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#128354  01/25/03 01:31 AM
Need help with Electrical Question


I need someone to explian how to solve this question.
What is the line amperage of a 3phase delta system if the coil amperages are: A=11 amps
Ia=____________Amps Ib=____________Amps Ic=____________Amps
thanks in advance



#128355  01/25/03 01:41 AM
Re: Need help with Electrical Question


Is this a school assignment?



#128356  01/25/03 01:49 AM
Re: Need help with Electrical Question


I am playing catch up. I took a masters 1 and 2 class in 96 and 97. Now I am trying to relearn everything. I am taking a class now and I am a little behind. I need help on finding the steps on how to solve a few problems.



#128357  01/25/03 02:03 AM
Re: Need help with Electrical Question


11 amps x 1.732
Do you undestand why it's not 22 amps?



#128358  01/25/03 02:12 AM
Re: Need help with Electrical Question


I don't understand any of this question.This one has me at a total loss.



#128359  01/25/03 12:17 PM
Re: Need help with Electrical Question


Delta
Phase voltage = line voltage Phase amperage * 1.732 = line amperage
Wye
Phase voltage * 1.732 = line voltage Phase amperage = line amperage
Draw your two diagrams of delta and wye, visually you can see why this is.
Write in all your numbersYou can see that in a delta that across one coil if you have 240 volts that the volts on the line will be the same.
Whereas in a wye you can see that if there is 120 volts across one coil(phase) that from line to line you are going through two coils and have to mutiply by 1.732 (not 2 because they are out of phase) to get the line voltage of 208.
Amperage in a delta gets increased because one line is hooked up to the corner of two coils. In a wye one line is hooked up to the corner of one coil only so amperage stays the same.
Hope this helps.
Remember a delta is a triangle and a wye is a Y.



#128360  01/25/03 03:59 PM
Re: Need help with Electrical Question


Use three 2wire heater elements that draw 11 amperes each at rated line voltage. If you connected all three in parallel to a 2wire source, the current on each line connection would be 11+11+11 or 33 amperes. Now with two heaters on the 2wire source, the line current is 11+11 or 22 amperes.
Expand the setup from a singlephase twowire source to three phase, and power the three heaters from a simple threephase threewire line. Voltage is immaterial for your answer, but most 3ø calculations assume [unless stated otherwise] balanced voltage and balanced current. Another way of describing it all is that everything’s symmetrical.
Doing only part of the connections and making some current checks along the way—rewire the heaters with one 2wire heater connected AøtoBø and the second BøtoCø. At this point only 2 heaters are connected, but at this point there is a trick in the line currents, although it isn’t a balanced3ø system right now. If the current in each 2wire heater is tested, [say, with a clampon ammeter checking either of the heater leads] they will measure 11 amperes in each connection, or 11 amperes on Aø, 11 amperes on Cø, OK? With both heaters connected to Bø, the current should intuitively be more than 11 amperes, for there’s two heaters running. For two 2wire heaters connected with one wire each on Bø, it may seem that the total Bø current might be 11+11 or 22 amperes.
But, through the magic of three phase, the current is only 86.6% of the expected 22 amperes, or about 19 amperes on Bø.
Continuing for a complete 3ø circuit, connecting the third 2wire heater on Cø and Aø, there’s now three symmetrical or balanced currents—about 19 amperes {11x1.732} on each phase, and not 22 {11+11}. The three heaters are connected in sort of a triangle or delta connection.
Ho’P, there you are for the short version. It’s a lot to absorb at once, and has taken years of installation and current checks on real systems for me to learn that all those books aren’t crazy or stupid, and that if you do it right, then you can predict the outcome—and most importantly—make money at it. Repetition, repetition, repetition… three 2wire heater elements that draw 11 amperes each at rated line voltage. If you connected all three in parallel to a 2wire source, the current on each line connection would be 11+11+11 or 33 amperes. Now with two heaters on the 2wire source, the line current is 11+11 or 22 amperes.
Expand the setup from a singlephase twowire source to three phase, and power the three heaters from a simple threephase threewire line. Voltage is immaterial for your answer, but most 3ø calculations assume [unless stated otherwise] balanced voltage and balanced current. Another way of describing it all is that everything’s symmetrical.
Doing only part of the connections and making some current checks along the way—rewire the heaters with one 2wire heater connected AøtoBø and the second BøtoCø. At this point only 2 heaters are connected, but at this point there is a trick in the line currents, although it isn’t a balanced3ø system right now. If the current in each 2wire heater is tested, [say, with a clampon ammeter checking either of the heater leads] they will measure 11 amperes in each connection, or 11 amperes on Aø, 11 amperes on Cø, OK? With both heaters connected to Bø, the current should intuitively be more than 11 amperes, for there’s two heaters running. For two 2wire heaters connected with one wire each on Bø, it may seem that the total Bø current might be 11+11 or 22 amperes.
But, through the magic of three phase, the current is only 86.6% of the expected 22 amperes, or about 19 amperes on Bø.
Continuing for a complete 3ø circuit, connecting the third 2wire heater on Cø and Aø, there’s now three symmetrical or balanced currents—about 19 amperes {11x1.732} on each phase, and not 22 {11+11}. The three heaters are connected in sort of a triangle or delta connection.
Ho’P, there you are for the short version. It’s a lot to absorb at once, and has taken years of installation and current checks on real systems for me to learn that all those books aren’t crazy or stupid, and that if you do it right, then you can predict the outcome—and most importantly—make money at it. Repetition, repetition, repetition…
[This message has been edited by Bjarney (edited 01252003).]



#128361  01/25/03 06:52 PM
Re: Need help with Electrical Question

Member

Joined: Jul 2002
Posts: 175
Canada


HouseOfPower, Perhaps these sketches will help illustrate what the others have explained. Ed



#128362  01/25/03 10:17 PM
Re: Need help with Electrical Question


Thanks for all the replys. It has been a big help and I need alot of it. I have been in business for 8 years now and have leased a licence. I am starting to study for my masters.



#128363  01/25/03 11:08 PM
Re: Need help with Electrical Question


Blueribbon illustrations, "Ed".




