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VFD confusion
11

VFD confusion

VFD confusion

(OP)
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?

RE: VFD confusion

7
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 .53, 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)

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RE: VFD confusion

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.

RE: VFD confusion

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: Eng-Tips.com Forum Policies

RE: VFD confusion

(OP)
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).

RE: VFD confusion

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

RE: VFD confusion

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.

RE: VFD confusion

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.

RE: VFD confusion

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.

RE: VFD confusion

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.


RE: VFD confusion

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.

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)

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RE: VFD confusion

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.


RE: VFD confusion

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.

RE: VFD confusion

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.

RE: VFD confusion

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.

RE: VFD confusion

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%.

RE: VFD confusion

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.

RE: VFD confusion

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.

RE: VFD confusion

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.

RE: VFD confusion

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.

RE: VFD confusion

Mike,

What you appear to be overlooking is some real fundamental stuff. I'm sure you understand it, you are just not registering it. Do you agree that if you need to do a certain amount of work then it requires a certain amount of energy input? I'll assume that you agree with that, because it is conservation of energy and if that's wrong then a whole lot of other things also start to fall apart. Can we further agree that the required energy to to do the work can be delivered at a fast rate, or at a slow rate, but it is nonetheless the same amount of energy for the same work?

Where I think you are going wrong is that your assumptions result in the the moving mass on the conveyor increasing, but you don't account for that in your maths. If you slow the conveyor to 50% speed then you must slow the loading rate to 50% speed, otherwise the mass on the conveyor will increase and the reduction in speed no longer has a linear relationship with the work being done, and you can not assume that the power consumed by the motor is reduced to 50% by reducing the speed to 50%.

RE: VFD confusion

Geeze, when the process requires 50% of the maximum material the conveyer can supply then you have to find the most efficient way to deliver that amount of material. You can't compare this power usage to the full 100% material delivery case. You only need 50% of the conveyer capacity. What are you going to do if you don't slow the conveyer, dump 50% of the material on the floor? No, you will cut back the feed onto the conveyer if you don't slow it down. So, you must compare 100% with a reduced fill vs a slower speed and 100% fill.

RE: VFD confusion

guys, you are mixing your metaphores. You realize that to so a certain amount of work requires a certain amount of energy. What you miss is why a vfd is used: it is not used to do the same amount of work in most applications, it is used to vary speed on a motor shaft to help a process. See my examples above; none use a vfd to put out the same energy at the end of the day. If you believe a vfd is only used for your narrow minded application of a conveyor that does not need different speeds for different amounts of product then so be it.
everyone realizes you all understand conservation of energy, no sense beating it; I see you will not accept that the most common use of a vfd is to vary the speed of a moor shaft, not to conserve someone's energy at the end of the day, so that is that. thanks.

RE: VFD confusion

by the way, as another example of why folks use a vfd on the conveyor, see two posts down in this forum:

http://www.eng-tips.com/viewthread.cfm?qid=327706

they have need for varying speed due to, as I suggested earlier in my example, varying product based on what they run that day. this is another example of why vfds are used.

RE: VFD confusion

they could have just left the vfd off the motor and ran it full speed all the time. but instead, they will reduce their energy bill each month by the amount of days they run it slower than full speed. energy savings by vfd.

RE: VFD confusion

Mike,

Your example is the one where we got into conservation of energy, with the assumption that halving the speed halves the current from the line. It's not impossible that this will happen, but it is highly improbable because most motors in industry drive either pumps or fans, neither of which would result in the change in line current which you seem to expect in your example.

You're right that the most common use of a VFD is to vary the speed of a shaft. One of the great misconceptions - and one for which the VFD manufacturers are largely to blame - is that using a VFD will inherently save energy. It won't: work is work, done quickly or slowly. A VFD can allow an over-sized motor to drive its load more efficiently in the right circumstances. For the addition of a VFD to allow energy to be saved, the constant speed application first has to be wasting energy. This is by no means a given.

RE: VFD confusion

ScottyUK ....the assumption that halving the speed halves the current from the line. It's not impossible that this will happen, but it is highly improbable because most motors in industry drive either pumps or fans, neither of which would result in the change in line current which you seem to expect in your example.

Scotty, we did seem to get off on an energy vs power tangent. I respectfully suggest that the real issue was the disbelief in the above fact, which is so, even for fans and pumps as well any and all applications where torque at the lower speed does not INCREASE. I would be happy to show the math again why this is 100% true if you'd like.

RE: VFD confusion

OK, I'm not arguing that reducing the speed of the motor won't show a reduction in the input power. I'm telling you that the reduction in input power doesn't necessarily mean you're saving money.

Here's a different example. You have to feed process water. You have 2 choices. Install a tank and use on-off control for the pump or install a VFD and vary the speed of the pump. The VFD will use less power during the low water demand times compared to the high water demand times, but it will use more energy at the end of the year compared to the tank. Sure, you can argue that the VFD saves energy compared to running the pump full-out all the time but the process doesn't require running the pump full-out all the time so it's a useless comparison.

The thread you linked is about changing the speed to suit a process. If the poster get paid by the pound for the cucumbers then you have to look at the energy used to move the cucumbers and the amount he gets paid for them. Sure, the conveyer will use less power and energy when running slower, but he likely will get paid less due to running less product weight per period of time. In this case, the $$ paid for the product vs the energy cost to run the product must be examined to see if money is being saved or not. I personally hope he's getting paid more per pound for the smaller cucumbers because he's also paying people to inspect X number of cucumbers per day and running smaller cucumbers means more manpower cost per pound of cucumber.

RE: VFD confusion

Lionel,

Of course we all can agree that if one has a process where they slow the conveyor down TOO slow to keep up with the cucumbers, they will loose money. OK. But, reality is that the numbers of cucumbers is not a constant: If it were, farmers could guarantee their income: they can't. Farmers who rent my land pay farmers insurance for this reason; if the crop is really bad, they get paid back the missing income. We rent approx. 300 acres to farmers of corn and soy beans here in Ohio..... these poor farmers are hurting this year due to the water drought. Reality is no one can predict exactly how many cucumbers will be made per unit time. The result is that farmers may produce LESS or MORE for a given crop per year. So, the conveyor moving them CAN CHANGE SPEED TO ACCOMODATE THE VARIABLE VOLUME. That means that yes, they don't have to have the conveyor running at base speed if this year is less cucumbers than last year. They can save money by reducing the speed of their conveyor. Simple math.

OK, off cucumbers. Onto widgets. See 2 posts down as I suggested: I propose that if you do a study and ask ALL VFD USERS WHY THEY BOUGHT A VFD you would find a high majority bought them to slow a process down - or speed it up - based on the product required. Yes, I give you there are a FEW processes that that will require the same amount of work (energy) to get the job done as you keep insisting. The village will use the same amount of water as you suggest per week whether it is sent slowly to the tank or fast to the tank. But I suggest to you that THAT is the EXCEPTION, not the rule.

If there is a way on this forum to make a survey, I will accept the majority answer: is a process using a VFD most likely to require constant work or constant power? Let the games begin!


Assuming you too, along with Scotty (beam me up Captain), require the math for why a VFD DOES 100% save power if turned down, I recorded my barn fan for you at 15hz, 30hz, and 60 hz. Rememember as you watch my video that a fan torque goes up by the CUBE of the speed; be that as it may, the video below clearly shows that turning a VFD down in speed SAVES POWER. I do not require the high air movement rate on a cooler day; hence I am allowed to SAVE MONEY BY TURNING IT DOWN! You cannot deny the imperical data, sorry. If you wish to post a similar real world test to show I am wrong, please do so. Here is the video I posted to something called U tOBe I think..... I am sorry, I am not an accountant, but an electrical engineer so do not know how some of this stuff works.

BTW, if someone is young enough to want to go for a patent on saving energy with a slight VFD modification that is patentable, let me know: I am old enough & well off enough after 40 years at this that I don't need it, but it would be a shame to leave it for another 10 years to be discovered...... back to topic, here ya go my friend. Actual data recorded off MY BARN FAN TODAY for you. :)

I will summarize: IT CLEARLY SHOWS THAT THE POWER USED ON A VFD DRIVEN MOTOR GOES DOWN LINEARLY WITH SPEED IT IS RUNNING: THAT TRANSLATES DIRECTLY TO $$ ENERGY SAVINGS IN YOUR ELECTRIC BILL.

http://www.youtube.com/watch?v=reo7CZLyHLg&fea...

]MOTOR LOAD SIDE OF VFD on faN
SPEED AMPS WATTS

60....... 5.3..... 805
30....... 4.6..... 427
15....... 4.5..... 284

INPUT SINGLE PHASE INPUT SIDE OF VFD (1.73x higher than 3phase)
SPEED AMPS WATTS

60...... 7.9...... 968
30...... 3.1...... 521
15...... 2.5...... 223

INPUT *IF 3PH* INPUT SIDE OF VFD (1.73x higher than 3phase)
SPEED AMPS WATTS

60...... 4.6
30...... 1.8
15...... 1.4

RE: VFD confusion

Your last post means nothing. You're stuck on this "a power reduction saves money" theory and haven't read or acknowledged anything about how you have to ensure the lower speed actually saves money. Saving power by running the motor slower doesn't automatically translate into saving money.

At this point, I've got nothing more to say. It's useless giving another example. The way you twist and change things and also the way you argue points I'm not even disagreeing with make it very clear you're not comprehending.

RE: VFD confusion

agreed. you will not accept putting 1/2 the power into a motor saves money. ok.

RE: VFD confusion

OK, I'll take up the cause now, you've worn out the others...

According to your argument then, this will equivocate to any process in which less work is done as long as less energy is consumed. So let's say I need to drive from San Francisco to Los Angeles with 50 people in a bus that gets 10 miles per gallon. If your logic holds true, I can get a smaller bus that only holds 25 people but gets 20 miles per gallon and I have saved energy! Forget the fact that 25 people stayed behind and didn't get to LA, that's irrelevant according to you. The fact is, I used a bus that consumed less fuel, ergo energy was saved. Right?

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)

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RE: VFD confusion

if you too agree that most vfd's are used in order to do the same amount of work as before they were applied, then ok. Truth be told, I think a survey would show that most vfd's are applied to help a process, not to do the same amount of work in a given time, so that argument is not valid. If I don't need the fan running 1800rpm, I can turn it down and be cool and save power, energy, and money. It is just sad this point is not understood. so be it.

RE: VFD confusion

Huh? You are thinking we are saying it will do the SAME amount of work as before the VFD is supplied???? That's exactly the opposite of what we are saying. If you turn down the speed, LESS work is being done. So the energy you "save" is invalid, because you are not doing the same work. Less energy with less work is just less energy USED, but is not energy SAVED; i.e. I think maybe you are equating energy REDUCTION with energy SAVINGS. There is a difference.

There is some sort of comprehension disconnect here that makes me think we are not saying different things, just approaching it in different ways.

"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: Eng-Tips.com Forum Policies

RE: VFD confusion

THE AMOUNT OF WORK BEING DONE DOES NOT MATTER TO THE ARGUMENT. I have even given at least 2 examples (I actually think there were 3) where the amount of work being done per day varies. Yet, you can't seem to comrehend these examples because you misquote and mis-reply to them.

With a process, you can either control the process by using a VFD or you could vary the process by some other means. BUT, YOU MUST CONTROL THE PROCESS BY SOME MEANS. Examples of other means to control a process;
Fan - damper
Pump - throttling valve
Conveyer - sliding gate on the feed hopper
Milling machine - gearbox
etc
etc

THE ENERGY SAVED (or the extra energy wasted) IS THE DIFFERENCE BETWEEN THE ENERGY USED BY THE VFD AND THE ENERGY USED BY THE OTHER PROCESS CONTROL MEANS. IT IS NOT THE DIFFERENCE BETWEEN THE ENERGY USED AT 100% THRUPUT AND THE ENERGY USED AT THE REDUCED SPEED.

If you can't comprehend the above 2 sentences then you shouldn't be telling anyone how a VFD will save energy. You're simply not qualified for that task.

RE: VFD confusion

(OP)
My very simplified understanding of the answer to my question is that you must change the amount of Power used. You can do this by either reducing the work done or by reducing the speed you are doing it at. In the end the process requirements for the conveyor line dictate the amount of material you need to move (i.e. the amount of work you need to do) and what speed you need to do it at. Simply using a VFD instead of gearbox won't save you energy, or very little if you want to look at VFD vs. gearbox energy losses.

RE: VFD confusion

Is it possible to replace gearbox with VFD for speed reduction in liquid ring vacuum pump application.

RE: VFD confusion

jraef (Electrical)
12 Aug 12 21:51
Huh? You are thinking we are saying it will do the SAME amount of work as before the VFD is supplied???? That's exactly the opposite of what we are saying. If you turn down the speed, LESS work is being done. So the energy you "save" is invalid, because you are not doing the same work. Less energy with less work is just less energy USED, but is not energy SAVED; i.e. I think maybe you are equating energy REDUCTION with energy SAVINGS. There is a difference.


that is exactly what I have been saying. Never have I said reducing motor speed by itself will reduce energy to do a given task; a given task needing x amount of energy will need x amount of energy to complete irregardless of speed.

these 2 points are the crux of it: if no vfd is in the picture, there is no chance to slow a motor down and save power and hence energy - assuming the process can benefit from it - or said in your words, assuming the process can benefit from the energy savings.

again, most processes where a vfd is used can benefit from, in your words, this energy savings, by turning the speed down. we can and have all given examples of this as well as processes where the energy required is a constant and a vfd will not save energy.

RE: VFD confusion

LionelHutz (Electrical)
12 Aug 12 22:38
THE ENERGY SAVED (or the extra energy wasted) IS THE DIFFERENCE BETWEEN THE ENERGY USED BY THE VFD AND THE ENERGY USED BY THE OTHER PROCESS CONTROL MEANS. IT IS NOT THE DIFFERENCE BETWEEN THE ENERGY USED AT 100% THRUPUT AND THE ENERGY USED AT THE REDUCED SPEED.


nonsense. energy saved by a vfd is certainly saved energy if you can turn the motor down and use less energy. You feel the extra money in your pocket at the end of the day is not real money?

RE: VFD confusion

Quote:

if no vfd is in the picture, there is no chance to slow a motor down and save power and hence energy

Is this a craftfully written statement to try and prove your point. Without a VFD you may not be able to slow the motor down, but there are still lots of ways to save power and energy.

What do you think happens to the conveyer power usage if you partially close the gate on the feed hopper?
What do you think happens to the power usage of a pump if you partially close a throttling valve on a pump?
What do you think happens to the power usage of a fan if you partially close the dampers or you lower the pitch of the blades?

Can you answer these 3 questions?

RE: VFD confusion

Quote (Mysterrose)

My very simplified understanding of the answer to my question is that you must change the amount of Power used. You can do this by either reducing the work done or by reducing the speed you are doing it at. ...
Sigh...

Almost, but I'm afraid that this side track has confused you a little.

... you must change the amount of Power used. You can do this by either reducing the work done or by reducing the speed you are doing it at reducing the wasted energy in the system. Reducing the speed WILL reduce the work done in many cases. If it does not, then there is no change in the energy consumed. Energy = Work. But reducing the work done results in energy reduction, not energy savings, because your process produces less of whatever it is intended to accomplish.

"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: Eng-Tips.com Forum Policies

RE: VFD confusion

I'll bite. It seems mikekilroy is making an argument for when VFDs do save energy by slowing down the speed and Jraef et al are making the argument that they don't ALWAYS save energy by slowing down the speed. I think we all agree that there are apps that work. Conveyors can fall into that category, but not always. I agree the hype is misleading, slowing the motor speed does not necessarily save energy/money at the end of the day.

RE: VFD confusion

How about this:
If you subscribe to the notion that REDUCING energy is the same as SAVING energy, then isn't the ultimate goal going to be to turn it off completely? You cannot save any more enegy than when the machine is off.

But of course, it is doing no work at all. So what have you "saved"?

"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: Eng-Tips.com Forum Policies

RE: VFD confusion

John2025 you are right on, yes. we all are saying the same thing now, except when carried to extreme as Jraef shows can be done by turning it off.... the anal issues being thrown out here are that some folks think there is no inbetween full on or full off work....

If I am working on the tractor in the hot barn and turn my fan on w/o a vfd, I will use .75hp (1kw) of power running my fan across the line full speed for the whole 1 hour it takes me to fix it, thus using 1kwh of energy.

Or, if I am working on the tractor in the hot barn and turn my fan on with a vfd at 1/2 speed (because it ain't as hot as it was yesterday and 1/2 speed is enough to cool me today), I will use .375hp (.5kw) of power for the whole 1 hour it takes me to fix it, thus using .5kwh of energy.

Lastly, If I am working on the tractor in the hot barn and turn my fan off to use no kwh of energy, I may not save a penny as the hospital bill for heat stroke will likely be much higher than the energy saved.

The moral of the story is there are processes as I have been saying that do not require the same work all the time, and thus can save money by turning the motor speed down.

RE: VFD confusion

Sorry guys, I should have admitted the horse was dead. I do wonder though, does beating the carcass slowly for a long time use less energy than beating it quickly for a short time? ;)

RE: VFD confusion

John2025 - No. mikekilroy is arguing that you use a VFD for process control and you will save the energy difference between full output and the reduced output. He's totally 100% ignoring the fact that without the VFD that another means to achieve process control is required. In fact, I now believe he thinks a 75kW motor will draw 75kW when running at full speed regardless of the load. Believing this fallacy would explain why he thinks the only way to reduce the power drawn by a motor is to reduce the motor speed and also would explain how to justify this 50% speed = 50% energy crap.

Personally, I think he's completely full of it. In his 11 Aug 12 7:42 post he claims that he designs motors and then in his 12 Aug 12 15:17 he claims he rents land to farmers and has also hinted that he is a farmer (fan in his barn, working on tractors). And to boot, he's got this great patentable VFD energy saving idea which he can't be bothered to persue because he's so well off. So, a motor designing land baron farmer who's to rich to be bothered to persue a new money making idea? Seriously?

It's easy to give an example of energy saving by using a VFD on a fan that was pumping excess waste air into the barn. The old system had no process control and the new system has process control. Using a waste energy application example is an easy way to cheat when proving that a VFD will always save huge amounts of energy. This is exactly the same as comparing a VFD controlled conveyer with a conveyer which is dumping the excess material onto the floor. This is exactly the same as comparing a VFD controlled pump with a pump which is dumping the excess water back to the inlet.

Try comparing a system with mechanical process control means vs updating the system with a VFD for process control. Do this - measure a the conveyer poer at 1/2 fill and full speed vs the conveyer power at 100% fill and 1/2 speed. You'll quickly find your 1/2 speed = 1/2 power & energy rule is complete crap.

RE: VFD confusion

I may be speaking from a glass house of past experience, but I agree. Horse is dead.

=====================================
(2B)+(2B)' ?

RE: VFD confusion

ha! that's funny loin! I'm glad you find my life experiences funny :) sometimes one can open the minds of closed minded folks, and sometimes one cannot.

yes, the horse died many days ago :)

RE: VFD confusion

My head is making mushy sounds. banghead

.

RE: VFD confusion

Well, sadly I fully expected you to ignore the 3 questions I posed. Looks like the horse is very dead.

John - sounds like we should apply for a research grant to study that one ;)

RE: VFD confusion

LionelHutz (Electrical)
13 Aug 12 16:03
Well, sadly I fully expected you to ignore the 3 questions I posed.


Sorry Loin, didn't think beating the dead horse anymore made any sense, but since you insist.... if one does something else to reduce the energy usage, guess what? energy usage goes down? Unfortunately, in my barn, I have none of those things on my fan; just a vfd. I get to reduce my energy usage by simply turning speed pot down. Someday you may open your mind enough to see the whole picture; there is more than one way to skin a cat. Saving energy dollars can be done many ways, including turning down a vfd speed pot if the excess motor energy usage is not required. Someday you may accept that, or not. If you want to continue this, for instance, your nonsense 75kw comment, off line, feel free to pm me and I will try to help open your mind to more ideas from the real world.

RE: VFD confusion

Geeze, you just must keep using your little barn fan example so lets pick on it a bit.

You keep going on about how your barn fan with the VFD uses 50% power at 50% speed. What would you think of your precious VFD if I presented a cheap easy to install product which would give you the same airflow but use 25% of the power? Would you still think your VFD is so wonderful? I bet not. Sure, the VFD is saving you 50% of the power compared to running at full speed, but it's now costing you 25% more power compared to my solution.

Now think on a grander scale. Lets scale up from a dinky little 1kW fan to a real fan, say 1000kW. I bet you'd want to buy my solution and throw the VFD in the scrap pile. After all, my solution is saving you 250kWh MORE energy every hour it runs at the reduced airflow and that amount of savings would add up to some serious cash at the end of the year.

RE: VFD confusion

3
mikekilroy - You're a new guy. Most people don't make such an entrance as you did here.

I think you should know that the three guys you've been conversing with (Lionel, Scotty and jraef) are among the most knowledgeable in this area (as in many others) on the forum. I'm not saying that means you have to agree with everything they're saying, but it's something you should know so you should consider before statements like "feel free to pm me and I will try to help open your mind".

And Lionel's name is Lionel.

Respect is a two way street. Give a little bit and you may get a little bit.

This thread is not heading anywhere productive, especially if you think the only thing left is to explain your point to the others (they understand it, believe me).

I'm not anyone different than anyone else here, and I don't really have a right to lecture anyone. That is not my intent, just trying to help out. If it were in-person vs forum I would have a private side conversation with you. We don't have PM's here, that's not an option. Really no other way to do it.

Personally, I hope you'll stick around and keep contributing in other threads.
Don't be put off by this thread, others will be different. I think you will grow to enjoy it.

=====================================
(2B)+(2B)' ?

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