03 transformer losses & efficiency (1)

7/27/2019 03 Transformer Losses & Efficiency (1)

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Electrical MachinesLSEGG216A

9080V


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TransformerLosses & Efficiency

Week 3


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Objectives1. Describe the power losses which occur in a

transformer

2. Describe the tests which allow the power losses of atransformer to be calculated

3. Calculate transformer losses and efficiency using testresults

4. Define the all day efficiency of a transformer

5. Calculate the all day efficiency of a transformer


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5. Calculate the all day efficiency of a transformer

6. Describe the relationship between transformer cooling andrating

7. Describe the methods of cooling

8. List the properties of transformer oil

9. Describe the tests conducted on transformer oil

Objective s


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Transformer RatingsTransformers are rated to supply a given output in

Volt Ampsor

VAat a specified frequency and terminal voltage.


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Transformer RatingsThey are NOT rated in Watts

The load power factor is unknown

I V S

PF S Power

PF

Power S


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Transformer RatingsThey are NOT rated in Watts

The load power factor is unknown


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ProblemV1 = 6,351 V V2 = 230 V

S = 2 kVA

Power output at unity PF ?

P = 2 kVA x 1

P = 2 kW

Power = S x PF


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Problem

PFVS

I

V1 = 6,351 V V2 = 230 V

S = 2 kVA

Full load secondary current at 0.8 PF ?

I = 10.87 A

0.82302000

I


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Student Exercise 1


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PFSPower

V2 = 200 VV1 = 1270 V

S = 20 kVA

0.1000,20 P

P = 20 kW

(a) Power output at unity power factor


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PFSPower

V2 = 200 VV1 = 1270 V

S = 20 kVA

8.0000,20 P

P = 16 kW

(b) Power output at 0.8 power factor


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VxPF

SI

V2 = 200 VV1 = 1270 V

S = 20 kVA

I = 100 A

(c) Full-load secondary current at unity power factor

200x1.020,000I


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VxPF

SI

V2 = 200 VV1 = 1270 V

S = 20 kVA

I = 62.5 A

(d) Secondary current when transformer supplies 10 kW at 0.8 power factor

200x0.810,000I


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EfficiencyPowerInput

PowerOutput

LossesOutput Input

LossesPowerOutputPowerOutput

Ratio between Input power and Output Power

PowerInputLossesPowerInput


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Efficiency

100PowerInput

PowerOutput%

Efficiency is normally expressed as a percentage


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Transformer EfficiencyPower

InPowerOut

OvercomeIron

Losses

OvercomeCopperLosses

Some Poweris used to:

= 100% = 95% = 90%


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Student Exercise 2


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PFSPower

V1 = 230 V V2 = 32 V

S = 20 kVA

= 90% PF = 0.85

(a) Power output of transformer

0.85100P

P = 85 W


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In

Out

V1 = 230 V V2 = 32 V

S = 20 kVA

= 90% PF = 0.85

(b) Power input

P = 94.4 W

Out

In

0.9W85

In


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LossesOutIn

V1 = 230 V V2 = 32 V

S = 20 kVA

= 90% PF = 0.85

(c) Losses

P = 9.4W

Losses8594.4


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Transformer LossesCopper Losses (Cu)

Varies with load current

Produces HEAT Created by resistance of windings

Short circuit test supplies copper losses


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Short Circuit TestCopper Losses (Cu)

SecondaryShort Circuited

LimitedSupplyVoltage 5-10 %

Wattmeter indicates Copper Losses (Cu)


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Short Circuit Test

1000.5lossCopper 2

Copper Losses (Cu) Finds Cooper losses at full load

Copper losses vary with the square of the load

Full load C u loss = 100 WTransformer loaded at 50%

PCu = 25 W

1000.25lossCopper


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010

2030

405060708090

100

110

120130140150

0 10 20 30 40 50 60 70 80 90 100 110

Copper Losses (Cu)

% Load

C u L o s s e s ( W )


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Transformer LossesIron Losses (Fe) Fixed

Always present

Related to transformers constructionEddy Currents

Reduced by laminations

Produces HEAT

Hysteresis

Reduced by using special

steels in laminations


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Open Circuit TestFinds Iron Losses (Fe)

Full SupplyVoltage

SecondaryOpen Circuit

Wattmeter indicates Iron Losses (Fe)


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Transformer EfficiencyStudent Exercise 3

100

PowerInput

PowerOutput%

LossesCu Load Full load 2%CuLosses


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100PowerInputPowerOutput

%

Sout = 30 kVAFe = 220 WCu FL= 840 W

Calculate %at Full Load

100LossesOutput PowerOutput%

100k0.22k84k30

k30%

.0

% = 96.6%


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100LossesOutput

PowerOutput%


Calculate %at 75%Load

1000.220.7522.522.5

% 2 84.0

% = 97%

1000.220.472522.5

22.5%

5.223075.0outS

5.47284075.0 275%Cu


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100LossesOutputPowerOutput

%



1000.220.51515

% 2 84.0

% = 97.21%

1000.220.2115

15%

15305.0outS


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100LossesOutputPowerOutput

%



1000.220.257.57.5

% 2 84.0

% = 96.49%

1000.220.05257.5

7.5%

5.73025.0outS


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100% = 96.6%75% = 97%50% = 97.21%25% = 96.49%


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0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

0 10 20 30 40 50 60 70 80 90 100 110

96.00

97.00

% Load

L o s s e s ( W )

%

Cu Losses

Fe Losses

%

Fe = Cu =Max


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Maximum Efficiency

Fe = Cu =Max

CuLoadFe 2

2LoadCuFe

Load CuFe

Load 840220

Fe = 220 Cu = 840

Load %= 51.18%

1000.220.5118300.5118300.5118

% 2 84.0

1000.22

15.35%

22.035.15

%= 97.21%


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0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

0 10 20 30 40 50 60 70 80 90 100 110

96.00

97.00


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All Day Efficiency

Most Transformers are connected permanently

The time that the transformer has to be calculated whendetermining efficiency

Able to determine the best transformer for the application by itsefficiency


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All Day EfficiencyTransformer A

Sout = 300 kVA Fe = 1.25 kVA Cu = 3.75 kVAHours Load kW kWh % Load Cu Loss Cu kWh Fe kWh Losses kWh Input kWh

1.00 6.00 5 100 500.0 33.33 0.42 2.08 6.25 8.33 508.336.00 7.00 1 200 200.0 66.67 1.67 1.67 1.25 2.92 202.927.00 8.00 1 300 300.0 100.00 3.75 3.75 1.25 5.00 305.008.00 9.00 1 360 360.0 120.00 5.40 5.40 1.25 6.65 366.659.00 12.00 3 300 900.0 100.00 3.75 11.25 3.75 15.00 915.00

12.00 14.00 2 280 560.0 93.33 3.27 6.53 2.50 9.03 569.0314.00 18.00 4 300 1200.0 100.00 3.75 15.00 5.00 20.00 1220.0018.00 20.00 2 360 720.0 120.00 5.40 10.80 2.50 13.30 733.3020.00 22.00 2 280 560.0 93.33 3.27 6.53 2.50 9.03 569.0322.00 1.00 3 200 600.0 66.67 1.67 5.00 3.75 8.75 608.75

5900.0 5998.02

98.37

Time Period

P out kWh = P in kWh =

% Eff =


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All Day EfficiencyTransformer B

Sout = 300 kVA Fe = 2.5 kVA Cu = 2.5 kVA

Hours Load kW kWh % Load Cu Loss Cu kWh Fe kWh Losses kWh Input kWh1.00 6.00 5 100 500.0 33.33 0.28 1.39 12.50 13.89 513.896.00 7.00 1 200 200.0 66.67 1.11 1.11 2.50 3.61 203.617.00 8.00 1 300 300.0 100.00 2.50 2.50 2.50 5.00 305.008.00 9.00 1 360 360.0 120.00 3.60 3.60 2.50 6.10 366.109.00 12.00 3 300 900.0 100.00 2.50 7.50 7.50 15.00 915.00

12.00 14.00 2 280 560.0 93.33 2.18 4.36 5.00 9.36 569.3614.00 18.00 4 300 1200.0 100.00 2.50 10.00 10.00 20.00 1220.0018.00 20.00 2 360 720.0 120.00 3.60 7.20 5.00 12.20 732.2020.00 22.00 2 280 560.0 93.33 2.18 4.36 5.00 9.36 569.3622.00 1.00 3 200 600.0 66.67 1.11 3.33 7.50 10.83 610.83

5900.0 6005.34

98.25% Eff =

Time Period

P out kWh = P in kWh =


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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/orAir Oil The most common methods of circulation are

Forced and/or Natural


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Transformer Classification

Transformers are allocated symbols which indicate the type of coolingused Can consist of up to 4 letters indicating the cooling system

1st Letter 2 nd Letter 3 rd Letter 4 th Letter

The cooling medium in contactwith the windings

The cooling medium in contactwith the external cooling system

Kind of Medium Circulation type Kind of Medium Circulation type


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Type ANDry Transformer with

Natural Air Flow

A ir N atural


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Type AFDry Transformer with

Forced Air Flow

A ir Forced


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Type ONAN

Oil Tank Cooling Natural Oil Flow - Natural Air Flow

O il N atural A ir N atural


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Type ONAF

Oil Tank Cooling Natural Oil Flow - Forced Air Flow

O il N atural A ir Forced


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Transformer ClassificationType OFAF

Oil Tank Cooling Forced Oil Flow Forced Air Flow

Oil

Forced

Air

Forced


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Transformer Oil

Low Viscosity

High Flash point

Chemically inert

Good insulator

Acts as Coolant & Insulator


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


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THE END

03 transformer losses & efficiency (1)

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