single vs. 3 phase

Status
Not open for further replies.
R

rattus

Senior Member
Easy as falling off a log.

Easy as falling off a log.

coulter said:
rattus -
Since it is easy, go head and show us the proof. Please include a derivation of the trig identify (difference of functions) you are going to use, specifically show your derivation for:
sinA - sinB = 2cos(1/2(A + B))sin(1/2(A-B))

I agree Vab is a sine - I just don't think the proof is that easy.

carl
Click to expand...

Quite simple Carl,

Let A = wt and B = wt - 120, then

sinA - sinB = 2cos(wt -60))sin(60) = cos(wt -60)1.732 = sin(wt + 30)1.732

QED, whatever that means.
 
highendtron

highendtron

Senior Member
single phase vs. 3 phase

single phase vs. 3 phase

I always thought that the distiction comes into play when the nuetral is considered. On a single phase system, a nuetral is shared only by each individual ungrounded conductor. On multi phase system the nuetral is shared by multiple ungrounded conductors. If a system is classifeid as two-phase, then, one might be tempted to share the grounged conductor between two ungrounded conductors. Among other things; heating, loading, and derating problems would occur.
 
R

rattus

Senior Member
coulter said:
Well that part certainly was simple.

I missed the simple part where you derived this magic.

carl
Click to expand...

All I did was substitute for A & B, evalutate sin(60), and convert cos to sin.

Only the first step is necessary for proof because sin(60) is a constant which leaves a sinusoidal function.

P.S. Smart will gladly fill in the details. Right Smart?
 
Last edited:
mivey

mivey

Senior Member
coulter said:
Well that part certainly was simple.

I missed the simple part where you derived this magic.

carl
Click to expand...
Now that's funny. He cheated and assumed they had the same frequency...oh that's right...they DO have the same frequency.:grin:
 
C

coulter

Senior Member
rattus said:
...Only the first step is necessary for proof because sin(60) is a constant which leaves a sinusoidal function.
Click to expand...
Smoke level 90
Mirror level 85:roll:

You certainly did a good job at dodging a clear answer.

carl
 
J

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
I am amazed at the number of people that get caught up on the exact meaning behind words that are actually used in a general manner. Our industry terminology is not as pure and universal as we think it is.

I believe most of the discussions of the number of phases are directly related to the problem of referring to "line conductors" by the term "phases" and by the idea that the math for solving system equations changes just because a "neutral" point becomes available.

What if the "phases" of a system were defined only by the number of different usable LINE to LINE voltages that were available? Can you think of a system that this method is not applicable to? The presence or absence of a "neutral" does not affect these descriptions, it only affects the number of wires in the circuit.

2 line conductors L1 and L2 = single phase; L1-L2
3 line conductors L1, L2, and L3 = three phase; L1-L2, L2-L3, L3-L1
4 line conductors L1, L2, L1', and L2' = two phase L1-L2, L1'-L2'

A standard single input 480V transformer with (2) 120V output coils can be wired to produce three different output systems 120V 2-wire, 240V 2-wire, and 120/240V 3-wire. The math for solving equations (i.e. current flow through the windings) for any of these systems should be consistent with the physical connection of the windings not how many wires are used.
 
mivey

mivey

Senior Member
if complicated makes you happy

if complicated makes you happy

Given:
sinA-sinB = 2*cos(1/2*(A+B))*sin(1/2*(A-B))

We are dealing with waves that have the same frequency so, let A = wt+@1 and B = wt+@2

sinA - sinB = sin(wt+@1) - sin(wt+@2)
= 2*cos(1/2*(wt+@1+wt+@2))*sin(1/2*(wt+@1-(wt+@2)))
= 2*cos(1/2*(2*wt+@1+@2))*sin(1/2*@1-@2)
= 2*cos(wt+(@1+@2)/2)*sin((@1-@2)/2) <--- the magic formula
we know sin(C) = cos(90+C) so the result
= 2*sin(90+wt+(@1+@2)/2)*sin((@1-@2)/2)
which is a sinusoidal. Case closed. (I've been waiting to say that):grin:

But rattus's proof used the specific wye case so let @1=0 and @2=-120 and we then have
sinA-sinB = 2*sin(90+wt+(0-120)/2)*sin((0--120)/2) (ooooooh a minus negative!!!:smile: )
= 2*sin(90+wt-60)*sin(60)
= 2*sin(wt+30)*sqrt(3)/2
= sqrt(3)*sin(wt+30)

[edit: identified the magic formula]
 
Last edited:
C

coulter

Senior Member
mivey said:
Given:
sinA-sinB = 2*cos(1/2*(A+B))*sin(1/2*(A-B)) ...
Click to expand...
I'll keep this simple.

mivey -
Cute - but useless. rattus said he could derive this formula you list as given (the very same) - simple as lifting a page from his freshman test. First it was easy, then it was trickey, finally it became boring.

Maybe he can - we will never know.

carl
 
C

coulter

Senior Member
rattus said:
OK Carl, let us hear your proof if you don't like mine.
Click to expand...
rattus -
I never offered a proof. You were the one that said it was easy, tricky, boring - lift a page from your freshman text.

If you can derive it fine - if not, that's okay too

carl
 
R

rattus

Senior Member
Maybe:

Maybe:

coulter said:
I'll keep this simple.

mivey -
Cute - but useless. rattus said he could derive this formula you list as given (the very same) - simple as lifting a page from his freshman test. First it was easy, then it was trickey, finally it became boring.

Maybe he can - we will never know.

carl
Click to expand...

Yes, we will never know because I choose not to waste my time on such a pointless exercise. Furthermore, none of this addresses the OP's question.

Quogue is satisfied, therefore I am satisfied.
 
R

rattus

Senior Member
coulter said:
rattus -
I never offered a proof. You were the one that said it was easy, tricky, boring - lift a page from your freshman text.

carl
Click to expand...

I never thought you would anyway.

Never said I would reinvent the wheel. The comments pertain to the sinusoidal voltage issue which was a response to Quogue's "prove it", and that issue has been resolved.
 
H

hardworkingstiff

Senior Member
Location
Wilmington, NC
Jon, aka winnie

Jon, aka winnie

winnie said:
A service supplied by two legs of a 208/120 wye is officially _named_ as a single phase service.
Click to expand...

Jon,
Mivey posted a definition (he said it was an IEEE reference) a few posts ago that seems to contradict this posting. Do you have a link or can you post the definition you are mentioning?

Thanks,
 
quogueelectric

quogueelectric

Senior Member
Location
new york
rattus said:
I never thought you would anyway.

Never said I would reinvent the wheel. The comments pertain to the sinusoidal voltage issue which was a response to Quogue's "prove it", and that issue has been resolved.
Click to expand...
Go ahead keep rubbing it in the cows nose.
 
mivey

mivey

Senior Member
hardworkingstiff said:
Jon,
Mivey posted a definition (he said it was an IEEE reference) a few posts ago that seems to contradict this posting. Do you have a link or can you post the definition you are mentioning?

Thanks,
Click to expand...
Referr to this thread:http://forums.mikeholt.com/showthread.php?t=82963

While I have not seen the red book text, jim mentioned the IEEE red book in post #54
jim dungar said:
This discussion has not been about motors but rather on the absolutely improper (according to the IEEE Red book and ANSI C84.1-1989) method of calling 120/208Y 3 wire circuits anything other than single phase.
Click to expand...

Also see post #66 in that thread where kingpb quotes C84.1
kingpb said:
Additionally, quoting from C84.1, in reference to secondary transformer configurations, ?Single-phase services and single-phase loads may be supplied from single-phase systems or from three-phase system. They are connected phase-to-phase when supplied from three-phase, three-wire systems and either phase-to-phase or phase-to-neutral from three-phase, four-wire systems."
Click to expand...
 
Status
Not open for further replies.
Top