High starting torque, low running power
High starting torque, low running power
(OP)
I have a process driving many screw conveyors in a mining facility. The material being conveyed is a fine powder. Generally we try to empty the conveyors at the end of a production run, but sometimed residual material is left in the conveyors. When this happens we need high torque to get them started again. Most of these conveyors have 10HP 460V 3ph induction motors on them. History has shown that a 7.5HP motor is not enough to start the screws, but when they are running 3HP is enough to keep them going. The PF of the 10HP motor is very low (.5) when running so lightly loaded. What methods have been used to solve this problem of the requirement for high starting torque but low running power? As you can tell we are oversizing the motor to avoid the problem but in the new super-energy-efficient world this option is becoming unacceptable. Could I use a part-winding motor in the opposite manner to that which it is usually applied? That is, disconnect some of the winding once the motor is started.
Tom Gilmartin, P.E.
Rutland, VT USA
Tom Gilmartin, P.E.
Rutland, VT USA





RE: High starting torque, low running power
Two alternatives come to mind:
1. Use a two speed, constant horspower motor. Say 5 Hp at 1745 and 5 Hp at 850 RPM. At the slow speed, you have the same torque as a 10 Hp.
2. Use some kind of a soft start device to reduce the line voltage to the motor once it is running. By reducing the line voltage, the PF and efficiency improves. Many soft starts have this feature.
RE: High starting torque, low running power
RE: High starting torque, low running power
Best regards,
Mark Empson
http://www.lmphotonics.com
RE: High starting torque, low running power
RE: High starting torque, low running power
VermontPE,
The bigger question I have for you is, why are you concerned about the low pf of the motor? Are you penalized for low pf? If you are, put PFC caps on and correct that problem. Far cheaper than VFDs. When lightly loaded like that you have only lost a few percent in efficiency, but remember the motor will only consume as much power as it needs, so it is a few percent of a much lower power value. The net efficiency losses are then actually minimal amounts of power. PFC caps can correct for any pf charges imposed by the utility.
Also, be aware that VFDs can give you full torque at startup, but that is full RUNNING torque, not necessarilly full STARTING torque. You may have discovered that you need the full Starting torque of a 10HP motor on a loaded restart, and even a vector VFD may not necessarilly give you that. Unfortunately you will probably not know that unless you buy one and try it. And in order to save energy on that application, your auger will need to run slower, which often causes other load problems with auger conveyors.
"Our virtues and our failings are inseparable, like force and matter. When they separate, man is no more." Nikola Tesla
Member, P3
LionelHutz (Electrical)
jraef is correct. Motors only draw the power they need and the efficiency of a motor only lowers slightly with less load. You're likely looking at something like a 2% change in motor efficiency in your application, compared to runing full load. I just don't think you'll find a different motor and starting means combination that will reduce your energy usage. For example, a VFD isn't 100% efficient and whater power you save in the motor may just be re-lost in the VFD.
VermontPE (Electrical)
(OP)
Thank you very much everyone for the replies.
CJCPE (Electrical)
Just finding the most efficient standard motor for the application may help a lot. The first part-load motor data I found online is for a Siemens TEFC NEMA motor. The 10Hp, 460V, 4P motor is listed at 91.7% eff for full load, 82.1% for 3/4 load and 91.3% for 1/2 load. Their 3Hp motor is listed at 89.5% for FL. It looks to me that the 10Hp would run at 3Hp with almost the same efficiency as a 3Hp motor at full load. Power factor for 10Hp 1/2 load is .66.
CJCPE (Electrical)
Oops, that 82.1% figure should be 92.1%.
LionelHutz (Electrical)
I hate to burst anyone's bubble, but applying a 10hp VFD onto your motor will likely reduce the overall system efficiency. A VFD has losses itself, which typically are around 2%-3% for a 10hp VFD. A VFD also causes higher losses in the motor. I just don't know right now how much but I know I've seen a significant temp difference in a motor running on a VFD vs directly on-line. So, lowering the output voltage of the VFD until you get best efficiency likely won't gain back these extra losses let alone save above these extra losses.
Marke (Electrical)
As you are only nterested in improving the efficiency of the motor, not changing the speed, a VSD would be a backward step.
MikeHalloran (Mechanical)
Could you double up on the motors? Say, a 3HP motor to provide running torque, as efficient as you can get it, and on the same shaft, a 7.5HP or 10HP high torque motor used just to get the screw started.
jraef (Electrical)
VermontPE,
Read the Eng-Tips Site Policies at FAQ731-376
RE: High starting torque, low running power
To make a guess, I'd say you're probably talking about, at most, a theoretical 200W of motor losses you could save in each motor application. This is assuming you could change from a 10hp motor to a 3hp motor running across the line. Obviously, if you have to make a larger motor compromise to still start the augers then it'll be even less. Then, if you add an active power device, such as a VFD, you're once again lowering this number due to losses of this device.
You'll have to find someone with some good measuring equipment, such as a 3-phase power analyzer, to perform before and after measurements to justify any changes.
RE: High starting torque, low running power
aolalde, your VFD reply seems a bit strange - we have thousands of VFDs in our company. Roughly 80% of the ~70MW worth of motors we have in NA are VFD. Of those I have never adjusted one to reduce the output voltage but maintain the same frequency. The only input I know of is to change the speed, and the drive reduces effective output voltage and frequency accordingly. Have you ever done what you suggest, and if so, what type of drive, manufacturer, settings, etc did you use?
Some numbers
10HP std motor at 30% load is about 80% effecient. This translates to 2.7975KW
3HP motor at 100% load is about 84% efficient This translates to 2.6643KW
If the motor runs 7500 hrs/yr, annual savings is 1000KW for the 3HP vs 10HP.
Not much you say, but consider in one plant I have 70 of these screws. That's 70MWh saved annually. Insert your own costs to see the $ savings.
I'm being pushed hard by upper management to save energy costs, so even though this is not that much I thought I would explore it.
Thanks again
Tom
RE: High starting torque, low running power
RE: High starting torque, low running power
RE: High starting torque, low running power
The efficiency or power curve for lowering the voltage looks like a U so if you lower it a little you'll get some gains but then when you go further your losses start going back up.
Did you know that when you operate a standard 1.15 serice factor motor with a VFD that the motor should be derated to a 1.0 service factor?
What about the obvious fact that your example is really void since you won't be able to use a 3hp motor anyways because you'll never start the auger with it.
The only realistic solution I'm seeing here is using a 2-speed motor as djs suggested.
As I already stated, do good before and after power measurements on a test auger to validate your power saving method.
RE: High starting torque, low running power
The motor has both Iron losses and copper losses and these are relatively small.
You can reduce the iron loss by reducing the applied voltage. When you reduce the voltage, there will be s amll increase in slip (increased slip loss) and an increase in copper loss (the work curent will increase).
You would need to measure the results, but the easiest answer is to run the motor in star (provided that the motor is delta connected for full output power).
Best regards,
Mark Empson
http://www.lmphotonics.com
RE: High starting torque, low running power
The auxiliary motor need not be electric; it could be an air motor.
OR,
I'm not sure if this exists: If you could find a fluid coupling that's actually a torque converter at high slip, that might enable you to start the screw with a motor smaller than 10HP.
Mike Halloran
Pembroke Pines, FL, USA
RE: High starting torque, low running power
I think that your answer lies within the posts by Marke, CJCPE and myself (to a lesser extent) for a very workable solution. As CJCPE pointed out, newer energy efficient motors are going to be a vast improvement over the horrible numbers you posted. 80% eff at 1/2 load is terrible. So terrible in fact, that you may very well qualify for an efficiency upgrade rebate form your local utility. It is worth checking into. It means changing out the motors, but that then provides you with an opportunity to make sure you get 12 lead motors so that they are capable of being connected in Star as Marke mentioned.
What he is suggesting is essentially using a Star-Delta (aka Wye-Delta) controller, and instead of letting it start in Star and run in Delta, you actually do it the other way around. That will mean that your 10HP motor will be reduced to about 3.33HP, but you already know that will work. Also important is the fact that it will still run at the same speed so your fines will not settle and increase the load, as they likely would with a 2 speed motor or VFD. Then if your power factor is still poor (although it probably won't be in this scenario), just add PFC caps as well and correct it to the utility line so that they don't penalize you.
Check with your utility to see if they offer any rebates based on the Federal Dept. of Energy "Motor Challenge" program for upgrading to energy efficient desgns. Here is a link to their website for more information DOE ITP program link In fact, I even stumbled across this old rebate form that appears to have been valid for Vermont. It has expired, but perhaps you can give them a call and se if anything else is available. New England motor rebate form
Even if your utility doesn't do it officially, approach them with the concept and they may be willing to contribute. And still, even without rebates, the energy savings alone will likely have a very attractive payback. Several of the major motor manufacturers offer free payback calculator programs that can put hard numbers to that concept.
Good luck.
"Our virtues and our failings are inseparable, like force and matter. When they separate, man is no more." Nikola Tesla
Member, P3
LionelHutz (Electrical)
I don't buy the Y-Delta solution. You are effectively lowering the motor terminal voltage to 58% of rated. This will lower the efficiency of the motor, not raise it.
VermontPE (Electrical)
(OP)
I agree Lionel, wye-delta is interesting but counter-intuitive to the idea of saving energy. I'm trying to get a real motor datasheet to show the efficiency running in wye vs delta.
jraef (Electrical)
Why do you think it will lower the efficiency? The only way to decrease efficiency is to increase losses. Where are the extra losses going to come from? They are exactly the same windings, just connected differently, and the rotor and stator is exactly the same. Remember, we know that the torque in a fixed frequency follows the square of the voltage, so at the effective 57.7% voltage, the torque will be .5772, or 33.33%. So if we apply the math to it, where the 10HP motor has 30 ft.lbs. (assuming 1750RPM), the new HP = Tq x RPM/5250, and we have the Tq reduced to 33% of normal, so 30 x .333 x 1750/5250 = 3.33HP. Nothing else has changed in this motor.
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RE: High starting torque, low running power
RE: High starting torque, low running power
RE: High starting torque, low running power
The only possiility would be in extra slip losses, which would make sense if the load remained the same, because of the loss in torque. But in this case we already know that the load only requires the torque of a 3HP motor to keep moving, a value slightly under the 3.33HP available from the motor connected in Wye. No additional slip = no additional losses. Plus, now you will be running closer to rated load, so your power factor will be better.
"Our virtues and our failings are inseparable, like force and matter. When they separate, man is no more." Nikola Tesla
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Laplacian (Electrical)
Before the two-winding motor idea is fully endorsed, better check to see feasibility of adding the second contactor to the existing motor starter. If the starters are in standard NEMA sized MCC cubicles, then you may have to completely re-arrange the cubicle positions and buy new MCC hardware. If they are on switchracks in explosion/dustproof enclosures, then you may need larger ones.
jraef (Electrical)
Good point.
CJCPE (Electrical)
To compare the efficiency of Y vs delta operation, 3 Hp, 9 lbs-ft torque is the assumed operating point for both connections. When the winding voltage is reduced to 57.7%, the torque at any given slip RPM is reduced to about 33.3% of the full-voltage value. That means that the slip at any given torque will increase to about 300% of the full-voltage value. Slip losses should should increase. Other losses will be reduced, but I think the net effect will be reduced efficiency.
jraef (Electrical)
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RE: High starting torque, low running power
Using the price our site pays for energy; approximately $14/KWH the savings would be nearly $1M/yr based on the energy savings you posted.
RE: High starting torque, low running power
RE: High starting torque, low running power
The 5 Hp two speed constant Hp alternative may be more efficient, but the two-speed design constraints may prevent a good high-efficiency design. The motor cost will probably be higher than a 10 Hp 4 pole motor. The NEMA frame size for a 5 Hp 8 pole motor is 254T vs 215T for a 10 Hp 4 pole motor. To that you must add the cost of converting to a two speed starter.
I still think that finding a new 10 Hp motor with the best available efficiency at the intended 3 Hp operating point will have the best chance of paying back the investment.
RE: High starting torque, low running power
Please explain this. Do you mean to say that if you have 4% slip at full voltage, you will have 12% in Wye? How so? The number of poles has not changed, the winding resistance has not changed, the frequency has not changed. What am I missing?
"Our virtues and our failings are inseparable, like force and matter. When they separate, man is no more." Nikola Tesla
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CJCPE (Electrical)
jraef
jraef (Electrical)
Huh. When you put it that way, I see your point. If you lower the entire torque values on the curve without changing speed, the angle of the curve from breakdown torque to synchronous is lower, which would increase the slip RPM percentage of the new curve, so in that case it would be additional slip losses. I was thinking of it in terms of what creates slip in the first place. It makes sense when I think about it in terms of another old motor trick, that of oversizing a motor in order to reduce slip speed loss. Thanks for the enlightenment. Have a Star.
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RE: High starting torque, low running power
Reducing the phase winding voltage reduces all of the torque values on the torque-speed curve without changing the speed values much. Between no-load and full-load, torque is pretty much proportional to slip. If the 10 Hp motor has 50 RPM slip at 30 lbs-ft, full load, it will have 15 RPM slip at 9 lbs-ft, 3 Hp load. For the Y connection, the torque for 50 RPM slip will drop to 1/3 or 10 lbs-ft. Slip for 9 lbs-ft will increase to from 15 RPM to 45 RPM. If you draw the curves, you can see what happens.
RE: High starting torque, low running power
"Our virtues and our failings are inseparable, like force and matter. When they separate, man is no more." Nikola Tesla
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aolalde (Electrical)
VermontPE:
.reliance. com/prodse rv/standri v/appnotes /d7726.pdf
aolalde (Electrical)
Marke & Jraef:
Marke (Electrical)
Hello aolalde
CJCPE (Electrical)
Reliant Energy of Houston has some information that discusses energy savings by motor voltage reduction.
p://www.re liant.com/ business/1 ,,CID43394 5,00.html? 2id=220037 &1id=2 20172& 3id=433947
aolalde (Electrical)
Good work CJCPE I agree. We should consider that those are typical figures and could change for different type and motor manufacturer.
electricpete (Electrical)
A brief discussion to explain the behavior discussed above. (nothing new to most of the particpants here).
itsmoked (Electrical)
I suggest VermontPE gets a delta-wye starter and a regular power meter and runs one of his of many conveyors for 30 days. Then block the starter in delta for a month.(or alternate every week) This is chump change for a plant of his size and could give a good handle on everything discussed in these great posts, including the delta-wye feasibility.
Read the Eng-Tips Site Policies at FAQ731-376
RE: High starting torque, low running power
See the following paper by Reliance Electric about Using AC PWM Drives on Non-Standard Voltages and Frequencies; http://www
RE: High starting torque, low running power
According to NEMA MG1 part 12.58.1 the following losses shall be included in determining an induction SC motor efficiency:
Stator I^2*R
Rotor I^2R
Core Loss
Stray Load Loss
Friction and Windage Loss
My question is; what is the meaning of the “extra slip losses” or “slip speed loss” that you mentioned?
RE: High starting torque, low running power
The slip losses are dissipated in the rotor resistance, so they are covered in the list that you included.
Best regards,
Mark Empson
http://www.lmphotonics.com
RE: High starting torque, low running power
htt
They provide a set of curves showing motor losses vs applied voltage for various load levels. I used data from the curves to calculate the efficiencies for the minimum loss points on the curves:
Full Load, 88.5% at 430V vs. 88.3% at 460V
3/4 Load, 89.1% at 408V vs. 88.4% at 460V
1/2 Load, 89.6% at 359V vs. 86.7% at 460V
1/4 Load, 89.2% at 279V vs. 80.4% at 460V
Note that, in this case, the optimum voltage at 30% load would be more than 57.7%.
If the existing motors are delta connected with terminals available for Y connection, the Y delta idea might provide good results. Although it would introduce additional losses and harmonics, a solid state voltage reduction device might provide significant savings also.
If the motors are replaced with new motors of the highest obtainable efficiency at the actual load point, adding a Y delta or solid state voltage reduction arrangement may not add so much efficiency improvement.
RE: High starting torque, low running power
Thanks for the clarification Marke. In reality the slip will be proportional to the shaft load ( very reduced for this aplication) and the magnetic strength of the motor.The induced voltage in the rotor is proportional to the slip and the rotor current proportional to the induced voltage. The resultant loss in the rotor winding (cage) is I^2*R.
Loading a 10 HP motor at reduced voltage but taking only around 3 HP load, will probably not increase the slip as compared to that of 10 HP load (nominal slip) at full voltage.
RE: High starting torque, low running power
Consider two major losses I^2*R and core losses. As we decrease voltage (with constant load), they go in opposite directions (I^2*R increases and core losses decrease).
Which one wins? Depends on load. I^2*R is dominant increases with load while core loss doesn’t. I^2*R effect wins at high load (perhaps above nameplate in the example above) and higher voltage is more efficient. Core effect wins at low load and lower voltage is more efficient at low load. You could draw a curve of most efficient voltage vs load and program the voltage accordingly... I guess that’s sort of what the Nola devices do.
==============================
I agree delta-wye start is an innovative approach. To play the devil’s advocate you might think about the followiong questions: Will the energy savings would be worth the cost of installation/material PLUS the increase in complexity and possible reduction in reliability. In my mind increased complexity almost always means reduced reliability... if nothing else there are just more components to fail. In this particular case we might think about possible trips occuring during switching if not set up exactly right... possible degradation of motor windings due to switching surges.
We don’t have any wye-delta starters so I am not familiar with them and just speaking in generalities. Others are welcome to comment from experience.
Also in my mind there is an element of uncertainty about the design assumptions – in the past you have only experienced problems during starting. But perhaps there are other temporary power overloads during conveyer operation which you can ride thru in the existing configuration but which would cause problems in the wye configuration.
One other possibility to examine would be a special high-starting torque motor of a lower horsepower rating. Still subject to the uncertainty mentioned above regarding momentary overloads during operation.
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RE: High starting torque, low running power
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