High starting torque, low running power 1

[VermontPE]

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

 

[djs]

Hello;
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.

 

[aolalde]

My personal suggestion is to leave the 10 HP motor and add a VFD driver. Since you need high torque only while starting, set your starting ramp at full load nameplate current (around 30 Lb-Ft starting torque and 14 amperes). For the normal running conditions, lower the voltage around 70% and leave the nominal frequency, that adjust will reduce the power of the motor to around 4.75 HP at full speed. With the magnetic flux reduction, the motor will increase the power factor and reduce the electric and magnetic losses, unfortunately the friction and windage losses will not be improved since the motor speed will not change that much but hopefully those are not the big loss contributors. Additional advantage will be the reduction of inrush or starting current.

 

[Marke]

Why not do the opposite of a star/delta start? start in delta and you will get the 10KW torque, then switch to star, using a closed transition, and you will get effectively a 3.3KW connection.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[itsmoked]

I like that Marke!

 

[jraef]

Good idea Marke, very creative, but assuming that he is is North America (from his nickname and use of HP instead of kW), it is not automatic that his motors are wound for that option as they would be elsewhere in the world. Part winding would work as well, but again, only if the motors are capable of it (i.e. wound for it). If not, you would need to chage out all the motors.

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
Read the Eng-Tips Site Policies at faq731-376
Member, [blue]P³[/blue]

 

[LionelHutz]

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.

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.

 

[VermontPE]

Thank you very much everyone for the replies.

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

 

[CJCPE]

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]

Oops, that 82.1% figure should be 92.1%.

 

[LionelHutz]

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.

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.

 

[Marke]

As you are only nterested in improving the efficiency of the motor, not changing the speed, a VSD would be a backward step.
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

 

[MikeHalloran]

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.

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

 

[jraef]

VermontPE,
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 http://www.oit.doe.gov/bestpractices/motors/] DOE ITP program link [/url] 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
Read the Eng-Tips Site Policies at faq731-376
Member, [blue]P³[/blue]

 

[LionelHutz]

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]

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]

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.

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
Read the Eng-Tips Site Policies at faq731-376
Member, [blue]P³[/blue]

 

[Laplacian]

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.

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.

 

[jraef]

Good point.

 

[CJCPE]

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.

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.