quickerthansilver
Member
- Location
- Highland, CA
During 3 phase calculations, where does the root 3 come from? For example, for 208V 3 phase we use sqrt(3)*208 to get 360 as a divider...
Root 3
- Thread starter quickerthansilver
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gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090213-2006 EST
It is related to sin 60 deg. The sin of 30 is 0.5, sin 60 = 0.866 = ( (3^0.5) /2).
You draw vector diagrams and from these detemine angles, and then ratios. Draw two equal right triangles with a 30 deg included angle. Make the two short sides common and one triangle the mirror image of the other. Treat the hypotenuse of one triangle as one line to neutral voltage and the base as 1/2 the line to line voltage. See if you can proceed from here.
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It is related to sin 60 deg. The sin of 30 is 0.5, sin 60 = 0.866 = ( (3^0.5) /2).
You draw vector diagrams and from these detemine angles, and then ratios. Draw two equal right triangles with a 30 deg included angle. Make the two short sides common and one triangle the mirror image of the other. Treat the hypotenuse of one triangle as one line to neutral voltage and the base as 1/2 the line to line voltage. See if you can proceed from here.
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Another way:
Another way:
Consider a balanced delta load. If we work through the complex math, we find that the line current splits into two load currents of equal magnitude with one leading the line current by 30 degrees, the other lagging by 30 degrees. The complex sum of these two currents must be the line current.
For example, let Ia be the line current; let Ild be the magnitude of the load currents.
Then,
Ild[cos(30) + jsin(30) +cos(-30) +jsin(-30)] = Ildx2x0.866 =1.732 = sqrt(3)xIld
The imaginary components cancel each other.
Another way:
Consider a balanced delta load. If we work through the complex math, we find that the line current splits into two load currents of equal magnitude with one leading the line current by 30 degrees, the other lagging by 30 degrees. The complex sum of these two currents must be the line current.
For example, let Ia be the line current; let Ild be the magnitude of the load currents.
Then,
Ild[cos(30) + jsin(30) +cos(-30) +jsin(-30)] = Ildx2x0.866 =1.732 = sqrt(3)xIld
The imaginary components cancel each other.
- Location
- New Jersey
- Occupation
- Journeyman Electrician
Three phase systems are generated with the phases 120 degrees apart. So think of it this way, if you walked in a straight line for 1 mile and then walked another mile on an angle of 120 degrees from the first line, your resultant distance (the distance between where you started and where you ended) would be in a straight line that would equal 1 mile * 1.73 or 1.73 miles.
If you use 120 volts as an example from the 3 phase WYE system you would end up with (120)(1.73)= 208 volts. Hence the term 208Y/120 volt system.
If you use 120 volts as an example from the 3 phase WYE system you would end up with (120)(1.73)= 208 volts. Hence the term 208Y/120 volt system.
quickerthansilver
Member
- Location
- Highland, CA
Thanks
Thanks
hmmm... Is there a diagram or picture somewhere?
Thanks
hmmm... Is there a diagram or picture somewhere?
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090214-2031 EST
An old book that covers this in substantial detail in chapters V and VI is
"Electrical Circuits and Machinery", by Hehre and Harness, John Wiley, 1942.
Google has a digitized version from the University of Michigan library. If you go to Google Books, search for this book, then pick Find A Library and you will be provided a list of librarys. I am not sure how Google determined my location, but they provided library information out to 148 miles.
We have one of Google's largest digitizing operations here in town.
Unless the digitized copies are in the public domain I have not found them useful. But that was because I searched for books already in my possession. The snippets are too small for me to reference anything to someone. But if you had no idea where to look for certain material, then being able to identify a book might be very useful. In the case of very rare books that are in the public domain the digitized version makes something available that otherwise would be unavailable. There are a large number of these here in our library.
I found some books on early electrical measuring instruments that were out of copyright, and therefore fully readable on line.
.
An old book that covers this in substantial detail in chapters V and VI is
"Electrical Circuits and Machinery", by Hehre and Harness, John Wiley, 1942.
Google has a digitized version from the University of Michigan library. If you go to Google Books, search for this book, then pick Find A Library and you will be provided a list of librarys. I am not sure how Google determined my location, but they provided library information out to 148 miles.
We have one of Google's largest digitizing operations here in town.
Unless the digitized copies are in the public domain I have not found them useful. But that was because I searched for books already in my possession. The snippets are too small for me to reference anything to someone. But if you had no idea where to look for certain material, then being able to identify a book might be very useful. In the case of very rare books that are in the public domain the digitized version makes something available that otherwise would be unavailable. There are a large number of these here in our library.
I found some books on early electrical measuring instruments that were out of copyright, and therefore fully readable on line.
.
Try this link: http://books.google.com/books?id=y-vq2G1lHUMC&printsec=frontcover&dq=electricians+Handbook#PPT72,M1
Depending on your level of understanding you can go forwards or backwards in the book.
What do you think about this one?
http://books.google.com/books?id=XM...q=Electrical+Circuits+and+Machinery#PPA152,M1
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