I told everyone I know about you and your Material. You are the best teacher I had. I'm a Mechanicall Engineer in Argentina and I KNOW YOU ARE THE BEST !. Hands down !. Luckyly , this hard lectures ARE SHORT ! :). Nice !. Thanks again !!!!!!!!!.
Hi Jim, just wanted to thank you for your work at Columbia Gorge Community College, as your 'Digital Electronics class' has provided me with the necessary information to tackle the problems I need solved. My educator has failed in his teachings to properly educate me, and I had to seek out additional information to fill in the gaps. You have helped many individuals learn topics you have covered, who have made contributions to the infrastructure of America due to your generosity. Educators like you make America, and other countries better. Don't ever make anyone put you down, as they don't know the sheer positive impact you've had on the world.
Good question. It's a unit conversion for 3 phase machines. The inrush code is sometimes called "kVA" note the k. k is 1000. Apparent power for a 3 phase AC machine is sqrt(3)*Line to Line Voltage*Line Current (see ruclips.net/video/bVjmy2vBj-0/видео.html ) 577*sqrt(3) is 1000.
Not just a Y start-Delta run can reduce inrush. There's a bunch of different types of reduced voltage starters. Check out this playlist: ruclips.net/p/PLdnqjKaksr8pQrEBTQ65GKlEmkb8AJRCE
If not specified on the name plate hopefully this data would be in the complete data sheet provided by the manufacturer. If not you could always put an amp clamp on during start to experimentally determine inrush.
Hi great Vid, but i am using this calculation on my goulds water pump and its not matching to the reading on my fluke clamp on. Dont know if am doing something wrong. The pump is 1/2 HP at 115 volts. The code is L. When i calculate using .1000 for single phase i get 4.12 Amps but my pump FLA is 10.8 Amps and my fluke meter reads 17.32 amps on start and 11.08 on run. Please help. Thanks much.
This might be an order of operations problem. Inrush is more than FLA. You should at least expect more than 10.8A. Assuming L=9.5 Voltage = 115 Mechanical Power (hp) = .5 Inrush = 9.5*.5*1000/115 = 41.3A Inrush is thankfully a brief phenomenon. Most likely the amp clamp is capturing the 17.3A after the fact and on the way down. Your measured current is very close to FLA.
@@bigbadtech And one more thing please. I do not see an efficiency or power factor on the pumps nameplate. I know it is required to calculate input power. is there another way to figure out the efficiency or power factor of the pump. Sorry if I ask too much, just want to learn this thing. Thanks again.
Easy way: there's got to be a data sheet for the pump in question somewhere on the internet. Hard way: Start at lecture 37 and watch to lecture 43 ruclips.net/video/flRocvQxQw4/видео.html
I told everyone I know about you and your Material. You are the best teacher I had. I'm a Mechanicall Engineer in Argentina and I KNOW YOU ARE THE BEST !. Hands down !. Luckyly , this hard lectures ARE SHORT ! :). Nice !. Thanks again !!!!!!!!!.
Hi Jim, just wanted to thank you for your work at
Columbia Gorge Community College, as your 'Digital Electronics class' has provided me with the necessary information to tackle the problems I need solved.
My educator has failed in his teachings to properly educate me, and I had to seek out additional information to fill in the gaps.
You have helped many individuals learn topics you have covered, who have made contributions to the infrastructure of America due to your generosity. Educators like you make America, and other countries better.
Don't ever make anyone put you down, as they don't know the sheer positive impact you've had on the world.
Glad you're making use of this material. Tell your friends!
Excellent, simple and time efficient explanation of inrush current, thank you.
Awesome video sir! Thank you!
Love it. Keep it
Hello, i'm sorry for the possibly naive question but where does the 577 in the equation come from?
Good question. It's a unit conversion for 3 phase machines. The inrush code is sometimes called "kVA" note the k. k is 1000. Apparent power for a 3 phase AC machine is sqrt(3)*Line to Line Voltage*Line Current (see ruclips.net/video/bVjmy2vBj-0/видео.html ) 577*sqrt(3) is 1000.
can we prevent in rush current by wye delta configuration start up?
Not just a Y start-Delta run can reduce inrush. There's a bunch of different types of reduced voltage starters. Check out this playlist: ruclips.net/p/PLdnqjKaksr8pQrEBTQ65GKlEmkb8AJRCE
@@bigbadtech ty sir, but all i need is yung cheapest one kasi de naman ganon ka critical yung gamit namin.
I work in the elevator industry and I found the code letter is not in the traction motor name plate. Then how do i determine the inrush current?
If not specified on the name plate hopefully this data would be in the complete data sheet provided by the manufacturer. If not you could always put an amp clamp on during start to experimentally determine inrush.
Hi great Vid, but i am using this calculation on my goulds water pump and its not matching to the reading on my fluke clamp on. Dont know if am doing something wrong. The pump is 1/2 HP at 115 volts. The code is L. When i calculate using .1000 for single phase i get 4.12 Amps but my pump FLA is 10.8 Amps and my fluke meter reads 17.32 amps on start and 11.08 on run. Please help. Thanks much.
This might be an order of operations problem. Inrush is more than FLA. You should at least expect more than 10.8A.
Assuming L=9.5
Voltage = 115
Mechanical Power (hp) = .5
Inrush = 9.5*.5*1000/115 = 41.3A
Inrush is thankfully a brief phenomenon. Most likely the amp clamp is capturing the 17.3A after the fact and on the way down. Your measured current is very close to FLA.
@@bigbadtech Okay thank u so much. I will also try another clamp on an see what happens. Thanks also for doing the calculation.
@@bigbadtech And one more thing please. I do not see an efficiency or power factor on the pumps nameplate. I know it is required to calculate input power. is there another way to figure out the efficiency or power factor of the pump. Sorry if I ask too much, just want to learn this thing. Thanks again.
Easy way: there's got to be a data sheet for the pump in question somewhere on the internet.
Hard way: Start at lecture 37 and watch to lecture 43 ruclips.net/video/flRocvQxQw4/видео.html
Stop your news negative crap
Indeed. Retire thee to thy fainting couch.