How VFDs save energy,
Power (P) = √3 x Voltage (V) x Current (I) x Power Factor (PF)
In an ideal VFD, the following would hold true:
Powerin = Powerout
But because a VFD has inefficiencies and requires a small amount of power consumption to power the brains of the drive, the input power will be slightly greater than the output power. For this, we will assume that this extra power draw is negligible.
With these two equations, we can then define the relationship between the VFD’s input and output:
Vin x Iin x PFin = Vout x Iout x PFout
Taking these equations into account, let’s use a 100-hp motor as an example with the following properties:
Power = 100 hp
Speed = 1,785 rpm
Voltage = 460 V
FLA = 115 A
Power factor = 0.86
Assume that the motor is running at 60 Hz on a VFD, drawing a no-load current of 40 A on the output of the VFD. With this, one would assume that the input current would also be the same, 40 A. However, using an ammeter on the drive’s input, a person is reading nearly zero amps! How is this possible? Is the drive creating power somehow? The answer simply is no, the drive is not creating power. The power factor causes this “discrepancy” in current When a motor is running at no load, the motor’s power factor can be assumed to be zero, not 0.86 86 as stated on the nameplate. The reason the power factor isn’t at 0.86 is because this is the motor power factor at full load. Alternatively, mechanical (friction) and electrical (resistive) losses in the motor prevent the power factor from being zero when running no load, but we’ll assume these losses to be zero just like we did for the VFD. Therefore, you would have the following:
Pout = 460V x 40A x 0
Pout = 0
Because the output power is zero, the input power also will be zero. With a fixed input voltage, the two variables would be current and power factor. Because current is needed for a power factor to exist, both current and power factor are zero, which means the low input current reading is indeed correct.
This explains why the input current to the VFD is so low when the motor is operating under no load conditions. But what about under load? The same concept still applies when the motor is under load. For example, assume the same motor is now operating at half speed, 30 Hz and producing full motor rated torque and drawing the motor’s full-load amps (FLA). This means that the electrical power that the motor is drawing is:
Pout = √3 x 230V x 115A x 0.86 = 39.4kW
Because the VFD is a power converter, this means that the input current is (assuming a 0.89 input power factor from a 3% line impedance):
Iin = (39.4kW) / (√3 x 460V x 0.89) = 55.6A
Because the VFD is operating at half speed and under full load, the input current is less than half of the output current.
In this example, the input current is less than half of the output, a result of having a higher power factor on the input side.
The difference in power factor between the input and output side of the motor is what makes it possible to have a higher output current than input current. Assuming the motor is now running at full load and using the same power factor values, your input current now becomes:
Iin = (√3 x 460V x 115A x 0.86) / (√3 x 460V x 0.89) = 111A
which is 4 A lower than the output current.
Powermechanical ∝ Speed x Torque
This means that if the motor is operating at half the speed and producing full torque, the motor is outputting half of its rated power. Consequently, if the motor is running at full speed and producing half torque, the motor is also outputting half of its rated power.\
Because of motor losses, the power relationship between the electrical power going into the motor and the mechanical power is:
PowerElectrical = (Powermechanical) / (EfficiencyMotor)
Revisiting the above example, if the motor is operating at 30 Hz, half the motor’s rated speed and producing full torque, then the mechanical power being produced is 50 hp. Assuming that the motor is 95% efficient, the electrical power that’s required is:
PowerElectrical = (50HP x 0.746) / 0.95 = 39.3kW
which means that the current on the input side of the VFD will be approximately 55 A. This same current will also hold true even if the motor is operating at full speed and producing half torque.
Table 1. Operating points for examples*
Ultimately, a VFD is merely a power conversion device that converts the fixed voltage and frequency of incoming power to a variable voltage and frequency output to provide the variable speed capabilities for which it was designed. Keep in mind the variables associated with electrical power (voltage, current and power factor) and their relationships when comparing the VFD’s input to its output. This also will hold true when using the motor’s mechanical power (speed and torque) to determine the amount of input power/current to the VFD. Taking all the variables into consideration, one can be pleasantly surprised to find the input current lower than the output current
Analysis and examples of power conversion by variable-frequency drives.
Variable frequency drives (VFD) are becoming more common place and more widely used in applications. They are capable of varying the output speed of a motor without the need for mechanical pulleys, thus reducing the number of mechanical components and overall maintenance. But the biggest advantage that a VFD has is the ability to save the user money through its inherit nature to save energy by consuming only the power that’s needed. The main question now is, How does a VFD accomplish this? The simple answer to this question is power conversion.
A VFD is similar to the motor to which it’s attached, they both convert power to a usable form. In the case of an induction motor, the electrical power supplied to it is converted to mechanical power through the rotation of the motor’s rotor and the torque that it produces through motor slip. A VFD, on the other hand, will convert its incoming power, a fixed voltage and frequency, to a variable voltage and frequency. This same concept is also the basis to vary the speed of the motor without the need of adjustable pulleys or gearing changes.
“A VFD will convert its incoming power, a fixed voltage and frequency, to a variable voltage and frequency.”
- Edward Tom, Yaskawa America, Inc.Electrical
Electrical power is defined as the following:Power (P) = √3 x Voltage (V) x Current (I) x Power Factor (PF)
In an ideal VFD, the following would hold true:
Powerin = Powerout
But because a VFD has inefficiencies and requires a small amount of power consumption to power the brains of the drive, the input power will be slightly greater than the output power. For this, we will assume that this extra power draw is negligible.
With these two equations, we can then define the relationship between the VFD’s input and output:
Vin x Iin x PFin = Vout x Iout x PFout
Taking these equations into account, let’s use a 100-hp motor as an example with the following properties:
Power = 100 hp
Speed = 1,785 rpm
Voltage = 460 V
FLA = 115 A
Power factor = 0.86
Assume that the motor is running at 60 Hz on a VFD, drawing a no-load current of 40 A on the output of the VFD. With this, one would assume that the input current would also be the same, 40 A. However, using an ammeter on the drive’s input, a person is reading nearly zero amps! How is this possible? Is the drive creating power somehow? The answer simply is no, the drive is not creating power. The power factor causes this “discrepancy” in current When a motor is running at no load, the motor’s power factor can be assumed to be zero, not 0.86 86 as stated on the nameplate. The reason the power factor isn’t at 0.86 is because this is the motor power factor at full load. Alternatively, mechanical (friction) and electrical (resistive) losses in the motor prevent the power factor from being zero when running no load, but we’ll assume these losses to be zero just like we did for the VFD. Therefore, you would have the following:
Pout = 460V x 40A x 0
Pout = 0
Because the output power is zero, the input power also will be zero. With a fixed input voltage, the two variables would be current and power factor. Because current is needed for a power factor to exist, both current and power factor are zero, which means the low input current reading is indeed correct.
This explains why the input current to the VFD is so low when the motor is operating under no load conditions. But what about under load? The same concept still applies when the motor is under load. For example, assume the same motor is now operating at half speed, 30 Hz and producing full motor rated torque and drawing the motor’s full-load amps (FLA). This means that the electrical power that the motor is drawing is:
Pout = √3 x 230V x 115A x 0.86 = 39.4kW
Because the VFD is a power converter, this means that the input current is (assuming a 0.89 input power factor from a 3% line impedance):
Iin = (39.4kW) / (√3 x 460V x 0.89) = 55.6A
Because the VFD is operating at half speed and under full load, the input current is less than half of the output current.
Power (kW) | Voltage (V) | Current (A) | Frequency (Hz) | Power Factor | |
Input | 39.4 | 460 | 55.6 | 60 | 0.89 |
Output | 39.4 | 230 | 115 | 30 | 0.86 |
In this example, the input current is less than half of the output, a result of having a higher power factor on the input side.
The difference in power factor between the input and output side of the motor is what makes it possible to have a higher output current than input current. Assuming the motor is now running at full load and using the same power factor values, your input current now becomes:
Iin = (√3 x 460V x 115A x 0.86) / (√3 x 460V x 0.89) = 111A
which is 4 A lower than the output current.
If… | Then… |
PFin < PFout | Iin < Iout |
PFin = PFout | Iin = Iout |
PFin > PFout | Iin > Iout |
Mechanical
The current a VFD draws on the input side also can be related to the mechanical power a motor is delivering. The basic relationship for motor power is:Powermechanical ∝ Speed x Torque
This means that if the motor is operating at half the speed and producing full torque, the motor is outputting half of its rated power. Consequently, if the motor is running at full speed and producing half torque, the motor is also outputting half of its rated power.\
Because of motor losses, the power relationship between the electrical power going into the motor and the mechanical power is:
PowerElectrical = (Powermechanical) / (EfficiencyMotor)
Revisiting the above example, if the motor is operating at 30 Hz, half the motor’s rated speed and producing full torque, then the mechanical power being produced is 50 hp. Assuming that the motor is 95% efficient, the electrical power that’s required is:
PowerElectrical = (50HP x 0.746) / 0.95 = 39.3kW
which means that the current on the input side of the VFD will be approximately 55 A. This same current will also hold true even if the motor is operating at full speed and producing half torque.
Table 1. Operating points for examples*
Motor speed (rpm) | Motor torque (ft-lb) | Motor power (hp) | Input power (kW) | Input current (A) |
0 | 0 | 0 | 0 | 0 |
0 | 100 | 0 | 0 | 0 |
900 | 0 | 0 | 0 | 0 |
900 | 295 | 50 | 39.3 | 55 |
1800 | 0 | 0 | 0 | 0 |
1800 | 295 | 100 | 78.8 | 111 |
*Motor is rated for 100 hp with a rated torque of 295 ft-lb. |
Ultimately, a VFD is merely a power conversion device that converts the fixed voltage and frequency of incoming power to a variable voltage and frequency output to provide the variable speed capabilities for which it was designed. Keep in mind the variables associated with electrical power (voltage, current and power factor) and their relationships when comparing the VFD’s input to its output. This also will hold true when using the motor’s mechanical power (speed and torque) to determine the amount of input power/current to the VFD. Taking all the variables into consideration, one can be pleasantly surprised to find the input current lower than the output current
Star Automations is one of the leading Supplier,support,service Provider of Variable frequency drives and industrial Automation Product.
Star automations specializes in development, manufacturing and sales of high-quality low-cost motion control products.
We are specialized in providing technical solutions and consultancy for these products.
We can provide third party services all over world and will delevery worldwide.
Star Automations Committed to Serves Its International Customers throughout the world
Abudhabi,Afghanistan,Albania,Algeria,Angola,Antigua,Argentina,Armenia,Australia,Austria,Azerbaijan,Bahamas,Bahrain,Bangladesh,Barbados,Belarus,Belgium, Benin, Bermuda, Bhutan, Bolivia, Bosnia-Herzegovina, Bostwana, Brazil, Brunei, Bulgaria, Burkina-Faso, Burundi, Cambodia, Cameroon, Canada, Cap,Verde, Cayman,Islands, Central,African,Republic, Chad, Chile, China, Colombia, Comoros, Congo, Cook,Islands,Costa,Rica,Croatia,Cuba,Cyprus,Czech,Republic,Democratic,Republic,of,Congo,Denmark,Djibouti,Dominican,Republic,Ecuador,Egypt,El,Salvador,Equatorial,Guinea,Erythrea,Estonia,Ethiopia,Fiji,Finland,France,French,Guyana,French,Polynesia,Gabon,Gambia,Georgia,Germany,Ghana,Greece,Guadeloupe, Guatemala,Guinea,Guinea,Bisso,Guyana,Haïti,Honduras, Hong-Kong,Hungary,Iceland,India,Indonesia,Iran,Iraq,Ireland,Israel,Italy,Ivory,Coast,Jamaïca,Japan,Jordan,Kazakhstan,Kenya,Kirghizistan,Kuwait,Laos,Latvia,Lebanon,Liberia,Libya,Liechtenstein,Lithuania,Luxembourg,Macedonia,Madagascar,Malawi,Malaysia,Maldives,Mali,Malta,Martinique,Mauritania,Mauritius,Mayotte,Mexico,Moldavia,Montenegro,Montserrat,Morocco,Mozambique,Myanmar,Namibia,Nepal, Netherlands,New Caledonia, New Zealand,Nicaragua,Niger,Nigeria,Norway,Oman,Pakistan,Palestine,Panama,Papua New Guinea,Paraguay,Peru,Philippines,Poland,Portugal,Puerto Rico,Qatar,Reunion Island,Romania,Russia,Rwanda,Saint Barthelemy,Sainte Lucie,Saint Martin,Saint Pierre and Miquelon,Saint Vincent and the Grenadines,Samoa,SaoTome and Principe,Saudi Arabia,Senegal,Serbia,Seychelles,Sierra Leone,Singapore,Slovakia,Slovenia,Solomon Islands,Somalia,South Africa,South Korea,Spain,Sri Lanka,Sudan,Surinam,Sweden,Switzerland,Syria,Taiwan,Tanzania,Thailand,Togo,Tonga,Trinidad and Tobago,Tunisia,Turkey,Turkmenistan,Turks and Caico,Tuvalu,Uganda,Ukraine,United Arab Emirates,United Kingdom,United States,Uruguay,Uzbekistan,Vanuatu,Venezuela,Vietnam,Virgin islands,Wallis and Futuna,Yemen,Zambia,Zimbabwe
service, repair, replace, built, rebuild, regrind, recondition, refurbish, remanufacture, retrofit and overhaul all VFD and MV drives.
For any other related information feel free to contact us.
I'm not sure where you are getting your info, but great topic. I needs to spend some time learning much more or understanding more. Thanks for wonderful information I was looking for this information for my mission.
ReplyDeleteFeel free to surf to my blog ... wordpress tutorial pdf
Also see my page :: wordpress grundkurs
I read this article completely regarding the comparison of hottest and previous technologies, it's amazing article.
ReplyDeletemy page - low carb erfolge
my web page: steinzeit diät
I've been exploring for a bit for any high quality articles or weblog posts in this kind of area . Exploring in Yahoo I at last stumbled upon this web site. Reading this information So i am happy to express that I've a very just right uncanny feeling I found out exactly what I
ReplyDeleteneeded. I most without a doubt will make certain to
do not forget this web site and provides it a look regularly.
Also visit my web blog proper diet
Heya superb blog! Does running a blog like this require a massive amount work?
ReplyDeleteI have absolutely no knowledge of coding but I was hoping to start my
own blog soon. Anyhow, if you have any suggestions or tips for new blog owners please share.
I know this is off topic nevertheless I simply needed to ask.
Many thanks!
Also visit my homepage: low carb diät buch
My site :: brot ohne kohlenhydrate kaufen
This is very interesting, You're a very skilled blogger. I have joined your rss feed and look forward to seeking more of your great post. Also, I have shared your site in my social networks!
ReplyDeleteMy page :: Finanzberatung Mainz
You're so awesome! I do not suppose I've truly read a single thing like that before.
ReplyDeleteSo nice to discover someone with some unique thoughts on this topic.
Really.. many thanks for starting this up. This web site is something that's needed on the internet, someone with some originality!
My blog post ... Steinzeit Diät Erfahrungen
Awesome! Its in fact awesome post, I have got much clear idea regarding
ReplyDeletefrom this paragraph.
Also visit my webpage :: paleo diät
I know this if off topic but I'm looking into starting my own blog and was wondering what all is required to get setup? I'm assuming having a blog like yours would cost a pretty
ReplyDeletepenny? I'm not very web smart so I'm not 100% sure. Any suggestions or advice would be greatly appreciated. Many thanks
Here is my web site; kohlenhydratarme Ernährung Rezept
Your style is very unique in comparison to other people I've read stuff from. Many thanks for posting when you've got the opportunity, Guess I'll just bookmark this site.
ReplyDeletemy web site ... anleitung lenkmatte
I think everything published was very reasonable.
ReplyDeleteHowever, think about this, what if you were to create a killer title?
I ain't suggesting your information is not solid., however suppose you added something that makes people want more? I mean "How VFDs save energy,vfd efficiency,VFD,vfd energy savings calculator,vfd basics,advantages of variable speed drives,what is vfd,vfd working principle,vfd fundamentals, variable frequency drive tutorial,variable frequency drive ,energy saving tips,energy saving ,vfd energy saving ," is a little plain. You might look at Yahoo's front page and note
how they create post titles to get viewers to open the links.
You might add a related video or a related pic or two to grab readers interested about what you've got to say. Just my opinion, it could bring your website a little livelier.
Feel free to visit my blog post ... finanzgericht
It's perfect time to make some plans for the longer term and it is time to be happy. I have learn this publish and if I may just I want to counsel you some interesting issues or tips. Perhaps you can write subsequent articles relating to this article. I wish to learn even more things about it!
ReplyDeleteHere is my web blog ... sony autoradio ausbauen