I have 20HP , 1440 RPM old motor running for blower with nameplate efficiency as 89%. When I checked out the power by applying load manager I got the power as 5.5kW. Now I want to replace this motor with proper rated motor and better efficiency. For ROI I am facing question of considering efficiency. What efficiency do I consider either 89% or less?? Is there any formula for calculating efficiency based on % loading ( especially below 50%)??
thanks and regards,
yogesh tapaswi
[email protected]

 

The effeciency of an electric motor is greatly affected by loading. Motor effeciency is about zero at no load (power in but no useful work out). The motor RPM is based on number if poles, frequency and the amount of actual load. The effeciency of motors is generally a rating at full load but all the information you are looking for is at two websites:

http://www.usmotors.com
http://www.marathonelectric.com

 

The motor is already properly rated. Do not change a motor just with a kw reading. What is the pick up time for the motor to reach full speed. What is the maximum current taken by the motor with inlet and discharge fully open. Is there any automatic system to close the dampers during starting? And there is a lot to know before you can try to change a blower motor.
regards,
Sekar

 

In manufacturer's catalogues, you will find graphs of efficiency versus load.
However, I do not think that effiency of a 15 kW motor at %35 load will be too much different than that of a 5.5 kW motor of the same kind at %100 load. As the motor physical dimension decreases, magnetisation current and magnetic losses (which are the reasons for inefficency) will increase. However, you may think to replace it with a high-efficient motor, where the losses are less.

 

Responding to Yogesh's Mon, Dec, 10:15 am, query:

Is your stated 5.5 kW value a) the measured input? Or b) the calculated output?

If a) and assuming a 4-pole, 50 Hz motor, the motor is operating at about 33% of its rating. This determination is based upon using a typical efficiency vs rating curve for "older" motors. Thus, the required motor capacity is about 6.6 Hp. Select the next commercially available motor size adequate to deliver at least 6.6 / 0.85 or 7.8 Hp. In the USA I would select 7.5 Hp. Note, the 0.85 figure is the load factor coincident with maximum efficiency.

If b) the motor is operating at about 35% of design. Thus, the required Hp is about 7.0 Hp. Select a motor size to deliver at least 7.0 / 0.85 or 8.2 Hp. If in the USA, I would select 10 Hp.

Two cautionary comments: 1) the criteria used in the calculations are based on my practice. 2) Motor selection should never be based solely on driven-load. Make sure that starting duty is adequately covered.

If additional detail is required contact me by e-mail or via the List.

Regards, Phil Corso, PE (Boca Raton, FL) [[email protected]]

 

Further to my earlier response, slip mesurement can determine acual output:

If an rpm meter is unavailable (and the location is unclassified, or determined to be non-hazardous) use a fluorescent lamp connected to the
same supply as the motor's. It is now a "strobe" light. The resultant shaft image will appear to slowly rotate backwards. Count the number of
rotations for one minute, and divide it by the number of poles to obtaiin actual slip. Then, calcuate the ratio of actual to design slip.
Multiply design Hp or kW by this ratio to determine actual output.

Note: The count is divided by the number of poles because the number of lamp flashes is usually (but, not always) 2 x line frequency!

Regards,
Phil Corso, PE
(Boca Raton, FL)
[[email protected]]

 

<P>Dear sekar/phil,
<P>following is present motor data:
<BR>HP -20, 1440 rpm
<BR>Rated Current - 28 Amps
<BR>Rated Efficiency - 89%
<BR>Rated Power Factor - 0.84
<BR>Time for attain full RPM = 15 Sec.

<P>Following is power monitor readings taken at output and with damper setting as 80% open.(fixed dampers )
<P>
<PRE>
Voltage Current P.F. kW
413 15.5 0.52 5.6
</PRE>
<P>(Infact , current should drop further as power factor is improved; I am applying PF capacitors to it)
<P>Can we Consider 9.3kW/11kW motor with efficiency as 92.5% as good replacement?

<P>With regards,
<BR>yogesh

 

1) Responding to Hakan Ozevin's Fri, Dec 27, 5:08pm comment regarding 'loss difference':

Using the original posted figure of 5.5 kW as the output value, then motor's operating at 1/3 of its rating. At this load the motor's efficiency is ca 63%, corresponding to an input of 8.7 Watts.

A new hi-efficiency motor rated 5.5 kW has an efficiency of ca 90 %, yielding an input of about 6.1 kW. Thus, the loss difference is 2600 Watts, or about an 47% energy savings. Nothing to sneeze at!

2) Responding to Sekar's Fri, Dec 27, 3:41pm, comment regarding 'rating':

I don't understand your statement that the motor is "already properly rated!" Do you mean that the saving doesn't justify the replacement engineering cost? But, isn't that the root of Yogesh's query?

Regards,
Phil Corso, PE
(Boca Raton, FL)
[[email protected]]

 

Responding to Yogesh's Mon, Dec 30, 10:25am transmittal of actual operating data:

1) The low PF is due to the low load-factor. A power-factor correction-capacitor will compensate for motor-current losses in the cable between the source and the motor. It will not reduce the motor's internal losses!

2) There is no doubt that a 9.3kW or 11kW motor appreciably reduces losses, thereby improving efficiency.

3) My major concern is the stated time-to-attain rated speed (runup-time)! What is the position of the damper during start? The value of 15-seconds may be the reason a 20 Hp motor was chosen in the first place. If true, the smaller motor will substantially increase runup-time. This will result in injurious temperature-rise to the motor,
thereby negatively impacting its life-expectancy!

You must determine if the 15-sec startup time is due to the additional load caused by the open damper? Or because the blower-inertia is greater
than the permissible value the motor is capable of handling?

Of course, the inertia value can be calculated using the parameters you provided. But this step may be unnecessary if runup-time can determined
with the damper closed!

Regards,
Phil Corso, PE
(Boca Raton, FL)
[[email protected]]

 

For A 20HP motor with 89% efficiency at full load -

when it is consuming 5.5kW
efficiency at load shall be 84.2% of consumption, which means the losses will be 5.5 * (1-0.842) = 869 Watts
Which means the mechanical work equiv. done is 5500-869 = 4631 Watts

for a 10HP motor (rated efficiency EFF3, for which the lower limit of efficiency is 90.1% at full load) delivering 4631 W, the load is 4631 / (10 * 746) = 62%, and at 62% load the efficiency of 10HP motors shall be 87,8% which means the loss will be 650 Watts.

Your savings shall be (869-650) = 219Watts.

Now as to saving 219W per hour of usage is worth the cost and bother of changing your motor. Who knows??

 

Further to previous discussions:

Those interested have probably noticed appreciable difference in loss calculations between Alpasian Pak's and mine. The reason, of course, is the source of pertinent data.

Alpasian's information may have come from a motor vendor, while mine came from the original USA Dept of Energy's (DOE) EPAct study. The point is that the results widely vary based on where/how data is obtained.

Also, upgrading to hi-efficiency motors for fans requires careful analysis. Fan horsepower varies as the cube of speed. Consder the following example. An older motor operates at 1435 rpm (slip = 4.33%) while the new one operates at 1450 rpm (slip = 3.33). The power of the new motor is (1450 / 1435)^3, or about 3.2% more than the older one. In other words the new motor is drawing more but doing no more useful work.

Regards,
Phil Corso, PE
(Boca Raton, FL)
[[email protected]]

 

How would I calculate the efficiencies of more than 7-8 years old motors? Is there drop in efficiency over time? Can anybody suggest me how to calculate drop in rated efficiencies of motors over time.
Thanks
my e mail is [email protected]

 

Responding to Subodh's Thu, Feb 6, 3:3pm query:

Please do not take offense at what I'm about to say, but... this is the perfect example of "catching the mice and letting the elephant's free!"

The Benefit-to-Cost Ratio (BCR) for the method proposed is nil. Instead, look for ways to improve the driven-machinery efficiency. This
approach will yield orders-of-magnitude higher BCRs!

The US government's edict to use "higher" energy-efficient motors has had zero-impact on our per-capita consumption of energy! I can't say the same about the predicted cost savings!

Regards,
Phil Corso, PE
(Boca Raton, FL)
[[email protected]]
{[email protected]}