Question on 3 phase........

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dionysius

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Consider a 3 phase 'Y' source.

V Phase to N is 120 VAC.

V Phase to Phase is 208 VAC.

Been a long time since trig and math.

Is V Phase to Phase a sinusoid that is same as adding V Phase + V (Phase + 120 deg)??????

In other words if I add Sin(x) to Sin(x+120 deg) do I get a Sine wave of same frequency (60 Hz) but amplitude with a mean of 208???

Anybody got this to show with real math so I can see it???
 
dionysius said:
Consider a 3 phase 'Y' source.

V Phase to N is 120 VAC.

V Phase to Phase is 208 VAC.

Been a long time since trig and math.

Is V Phase to Phase a sinusoid that is same as adding V Phase + V (Phase + 120 deg)??????

In other words if I add Sin(x) to Sin(x+120 deg) do I get a Sine wave of same frequency (60 Hz) but amplitude with a mean of 208???

Anybody got this to show with real math so I can see it???
Click to expand...
It's the difference.

Wye L-N & L-L waveforms.jpg

Solid lines represent L-N measurements of 0° and -120° Lines (aka "A" & "B"). The Neutral reference is not depicted. Dashed lines represent L-L measurement with the "B" Line as the reference.
 
dionysius said:
Consider a 3 phase 'Y' source.

V Phase to N is 120 VAC.

V Phase to Phase is 208 VAC.

Been a long time since trig and math.

Is V Phase to Phase a sinusoid that is same as adding V Phase + V (Phase + 120 deg)??????

In other words if I add Sin(x) to Sin(x+120 deg) do I get a Sine wave of same frequency (60 Hz) but amplitude with a mean of 208???

Anybody got this to show with real math so I can see it???
Click to expand...
PS: It's root mean square of 208.
 
Smart $ said:
It's the difference.

View attachment 15361

Solid lines represent L-N measurements of 0° and -120° Lines (aka "A" & "B"). The Neutral reference is not depicted. Dashed lines represent L-L measurement with the "B" Line as the reference.
Click to expand...

I think I see it now from looking at the graphs......

Use the old identity: Sin(A+B) = Sin(A)Cos(B) + Cos(A)Sin(B) which I still remember!!!

Sin(x) - Sin(x+120deg) = Sin(x) - [Sin(x)*Cos(120) + Cos(x)*Sin(120)]

= Sin(x) -[-(1/2)*Sin(x) + (1.732/2)*Cos(x)]

= (3/2)*Sin(x) + (1.732/2)*Cos(x)

= 1.732[(1.732/2)*Sin(x) + (1/2)*Cos(x)]

= 1.732[Cos(30)*Sin(x) + Sin(30)*Cos(x)]

= 1.732*Sin(x + 30)

Which does look like the composite....amplitude is times root(3) and phase angle is shifted 30 deg if my math is correct

The 30 deg does not look right however......getting late.

It does prove the V L-L is single phase 60 Hz also which I had from another posting somewhere
 
Smart $ said:
= (3/2)*Sin(x) – (1.732/2)*Cos(x)
Click to expand...

Thank you for this correction.....

So here is the corrected version:

Use the old identity: Sin(A+B) = Sin(A)Cos(B) + Cos(A)Sin(B) which I still remember!!!

Sin(x) - Sin(x+120deg) = Sin(x) - [Sin(x)*Cos(120) + Cos(x)*Sin(120)]

= Sin(x) -[-(1/2)*Sin(x) + (1.732/2)*Cos(x)]

= (3/2)*Sin(x) - (1.732/2)*Cos(x)

= 1.732[(1.732/2)*Sin(x) - (1/2)*Cos(x)]

= 1.732[Cos(330)*Sin(x) + Sin(330)*Cos(x)]

= 1.732*Sin(x + 330)

This does look correct per the curves......
 
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