question!

3 phase complex power (S3Φ)

Line-to-line voltage (VLL)

Approach

Convert 3Φ values into single phase (also called per-phase) values

This is because balanced three phase circuits are solved as a single line with a neutral return.

E.g. Vbase =Vφ=VLL

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Per unit system cont.

Worked example 1

A balanced - Y – connected voltage source with Eab = 480∠0° volts is applied to a balanced - ∆ load with Z∆ = 30∠40°Ω. The line impedance between the source and the load is ZL = 1∠85°Ω for each phase. Calculate the per-unit and actual current in phase a of the line using Sbase3φ =10kVA and VbaseLL = 480 volts and draw the circuit.

Per-Unit Values in AC

Remembering the general formulae

actual value of that quantity

quantity in p.u.= base value of that quantity

The per-unit AC values are

V inp.u.= Vactual
Vbase / I inp.u.= Iactual
Ibase
Z inp.u.= Zactual
Zbase / S inp.u.= Sactual
Sbase

Per-Unit Values Cont..

Worked Example 2

A single - phase two-winding transformer is rated 20kVA, 480/120 volts, 60Hz. The equivalent leakage impedance of the transformer referred to the 120-volt winding, denoted winding 2, is Zeq2 =0.0525∠78°Ω. Using the transformer ratings as base values, determine the per-unit leakage impedance referred to winding 2 and referred to winding 1.

Class Example 2

An electrical lamp is rated 240 Volts, 100 watts. Determine the per-unit and percent impedance of the lamp. Draw the p.u. equivalent circuit

Class Example 3

Solve the following circuit using the per-unit method. Determine Ztotal, I, and the power consumed all in p.u for the circuit. Vbase is specified as 100 volts and Sbase as 1000 VA.

Transformation between bases

When performing calculations in a power system, all per-unit values must be converted to the same base

Transformation between bases is a means of converting per-unit impedances from one base to another

Z pu,new =Z pu,old VVbasebase,,newold 2 ×SSbasebase,,newold 

Example 5: The old reactance of a generator is given as 0.25 per unit based on the generator's nameplate rating of 18kV, 500 MVA. Using the new base value 20kV and 100MVA, find the new p.u. impedance.

V2Sbase,new = 0.2518k 2 ×100M = 0.0405 pu

Z pu,new =Z pu,old Vbasebase,,newold  ×Sbase,old   20k 500M

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