APPENDIX 5B. SCALING RELATIONSHIPS IN TRANSFORMER

MANUFACTURING

TABLE O F CONTENTS

5B.1 I NTRODUCTION

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5B­1

5B.2 T HEORY AND BASIS FOR SCALING RULES

. . . . . . . . . . . . . . . . . . . . . . . . . . . . 5B­4

5 B.2.1 Power and Voltage Equations

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5B­4

5B.2.2 Losses

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5B­5

5B.2.3 Output and Winding Coefficients

. . . . . . . . . . . . . . . . . . . . . . . . . . . 5B­10

5B.2.4 Scaling Laws

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5B­11

LIST O F TABLES

Table 5 B.1 Common Scaling Relationships in Transformers

. . . . . . . . . . . . . . . . . . . . . . . . . 5B­2

Table 5B.2 Nominal 60 Hz, Core­Type, Liquid­Filled, 12 kV Distribution Transformers

. . 5B­11

LIST O F FIGURES

Figure 5 B.1 Size and Performance Relationships by kVA Rating

. . . . . . . . . . . . . . . . . . . . . 5B­3

Figure 5B.2 Basic Three­Phase Transformer Dimensions

. . . . . . . . . . . . . . . . . . . . . . . . . . . 5B­12

5B­iAPPENDIX 5B. SCALING RELATIONSHIPS IN TRANSFORMER

MANUFACTURING

5B.1 I NTRODUCTION

There e xist certain fundamental relationships between the ratings in kilovolt­amperes

(kVA) of transformers and their physical size and performance. A rather obvious such

relationship is the fact that large transformers of the same voltage have lower percentage losses

than small units, i.e., large transformers are more efficient. These size­versus­performance

relationships arise from fundamental equations describing a transformer's voltage and kVA

rating. For example, by fixing the kVA rating and voltage frequency, the product of the

conductor current density, core flux density, core cross­sectional area, and total conductor

cross­sectional area is constant.

To i llustrate this point, consider a transformer with frequency, magnetic flux density,

current density, and basic impulse insulation levels (BIL) all fixed. If one enlarges (or decreases)

the kVA rating, then the only free parameters are the core cross­section and the core window area

through which the windings pass. Thus, to increase (or decrease) the kVA rating, the dimensions

for height, width, and depth of the core/coil assembly may be scaled equally in all directions.

Careful examination reveals that linear dimensions vary as the ratio of kVA ratings to the 1⁄4

power. Similarly, areas vary as the ratios of kVA ratings to the 1⁄2 power and volumes vary as the

ratio of the kVA ratings to the 3⁄4 or 0.75 power. Hence the term "0.75 scaling rule." Table 5B.1

depicts the most common scaling relationships in transformers.

5B­1Table 5B.1 Common Scaling Relationships in Transformers

Parameter B eing Scaled Relationship t o kVA Rating

(varies w ith ratio of kVA x )

Weight (kVA

1 /kVA

0 ) 3/4

Cost (kVA

1 /kVA

0 ) 3/4

Length (kVA

1 /kVA

0 ) 1/4

Width (kVA

1 /kVA

0 ) 1/4

Height (kVA

1 /kVA

0 ) 1/4

Total Losses (kVA

1 /kVA

0 ) 3/4

No­load Losses (kVA

1 /kVA

0 ) 3/4

Exciting Current (kVA

1 /kVA

0 ) 3/4

% Total Loss (kVA

1 /kVA

0 ) ­1/4

% No Load Loss (kVA

1 /kVA

0 ) ­1/4

% Exciting Current (kVA

1 /kVA

0 ) ­1/4

% Resistance (R) (kVA

1 /kVA

0 ) ­1/4

% Reactance (X) (kVA

1 /kVA

0 ) 1/4

Volts/Turn (kVA

1 /kVA

0 ) 1/2

The three elements listed below are true as the kVA rating increases or decreases, if the

following factors are held constant: the type of transformer (distribution or power transformer,

liquid­filled or dry­type, single­phase or three­phase), the primary voltage, the core configuration,

the core material, the core flux density, and the current density (amperes per square inch of

conductor cross­section) in both the primary and secondary windings.

1. The p hysical proportions are constant (same relative shape),

2. The e ddy loss proportion is essentially constant, and

3. The i nsulation space factor (voltage or BIL) is constant.

In p ractical applications, it is rare to find that all of the above are constant over even

limited ranges; however, over a range of one order of magnitude in both directions (e.g., from

50kVA to 5kVA or from 50kVA to 500kVA), the scaling rules shown in Table 5B.1 can be used

to establish reasonable estimates of performance, dimensions, costs, and losses. In practice,

these rules can be applied over even wider ranges to estimate general performance levels. The

same quantities are depicted graphically in Figure 5B.1 for reference.

5B­2S ize/Performance Relations vs. Rating

10.00

1.00

0.10

0.1 1 10

Ratio of kVA's

Figure 5B.1 Size and Performance Relationships by kVA Rating

To i llustrate how the scaling laws are used, consider two transformers with kVA ratings

of S

0 and S

1 . The no­load losses (NL) and total losses (TL) of these two transformers would be

depicted as NL

0 and TL

0 , and NL

1 and TL

1 . Then the relationships between the NL and TL of the

two transformers could be shown as follows:

0 .75 0.75 

### ⎛ S ⎞ ⎛ S ⎞

NL = NL

0 

### ⎜ 1 

### ⎟ and T L

= TL

. 0 

### ⎜ 1 

### ⎟

1 1 

### ⎝ S

0 

### ⎠ ⎝ S

0 

### ⎠

These two equations can be manipulated algebraically to show that the load loss also

varies to the 0.75 power. Starting with the concept that total losses equals no­load losses plus

load losses, one can derive the relationship for load loss (LL), and show that it also scales to the

0.75 power. S pecifically:



### LL

1 

### = TL

1 

### ­ NL

1

Plugging the TL1 and NL1 terms into this equation:Ratio of Quantity ­

­

­

­ Weight, Cost, Total

Loss, No­load loss,

Exciting Current

0.75 power

Volts/ turn 0.5

power

Dimensions Percent

Reactance 0.25 power

Percentage: No­load

Loss, Total Loss,

Resistance ­0.25

power

0 .75 0.75 

### ⎛ S ⎞ ⎛ S ⎞

LL = TL

0 

### ⎜ 1 

### ⎟ ­ NL

0 

### ⎜ 1 

### ⎟

1 

### ⎝ S

0 

### ⎠ ⎝ S

0 

### ⎠

5B­3

### . 0 .75

1



### = 

## ( TL − NL

0 

## ) ⎛ ⎜

S ⎞

0 

### ⎟ 

### ⎝ S

0 

### ⎠

That is,

0 .75 

### ⎛ S ⎞



### L

L = LL

0 

### ⎜ 1 

### ⎟

1 

### ⎝ S

0 

### ⎠

In this way, the 0.75 scaling rule can be used to derive the losses of a transformer,

knowing the losses of a reference unit, if the specified type of transformer is held constant, and

key parameters are fixed—such as the type of core material, core flux density, and conductor

current density in the high and low voltage windings.

5B.2 T HEORY AND BASIS FOR SCALING RULES

To u nderstand the origins of winding and output coefficients and related scaling laws, it

is necessary to review some basic equations and definitions. Most are lifted freely or derived

from similar material in Modern Power Transformer Practice

, Wiley 1979, edited by R.

Feinberg. 1 No mathematics beyond elementary algebra is required, but a good deal of implied

physics and electrical engineering is required to fully appreciate these derivations.

5B.2.1 P ower and Voltage Equations

The m achine equation relates the induced volts, V, per phase to the number of turns (N)

the frequency (f) in Hertz, the peak core flux density B

m in Tesla, and the cross­sectional area of

the core steel (A

Fe ) in square meters. The units are mixed to simplify the basic equations, a

common practice in transformer design texts. The machine equation is derived from Faraday's

law, which is expressed as



### ∂

φ

v = − N



### ∂ t

∂ φ

where v is the instantaneous value of V, and is the derivative of changing magnetic flux



### ∂ t with respect to time.

Considering V as the root­mean­square (RMS) value of a sine­wave alternating current

voltage, the above equation can be converted into:

V/N = 4 .44 f B

m Eq. 1 A

Fe

5B­4The voltage and turns may apply to either the primary or the secondary winding and, for the ideal

transformer with no losses and no­leakage flux,

V

1 /V

2 = N

1 /N

2 = n = I

2 /I

1

where V

1 and V

2 represent primary and secondary voltages respectively, N

1 and N

2 primary and

secondary turns, and I

1 and I

2 primary and secondary currents in amperes (amps). The quantity n

is referred to as the “turns ratio.” With the parameters defined, and using Equation 1, the output

or transformer capacity (S) in megavolt­amperes (MVA) per phase can be expressed as:

S

= 4.44 f B

m A

Fe N I Eq. 1a

The o verall cross­section of primary plus secondary conductors in square meters is

A

Cu = (N

1 a 1 + N

2 a 2 ) x 10 ­6

and, assuming current densities for primary and secondary windings to be equal, then

A

Cu = 2 x 10 ­6 Na

where “ a” is the conductor cross­section in square millimeters (mm 2 ) of an individual turn

referred to the winding with N turns, and a 1 and a 2 are conductor cross­sections of primary and

secondary turns, respectively. As long as the winding current densities are equal, either winding

may be used as reference, provided the choice of primary or secondary is consistent. Starting

with Equation 1a, using the A

Cu relationship explained above, and letting J represent current

density in amps per mm 2 :

S

= 2.22 f B J A

Fe Eq. 2

m A

Cu

Let A

w be the core window area in square meters, and k

w the window space factor, as

given by 2 A

Cu /A

w . (Refer to Figure 5B.2 and note that, in a three­phase transformer, there are

two coil phases occupying a given core window). This fraction is indicative of the insulation and

cooling channel requirements. For distribution transformers, k

w is found to be about 0.3–0.4 for

nominal 12 kV systems. Using these definitions,

S

= 1.11 f B J A

Fe k A Eq. 2a

m w w

Note that, for a given MVA rating, and specified flux and current densities, the product of

conductor and core cross­section is constant and inversely related; i.e. A

Fe α 1/A

Cu .

5B.2.2 L osses

Ideally, i f the values of energy loss in Watts per kilogram (W/kg) of unit mass of the core

and windings are known, the total core and load losses (P

Fe and P

Cu ) can be readily obtained.

5B­5

### . These results are accomplished by multiplying the W/kg for both core and windings by the core

mass and the conductor mass respectively (or by their volumes times material densities).

The D epartment uses the convention that lower case corresponds to per­unit quantities

and upper case corresponds to total or total­per­phase quantities. Load losses consist of resistive

(p

R ) and eddy (p

i ) components. Expressions can be derived that express each in terms of the

conductor properties and geometry. The fraction of eddy losses plays an important role and can

be expressed as



### % P 

### ⎛ ⎞

i %P

i = 100 P

i /P

R , or 

### P = P

R 

### ⎜ ⎟

i 

### 1 00 

### ⎝ ⎠

Ignoring stray loss, (which is associated with eddy losses), let P

t represent total load loss

for a three­phase transformer. That is,

P = 3 P

Cu

t

Also assume the same eddy loss fraction in primary and secondary windings.



### %P ⎛ ⎞ ⎛ % P

i 

### ⎞

i 

### P

= k P P

Cu = P

R + P

i = P

R + 

### = 1 + ⎟

R i R 

### P

R 

### ⎜ ⎟ ⎜

100 ⎝ ⎠ ⎝ 1 00 ⎠

Closely associated with the load loss of a transformer is its impedance. When the load

loss of a given transformer is determined by test (the watt­meter reading in the test circuit), that

same test also provides the value of the impedance (the voltmeter reading in the test circuit).

Impedance in a transformer is expressed in terms of the “impedance voltage,” which is defined as

“the voltage required to circulate rated current through one of two specified windings of a

transformer when the other winding is short­circuited, with the windings connected as for rated

voltage operation” (IEEE C57.12.80).

For c onvenience, “percent impedance,” %Z, is used to describe the impedance voltage of

a transformer. In accordance with the definition given above,

%Z

= 

### I Z 

### × 100



### V

that is, when related to the primary or secondary winding of a transformer, the percent impedance

is the percent voltage drop due to impedance when rated current flows through the respective

primary or secondary winding of the transformer.

The %Z may be represented by its resistive and reactive components, %R and %X, as

2 2 %Z

= 

# ( % R



# ) + 

# ( %X 

# )

5B­6

### . Therefore, one can express percent resistance (%R) as follows:



### I R × 10 2



### %

R =



### V

Note that R in the numerator must represent the total resistance in the transformer

windings. Therefore, if the transformer is being viewed from the primary terminals, the value of

R would be the total resistance of the primary winding, plus the total resistance of the secondary

winding referred to the primary winding, (R

2 (N

1 /N

2 ) 2 ).

Where t he percent impedance, percent reactance, and percent resistance are related to the

voltage across the primary or secondary winding of a transformer, the percent load loss (%I 2 R) is

related to the MVA capacity of the transformer, stray loss being ignored as stated previously.

Multiplying n umerator and denominator in the above equation by I, and letting P

t represent total load loss in watts and S represent the MVA per phase rating, one can determine

the percent load loss as:

2 2 

### I R × 10 2 

### I R × 10 2



### = Percent load loss

= 

### I × V 

### 3 S 10 6 

### ×

− 4 

### 1 0 P

t 

### ∴ %R =



### 3 S

Thus, an expression of %R is equivalent to indicating the transformer's load loss.

From E quation 2, it is evident that, once the core flux density and current density are

fixed, the transformer rating is dependent on the core cross­section and window area. Next, one

can derive information about the window shape.

In a detailed discussion of the reactance, the electrical characteristics would depend on:

• the r atio of winding height (h) to the winding mean turn (s), and

• the r atio of the cross­sectional areas of the core and conductor (A

Fe /A

Cu ).

The m ean value of s (a linear measurement, recording the circumference), is given by the

equation s = (s

1 + s 2 )/2, where s 1 is the mean turn of the primary winding and s 2 is the mean turn

of the secondary winding.

These r atios, together with the necessary space factors for insulating and cooling

clearances, establish the relative volumes of the core and conductor. Consequently, if fixed

values for the specific loadings and, therefore specific losses for core and conductor can be

assumed, the ratios of core loss and load loss are established.

5B­7. The following application of relationships derives an expression relating the flux and

current densities. The expression starts with:



### ⎛ %P

i 

### ⎞

P

= k P P

Cu = 

### ⎜ 

### 1 +

100 ⎠ ⎟

R i R 

### ⎝

2 2 

## P

= (I R + I R )k

,

C

u 1 1 2 2 i

where subscripts 1 and 2 indicate primary and secondary windings, respectively. The resistance

per phase of the primary winding is given by



### ρ

N s 

### R = 1 1 

### ohms,

1 

### a 1

where a 1 is the cross­sectional area of the primary copper conductor, and ρ is the resistivity at full

load operating temperature of the conductor, 21.4 x 10 ­3 ohm­meters. The value of R

2 is similarly

obtained:

2 2 

### ⎛ I ρ N s I ρ N s ⎞

C u 1 1 

### + 2 2 2 

### ⎟

i 

### ∴ P

= 

### ⎜ 1 

### k

a 

### ⎝

1 

### a 2 

### ⎠



### ⎛ I s I s ⎞

2 2 

### ∴ P

= IN 

### ⎜ 1 1 

### + ⎟ ρ k

C u i



### a 

### ⎝

1 

### a 2 

### ⎠

where IN is the ampere­turns in either winding. As before, the assumption of equal current

densities in the windings is made, driven by the condition for minimum I 2 R loss. Accordingly,

= 2 INJsρ k

i P

Cu



### P

Cu

, the current density equation.

, 

### J ∴ =



### 2 INs 

### ρ k

i

Multiplying Equation 1 by I and rearranging algebraically, one gets:



### VI IN =



### 4 .44fB A

m Fe

5B­8It was established earlier that S is the rating per phase in MVA, i.e., VI

= 10 6 S. Thus:

6 

### 10 S

∴ I N =



### 4 .44fB A

m Fe

Using the current density equation, substituting the resistivity value for ρ, and the above

value for IN, one can derive that:



### 1 04 × 10

− 6 

### fB A P J = m Fe Cu

. 

### k sS

i

The watts of conductor loss (for copper) can be expressed as a percentage of the

transformer MVA rating:



### P × 10 2



### %P = Cu

Cu 

### S

or, in kilowatts:



### P × 10 2 

### 0.1P

C

u 

### = Cu %P

Cu 

### =



### S 10 3 

### S × .

By substituting in the revised equation for J (amperes per square meter), one gets



### × − 6 

### × − 6 

### 104 10 fB A S %P 1040 10 fB A 

### J = m Fe 

### × Cu m Fe



### = × %P Eq. 3

C u 

### k sS 0.1 k s

i i

If aluminum windings were used instead of copper, a value of 655 would be substituted

for 1040. The expression assumes equal J in both windings, and that both windings are made of

the same material. The losses are expressed at operating temperature.

If J and B

m are chosen independently, the transformer will have a natural value of

conductor loss depending on the ratio A

Fe /s. Conversely, if losses are specified, the choice of J is

determined by B

m and A

Fe /s. Note that this relationship gives no information about the other

transformer dimensions. The impedance, voltage, and other space requirements provide the

majority of this information.

5B­95B.2.3 Output and Winding Coefficients

Starting w ith the output or power Equation 2, one can write:



### S

S

= 2.22 f B J A

Fe or A

Fe =

m A

Cu 

### 2 .22 f B J A

m Cu

Then, without changing the value, one can state:



## ( ) 2

2

m 

### S



## ( ) m

2

m 

### ) S A =

=

Cu 

### 2.22f B J A Fe Cu

Cu 

### 2.22f B J (A )(A

2.22f B J A Fe or

m 

### A S 

### ) A = Fe

Cu 

### (2.22f B J)(A Fe Eq. 4

Use K

AS to represent the portion of Equation 4 to the right of

. 

### S

The expression K

AS is essentially constant for a wide range of transformer classes and is

called the output coefficient. For three­phase, liquid­filled distribution transformers at 60 Hz, the

value of K

AS ranges from 0.050 to 0.055, with a nominal median value of 0.052. For singlephase, wound­core, liquid­filled units at 60 Hz, the median value is about 0.040.

In a similar fashion, making use of Equation 4, we can restate Equation 1 as follows:

2 

### V 

### ( 4.44fB ) SA



### = 4.44fB A 

### = m Fe

m Fe 

### N 2.22fB JA

m Cu



### ⎛ 8.88fB

m 

### ⎞ 

### ⎛ A

Fe 

### ⎞ ⎟ 

## ( S 

## )

= 

### K S Eq. 5

= 

### ⎜ ⎝ J ⎟ ⎠ 

### ⎜

⎝ A

Cu 

### ⎠ VS

The expression K

VS is also essentially constant for a wide range of transformer classes

and is called the winding coefficient. One can also express K

VS in terms of K

AS :

= 4 .44 f B

m K

AS K

VS

5B­10For 60 Hz systems, this may be rewritten as K

VS = 266.4 B Thus the median values

m K

AS . for K

VS become 21.5 for three­phase and 17.0 for single­phase, wound­core distribution

transformers at 60 Hz with B

= 1.55 Tesla. Equations 4–5 provide initial estimates for

m transformer dimensions in studies. They are the starting basis for the scaling laws used to scale

designs and performance. Typical values are given in Table 5B.2 for core type, liquid­filled, 60

Hz distribution transformers at 12 kV, 95 kV BIL.

Table 5 B.2 Nominal 60 Hz, Core­Type, Liquid­Filled, 12 kV Distribution Transformers

Class B

m of Dist. J(A/mm 2 ) (Tesla) A

Fe /A

Cu K

AS K

VS %X

Transf. Range Nominal Nominal Range Nominal Range Nominal

3­Phase 2.4­3.2 2.7 1.55 1.4­2.8 1.6 0.050­ 0.052 21.5 4.75

0.055

1­Phase 2.0­2.5 2.3 1.55 0.65­0.85 0.8 0.038­ 0.041 17.0 4.75

0.043

5B.2.4 Scaling Laws

Having e stablished the output and winding coefficients, it is instructive to examine the

origin of the 0.75 rules for scaling transformer losses. To illustrate, first of all, one needs to set

relationships as follows:



### V

= 

### K S

VS 

### N



## A

= K 

## S

Fe AS



### 1 A

Cu 

### = K

CS 

### S

, (where 

### K = )

CS 

### K

A

S

s ~ (A

Fe 0.5 + 

### b

w )

~ S 0.25



### 4

The shape of the window is set by voltage and the ratio h/s, which is essentially constant

for a given voltage and size, thus setting b

w . Refer to Figure 5B.2 for dimensional definitions.

5B­115B­12 2 

### I R

1000 2 

### S R

V 1000 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 2 

### S R

V 1000 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠

17 2 2

2

Cu 

### 4.28 10 S sN

A V − 

### × 0.75 

## K S s K'S . 

## × =

b

can

S

2 π

S 1 2 π

S 2 2 π 1

4 b

w

1 2

n 1

2 b

can

1

2 b

w

1 2 d –

b 1 b 02 b 01 t 0

b 2

n 01 n 02 4 3 n

w b

w

2b Fe

A FeA

Note:  This diagram is not to scale b

can

S

2 π

S 1 2 π

S 2 2 π 1

4 b

w

1 2

n 1

2 b

can

1

2 b

w

1 2 d –

b 1 b 02 b 01 t 0

b 2

n 01 n 02 4 3 n

w b

w

2b Fe

A FeA

Note:  This diagram is not to scaleSource: R. Feinberg, Modern PowerTransformer Practice Now, one considers the load losses, P

Cu (in kW/phase):

P

Cu =   = =

= =

The other scaling laws are derived in a similar fashion.

Figure 5B.2 Basic Three­Phase Transformer DimensionsREFERENCES

1. Modern p ower transformer practice, Edited by R Feinberg, Wiley Publishers, New York, NY,

1979. ISBN: 047026344X

5B­13