Electrical Machines LSEGG216A 9080V

Transformer

Basic of Transformer A basic transformer consists of two sets of coils or windings. Each set of windings is simply an inductor. AC voltage is applied to one of the windings, called the primary winding. The other winding, called the secondary winding, is positioned in close proximity to the primary winding, but is electrically isolated from it.

Transformer Losses & Efficiency

Objectives 1.Describe the power losses which occur in a transformer 2.Describe the tests which allow the power losses of a transformer to be calculated 3.Calculate transformer losses and efficiency using test results 4.Define the all day efficiency of a transformer 5.Calculate the all day efficiency of a transformer 6.Describe the relationship between transformer cooling and rating

5.Calculate the all day efficiency of a transformer 6.Describe the relationship between transformer cooling and rating 7.Describe the methods of cooling 8.List the properties of transformer oil 9.Describe the tests conducted on transformer oil Objective s

Transformer Ratings Transformers are rated to supply a given output in Volt Amps or VA at a specified frequency and terminal voltage.

They are NOT rated in Watts The load power factor is unknown

Transformer Ratings They are NOT rated in Watts The load power factor is unknown

Problem V 1 = 6,351 V V 2 = 230 V S = 2 kVA Power output at unity PF ? P = 2 kVA x 1 P = 2 kW Power = S x PF

Student Exercise 1

Efficiency Ratio between Input power and Output Power

Efficiency Efficiency is normally expressed as a percentage

Transformer Efficiency Power In Power Out Overcome Iron Losses Overcome Copper Losses Some Power is used to:

Transformer Losses Copper Losses (Cu) Varies with load current Produces H EAT Created by resistance of windings Short circuit test supplies copper losses

Short Circuit Test Copper Losses (Cu) Finds Cooper losses at full load Copper losses vary with the square of the load Full load C u loss = 100 W Transformer loaded at 50% P Cu = 25 W

Copper Losses (Cu) % Load Cu Losses (W)

Transformer Losses Iron Losses (F e ) Fixed Always present Related to transformers construction Eddy Currents Reduced by laminations Produces HEAT Hysteresis Reduced by using special steels in laminations

Open Circuit Test Finds Iron Losses (Fe) Full Supply Voltage Secondary Open Circuit Wattmeter indicates Iron Losses (Fe)

S out = 30 kVA Fe = 220 WCu FL = 840 W Calculate η%at 75%Load η% = 97%

% Load Losses (W) η%η% Cu Losses Fe Losses η%η% Fe = Cu =Max η

All Day Efficiency Most Transformers are connected permanently The time that the transformer has to be calculated when determining efficiency Able to determine the best transformer for the application by its efficiency

Transformer Cooling Transformer ratings can be increased if their windings are cooled by some external means The most common cooling mediums are in direct with transformer windings; and/or AirOil The most common methods of circulation are Forced and/or Natural

Transformer Classification Transformers are allocated symbols which indicate the type of cooling used Can consist of up to 4 letters indicating the cooling system 1 st Letter2 nd Letter3 rd Letter4 th Letter The cooling medium in contact with the windings The cooling medium in contact with the external cooling system Kind of MediumCirculation typeKind of MediumCirculation type

Transformer Classification Type AN Dry Transformer with Natural Air Flow A ir N atural

Transformer Classification Type AF Dry Transformer with Forced Air Flow A ir F orced

Transformer Classification Type ONAF Oil Tank Cooling Natural Oil Flow - Forced Air Flow O il N atural A ir F orced

Transformer Classification Type OFAF Oil Tank Cooling Forced Oil Flow – Forced Air Flow O il F orced A ir F orced

Transformer Oil Low Viscosity High Flash point Chemically inert Good insulator Acts as Coolant & Insulator

Transformer Oil Tests Dielectric Strength Acidity Power factor Interfacial tension Dissolved Gas

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