03 Transformer Losses & Efficiency

Electrical Machines LSEGG216A  9080V 

Transformer Losses & Efficiency

Week 3

Objectives 1.

Desc Descri ribe be the the pow power er loss losses es wh whic ich h occ occur ur in a tra trans nsfo form rmer  er 

2.

Desc Descri ribe be the the tes tests ts wh whic ich h all allow ow the powe powerr los losse ses s of of a transformer to be calculated

3.

Calc Calcul ulat ate e tran transf sfor orme merr loss losses es and and effi effici cien ency cy usi using ng tes testt resu result lts s

4. 5.

Defi Define ne the the all all day day eff effic icie ienc ncy y of of a tra trans nsfo form rmer  er  Calc Calcul ulat ate e the the all all day day effi effici cien ency cy of of a tran transf sfor orme mer  r 

6.

Desc Descri ribe be the the rela relatio tions nshi hip p betwe between en tra trans nsfor forme merr cool coolin ing g and and rating

Objectives 5.

Calc Calcul ulat ate e the the all all day day effi effici cie ency ncy of a tran transf sfor orme mer r

6.

Desc Descri ribe be the the rel relat atio ions nshi hip p betw betwee een n tran transf sfor orme mer r cooling and rating

7.

Descri scribe be the the metho thods of cooli oling

8.

List List the the prop proper erti ties es of tran transf sfor orme mer r oil oil

9.

Desc Descri ribe be the the test tests s con condu duct cte ed on tra trans nsfo form rmer er oil oil

Transformer 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  S  =

 Power   PF 

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

Problem

S = 2 kVA

V1 = 6,351 V

V2 = 230 V

Power output at unity PF ?

Power = S x PF P = 2 kVA x 1 P = 2 kW

Problem

S = 2 kVA

V1 = 6,351 V

V2 = 230 V

Full load secondary current at 0.8 PF ?

I

I

=

=

S V × PF

2000 230× 0.8

I = 10.87 A

Student Exercise 1

S = 20 kVA V1 = 1270 V (a)

V2 = 200 V

Power output at unity power factor

Power= S× PF

 P  = 20,000 ×1.0

P = 20 kW

S = 20 kVA V1 = 1270 V (b)

V2 = 200 V

Power output at 0.8 power factor

Power= S× PF  P  = 20,000 × 0.8

P = 16 kW

S = 20 kVA V1 = 1270 V (c)

V2 = 200 V

Full-load secondary current at unity power factorS

I

I

=

=

VxPF

20,000 200x1.0

I = 100 A

S = 20 kVA V1 = 1270 V (d)

V2 = 200 V

Secondary current when transformer supplies 10 kW at 0.8 power factor

I I

=

=

S VxPF 10,000 200x0.8

I = 62.5 A

Efficiency Ratio between Input power and Output Power Output Power η= Input Power

 Input  = Output  + Losses

η=

OutputPower OutputPower+ Losses

η=

Input Power − Losses Input Power

Efficiency Efficiency is normally expressed as a percentage

η% =

Output Power × 100 Input Power

Transformer Efficiency Power In

Power Out

η = 100% 90% 95%

Some Power is used to:

Overcome Copper Losses

Overcome Iron Losses

Student Exercise 2

S = 20 kVA η = 90% V1 = 230 V (a)

V2 = 32 V PF = 0.85

Power output of transformer

Power = S× PF

P = 100× 0.85

P = 85 W

S = 20 kVA η = 90% V1 = 230 V (b)

η=

Out In In =

V2 = 32 V PF = 0.85

Power input

Out η

In =

85W 0.9

P = 94.4 W

S = 20 kVA η = 90% V1 = 230 V (c)

V2 = 32 V PF = 0.85 Losses

In − Out = Losses

94.4− 85= Losses P = 9.4W

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)

Limite d Supply Voltag e ≈ 510 %

Secondary Short Circuited

Wattmeter indicates Copper

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

Copper loss = ( 0.5)

2

×

100

Copper loss = 0.25× 100

PCu = 25 W

Copper Losses (Cu)

e s s o L u C

W( s

)

150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 0

10

20

30

40

50

60

70

80

% Load

90

100

110

Transformer Losses Iron Losses (Fe) •Always present •Fixed

•Related to transformers construction Eddy Currents Hysteresis Reduced by laminations Produces HEAT

Reduced by using special steels in laminations

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

Secondary Open Circui

Wattmeter indicates Iron Losses (Fe)

Transformer Efficiency Student Exercise 3 Output Power η% = × 100 Input Power

LossesCu = ( load % )

2

 Full   Load  Cu Losses

×

Sout = 30 kVA

Cu

FL

= 840 W Fe = 220 W

Calculate η%at Full Load Output Power η% = × 100 Input Power

η% =

Output Power × 100 Output + Losses

30k η%= × 100 30k + 0.84k + 0.22k

η% = 96.6%

Cu

Sout = 30 kVA

FL

= 840 W Fe = 220 W

Calculate η%at 75%Load Sout

Output Power × 100 η% = Output + Losses

η% =

η% =

22.5+ ( ( 0.75)

)

× 0.84 +

×

0.22

22.5 × 100 22.5+ 0.4725+ 0.22

η% = 97%

0.75 × 30

Cu 75% = ( 0.75)

22.5 2

=

100

2

=

22.5

× 840 =

472.5

Cu

Sout = 30 kVA

FL

= 840 W Fe = 220 W

Calculate η%at 50%Load Output Power η% = × 100 Output + Losses

η% =

15 15+ ( ( 0.5)

2

)

× 0.84 +

×

0.22

15 η% = × 100 15+ 0.21+ 0.22

η% = 97.21%

Sout 100

=

0.5 × 30

= 15

Cu

Sout = 30 kVA

FL

= 840 W Fe = 220 W

Calculate η%at 25%Load η% =

η% =

Output Power × 100 Output + Losses

7.5 7.5+ ( ( 0.25)

2

)

× 0.84 +

×

0.22

7.5 × 100 η% = 7.5+ 0.0525+ 0.22

η% = 96.49%

Sout 100

=

0.25 × 30

=

7.5

100% 75% 50% 25%

η = 96.6% η = 97% η = 97.21% η = 96.49%

1.4 1.2

W( s e s s o L

)

η%

Cu Losses Fe Losses

1.0 0.8 0.6

97.00

0.4 0.2 0.0

96.00 0

10

20

30

40

50

60

70

80

% Load Fe = Cu =Max η

90

100

110

η %

Maximum Efficiency Fe = 220

Cu = 840

220 =  Load  840

Fe = Cu =Max η

Fe = ( Load) Fe Cu

Fe Cu

=

2

( Load)

×

2

Load %= 51.18%

Cu η% =

0.5118 × 30

( 0.5118 ) × 0.84) + 0.22 × 30) + ( ( 0.5118 η%=

=  Load 

×

2

15.35

×

15.35 + 0.22 + 0.22

100

η%= 97.21%

10

1.4 1.2 1.0 97.00 0.8 0.6 0.4 0.2 0.0

96.00 0

10

20

30

40

50

60

70

80

90

100

110

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

All Day Efficiency Transformer A

Sout = 300 kVA Time Period Hours Load kW 1.00 6.00 5 100 6.00 7.00 1 200 7.00 8.00 1 300 8.00 9.00 1 360 9.00 12.00 3 300 12.00 14.00 2 280 14.00 18.00 4 300 18.00 20.00 2 360 20.00 22.00 2 280 22.00 1.00 3 200

Fe = 1.25 kVA

Cu = 3.75 kVA

kWh % Load Cu Loss Cu kWh Fe kWh Losses kWh Input kWh 500.0 33.33 0.42 2.08 6.25 8.33 508.33 200.0 66.67 1.67 1.67 1.25 2.92 202.92 300.0 100.00 3.75 3.75 1.25 5.00 305.00 360.0 120.00 5.40 5.40 1.25 6.65 366.65 900.0 100.00 3.75 11.25 3.75 15.00 915.00 560.0 93.33 3.27 6.53 2.50 9.03 569.03 1200.0 100.00 3.75 15.00 5.00 20.00 1220.00 720.0 120.00 5.40 10.80 2.50 13.30 733.30 560.0 93.33 3.27 6.53 2.50 9.03 569.03 600.0 66.67 1.67 5.00 3.75 8.75 608.75

P out kWh = 5900.0

P in kWh = 5998.02

All Day Efficiency Transformer B

Sout = 300 kVA Time Period Hours Load kW 1.00 6.00 5 100 6.00 7.00 1 200 7.00 8.00 1 300 8.00 9.00 1 360 9.00 12.00 3 300 12.00 14.00 2 280 14.00 18.00 4 300 18.00 20.00 2 360 20.00 22.00 2 280 22.00 1.00 3 200

Fe = 2.5 kVA

Cu = 2.5 kVA

kWh % Load Cu Loss Cu kWh Fe kWh Losses kWh Input kWh 500.0 33.33 0.28 1.39 12.50 13.89 513.89 200.0 66.67 1.11 1.11 2.50 3.61 203.61 300.0 100.00 2.50 2.50 2.50 5.00 305.00 360.0 120.00 3.60 3.60 2.50 6.10 366.10 900.0 100.00 2.50 7.50 7.50 15.00 915.00 560.0 93.33 2.18 4.36 5.00 9.36 569.36 1200.0 100.00 2.50 10.00 10.00 20.00 1220.00 720.0 120.00 3.60 7.20 5.00 12.20 732.20 560.0 93.33 2.18 4.36 5.00 9.36 569.36 600.0 66.67 1.11 3.33 7.50 10.83 610.83

P out kWh = 5900.0

P in kWh = 6005.34

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

Force

and/or

Natur

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

3rd Letter

4th Letter

 The cooling medium in contact  The cooling medium in contact with the windings 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

ONAN Oil Natural Air Natural  Type

Oil Tank Cooling Natural Oil Flow - Natural Air Flow

Transformer Classification

ONAF Oil Natural Air Forced  Type

Oil Tank Cooling Natural Oil Flow - Forced Air Flow

OFAF

Transformer Classification  Type

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

Transformer Oil Tests • • • • •

Dielectric Strength Acidity Power factor Interfacial tension Dissolved Gas