DRAFT

5B-i Transformer_DRAFT_TSD_Appendix5B_version08-01-05.wpd APPENDIX 5B. SCALING RELATIONSHIPS IN TRANSFORMER MANUFACTURING TABLE OF CONTENTS

5B.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5B-1 5B.2 THEORY AND BASIS FOR SCALING RULES . . . . . . . . . . . . . . . . . . . . . . . . . . 5B-4 5B.2.1 Power and Voltage Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5B-4 5B.2.2 Losses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5B-5 5B.2.3 Output and Winding Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 5B-10 5B.2.4 Scaling Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5B-11

LIST OF TABLES Table 5B.1 Common Scaling Relationships in Transformers . . . . . . . . . . . . . . . . . . . . . . . . . 5B-2 Table 5B.2 Nominal 60 Hz, Core-Type, Liquid-Filled, 12 kV Distribution Transformers . . 5B-11

LIST OF FIGURES Figure 5B.1 Size and Performance Relationships by kVA Rating . . . . . . . . . . . . . . . . . . . . . 5B-3 Figure 5B.2 Basic Three-Phase Transformer Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . 5B-12DRAFT

5B-1 Transformer_DRAFT_TSD_Appendix5B_version08-01-05.wpd APPENDIX 5B: SCALING RELATIONSHIPS IN TRANSFORMER MANUFACTURING

5B.1 INTRODUCTION There exist certain fundamental relationships between the ratings in 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 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, there is the term "0.75 scaling rule." Table 5B.1 depicts the most common scaling relationships in transformers.DRAFT

5B-2 Transformer_DRAFT_TSD_Appendix5B_version08-01-05.wpd Table 5B.1 Common Scaling Relationships in Transformers Parameter Being Scaled Relationship to kVA Rating (varies with 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 No-load 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 % R (kVA 1 /kVA 0 ) -1/4 %X (kVA 1 /kVA 0 ) 1/4 Volts/Turn (kVA 1 /kVA 0 ) 1/2 The following three elements are true as the kVA rating increases or decreases if the following conditions are met: holding constant the type of transformer (distribution or power transformer, 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 (say 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.DRAFT

5B-3 Transformer_DRAFT_TSD_Appendix5B_version08-01-05.wpd Size/Performance Relations vs. Rating

0.10 1.00 10.00

0.1 1 10 Ratio of kVA'sRatio of Quantity Weight, Cost, Total Loss, No-load loss, Exciting Current - 0.75 power Volts/ turn - 0.5 power

Dimensions - Percent Reactance 0.25 power

Percentage: No-load Loss, Total Loss, Resistance - -0.25 power

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

. 0.75 0.75 1 1 1 0 1 0 0 0 

### S S NL NL and TL TL S S ⎛ ⎞ ⎛ ⎞
= = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

0.75 0.75 1 1 10 0 0 0 

### S S LL = TL - NL S S ⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ To illustrate how the scaling laws are used, consider two transformers with kVA ratings of S 0 and S 1 . The no-load 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:

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, we can derive the relationship for load loss (LL), and show that it also scales to the 0.75 power. Specifically: 

### LL 1 

### = TL 1 

### - NL 1 Plugging the TL1 and NL1 terms into this equation:DRAFT

5B-4 Transformer_DRAFT_TSD_Appendix5B_version08-01-05.wpd 

## ( ) 0.75 1 0 0 0 

### S TL NL S ⎛ ⎞
= − ⎜ ⎟ ⎝ ⎠

0.75 1 1 0 0 

### S LL = LL . S ⎛ ⎞ ⎜ ⎟ ⎝ ⎠



### v N t ∂φ = − ∂

t ∂φ ∂ That is,

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 In order 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 the text, 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 Power and Voltage Equations 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 B m in Tesla, and the cross-sectional 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

, where v is the instantaneous value of V, and 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 = 4.44 f B m AFe Eq. 1DRAFT

5B-5 Transformer_DRAFT_TSD_Appendix5B_version08-01-05.wpd The voltage and turns may apply to either primary or secondary winding and, for the ideal transformer with no losses and no-leakage 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 MVA per phase can be expressed as: S = 4.44 f B m AFe N I Eq. 1a The overall cross-section 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 cross-section in square millimeters (mm 2 ) of an individual turn referred to the winding with N turns and a 1 and a 2 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 1a), using the A Cu relationship explained above, and letting J represent current density in amps per mm 2 : S = 2.22 f B m J AFe A Cu Eq. 2 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 three-phase 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 m J AFe k w A w Eq. 2a 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. A Fe 

### " 1/A Cu . 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 (P Fe and P Cu ) can be readily obtained.DRAFT

5B-6 Transformer_DRAFT_TSD_Appendix5B_version08-01-05.wpd i i R 

### %P P = P 100 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠

i i R R iR 

### %P %P P 1 P kP. 100 100 ⎛ ⎞⎛ ⎞ =+ = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠



### IZ 100 V ×



# ( )( ) 2 2 

### %R %X + 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 to total-per-phase 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 = 100 P i /P R , or Ignoring stray loss, (which is associated with eddy losses), let P t represent total load loss for a three-phase transformer. That is, P t

= 3P Cu Also assume the same eddy loss fraction in primary and secondary windings.

P Cu = P R + P i = P R + 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, %Z = 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 =DRAFT

5B-7 Transformer_DRAFT_TSD_Appendix5B_version08-01-05.wpd 2 22 2 6 

### I R 10 I R 10

I V 3S 10

× × = × ×
4 t 

### 10 P %R 3S − 

### ∴ = 2 

### IR 10 %R . V × = Therefore, we can express percent resistance (%R) as follows:

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 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 (%I 2 R) 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 P t represent total load loss in watts and S represent the megavoltamperes (MVA per phase) rating, we determine the percent load loss as:

Percent load loss =

Thus, an expression of %R is equivalent to indicating the transformer's load loss. From equation (2), it is evident that once the core flux density and current density are fixed, the transformer rating is dependent upon the core cross-section and window area. Next, we will 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 • the ratio of the cross-sectional areas of the core and conductor (A Fe /A Cu ). The mean 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 ratios, together with the necessary space factors for insulating and cooling clearances, establish the relative volumes of the core and conductor. Consequently, if fixed valuesDRAFT

5B-8 Transformer_DRAFT_TSD_Appendix5B_version08-01-05.wpd i R iR 

### %P 1 P kP 100 ⎛ ⎞ + = ⎜ ⎟ ⎝ ⎠

2 2 Cu 1 1 2 2 i 

## P (I R I R )k , = +

1 1 1 1 

### N s R ohms, a ρ =

2 2 1 11 2 2 2 Cu i 1 2 

### I Ns I Ns P k a a ⎛ ⎞ ρ ρ ∴= + ⎜ ⎟ ⎝ ⎠

11 2 2 Cu i 1 2 

### Is Is P IN k a a ⎛ ⎞ ∴= + ρ ⎜ ⎟ ⎝ ⎠

Cu i 

### P J 2INs k ∴ = ρ 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 derive an expression relating the flux and current densities. The expression starts with:

P Cu =

where subscripts 1 and 2 indicate primary and secondary windings, respectively. The resistance per phase of the primary winding is given by

where a 1 is the cross-sectional area of the primary copper conductor, and 

### D is the resistivity at full load operating temperature of the conductor, 21.4 x 10 -3 ohm - meters. The value of R 2 is similarly obtained:

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 I 2 R loss. Accordingly, P Cu= 2INJs 

### D k i ,the current density equation. ,
Multiplying equation (1) by I and rearranging algebraically we getDRAFT

5B-9 Transformer_DRAFT_TSD_Appendix5B_version08-01-05.wpd m Fe 

### VI IN 4.44fB A =

.

6

m Fe 

### 10 S IN 4.44fB A ∴ =

6 m Fe Cu i 

### 104 10 fB A P J k sS − 

### × =

.

2 Cu Cu 

### P 10 %P S × =

2 Cu Cu 3 

### P 10 0.1P S S 10 × = = × .

6 6 m Fe Cu m Fe Cu i i 

### 104 10 fB A S %P 1040 10 fB A J % P ksS 0.1 ks − − 

### × × = ×= × It was established earlier that S is the rating per phase in MVA, that is VI = 10 6 S. Thus:

Using the current density equation, substituting the resistivity value for 

### D , and the above value for IN, we can derive that:

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

or in kilowatts:

%P Cu By substituting in the revised equation for J (amperes per square meter), we get
Eq. 3

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 upon the ratio A Fe

/s. Conversely, if losses are specified, the choice of J is determined by B m and AFe /s. Note that this relationship gives no information about the otherDRAFT

5B-10 Transformer_DRAFT_TSD_Appendix5B_version08-01-05.wpd m Cu 

### S 2.22 f B J A



## ( ) 2 2 m Cu 

### S 2.22f B J A 

## ( ) m Fe Cu 2 m Cu 

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

Fe m Cu 

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

### S

2 m Fe m Fe m Cu 

### (4.44fB ) SA V 4.44fB A N 2.22fB JA = =

VS 

### K S 

## ( ) Fe m Cu 

### A 8.88fB S J A ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 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 (2), we can write:

S = 2.22 f B m J AFeA Cu or AFe= Then, without changing the value, we can state:

AFe= = or

AFe= Eq. 4

Use K AS to represent the portion of Equation (4) to the right of

.

The expression K AS 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 K AS ranges from 0.050-0.055 with a nominal median value of 0.052. For single-phase, wound core, liquid filled 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:

= = Eq. 5 The expression K VS is also essentially constant for a wide range of transformer classes and is called the winding coefficient. We can also express K VS in terms of K AS :DRAFT

5B-11 Transformer_DRAFT_TSD_Appendix5B_version08-01-05.wpd VS 

### V K S N =

Fe AS 

## A KS =

Cu CS A KS = CS AS 

### 1 K K =

w 

### b

4 K

VS = 4.44 f B m K AS For 60 Hz systems this may be rewritten as K VS = 266.4 B m K AS . Thus the median values for K VS become 21.5 for three-phase and 17.0 for single-phase wound core distribution transformers at 60 Hz with B m = 1.55 Tesla. Equations (4)-(5) 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 for core type, liquid filled, 60 Hz distribution transformers at 12 kV, 95 kV BIL.
Table 5B.2 Nominal 60 Hz, Core-Type, Liquid-Filled, 12 kV Distribution Transformers

Class of Dist. J(A/mm 2 ) B m (Tesla) A Fe /A Cu K AS K VS %X Transf. Range Nominal Nominal Range Nominal Range Nominal 3-Phase 2.4- 3.2 2.7 1.55 1.4-2.8 1.6 0.050- 0.055 0.052 21.5 4.75
1-Phase 2.0- 2.5 2.3 1.55 0.65-0.85 0.8 0.038- 0.043 0.041 17.0 4.75 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, we need to set relationships as follows:

, (where )
s ~ (A Fe 0.5 + ) ~ S 0.25
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 b w . Refer to Figure 5B.2 for dimensional definitions.DRAFT

5B-12 Transformer_DRAFT_TSD_Appendix5B_version08-01-05.wpd 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 Power Transformer Practice Now, we consider the load losses, P Cu (in kW/phase):

P Cu = = =

= =
The other scaling laws are derived in a similar fashion.

Figure 5B.2 Basic Three-Phase Transformer DimensionsDRAFT

5B-13 Transformer_DRAFT_TSD_Appendix5B_version08-01-05.wpd 1. Modern power transformer practice, Edited by R Feinberg, Wiley Publishers, New York, NY, 1979. ISBN: 047026344X REFERENCES