Pole mounted utility transformer

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Re: Pole mounted utility transformer

Do you mean the axis x? Because we don't have x defined here.
 
Re: Pole mounted utility transformer

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You already know that x can be found by:

x = cos (angle)

The arrows between the x and y axes can represent vectors. You can also find the vector components (the lengths of x and y) using:

Ax = A cos (angle)
Ay = A sin (angle)

Ax is the length in the x axis
Ay is the length in the y axis
A is the value of the vector.

A (the vector) = 120
angle (from x axis to A) = 60?

Ax = A cos (angle)
60 = 120 cos 60
 
Re: Pole mounted utility transformer

Sam, be careful what you say.

The x and y axes are infinite in length. The extremes of both are +/- infinity.

The convention is to describe the sides of a right triangle with A, B, and C. The letters a and b denote the angles opposite sides A & B.

C is the hypotenuse,

B is the horizontal leg,

B = C*cos(a) = C*sin(b)

A is the vertical leg,

A = C*sin(a) = C*cos(b)

Don't clutter things up by renaming variables unless there is a valid reason to do so.

Just had a thought, "Irish coffee at Fisherman's Wharf".
 
Re: Pole mounted utility transformer

I've seen so many differnt conventions used. I'm trying to keep it standard. This notation is from my nearest physics text. That's why I define everything.

Is that San Francisco's Fisherman's Warf?
 
Re: Pole mounted utility transformer

Are you from my side of the universe Rattus?

And, are you wireless right now? :cool:
 
Re: Pole mounted utility transformer

Rattus, are you talking about C*, the extended complex plane? :confused:

Edit: Because if you are, Charlie B. and I have an agreement that I wont use the extended complex plane in the application of elctrical mathmatics. (Although I did leave an escape clause)

Edit: Or is * squared?

[ February 09, 2005, 05:26 AM: Message edited by: physis ]
 
Re: Pole mounted utility transformer

Naw Sam,

The asterisk is used as a symbol for multiplication in some software to avoid confusing it with the variable "x".

Complex infinity has no application in this forum.

[ February 09, 2005, 10:10 AM: Message edited by: rattus ]
 
Re: Pole mounted utility transformer

Oh yeah, could be that too. That's why you shouldn't stay up so late.

Edit: You had infinity and C* in there, I was having flash backs from another thread. :D

[ February 09, 2005, 03:28 PM: Message edited by: physis ]
 
Re: Pole mounted utility transformer

Physis,
Consider this, the 60 degree delta vector angle in Ed's diagram is in the second quadrant.This angle is referenced to -x axis at 180 degree. We know this because the phase voltages in a three phase Delta secondary are 120 degrees out of phase.Therefore,180-60=Angle Theta 120 degrees.
Angle Theta is the angle of reference.
 
Re: Pole mounted utility transformer

That's not really the issue but:

What do you get if you put it in the second quadrant?
 
Re: Pole mounted utility transformer

Sam,

Ed's diagram does not have arrow heads therefore it is not a true vector diagram.

Draw the delta diagram with the arrows connected head to tail at 0, 120, and -120 to form an equilateral triangle. Some are confusing the complement of 120 degrees (60) with the angle of separation. Not so! The phase angles must be referenced to the positive x axis.

60 degrees of separation would fry the tranformer first before the load could be fried.
 
Re: Pole mounted utility transformer

Rattus,

I'm not sure you're following the discussion between Sparks1 and myself.

This starts at a question regarding a triangle I kept encountering.

Sparks1 came up with Ax (or "b") = 60.005.

I used, I'm sure, a different formula, and get exactly 60.

I'm trying to stay referent.

If you go back a ways you'll see where the 60? becomes the topic.

Edit: I said tringle instead of triangle :D

[ February 10, 2005, 08:22 PM: Message edited by: physis ]
 
Re: Pole mounted utility transformer

Sam, if you are talking about the x and y values of the + 120 degree vector, it would be:

V*[cos(120) + jsin(120)] = V[0.5 + j*sqrt(3)/2]

For V = 120Vrms, the voltage vector is:

-60 + j105, the 105 is not exact

The extra 0.005 may come from computational error in using the Pythagorean theorem. Better to use trig. Besides, you cannot obtain five significant figures when your inputs are only three figures.

[ February 10, 2005, 10:35 PM: Message edited by: rattus ]
 
Re: Pole mounted utility transformer

That's exactly my guess my Rattus. More accurately, a lack of decimal places precision. Although I haven't done the math to verify the hypothesis.

Edited cause I can't spell

[ February 11, 2005, 12:00 PM: Message edited by: physis ]
 
Re: Pole mounted utility transformer

I never was trying to implicate anybody. Only demonstrate the meathod I used.
 
Re: Pole mounted utility transformer

Rattus,
Thank you for using the trig relationship in explaining why Ed's triangle would burn out the secondary side of a Delta transformer!
The .005 was computational,I carried out three places using the Pythagorean Theorem.

I think I'll have that Irish Whiskey now! :eek
 
Re: Pole mounted utility transformer

Ed wasn't gonna burn up any transformers. :D :D

I think a beer sounds good.
 
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