Monday, November 28, 2011

RAPID ASSESSMENT OF THE VIABILITY OF MOTOR REPLACEMENT FOR ENERGY EFFICIENCY

Motors driving centrifugal pumps at a water purification plant
Induction motors are prevalent on every industrial site, and are typically responsible for a sizable portion of the total electrical energy consumed (reportedly up to 40%, but it depends heavily on the industry concerned). As technology regarding their construction has moved forward, they have become more efficient, meaning that the amount of energy supplied to an induction motor for the equivalent amount of work performed has reduced. It is widely accepted that most of the life-cycle costs associated with owning an induction motor are due to energy consumption rather than the capital and maintenance costs of the motor itself. Sites with older motor populations can therefore benefit from replacement of older motors with newer, more efficient designs.
Motors are generally classified as being either of standard, high/improved or premium efficiency, with various different standards in use which are not necessarily equivalent (if interested, you can find out more here).

In some parts of the world, minimum efficiency levels are regulated. In the European Community, high efficiency motors became mandatory for new installations in 2011, for example, with stricter standards planned for 2015 and beyond. Various financial incentives are also in place in different parts of the world – in South Africa, the national power utility provides incentives for motor replacement on condition that the motor being replaced is destroyed.

The issues around motor efficiency are diverse, and extend beyond their design characteristics. To begin with, on-site measurement of efficiency is not typically done, and hence the assessment of an existing motor’s efficiency is in most instances an intelligent guess. While the design efficiency of an existing motor is generally available from manuals, and is sometimes displayed on the motor’s nameplate, this efficiency could have changed over time, particularly if the motor has been re-wound. Since in general we are talking about a decrease in efficiency here, using the design efficiency is nevertheless a conservative measure when calculating the benefits of replacement. Note also that the efficiencies quoted are based on full-load conditions. If a motor is poorly loaded, the impacts on efficiency can be very significant, as shown by the typical load-efficiency curve illustrated below.


It is clear that in this example, at loadings of less than 50% efficiency levels fall dramatically. While such loadings are uncommon, I have observed them on occasion, particularly for larger motors. It should however be considered that the illustrated relationship does depend also on motor size, with larger motors able to maintain higher efficiencies at lower loadings.
Use of variable speed drives complicates the efficiency issue further. VSD’s (also called variable frequency drives) contribute to losses directly themselves (these are of the order of 1-3%), but motor speed reduction (caused by the motor operating at lower frequencies than rated) also causes a reduction in motor efficiency. You can find more information in this article. The full-load efficiency levels of new motors are determined according to standard tests and manufacturers are expected to meet the defined efficiency standards in order to classify a motor as being within a specific efficiency class.
Let’s explore a few simple (understanding from the above that it’s not that simple) relationships to illustrate how efficiency increases contribute towards reduced energy consumption.

Motor efficiency% = η = output power (at the shaft) / input power (from the energy source) x 100%. For a given level of output power (OP), an energy-efficient motor requires less input power (IP) than a standard-efficiency motor.  

For any motor, OP = η x IP / 100 (with efficiency in %) and therefore, since we can assume that in evaluating opportunities to install energy-efficient motors the process being driven requires a specific amount of power,  ηstandard x IPstandard / 100 = ηnew x IPnew / 100

From IP = OP x 100 / η, the change in input power requirements arising from the use of energy efficient motors, ΔIP, is given by: 

ΔIP = OP x 100 x (1/ ηstandard – 1/ ηnew)               
    = ηstandard x IPstandard x (1/ ηstandard – 1/ ηnew)

So, for a motor drawing an input power of 50 kW, with an efficiency of 92.7% and with a replacement motor of 95% efficiency available, the power savings would be of the order of ΔIP = 92.7% x 50 x (1/92.7 – 1/95) = 1.21 kW.

This is the input power under average conditions, since the power draw of a motor varies with time, particularly under transient conditions like start-up, and also as a function of the specific processes in which the motor is employed. A motor driving a bandsaw will draw different amounts of power depending on the material being cut and the condition of the blade, for example.

The amount of energy saved is the power saving multiplied by the time over which the savings are realised. While motors run continuously in some plants, in most cases you will need to determine the running hours. The energy savings in the case of our example, were this motor to run continuously, would be of the order of 1.21 kW x 24 hours/day =  29 kWh/day. The value of these energy savings would depend on local tariff structures, and there may also be demand savings associated with replacement. 

A few things become apparent (or even self-evident) from the preceding analysis:
·  The bigger the efficiency differential, the better – generally,  the larger a motor is, the smaller is the efficiency differential, since larger motors tend to be more efficient than smaller ones;
·  The longer the running hours, the greater the amount of energy saved. Hence if you run a dayshift operation, expect it to be tougher to justify replacements, even for motors that run continuously;
·   Some measurement is needed to determine input power. While you can log the power drawn, simple spot checks are a useful starting point. If you measure only current and voltage, you will need to make assumptions regarding power factor in calculating the power drawn. This is the least-preferred approach, since like efficiency, power factor varies with motor loading. Alternatively you could measure power factor directly, together with the voltage and current. This would give an accurate indication of power consumption. Better still, hand-held clamp meters are available which measure power directly. 
Conducting such an exercise on a site with a large motor population is possible if you chip away at it over time. It is useful to have such information on record, and to track changes through periodic measurements. These checks could even be incorporated into your preventive maintenance programme, with input power draw triggering further investigations should there be an upward trend. If however you would like a first-order overview of energy-saving opportunities and their potential viability, here is an approach I use to rapidly review opportunities on industrial sites.
·    Construct a record of every motor on the site in terms of its power rating, full-load efficiency level (if available), full-load power factor and rated speed;
o If efficiency is not known, estimate it, either using a standard efficiency value, or correlations for older motors;
o The speed is required in order to calculate the number of poles. This is often indicated on the nameplate. If not, measure it with a tachometer. Often the poles are directly indicated, in which case you do not require the speed value;
·     Assume a loading value - I generally use 80%. At these loading levels, efficiency and power factor levels are typically close to what they would be at full load. Of course, actual loadings could be far lower or even higher;
·  Where a given motor size/speed combination is prevalent for a number of motors, conduct this analysis for just one, which will represent all of them;
·  Find the equivalent high and premium efficiency motors for each individual size/speed combination, and obtain rough prices for the replacement motors. You can generally get a price list from a motor manufacturer - take care to note if the motors are foot mounted or flange-mounted, since there is typically a fairly significant difference in price between equivalent-capacity motors with different mounting arrangements;
·   Determine your unique payback requirements – for example, do you require a payback within 3 years? Bear in mind that the average induction motor can be expected to have a useful life of 100,000 hours.
·  For each motor/speed combination, determine the annual running hours that will lead to a break-even situation over your payback time horizon, based on energy tariffs, energy savings arising from the efficiency differential and the costs of the motor. I use rough costs at this stage – it’s quite simple to construct a correlation between motor price and power rating for premium efficiency motors once you have accumulated a sufficient number of quotations. For simplicity, I don’t include maintenance costs at this stage, but you could do so.
·  Now compare these running hours to the running hours of your facility, the running hours of your individual processes and finally the running hours of individual motors within those processes. If the breakeven running hours exceed the motor’s expected running hours by a significant amount, further investigation is probably not promising. If however the breakeven hours are far lower, investigate further with measurements to determine actual loading and analysis. A firmer motor price could also be obtained by asking a supplier to quote for the specific motor concerned. Recalculate with this firmer data and make a decision.
Wondering how to calculate those breakeven hours?
Annual Energy savings = ΔIP x Running hours/annum x Cost/unit energy
 =ηstandardxIPstandardx(1/ηstandard–1/ηnew)x Running hours/annum x Cost/unit
By setting these annual energy savings to a value which offsets the cost of the replacement motor over the chosen time horizon (e.g. 5 years), the running hours required annually to achieve this can be calculated. Use of a spreadsheet and the goal-seek function is a simple way to do this for multiple motors. An example is outlined in the table below for a sample of 9 power/speed combinations.

RATED MOTOR SIZE (KW)
NO. OF POLES
EXISTING MOTOR   η %
FULL LOAD POWER DRAW (KW)
ESTIMATED AVERAGE INPUT POWER @ 80% LOADING (KW)
PREMIUM MOTOR   η %
SAVINGS OVER 5 YEARS (RANDS)
ESTIMATED MOTOR REPLACEMENT COST (RANDS)
ANNUAL RUNNING HOURS REQUIRED (HRS)
0.75
4
72.1
1.04
0.83
82.5
3900
3900
12392
1.1
4
75
1.47
1.17
84.1
4100
4100
10765
2.2
4
79.7
2.76
2.21
86.7
4700
4700
8787
2.2
6
77.7
2.83
2.27
84.3
4700
4700
8834
3
6
79.7
3.76
3.01
85.6
5240
5240
8415
3
4
81.5
3.68
2.94
87.7
5240
5240
8390
7.5
4
86
8.72
6.98
90.4
7590
7590
7450
11
4
87.6
12.56
10.05
91.4
9500
9500
7582
15
4
88.7
16.91
13.53
92.1
11800
11800
7876



In the calculations in the table, the annual running hours are varied such that for each motor, the savings achieved over a 5-year period are equal to the costs of replacement with an energy-efficient motor, assuming a motor loading of 80% for the existing motor. 

If for example, this site operated for 24 hours a day and for 340 days of the year, total operating hours would be 24 x 340 = 8,160 hours/annum. A glance at the required running hours quickly shows that motors of 7.5 kW, 11 kW and 15 kW should be investigated further. These investigations should involve determination of actual power consumption through measurement, a determination of actual running hours and an assessment of financial viability based on these measurements. In fact, this analysis could be extended to include motors from 2.2 kW in size given that these are relatively close in terms of breakeven running hours.   

As regards the other motors, the overall situation could change (assuming the efficiency differential remained constant) if power prices increased to a higher level, if motor prices decreased and/or if running hours increased. The exercise should also be periodically repeated to account for improvements in motor efficiency arising from technological advances over time.

You could make your approach more comprehensive than the one proposed by assuming a loading level of 100%. This would increase the savings potential due to motor replacement and widen the net in terms of the number of motors that would require detailed measurement. In my experience, motors in most factories tend to be loaded at levels below 80%, so this is just an arbitrary value I have chosen.
A final point: motors form part of a larger system, and before pursuing energy-efficient motor options, don’t lose sight of opportunities such as reductions in running time, installation of more efficient drive systems (belts, chains, gearboxes etc.) where applicable, and increases in the efficiency of the equipment being driven e.g. a different mixer design may require less power to achieve the same degree of mixing. Each system being driven should be examined in its entirety, with energy-efficient motors being a component of that analysis.