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.

Re: Need help with Electrical Question#128357 01/25/0301:03 AM01/25/0301:03 AM

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 numbers--You 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.

Re: Need help with Electrical Question#128360 01/25/0302:59 PM01/25/0302:59 PM

Use three 2-wire heater elements that draw 11 amperes each at rated line voltage. If you connected all three in parallel to a 2-wire source, the current on each line connection would be 11+11+11 or 33 amperes. Now with two heaters on the 2-wire source, the line current is 11+11 or 22 amperes.

Expand the setup from a single-phase two-wire source to three phase, and power the three heaters from a simple three-phase three-wire 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 2-wire heater connected Aø-to-Bø and the second Bø-to-Cø. 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 balanced-3ø system right now. If the current in each 2-wire heater is tested, [say, with a clamp-on 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 2-wire 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 2-wire 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.

H-o’-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 2-wire heater elements that draw 11 amperes each at rated line voltage. If you connected all three in parallel to a 2-wire source, the current on each line connection would be 11+11+11 or 33 amperes. Now with two heaters on the 2-wire source, the line current is 11+11 or 22 amperes.

Expand the setup from a single-phase two-wire source to three phase, and power the three heaters from a simple three-phase three-wire 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 2-wire heater connected Aø-to-Bø and the second Bø-to-Cø. 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 balanced-3ø system right now. If the current in each 2-wire heater is tested, [say, with a clamp-on 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 2-wire 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 2-wire 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.

H-o’-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 01-25-2003).]

Re: Need help with Electrical Question#128361 01/25/0305:52 PM01/25/0305:52 PM

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.

Re: Need help with Electrical Question#128363 01/25/0310:08 PM01/25/0310:08 PM