03 Transformer Losses & Efficiency

(1)

Electrical Machines

LSEGG216A

9080V

(2)

(3)

Transformer

Losses & Efficiency

Week 3

(4)

Objective

s

_sss

1.

1. DeDescscriribe be ththe pe powower er lolosssses es whwhicich oh occccur ur in in a ta traransnsfoformrmer er 2.

2. DeDescscriribe be ththe te tesests ts whwhicich ah allllow ow the the popowewer lr lososseses os of af a transformer to be calculated

transformer to be calculated 3.

3. CaCalclcululatate tre tranansfsforormemer lor losssses aes and end effifficicienency ucy usising tng tesest ret resusultltss 4.

4. DeDefifine ne ththe ae all ll daday ey efffficicieiencncy oy of a tf a traransnsfoformrmer er 5.

5. CaCalclcululatate the the ale all dal day efy effificicienency ocy of a trf a tranansfsforormemer r 6.

6. DeDescscriribe tbe the rhe relaelatiotionsnshihip bep betwetween ten traransnsforformemer cor coololining ang andd rating

(5)

5.

5. CaCalclcululatate the the ale all dal day efy effificicieencncy of y of a tra tranansfsforormemerr 6.

6. DeDescscriribe tbe the rhe relelatatioionsnshihip bep betwtweeeen trn tranansfsforormemerr cooling and rating

cooling and rating 7.

7. DDeescscriribe be ththe e mmeeththoods ds oof f ccooololiingng 8.

8. LiList st ththe e prpropoperertities es of of trtranansfsforormemer r oioill 9.

9. DeDescscriribe tbe the he teteststs cs cononduductcteed on td on traransnsfoformrmer er oioill

Objective

(6)

Transform

Transformer

er Ratings

Ratings

Transformer Ratings

Transformers are rated to supply a given

Transformers are rated to supply a given output

output

in

Volt Amps

or

VA

at a specified frequency and

(7)

Transformer Ratings

They are NOT rated in Watts

The load power factor is unknown

I

V

S

=

×

PF

S

Power

₌

_×

PF

Power

S

₌

(8)

Transformer Ratings

They are NOT rated in Watts

(9)

Problem

V

₁

= 6,351

V

₂

= 230 V

S = 2 kVA

Power output at unity

PF ?

P = 2 kVA x 1

P = 2 kW

(10)

Problem

PF

V

S

I

× =

V

₁

= 6,351

V

₂

= 230 V

S = 2 kVA

Full load secondary current at 0.8 PF ?

I = 10.87 A

0.8

230 2000

I

× =

(11)

(12)

PF

S

Power

= ×

V

₂

= 200 V

V

₁

= 1270

V

S = 20 kVA

0 .

1

000 ,

20

× =

P

P = 20 kW

(a)

Power output at unity power

factor

(13)

PF

S

Power

= ×

V

₂

= 200 V

V

₁

= 1270

V

S = 20 kVA

8 .

0

000 ,

20

× =

P

P = 16 kW

(14)

VxPF

S

I

=

V

₂

= 200 V

V

₁

= 1270

V

S = 20 kVA

I = 100 A

(c)

Full-load secondary current at unity power

factor

200x1.0

20,000

I

=

(15)

VxPF

S

I

=

V

₂

= 200 V

V

₁

= 1270

V

S = 20 kVA

I = 62.5 A

(d)

Secondary current when transformer

supplies 10 kW at 0.8 power factor

200x0.8

10,000

I

=

(16)

Efficiency

Power

Input

Power

Output

η

=

Losses

Output

Input

= +

Losses

Power

Output

Power

Output

η

+ =

Ratio between Input power and Output

Power

Input

Losses

Power

Input

η

= −

(17)

Efficiency

100 Power

Input

Power

Output

η%

=

×

Efficiency is normally expressed as a

percentage

(18)

Transformer

Efficiency

Power

In

Power

Out

Overcome

Iron

Losses

Overcome

Copper

Losses

Some Power

is used to:

η = 100%

η = 95%

η = 90%

(19)

(20)

PF

S

Power

= ×

V

₁

= 230 V

_V

2

= 32 V

S = 20 kVA

η

=

90%

PF

=

0.85 (a)

Power output of transformer

0.85

100 P

= ×

(21)

In

Out

η

=

V

₁

= 230 V

_V

2

= 32 V

S = 20 kVA

η

=

90%

PF

=

0.85 (b)

Power input

P = 94.4 W

η

Out

In

=

0.9 W

85 In

=

(22)

Losses

Out

In

− =

V

₁

= 230 V

_V

2

= 32 V

S = 20 kVA

η

=

90%

PF

=

0.85 (c)

Losses

P = 9.4W

Losses

85

94.4

− =

(23)

Transformer

Losses

Copper Losses

(Cu)

•Varies with load current

•Produces HEAT

•Created by resistance of windings

•Short circuit test supplies copper

losses

(24)

Short Circuit Test

Copper Losses

(Cu)

Secondary

Short

Circuited

Limite

d

Supply

Voltag

e ≈

5-10 %

(25)

Short Circuit Test

(

0.5 )

100 loss

Copper

= 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

W

Transformer loaded at

50%

P

_Cu

= 25 W

100

0.25 loss

Copper

= ×

(26)

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

Copper Losses

(Cu)

% Load

C

u

Lo

ss

es

(W

)

(27)

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

(28)

Open Circuit Test

Finds Iron Losses

(Fe)

Full

Supply

Voltage

Secondary

Open Circui

Wattmeter indicates Iron

Losses (Fe)

(29)

Transformer

Efficiency

_{Student Exercise 3}

100 Power

Input

Power

Output

η%

= ×

(

load

)

×

Full

Load

Cu

Losses

=

%

2

Cu

Losses

(30)

100 Power

Input

Power

Output

η%

= ×

S

_out

= 30 kVA

Cu

FL

= 840 W

Fe = 220 W

Calculate η%at Full Load

100 Losses Output Power Output η% × + =

100 k

0.22 k

84 k

30 k

30 η%

× + + =

.

0 η% =

96.6%

(31)

S

_out

= 30 kVA

Cu

FL

= 840 W

Fe = 220 W

Calculate η%at 75%Load

(

)

(

0.75 )

0.22

100

22.5

22.5 η%

₂ × + × + = 84 . 0

η% = 97%

100 0.22 0.4725 22.5 22.5 η% × + + = 5 . 22 30 75 . 0 _× ₌ = out

S

(

0.75

)

2 _×840 ₌ 472.5 = 75%

Cu

(32)

S

_out

= 30 kVA

Cu

FL

= 840 W

Fe = 220 W

Calculate η%at 50%Load

( )

(

0.5 )

0.22

100

15

15 η%

₂ × + × + = 84 . 0

η% = 97.21%

100 0.22 0.21 15 15 η% × + + = 15 30 5 . 0 _× ₌ = out

S

(33)

S

_out

= 30 kVA

Cu

FL

= 840 W

Fe = 220 W

Calculate η%at 25%Load

(

)

(

0.25 )

0.22

100

7.5

7.5 η%

₂ × + × + = 84 . 0

η% = 96.49%

100

0.22 0.0525

7.5

7.5 η%

× + + = 5 . 7 30 25 . 0 _× ₌ = out

S

(34)

100%

η = 96.6%

75%

η = 97%

50%

η = 97.21%

25%

η = 96.49%

(35)

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

Lo

ss

es

(W

)

η

%

Cu

Losses

Fe

Losses

η%

Fe = Cu =Max

η

(36)

Maximum Efficiency

Fe = Cu =Max η

(

Load

)

Cu

Fe

= 2 ×

(

)

2

Load

Cu

Fe

= Load =

Cu

Fe

Load =

840

220 Fe =

220 Cu =

840 Load %=

51.18%

(

0.5118

30 )

(

0.5118

)

0.22

10

30 0.5118

η%

₂ × + × + × × = 84 . 0

100

0.22

15.35 η%

× + + = 22 . 0 35 . 15

η%= 97.21%

(37)

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

(38)

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}

(39)

All Day Efficiency

Transformer A

S

_out

= 300 kVA

Fe = 1.25 kVA

Cu = 3.75 kVA

Hours 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 8.33 508.33 6.00 7.00 1 200 200.0 66.67 1.67 1.67 1.25 2.92 202.92 7.00 8.00 1 300 300.0 100.00 3.75 3.75 1.25 5.00 305.00 8.00 9.00 1 360 360.0 120.00 5.40 5.40 1.25 6.65 366.65 9.00 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 9.03 569.03 14.00 18.00 4 300 1200.0 100.00 3.75 15.00 5.00 20.00 1220.00 18.00 20.00 2 360 720.0 120.00 5.40 10.80 2.50 13.30 733.30 20.00 22.00 2 280 560.0 93.33 3.27 6.53 2.50 9.03 569.03 22.00 1.00 3 200 600.0 66.67 1.67 5.00 3.75 8.75 608.75 5900.0 5998.02 Time Period PoutkWh= PinkWh=

(40)

All Day Efficiency

Transformer B

S

_out

= 300 kVA

Fe = 2.5 kVA

Cu = 2.5 kVA

Hours Load kW kWh % Load Cu Loss Cu kWh Fe kWh Losses kWh Input kWh 1.00 6.00 5 100 500.0 33.33 0.28 1.39 12.50 13.89 513.89 6.00 7.00 1 200 200.0 66.67 1.11 1.11 2.50 3.61 203.61 7.00 8.00 1 300 300.0 100.00 2.50 2.50 2.50 5.00 305.00 8.00 9.00 1 360 360.0 120.00 3.60 3.60 2.50 6.10 366.10 9.00 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 9.36 569.36 14.00 18.00 4 300 1200.0 100.00 2.50 10.00 10.00 20.00 1220.00 18.00 20.00 2 360 720.0 120.00 3.60 7.20 5.00 12.20 732.20 20.00 22.00 2 280 560.0 93.33 2.18 4.36 5.00 9.36 569.36 22.00 1.00 3 200 600.0 66.67 1.11 3.33 7.50 10.83 610.83 5900.0 6005.34 Time Period PoutkWh= PinkWh=

(41)

Transformer Cooling

• Transformer ratings can be increased if their windings are cooled by some external means