Graffeo - Single Phase Problem

[applepieordie]

Hi,

can someone explain to me the whole 'two wires' thing in single phase circuit and why the impedance is 2 times.

Thanks in advance

 

[applepieordie]

Hi,

can someone explain to me the whole 'two wires' thing in single phase circuit and why the impedance is 2 times.

Thanks in advance

Here is the problem im talking about. doesnt look like it attached properly in my first post

View attachment 9785View attachment 9786

 

[jwholla]

The reason it's two wires is because you have one wire going to the load and one wire coming back. The length given in the problem is the length only in one direction; therefore, to account for the total impedance of the wires, you must multiply the distance by 2. Hopefully this makes sense. 

 

[cupojoe PE PMP]

Like jwholla said, you have two directions. The voltage across the load terminals will be lower due to impedance on both sides of the load.

 

[rg1]

In applying Ohms law V=IR or V=IZ we have to take current and the resistance of the path it follows. In single phase it follows two conductors, so we take 2 times of the one conductor.  In the attached figure of single phase the voltmeters V2 and V4 read the voltage drops in conductors, V3 is load voltage and V1 is source voltage. So V1= (V2+V3)+V4. Since V3 and V4 are Voltage drops on conductors having same impedance and same current, they are equal. So we take double of  the one. I have taken one side of source as zero voltage and shown the voltages of the other nodes with reference to it to give better appreciation. Does this make sense?

View attachment 9790

 

[applepieordie]

In applying Ohms law V=IR or V=IZ we have to take current and the resistance of the path it follows. In single phase it follows two conductors, so we take 2 times of the one conductor.  In the attached figure of single phase the voltmeters V2 and V4 read the voltage drops in conductors, V3 is load voltage and V1 is source voltage. So V1= (V2+V3)+V4. Since V3 and V4 are Voltage drops on conductors having same impedance and same current, they are equal. So we take double of  the one. I have taken one side of source as zero voltage and shown the voltages of the other nodes with reference to it to give better appreciation. Does this make sense?

View attachment 9790

yes this makes sense and is much easier to see when the circuit is drawn out as is in your reply rather than the one line diagram in the solution. thank you all for the replies.

 

[BigWheel]

All of these explanations are great and completely valid.

To make this easy to remember for instant recall during the exam, use this:

"Single-Phase is Two Times, and Three-Pase is One Times."

...or, "1 is 2 and 3 is 1."

Kinda clumsy, but it got me through Circuits II. ?

 

[TNPE]

Actually, 3-phase is not 1, rather it is 1.732.  It still has to return to the source.  It doesn't hit an arbitrary load and stop there.  How does it return on a 3-phase system, you ask?  On another phase (although, the imbalance returns on the neutral, if in fact it is a 3-phase, 4-wire system).  

 

[BigWheel]

Actually, 3-phase is not 1, rather it is 1.732.  It still has to return to the source.  It doesn't hit an arbitrary load and stop there.  How does it return on a 3-phase system, you ask?  On another phase (although, the imbalance returns on the neutral, if in fact it is a 3-phase, 4-wire system).  

The "1 is 2 and 3 is 1" monicker means "single phase counts 2 times the conductor length and 3 phase counts 1 times the conductor length."

The OP was confused about why the conductor length was doubled.

 

[supra33202]

Why count 1 transmission line for 3-phase system?

One phase conductor connecting the load

Another phase conductor/neutral conductor connecting the load

Should we count 2 transmission line (2 conductors) for 3-phase system?

 

[BigWheel]

Why count 1 transmission line for 3-phase system?

One phase conductor connecting the load

Another phase conductor/neutral conductor connecting the load

Should we count 2 transmission line (2 conductors) for 3-phase system?

In a single-phase system, current travels to the load on the line conductor and returns on the neutral conductor, so this "series" circuit voltage is dropped across both the line and neutral conductors.

In a three-phase system, you have two conditions: balanced and unbalanced. In a balanced three-phase system, there is no current in the neutral conductor (each phase's current cancels the other two phase's currents). Some examples of balanced 3-phase loads include 3-phase motors, pumps, and VFDs.

Conversely, with an unbalanced 3-phase system, current is produced in the neutral; however an unbalanced 3-phase system is rarely (if ever) calculated considering the line and neutral impedances because the effort to perform the calculations isn't worth the results.

I'm happy to talk to this and prove what I just claimed, but I'll spare the community until pressed to do so, but I will conclude with this: a three-phase system is not a series system; it's a parallel system, so the math is considerably more complicated than a single phase analysis, and the kicker is that you can do the math and calculate the true 3-phase voltage drop of an unbalanced 3-phase system, but the results of these calculations will leave you wondering why you even bothered.

Finally, it helps to understand why "Voltage Drop" is such a hot topic. Is it because it's a code requirement? It's not a code requirement, but rather a recommendation. It is absolutely true that too much voltage drop will cause the load to behave erratically, and can even be a very dangerous (and even deadly) situation for craftspersons working at the load; nevertheless, it's not a "CODE REQUIREMENT." The NEC recommends no more than a 3% voltage drop on the branch circuit and no more than 2% on the feeder. Voltage drop calculations are essential to the difference between a true engineered design and a classroom assignment.

 

[TNPE]

But you failed to mention how you have a "circuit" in a 3-phase, 3 wire system.  As mentioned above, this is where sqrt 3 rears itself yet again, and the circuits are made on another phase (cyclically alternating...i.e. A-B-B-C-C-A, for example, where A returns on B, B returns on C etc.)

 

[BigWheel]

Yep, that's true. The three phases serve as return paths for each other, but I think dwelling on this too much will just create more confusion than necessary for people trying to prepare for the exam when all they want to know is why a single phase VD calculation counts the conductor length twice while the 3-phase VD calculation only counts it once.

There are two types of 3-phase configurations: delta and wye. In a delta configuration, there is no neutral, so that was easy - just one conductor length is considered.

In a wye configuration, there is a neutral. The return current on the neutral in a grounded-wye three-phase system is a vector sum of the three (3) phase currents. 

In a balanced, grounded-wye configuration, the vector sum of the currents will result in 0 amps in the neutral, so no voltage drop in the neutral; therefore, only one conductor length is considered.

In an unbalanced, grounded-wye configuration, the current in the neutral depends on what the unbalanced conditions are. It can be very close to zero, meaning we're almost (but not quite) balanced (typical), or it can be as high as one of the phase currents alone (atypical, and is just one of the reasons why the NEC requires that the neutral conductor in a grounded-wye configuration is the same gauge/size as the phase conductors).

But the total three-phase voltage drop is a vector sum of the voltage drop across each phase and the neutral. The sqrt3 simply falls out of the 3-phase formulae derivations.

When performing 3-phase voltage drop calculations in practice, a balanced three-phase load is usually assumed because you're working with something like a motor, pump, transformer, VFD (intrinsically balanced), or you're hanging a new three-phase panelboard, and if you're doing that (hanging a new panelboard), a good engineer recognizes that it's good practice to arrange the circuits such that the anticipated loads are balanced across all three phases of the new panelboard.

 

[TNPE]

No confusion, but it's a fundamental concept that has to be understood to move forward with -why?- you don't account for a return path in 3-phase applications of this type.  Forget about distance, I'm not talking about distance -or counting something twice-, purely concepts.

 

[rg1]

No confusion, but it's a fundamental concept that has to be understood to move forward with -why?- you don't account for a return path in 3-phase applications of this type.  Forget about distance, I'm not talking about distance -or counting something twice-, purely concepts.

The issue is only and only of application of KVL. You trace your path of current and apply KVL whether three phase or single pahse, you get the answer. But all may not understand the concept properly and so for them thumb rules are handy to solve the problem to pass PE exam. So once we say that unbalanced (slightly) three phase problem is not asked in PE axam the rules of thumb being followed and being reiterated by @BigWheel are good enough. 

 

[TNPE]

True, rg1.  But I find it more advantageous to be successful if one understands concepts more so than a generic memory trigger.  For those who have sat for this test can agree that understanding concepts, more times than not, will determine any given candidate's ability to pass.  Those who know them and fully understand -why?- will be successful more than those who look for an equation to plug and chug.  This exam is not designed for that!  I'm all for memory triggers that will always be right (for example, eli the ice man, which I just gave to another visitor to this site a few mins ago), but with subtle nuances in presentation or what's being asked for, and it's a totally different problem that doesn't work by cute slogans.

 

[supra33202]

In a single-phase system, current travels to the load on the line conductor and returns on the neutral conductor, so this "series" circuit voltage is dropped across both the line and neutral conductors.

In a three-phase system, you have two conditions: balanced and unbalanced. In a balanced three-phase system, there is no current in the neutral conductor (each phase's current cancels the other two phase's currents). Some examples of balanced 3-phase loads include 3-phase motors, pumps, and VFDs.

Conversely, with an unbalanced 3-phase system, current is produced in the neutral; however an unbalanced 3-phase system is rarely (if ever) calculated considering the line and neutral impedances because the effort to perform the calculations isn't worth the results.

I'm happy to talk to this and prove what I just claimed, but I'll spare the community until pressed to do so, but I will conclude with this: a three-phase system is not a series system; it's a parallel system, so the math is considerably more complicated than a single phase analysis, and the kicker is that you can do the math and calculate the true 3-phase voltage drop of an unbalanced 3-phase system, but the results of these calculations will leave you wondering why you even bothered.

Finally, it helps to understand why "Voltage Drop" is such a hot topic. Is it because it's a code requirement? It's not a code requirement, but rather a recommendation. It is absolutely true that too much voltage drop will cause the load to behave erratically, and can even be a very dangerous (and even deadly) situation for craftspersons working at the load; nevertheless, it's not a "CODE REQUIREMENT." The NEC recommends no more than a 3% voltage drop on the branch circuit and no more than 2% on the feeder. Voltage drop calculations are essential to the difference between a true engineered design and a classroom assignment.

Thanks for spending time on answering my questions!

in a single-phase system, current travels to the load on the line conductor and returns on the neutral conductor, so this "series" circuit voltage is dropped across both the line and neutral conductors.

I agree.

In a three-phase system, you have two conditions: balanced and unbalanced. In a balanced three-phase system, there is no current in the neutral conductor.

I agree.

(each phase's current cancels the other two phase's currents)

Could you elaborate?

I'm happy to talk to this and prove what I just claimed, but I'll spare the community until pressed to do so, but I will conclude with this: a three-phase system is not a series system; it's a parallel system, so the math is considerably more complicated than a single phase analysis, and the kicker is that you can do the math and calculate the true 3-phase voltage drop of an unbalanced 3-phase system, but the results of these calculations will leave you wondering why you even bothered.

If you time, could you tell me more?

three-phase system is not a series system; it's a parallel system

Could you elaborate?

My thoughts: 3-phase system is kind of like 3 separate 1-phase systems. Regardless of delta or wye connections, the load will connect to 2 conductors (phase and phase or phase and neutral). When we do voltage calculation, I still don't see why we don't use the same equation for 1-phase and 3-phase system.

To complete a circuit. Current flows from the source through one conductor into the load and flow out of the load through another conductor. When we do voltage drop calculation, I still don't understand why we don't account for the return path conductor for 3-phase system.

For a typical 3-phase PE  circuit analysis problems, we look at one phase and solve the problem, and convert our 1-phase answer to 3-phase answer. Why is it different when we do voltage drop problems?

For the PE exam's voltage drop questions, I know I should count 1 transmission line for 3-phase system, and 2 transmission lines for 1-phase system. But I still want to know why.

Thanks!

 

[TNPE]

To eliminate confusion, the sqrt3 is not a product of the run-of-the-mill use that is seen in power calculations.  It is actually a factor of the distance, but due to phase shifts between phases in 3-phase systems, this is why it is sqrt3 and not 2.  Remember, this is under balanced conditions; if unbalanced, then the method of symmetrical components has to be implemented.