Parallel conductors

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Carultch

Senior Member
Location
Massachusetts
Of course that's picking one failure mode that makes it worse when tapping all 3 - But I can't think offhand of a failure mode that's worse when tapping only one

Let's start with a 500A service, made up of two parallel sets of 250 kcmil. Suppose we desire to connect 200A of PV, which is 3/0. Ignore the 1.25 safety factor, for simplicity, i.e. assume that this is already included in all applicable ampere figures. The 250 kcmil service conductors run a distance of 500 ft from the transformer to the service disconnect, and the PV is tapped onto set #1, within 1 ft of wire length from the service disconnect. That's a voltage drop of about 10 volts, under full load.

Under full load/source, the PV system produces 200A, and the building loads use 500A. This means 300A comes from the utility, and must be divided between the 2 sets of 250 kcmil.

Voltage drop between the two sets must be equal. Current will adjust between both sets, until voltage drop is equal.

The voltage difference from the tap point to the service disconnect is negligible. Most of the voltage drop happens in the distance from the transformer. The 300A from the utility will divide between the two sets, at 150A each. And it will divide this way, in order to make the voltage drop equal in both sets. Resistance of 499 ft and 500 ft of the same wire is nearly the same.

Set #2 experiences 150A. No problem for a 250 kcmil wire.
Set #1 for most of the distance experiences 150A. And in the final foot, 200A are added to it. That ends up being 350A.

The service disconnect sees 500A, which it is rated to withstand. But there is no overcurrent system to protect the final foot of set #1, from this local overload condition.
 

Carultch

Senior Member
Location
Massachusetts
Let's start with a 500A service, made up of two parallel sets of 250 kcmil. Suppose we desire to connect 200A of PV, which is 3/0. Ignore the 1.25 safety factor, for simplicity, i.e. assume that this is already included in all applicable ampere figures. The 250 kcmil service conductors run a distance of 500 ft from the transformer to the service disconnect, and the PV is tapped onto set #1, within 1 ft of wire length from the service disconnect. That's a voltage drop of about 10 volts, under full load.

Solve this circuit for current in R1, R2, and R1'. Each of the given ohm values is the effective round-trip AC resistance of the length of the 250 kcmil feeder. 490V is the voltage at the transformer, and 480V is the voltage at the loads. Answers should be close to my above example, to demonstrate the issue with breaking the symmetry.
 

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GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
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Retired PV System Designer
Thanks Carultch,
I felt strongly that there was a way that current could increase in the tapped conductor but just could not put my finger on it.
Nice work!
 

Smart $

Esteemed Member
Location
Ohio
This has been covered... but I don't think it has set in. :blink:

In order to see the non-compliance, one has to look at it from the perspective of parallel conductors from PV to POCO source. Starting at the PV system disconnect, the parallel conductors start at the tap. When only tapping one set of parallel service conductors, the service conductor on the line side and the service conductor conductor on the load side of the tap become the parallel sets of the PV system [red vs. blue and 310.10(H)].

unequal%20parallel.gif
 

kwired

Electron manager
Location
NE Nebraska
Thanks Carultch,
I felt strongly that there was a way that current could increase in the tapped conductor but just could not put my finger on it.
Nice work!

So did I, it may be that in many cases it isn't significant enough to matter, but everything ever taught about parallel conductors says to ensure impedance of each element of the run has as close as possible to same impedance or you will get imbalances in current flow, so logic tells me why intentionally introduce something different in only one element and disturb the balance you tried to achieve when you selected same size, length and other same physical characteristics of those conductors? Adding more current to just a part of one of the conductors is going to change the voltage drop across that conductor = it changed impedance, doesn't matter if impedance went up or down - it is still not same as the others of the set and depending on load conditions one element will have different current then the others for certain.
 

Carultch

Senior Member
Location
Massachusetts
So did I, it may be that in many cases it isn't significant enough to matter, but everything ever taught about parallel conductors says to ensure impedance of each element of the run has as close as possible to same impedance or you will get imbalances in current flow, so logic tells me why intentionally introduce something different in only one element and disturb the balance you tried to achieve when you selected same size, length and other same physical characteristics of those conductors? Adding more current to just a part of one of the conductors is going to change the voltage drop across that conductor = it changed impedance, doesn't matter if impedance went up or down - it is still not same as the others of the set and depending on load conditions one element will have different current then the others for certain.

For linear ohmic components, impedance doesn't depend on voltage or current. It's a function of the geometry/materials/EM field spaces of the component, and the AC frequency. So that would mean size, length, type, temperature, conditions of use, and conduit type.

It is the voltage drops and imbalance of current distribution that causes the issue.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Why does it decrease current?
Kirchoff's Law. When the PV is operating the current between the tap and the load stays the same but the current from the service to the tap is reduced by however much current the PV system is injecting. That's for a single conductor, of course, but I do not see how if it were one of a set of parallel conductors it could result in more current in the non-tapped conductors.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
Kirchoff's Law. When the PV is operating the current between the tap and the load stays the same but the current from the service to the tap is reduced by however much current the PV system is injecting. That's for a single conductor, of course, but I do not see how if it were one of a set of parallel conductors it could result in more current in the non-tapped conductors.

You did not follow the details of C's analysis.
If the current from the service to the tap point is reduced, then the voltage drop too.
But the voltage drop of each of the parallel sets, end to end, must be the same

The only way this can balance is for the current in the tapped conductor to increase and that in the untapped conductors to decrease.
The end result is that less current still flows in the service to tap segment, but more current flows in the tsp to load segment.
Consider the extreme case in which the inverter supplies the (or is capable of supplying) the entire load current for that parallel segment.
The voltage drop from service to load will, to first approximation, now be entirely frop tsp to load. This cannot balance the other parallel segments so there will still be current (and substantial current at that) flowing from service past the tap to the load.
QED.
 

Carultch

Senior Member
Location
Massachusetts
You did not follow the details of C's analysis.

If anyone is lost with my explanation, I recommend downloading a trial version of P-Spice, and drawing out my network of resistances, voltage sources and current sources. Then simulate the current in each resistor (representing each wire segment in the situation of question). And play with the resistances, voltages and currents, to see how this loophole results in current exceeding the wire ampacity.
 

Smart $

Esteemed Member
Location
Ohio
It can. It overloads the segment of the tapped conductor, between the loads and the tap point.
Not convinced. I've worked with SPICE-based apps before (not for several years now, though). They all require proper setup to get true-to-life results. rather than getting into the pros and cons of SPICE-based apps, suffice it to say in many to most cases regarding power circuits that we discuss here, set up will fall short of mimicking a true-to-life scenario. (And just to add, I do not have a SPICE-based app installed on my computers and I'm not going to install one just to make a point).

Anyway, I am curious as to what your actual results were (or are). Let's say POCO 490V 1?, 250kcmil parallel conductors as depicted earlier. PV set up as constant current supply of 200A. One leg of supply system grounded... as I recall SPICE requires some point on the system to be grounded. System load set up to sink 500A or anything close (excluding conductor resistance simulating resistances... I know it is difficult to setup up the load to be dynamic enough to have an exact 500A draw, anticipate 480V at load end of service conductor sets). Don't forget to put in matching resistances on the return legs.

What is the current level on each section of the service conductors?
 

jaggedben

Senior Member
Location
Northern California
Occupation
Solar and Energy Storage Installer
Voltage drop between the two sets must be equal. Current will adjust between both sets, until voltage drop is equal.

Since the inverter is a current source, why can't we say that voltage will adjust between all nodes, until both parallel runs are equal over the total length.

Or in other words...

Solve this circuit for current in R1, R2, and R1'. Each of the given ohm values is the effective round-trip AC resistance of the length of the 250 kcmil feeder. 490V is the voltage at the transformer, and 480V is the voltage at the loads. Answers should be close to my above example, to demonstrate the issue with breaking the symmetry.

In the diagram, why can't I posit currents and calculate the voltage instead of positing the voltage and calculating the currents?
 

jaggedben

Senior Member
Location
Northern California
Occupation
Solar and Energy Storage Installer
This has been covered... but I don't think it has set in. :blink:

In order to see the non-compliance, one has to look at it from the perspective of parallel conductors from PV to POCO source. Starting at the PV system disconnect, the parallel conductors start at the tap. When only tapping one set of parallel service conductors, the service conductor on the line side and the service conductor conductor on the load side of the tap become the parallel sets of the PV system [red vs. blue and 310.10(H)].

unequal%20parallel.gif

I just re-read this post and finally understand it. I would agree - you have a non-compliant inverter output circuit, no matter which way the current is going.
 

jaggedben

Senior Member
Location
Northern California
Occupation
Solar and Energy Storage Installer
Since the inverter is a current source, why can't we say that voltage will adjust between all nodes, until both parallel runs are equal over the total length.

Or in other words...

In the diagram, why can't I posit currents and calculate the voltage instead of positing the voltage and calculating the currents?

Nevermind these questions. After working out some calculations on paper I see that it doesn't affect the issue. No matter how you try to do it you end up with the current from the utility basically splitting between the parallel conductors, and adding to the PV current between the tap and load. So for the 500A example you end up with 150A on the untapped line and 350 between the tap and load. The effect of the PV current amounted to less than a rounding error (i.e. milliamps).

The one thing worth mentioning is that of course the Vdrop will be less between the utility and load when the PV is operating. Assuming the service end voltage remains constant, the voltage at the load will be higher. I was trying to get at that with my questions, but I think I see how that works now.
 

Carultch

Senior Member
Location
Massachusetts
Since the inverter is a current source, why can't we say that voltage will adjust between all nodes, until both parallel runs are equal over the total length.

Or in other words, In the diagram, why can't I posit currents and calculate the voltage instead of positing the voltage and calculating the currents?


You certainly can, and that could've been a better way to propose the problem. Maybe an even more realistic representation is to prescribe KVA values, but that would just add calculations that distract from the concept.

A real utility network is very complicated, because all the other neighboring services affect the voltage at your service point. They have a standard to supply power within a given percentage tolerance above and below nominal at your service point, and the utility-owned transformers are tuned to get as close as possible to nominal for each customer. It is a moving target to maintain this.

Inverters work as a current source. They work by measuring the grid voltage, and supplying the current that matches the KVA generated. And adjust their own voltage to be ever so slightly higher than voltage at the point of interconnection. Simply high enough to drive that current through the wiring to the point of interconnection.

The takeaway concept is that the current divides as needed, in order for the voltage to be equal at the termination of all paths to each node. I set this problem up, in order for main service conductors to dominate the resistance, and make it easy to find the weak link without too much algebra. You know the 300A is shared nearly equally, and then you add 200A to one of the 150A. Now you see the problem.
 

Carultch

Senior Member
Location
Massachusetts
Not convinced. I've worked with SPICE-based apps before (not for several years now, though). They all require proper setup to get true-to-life results. rather than getting into the pros and cons of SPICE-based apps, suffice it to say in many to most cases regarding power circuits that we discuss here, set up will fall short of mimicking a true-to-life scenario. (And just to add, I do not have a SPICE-based app installed on my computers and I'm not going to install one just to make a point).

Anyway, I am curious as to what your actual results were (or are). Let's say POCO 490V 1?, 250kcmil parallel conductors as depicted earlier. PV set up as constant current supply of 200A. One leg of supply system grounded... as I recall SPICE requires some point on the system to be grounded. System load set up to sink 500A or anything close (excluding conductor resistance simulating resistances... I know it is difficult to setup up the load to be dynamic enough to have an exact 500A draw, anticipate 480V at load end of service conductor sets). Don't forget to put in matching resistances on the return legs.

What is the current level on each section of the service conductors?


This is a "lumped system", where the resistances represent a round-trip feeder resistance. A proper setup would have a lot of duplicate resistors, to represent the return path and all the phases.

The ground point in a PSPICE model is more analogous to neutral, the way we use it. It is a point at which we define voltage to equal zero, and all connections to it to be electrically continuous with it. Unlike ground for us, which shouldn't carry current under ordinary circumstances, current can flow to and from a ground point in a spice model.

If you only define one ground point in the PSpice model, there will be no current in the connection to it.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
Although it is perfectly valid to start with the current values, you will at the end have to modify those current values to satisfy the constraint equation that the voltage drops on all of the parallel segments (end-to-end) are equal. (That is what ggunn left out of his analysis.)
You cannot just end up with two or three different voltages on parallel components. :)
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Inverters work as a current source. They work by measuring the grid voltage, and supplying the current that matches the KVA generated. And adjust their own voltage to be ever so slightly higher than voltage at the point of interconnection. Simply high enough to drive that current through the wiring to the point of interconnection.
A current source is perfectly happy if the voltage is the same. It would work just fine with superconductors with no voltage drop whatsoever.
 
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