AC Motor Efficiency - A Guide to Energy Savings

What we have done, and what we are doing
to help reduce energy consumption


Part 6: EVALUATING MOTOR EFFICIENCY

| Price Premium | Simple Payback | Present Value Analysis | Energy Efficient Payback Analysis | Energy Costs |

Today’s motor user faces the evaluation of higher efficiency motors at premium prices with standard designs. In many applications, an initial purchase price premium can be justified based on energy cost savings. The basis for this justification depends on the individual user’s situation. Factors such as running hours, cost of electricity, payback period, tax rate, cost of capital and service life affect the premium price justification, and will vary with the individual user.

Price Premium Per Kilowatt Saved

There are two methods for user determination as to how much additional capital investment or price premium is justified per kilowatt (KW) of power load reduction.

If, for example, an additional capital investment of $1,000/KW is determined, the motor user would be justified in paying up to a $1,000 price premium for every KW saved because of the watts loss reduction by using a more efficient motor.

Motor efficiency is a measure of the power loss in the conversion of electrical power to mechanical power. An example of the power or watts loss reduction provided by a higher efficiency 25 HP motor follows:

Watts Loss + Input - Output =

HP x 746

- (HP x 746)
 

EFF

 
     
Watts Loss 88.3% Eff. Motor =

25 x 746

- (25 x 746) = 2471 Watts
 

.883

 
     
Watts Loss 93% Eff. Motor =

25 x 746

- (25 x 746) = 1275 Watts
 

.936

 

Figure 15
Watts Loss 25 HP Motor

Operating the 93.6% efficiency motor vs. the 88.3% efficiency motor thus saves 1196 watts. In other words, the power loss reduction is 1.196 kilowatts.

Kilowatts saved can also be calculated using the simple formula below.

Kilowatts Saved = HP x .746 [1   1]
 
-
  Std Motor Eff.   Prem. Motor Eff.

Figure 16
Kilowatts Saved Formula

Two methods of calculating the additional capital investment to justify the price premium per KW of loss reduction are shown next:

Simple Payback

The first method is a simple payback analysis where the user specifies how long he is willing to wait for his after tax power cost savings to equal the premium paid. The user must specify:

Example:

HRS/YR = Hours of motor operation per year
$/KWH = Cost of electricity (dollars per kilowatt hour)
YRS = Maximum acceptable years to payback or break even
T = Tax rate
Then:    
P = Justified Price Premium/KW Saved =

HRS/YR x $/KWH x YRS x (1–T)

A motor user that averages three - 8 hour shifts per day; 6 days a week; 50 weeks per year has a power cost of $.04/KWH and considers 3 years breakeven or payback period and has a tax rate of 38%.

P

=

HRS/YR x 4/KWH x YRS x (1–T)
 

=

7200 x $.04 x 3 x (1–.38)
 

=

$536.00

This user could justify up to a $536 price premium for a motor that saved or reduced the KW load by 1 KW.

Using the two alternative efficiencies in the previous 25 HP motor example, this user could justify paying an initial price premium for the high efficiency motor of up to $641 (P x loss reduction = $536/KW X 1.196KW = $641).


Present Value Analysis

The second method for determining the justified price premium per kilowatt of power loss reduction involves a formal present value analysis. It allows the motor user to specify required rate of return (minimum acceptable lRR) rather than payback period.

The user must specify all of the following:

L = Expected service life of motor (5, 10, or 15 years)
R = Required rate of return
HRS/YR = Hours of motor operation per year
$/KWH = Current cost of electricity (dollars per kilowatt hour)
T = Tax Rate

In this analysis, the user can also specify his projection for the annual growth in the cost of electricity: zero (from Figure 17), 5% (from Figure 18) or 10% (from Figure 19). The justified premium/KW of power loss reduction (P) consists of two components: one due to energy savings (PE) and one due to depreciation benefits. (PD).

P = PE + PD

To determine PE, select a present value factor (PVF) from Figure 17 (0% power cost growth) or Figure 18 (5% power cost growth) or Figure 19 (10% power cost growth). The PVF is selected according to the user specified values for R and L. The following equation can then be solved for PE:

PE = PVF x $/KWH x HRS/YR x (1–T)

To determine PD, select a present value factor (PVF) from Figure 17 according to the same user specified values for R and L.

The following equation can then be solved for PD :

PD = PVF x PE x T
L

The total justified premium per KW of power load reduction is the sum of PE and PD.

P = PE + PD

Example:The motor user specifies the following:

10 YRS = L = Expected service life of motor
30% = R = Required rate of return
7200 = HRS/YR = Hours of motor operation per year
$.04 = $/KWH = Current cost of electricity
38% = T = Tax rate

Also the user states that he expects real electrical costs to increase at 5% per year.


The user could justify up to a $740 price premium for every KW of power loss reduction. Using this analysis in the 25 HP example above, the user could justify paying the initial price premium of $657 for the highly efficient motor (P x loss reduction = $740/KW X 1.196KW = $885).

It should be noted that this analysis method is moderately conservative for two reasons:

It assumes a straight line depreciation and the depreciable base (PE) is understated due to ignoring depreciation benefits in the PE calculation.


L = Service Life
    5 Years 10 Years 15 Years
R =
Return
On
Investment

10%
15%
20%
25%
30%

3.79
3.35
2.99
2.69
2.44

6.14
5.02
4.19
3.57
3.09

7.61
5.85
4.68
3.86
3.27

Figure 17
PVF Constant Energy Costs

L = Service Life
    5 Years 10 Years 15 Years
R =
Return
On
Investment

10%
15%
20%
25%
30%

4.36
3.84
3.41
3.05
2.76

7.82
6.28
5.16
4.34
3.71

10.56
7.82
6.06
4.87
4.03

Figure 18
PVF Energy Costs Increasing At 5% Year

L = Service Life
    5 Years 10 Years 15 Years
R =
Return
On
Investment

10%
15%
20%
25%
30%

5.00
4.38
3.88
3.46
3.11

10.00
7.90
6.39
5.29
4.47

15.00
10.71
8.02
6.26
5.05

Figure 19
PVF Energy Costs Increasing At 10% Year


Energy Efficient Motor Payback Analysis

Many motor users have developed various methods of evaluating efficiency vs. price premium. The most common method is to develop a price penalty per unfavorable kilowatt. The KW penalty takes into account the payback period, tax rate, hours of operation, cost of capital, cost of power, and other variables significant to this specific operation.

To evaluate price vs. value, the KW penalty is added to the initial price of the motor with the lowest efficiency. The 25 HP motor @ 88.5% depicted in Figure 15 would have a penalty of 1.196 KW x the developed penalty. Some users have calculated this penalty to be as high as $1,500 to $2,000 per KW difference.

MOTOR HORSE POWER: 75 ENERGY COST: $0.050 KW/HR
LOAD HORSEPOWER: 75.00 HOURS OF OPERATION: 4380/YR
SYNCHRONOUS SPEED: 1800 LIFE CYCLES OF MOTOR 10.00 YEARS
ENCLOSURE: TEFC NOMINAL XE MOTOR EFF: 95.0%
NOMINAL STD MOTOR EFF: 90.2% XE MOTOR NET PRICE $3608
STD EFF MOTOR NET PRICE: $2864    
SIMPLE SAVING PER YEAR: $686.00
SIMPLE INVESTMENT PAYBACK:  
(BASED ON XE MOTOR PRICE PREMIUM OF $744)  
WITHOUT REBATE 1.0800 YEARS
WITH REBATE -0.0100 YEARS
KILOWATT CE,AND SAVED PER MOTOR: 3.134 KW
CASH FLOW
INCOME TAX RATE: 38.00% TAX LIFE: 5.00 YEARS
DEPRECIATION METHOD: DOUBLE DECLINING BALANCE WITH INTEREST RATE OF: 8.0%
FIRST YEAR SAVINGS DOES NOT INCLUDE REBATE OF $750
YEAR SAVING DEPREC. ON PREM TAXABLE INCOME INCOME TAX AFTER TAX
CASH FLOW
ACUM’D
CASH FLOW
1 $686.00 $297.60 $388.40 $147.59 $538.41 $ 538.41
2 $713.44 $178.56 $534.88 $203.25 $510.19 $1048.60
3 $741.98 $107.14 $634.84 $241.24 $500.74 $1549.34
4 $771.66 $ 64.28 $707.38 $268.80 $502.86 $2052.20
5 $802.53 $ 38.57 $763.96 $290.30 $512.23 $2564.43
6 $834.63 $ 0.00 $834.63 $317.16 $517.47 $3081.90
7 $868.02 $ 0.00 $868.02 $329.85 $538.17 $3620.07
8 $902.74 $ 0.00 $902.74 $343.04 $559.70 $4179.77
9 $938.85 $ 0.00 $938.85 $356.76 $582.09 $4761.86
10 $976.40 $ 0.00 $976.40 $371.03 $605.37 $5367.23

Figure 20

Reliance has developed a series of efficiency analysis computer programs (Figure 20) that are available through most Reliance sales offices. These programs are available to assist the user in determining the value of an energy efficient motor versus a standard motor. And the analysis can be made for either a single unit or a multi–unit analysis. The programs consider many variables to include hours of operation per year, cost of power, tax rates, utility rebates, interest rates, motor tax life, operation methods, inflation rates, and the life cycle of the motor. These factors are all considered variables to be determined by the User and the programs take them into account when evaluating the price premium compared to efficiency savings. Many users have found these aides invaluable in evaluating the value of energy efficient motors.


Energy Costs

Energy costs are forecast to continue rising steadily. Disallowing for inflation, a 3-5% increase rate in electricity costs is forecast by most conservative sources.




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All Rights Reserved.