NCEES Power Problem 108

[BebeshKing PE]

How did they get the formula Ia+Ib= -In?

What I know is In=Ia+Ib+Ic. And if Ic=0, should it be Ia+Ib=In, right?

What am I missing here?

Thank you,

 

[DLD PE]

Yes for unbalanced loads, Ia+Ib+In = 0

Maybe this will help:

 

[DLD PE]

Oh no I'm having a brain fart:  How do we know Ic=0?  

I like how everyone posts these problems.  When I think I know something that I used to know, all of a sudden I realize it hasn't stuck yet.

 

[BebeshKing PE]

Oh no I'm having a brain fart:  How do we know Ic=0?  

I like how everyone posts these problems.  When I think I know something that I used to know, all of a sudden I realize it hasn't stuck yet.

I did the same solution Based on @supra33202 's solution per your attached comment above. I just don't know why the NCEES solution have the minus neutral current.

I know the problem was just asking about the magnitude, but I would also want to know if my concept is correct which is In=Ia+Ib+Ic. 'coz maybe in the future they will also asked for the vector of the neutral current.

In my thought, I assumed Ic=0 since this is a 3 phase system but there is no load in phase c and assumed to be open circuited. Hence, In=Ia+Ib

 

[Orchid PE]

It doesn't really matter which side I neutral is on.

With Ia+In+Ic=In or Ia+Ib+Ic+In=0 they will work out to the same answer because of KCL.

 

[BebeshKing PE]

It doesn't really matter which side I neutral is on.

With Ia+In+Ic=In or Ia+Ib+Ic+In=0 they will work out to the same answer because of KCL.

@Chattaneer PE, will the Neutral current displaced by 180 degrees if we use the other way around?

Ia+Ib+Ic+In=0

transpose In to the right of the equation:

Ia+Ib+Ic=-In  --> 180 degrees displaced

 

[Orchid PE]

@Chattaneer PE, will the Neutral current displaced by 180 degrees if we use the other way around?

Ia+Ib+Ic+In=0

transpose In to the right of the equation:

Ia+Ib+Ic=-In  --> 180 degrees displaced

The angle will tell you the direction. All depends if it's included at summing into the node or leaving the node.

We know phase A and B will be entering the node, and because of that we know the neutral current will be leaving the node and of equal magnitude of A+B, but the angle will be 180 the opposite direction.

 

[BebeshKing PE]

The angle will tell you the direction. All depends if it's included at summing into the node or leaving the node.

Thanks @Chattaneer PE.

So on this problem, the utility Y connected source is sending a current to the loads. And if the reference node is at the neutral, the 3 phase ungrounded currents are leaving the nodes and the  grounded neutral current is entering the node. Hence,

entering the node = leaving the node

In= Ia+Ib+Ic??

Please correct me if I'm wrong.

Thank you,

 

[Orchid PE]

I would say phase A and B currents are entering the node, and neutral current is leaving the node.

 

[Orchid PE]

Here's how I would draw the diagram:



From this we can make the equation Ia + Ib = In.

Ia = (200+100j) kVA / (7.62 ∠ 0 kV) = 29.3 ∠ 26.56

Ib = (200+100j) kVA / (7.62 ∠ 120 kV) = 29.3 ∠ 86.56

In = Ia + Ib = 29.3 ∠ 26.56 + 29.3 ∠ 86.56 = 29.3 ∠ -33.4

----

If we drew In going into the node (->, towards the right), the equation would be Ia + Ib + In = 0 (summing all entering the node). Transform to Ia + Ib = -In.

When we do all the calculations again, we get In = -29.3∠ -33.4. Since this is negative, we know In has to be leaving the node.

----

Just like in circuits when using KCL, the sign of the currents tells use the direction of flow based on how we draw the diagram.

 

[BebeshKing PE]

Here's how I would draw the diagram:

View attachment 16225

From this we can make the equation Ia + Ib = In.

Ia = (200+100j) kVA / (7.62 ∠ 0 kV) = 29.3 ∠ 26.56

Ib = (200+100j) kVA / (7.62 ∠ 120 kV) = 29.3 ∠ 86.56

In = Ia + Ib = 29.3 ∠ 26.56 + 29.3 ∠ 86.56 = 29.3 ∠ -33.4

----

If we drew In going into the node (->, towards the right), the equation would be Ia + Ib + In = 0 (summing all entering the node). Transform to Ia + Ib = -In.

When we do all the calculations again, we get In = -29.3∠ -33.4. Since this is negative, we know In has to be leaving the node.

----

Just like in circuits when using KCL, the sign of the currents tells use the direction of flow based on how we draw the diagram.

@Chattaneer PE, great explanation! Quick question though, is there gonna be a way in theory(or maybe in reality) that all the phase currents(Ia,Ib,Ic) including the neutral current(In) will all be entering a common node? If yes, in what scenario? 

Thanks,

 

[Orchid PE]

@Chattaneer PE, great explanation! Quick question though, is there gonna be a way in theory(or maybe in reality) that all the phase currents(Ia,Ib,Ic) including the neutral current(In) will all be entering a common node? If yes, in what scenario? 

Thanks,

The only time current flows on a neutral is when the load is not balanced.

Let's say phase A and B currents are the same, but greater than phase C current. In this scenario, the "excess" current from phases A and B that is not cancelled out by phase C will flow on the neutral, back towards the source.

----

Technically speaking, not to confuse you, since all this is AC the current alternates directions (60 times a second). 

 

[Zach Stone P.E.]

How did they get the formula Ia+Ib= -In?

What I know is In=Ia+Ib+Ic. And if Ic=0, should it be Ia+Ib=In, right?

Here's the trick:

Most books do not explain this well and gloss over it entirely. If you use a KCL equation to solve for the neutral current (In) from the three line currents (Ia, Ib, Ic) of a wye wye system, it looks like this:

• In = Ia + Ib + Ic
  

However, the direction of the neutral current is typically shown from the source to the load, just like the line currents. This is where the negative sign comes from. How do we change the direction of the neutral current to express it mathematically as flowing from the source to the load? We multiply it by negative one, or lead/lag the phase angle by 180 degrees (both have the same affect). 

The neutral current formula then becomes:

• In = -(Ia + Ib + Ic)   or   In = -Ia - Ib - Ic    or   -In = Ia + Ib + Ic 
  

In this particular problem since there is no C line current, the following happens once we substitute Ic = 0 into the previous formula:

• In = -(Ia + Ib + 0)
  
• In = -(Ia + Ib)
  

However, since the problem is only asking for the magnitude of the neutral current, and not the complex neutral current with magntidue and angle, the negative sign does not matter. 

Multiplying a complex number by negative one is really just leading/lagging the phase angle by 180 degrees, it has no affect on the magnitude.

So even if you solved it without the negative sign for the neutral current, you'll still end up with the same magnitude of 29.3A that the author did.

Try it:

 

[BebeshKing PE]

Here's the trick:

Most books do not explain this well and gloss over it entirely. If you use a KCL equation to solve for the neutral current (In) from the three line currents (Ia, Ib, Ic) of a wye wye system, it looks like this:

• In = Ia + Ib + Ic
  

However, the direction of the neutral current is typically shown from the source to the load, just like the line currents. This is where the negative sign comes from. How do we change the direction of the neutral current to express it mathematically as flowing from the source to the load? We multiply it by negative one, or lead/lag the phase angle by 180 degrees (both have the same affect). 

The neutral current formula then becomes:

• In = -(Ia + Ib + Ic)   or   In = -Ia - Ib - Ic    or   -In = Ia + Ib + Ic 
  

In this particular problem since there is no C line current, the following happens once we substitute Ic = 0 into the previous formula:

• In = -(Ia + Ib + 0)
  
• In = -(Ia + Ib)
  

However, since the problem is only asking for the magnitude of the neutral current, and not the complex neutral current with magntidue and angle, the negative sign does not matter. 

Multiplying a complex number by negative one is really just leading/lagging the phase angle by 180 degrees, it has no affect on the magnitude.

So even if you solved it without the negative sign for the neutral current, you'll still end up with the same magnitude of 29.3A that the author did.

Try it:

View attachment 16241

Nice. This is what I thought, that typically the neutral current is also flowing from source to the load. and since it will be negative, the actual flow will be opposite.

Thank you @Zach Stone, P.E.