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  1. #14
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    Quote Originally Posted by Achicagoperator View Post
    On a side note has anyone ever noticed while physically amping a load with a vfd on it that the line sides(inlet to the drive)amperage is lower than the loads(outlet of the drive)side.Could it be because of the capacitor storage banks?I have ABB drives.
    you are seeing the effects of Power Factor. a drive has a nearly 100% (or 1.0) power factor. the motor, however, still has its 35-80+ power factor. the amp draw on the motor will need to be higher in order to compensate for the lower power factor.

    keep in mind that the 'Power' (measured in kW) will be higher on the 'Line' side than the 'Load' side due to the inefficiency of the drive itself.

    Attached is a picture of 2 screen shots that I took on a York Drive on a YT chiller. The left screen shot is taken between the line disconnect and the 3 phase inductor (which is before the drive) and the right screen shot is taken between the drive and the motor. You can clearly see the power factor changes quite a bit between the line and the motor, but the Power on the motor is much less than the line side.
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  2. #15
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    You could check the accuracy of your meter by simply comparing it to the displayed amp draw which should be fairly accurate .

    Your basic meter is set up and calibrated for a AC sinosoidal wave form. VFDs output a frequency modulated square wave.

    Its actually alternating DC fired from the 800 volt internal DC bus via IGBT switching.

    Explains your capacitor bank. Never get near that dc bus if you like breathing.

  3. #16
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    When measuring the output to the motor it is usually not running at the same frequency as the input current. Let's take the losses of the VSD out of the equation. The input kVA has to equal the output kVA. If the input current is 100A @ 460 VAC then if the output voltage is 330 VAC (43 Hz) the output current has to be 139 A.

  4. #17
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    Back to the question at hand. If you are using a centrifugal chiller with a VFD. The lower the lift of the compressor, the more the VFD will drop. The only way that you can reduce lift is by removing heat. That is done with condenser water; reduce the flow and you will reduce the effectivness of the condenser.The water flow should be kept at the flowrate that the manufacturer has designed it for. Not to mention that when you lower water flowrate, the tubes will foul at a quicker rate. Lower condenser flowrate is a marketing strategy that some manufacturers uses. I would rather save money on a 500 amp chiller than on a 30 amp pump motor. The math just works out better for the customer.

  5. #18
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    Quote Originally Posted by KnewYork View Post
    When measuring the output to the motor it is usually not running at the same frequency as the input current. Let's take the losses of the VSD out of the equation. The input kVA has to equal the output kVA. If the input current is 100A @ 460 VAC then if the output voltage is 330 VAC (43 Hz) the output current has to be 139 A.
    Lol..... 20 bux its the meter. There are no appreciable losses with running a VFD with the exception of harmonics that they can produce that can cause overheating.

    Its a PWM non- sinasoidal DC square wave form hes measuring most likely with a hand held standard clamp meter designed for 60hz AC voltage.


    I went to Seimens VFD class many years ago when I worked for a manfacturer
    and baxk then there were only a couple of meters available that could give a accurate output represemtation of amps or voltage.

    Easy test for meter guy. Checks meters reading to VFDs reading.

  6. #19
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    Wow...This thread has gotten rather deep in VSD readings as well as proclaimed savings here and there...
    First off using any field meter, clamp-on etc. be it "True RMS" or not to attempt to get any semblance of KW in or out of "Any" VFD is a total waste of time and effort.
    The reasons are simple.....
    There ain't "Nothing" resembling a Sinosudal waveform anywhere near a VFD...
    Even the Hi-tech VFD's using "AFE" technology in lieu of diode rectifiers rip the power supply to shreds.( Active Front End - draws a pretty good sinusoid from the mains using an active switching input stage instead the harmonic-rich waveform of a diode rectifier)
    Sooooooo....Unless you have access to the toys Jayguy has to play with or can obtain a watt-hour meter to submeter the input to a VFD just trust the readings that the particular VFD is giving you on its readout.
    As Jayguy has shown it is the "Power" that is being sent to the VFD that one will be billed for and "Not" the "Apparent" power of the VFD output.
    Ain't "None" of us as smart as "All" of us..

  7. #20
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    Quote Originally Posted by jayguy View Post
    you are seeing the effects of Power Factor. a drive has a nearly 100% (or 1.0) power factor. the motor, however, still has its 35-80+ power factor. the amp draw on the motor will need to be higher in order to compensate for the lower power factor.

    keep in mind that the 'Power' (measured in kW) will be higher on the 'Line' side than the 'Load' side due to the inefficiency of the drive itself.

    Attached is a picture of 2 screen shots that I took on a York Drive on a YT chiller. The left screen shot is taken between the line disconnect and the 3 phase inductor (which is before the drive) and the right screen shot is taken between the drive and the motor. You can clearly see the power factor changes quite a bit between the line and the motor, but the Power on the motor is much less than the line side.
    thanks for the info and pics,makes a bit more sense now.

  8. #21
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    Quote Originally Posted by Commtec77
    Lol..... 20 bux its the meter. There are no appreciable losses with running a VFD with the exception of harmonics that they can produce that can cause overheating.

    Its a PWM non- sinasoidal DC square wave form hes measuring most likely with a hand held standard clamp meter designed for 60hz AC voltage.
    I disagree with you on two points. There are losses through a VSD. Converting AC to DC and back to AC costs money. At one time it was as much as 6%. I think it's probably less than that now, but the conversion still costs money.

    Secondly, the output waveform of 99% of the drives today are sinusoidal like PWM A/C (not DC) waveforms. Square wave VSDs went out with leisure suits. Unfortunately I am not at a computer where I can attach a drawing of a typical PWM waveform.

    One can measure current going to the motor with a Fluke 87 meter. As is being pointed out the customer is getting charged for the input current to the drive.

  9. #22
    Quote Originally Posted by JEF1980 View Post
    Really need more info on you system, but if you have drives now they most likley opened your triple duty valves to 100% and at lower loads depending on control system they lower the flow to a point at lower load conditions. Is your building automation controling the drives now?
    Yes that is what they have done using differental pressure switch's to control the VFD on the pumps. There is what they call a sales screen it says 1400 gpm for the condensor water is that a constant flow or a max flow?
    The chillers are York's two 550 ton and one 400 ton with three cell 500 ton towers (see the problem) I am 50 tons short on the tower when one chiller runs.

  10. #23
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    Quote Originally Posted by engineer12
    There is what they call a sales screen it says 1400 gpm for the condensor water is that a constant flow or a max flow? The chillers are York's two 550 ton and one 400 ton with three cell 500 ton towers (see the problem) I am 50 tons short on the tower when one chiller runs.
    The 1400 GPM shown on the sales order screen is the design flow. You can give it more or less within reason. Running one 500 T. tower with a 550 T. Chiller won't hurt much of the time. Even at 90% load you will be doing 495 T. at design flow and delta T. Under normal circumstances one would be bringing on a second chiller once the lead chiller was at 90%.

  11. #24
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    Quote Originally Posted by KnewYork View Post
    I disagree with you on two points. There are losses through a VSD. Converting AC to DC and back to AC costs money. At one time it was as much as 6%. I think it's probably less than that now, but the conversion still costs money.

    Secondly, the output waveform of 99% of the drives today are sinusoidal like PWM A/C (not DC) waveforms. Square wave VSDs went out with leisure suits. Unfortunately I am not at a computer where I can attach a drawing of a typical PWM waveform.

    One can measure current going to the motor with a Fluke 87 meter. As is being pointed out the customer is getting charged for the input current to the drive.

    Good points NY.

    First I said no appreciable losses not no losses at all. Also Ive spent a lot of time with Seimens 100hp VFDs in a test R and D setting with a oscilloscope monitoring output wave fors from VFDs. Modulated square wave dc and it wasnt that long ago.

    I installed a Yaskawa VFD a couple of years ago that was PWM DC square wave out put. Still have the manual. It contructs a pseudo AC sign wave using 6 IGBTs firing out MODULATING DC.

    The 6 step waveform inverters are obsolete but I havent gotten the memo about pwm dc square wave being obsolete.

    A sinusoidal wave form by its definition is constant and non changing in frequency. I dont see the value of going to the " ac pwm " waveform" and cant find any reference to it. The mptor "sees" the current waveform as ac sinusoidal even if its DC modulated.

    Maybe you can post a link so I can get up to snuff on modern drives. As far as meters Fluke had a couple that could measure accurate amp draw. Fluke 87 wouls need a cllamp attachment obviously to measure VFD current. I use to use a old analog amprobe that was prerry accurate.

  12. #25
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    We have a difference of opinion. IGBTs convert straight line DC to PWM AC. If the output was DC then you would have to have a DC motor. I can't see how you can get square wave DC when DC by nature is straight line. I guess I'd have to read their manual. I disagreed with a manufacturer when their national sales rep said their motor was a DC motor. Their literature showed IGBTs and a three lead motor...DC? Really? It is an AC motor.

    I guess we'll have to agree to disagree.

  13. #26
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    Talking

    Quote Originally Posted by KnewYork View Post
    We have a difference of opinion. IGBTs convert straight line DC to PWM AC. If the output was DC then you would have to have a DC motor. I can't see how you can get square wave DC when DC by nature is straight line. I guess I'd have to read their manual. I disagreed with a manufacturer when their national sales rep said their motor was a DC motor. Their literature showed IGBTs and a three lead motor...DC? Really? It is an AC motor.

    I guess we'll have to agree to disagree.
    Or maybe I can convince you. :-)

    First your observing a voltage PWM dc square wave. Because of the innate inductive properties of the AC motor it "sees"current waveform as a sinusoidal ac waveform.

    AC wave forms alternate peak to peak changing polarity and so does a PWM DC square wave. The properties of the modulating square wave allow for very specific control over the modulating frequency.

    If you saw a dc square wave on a oscilloscope it would have the same electrical properties as its AC counter part. The waveform just looks different. The motor is still a AC motor.

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