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Discussion Starter #1
A few years ago we looked at a VFD system for our pumps that was solely to save money by throttling the pump to follow demand more closely. And it might have done other tricks. In the end we never bought it because the cost was nearly $4000, and the savings would take many years to pay for itself.

However, as many of you know, electrical line charges are getting out of control in Alberta, going up at about 10% a year (over the last four years or so). If this continues, within about ten years we'll be paying more line charges than we do for the actual electricity. Fortis also now calculates your fixed charge based on the peak use. This means if you start your 100 HP pump, the demand during start peaks at about 130 HP, and that is what Fortis bases your fixed charge rate on. So I've been running some numbers and suddenly the VFDs are looking to be more economical. If I can reduce my peak demand to my actual use (slow start the motor), that's a reduction of about 30% a year of my line charges, which will soon be a fairly significant thing. I heard of a guy somewhere that reduces his charges by spinning up the pumps with a pto before turning on the electricity!

Have any of you used VFD in your operation? Prices have come down somewhat for the technology. I figure if a VFD for a 100 HP pump could probably pay for itself in just a few years, if the rates continue to spiral upwards. VFD also opens up the way for variable-rate irrigation some day when that becomes affordable.

Any advice or experience you want to share?
 

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Peak Electrical demand is Calculated on the energy meter over a 15 minute period. If your motor takes 15mins to get up to speed then the result would be a larger demand reading on the demand value of the meter. My experience with these units would tell me that the motor will use whatever its need as far as power is required. A VFD unit is solely used for speed control for AC motors. ( Done by modifying the frequency applied to the motor.) One of the side benefits are a slight reduction of power used. I don't think it would equate to 30%. As a note, if you reduce the frequency to the motor the current through it will go up possibly leading to uneven heating within the motor and risking burning it out. These values can be calculated to see if it works in the field but probably can be a close approximation. If you are using it for irrigation, Your tolerances for a VFD to work will be tight if your motor is sized to your load. If your motor is oversized, the VFD will be a better fit as it will right size the unit to the load giving you a better power factor and a smaller bill.
 

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Discussion Starter #3 (Edited)
I agree it won't save 30% in energy use. At most it's just a few percent if the motor is sized correctly, as you say. So I'm not looking at it to save drastic amounts of net electricity.

That's interesting about how you say they calculate the peak demand. Perhaps we're talking about different things. I'm talking about what Fortis calls "demand KW max." I personally haven't been looking at the bills (my father has been the one analysing them), but I understand that the bills state this. And for all our pumps Fortis has a number 10-30% higher than the normal runtime KW usage. This number can come from nowhere else but motor start. EDIT: this number also shows up on the meter itself as well as the bill, if I'm not mistaken.
 

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Discussion Starter #4
I was wrong about the fixed charges being greater than the actual net electrical rates in ten years. It's actually now. Last year our fixed charges were over 56% of our total electrical bill (and rising). And a chunk of that is based on peak demand KW, which is why I'm analyzing the effect of VFD on this. However, if you're correct, then VFD wouldn't influence it at all. But our bills suggest otherwise.
 

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My understanding is peak charge is the shock load caused by the motor starting, that's how Fortis has explained it to me. I've been doing a little shopping for VFDs myself and doing calculations for our hundred horse pumps. Like you I'm on the fence and still trying to determine if it will pay. I really like the idea of variable rate water application also. We have a few older systems that are acting up and they would be ideal candidates for replacement. I'm shopping for US suppliers for components to see what I can save.
 

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I know the saskpower rate structure better and you could look up saskpower website to confirm. I will use saskpower rate structure to explain the general idea.

Saskpower has a component called "kva demand", not kw demand. As jd says, the demand is set over a period of time, 20 minutes I believe for saskpower. If the you run a max load of 75 kva for 20 min or longer, then thats your demand for that month. Doesn't matter if you drop down to 10 kva for the next 23.5 hours and 29 days of that month. You can peak at 150 kva for 30 seconds and drop down to 75 kva for the next 19.5 minutes and your kva demand will be around 75 kva.

For saskpower farm accounts we don't get billed a demand charge if our kva demand is under 50. Only above that do we need to worry.

I can get into a long explanation of kva vs kw, but best to read up a bit. Basically a kva is the amount of power drawn by a motor, kw is the actual work done by a motor. Kw = kva x power factor x efficiency. If I had your motor nameplate data I could calculate for you. A general guideline is 1 hp = 1 kva and then you can improve from there. I have always seen demand as kva, not kw so you should read your bill closer.

Where vfds save energy depends on how you normally control flow. If your pumps normally run at full output then a vfd at full output won't save you much. If you close your valves to reduce flow by 50% and the motor runs that way 70 % of the year let's say, and so your motor is working against that head pressure, then you could save by putting in a vfd.

And a regular motor will work fine with a vfd over a range of 50 to 150% of the motor nameplate speed.

There is more to it. I've done a lot of power studies and supplied equipment to reduce demand and a short note doesn't cover everything. But if you have a 100 hp motor and it is needed to do 100 hp of work then you will always have a demand charge. We can get your demand as low as possible, perhaps 85 kva for that 100 hp motor but can't make it go away completely.
 

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I read the rate structures for fortis and it is kw demand and over 15 minutes so that makes it easy. A 100 hp motor loaded to 100 hp and running steady for 15 minutes will take around 81 kw demand (100 hp x 746w / 0.92 efficiency). Motor starting is less usually under 10 seconds so you won't impact your demand with startup.

You get your first 40 kw demand for free so the only ways to really reduce demand is to use a smaller motor some how. Can you run a 50 hp for twice as long, run two 50's from two different services maybe. If you could do that you wouldn't have any demand charges.

A vfd supplier should be able to give you some numbers on kwh savings with variable pump speeds.
 

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Discussion Starter #8
I agree with offroadnt here. Despite what Fortis claims, their peak demand kw is not based on a 15 minute average. Like I say, my peaks are all 10-30% (or more) higher than what's on the label of the pump (and what I have measured with my amp clamp meter). And I'm not sitting there turning pumps on and off to get this peak either.

And I'm not that interested in kwh savings during runtime, though if there were savings there, that'd be a bonus. But where the VFD could pay for itself is in reducing this shock load peak kw fixed rate charge.

I'm also looking into US (or international) suppliers. There are issues with importing equipment though. They need to meet Canadian electrical code. My cousin bought a couple of large units from the US to run turbines with (8 pivots or so off of the two turbines). I need to ask him more about this. His turbine project is going in sometime next month.
 

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Not Sure what you mean by Shock loading. The meter will not capture this unless it has a duration of longer than 15 Mins. I have worked in the Utility industry for 28+ years and the meters in AB are set up and programmed by Either Fortis Alberta or the Meter Manufacturer to Industry Canada Specs. On the older meters demand was posted on the face plate to register ( Capture ) 90 % after 10 Mins of use and 99% after a 15 Min Period. This was on the older Thermal Demand Meters ( Glass Face ). The industry now uses mostly Electronic meters which mimic the old thermal demand meters using electronic algorithims to measure demand.

When the industry used the old glass thermal meters you as a customer were either billed on Killowatts ( KW, real Power ) or Kilovolt Amperes ( KVA Apparent Power ) Each meter had a different item number and was placed onto the service dependent on how efficiently the customer was using the power service. If your power factor ( defined as the ratio of KW/KVA) was poor or less than 0.9 or 90 % you were issued a KVA meter onto your service. If you did some work by installing power factor correction by adding capacitors, you could bring the power factor under control and save quite a bit of money.

Nowadays with the electronic meters, the utility only employs 1 style of meter which can do either KW or KVA demand DEPENDING ON THE POWER FACTOR the meter sees.

The meter read is sent every day to the utility server and occasionally the demand is read and reset once a month. The date and time of the maximum demand is recorded in the system, if everything is working.

Sorry for the long explanation. This cannot be explained in 1-2 lines.

During my work day ( Job # 2 ) I can see many services with poor power factors while field testing meters as low as 40% or .4.

I believe with motors of this size a VFD can help correct the power factor if you are not employing Power factor correction. I have seen many Oilfield sites that employ many VFD's for lifting , Fan Speed control or pipeline pumps run Power Factors in the range of 95-98 % as they like to optimize the power consumption of their facilities. This would save you money no doubt and would pay for the VFD in short order.

Get an electrical expert on-site who you can trust not someone selling you expensive hard to run crap that won't pay back and analyse your system. If its one motors this would not be a lot of work.

My feeling is that your motor as it sits is not optimized for your service, Motors can draw more than nameplate data, and the utility meter seldom is wrong. Seen that less than 5 times over the years.
 

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Peak Electrical demand is Calculated on the energy meter over a 15 minute period. If your motor takes 15mins to get up to speed then the result would be a larger demand reading on the demand value of the meter. My experience with these units would tell me that the motor will use whatever its need as far as power is required. A VFD unit is solely used for speed control for AC motors. ( Done by modifying the frequency applied to the motor.) One of the side benefits are a slight reduction of power used. I don't think it would equate to 30%. As a note, if you reduce the frequency to the motor the current through it will go up possibly leading to uneven heating within the motor and risking burning it out. These values can be calculated to see if it works in the field but probably can be a close approximation. If you are using it for irrigation, Your tolerances for a VFD to work will be tight if your motor is sized to your load. If your motor is oversized, the VFD will be a better fit as it will right size the unit to the load giving you a better power factor and a smaller bill.
A VFD not only changes the frequency to the motor, but adjust the voltage as well, in order to a maintain the the V/Hz ratio, at 7.67V/Hz. So at 45 Hz, the output voltage will be 345 volts, and at 30 Hz, the output voltage will be 230 volts. With lower voltage, there is less force to overcome the BEMF of the motor, so there is less current draw by the motor. The drop in current in not linear either, in other words, at 30 Hz the current draw in not going to be half. Let's take a 100 Hp motor, that has a full load amps of 111 amps, at 60 Hz with a power factor of .88. At 45 Hz (75%), the motor will draw 63 amps, and produce 43 Hp, already at 45 Hz, the reduction in amps is nearly 50%. At 30 Hz, (50%) this same motor, will draw only 40 amps, and produce 18 Hp. So yes, the savings can be significant. We have several on corner systems, when the corner system is shut off, the drive will operate in the 40 to 45 Hz range, it is only when all of the sprinklers on the corner system are on, is when the drive will operate at 60 Hz.
 

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All good info guys, and nice to know how the utility elects to use KW or KVA demand billing, in Sask it is only kVA demand and so power factor correction is a big component of reducing kVA demand. But we need to keep a lot of things in context, for example, when that VFD is running at 60Hz and that 100 HP motor is the right sized motor for the job and it runs for more than 15 minutes straight in that month, then you have set the new demand. The VFD won't reduce the demand, the size of the motor and how much it is loaded will set the kW demand. Also, if the pump runs two months a year you may find it harder to justify the kWh reducition by being able to run the motor at say 60 HP instead of at 100HP with a throttled output (non-VFD) operation. For industrial sites where the motor runs 24h x 365 days at a range of 50 to 100 HP at any given time then the savings in kwh add up because the amount of hours run at less than 100HP load. Numbers need to be crunched before justifying vfd for kwh savings. VFDs can also be used solely to reduce impact on starting to mechanical drives, but a pump/fan not so much because they start unloaded - only as speed and thus flow build up does the torque on the shafts increase.

Keep sight of what you want to accomplish, know the theory and application and decide from there what will work. I can think of many applications where the VFD is definitely the way to go. For KW demand reduction as defined by the Fortis site and what industry knows it as, then a VFD won't help there. And would need to know more info before saying the kwh reduction will pay for the VFD also.
 

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Discussion Starter #12
Wow, good information, Ken, ncrc5315, jdhrasko. Thank you for sharing your knowledge and getting me, and hopefully others, down a correct track.

Is there a way to (safely) measure power factor on a pump with things like a volt meter and a clamp ammeter? I can measure volts and amps, but no idea how to measure watts (I have a good handle on DC but certainly not AC). Is this something a normal electrician can measure, or is it more in the purview of folks like Ken who do measurements and service optimization for a living?
 

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Wow, good information, Ken, ncrc5315, jdhrasko. Thank you for sharing your knowledge and getting me, and hopefully others, down a correct track.

Is there a way to (safely) measure power factor on a pump with things like a volt meter and a clamp ammeter? I can measure volts and amps, but no idea how to measure watts (I have a good handle on DC but certainly not AC). Is this something a normal electrician can measure, or is it more in the purview of folks like Ken who do measurements and service optimization for a living?
A fluke 199 scopemeter, will measure PF. It can also be calculated, but I would have to dig up the calculation. To calculate watts, measure the voltage x amps = watts. 460 x 30 amps =13800 watts, or 13.8Kw. To calculate Hp: (Amps x volts x (sqrt of 3) X PF)/765 or
(30 amps x 460 volts x (sqrt or 3) X .88 (PF))/765 = 27.5 Hp, approximately.
 

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Yeah NCRC, there are probably some reasonably priced scope meters these days and they would give a power factor and actual kw. What you need to know when you measure amps and volts with regular multimeter that you are actually reading the VA, not the watts. You apply you power factor calculation to your VA reading to get watts. That is why VA is also known as "apparent power" as it is easily measured and is also what goes into a motor, or other load with a power factor, it doesnt actually indicate how much work is being done. So, for a three phase load, you measure phase to phase voltage and one of your phase currents and get something like
Kva = 600v x 100a x root 3 (use 1.73) = 103.9 kva

Then to get kw you take the 0.88 pf example and go 103.9 x 0.88 = 91.5 kw. A good double check is that what ever you measure (kva) will be greater than the kw.

Or if you know the kva and kw then your pf = kw/kva
 

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Discussion Starter #15
Thanks everyone for setting me straight here. I wasn't questioning Fortis' accuracy, rather I didn't understand where their numbers were coming from, how they were calculated, or what they meant really. With what I've learned, here are some numbers I ran, based on some amp readings I took on my pump this fall:

48 amps/leg * 3 leg * 277 volts = 39900 VA (39.9 KVA)

Now if I calculate how much power is actually getting transferred to the motor shaft (well not counting friction and other losses), using an estimated power factor of 0.8:

39.9 KVA * 0.8 (PF) = 31.9 KW * 1.34 KW/horsepower = 42.8 hp

That's not far off the sticker rating. The nozzles are all worn, so I'm not surprised the pump is running a tiny bit over spec.

If I add in estimated maximum pivot amps, then I get a KVA that's not far off of Fortis' peak demand KVA.

So, conclusion, VFD will not help my fixed charges (in some cases I've read it can make the PF worse).
 

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Yeah NCRC, there are probably some reasonably priced scope meters these days and they would give a power factor and actual kw. What you need to know when you measure amps and volts with regular multimeter that you are actually reading the VA, not the watts. You apply you power factor calculation to your VA reading to get watts. That is why VA is also known as "apparent power" as it is easily measured and is also what goes into a motor, or other load with a power factor, it doesnt actually indicate how much work is being done. So, for a three phase load, you measure phase to phase voltage and one of your phase currents and get something like
Kva = 600v x 100a x root 3 (use 1.73) = 103.9 kva

Then to get kw you take the 0.88 pf example and go 103.9 x 0.88 = 91.5 kw. A good double check is that what ever you measure (kva) will be greater than the kw.


Or if you know the kva and kw then your pf = kw/kva
My Flukes are true RMS meters, so I have the capability to read actual VFD output, I forget that most meters can't read this.
 

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Wow, good information, Ken, ncrc5315, jdhrasko. Thank you for sharing your knowledge and getting me, and hopefully others, down a correct track.

Is there a way to (safely) measure power factor on a pump with things like a volt meter and a clamp ammeter? I can measure volts and amps, but no idea how to measure watts (I have a good handle on DC but certainly not AC). Is this something a normal electrician can measure, or is it more in the purview of folks like Ken who do measurements and service optimization for a living?
Take a strong magnet on top of the Landis and Gyr Power Meter you have , swipe the magnet over the 12:00 Position on top of the meter. This will temporarily put the meter in an alternate mode. This will give you at the display instantaneous values the meter is seeing along with the Power factor. No need to buy one, A Fluke 430 or Fluke 1750 Power analyser meter can be rented from some of the larger rental shops in Alberta. I have come across them during my days work. No idea what they rent for.

I Have attached some reference for the values you will see in the alternate mode. There is no issues for you to do this as a customer. TOU in the table stands for "time of use" Rates which at the moment are not in play in Alberta.
 

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You are ok to do this , only works with the black demand meters, we use this at work all the time, I also have software to read an query the meters on a more technical nature. By doing this there is no way to affect the meter in any way.
 

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Discussion Starter #20
A few more questions, possibly naive questions, for the experts.

1. Power Factor in essence means that a certain percentage of the apparent power that is entering my pump is returning to the grid, correct? (if current leads or follows voltage, then the power waveform is negative during some period of time each cycle.) Thus the pump is demanding a certain amount of power, but the grid has to deliver that power plus a little bit extra that gets returned.

2. What happens to this power that returns to the grid? Is it lost to everyone or is it used by another load on the system?

I realize as I type this that while transmission charges are based on apparent power, the generation charge itself is purely kwh, so in that sense I'm not paying for electrical generation my motor is returning to the grid. However the transmission charges take into account this extra power being delivered, though it is returned.

3. Finally, is it feasible, to install capacitors to raise the power factor closer to one? Just for information sake at this point.
 
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