New NCEES #501

[Morz58]

I have been reviewing this problem with no success, can anyone tell me why in the solution you multiply by 2, why you are adding two currents instead of one?

 

[Flyer_PE]

^If I understand your question correctly, they are multiplying the distance by 2 since you have 200 ft of wire going out, and another 200 ft of wire coming back.

 

[Morz58]

I understand the multiplying by 200ft for the distance, but in the solution they set it up as

(Iline)(Z)+Vload+ (Iline)(Z), what are there two (Iline)(Z)’s? I hope that I am explaining it well enough.

 

[niurou]

I understand the multiplying by 200ft for the distance, but in the solution they set it up as (Iline)(Z)+Vload+ (Iline)(Z), what are there two (Iline)(Z)’s? I hope that I am explaining it well enough.

because you have 200ft conductor to the load, and another 200ft conductor from the load to the source, so that it can be a loop and current can flow. these two conductors all carry currents. so you need to count 200ft twice. This is true for the single phase.

For balanced three phase, the conductor from the load back to source is the Neutral conductor, but balanced three phase system has no current flows in the neutral conductor, so in this case, only one way conductor is calculated for voltage drop.

I hope that you are preparing for the Oct exam not the April one .

 

[Morz58]

because you have 200ft conductor to the load, and another 200ft conductor from the load to the source, so that it can be a loop and current can flow. these two conductors all carry currents. so you need to count 200ft twice. This is true for the single phase.
For balanced three phase, the conductor from the load back to source is the Neutral conductor, but balanced three phase system has no current flows in the neutral conductor, so in this case, only one way conductor is calculated for voltage drop.

I hope that you are preparing for the Oct exam not the April one .

I am sure if you aren’t explaining it well or not but your not correct. It is not a loop, both conductors are carrying current to the load, just because it is single phase does not mean there is no neutral, just look at your house panel, the white wire that’s the neutral. Think of this as the dryer or range in your house, ie: 240volt 50amps they are typically wired with two hot leggs and ground. Where the 50amps are split between the two wires (mostly) at full load, there is no conductor “looping” back to the source. Thanks for helping me figure it out anyway, and yes btw I am taking the test in April, maybe you shouldn’t.

 

[niurou]

I am sure if you aren’t explaining it well or not but your not correct. It is not a loop, both conductors are carrying current to the load, just because it is single phase does not mean there is no neutral, just look at your house panel, the white wire that’s the neutral. Think of this as the dryer or range in your house, ie: 240volt 50amps they are typically wired with two hot leggs and ground. Where the 50amps are split between the two wires (mostly) at full load, there is no conductor “looping” back to the source. Thanks for helping me figure it out anyway, and yes btw I am taking the test in April, maybe you shouldn’t.

240v/120v is different story. each line voltage is 120v, and they are 180 degree out of phase, so add up to 240v. if it's balanced, for 120v load line to neutral voltage drop, the "loop back" conductor is the neutral, since neutral has no current, you calculate only one way conductor.

but for 240v load voltage drop, you still need to count two way conductors. because now you are not using the neutral and both line conductor are carrying current.

of course there must be a conductor loop back to the source, no matter AC or DC, circuit is a loop. no loop no circuit.

 

[cabby]

Attached is how I approached the problem.

 

[geofs_PE]

240v/120v is different story. each line voltage is 120v, and they are 180 degree out of phase, so add up to 240v. if it's balanced, for 120v load line to neutral voltage drop, the "loop back" conductor is the neutral, since neutral has no current, you calculate only one way conductor.
but for 240v load voltage drop, you still need to count two way conductors. because now you are not using the neutral and both line conductor are carrying current.

of course there must be a conductor loop back to the source, no matter AC or DC, circuit is a loop. no loop no circuit.

The neutral of a 120V, 1-phase circuit carries the same current as the line under normal conditions and the voltage drop is calculated based on conductor length of twice the distance of the load from the source. The statement "since neutral has no current, you calculate only one way conductor" is incorrect.

 

[niurou]

The neutral of a 120V, 1-phase circuit carries the same current as the line under normal conditions and the voltage drop is calculated based on conductor length of twice the distance of the load from the source. The statement "since neutral has no current, you calculate only one way conductor" is incorrect.

The OP mentioned the 240v water heater and ranger, so I assumed he is talking about the split phase 240/120v in American. In this case, for a ideal balanced 120v load, neutral has no current, and the theoretic voltage drop is only one way the distance for 120v load. Pretty much the same as the balanced three phase voltage drop.

 

[Art]

first, the current doesn't 'cancel' in a 240 circuit...let's say the load is 10 ohm, the current is 24A in each leg...

you MUST have a loop or circuit path to conduct current...

EACH leg carries 24A, it's ALTERNATING current...

so the loss would be (Iline)(Zleg1)+Vload+ (Iline)(Zleg2), we assume that Z is the same in each leg...ie, same wire, same length

(Iline)(Z)+Vload+ (Iline)(Z) or Iline(Z+Z) + Vload or 2ZI line + Vload...

this is elementary stuff guys...

 

[Morz58]

first, the current doesn't 'cancel' in a 240 circuit...let's say the load is 10 ohm, the current is 24A in each leg...you MUST have a loop or circuit path to conduct current...

EACH leg carries 24A, it's ALTERNATING current...

so the loss would be (Iline)(Zleg1)+Vload+ (Iline)(Zleg2), we assume that Z is the same in each leg...ie, same wire, same length

(Iline)(Z)+Vload+ (Iline)(Z) or Iline(Z+Z) + Vload or 2ZI line + Vload...

this is elementary stuff guys...

This question is quite simple once you realize what they are asking (my problem) it is a simple voltage drop, if anyone has the NEC 2008 handbook, Article 215 on page 108 works out a comparable example. The main point to take away from this is that when you have a single phase circuit multiply by 2.