Calculations

Status
Not open for further replies.
E

Eric1119

Member
Calculate the neutral current in a 208Y/120 volt 3 phase, 4 wire system when he current in phase A is 40Amps, in phase B is 20 Amps and in phase C is 30 Amps.:)
 
You are asking homework or study questions and we will not just provide you with answers, you must show your answers first and how you arrived at them then we will point you in the right direction.


Roger
 
Eric1119 said:
Calculate the neutral current in a 208Y/120 volt 3 phase, 4 wire system when he current in phase A is 40Amps, in phase B is 20 Amps and in phase C is 30 Amps.:)
Click to expand...

1 answer I can come up with by manipulating the loads is

1- 20 amp 3 phase load

1- 10 amp 208 volt single phase load connected A to C

1- 10 amp 120 volt single phase load

gives you 10 amps neutral current.

My guess is that is not the answer you were looking for.
 
masterinbama said:
1 answer I can come up with by manipulating the loads is

1- 20 amp 3 phase load

1- 10 amp 208 volt single phase load connected A to C

1- 10 amp 120 volt single phase load

gives you 10 amps neutral current.

My guess is that is not the answer you were looking for.
Click to expand...

I think that your logic is incorrect. I came up with 17.3 amps.
 
Squrt ( Ia^2+Ib^2+Ic^2)-(Ia*Ib+Ib*Ic+Ic*Ia)

I=amps
a, b, and c = the 1st, 2nd, and 3rd phase amperage respectively
^2 = squared
Sqrt = after all is done you take the square root of the answer.

I get the square root of 300 = 17.3
 
I agree with the 17.3A given by both Dexie and Infinity.
But only for the specific case where the power factor on all three phases is exactly the same i.e. the currents are phase displaced from each other by 120degE.
In the more general case where the power factors are not the same you could get a quite different answer. The example below still has 40A, 20A and 30A for A,B, and C respectively but the neutral current (black) is a little over 50A rms.

three-phase02.jpg
 
1 answer I can come up with by manipulating the loads is

1- 20 amp 3 phase load

1- 10 amp 208 volt single phase load connected A to C

1- 10 amp 120 volt single phase load


I should have stipulated the first 2 loads have no neutral connection, only the 10 amp 120 volt load
 
masterinbama said:
I should have stipulated
the first 2 loads have no neutral connection,
only the 10 amp 120 volt load
Click to expand...

MasterInBama,
alias Perry Mason, "stipulate"!

The OP "stipulated" a WYE circuit.

Try this on any combo:

Square root of
( (Ia^2)+(Ib^2)+(Ic^2) )-( (Ia*Ib)+(Ib*Ic)+(Ic*Ia) )

with the "stipulation" that
if a line is not connected to neutral,
then that line current is zero.

:)
 
Last edited:
masterinbama said:
1 answer I can come up with by manipulating the loads is

1- 20 amp 3 phase load

1- 10 amp 208 volt single phase load connected A to C

1- 10 amp 120 volt single phase load


I should have stipulated the first 2 loads have no neutral connection, only the 10 amp 120 volt load
Click to expand...

That still doesn't work.
For example, the current in the 120V single phase load won't be in phase with the current in the 208V single phase load so you can't just add them arithmetically.
 
Besoeker said:
That still doesn't work.
For example, the current in the 120V single phase load won't be in phase with the current in the 208V single phase load so you can't just add them arithmetically.
Click to expand...
There isn't any adding required for the neutral conductor. The loads only add, vectorially, at the line connection. ;)
 
Smart $ said:
There isn't any adding required for the neutral conductor. The loads only add, vectorially, at the line connection. ;)
Click to expand...
The loads also add vectorially in the neutral. That's how you end up with zero in the neutral if you have a balanced load. Suppose you have 1A at zero, 1A at 2Pi/3, and 1A at 4Pi/3.

Then:
Ia = 1
Ib = -0.5 - jsqrt(3)/2
Ic = -0.5 + jsqrt(3)/2

In = Ia + Ib + Ic
= 0

QED
;)

But my point was that masterinbama could only have arrived at the 40A for Ia by arithmetic addition of the currents he gave (20+10+10)
 
Smart $ said:
If anyone reading that thread wants to download my Excel neutral calculator, which allows for entry of and accounts for power factor
Click to expand...
I already have one I made earlier, thanks. That's how I generated the waveforms I put in post #9.
:)
 
Besoeker said:
The loads also add vectorially in the neutral. That's how you end up with zero in the neutral if you have a balanced load. Suppose you have 1A at zero, 1A at 2Pi/3, and 1A at 4Pi/3.

Then:
Ia = 1
Ib = -0.5 - jsqrt(3)/2
Ic = -0.5 + jsqrt(3)/2

In = Ia + Ib + Ic
= 0

QED
;)

But my point was that masterinbama could only have arrived at the 40A for Ia by arithmetic addition of the currents he gave (20+10+10)
Click to expand...

My first post was a little vague. Allow me to add to it

1 answer I can come up with by manipulating the loads is

1- 20 amp 3 phase load with no neutral connection

1- 10 amp 208 volt single phase load connected A to C with no neutral connection

1- 10 amp 120 volt single phase load

gives you 10 amps neutral current.
 
Besoeker said:
I agree with the 17.3A given by both Dexie and Infinity.
But only for the specific case where the power factor on all three phases is exactly the same i.e. the currents are phase displaced from each other by 120degE.
In the more general case where the power factors are not the same you could get a quite different answer. The example below still has 40A, 20A and 30A for A,B, and C respectively but the neutral current (black) is a little over 50A rms.

three-phase02.jpg
Click to expand...

When I first read the OP, and did a quick calculation in my head, my mind said....worst case, 50A on the neutral.

Try as I might, I cannot repeat the calculation.:roll:

I had either a flash of brilliance, or a moment of delusion.:D

steve
 
Status
Not open for further replies.
Top