UNITED STATES DEPARTMENT OF AGRICULTURE 

Rural Utilities Service 

RUS Bulletin 1724E-301 

RD-GD-2009-57

SUBJECT: · Guide for the Evaluation of Large Power Transformer Losses 

TO: All Electric Borrowers, Consulting Engineers, and Rural Utilities Service Staff

EFFECTIVE· DATE: Date of Approval 

OFFICE OF PRIMARY INTEREST: Transmission Branch, Electric _Staff Division. 

FILING INSTRUCTION: This bulletin replaces RUS Bulletin 1724E-301, "Guide forthe 

Evaluation of Large Power Transfonner'Losses," issued in December, 1997.

AVAILABILITY: This bulletin can be accessed on the Internet at:

http://www.usda.gov/rus/electric/bulletins.htm

PURPOSE: This bulletin provides guidelinesfor evaluating transformer.losses and including 

such losses in bid evaluations.

CONTRIBUTORS: The following current and former members of the Substation 

Subcommittee oftheTransmission and Distribution (T&D) Engineering Committee ofthe 

N tional Rural Electric Cooperative Association.

Pehosh, Mike, NRECA, Washington, DC 

Metro, Patti, NRECA, Washington, DC

- Saint, Bob, NRECA, Washington, DC

Eskandary, Mike; USDA, RUS

Malone, Ken, Middle TN EM(:, Murfreesboro, TN 

Kahanek, Bil, Bluebonnet Electric Cooperative, TX 

Ruppard, Paul, East Kentucky Power Coop., KY 

Powell, k - Burns & McDonnell Engineering, TX

Groves, Kevin, West Plains Engineering, MO 

Barnette, Thomas, Berkeley Electric Coop., SC

· Seesholtz, Kevin, South Central Power Co., OH

Newton, Jarod, United Services Group, MN

Higbe, Shawn, McCallThomas Engineering, SC

White, Kevin, N.E. MO Electric Power, MO

Gieger, Dan, Heartland Eng Services,MN

Disclaimer: The contents of this guidance document does not have the force and effect of law and is not meant to bind the 

public in any way. This document is intended only to provide clarity to the public regarding existing requirements under the

law or agency policies.

Acting Administrator 

Rural Utilities Service
Bulletin 1724E-301

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BLANK PAGE
Bulletin 1724E-301

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

2. FORMULAE...........................................................................................................................................................6

3. VALUES FOR FORMULAE ................................................................................................................................8

4. INFLATION .........................................................................................................................................................11

5. EXAMPLE............................................................................................................................................................14

6. CONCLUSIONS...................................................................................................................................................18

APPENDIX A: EQUIVALENT LEVEL LOAD FORMULA ..............................................................................19

APPENDIX B: EQUIVALENT LEVEL COST FORMULA................................................................................23

INDEX:

CONTRACTS:

Evaluation of Large Power Transformer Losses 

DESIGN, SYSTEM:

Evaluation of Large Power Transformer Losses 

MATERIALS AND EQUIPMENT:

Evaluation of Large Power Transformer Losses

POWER LOSS:

Evaluation of Large Power Transformer Losses

ABBREVIATIONS

A Base cost before inflation 

A′ Cost adjusted for inflation

FRC Fixed charge rate for capital investment expressed as a decimal in dollars per dollar 

of investment

EC Cost of energy in dollars per kilowatt-hour 

EC′ Energy charge adjusted for inflation

G Peak ratio which is the ratio of peak load to full rated load 

kVA kilovoltampere

kW kilowatt

kWh kilowatt-hour

K Peak responsibility factor which is the ratio of transformer load at the time of the 

system peak to the transformer peak load

LFA Loss factor for the auxiliary equipment

LFT Transformer loss factor which is the ratio of average transformer losses to peak 

transformer losses

LS Equivalent level losses; value that results in the same total losses as the yearly 

increasing load

MVA Transformer load, megavoltamperes
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OA Self-cooled rating for oil-filled power transformers

P The rate of the increase in costs per kWh associated with power generation and 

transmission expressed as a decimal

PW Present worth

SI The system capital investment in dollars per kilowatt required to supply the power 

losses of the transformer

TAL Losses due to transformer auxiliary equipment in kilowatts 

TLL Transformer’s guaranteed load losses in kilowatts

TNLL Transformer’s guaranteed no-load losses in kilowatts

Y1 The ratio of load to capacity when transformer is installed 

Y2 The ratio of load to capacity when transformer is removed 

r′ Equivalent inflation rate
Bulletin 1724E-301

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1. INTRODUCTION

Losses and purchase price should be considered when deciding which transformer to purchase. 

The purpose of this bulletin is to present a uniform approach that can be used to determine the 

dollar value of these losses over the life of the transformer. Below is typical wording of a 

transformer loss evaluation clause for insertion into bidding documents that specifies how losses 

will be evaluated.

“Load, no-load and auxiliary losses at 50 MVA for the 

30/40/50 MVA transformer will be evaluated as follows:

No-Load Load Auxiliary

Losses Losses Losses

$/kW 2450 $/kW 1304 $/kW 756

The cost of losses for each transformer will be calculated 

by multiplying the appropriate dollars/kW values above by 

the guaranteed load losses at 55°C rating and no-load 

losses at 100% voltages. This cost will be added to the bid 

price for evaluation.”

1.1 Example: Using the loss evaluation factors given above, determine which manufacturer’s 

transformer has the lowest evaluated cost including losses.

161/34.5 kV, 30/40/50 MVA Transformer

Manufacturer A’s 

Transformer

Manufacturer B’s 

Transformer

Bid price $424,500 $436,000

No -load losses 59 kW 53 kW

Load losses at 50 MVA, 

55°C temperature rise

224 kW 218 kW

Auxiliary losses at 

50 MVA 55°C

2.0 kW 2.5 kW

temperature rise
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1.2 Solution

A B

Bid Price = $424,500 = $436,000

Total cost of no-load losses 59 kW (2450 $/kW) = $144,550 53 kW (2450 $/kW) = $129,850

Total cost of load losses 224 kW (1304 $/kW) = $292,096 218 kW (1304 $/kW) = $284,272

Total cost of auxiliary losses 2.0 kW ( 756 $/kW) = $ 1,512 2.5 kW (756 $/kW) = $1,890

TOTAL COST = $862,658 = $852,012

Although the transformer from Manufacturer A has the lowest bid price, the transformer from 

Manufacturer B has the lowest evaluated total cost.

In addition to giving loss evaluation values, the bid documents should also have penalty values 

that the manufacturer is to be charged for every kilowatt by which the actual tested transformer 

losses exceed the guaranteed losses upon which the bids are evaluated. It is important to have 

such penalty values in order to give an incentive to the manufacturers to provide the most 

accurate guaranteed loss values possible. The penalty values should be expressed in the same 

dollars per kW manner as the bid evaluation values but should be somewhat higher. An 

increment of approximately 20 percent is recommended.

2. FORMULAE

The three different types of transformer losses that should be evaluated separately are:

a. Load losses (sometimes called copper or coil losses);

b. No-load losses (sometimes called core or iron losses); and

c. Auxiliary losses (electric fan losses, other such equipment losses).

Load losses are primarily from the I2

R losses in the transformer windings and eddy current 

losses. If a value of load losses is not directly given, load losses can be determined by 

subtracting no-load losses from total losses.∗ No-load losses consist of the hysteresis and the 

eddy current losses in the iron core of the transformer and the I2

R losses in the windings due to 

the excitation current. Auxiliary losses consist of the power necessary to drive the auxiliary

cooling pumps and fans.

∗

If the total losses at full load are 100 kW and the no-load losses are 10 kW, then the load (or 

copper or coil) losses are 90 kW.
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The formulae below yield the total costs of the losses that should be added to the purchase price 

of the transformer as shown in the Example 1.1:

Costof no −load Υ 8760 • (EC)⁄

  =

 lossesindollars  ′

≤

SI +

FCR ∞ƒ

•TNLL (Eq. 1)

 Costof load  Υ 2 8760 • (EC)(LFT )(G)⁄

lossesindollars  = ′

≤

SI(K )(G)+ FCR

∞ƒ

• TLL (Eq. 2)

 

(Eq. 3)

where:

G = peak ratio

K = peak responsibility factor

SI = the system capital investment in dollars per kilowatt required to supply the 

power losses of the transformer;

8760 = the number of hours in a year;

EC = the cost of energy in dollars per kilowatt-hour;

FCR = fixed charge rate for capital investment expressed as a decimal in dollars per 

dollar of investment;

LFA = the loss factor for auxiliary equipment;

LFT = the transformer loss factor which is the ratio of average transformer losses to 

peak transformer losses;

TNLL = the transformer’s guaranteed no-load losses in kilowatts; 

TLL = the transformer’s guaranteed load losses in kilowatts;

TAL = the losses due to transformer auxiliary equipment in kilowatts

A detailed discussion of the factors in Equations 1 through 3 follows in Section 3.

∞

ƒ

•TAL

FCR

2 8760 • (EC)(LFA)⁄ ′

≤

S I(K )+

Υ  =

 lossesindollars 



Costof auxiliary
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3. VALUES FOR FORMULAE

3.1 SI: The System Investment (SI) charge is the cost of generation and transmission facilities 

per kilowatt necessary to supply the additional demand resulting from the transformer losses at 

the system peak. Since a transformer located directly at a generating station does not require an 

investment in transmission facilities, the SI value used to evaluate the losses in the generating 

station transformer should be less than the SI of a transformer to be located at the receiving end 

of a transmission line.

One method for determining the SI value involves adding the construction cost (dollars per 

kilowatt) of a recently completed or soon to be completed generating station to the cost of the 

transmission facilities (dollars per kilowatt) required to connect the transformer to the plant. If 

power is purchased rather than self-generated, the SI value can be determined by dividing the 

demand charge in dollars per kW per year by the fixed charge rate (FCR). Since there is more 

than one method of evaluating the SI value, the method that is judged to yield the most realistic 

results should be used.

3.2 FCR: The fixed charge rate (FCR) represents the yearly income necessary to pay for a 

capital investment. FCR is expressed as a percentage of capital investment. The rate covers all 

costs that are fixed and do not vary with the amount of energy produced. The rate includes 

interest, depreciation, taxes, insurance, and those operations and maintenance expenses that do 

not depend on system kilowatt-hours sold.

The interest rate used should be the same as the interest rate of the loan acquired to purchase the 

transformers. If loan funds are not used, a blended rate of the interest earned on deposited funds 

should be used.

The practice of including some operations and maintenance expenses in the fixed charge rate is a 

matter of judgment. Some typical values for the components of the carrying charge rate are as 

follows:

Interest 7.50%

Depreciation 2.75%

Insurance 0.60%

Taxes 1.00%

Operations and Maintenance 2.76%

Carrying Charge Rate 14.61%

3.3 EC: The energy charge (EC) is the cost per kilowatt-hour for fuel and other expenses that 

are directly related to the production of electrical energy. Although the costs per kilowatt-hour 

will vary with the level of demand, a single energy charge representing an average cost per 

kilowatt-hour throughout the load cycle should be used for the sake of simplicity. Equations 1 

and 2 are based on the assumption that the energy charge remains constant throughout the life of 

the transformer. See Section 4 for a discussion of the effects of inflation and increasing costs on 

the energy charge. If power is purchased, EC will be the kWh (or energy) cost of power.
Bulletin 1724E-301

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3.4 K: The peak responsibility factor (K) is intended to compensate for the transformer peak 

load losses not occurring at the system peak losses. This means that only a fraction of the peak 

transformer losses will contribute to the system peak demand. The value of K can be determined 

from:

Peak responsibility Transformer load at time of system peak

  =

 factor(K)  Transformer peak load (Eq. 4)

It should be pointed out that K is squared in Equations 2 and 3 because K is a ratio of loads 

while losses are proportional to the load squared. Any value of K that seems appropriate can be 

used. The following are recommended values that appear to be reasonable.

Transformer Type K K2

Generator step-up 1.0 1.00

Transmission substation 0.9 0.81

Distribution substation 0.8 0.64

3.5 LFT: The transformer loss factor is defined as the ratio of the average transformer losses to 

the peak transformer losses during a specific period of time. For the sake of simplicity, the 

equations assume that the transformer loss factor is a constant and that it does not change 

significantly over the life of the transformer.

The transformer loss factor can be determined directly using the equation:

Transformer loss = kW − hours of loss during a specified time period (Eq. 5)

 factor (LFT)  (Hours) (Peak loss in kW in this period)

LFT can also be approximated from the load factor (the average load divided by the peak load 

for a specified time period) using the empirical equation below:

Transformer loss = 0.8 • (load factor) 2 + 0.2 • (load factor) (Eq. 6)

 factor (LFT) 

Where:

Load Factor kWh per year 

8760 • peak kW =
Bulletin 1724E-301

Page 10

′

Load factor is the ratio of the average load over a period of time to the peak load occurring in 

that period. The load factor is a commonly available system parameter. The one-hour integrated 

peak value should be used.

3.5.1 Example: Determine the transformer loss factor for a substation transformer that has a 

load factor of 47 percent.

3.5.2 Solution:

Transformer loss factor = 0.8•(0.47)2 + 0.2•(0.47) 

Transformer loss factor = 0.271

3.6 G: The peak ratio is defined by the equation:

Υ Peak annual transformer load ⁄

2

PeakratioG = ∞

≤ Full rated transformer load ƒ

(Eq. 7)

For the peak annual transformer load, the one hour integrated peak value should be used.

The purpose of the peak ratio is to relate the value of Equation 2 to the full rated transformer 

load and not to the peak transformer load that would otherwise result if G were not in the 

equation.

If the total kVA of all transformers is known for your system and the peak kW (or kVA) load is 

known, then the average peak ratio for your system would be:

If the peak kW is known, but the peak kVA is unknown, assume a reasonable power factor on 

peak and calculate peak kVA as follows:

kVA = kW

power factor

If the transformer being purchased has a peak ratio different from the average, use that value. If 

the transformer will be installed at a known substation, use the billing data and assumed load 

growth for that substation.

The equations above are based on the assumption that the peak annual transformer load remains 

the same throughout the life of the transformer. If the load on the transformer is expected to 

increase annually, then use a reasonable equivalent level yearly peak load value based on

⁄

2

Υ Peak kVAload =′

≤TotalkVAof alltransformers ∞ƒ

PeakratioG
Bulletin 1724E-301

Page 11

experience even though the expected peak loading value will increase every year. Another 

method is to calculate a value using Equation A-4 which is explained and derived in 

Appendix A. Equation A-4 yields an equivalent level load that will result in the same total 

losses as the actual non-level loading pattern.

3.7 LFA: The auxiliary loss factor compensates for the transformer auxiliary equipment that 

operates during only part of the transformer’s load cycle. For a transformer with two stages of 

cooling:

LFA = (0.5)• (probability first stage of cooling will be on at any given time) + (Eq. 8)

(0.5)• (probability second stage of cooling will be on at any given time)

The choice of the proper probabilities in the above equation is a matter of judgment based on 

historical system loading patterns. It is expected that the above probabilities under normal 

loading patterns will be extremely low. Since energy use and losses associated with transformer 

auxiliaries are extremely small over the life of the transformer, they could be ignored. The 

capital cost associated with auxiliaries are significant and should be considered.

4. INFLATION

The problem of dealing with inflation in economic studies is a difficult and complex topic that is 

frequently misunderstood. The purpose of this section is not to provide an in-depth analysis of 

the subject but rather to provide some general guidelines.

One method of handling inflation would be to increase future variable costs, such as the costs of 

losses, by the percentage represented by the general inflation rate. Since Equations 1, 2, and 3 do 

not have any provisions for costs that increase over the years, an equivalent level cost that takes 

into account future cost increases should be used. Equation 9 will yield such a value and can be 

used to adjust for inflation. (See Appendix B for derivation.)

(Eq. 9)

where:

(1+ r) X =

(1 + i) 

 

 

for r≠i (Eq. 10)

A′ = the cost adjusted for inflation

≤ ∞ƒ≤ ƒ

Υ1−X ⁄ n ⁄Υ i(1+i) n

= A•X•′

1−X

∞′

(1+i)n −1

A' ∞
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A = the base cost before inflation

n = the number of years in the inflation period.

It is recommended that “n” be taken as 35 years, which is the assumed transformer life. 

By assuming an “n” equal to the life of the transformer, an implicit assumption is being 

made that inflation will continue throughout the life of the transformer.

i = money interest rate per year expresses as a decimal, e.g., for 7%, i =.07)

r = the rate of inflation per year expressed as decimal; e.g., for 3% inflation, r =.03)

Υ i • (1 + i) n ⁄

The term ′

(1 + i) n − 1

∞ is called the capital recovery factor and tables for determining it are

≤ ƒ

easily available in most standard engineering economy texts and computer software. While the 

above method increases future costs, it fails to take into account the value of money decreasing 

with inflation.

Another method of handling inflation is to assume that the increase of costs in the future will be 

balanced out by the decrease in the value of money, thus allowing us to ignore inflation 

altogether. The problem with this approach is that the assumption does not always hold true 

because costs of certain items may increase faster than the inflation rate.

A third method of treating inflation that appears more realistic than the two methods mentioned 

previously is to compensate both for the increase in costs associated with the generation and 

transmission of electric power and for the decrease in the value of the dollar due to the generally 

prevailing inflation rate. This compensation can be accomplished by coming up with an 

“equivalent inflation rate” (r′) that could be used in Equations 9 and 10. The formula for the 

equivalent inflation rate follows:

Υ  1+ P ⁄ r' = − for P ≥ ig (Eq. 11) ′  1∞

≤1+ ig ƒ

where:

r′ = the equivalent inflation rate

P = the rate of the increase in costs per kWh associated with power generation and 

transmission expressed as a decimal.

ig = the inflation rate for the economy as a whole expressed as a decimal.

The approximate form of the equation above is:

r' = P − ig (Eq. 12)
Bulletin 1724E-301

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′ ∞

′

≤

4.1 Example 1: Find the equivalent inflation rate factor by which the energy charge rate should 

be adjusted to compensate for inflation if the following factors apply:

• General inflation rate (ig) = 5%

• Rate of increase of generating costs per kWh (P) = 5%

• Time value of money (i) = 7%

4.2 Solution for Example 1:

Solving for the equivalent inflation rate:

r' = Υ1 + P ⁄− 1 1 + ig

≤ ƒ

r' = Υ1+.05⁄− 1 = 0

≤1+.05ƒ

Solve Equation 10 for r′ = 0, we get:

X = Υ1+ 0⁄ = Υ 1 ⁄

′

≤ 1 + i ∞ƒ ′≤1 + i∞ƒ

Then solving Equation 9, we get: Υ  1 

n ⁄

 1  ′1 − 

 1 + i





∞ Υ i(1 + i) n ⁄

A' = A•  ′ ∞ ′ n ∞  1 + i ′ 1− 1  ∞ ′

≤(1 +i) − 1∞ƒ

′

′ 

 1 + i



 ∞

 1  Υ (1 + i)[ (1 + i) n − 1]⁄ Υ i (1 + i) n ⁄

A' = A•

1+ i





′

(1 + i) n[(1 + i) − 1] 

∞ ′

(1 + i) n − 1

∞

′≤

A' = A

∞ƒ′

≤ ∞ƒ

Thus, the cost adjusted for inflation is the same as the base cost. The above result was to be 

expected. Since the general inflation rate can be taken to be the rate at which the value of money 

decreases and since for this case it is equal to the rate of increase in costs, the two factors can be 

considered to cancel one another out and inflation can be ignored.

4.3 Example 2: Find the factor by which the energy charge rate should be adjusted to

ƒ
Bulletin 1724E-301

Page 14

compensate for inflation if the following factors apply:

• General inflation rate (ig) = 5%

• Rate of increase of generating costs per kWh(P) = 6.5%

• Time value of money (i) = 8%

4.4 Solution for Example 2: Solving for the equivalent inflation rate:

Equation11(exact) r' = Υ 1 + P ⁄

− 1= Υ1 + 0.065⁄− 1=0.0143 ′

1+ig∞ ′

≤ 1 + 0.05∞ƒ

≤ ƒ

Equation 12 (approximate): r' = P − ig = 0.065 − 0.05= 0.015

Solving Equation 9 for r = 0.0143, and n = 35 years:

x = 1 + 0.0143= 1 + 0.08

0.9392

From engineering economics tables, the capital recovery factor for i = 8% and n = 35 years, is 

0.08580. Therefore:

Υ1 − (0.9392)35 ⁄

A = A• (0.9392) ′

′

≤

A' = A • (1.178)

1 − 0.9392

∞ (0.08580)

∞ƒ

In general, the key to properly accounting for inflation in economic studies is to realize that 

inflation increases not only dollar costs, but decreases the value of the dollar and affects all other 

factors that are related to money and the time value of money.

5. EXAMPLE

A 161/34.5 kV transformer rated at 30/40/50 MVA is to be purchased. This transformer is to be 

installed in a substation located at the end of an 80 mile (128 km) transmission line. Determine 

the load, no-load, and auxiliary loss evaluation values in dollars per kilowatt of the guaranteed 

losses at the 50 MVA rating.

Assume:

• Capital cost of power plant is $1,000/kW.
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′

′

′

• Capital cost of line and associated facilities is $130/kW.

• Average energy cost is $0.02/kWh.

• Carrying charge rate is 14.6%.

• Time value of money is 9%.

• Load factor will stay at a constant value of 53% throughout the life of the transformer.

• Annual peak load will remain constant at a value of 53 MVA.

• Non-capital costs associated with generation and transmission increase at 5% per year.

• General inflation rate is 4%.

5.1 The first solution step is to adjust the energy charge for the difference between the general 

inflation rate and the inflation of costs.

Thus, the energy charge adjusted for inflation:

Υ1 − X n ⁄ Υ i (1 + i) n ⁄

EC' = EC• X• ′ 1 − X

∞ ′

(1 + i) n − 1

∞

≤

X =Υ1+r⁄

≤1 + iƒ

ƒ ′≤ ∞ƒ

Adjusted inflation rate:

r' = P −ig

X =Υ1+.01⁄

≤1+.09ƒ

= 5 − 4 = 1%

= 0.927

Assuming that inflation will continue into the unforeseeable future, a value of n=35 will be used, 

as 35 years is the assumed life of a transformer.

EC' = EC • (0.927) (12.73) (0.0946)

EC' = Υ$0.02 ⁄

(0.927) (12.73) (0.0946)

≤ kWh ƒ

EC' = $0.022 / kWh

∞

∞ƒ ′

≤(1+ 0.09)35 −1

∞ƒ

∞′ ′

≤

1−0.927

Υ1−0.92735 ⁄Υ0.09(1+ 0.09)35 ⁄

EC' =EC•(0.927)′
Bulletin 1724E-301

Page 16

ƒ

5.2 The system capital investment is equal to the cost of the plant plus the cost of transmission 

and associated facilities, all per kW; thus:

SI =$1,000 / kW + $130 / kW = $1,130 kW

5.3 Solving Equation 1 for the cost of no-load losses in dollars per kilowatt of losses:

5.4 Solving Equation 2 for the cost of the load losses in dollars per kilowatt of losses:

According to Section 3.4, a peak responsibility factor (K) of 0.8 would be appropriate.

The peak ratio:

Υ peak annual transformer load ⁄

2

G =′

≤ full rated transformer load ∞

Υ53 MVA ⁄

2

G =′

50MVA∞ = 1.124

≤ ƒ

=$2,450/kWof no − loadloss

= SI+

8760• EC'

FCR

=1130+

8760(0.022)

0.146

∞

ƒ perkWof loss

′

≤

′ lossesindollars ∞

ΥCostofno−load⁄

=(SI)(K2)(G)+

8760 • EC'•(LFT)(G)

∞ FCR

ƒ ofloss

′

≤

′ indollarsperkW ∞

ΥCostofloadlosses⁄
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Page 17

Transformer loss factor:

5.5 Solving Equation 3 for the cost of the auxiliary losses:

From the system loading pattern, it is judged that the probability that the first stage of cooling 

will be on at any one time is 0.04 and that the probability that the second stage of cooling at any 

one time is 0.01. Thus, the loss factor for all auxiliary equipment operating is as follows:

LFA = 0.5(0.04) + 0.5(0.01)

LFA = 0.025

ΥCost of auxiliary⁄ ′

lossesall running∞ = 1130 (0.82 ) + 8760 (0.022)(0.025) 0.146

≤ ƒ

ΥCost of auxiliary⁄ ′

lossesall running∞ = $723 + 33.0

≤ ƒ

ΥCost of auxiliary⁄ ′

lossesall running∞ = $756

≤ ƒ

The three loss values are:

No-Load Core Losses Load(Copper) Losses Auxiliary Losses

$2,450/kW $1,304/kW $756

=$1,304/kWofloadlossesat50MVA

0.146 ∞

≤ perkWofloss ƒ

′ =1130(0.8)2 (1.124) +

8760(0.022)(0.331)(1.124)

Costof load ⁄ ′

lossesindollars

∞

Υ

LFT =0.331

LFT = 0.8(0.53)2+0.2(0.53)

≤ ƒ FCR ′

lossesperkWofloss∞

Υ Costof auxiliary ⁄

=SI(K 2 )+

8760 • EC'•LFA
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6. CONCLUSIONS

This bulletin provides formulae, energy loss concerns, and examples of a suggested method for 

evaluating transformers for purchase. Borrowers and others could use the method presented as 

part of a standard procurement practice to ensure that the most economical. long-term, 

purchasing decision is achieved.

However, because of the many variables involved, such as inflation rates, peak loading times, 

investment costs, etc., users of this evaluation method should exercise judgment when using the 

formulae
Bulletin 1724E-301

Appendix A

Page 19

APPENDIX A: EQUIVALENT LEVEL LOAD FORMULA

For the equivalent level load formula, it is assumed that the load on the transformer is initially 

Y1, load on the transformer increases at a rate of k percent per year, and when the load reaches 

Y2, the transformer is either changed out or a second transformer is installed. It is further 

assumed that this pattern continues for the life of the transformer.

The equivalent level load is an equivalent level loss value that results in the same total losses as 

the yearly increasing load. This loss value can then be equated to a load.

Let:

k = the rate of growth of the load, expressed as a decimal

Y1 = the ratio of load to capacity when transformer is installed

Y2 = the ratio of load to capacity when transformer is removed or supplemented 

t = time in years that it takes for load to grow from Y1 to Y2.

If it is assumed that the load growth cycle on the transformer is repeated throughout its life, then 

only the equivalent level load for one cycle is required.

Since losses are proportional to the square of the load:

a ∫

t Y

2 (1 + K) 2n dn

LS = o 1

t (Eq. A-1)

Where:

LS = equivalent level losses 

n = time in years

a = proportionality factor between losses and load such that:

LS = a(load)2 (Eq. A-2)
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Page 20

≤ 

Completing the integral and solving Equation A-1 results in the following for LS:

LS =aY 2 (1+k)2t −1 (Eq. A-3)

1 2tln(1+k)

Setting Equation A-2 equal to Equation A-3, we solve for “load”, the equivalent level peak load 

and end up with:

ΥEquivalent level⁄ 

′ peak load ∞ = Y

ƒ 

(Eq. A-4)

The derivation of time “t” is as follows:

Y2

Y2 

Y1

log

Y2

Y1

= Y1 (1 + k) t

= (1 + k) t

= t log (1 + k)

 Y2 

log Y  (Eq. A-5)

t =  1

log (1 + k)

Where it cannot be assumed that the loading pattern on a transformer will be repeated throughout 

its life, the rationale used in the derivation above can still be used to determine the equivalent 

level loss value. The equivalent level loss value can be determined by dividing the integral of 

the square of the load curve by the time period involved.

Example

On a particular system a triple-rated oil-filled transformer is to be installed where the peak load 

is 0.95 of the self-cooled (OA) rating. The load is assumed to grow at 8 percent per year. When 

the peak load reaches 1.9 times the OA rating, an additional unit will be installed.

(1 + k) 2t − 1 

ln (1 + k) 1 2t
Bulletin 1724E-301



≤ ƒ

≤ ƒ

Appendix A

Page 21

Assuming the above pattern will continue throughout the life of the transformer, determine the 

equivalent level peak load.

Solution

Substituting known values into Equation A-5, we find:

log 1.9 

0.95

t = log(1 + 0.08) = 9.0 years

Then, substituting k= 0.08, t= 9 years, and Y1= 0.95 into Equation A-4, we find:

ΥEquivalentlevel⁄ = ′ peak load ∞ 0.95

ΥEquivalent level⁄ = 1.40 of OA rating ′ peak load ∞

. (1 + 0.08)2(9.0) − 1

(2) (9.0) ln (1+ 0.08)
Bulletin 1724E-301

Page 22

BLANK PAGE
Bulletin 1724E-301

(1+i)n∞

∞

Appendix B

Page 23

APPENDIX B: EQUIVALENT LEVEL COST FORMULA

The following provides a derivation of Equation 9 which is used to adjust cost evaluations for 

inflation. Equation 9 will produce a level cost (or energy charge rate) that will yield the same 

total present worth value as a cost (or energy charge rate) that is increasing at “r” percent per 

year.

The present worth of a cost increasing at “r” percent per year is:

Υ(1 + r) PW = A •′

′

≤

(1+i)

+ (1 + r) 2

(1 + i)2 ⋅⋅

(1+ r)n ⁄

ƒ

(Eq. B-1)

where:

(Eq. 10)

PW = A • X • (1 + X + X2KXn−1)

Algebraic manipulation yields the following:

Υ1 − Xn ⁄

PW =AX′

≤ 1 − X ∞ (Eq. B-2)ƒ

Multiplying the above by the capital recovery factor to get an equivalent level yearly cost yields:

Υ1 − Xn ⁄ Υ i(1+ i)n ⁄

A' =AX′ 1 − X

∞′

(1+i)n −1

∞ (Eq. 9)

≤ ƒ ≤ ƒ

i = the time value of money

r = the rate of inflation

A = the cost before inflation

PW = the present worth

n = the time period in years

If we let X =

1+r

1+i



 

then: