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Single phase to three phase calculation relation

Bjtpower_magic

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Hi Everyone,

Could you please help me with the relationship between Single phase to 3 Phase.
I have
L1-N=230V
L2-N=50V
L3-N=70V
N=0V

From above data how i could derive voltage between phases like L1-L2, L2-L3 and L3-L1
 

wwfeldman

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3 phase is three single phase systems
one phase is used as a reference and consequently its phase angle is 0 degrees
one phase is shifted 120 degrees from the reference
one phase is shifted (-120 degrees) from the reference
hence, there is 120 degrees from any phase to any other phase
(-120 is the same as +240)

L-L is line to line, as in a delta arrangement and there is no neutral
L-N is line to neutral, as in a Y arrangement

see these
 

FvM

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The voltages can be calculated under the assumption, that the voltages have 3x120° phase relation. But this must not necessarily the case, it depends on how the asymmetrical three phase system is generated. In other words, the problem specification is incomplete.
 

c_mitra

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Could you please help me with the relationship between Single phase to 3 Phase.
I have
L1-N=230V
L2-N=50V
L3-N=70V
N=0V

From above data how i could derive voltage between phases like L1-L2, L2-L3 and L3-L1
Begin at the end; N=0V does not convey much; it was measured against what?
Anyway, the neutral (as the name says) is usually at the ground potential ...

Under normal conditions, all the three phases (L1, L2 and L3) are symmetrically placed with respect to the neutral. So all the three voltages should be the same.

It is not possible that all the three phases show positive voltage at any given instant. Also we need to know the peak voltage so that the waveforms can be plotted.
 

Bjtpower_magic

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Hi,

Are the given values "RMS" values, or amplitude, or instantaneous?

Klaus
Values are given in RMS
I want to calculate in Star connection not delta
--- Updated ---


Begin at the end; N=0V does not convey much; it was measured against what?
Anyway, the neutral (as the name says) is usually at the ground potential ...

Under normal conditions, all the three phases (L1, L2 and L3) are symmetrically placed with respect to the neutral. So all the three voltages should be the same.

It is not possible that all the three phases show positive voltage at any given instant. Also we need to know the peak voltage so that the waveforms can be plotted.
Measured against N.. system is Star connection
 
Last edited:

KlausST

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From above data how i could derive voltage between phases like L1-L2, L2-L3 and L3-L1
And
I want to calculate in Star connection not delta
How do both match? Since L1-L2 is delta...
****
As far as I can see there is no solution if you expect equal delta voltages.(since 50V + 70V is smaller than 230V)
If you don't expect equal delta voltages, then there are infinite solution, or in other words: it can't be solved.

Klaus
 

c_mitra

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Measured against N.. system is Star connection
Values are given in RMS

I want to calculate in Star connection not delta

--- Updated ---

The voltages on each of the three phases (lines in this context) d
Values are given in RMS

I want to calculate in Star connection not delta

--- Updated ---

Voltages does not depend on the way you connect it.

Star or delta refers to different way the three phases are connected to a load. But that does not change the voltage on the lines.

Now coming to your comment that the three reported values are RMS values:

Three phases are identical except in their phase (they are phase shifted from each other by +/- 120 deg). They usually have equal amplitude.

If due to some "fault" the voltages have changed from the "common" value, you need to confirm that the phases have not been changed.

The RMS value is closely related to the peak value: RMS voltage = peak voltage / (sqrt (2)); peak value = 1.414 * RMS voltage

Because the three phases are not the same amplitude, you cannot use the same idea. Take L1 as reference (230*1.414*sin(t)) and L2 as +120 deg shifted (50*1.414*sin(t+120)) and L3 as +240 deg shifted (70*1.414*sin(t+240)). Now you can calculate RMS (L1-L2) etc.

I do not want to do that calculation!!
 

Bjtpower_magic

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Because the three phases are not the same amplitude, you cannot use the same idea. Take L1 as reference (230*1.414*sin(t)) and L2 as +120 deg shifted (50*1.414*sin(t+120)) and L3 as +240 deg shifted (70*1.414*sin(t+240)). Now you can calculate RMS (L1-L2) etc.

Based on above.
What is "t"

If i considered t=0, then sin (0)=0, L1=0V

Sin (120)=0.86, L2=50*1.414*0.86=60.8V
Sin (240)=-0.86, L3=70*1.414*0.86=-85V

Can you please comment on L1 Calculation?
 

c_mitra

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Based on above.
What is "t"

If i considered t=0, then sin (0)=0, L1=0V

Sin (120)=0.86, L2=50*1.414*0.86=60.8V
Sin (240)=-0.86, L3=70*1.414*0.86=-85V

Can you please comment on L1 Calculation?
Because you have not mentioned the frequency, let us assume it is 50Hz. The waveform is be given (for L1) as 230*1.414*sin(2*pi*50*t)

Here t is the time and you need to consider that for one period (complete cycle).

Because sine function has a period of 2*pi, the above equation has a period with t corresponding to t=1/(50) or 20 ms.

Here pi is the irrational number; 120 is in degrees; it is actually 2*pi/3 and 240 degrees is to be put as 4*pi/3. The argument of the sine function has to be in radian.

Your calculations are correct but they represent instantaneous values. You have to calculate the RMS.

Doing that needs calculus (integration). I shall tell you a simpler method.

Use an excel sheet and put the first col as t: start from zero and go to 20 in steps of 0.1; you will have 200 rows of number.
second col use 2*pi*50*col(1)/1000; you get it? It is quite simple. This col will run from 0 to 2*pi (div by 1000 is for sec to ms)
Third col use 230*1.414*sin(col(1)); fourth col use 50*1.414*sin(col(1)+2*pi/3); fifth col use 70*1.414*sin(col(1)+4*pi/3)
3rd 4th and 5th col are giving L1, L2 and L3. To calculate L1-L2; do the following:

6th col = (col(3)-col(4)); 7th col =col(6)*col(6); 8th col = sqrt(sum (col(8)))
 

KlausST

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Hi,

Without additional information one can not solve the maths.

When one thinks about 3 phase grid voltage, all three delta voltages usally are almost identical and the phases are 0°, 120° and 240°.
But when the star of the source is not conected to the star of the load...and the load is not symmetric, then you see different star_voltage referred to load_star.

So tell us - no, draw a sketch - of your situation. With source, load, star points..voltages, phases...

Klaus
 

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