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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. 5. Define the all day efficiency of a transformer Calculate the all day efficiency of a transformer Objectives 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 Transformer Ratings Transformers are rated to supply a given output in Volt Amps or VA at a specified frequency and terminal voltage. Transformer Ratings They are NOT rated in Watts The load power factor is unknown S V I Power S PF Power S PF Transformer Ratings They are NOT rated in Watts The load power factor is unknown Student Exercise 1 Efficiency Ratio between Input power and Output Power Output Power η Input Power Input Output Losses Output Power η Output Power Losses Input Power Losses η Input Power Efficiency Efficiency is normally expressed as a percentage Output Power η% 100 Input Power Transformer Efficiency Power In Some Power is used to: Power Out Overcome Copper Losses Overcome Iron Losses Transformer Losses Copper Losses (Cu) •Varies with load current •Produces HEAT •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 Cu loss = 100 W Transformer loaded at 50% Copper loss 0.5 100 2 Copper loss 0.25 100 PCu = 25 W Cu Losses (W) Copper Losses (Cu) 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 0 10 20 30 40 50 60 % Load 70 80 90 100 110 Transformer Losses •Fixed Iron Losses (Fe) •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) Cu FL= 840 W Sout = 30 kVA Fe = 220 W Calculate η%at 75%Load Output Power η% 100 Output Losses 22.5 η% 100 2 22.5 0.75 0.84 0.22 η% 22.5 100 22.5 0.4725 0.22 η% = 97% S out 0.75 30 22 .5 Cu75% 0.75 840 472.5 2 1.4 Losses (W) 1.0 0.8 η% Cu Losses 1.2 Fe Losses 97.00 0.6 η% 0.4 0.2 0.0 96.00 0 10 20 30 40 50 60 % Load Fe = Cu =Max η 70 80 90 100 110 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; Air and/or Oil • 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 1st Letter 2nd Letter The cooling medium in contact with the windings 3rd Letter 4th Letter The cooling medium in contact with the external cooling system Kind of Medium Circulation type Kind of Medium Circulation type Transformer Classification Type AN Air Natural Dry Transformer with Natural Air Flow Transformer Classification Type AF Air Forced Dry Transformer with Forced Air Flow Transformer Classification ONAF Oil Natural Air Forced Type Oil Tank Cooling Natural Oil Flow - Forced Air Flow Transformer Classification Type OFAF Oil Forced Air Forced Oil Tank Cooling Forced Oil Flow – Forced Air Flow Transformer Oil Acts as Coolant & Insulator • Low Viscosity • High Flash point • Chemically inert • Good insulator THE END