3phase volt drop.

[Simonslimline]

This is one of a weekly question by sparks magazine posted on Facebook to win free tools each week. Everyone has answered B. 8Volts. I am not sure this is correct. Please could anyone help me with this. I have posted it up in here rather than just the trainee section so more people will see it. Not bothered about the prize. I just have a feeling that 8volts is wrong, but would happily be corrected on this. Thanks.

To comply with IEE Wiring Regulations the voltage drop on a three phase 400v Circuit should not exceed:
a) 4v
b) 8v
c) 20v
d) 32v

 

[Silly Sausage]

I think I've sussed it!!!

 

[Chr!s]

I think I've sussed it!!!

Which bit?

If we take 20 amps at a reference of (-240), whats the angle to the reference phase?

cheers

 

[Simonslimline]

I think I've sussed it!!!

The meaning of life or the volt drop question!:teeth_smile:

 

[Silly Sausage]

Look at the 3 phase waveforms...
When L1 is at V[SUB]pk[/SUB], L2 & L3 are at -0.5V[SUB]pk[/SUB]'

so... assuming 5% drop and balanced load,
vd on L1 is 0.05*sqrt(2)*230 = 16.3V
vd on L2/3 0.05*sqrt(2)*-230/2 = -8.13V

so total vd is 16.3 + (-8.13) = 8.13Volts.



Whether there's any truth in the above, I'll wait for higher beings to comment!!!

 

[Silly Sausage]

The meaning of life or the volt drop question!:teeth_smile:

Oh, that's easy, lots of beer!

 

[Chr!s]

Look at the 3 phase waveforms...
When L1 is at V[SUB]pk[/SUB], L2 & L3 are at -0.5V[SUB]pk[/SUB]'

so... assuming 5% drop and balanced load,
vd on L1 is 0.05*sqrt(2)*230 = 16.3V
vd on L2/3 0.05*sqrt(2)*-230/2 = -8.13V

so total vd is 16.3 + (-8.13) = 8.13Volts.



Whether there's any truth in the above, I'll wait for higher beings to comment!!!

None what so ever

 

[Simonslimline]

Look at the 3 phase waveforms...
When L1 is at V[SUB]pk[/SUB], L2 & L3 are at -0.5V[SUB]pk[/SUB]'

so... assuming 5% drop and balanced load,
vd on L1 is 0.05*sqrt(2)*230 = 16.3V
vd on L2/3 0.05*sqrt(2)*-230/2 = -8.13V

so total vd is 16.3 + (-8.13) = 8.13Volts.



Whether there's any truth in the above, I'll wait for higher beings to comment!!!

Definitely something i have not done at college. Very interesting though.

 

[Chr!s]

Archy, there is a nice easy one in post 62

 

[Simonslimline]

Oh, that's easy, lots of beer!

Yep i agree. Drinking beer is much simpler and easier than electrics aswell.

 

[Chr!s]

Look at the 3 phase waveforms...
When L1 is at V[SUB]pk[/SUB], L2 & L3 are at -0.5V[SUB]pk[/SUB]'

so... assuming 5% drop and balanced load,
vd on L1 is 0.05*sqrt(2)*230 = 16.3V
vd on L2/3 0.05*sqrt(2)*-230/2 = -8.13V

so total vd is 16.3 + (-8.13) = 8.13Volts.



Whether there's any truth in the above, I'll wait for higher beings to comment!!!

Just remember V = I x Z, so without I and Z there is no V

Cheers

 

[Chr!s]

I can see now said the blind man

Ive not seen that method Simon provided

Okay i used polar coordinates to arrive at my 15 amps at a angle of 60 degrees.

Reference L1 = 40, L2 =30 and L3 =20

So in rectangular form we have 15 + j8.66, if we Sqroot the rectangular form you get the magnitude 17.32 amps.

My phasor diagrams back this up, what happened to my earlier diagrams, dodgy ruler

Ill crawl back under my stone now.

Chris

 

[D Skelton]

Do what?!?!
Workings please!

No need for workings, people are looking way too much into this.

Imagine a TP+N board. On this board is one single balanced three phase load pulling 10A per phase, what is the current in the supply neutral? 0A right? Add a 5A single phase load onto that board, what is the current in the supply neutral then? 5A. This would be both the total, and the maximum amount of current in the neutral.

Add another 5A single phase load to a different phase as the first 5A load, the total neutral current is now 10A, but the maximum amount of current in the neutral at any one time is still 5A.

 

[Simonslimline]

Edit - Found info

 

[davesparks]

No need for workings, people are looking way too much into this.

Imagine a TP+N board. On this board is one single balanced three phase load pulling 10A per phase, what is the current in the supply neutral? 0A right? Add a 5A single phase load onto that board, what is the current in the supply neutral then? 5A. This would be both the total, and the maximum amount of current in the neutral.

Add another 5A single phase load to a different phase as the first 5A load, the total neutral current is now 10A, but the maximum amount of current in the neutral at any one time is still 5A.

The neutral current won't be 5 amps in your final scenario there.

 

[D Skelton]

The neutral current won't be 5 amps in your final scenario there.

It would be. Do the math