Maybe it's just me. Is there anyone else out there? Is this a true statement about what is happening in America today? There is a glut of electric motors in America because no one needs them any more. The nice people in other countries will do those so much troublesome taskes for us and all we need to do is buy them. True, they may be marginally the quality we are used to but they are so inexpensive that we can simply buy another and another when they fail. John.
Hello,
Junking out electric motors is how my buddy and I pay for most of our shop supplies! He finds them and brings them home and cuts what ever they're attatched to off the motor, then gives it to me. I then take a chisel I forged from a car coil spring, and 1 or 2 swipes per screw (usually 3 or 4 per motor), with the chisel and hammer and a few taps here and there with the hammer and I have the end caps off. (they're alum) I then give him the end caps (he knocks the bushing/bearings/seals out) and they're ready for the alum. scrap pile. I then take said chisel and cut the copper off of one end of the laminates, Then I take a big screw driver and and pry the copper loops out the other end. The copper is then "cleaned" of any insulated wire and put in the copper barrel (copper's usually over $1 a pound) The armatures and the laminates go in a scrap steel barrel, to also be scrapped. My buddy brings in about 10-50 electric motors a week. Most are from AC/HVAC units, plus some misc motors. We also do alot of other scrapping, batteries, copper pipe, alum cans, anything folks want hauled off. My buddy drives a truck for a local freight company that gives him a regular route, I joke with him that he hauls in more than he hauls out!!!! This is how we do it and provide some folks a way to get rid of stuff that they don't want to mess with. We love to hear, "You can have it, if you haul it off"!!!!
Thanks
Richard
http://www.metalworld.com/a/.html
Electric motors are scrap, just as anything else you don't have a need for, that has a recycle value.
Small motors are higher because there is relatively more copper in them. Large motors have copper, but the bulk of the weight is scrap steel/iron.
Kudos to Meco, above. Look at his system. You want more, you gotta work a little more.Fill your pickup with 1/2 HP motors, sell them for 100 a ton.
Strip the copper. Sell for 2 bucks a pound, aluminum endbells, what, 70 cents a pound? Sell the scrap steel for 150 to 200 a ton.
Making money is not easy. You don't wanna work at it, I don't care what machines you got, if you ain't a little bit hungry, you ain't gonna get rich.
Cheers,
George
Bob-
Sounds about right for the price. For a while, back in the early spring of '05, the price had dipped because China had run out of credit with some of the banks, but once they got back in the game in late summer the prices went back up. Everything- copper, aluminum, brass, stainless- they're buying it by the shipload because it's cheaper than mining it themselves.
It's good for us recyclers. I am a trustee of a Masonic program that buys kids from needy families essential things through the school systems, and 90 percent of our local lodge's funding comes through our recycling program. We gave away almost $.00 last year, and are on target to do the same this year. A bunch of us are in various businesses, and you would not believe what people put in the dumpster. We all get together a couple Saturday mornings a month to have coffee, and break stuff down, as a "pure" piece is worth more than one that has other metals, etc. still attached.
Up here, dead car batteries are a buck and a half each, scrap value. Motors are worth 27-1/2 cents a pound, and Carl is right, it's just not economically feasible to rebuild anything under 10 horse. It's cheaper new.
Billy
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Maybe it's just me. Is there anyone else out there? Is this a true statement about what is happening in America today? There is a glut of electric motors in America because no one needs them any more. The nice people in other countries will do those so much troublesome taskes for us and all we need to do is buy them. True, they may be marginally the quality we are used to but they are so inexpensive that we can simply buy another and another when they fail. John.
I could look up the numbers at my job however with a lot of the older motors usually if you rip them out scrap them and replace with newer higher efficiency the motor can pay for itself very quickly depending how many hours a year you run the motor. Motors are one of those items where the yearly cost to run them if say run 2-3shifts a day 50 months a year can easily cost you far more then they cost to run. With newer technologies and more modern extra strong magnets that have only come around in the last few years I could easily see where that old humongeous 20-30yr old 5hp motor can be replaced with a brand spanking new high efficency motor and have a payoff period of less then a year or so.
These days to stay competitive with the rest of the world you can't afford to be wasting money on energy. Efficiency is one of the best ways for a manufacturer to lower costs with out having to cut corners or lower the end quality.
Adam
Hello,
Thanks for the kudos George! Yep the system we use on scraping out electric motors could use some improvements, a air chisel with a custom forged chisel could do it a lot faster, and easier than a hammer and hand chisel. I've just been too busy to make a chisel like what I need for these motors that'll work in a air chisel.
On the bigger motors(5hp+), we get the end caps(Cast iron mostly) off of the motor and the armature out. If they have a bearing thats over 3" I scrap the balls out of the bearing for other uses. Then we take a electric concrete saw with a metal abrasive blade on it and cut the outer shell off and then take the saw and cut the one end off of the copper loops. Now these bigger motors have alot of shellac on them so we have to burn the shellac off. Once they cool, the copper pulls out easily, nice and clean. This all goes for various clean scrap. The best money you can make scrapping this stuff out. If you have a band saw you could cut the motor in half after the end caps come off, and the armature is out, then you wouldn't have to chisel the copper ends, just pull out of each half.
Some times to make the best money ya got to work a little harder!
Thanks!
Richard
I was asked to give an estimated running cost of operating a 3 phase 75 HP motor for 80 hrs. I assumed at 480 Vac that the motor would have roughly 100 FLA.
My equation then became KW = 480 V * 100 A * 1.732 / (ignored power factor and demand) and came to roughly 83 KW.
I had a coworker check to verify my answer and he used a different method. He used 1 HP = .746 KW and came to roughly 56 KW. Does motor phase affect this equation?
Which method is correct when solving for KW in a three phase system and why?
Any insight would be apprecciated, Thanks.
Your "cost of operating" has the word "cost" in there. Cost of electricity is for kW, not Amps. Amps is essentially irrelevant.
So yes, there are 746 watts/HP, that has nothing to do with amps, volts, phase etc. That is the power that is used. You are metered on kWH, (kilo
Watt Hours), so to determine the cost of operating something electrical, simply take the watts (kW) x operating time (Hours).
Now, without measuring anything all you can REALLY estimate is the
MAXIMUM POSSIBLE cost of operating that motor, because all you have is the RATING of that motor, not a measurement of how much power it is actually consuming. So 75HP x .746 = 55.95kW, x 80 hours = 4,476 kWH, x whatever your electricity rate is will give you the worst case operating cost. It will probably be less than that however, you will only be able to tell the exact amount by measuring.
- EST
You need to know what is the motor load, how constant, and duty cycle.
A useful method would be to measure the KWH for one typical 24 hour period.
If you want a wild guess at full mechanical load, then maybe use 1 KW/HP as an estimate, or 75 KW. Then you need to estimate the number of full load KW per day, possibly 16 hours. At 16 hours per day the consumption would be 16*1*75 = KWH/day, and if cost is $0.10/KWH the daily cost is $120.
You can change the cost all over the place based upon assumption you make.
.
Note that the stated horsepower of a motor is normally the mechanical output of which the motor is capable.
In this case it is stated to be 75 HP which is indeed about 56KW.
The input must be more than this owing to the losses in the motor, in the absence of detailed information on the motor efficiency, I would allow approaching 10% for losses giving an input of about 61KW.
Multipying 61 by the cost of power per KWH would give the hourly operating cost.
This assumes that the motor is fully loaded, the actual loading may be much less and can only be determined by measurement.
- EST
You need to know what is the motor load, how constant, and duty cycle.
A useful method would be to measure the KWH for one typical 24 hour period.
If you want a wild guess at full mechanical load, then maybe use 1 KW/HP as an estimate, or 75 KW.
I agree with you points on motor loading and duty cycle. But think that taking 1kW for 1 hp is an unduly pessimistic assumption about motor efficiency at the rating being discussed here.
- EST
Besoeker:
I agree that 1 KW is probably high for a motor of that size, but it is an easy number to use for a quick estimate.
Quite.
Of course, if the motors were rated in kW instead of the archaic HP, that would then eliminate the need for conversion factors.......
Runs for cover........
There are too many other undefined parameters that a high estimate won't make much difference. Actual KWH measurements over a reasonable time are needed to get good information.
I agree. With just the motor rating alone, all you can calculate is the maximum energy that could be consumed over the 80 hour period.
Once again
Once again
So everyone missed my point. I guess i will spell out my question.
This is all theoretical. Only want Ballpark answers.
A 75 HP motor at 480 find KW.
If single phase 75HP * .746KW/HP = roughly 56 KW
If three phase 75 HP * .746KW/HP = roughly 56 KW
If three phase and assuming 100 FLA at 480V
KW = Volts *Amperes * Power Factor * 1.732
KW = 480 * 100 *Assume 1 (reality .8) *1.732 = 83 KW or (100 KW if power factor is used)
All i want to know is if you generically use the formual for 1 hp = 746 watts do you need to account for phree phase situation by multiplying by 1.732.
I understand you are only charged for real power usage (KW) unless you are on a commerial site where your rate structure can ding you for you KVA usage (Low power factor = high ding). Amps doesn't matter you just want KW (I was using a generic voltage and Current to find KW).
- EST
JUSTEINEE:
Read broadgage's comments very carefully and understand what is being said.
You have a hang-up on some point that is not clear to most of us.
Power output = power input * efficiency
This has nothing to do with voltage, power factor, three phase, room temperature, speed, the phase of the moon, sunspot activity, or anything else other than as these factors influence the efficiency.
KW and HP are both units of power, the rate of doing work, and there is a linear equation that relates one to the other. KWH and horsepower-hours are units of energy, or work.
.
"1 HP=746 Watts regardless of single phase or three phase.
In many cases though the HP is the mechanical output, not the electrical input which must be greater since the motor must be less than 100% efficient.
As a rough estimate I would add about 10% to the output in order to estimate the input.
If the motor is not fully loaded then measurement by use of a KWH meter is the best option."
Thanks. Its kinda clearer now.
My hang up was if 1 hp = .746 kw and 75 HP=56 KW and 56 KW = V*I*PF*1.732 then why does V*I*PF*1.732 not equal 56KW. Equations should be interchangable.
Single Phase
KW = I x E x PF/
HP = I x E x Eff x PF/746
HP = KW x Eff/746
Three Phase
KW = I x E x 1.73 PF/
HP = I x E x Eff x 1.73 x PF/746
HP = KW x Eff/746
Whether Single phase or three phase HP = KW x EFF/746
I suppose the differences in our assumed numbers were due to Efficiency and Power factor. If i used a power factor of .67 i would arrive at the same answer of 56 KW.
I suppose from now on i will just use the 1 hp = .746 KW for my calcs.
- EST
JUSTEINEE:
I think I now see your hang-up.
From your post #13:
HP = KW x Eff/746
You continue to relate HP to the motor output power and WATTS to motor input power. Erase this concept. Horsepower and watts are the same kind of animal, they are are both power and related to each other by the unit conversion equation
W = 746 * HP .
Your equation is wrong unless you write it as
Motor Output HP = Input Electrical power * Efficiency / 746
If you do not explicitly define your variables, then the equation can mean all sorts of things.
When you use a variable, such as HP, that has the implication of power, and relate it to some other variable, such as W, that also implies power, then this implies a conversion factor formula relating the two units.
At either the output or input of the motor you can express the power at that point in WATTS or HP or ERGS/second or many other units.
Your confusion seems to arise from standard usage in the US where we typically describe the power output of a motor in HP and the motor input in WATTS ( more likely VA * power factor ) and thus you are confused in the meaning of these units.
.
That footnote applies to the columns for the synchronus motors not a squirrel cage induction motor.
Thanks kwired, I whiffed that eh?
It's also easy for me to miss what gar posted when I don't slow down;
"When you use a variable, such as HP, that has the implication of power, and relate it to some other variable, such as W, that also implies power, then this implies a conversion factor formula relating the two units."
What's a good field book for motors? I hear Dewalt does a good job but never used one. Ugly's style is a little hard to find things in, but it is compact for the tool bag.
Thanks kwired, I whiffed that eh?
It's also easy for me to miss what gar posted when I don't slow down;
"When you use a variable, such as HP, that has the implication of power, and relate it to some other variable, such as W, that also implies power, then this implies a conversion factor formula relating the two units."
What's a good field book for motors? I hear Dewalt does a good job but never used one. Ugly's style is a little hard to find things in, but it is compact for the tool bag.
I use mostly Square switchgear, breakers, controls. Ask a distributor for a motor slide chart calculator. I don't have any new ones but would guess they still have them.
You slide the chart to match the motor horsepower and then follow down to the correct voltage. Here it tells you full load current (based on NEC tables), minimum copper wire size (75 deg), recommended thermal magnetic breaker trip setting, Catalog number of of suggested fusible switch, size of dual element time delay fuses needed, catalog number of NEMA magnetic starter, and catalog of thermal overload elements. (I don't like to use the overload selection because using the actual load marked on the motor is the best way to protect it) It has all standard motor sizes from 1/2 - 200 hp three phase and turn it over and the other side has 1/6 to 10 hp single phase.
I have seen similar slide charts from other manufactureres in the past. Is faster than using any book I have seen. Especially if designing a single circuit install.