VFD confusion 11

[Mysterrose]

Trying to understand how VFD's save you money, and I've seen people throw around that a x% reduction in speed is equal x% cubed in energy reduction. From what I can tell this is only true for pumping or air handling type applications where speed and flow are roughly a 1:1 ratio but speed to energy consumption has a cubed factor.

What about applications like a conveyor system? I'm thinking processing plants, shipping plants, heck even an escalator. Right now say they use a gearbox or belts to reduce the speed of the conveyor. Going to a VFD would eliminate the gearbox/belts and allow you to direct couple motor to conveyor just driving the motor at required speed. What I don't understand is if in a situation as described does the VFD save any money?

To me a gearbox/belts is a zero energy device in theory. I'm assuming no losses due to friction, heat, etc. So removing the gearbox/belts does not reduce amount of energy used. If you're not reducing the amount of energy used, how do you save money?

In the above assuming that nothing changes in terms of speed of conveyor or loading on conveyor. You're still processing or moving same amount of "stuff". So aren't you still doing the same amount of work?

 

[jraef]

You are the victim of a very common marketing trick: say something that is true but IMPLY that it is something that is EXCLUSIVE.

In ALL centrifugal loads like pumps and fans (although not all pumps and fans are centrifugal), there is a physics property that is called the Affinity Law that states that for any given change in FLOW, the power required to make that change varies by the cube of the speed. So for example at 1/2 flow, the power needed to accomplish that is .5[sup]3[/sup], or 12.5%. What the VFD marketing people don't tell you, is that this is the case REGARDLESS of whether or not the VFD is there. In other words if you just choke off the output of a fan to 50% flow, it ALSO reduces the load on the motor to 12.5% of what it was. That alone is not exclusive to VFDs. It's a very common misconception that the VFD marketing people tend to leave alone. I work for a VFD mfr, my own company's literature and training courses say it. I decry that, but it goes unheeded because "everyone else says it and if we don't, people will think our drives don't offer that". And you know what? They are correct in that perception!

But that does not mean that VFDs do not save energy, they do, just not because of that issue alone. The correct statement is to say that in centrifugal machines when flow reduction is necessary to the process, VFDs SAVE ENERGY COMPARED TO OTHER METHODS OF SPEED REDUCTION. The problem is, that's not a nifty little sound bite...

Where the energy savings come from is in COMPARISON to throttling valves and vane controls etc. In a pump for example, if you use a valve to throttle back the flow, and run the motor at full speed, the motor load will drop by the Affinity Law, but there will be losses in the valve itself; a pressure drop, heat and turbulence all extract energy from the system. Using a VFD to reduce the output INSTEAD of using that valve saves on the energy LOST in that valve. In addition, running a motor at full speed, but at 12.5% load (from our example), also has additional losses associated with the motor itself. This is because there are fixed losses in a motor that do not change with load, they are associated with it just BEING a motor. A decent proportion of THOSE losses are associated with magnetic energy being consumed, and that varies with the voltage applied. So when you reduce speed with a VFD, you are also reducing the voltage, and thereby also reducing those losses in the motor. So between the reduced motor losses and the reduced losses across the flow restriction, there are significant energy savings available to take advantage of. As a general rule, most VFDs will pay for themselves in anywhere from 12-24 months, depending on how much of the time they operate at reduced flow. But on the other hand if you DON'T operate at reduced flow very often, it may not be worth doing at all.

Now apply this to your conveyor example. Does our conveyor NEED to operate at varying speeds? Not reduced, but VARYING. Because if it's just that you need a REDUCED speed, change the pulley or gear ratio. If you need to have DIFFERENT speeds all of the time, as in more than two (because 2 speed motors are easy), then you can compare a VFD to some sort of MECHANICAL variable speed drive. If that's the case, a VFD may indeed save on energy. but it's a lot more complicated than that. With MECHANICAL methods of changing speed, you usually end up with LESS speed but MORE torque. With a VFD, you get less speed but THE SAME torque. In some cases, this may be a problem, i.e. the original designer had selected a motor speed and pulley ratio that was based on the FINAL torque at the conveyor belt. If you replace the pulleys with a VFD and end up with LESS torque, you stall the motor. Lot's of energy saved , but no work performed! So can you save energy on a non-centrifugal load? The answer is maybe, maybe not, which marketing people HATE.

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376

 

[zdas04]

jref,
Outstanding explanation. As a flangehead I just have to clarify one point. Everywhere you said "centrifugal" you should have said "dynamic". There are a number of dynamic fluid-transfer devices that operate under the affinity laws, centrifugal is just one of them (albeit the most common). Axial pumps and compressors exhibit the same speed vs. power behavior and centrifugal pumps and compressors for example. I know that you were trying to differentiate dynamic pumps which operate within the affinity laws from positive displacement pumps that don't, but I keep getting into arguments with EE's who don't have your understanding of the subject and think that only centrifugal pumps get the reciprocal n cubed bang for the buck.

Your point about gears and pulleys is a good one. I can do a transmission and a soft start for slightly more cost than a VFD without the torque problems you mention. End users really need to think about what they are doing and how much flexibility it really needs.

Like you implied, VFD equipment is often very effective, but it is not a universal answer to all problems.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.

 

[jraef]

Yes, I know I was being guilty of oversimplification by using "centrifugal", but not being a flangehead, I wasn't sure if there was another term (such as "dynamic") that would have been universally suitable without more lengthy explanation (and my diatribe was already long). Outside of the US engineering community people use the term "quadratic" loads, which I like because it really describes the dynamic involved, but again, I wanted to keep it as clean as possible. I've used quadratic loads in talks in the past and had been greeted by blank stares, then had to side track and explain quadratic formulas and how they relate... lost everyone.

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376

 

[Mysterrose]

I think this makes more sense when you mentioned that mechanical means gives less speed but more torque, but going to a VFD will give less speed, but same Torque output by motor. I was going through all the mechanical equations and what happens when you simply replace gearbox with motor running at slower speed and kept arriving at that same thing. This is what bogged me down I think. When I was looking at it I was making assumption no conveyor speed and no conveyor loading changes which meant no changes to amount of torque required to drive it at same speed. This meant no changes to amount of work I had to do in a given time (i.e. amount of power required stayed the same).

If I am understanding this correctly from a very high level. VFD's save money only if you can reduce amount of work the motor needs to do in a given time, which in essence is saying you need to reduce amount of power required. Centrifugal loading applications leverage affinity law to reduce the amount of work the motor needs to do. Non-cetrifugal applications you would either need to change the loading or distance (i.e. change amount of work you need to do), or reduce speed (i.e. change how fast you are doing the work).

 

[LionelHutz]

When looking at efficiency, you have to look at the complete picture. Efficiency of a pump would be measured by looking at the gallons pumped per kWh used. Too many people focus on the kWh used per unit of time (day or hour or month) because this is a easy number to see simply by using a power meter and observing the power drop at slower speeds.

On a conveyer, you are typically moving X amount of material per unit time. It will take the same energy to move that material from one end of the conveyer to the other, regardless of the speed (within practical limits). So, it comes down to which speed makes the conveyer parts more efficient. Does running slower with more load cause more or less friction in the belt and rollers compared to running faster with less load?

Here is a good thread on the subject in case you haven't seen it. The very generalized conclusion is that a system that can allow a varying head pressure is required to save energy, which is not the case in many pump applications. It can be the case for many fan applications though.

http://www.eng-tips.com/viewthread.cfm?qid=260203

 

[mikekilroy]

good points on load changes on various types of equipment with speed changes....

good points on vfd being a constant torque device at below base speed.

but two other points to keep in mind:

1) vfd typically goes FASTER than base speed at constant HP; so if one needs higher torque at lower speed (very few processes are like this!) they Can get double the bang by adding gearing and a vfd; run the motor over base speed at constant HP to double the overall speed range at double the torque if this is significant.....

2) and missed in replies above, YES, vfd DOES save money if run at less than base speed all by itself. Again, very very few processes need more torque at lower speed and the majority will need less torque. so just by adding the vfd, the torque probably would go down, so the current draw will also. BUT there is a 1:1 savings by going slower: run 1/2 speed, you output 1/2 voltage to the motor at that point. Since the ac input to the vfd is constant voltage, the input current goes to 1/2 the load current - hence you are now using 1/2 the input power. Ditto at lower speeds: example: go 1/4 base speed outputing 20 amps into the motor, and the input current will be 20/4 or 5 amps. This is indeed straightforward power savings.

 

[LionelHutz]

mikekilroy - your second point is extremely oversimplified. If I run a water pump at 1/2 speed then the input power will be much less than 1/2 of the full speed power. However, I'm not saving any energy with a pump that is running dead-headed with no flow.

 

[mikekilroy]

oversimplified? not at all.... question was not what if someone turns a vfd down so far that its output does no work anymore.... it was:

Mysterrose (Mechanical)
30 Jul 12 11:51
Trying to understand how VFD's save you money.....
What about applications like a conveyor system? ...... What I don't understand is if in a situation as described does the VFD save any money?

so you see, the OP asked a specific question about slowing something like a conveyor down; do it & the torque load remains the same or goes down slightly. the input power at 1/2 speed will be 1/2 the power that was pulled by the vfd at 100% speed. question answered. yes, power saved. no over simplification here at all. sometimes a simple straightforward answer to a direct question is the best answer.

 

[LionelHutz]

LOL, yes your first answer was oversimplified. You basically posted that reducing the speed of a motor will cause a reduction in power used. There were no conditions given where this would allow for energy savings. The probem is that no-one should care only about the power into the motor. Everyone should be concerned with the work done per unit of energy used.

You also answered the question out of context. If you quoted the complete post, then you'd realize the OP asks about removing a gearbox and keeping the conveyer at the same speed with the motor at a reduced speed. In that case, the conveyer is still moving the same amount of material and that will require the same amount of power. There can be some savings in the removal of the gearbox, but they might be lost again due to the extra losses in the VFD and the motor driven by the VFD.

Also, now you have posted that a conveyer load will show reduced power usage with reduced speed. This is flawed.

I certainly would not expect the power to reduce directly with the motor & conveyer speed if feeding a constant material supply onto the conveyer. It requires a certain amount of energy to move the material from point A to point B so you can't reduce the energy used in 1/2 by reducing the conveyer speed in 1/2 and doubling the conveyer load. This case may or may not save some energy depending on the conveyer and load, but it certainly isn't a simple savings like you posted.

Lets look at the other case. Reduce the speed by 1/2 and also reduce the feed by 1/2 so the conveyer now moves 1/2 the material in a given period of time. I see 1/2 the power and twice the time which is the same energy used to move a given amount of material. No energy savings in this case.

 

[jraef]

LOL, yes your first answer was oversimplified. You basically posted that reducing the speed of a motor will cause a reduction in power used. There were no conditions given where this would allow for energy savings. The probem is that no-one should care only about the power into the motor. Everyone should be concerned with the work done per unit of energy used.

Bingo, but this may be an issue of the semantics.

I think the salient point here is the difference between power and energy. Yes, technically mikekilroy is correct in that when you reduce the speed of something like a conveyor, you are reducing the power consumed at that moment in time, but ENERGY is a function of power AND time. So if you are doing less work but for a longer period of time, the base energy per unit of work will not change, which is LH's point. But you may actually be consuming slightly MORE energy in reality, because there are fixed losses that have only to do with operating a motor in the first place, regardless of the load on it. So the longer you have to operate that motor to accomplish the same net amount of work, the more those losses add up over time.

Side note: I've had to have this discussion (argument) with civil engineers on the value, or lack thereof, in putting VFDs on pumps to "save energy" that are just filling water tanks. There are other reasons to do it, but it does not save energy unless there is some other compelling reason to vary the flow, in which case it comes back to what I said earlier about the DIFFERENCE in changing flow with a VFD vs some other method.

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376

 

[mikekilroy]

huh? first you say my 2) answer was oversimplified, which I then explained was not so, now you say my 1) answer is also oversimplified? huh?

my 1) answer for reference:

1) vfd typically goes FASTER than base speed at constant HP; so if one needs higher torque at lower speed (very few processes are like this!) they Can get double the bang by adding gearing and a vfd; run the motor over base speed at constant HP to double the overall speed range at double the torque if this is significant.....

What part of this don't you agree with?

You then say I answered the question 'out of context." How, I quoted the OP question? Then directly answered it.

You also say a that a conveyor running at reduced speed will not result in less power required? huh? you should try to explain this comment..... First let me reiterate fact: a conveyor running at 1/2 speed will require 1/2 the power to do so since it will be running at 1/2 speed N, and the friction torque elements (Tfstatic+ Tfviscous) will result in a new torque of Tffriction that is the same but Tfviscous will be less. So even torque will decrease in addition to speed. But ignore that if you will. Speed is 1/2. Look at power. P=N*T/5252 where P=hp, N is in rpm, T is in #-ft. So even leaving T at same level, with N=1/2 full load speed, P is still 1/2. What part of this do you disagree with?

Help me also understand your last statement.... If someone has a conveyor and they need to cut to 1/2 to stay even with the process, why do you think the material will double? One does not cut a conveyor speed in halve to make it back up on the conveyor.

 

[mikekilroy]

jraef (Electrical)
10 Aug 12 14:22
Quote (LionelHutz)
LOL, yes your first answer was oversimplified. You basically posted that reducing the speed of a motor will cause a reduction in power used. There were no conditions given where this would allow for energy savings. The probem is that no-one should care only about the power into the motor. Everyone should be concerned with the work done per unit of energy used."

I am too dumb to know how to copy previous posts for reference on this forum: I am an electral engineer, not a poet (as Dr. McCoy would say). sorry.

As Jraef states here, why would anyone want to slow down a conveyor to 1/2 speed to do the same amount of work? They would not. Reason for slowing down a conveyor is because that higher speed is not required to keep up with the process. So you slow down the conveyor TO SAVE POWER AND THUS ENERGY. Let's stay with reality folks. Playing what-ifs with nonsense work vs energy statements does not help the OP understand why vfd's can significantly save money on one's energy bill.

 

[ScusaMe]

Lionel -

When Abraham Lincoln stated "...that All men are created equal...." he wasn't referring to comprehension abilities.

Ergo, One man's "extremely oversimplified" is another man's Clarity.

We all don't share the same yardstick !

It was ever thus.

 

[LionelHutz]

OK I was referring to your first response second point.

As for the power consumed. You have to consider all the variables for the speed where you want to operate. In other words, don't use blanket rules for this kind of thing.

You also have to read the complete reply. I never posted that running a conveyer at a lower speed doesn't reduce the power. In fact, I did post that running at a slower speed can reduce the power in the case where the amount of material on the conveyer remains the same.

And once again. The OP asked about removing a gearbox and keeping the conveyer speed the same. You can't be claiming you have the one and only true answer to the question when you never answered the question. He didn't ask about reducing the belt speed at all or reducing the amount of material used. I never answered the question either since it was already answered. I just added some comments in case the OP was wondering about conveyer speed.

Last I checked, the hydro company bills in kWh. I could care less what the instantaneous kW is. The kWh to run the process for the month is what I pay for and what I care about.

As for you point #1 in your first post. The motor is not simply a constant HP device as you exceed the base voltage or once you begin to reduce the V/Hz applied to the motor. It will be for a bit, but then you use up the "excess" torque capability and the motor begins to lose HP. I've seen motors that can't accelerate past 75Hz with an open shaft. Not much of a constant HP device there. You have to remember, as you increase speed in the constant HP range, the torque capability of the motor reduces by the square of the voltage reduction while the required torque reduces linearly by the overspeed. At some point, the motor will not be able to produce the torque required to maintain the constant HP.

As for you last point in your last post. Now you're beginning to give specifics where energy savings are possible. Now you're talking running at a slower speed to match a process. Big difference compared to running a stockpiling conveyer at a slower speed just to hopefully save energy. Same applies to a water pump. Jraef has posted about this already. Big differeence between using a VFD to reduce the pump pressure and flow for a process compared to using a VFD to reduce the flow when filling a tank. The first case can save energy compared to using other methods to reduce pressure and flow. The second case is an energy waster.

 

[LionelHutz]

I forgot this point. Running the conveyer at a reduced speed even for process control will not save energy directly based on the motor speed.

Here's an example. Say you have a process where there are 2 possible operating conditions.
#1 - requires 100% speed and 100% load on the conveyer.
#2 - requires 50% of the material delivered in condition #1.

OK. So condition #1 should be obvious. A VFD running at full speed causes extra losses. No savings in this case.

For condition #2. There are 2 choices. Run the conveyer at 100% speed and 50% load or run the conveyer at 50% speed and 100% load. It should be pretty obvious that running at 50% speed and 100% load will not give a 50% energy savings compared to running at 100% speed and 50% load. You might save some power, but certainly not 50%.

 

[mikekilroy]

ok lionel, you got a point; I did not address the energy side of the equation just the power side since that is how I read OP question. I reread it and you are correct, it clearly asks on the conveyor example for replacing the gears with vfd, which of course would require a higher torque motor so no saving there, and asked about energy savings if you ended up having to do the same amount of work at the end of the day, in which case vfd again would not save energy.

But you miss the point of energy savings by reducing the motor speed. Ask yourself an honest question: how often does someone slow the speed of a motor down to make it run longer? a vfd is normally used to tweak a process:

- a conveyor master slaved to a machine that runs anywhere from 100 units/min to 1000 units/min depending on the product run that day; so slow the motor down to match the line
- a fan blowing cooling air with a temp sensor modulating it
- a machine tool spindle; its speed is determined by the material removal rate possible with a given tool
- etc
- etc
- etc

So you see that slowing a vfd is very common due to many many reasons and - I would suggest - SELDOM to do the same amount of work in more time, so most applications of vfd's DO care about the instananeous KW usage.

You go on to suggest that it is unfair to also say a vfd will allow a motor to run in contant HP above its base speed. I understand you are trying to be picky, but you should not confuse folks with that: the fact remains it is true, even as you pointed out. I did not get into breakdown torque as that was not pertinent here. Of course a point is reached eventually where torque along the constant HP curve intersects breakdown torque. I don't recall stating anything about how MUCH higher than base speed a motor will go - that would not have made sense as it is motor design dependent. FYI, your motor that reached breakdown torque at only 75hz was a very unique motor; since you brought it up, most will reach easily 120, most at least 200hz at constant HP. In addition it is quite common for applications that use a vfd in this high speed mode require LESS torque the higher they go (such as high speed machining aluminum) so even passing breakdown torque does not mean the higher speed is not applicable to a process. we often rate motor speed torque curves showing this reduction in HP at very high speeds of 400, 600, 800hz output. Quite often on special motors we design them to run at less than vfd input voltage at base speed to allow us to conntinue the v/hz curve linear above base speed to overcome these very high speed limits, but that is another discussion. Anyway, this point of breakdown meeting rated torque is easily calculatable before applying a certain motor to a process that requires those speeds.

As to your last post you should go back and read my #2 point again: Yes, running a motor at same output torque & 1/2 speed will pull 1/2 the power from the vfd input. This means 1/2 the energy. Your post is confusing this fact by making the load torque variable. Let me know if you would like me to explain that in more detail.

 

[LionelHutz]

I have no problem understanding that a motor run at 1/2 speed is running at 1/2 power if the load torque remains the same. However, my last point didn't confuse or introduce anything that should not be considered. Comparing the power draw of the 100% speed and 100% load case to the power draw of the 50% speed and 100% load case is 100% wrong.

 

[mikekilroy]

LionelHutz (Electrical)
11 Aug 12 12:01
....Comparing the power draw of the 100% speed and 100% load case to the power draw of the 50% speed and 100% load case is 100% wrong.

sorry, you are wrong for most cases. see some of my examples FYI. so be it. we will agree to disagree.

 

[mikekilroy]

417LionelHutz (Electrical)
11 Aug 12 12:01
I have no problem understanding that a motor run at 1/2 speed is running at 1/2 power if the load torque remains the same. However, my last point didn't confuse or introduce anything that should not be considered. Comparing the power draw of the 100% speed and 100% load case to the power draw of the 50% speed and 100% load case is 100% wrong.

Lionel, what you said in previous post of 10 Aug 22:40 was acceptable although potentially misdirected as I said since most processes don't work that way. But this post is flat wrong.

Let me do the math on this one once more, using 230v voltages and 10 amp motor current for example:

100% speed means 230v input vfd voltage to output 230v plus 10 amps load means 10amps input current so power used is 10*230*1.73= 3,979 watts. Hence, power in = power out (plus small losses). We agree.

But 50% speed means 230v input vfd voltage to 115v output plus 100% load means 10 amps output. so power out = 115v*1.73*10amp = 1,989 watts out. So power of 1,989 watts/1.73/230v= 5 amps draw, not 10. 1/2 power in for 1/2 speed. No arguing this point I am afraid.

Let's agree to disagree: You say most times when a vfd is used to run slower than the motor's base speed the motor will have to run longer to make up for the missing work. I say nonsense and gave examples.