5B-i 

APPENDIX 5B. SCALING RELATIONSHIPS IN TRANSFORMER 

MANUFACTURING

TABLE OF CONTENTS

5B.1 INTRODUCTION ...................................................................................................1

5B.2 THEORY AND BASIS FOR SCALING RULES ..................................................3

5B.2.1 Power and Voltage Equations ..................................................................................3

5B.2.2 Losses .......................................................................................................................5

5B.2.3 Output and Winding Coefficients ............................................................................8

5B.2.4 Scaling Laws ..........................................................................................................10

5B.2.5 Devitations from .75 ..............................................................................................10

LIST OF TABLES

Table 5B.1 Common Scaling Relationships in Transformers ........................................... 1

Table 5B.2 Nominal 60 Hz, Core-Type, Liquid-Filled, 12 kV Distribution Transformer

........................................................................................................................................... 10

Table 5B.3 Scaling Exponents By Equipment Class ........................................................ 12

LIST OF FIGURES

Figure 5B.1 Size and Performance Relationships by kVA Rating .................................... 2

Figure 5B.2 Efficiency Levels within an Equipment Class (Logarithmic) ...................... 12

Figure 5B.3 Basic Three-Phase Transformer Dimensions .............................................. 13
5B-1 

APPENDIX 5B. SCALING RELATIONSHIPS IN TRANSFORMER 

MANUFACTURING

5B.1 INTRODUCTION

There exist certain fundamental relationships between the ratings in kilovolt-amperes 

(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 size versus performance 

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 cross sectional area, and total conductor cross 

sectional area is constant. 

To illustrate 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 cross section 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.

Table 5B.1 Common Scaling Relationships in Transformers

Parameter Being Scaled Relationship to kVA Rating 

(varies with ratio of kVAx)

Weight (kVA1/kVA0)

3/4

Cost (kVA1/kVA0)

3/4

Length (kVA1/kVA0)

1/4

Width (kVA1/kVA0)

1/4

Height (kVA1/kVA0)

1/4

Total Losses (kVA1/kVA0)

3/4

No-load Losses (kVA1/kVA0)

3/4

Exciting Current (kVA1/kVA0)

3/4

% Total Loss (kVA1/kVA0)

-1/4

% No Load Loss (kVA1/kVA0)

-1/4

% Exciting Current (kVA1/kVA0)

-1/4

% Resistance (R) (kVA1/kVA0)

-1/4

% Reactance (X) (kVA1/kVA0)

1/4

Volts/Turn (kVA1/kVA0)

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, 
5B-2 

liquid filled or dry-type, single-phase or three-phase), the primary voltage, the core 

configuration, the core material, the core flux density, and the current density (amperes per 

square inch of conductor cross section) in both the primary and secondary windings. 

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

2. The eddy loss proportion is essentially constant, and 

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

In practical 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.

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

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

of S0 and S1. The no-load losses (NL) and total losses (TL) of these two transformers would be 

depicted as NL0 and TL0, and NL1 and TL1. Then the relationships between the NL and TL of the 

two transformers could be shown as follows: 

NL1 = NL0 �

S1

S0

�

0.75

 and TL1 = TL0 �

S1

S0

�

0.75
5B-3 

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 no-load losses plus 

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

0.75 power. Specifically: 

LL1 = TL1 − NL1

Plugging the TL1 and NL1 terms into this equation: 

LL1 = TL0 �

S1

S0

�

0.75

− NL0 �

S1

S0

�

0.75

= (TL0 − NL0) �

S1

S0

�

0.75

That is,

LL1 = LL0 �

S1

S0

�

0.75

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 THEORY AND BASIS FOR SCALING RULES

To understand 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.i

5B.2.1 Power and Voltage Equations 

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

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

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

the frequency (f) in Hertz, the peak core flux density Bm in Tesla, and the cross-sectional area of 

the core steel (AFe) 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 

v = −N ∂∅

∂t
5B-4 

where v is the instantaneous value of V, and ∂∅

∂t is the derivative of changing magnetic flux with 

respect to time. 

Considering V as the root-mean-square (RMS) value of a sine-wave alternating current 

voltage, the above equation can be converted into: 

V N = 4. ⁄ 44fBmAFe

Equation 5B.1 

The voltage and turns may apply to either the primary or the secondary winding and, for 

the ideal transformer with no losses and no-leakage flux,

V1⁄V2 = N1⁄N2 = n = I2⁄I1

where V1 and V2 represent primary and secondary voltages respectively, N1 and N2 primary and 

secondary turns, and I1 and I2 primary and secondary currents in amperes (amps). The quantity n 

is referred to as the “turns ratio.” With the parameters defined, and using Equation 5B.1, the 

output or transformer capacity (S) in megavolt-amperes (MVA) per phase can be expressed as:

S = 4.44fBmAFeNI

Equation 5B.2 

The overall cross-section of primary plus secondary conductors in square meters is 

ACu = (N1a1 + N2a2) × 10−6

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

ACu = 2 × 10−6 Na

where “a” is the conductor cross-section in square millimeters (mm2) of an individual turn 

referred to the winding with N turns, and a1 and a2 are conductor cross-sections 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 5B.2, using the ACu relationship explained above, and letting J represent current 

density in amps per mm2: 

S = 2.22 f BmJAFeACu

Equation 5B.3 

Let Aw be the core window area in square meters, and kw the window space factor, as given 

by 2 ACu⁄Aw. (Refer to Figure 5B.3 and note that, in a three-phase transformer, there are two coil 

phases occupying a given core window). This fraction is indicative of the insulation and cooling 
5B-5 

channel requirements. For distribution transformers, kw is found to be about 0.3–0.4 for nominal 12 

kV systems. Using these definitions,

S = 1.11 f BmJAFekwAw

Equation 5B.4 

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

conductor and core cross-section is constant and inversely related; i.e. AFe α 1/ACu. 

5B.2.2 Losses

Ideally, if 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 (PFe and PCu) can be readily obtained. 

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 Department uses the convention that lower case corresponds to per-unit quantities 

and upper case corresponds to total or total-per-phase quantities. Load losses consist of resistive 

(PR) and eddy (Pi) 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

%Pi = 100 Pi⁄PR, or Pi = PR �

%Pi

100�

Ignoring stray loss, (which is associated with eddy losses), let Pt represent total load loss 

for a three-phase transformer. That is,

Pt = 3PCu

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

PCu = PR + Pi = PR + PR �

%Pi

100� = �1 +

%Pi

100� PR = kiPR

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 short-circuited, with the windings connected as for rated 

voltage operation” (IEEE C57.12.80). 

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

a transformer. In accordance with the definition given above, 
5B-6 

%Z =

IZ × 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 

%Z = �(%R)2 + (%X)2

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

%R =

IR × 102

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, (R2(N1/N2)

2

). 

Where the 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 (%I2R) is 

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

Multiplying numerator and denominator in the above equation by I, and letting Pt represent total 

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

loss as:

Percent load loss = I

2R × 102

I × V = I

2R × 102

3S × 106

∴ %R =

10−4Pt

3S

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

From Equation 5B.3, it is evident that, once the core flux density and current density are 

fixed, the transformer rating is dependent on the core cross-section 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 ratio of winding height (h) to the winding mean turn(s), and 
5B-7 

• The ratio of the cross-sectional areas of the core and conductor (AFe /ACu). 

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

equation s = (s1 + s2)/2, where s1 is the mean turn of the primary winding and s2 is the mean turn 

of the secondary winding. 

These ratios, 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. 

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

current densities. The expression starts with: 

PCu = �1 +

%Pi

100� PR = kiPR

PCu = �I1

2R1 + I22R2�ki,

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

per phase of the primary winding is given by 

R1 = ρN1s1

a1

ohms,

where a1 is the cross-sectional area of the primary copper conductor, and ρ is the resistivity at full 

load operating temperature of the conductor, 21.4 x 10-3 ohm-meters. The value of R2 is similarly 

obtained: 

∴ PCu = �

I1

2ρN1s1

a1

+

I2

2ρN2s2

a2

� ki

∴ PCu = IN�

I1s1

a1

+

I2s2

a2

� ρki

where IN is the ampere-turns in either winding. As before, the assumption of equal current 

densities in the windings is made, driven by the condition for minimum I2R loss. Accordingly, 

PCu = 2INJsρki

∴ J = PCu

2INsρki

, the current density equation. 

Multiplying Equation 5B.1 by I and rearranging algebraically, one gets:
5B-8 

IN = VI

4.44fBmAFe

It was established earlier that S is the rating per phase in MVA, i.e., VI = 106S. Thus: 

∴ IN = 106S

4.44fBmAFe

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

value for IN, one can derive that: 

J =

104 × 10−6fBmAFePCu

kisS

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

transformer MVA rating:

%PCu = PCu × 102

S

or, in kilowatts: 

%PCu = PCu × 102

S × 103 = 0.1PCuS .

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

J =

104 × 10−6fBmAFeS

kisS

×

%PCu

0.1 = 1040 × 10−6fBmAFekis

× %PCu

Equation 5B.5 

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

conductor loss depending on the ratio AFe/s. Conversely, if losses are specified, the choice of J is 

determined by Bm and AFe/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.

5B.2.3 Output and Winding Coefficients

Starting with the output or power Equation 5B.3, one can write:
5B-9 

S = 2.22fBmJAFeACu or AFe = S

2.22fBmJACu

Then, without changing the value, one can state: 

AFe = � S2

(2.22fBmJACu)2 = √S�2.22fBmJ(AFe)(ACu)

(2.22fBmJACu)2 or

AFe = √S� AFe

(2.22fBmJ)(ACu)

Equation 5B.6 

Use KAS to represent the portion of Equation 5B.6 to the right of √S

The expression KAS is essentially constant for a wide range of transformer classes and is 

called the output coefficient. For three-phase, liquid-filled distribution transformers at 60 Hz, the 

value of KAS ranges from 0.050 to 0.055, with a nominal median value of 0.052. For singlephase, wound-core, liquid-filled units at 60 Hz, the median value is about 0.040. 

In a similar fashion, making use of Equation 5B.6, we can restate Equation 5B.1 as follows: 

V

N

= 4.44fBmAFe = �(4.44fBm)2SAFe

2.22fBmJACu

= ��

8.88fBm

J � �AFeACu

� (S) = KVS√S

Equation 5B.7 

The expression KVS is also essentially constant for a wide range of transformer classes 

and is called the winding coefficient. One can also express KVS in terms of KAS: 

KVS = 4.44fBmKAS

For 60 Hz systems, this may be rewritten as KVS= 266.4 BmKAS. Thus the median values 

for KVS become 21.5 for three-phase and 17.0 for single-phase, wound-core distribution 

transformers at 60 Hz with Bm = 1.55 Tesla. Equation 5B.6 and Equation 5B.7 provide initial 

estimates for 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. 
5B-10

Table 5B.2 Nominal 60 Hz, Core-Type, Liquid-Filled, 12 kV Distribution Transformer 

5B.2.4 Scaling Laws 

Having established 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

N

= KVS√S

AFe = KAS√S

ACu = KCS√S, �where KCS = 1

KAS

�

s~ �AFe

0.5 +

bw

4 � ~s0.25

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 bw

. Refer to Figure 5B.3 for dimensional definitions.

Now, one considers the load losses, PCu (in kW/phase):

PCu = I

2R

1000 = �

S

V

�

2 R

1000

= 4.28 × 10−17S2sN2

ACuV2 = K√S × s = K′

S0.75

The other scaling laws are derived in a similar fashion. 

5B.2.5 Deviations from .75 

Although these laws dictate that an ideal transformer will yield a scaling exponent of 

0.75, DOE recognizes that a different exponent may produce better behaved results based on 

real-world engineering. For the Final Rule, DOE used unique scaling exponents for each 
5B-11

equipment class. For each equipment class DOE derived an exponent to scale relative kVA 

rating by examining the proposals discussed during the negotiations. Because the proposals 

discussed during the negotiations included efficiency levels across multiple designs lines, a 

scaling relationship was implied by the proposal. The exponents used for each equipment class 

are shown below in Table 5B.3. 

If one imagines the standard for a particular equipment class as a function on a plot of 

efficiency (y-axis) versus kVA (x-axis), then the efficiency levels in each design line are a series 

of points along an imaginary vertical line that intersects the x-axis at the design line’s kVA. If 

there is more than one design line in a given equipment class, there will be more than one series 

of points. Because exponential scaling is performed on losses and because exponential function 

will appear as straight lines on logarithmic plots, the concept is more tractable if illustrated that 

way, as is done in Figure 5B.2 below. Note that efficiency and loss values have a one-to-one 

correspondence with each other, so one can use whichever coordinate is easier to illustrate 

identical information. Although standards are ultimately given in terms of efficiency, DOE 

performs the scaling in loss coordinates. Also note that the following figures are given to 

illustrate the scaling concept, and have no relation to actual transformer data.

If one is to select efficiency levels for each design line, as was done by the negotiating 

committee for MVDT transformers, the task remains to scale those chosen efficiencies at certain 

kVA ratings to all of the other kVA ratings that DOE covers. Drawing a straight line1

 through 

the chosen points accomplishes that goal, but may produce a slope different from .75. 

Deriving the .75 rule requires a number of assumptions to be made, among them that the 

overall form and proportions of the transformer remain intact as it changes in size. This 

assumption may break down in a number of ways. For example, MVDT BIL ratings require 

fixed spacings between the edge of a winding and the window of a core. Proportionally, these 

fixed values will be much larger for smaller transformers than for larger units. Thus, while the 

rest of the transformer may behave closer to what the .75 rule would predict, the “fixed” portion 

will cause losses to fall more slowly with decreasing kVA. Stated alternatively, losses will grow 

more slowly with increasing kVA and imply a scaling behavior of less than .75. 

 

1 A straight line in logarithmic space is an exponential in the original dimensions, which is the logical scaling 

behavior for transformers to exhibit.
5B-12

Figure 5B.2 Efficiency Levels within an Equipment Class (Logarithmic)

Table 5B.3 Scaling Exponents By Equipment Class 

Distribution Transformer Equipment Class Scaling Exponent 

1. Liquid-immersed, medium-voltage, single-phase .76

2. Liquid-immersed, medium-voltage, three-phase .79

3. Dry-type, low-voltage, single-phase .75

4. Dry-type, low-voltage, three-phase .74

5. Dry-type, medium-voltage, single-phase, 20-45 kV BIL .67

6. Dry-type, medium-voltage, three-phase, 20-45 kV BIL .67

7. Dry-type, medium-voltage, single-phase, 46-95 kV BIL .67

8. Dry-type, medium-voltage, three-phase, 46-95 kV BIL .67

9. Dry-type, medium-voltage, single-phase, > 96 kV BIL .68

10. Dry-type, medium-voltage, three-phase, > 96 kV BIL .68
5B-13

Figure 5B.3 Basic Three-Phase Transformer Dimensions

 

i

Modern power transformer practice, Edited by R Feinberg, Wiley Publishers, New York, NY, 1979. ISBN: 

047026344X