Single phase parallel conductor line modeling
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Hi,
I am new to OpenDSS
Can anyone show an example of how to define linegeometry for a single phase line two parallel condcutors and a neutral. The following figure shows an exapmle of spacing for such type of configuration.
Any help will be greatly appreciated.
Thanks,
Divya
Last edit: Divya Vedullapalli 2017-03-08
Last edit: Divya Vedullapalli 2017-03-08
Hello Divya,
Celso and I have a code that you can use as an example. Find attached the figure.
// -----------------------------------------------------------------------//
// Name: Paulo Radatz and Celso Rocha
// e-mail: paulo.radatz@gmail.com and celsorocha11@gmail.com
// -----------------------------------------------------------------------//
// -----------------------------------------------------------------------//
// ------------Line -> Wiredata and Linegeometry------------//
// -----------------------------------------------------------------------//
Clear
New Circuit.TheveninEquivalent phases=3 basekv=13.8 bus1=K
// -----------------------------------------------------------------------//
// Example from Kersting's Book
// -----------------------------------------------------------------------//
New Wiredata.Phase GMR=0.0244 DIAM=0.721 RAC=0.306
~ NormAmps=530
~ Runits=mi radunits=in gmrunits=ft
New Wiredata.Neutral GMR=0.00814 DIAM=0.563 RAC=0.592
~ NormAmps=340
~ Runits=mi radunits=in gmrunits=ft
New Linegeometry.PoleExample nconds=4 nphases=3 reduce=No
~ cond=1 Wire=Phase x= -4 h=29 units=ft
~ cond=2 Wire=Phase x= -1.5 h=29 units=ft
~ cond=3 Wire=Phase x= 3 h=29 units=ft
~ cond=4 Wire=Neutral x= 0 h=25 units=ft
New Line.LineExample bus1=K.1.2.3.0 bus2=L.1.2.3.4
~ Geometry=PoleExample
~ Length=1 units=mi
~ EarthModel=Carson
Solve
Dump Line.LineExample debug
// -----------------------------------------------------------------------//
Best regards
Paulo Radatz
Hello Paulo,
Thank you for your reply. The example you provided is for a 3 phase 4 conductor line.
But I need an example for single phase 3 conductor (2 parallel phase conductors and a neutral conductor) line connection.
Thanks.
Divya
Attached is the figure.
Hello Divya,
You can use the code I sent to you as an example.
So, you mean that 1 phase uses two conductors. I am not sure if you can define the buses of the line as, for example, buses=[ A.1.1.4 B.1.1.4]. I am going to see if it works, I've never done it before. However, I think you can create the buses as Buses=[A.1.2.4 B.1.2.4] and include two reactors with low impedance (short-circuit). New reactor.shortCircuitBusA Bus1=A.1.2 R=0.00001 X=0.00001
I would recommed you to have a look at those ideas and see if they work.
Best Regards
Paulo Radatz
Thank you Paulo. I will check them.
Best regards,
Divya
Hello Divya,
The first idea worked just fine, look at the example below, I just deleted phase c.
// -----------------------------------------------------------------------//
// Name: Paulo Radatz and Celso Rocha
// e-mail: paulo.radatz@gmail.com and celsorocha11@gmail.com
// -----------------------------------------------------------------------//
// -----------------------------------------------------------------------//
// ------------Line -> Wiredata and Linegeometry------------//
// -----------------------------------------------------------------------//
Clear
New Circuit.TheveninEquivalent phases=3 basekv=13.8 bus1=K
// -----------------------------------------------------------------------//
// Example from Kersting's Book - Kron Reduction
// -----------------------------------------------------------------------//
New Wiredata.Phase GMR=0.0244 DIAM=0.721 RAC=0.306
~ NormAmps=530
~ Runits=mi radunits=in gmrunits=ft
New Wiredata.Neutral GMR=0.00814 DIAM=0.563 RAC=0.592
~ NormAmps=340
~ Runits=mi radunits=in gmrunits=ft
New Linegeometry.PoleExample nconds=3 nphases=2 reduce=No
~ cond=1 Wire=Phase x= -4 h=29 units=ft
~ cond=2 Wire=Phase x= -1.5 h=29 units=ft
~ cond=3 Wire=Neutral x= 0 h=25 units=ft
New Line.LineExample bus1=K.1.1.0 bus2=L.1.1.4
~ Geometry=PoleExample
~ Length=1 units=mi
~ EarthModel=Carson
New load.teste phases=1 bus1=L.1.4 kw=1000 kv=13.8
Solve
Dump Line.LineExample debug
Looking at the Current elements report, we got:
ELEMENT = "Line.LINEEXAMPLE"
K 1 23.132 / -30.1 = 20.017 +j -11.593
K 1 25.402 / -27.2 = 22.603 +j -11.593
K 0 48.538 /_ 151.4 = -42.62 +j 23.226
L 1 23.146 / 149.9 = -20.017 +j 11.621
L 1 25.416 / 152.8 = -22.603 +j 11.623
L 4 48.546 /_ -28.6 = 42.62 +j -23.244
As you can see there are two inject currents flowing into the terminal 1. That's because we have two conductors for the phase which we connected between nodes 1.
I've never thought about it before, but as I expected OpenDSS can handle it as well.
Now, you just need to set your values into this code.
Best Regards.
Paulo Radatz