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#1

# Current distribution in sub-transmission network

What factors determine current distribution between two 33kV feeders feeding a 33/11kV Substation within a sub-transmission network.
09-19-2013 03:15 PM
Top #2
cable type and size and the location of each feeder
09-19-2013 06:13 PM
Top #3
Assuming the two 33kV feeders are of the same type and size but they are originating from 2 different 33kV buses like in a mesh network. Doesn't the source potential of each of the feeders affect the current distribution on the feeders?
09-19-2013 08:23 PM
Top #4
Try the voltage divider rule. Take the R + X of each feeder (resistance and reactance) and find the Z. Remember that square root of R² + X² = Z. Now that you have the Z of each feeder, now find the Z of the two in parallel. To do this we have Z = 1/(1/Z1 +1/Z2). So, if Z1 = 2.16 ohms and Z2 = 1.67 ohms, then our Z of the two in parallel is 0.94 ohms. Now we pass the current of the entire substation between these two feeders. Let’s say that the current is 240 amps. Now we have 240Ax0.94ohms = 226 volts (IxZ=V). And since voltage divided by impedance gives us current (V/Z = I) we now take the voltage drop across the two feeders in parallel and divide each of the feeder impedances to get the separate feeder current. So we get 226V/2.16ohms = 105 amps (feeder 1) and 226V/1.67ohms = 135 amps (feeder 2). I have not tried this with your exact situation. Having different voltages from two different substations will change things, but at least this way you have a good start on the problem.
09-19-2013 10:55 PM
Top #5
Power angle between the two location and voltage level.

Power angle to determine power flow and voltage for var flow, the resultant current is you total less the line charging current requirements.
09-20-2013 01:02 AM
Top #6
If both feeders are connected at both ends current will divide according to impedance ratio but if not connected together the pf of each load will also play a role.
09-20-2013 03:39 AM
Top #7
I see where I went wrong. Sinnadurai and Raymond are correct. Since one end is tied together and the two other ends are from different substations, then you will have the classic voltage sending and receiving formula. Since the load is the one substation, then their will only be one power factor of the one load, so I would think this formula would apply: Es = Square Root of ((ErCosƟ + IR)² +(ErSinƟ +IX)²), which is square root of ((Receiving voltage times the cosine of the current phase angle plus current times resistance of the line)² + (Receiving voltage times Sine of the current phase angle plus current times reactance of the line)²). The voltage drop across each line would be VD=I(RcosƟ +XsinƟ) where R is the line resistance and X is the line reactance and the Ɵ is the phase angle of the load. Seems like a lot of trouble when we have computer programs to do this work for us.
09-20-2013 06:36 AM
Top #8
Sinnadurai, your point is well noted however the loads are tied together at the receiving end of the feeders. So it is assumed their power factor of load to both feeders will be the same.
09-20-2013 08:52 AM
Top #9
Let say we modify the original question an that a capacitor bank was in service at the 33kv/11Kv substation. The actual voltage on the bus is meaured at 34.0 kv.

The remote (source) ends where the 2 subs are are fed from 2, call it sub 1 and sub 2, Its bus voltage is measured at 34.0 kv for both subs. The line length is 10Km for both.

Question:
Will there be current flowing in the lines since the voltage are identical at both ends ?

If there is zero voltage drop on the line does it mean that there is zero current flowing in the line.

When we look at the incoming metering, Line 1 shows 10MW and 1 Mvar and line 2 show 5MW and 1Mvar.

How is that possible when there should be 0 amps flowing in both lines ?

Hope the scenario is correct !!!!
09-20-2013 11:04 AM
Top #10
Raymond. The fact that both ends of the feeder have the same potential does not mean no current will flow. The current that will flow will be based on the load the receiving end of the bus with voltage 34.kV. Thus I = S/(rt 3* 34.0kV). The potential difference on the line just influences the magnitude of the current that will flow. An iterative process determines the resultant receiving end voltage and the eventual current to the drawn by the load. Read on Newton Raphson or Gauss - Seidal Method of Load flows.
09-20-2013 01:27 PM
Top #11
I think Ekow must have experienced unequal loads on the two incoming cable feeders from different substations.

The unequal loads could be caused by different voltage levels at the upstream substations and the difference in cable impedance (due to lengths or sizes). Different voltage levels at the upstream substations could be the major influence. In a more serious situation, one cable feeder may be importing high current and the other cable feeder may be exporting. The cable feeder that imports high current may be tripped by overcurrent protection relay.

Check how you can change the upstream substations’ voltage levels so that they can be as close as possible. This may be difficult as some intermittent loads may swing the voltages that will cause the voltage differences at different times.

The easier way is to avoid operating the 2 incoming feeders in parallel, ie, by operating an open point at one of the 3 circuit breakers (bus-coupler or the incomers).
09-20-2013 04:19 PM
Top #12
IF 2 different feeders supplies power to a single load, the total impedance of each line will be the factor that will determine how much current each feeder will share. This is true if the 2 feeders comes from the same source or same bus.

If the 2 feeders came from different sources, then the sending end voltage of each source will determine how much each feeder will share, considering also the line impedances of each feeder affecting the voltage loss along the the line down to the receiving end (load)