annual cost of electric motor

[maytag]

I'm trying to justify energy savings on an electric motor that runs 3 hydraulic pumps. The motor is 125 hp, 480 VAC and amp load will vary according to how many of the pumps are loaded ( amp draw is approx. 50 amps with no pumps loaded and increases to approx 140 as work is done)probably averaging 75 amps. This is in a cotton gin so it is very seasonal about 600-8oo hours a year. Can anyone help me calculate the annual kilowatt hours?
thanks,
Maytag

 

[davidbeach]

Figure out the annual horsepower hours; the conversion to kilowatt hours will require a few assumptions about efficiency and so forth, but without a good picture of the load profile it will be impossible to calculate kwhr.

 

[dpc]

Not really. If you really want an accurate calculation, you will need to know the power factor. You may have motor data giving power factor at various loads.

The best way is to connect a portable power meter and monitor the actual POWER (not amps) being used.

kW x time = energy (in kWh, generally). But if you need a calculation accurate enough to compare energy usage of two different motors, you'll really need the power factor or an actual kW measurement.

 

[maytag]

Thank you both for your replies. The energy savings would be from an accumulator bank and associated vavles recouping wasted hydraulic energy. I could shut 1 motor/pump group off and use it for a spare. With the very short yearly use I see years paying back on the initial investment-just trying to get a ballpark figure for the cost of the 1 motor/pump group.
Maytag

 

[ozmosis]

There are software packages that can be used to determine these calculations and use them to determine the use of more efficient motors. However, only 600hrs/yr use would not really justify changing a motor unless it failed and the introduction of any control on the motors (VFD) may also be limited with a viable ROI.
An interesting and independant tool is:

http://re.jrc.ec.europa.eu/energyefficiency/eurodeem/imssa.htm

NB: Big download (approx 25MB) for the tool itself and I would read the pdf first.

 

[waross]

Well you haven't given quite enough information for a meaningful answer.
I just spent 1/2 hour running assumed power factors and getting unrealistic answers.
It appears that your motor may have power factor correction installed. That makes the numbers realistic.
That makes it easy.

50 Amps x 480 Volts x 1.73 / 1000 = 41.6 kW
75 Amps x 480 Volts x 1.73 / 1000 = 62.4 kW
140 Amps x 480 Volts x 1.73 / 1000 = 116 kW

Now it's back to you to estimate the number of hours at each loading level and simply multiply and add.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[waross]

Your scheme may cost the same to run. You need the same number of horsepower-hours regardless of whether you use 1 or 2 motors. With one motor you will save the losses of one motor, but these losses will be replaced by the losses in the accumulator and extra valves.
We really need to see the load profile.
You may benefit from an RMS horsepower survey, if your maximum loading is not more than about 80% over the nominal capacity of one motor/pump set and there are intervening periods of light loading.
For more information on RMS horsepower loading see the Cowern papers.

RMS HORSEPOWER LOADING
There are a great many applications especially in hydraulics and hydraulically-driven machines that have
greatly fluctuating load requirements. In some cases, the peak loads last for relatively short periods
during the normal cycle of the machine. At first glance, it might seem that a motor would have to be
sized to handle the worst part of the load cycle. For example, if a cycle included a period of time where
18 HP is required, then the natural approach would be to utilize a 20 HP motor. A more practical
approach to these types of “duty cycle loads” takes advantage of an electric motor’s ability to handle
substantial overload conditions as long as the period of overload is relatively short compared to the total
time involved in the cycle.
The method of calculating whether or not the motor will be suitable for a particular cycling application is
called the RMS (root mean squared) horsepower loading method. The calculations required to properly
size a motor for this type of application are relatively simple and are presented in this paper.

http://www.baldor.com/support/literature_load.asp?LitNumber=PR2525

You may be able to safely overload one motor for short periods of time.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[maytag]

All hydraulic energy does one of two functions-it either does work or converts to heat. I have enough wasted hydraulic energy with this process to idle one motor/pump group. If this was a 24/7/365 operation it would be a
no-brainer but with the "seasonal" use it is a hard sell as the capital investment for the energy recovery is the same regardless of the usage. Thank all of you for your responses.
Maytag

 

[jasonf987]

are you charged different tariffs for different periods throughout the year? (summer vs winter tariff) if so then it might be an easier "sell" to justify running it in the period where your electricity costs would most benefit your goal.

 

[tygerdawg]

A few years ago I had to do energy calculations to justify an issue about running / not running a plant air compressor. Somewhere on the US Department Of Energy website

http://www.doe.gov

I found some useful calculations for energy used by an electric motor. If you dig around on that website, you may find something useful.


TygerDawg
Blue Technik LLC
Virtuoso Robotics Engineering

http://www.bluetechnik.com

 

[LionelHutz]

OK, can you calculate the hydraulic energy in an idling motor/pump?

If so, the input power to the motor will just be the losses of the motor which can just be estimated as around 10% more energy.