Calculating energy savings for replacing an electric motor

[KP2]

Hello, I am looking into replacing an old air compressor motor with a new energy efficient motor and I have some confusion on how to obtain the Brake Horsepower rating, (BHP).



Reminder:
N = The total number of defined combinations of load and Hours. (e.g., 1=10% of the hours at 100% load; 2=20% of
the hours at 83% load, etc.)
BHPN = the HP required by the load at combination N
hoursN=the time, in hours, that is spent at combination N


I have this formula, BHP = E x √3 x I x Percent Efficiency x PF/746, but the utility reminder is confusing me, and when I research the BHP it is refered to the demand of the compressor.

Any help would be appreciated.
Thanks
Kevin

 

[gar]

121101-1340 EDT

Brake horsepower is real actual horsepower as distinguished from horsepower calculated theoretically from engine displacement, compression ratio, and possibly other factors as was done in the early days of automobiles.

Brake horsepower is the power measured at the engine output using a dynamometer. Edison used a belt scheme with a weight resting on a scale to measure in-line power from the steam engine to a dynamo under test to compare with the electrical power output. This mechanism can be seen in the Menlo Park Machine Shop in Greenfield Village, Dearborn, Michigan. In those days they did not have straingage torque transducers.

Another means to measure brake horsepower is to float the load or brake on low friction bearings, attach a lever arm to the load, and at a known radial distance on the lever arm connect to a restraining force measuring scale. In addition you need to know RPM.

All of the above is for background.

What you need to know is the mechanical power input to the compressor and the electrical power input to the motor at the same time. The ratio gives you the efficiency of the motor at that operating point. This is needed for both motors.

If your desire is to calculate the energy saving, then you need a load profile over time. See my photo 29 near the end at http://beta-a2.com/EE-photos.html

If you have the load profile for the compressor input and the motor input for both motors, then you can determine the energy saving of the more efficient motor compared to the present motor.

.

 

[gar]

121101-1423 EDT

My refrigeration compressor curve probably shows less power variation thru a cycle than does an air compressor. My quick measurement below may put them in the same ballpark, but a different shape. Note: initial inrush power is of no concern because motor on time is very long in comparison.

The equation you presented needs some explanation.

Basically it assumes a constant horsepower load which the load is not. However, an average horsepower while the motor is running may be satisfactory.

Horsepower is converted to kW by the 0.746 factor. This is assumed to be the mechanical power output of the motor measured in kW, and the equation assumes power is constant.

Dividing output power by motor efficiency gives you input power. Input power times run time produces input energy. Since power is in kW and time is in hours the result is in kWh.

The energy consumption over some time period for one motor system is subtracted from the other to determine the energy saving.

In your case if you lack adequate instrumentation to record power vs time, then you might measure power input to the present motor of the compressor at several points in the cycle. Should be done at equal timed intervals. Determine an average power. Calculate an average BHP from this.

Really you probably don't need to use the equation given, and thus you would not need the BHP.

On a small air compressor from 0 to full pressure some readings were made in watts, but not at controlled time points:
828, 902 930, 945, 958, 978, 960, 940, 935, 931, 930.

I would probably average this at 945 W as a reasonable estimate. Thus, input horsepower on average is about 1.27 HP. Multiply by decimal efficiency to get output BHP.

Can you define what your goal is for this project? Do you want to estimate the energy saving per year, and therefore cost saving? Is the goal to determine whether it is cost justified to buy and install a new motor?

Suppose you had an energy saving of 5% on an average power consumption of 10 kW. If power costs $0.10 per kWh, then the saving is $0.005 per kWh. 10 kW for a year is 10*8766 = 87,660 kWh per year. The saving would be $438/year.

We have a 5 HP air compressor and under normal air use requirements probably has a duty cycle of under 10%. When more air is needed, meaning almost continuous flow blowing off parts, then the 20 HP unit is started. A guesstimate for us is a more efficient motor might save $20 per year. A much better place to spend money is on a reduction of lighting costs.

.

 

[KP2]

The formula comes directly from the Utility Company, as well as the reminder, pg 41,42;

http://www.ctenergyinfo.com/2012 CT Program Savings Documentation FINAL(1).pdf

The part of the calculation I don't understand is;
N = The total number of defined combinations of load and Hours. (e.g., 1=10% of the hours at 100% load; 2=20% of
the hours at 83% load, etc.)

I understand the N of the summation to indicate adding together each seperate calculation for the individual motors.

Does the example given; 1=10% of the hours at 100% load, simplifiy the equation by stating motor #1 of the equation runs at full HP for 10% of the measured time period.

121101-1423 EDT

Really you probably don't need to use the equation given, and thus you would not need the BHP.

On a small air compressor from 0 to full pressure some readings were made in watts, but not at controlled time points:
828, 902 930, 945, 958, 978, 960, 940, 935, 931, 930.

I would probably average this at 945 W as a reasonable estimate. Thus, input horsepower on average is about 1.27 HP. Multiply by decimal efficiency to get output BHP.

Can you define what your goal is for this project? Do you want to estimate the energy saving per year, and therefore cost saving? Is the goal to determine whether it is cost justified to buy and install a new motor?
.

Yes, Yes, and Yes. I am interested in replacing 2 electric motors on 2 different air compressors for this project; where we are already upgrading the lighting system, and the customer has requested we add this to the project.

121101-1423 EDT

Suppose you had an energy saving of 5% on an average power consumption of 10 kW. If power costs $0.10 per kWh, then the saving is $0.005 per kWh. 10 kW for a year is 10*8766 = 87,660 kWh per year. The saving would be $438/year.

We have a 5 HP air compressor and under normal air use requirements probably has a duty cycle of under 10%. When more air is needed, meaning almost continuous flow blowing off parts, then the 20 HP unit is started. A guesstimate for us is a more efficient motor might save $20 per year. A much better place to spend money is on a reduction of lighting costs.

.

I see, the best option is to get a power logger, and record at least a 24 hour cycle, then calculate the average horsepower. Is this also the duty cycle?

Once I determine the average HP needed I can complete the cost savings analysis.

Just to be clear, I can not accurately complete this projection based on nameplate information alone?

Thanks Again

 

[John120/240]

The formula comes directly from the Utility Company, as well as the reminder, pg 41,42;

http://www.ctenergyinfo.com/2012 CT Program Savings Documentation FINAL(1).pdf

The part of the calculation I don't understand is;
N = The total number of defined combinations of load and Hours. (e.g., 1=10% of the hours at 100% load; 2=20% of
the hours at 83% load, etc.)

I understand the N of the summation to indicate adding together each seperate calculation for the individual motors.

Does the example given; 1=10% of the hours at 100% load, simplifiy the equation by stating motor #1 of the equation runs at full HP for 10% of the measured time period.




Yes, Yes, and Yes. I am interested in replacing 2 electric motors on 2 different air compressors for this project; where we are already upgrading the lighting system, and the customer has requested we add this to the project.



I see, the best option is to get a power logger, and record at least a 24 hour cycle, then calculate the average horsepower. Is this also the duty cycle?

Once I determine the average HP needed I can complete the cost savings analysis.

Just to be clear, I can not accurately complete this projection based on nameplate information alone?

Thanks Again

Welders often have a duty cycle rating. It relates to "On" time vs "Off" time. Duty cycle

is a rating that lets you work your compressor for so long, then you are supposed to let

the motor rest so that it can cool down. Hope I said that right, if not correct me

 

[gar]

121101-1151 EDT

KP2:

Unless there is a large variation in the power profile from one pump cycle to the next, there is no reason to study more than one cycle for energy consumption per cycle for each present motor pump combination energy consumption.

From my prior measurement of the small air compressor average power was 945 W and pump up time was approximately 1 minute. I calculate 0.945 * 1 /60 = 0.016 kWh of energy per pump cycle. Also I calculated electrical input power at 1.27 HP. This pump is very seldom used. So per year it costs about 0.016*4*365 = 23 kWh and at $0.16/kWh about $3.68 per year.

Using input power, which can be measured with moderate ease, you need to know the efficiency of the present motor at the operating point to determine the motor output mechanical HP. This provides the BPH value for the equation. I would rather get the HP value from measurement of the motor output torque and RPM and from those values determine the motor efficiency. But that is a difficult measurement for you.

Let's assume 75 % efficiency for the present motor at the load operating point. Then output mechanical power is 1.27*0.75 = 0.95 HP.

It is highly likely that all pump up cycles from 70 PSI to 100 PSI are nearly the same and of the same duration unless the pump runs continuously, or there is a lot of variable consumption during pump up. The 70 and 100 values are just assumed as possible on and off values.

You can calculate the input energy per pump up cycle for the present motor, and for the new motor. Determine the energy difference per pump up cycle. Multiply this by the number of pump up cycles per year for the total saving.

If you can not reasonably determine the number of cycles, or there is a large variation in air usage vs time, then you would need to do input energy measurement over an adequate time period to average these variations.

Needs to be done separately for both air compressors.

Duty cycle is the ratio of on time to total cycle time. If on 10% of the time, then the duty cycle is 1/10.

.

 

[KP2]

Gar, thanks so much for your input on this thread. The first system, has 4 - 6 workers running pneumatic tools, grinding and polishing stone, for about 3744 hours per year, so the best way will be to collect some data, otherwise I feel I'm still only approximating. I will plan do as you have described using the average iinput power and current efficiency to determine the BHP.

Thanks
Kevin

 

[gar]

121102-0956 EDT

I am going to try to simplify my discussion.

The goal is to determine the energy saving produced by changing from a standard efficiency motor to a high efficiency motor.

The efficiency of each motor needs to be known for the operating load.

The operating mechanical load on the motor is unknown. If an efficiency curve for the present motor is available, then by measuring electrical power input to the present motor under operating conditions the output mechanical power can be determined. This can be converted from watts to HP if needed. But there is really no need to convert to HP.

It will be assumed that load power is constant when driving the compressor, and the same load power applies for both motors.

The reduction in power obtained by changing from the standard motor to the high efficiency motor is:

P_diff = P1 * Eff₁ * ( 1/Eff₁ - 1/Eff₂ ) or
P_diff = P1 * ( 1 - Eff₁/Eff₂ )

P1 is the electrical power input to the standard motor under average load conditions.

Eff₁ is the efficiency of the standard motor at that load condition.
Eff₂ is the efficiency of the high efficiency motor at that load condition.

Multiply the power difference by Average duty cycle * (8766/1000) for the kWh saved per year.

Do this separately for each compressor.

8766 is the number of hours in a 365.25 day year, and 1000 is watts to kilowatt.

.

 

[gar]

121102-1048 EDT

KP2:

My guess is that you have a 25 to 100 HP compressor, and it operates nearly continuously for your 3744 hours. You might question if a completely new scroll type compressor system could provide greater savings.

Also air tools are very expensive from an energy perspective. Could you change to electric tools?

.

 

[weressl]

Hello, I am looking into replacing an old air compressor motor with a new energy efficient motor and I have some confusion on how to obtain the Brake Horsepower rating, (BHP).

View attachment 7647

Reminder:
N = The total number of defined combinations of load and Hours. (e.g., 1=10% of the hours at 100% load; 2=20% of
the hours at 83% load, etc.)
BHPN = the HP required by the load at combination N
hoursN=the time, in hours, that is spent at combination N


I have this formula, BHP = E x √3 x I x Percent Efficiency x PF/746, but the utility reminder is confusing me, and when I research the BHP it is refered to the demand of the compressor.

Any help would be appreciated.
Thanks
Kevin

To determine the savings of motor replacements; compressors are probably the most difficult to determine. As compressors load/unload they go through a load profile that goes from <50% to 130% or even higher. Efficiency is variable so it would be different for each operating point along that curve. Depends on the load and the size of the buffer tank, the cycle time also varies. So while the shaft HP during the cycle maybe a squared curve, the electric input curve will be biased by the efficiency curve. You can imagine of generating the curves for two motors and trying to calculated the differential not being an easy task.

 

[KP2]

Hello Gentelmen, I wanted to thank you again for the help.

This is the energy consumption fo the motor for one hour, I used a power logger to get these readings.

Energy
Eat= 3.609 kWh
Est= 4.126 kVAh
Pt= 8.3 kW
Qt= 4.05 kVar
St= 9.24 kVA
"Measure
Time" 1:00:02

So this is what I calculated;


Sorry I dropped off on this post for a bit, but I was on the road traveling or almost two weeks.


Thanks Again & Happy Holidays
Kevin

 

[gar]

121130-1111 EST

KP2:

What are the definitions of the values you provided?

I will assume Pt = 8.3 kW is the running input power to the motor. Thus, input HP = 11.13 . If I assume 88.5% is the efficiency of the present motor, then the mechanical output HP = 9.85 . But I do not need this information other than to size the motor.

If a new more efficient motor of 91.7% is used, then the input power changes from 8.3 to 8.3*0.885/0.917 = 8.01 kW. Your saving in energy per hour of run time is 0.29 kWH.

Your duty cycle is about 3.609/8.3 = 0.435 . Thus, your saving is 0.29 * 0.435 = 0.126 kWH per hour that AC power is provided during working hours. If you ran continuously for 1 year (8766 hours) the saving would be 8766 * 0.126 = 1105 kWH. At $0.10 / kWH this is $110.50 saving per year.

You are estimating 936 hours. My figure is 43.5% where you have 45%. For 2080 hours my result is 2080 * 0.126 = 262 kWH or at $0.10 the saving is $26.20 per year.

The change is not justified unless I made a mistake.

.

 

[weressl]

121130-1111 EST

KP2:

What are the definitions of the values you provided?

I will assume Pt = 8.3 kW is the running input power to the motor. Thus, input HP = 11.13 . If I assume 88.5% is the efficiency of the present motor, then the mechanical output HP = 9.85 . But I do not need this information other than to size the motor.

If a new more efficient motor of 91.7% is used, then the input power changes from 8.3 to 8.3*0.885/0.917 = 8.01 kW. Your saving in energy per hour of run time is 0.29 kWH.

Your duty cycle is about 3.609/8.3 = 0.435 . Thus, your saving is 0.29 * 0.435 = 0.126 kWH per hour that AC power is provided during working hours. If you ran continuously for 1 year (8766 hours) the saving would be 8766 * 0.126 = 1105 kWH. At $0.10 / kWH this is $110.50 saving per year.

You are estimating 936 hours. My figure is 43.5% where you have 45%. For 2080 hours my result is 2080 * 0.126 = 262 kWH or at $0.10 the saving is $26.20 per year.

The change is not justified unless I made a mistake.

.

Not to mention that a higher efficiency motor will run faster and if it is a centrifugal load, it will consume more power. It will provide more 'work'. eg. higher pressure, flow etc. what you may not even need, but your cost will be even higher than Gar calculated it and your return; less.

 

[G._S._Ohm]

Calculating energy savings for replacing an electric motor


There are statistical methods for determining the new motor kwh/old motor kwh to some level of confidence (75%, 90%, 95%) providing you have data for the old motor.

ANOVA is one way, IIRC. I took a 40 hr course on stats at a NASA facility and this was the best 40 hrs I ever spent. The guy gave us look-up tables that I never would have figured out how to use.

He looked like
http://www.google.com/imgres?imgurl=http://www.longislandpress.com/wp-
content/uploads/2010/10/sarducci.jpg&imgrefurl=http://www.longislandpress.com/2010/10/30/father-guido-sarducci-father-guido-sarducci-gives-benediction-at-rally-to-restore-sanity/sarducci/&h=359&w=500&sz=136&tbnid=Q1qRZo_BpKP4XM:&tbnh=90&tbnw=125&prev=/search%3Fq%3Dfather%2Bsarducci%26tbm%3Disch%26tbo%3Du&zoom=1&q=father+sarducci&usg=__mGst9fnK4rI_KpujX7s-5cmEzKA=&docid=lmZHO2JvGmyoXM&sa=X&ei=nfy4UJAB8IfRAcz6gNAE&ved=0CE0Q9QEwBQ&dur=2027
except for the hat. I forget whether he had a mustache.

Stats summarize and can prevent us from getting bogged down in a swamp of details.

 

[iceworm]

Not to mention that a higher efficiency motor will run faster and if it is a centrifugal load, it will consume more power. It will provide more 'work'. eg. higher pressure, flow etc. what you may not even need, but your cost will be even higher than Gar calculated it and your return; less.

LZ -
I not getting this. Why would it run faster? How would one calculate how much faster?

ice

 

[gar]

121201-1052 EST

iceworm:

With other parameters fixed the slip at a given torque in an induction motor is a function of rotor resistance. So if part of the criteria of improving motor efficiency was to use a lower resistance rotor, then speed would increase at a given load.


weressl probably has field experience with the differences between normal older motors and new high efficiency units. I don't.

For theory on induction motors one reference is "Alternating-Current Machinery", by Bailey and Gault, McGraw-Hill, 1951. See chapter 11. I used the book, but never met either of the authors. J. G. Tarboux was my professor for that course, and an outstanding teacher. I did know Gault's wife because after his death she was a secretary at the Electronics Defense Group where, as a student, I was employed part time.

 

[iceworm]

121201-1052 EST
... For theory on induction motors one reference is "Alternating-Current Machinery", by Bailey and Gault, McGraw-Hill, 1951. See chapter 11. I used the book, but never met either of the authors. J. G. Tarboux was my professor for that course, and an outstanding teacher. I did know Gault's wife because after his death she was a secretary at the Electronics Defense Group where, as a student, I was employed part time.

Thanks for the reference, but I really don't have access to a 6 decade old college text. I current use Chapman, "Electric Machinery Fundamentals", 2005 (and I have a 1985 version - a bit different). I did not use it in school, don't know any of the authors or their wives, and I don't remember the name of my instructor or the text he used - except he was an old crotchety guy with a bad arm, and really good with explaining the machinery in the labs.

I know, not nearly the depth and magnitude of your Bailey and Gault reference, but perhaps all that us inept kids can handle.

121201-1052 EST
... With other parameters fixed the slip at a given torque in an induction motor is a function of rotor resistance.

yes, that's true. Although rotor reactance matters, but to a lesser extent - depending on the rotor cage design

121201-1052 EST
... So if part of the criteria of improving motor efficiency was to use a lower resistance rotor, then speed would increase at a given load.

Well, first off "if". Second, as I understand (very limited, this is not my area), control of motor characteristics by cage rotor design is quite prevalent, particularly in energy efficiency motors. Too low rotor resistance ruins the starting torque.

For example, one might look at the current design characteristics of a Design A motor. The slip over the normal operating range is nearly indistinguishable from a Design B. (again not my area of expertise) And yes, Design A is getting more common.

Chapman discusses this, but perhaps not prevalent in 1951.

121201-1052 EST
... weressl probably has field experience with the differences between normal older motors and new high efficiency units. .

I hope so. That is why I am asking him. But if someone else knows, I certainly encourage them to respond. Cause I don't get it clearly.

My view (between normal efficiency - high effficiency as defined by Nema MG-1 12-12):
The No-Load speed is the same - both are very near synchronous.
The Full-Load speed is nearly the same - I don't have a catalog handy so i can't verify.

So that leaves the speed differences in the normal operating range. A normal Design B has a bit of a belly (a paunch - so to speak). High efficiency tend to be more linear. (reference: Nema MG-1, Chapman) I'm not so sure a high efficiency couldn't have more slip at part load. But if it did, I would not expect more that a few hundreths of a percent. But then again - not my area.

ice

 

[gar]

121201-1303 EST

iceworm:

Using a different thought process consider the following:

The motor is a black box. Electrical energy goes in, and mechanical and heat energy come out.

Assume electrical energy going in and the load torque were constant. If the efficiency goes up, meaning less heat loss, and therefore more mechanical energy, then speed must increase.


If by some means the the output speed and torque were constant at the same operating point for the two motors, then the more efficient motor would need to be provided with a lower input voltage. Or by effectively changing the transformer ratio in the motor the voltage could remain constant, but input current would be less.


What design criteria are used for the more efficient motors I do not know.

.

 

[gar]

121201-1320 EDT

iceworm:

Yes, rotor resistance has a big effect on starting torque. Basic motor theory has not changed over the last 50 years.

I would agree with you that it is not obvious whether slip would increase, stay the same, or decrease as the designed motor efficiency was increased. Practical experience with real present designs would determine what happens.

A side point in the air compressor application. I don't think it would matter much whether slip was different between the two motors or not. In the bang-bang (on-off) air compressor application a certain amount of energy is removed by use of air. This energy has to be replaced by the motor and pump. Whether it is input a little faster or slower probably does not make much difference.

The general experience I had with teachers and books was that classes that were taught by the book author for the class were presented in a much more intuitive and clear way than when just a teaching assistant was the teacher.

.

 

[iceworm]

121201-1320 EDT
I would agree with you that it is not obvious whether slip would increase, stay the same, or decrease as the designed motor efficiency was increased. Practical experience with real present designs would determine what happens. ...

Yes

121201-1320 EDT
...A side point in the air compressor application. I don't think it would matter much whether slip was different between the two motors or not. In the bang-bang (on-off) air compressor application a certain amount of energy is removed by use of air. This energy has to be replaced by the motor and pump. Whether it is input a little faster or slower probably does not make much difference. ...

Yes. Excellent phrasing. That's where I was trying to go.

121201-1320 EDT
...The general experience I had with teachers and books was that classes that were taught by the book author for the class were presented in a much more intuitive and clear way than when just a teaching assistant was the teacher. ...

Could be. But that was 40 years ago and my recollection is a bit hazy. But I must have. That's what college professors do - write books and then require the text to be used for their classes.

ice