What is the effect of capacitor voltage rise effect when load shedding 1

[yyoungengineers]

HI everyone,
I am studying how to place capacitors in a 12kv distribution network and i have a couple of questions that i can't seem to answer with the literature i read.

Lets consider a simple example,suppose you place a fixed capacitor on a 12KV line that is compensating 3 loads. see attachemnt.
now if you want to determine the voltage rise that the capacitor causes , the literature says :
Voltage rise = Capacitor current * system reactance
Where sytem reactance is the reactance between the voltage source and the capacitor.
What i don't understand is : When one or 2 of the loads are put out of service. How does the load that is still in service expercience a higher voltage rise? Because the system reactance didn't change! and the capacitor current didn't not change also. How am i seeing this wrong?
The reason for his question is that the utility has placed a fixed capacitor in an area with residential loads, but when the network was reconfigured, where some load behind the capacitor where shed, some customers claimed to have experienced high voltages.As a result the banks where put out of service.
I want to find out an explanation dor this.

Thanks in advance

 

[davidbeach]

The magnitude of the voltage rise due to the capacitor may not change much, but the voltage drop due to the loads certainly changes. With the higher voltage due to less load induced voltage drop you will have more capacitor current.

 

[yyoungengineers]

I can't understand this a 100% , could you elaborate on the part of the voltage drop? Can i prove this with a equation.

 

[davidbeach]

V = I * Z where all three are complex quantities. I is the current in the line, Z is the line impedance and V is the voltage drop.

 

[yyoungengineers]

How would you know to what extent the voltage rises when the inductive load decreases? What could e a solution when network restoring?changing the tap of the transformer where the customers are overcompensated ?putting the capacitor out of service?

 

[davidbeach]

Calculate the voltage drop due to load both with and without the specific load in question. Pretty basic stuff, Ohm's Law really.

 

[pwrtran]

yyoungengineers -

Knowing the sending end voltage and the receiving end real and reactive power, the voltage drop is calculated as:
ΔV = V1-V2 ≈ (P*R+Q*X)/V1
Now you have Qc, contributed by the cap banks, then
ΔV' = V1-V2 ≈ [P*R+(Q-Qc)*X]/V1

The difference on ΔV will be the voltage rise caused by the capacitors.

Regards,

 

[ijl]

Yes, Ohm's law, but not necessarily as simple as V = I * Z. If the values of the capacitor and the network components are such that the net frequency is near the resonance frequency of the network, then the change in the voltage may be much larger than that.

yyoungengineers, is it possible to get the parameters of loads and the network and the capacitor(s) for an actual calculation of the voltage drop in the network?


Actually, the problem may be still more difficult, because, as yyoungengineers wrote, the capacitor current did not change !?

 

[yyoungengineers]

Thanks for the comments.

It is not possible to get the load precisely.
The data that is available includes transformer rating , line parameters, but the load is not monitored.
Several moment measurement are taken with the use of high voltage meters.
However, the loads are constantly fluctuating , therefore these measurements should be considered as approximations.
When i simulted different scenarios in ETAP i only got a voltage rise of a few hundredth.


Can you provide me some literature on the resonant phenomena? I don't wanna exclude nothing.

Thanks again

 

[ijl]

What I mean with the resonance is the following: The network can be simplified as a RLC-circuit, see the attached picture. Let's assume for simplicity that the load is purely resistive. If it happens that wL is nearly the same as 1/wC, i.e. the resonance frequency of the circuit is nearly the same as the frequency of the network (50 or 60Hz), then the voltage over the load can change considerably, when the load changes. (You can work out the details yourself).

But if you already have carried out the calculations, then this guess of the resonance as the cause of the voltage rise may not be correct.

 

[yyoungengineers]

You have misunderstood me....What i meant is that when i simulate the network, the voltage rise is not significant. The calculated voltage rise would not have caused the overvoltage that were experienced by customers.
Now that i am trying to place capacitors into the network, i want to prevent such phenomena. Therefor i want to know what considerations should be made in such situations...

 

[pwrtran]

When i simulted different scenarios in ETAP i only got a voltage rise of a few hundredth.

How big is your capacitor and how much voltage rise you think would make sense to you?

I suggest you check your local distribution code or standards apply to your area to find out the maximum voltage fluctuations is allowed!

A few hundreds, L-L or L-G? Say you have L-G 200V rise on your 12kV system, that is almost 3% voltage rise! I wouldn't be surprised if there are some customer's complaints, that's all depends on what tx tap set to and how sensitive of their equipment.

 

[yyoungengineers]

No that is what i expected such a voltage rise would cause considerable high voltages. But in my simulations the voltage rises from 12.027 KV to 12.06 KV, thus a voltage rise of 33 V only. The voltage on the line are 12KV L-L and the capacitor are 150 kvar (50 KVAR per phase). I don't want to focus on the maximum allowable voltages yet, i want to put the scope on the reason of the voltage rise.

 

[jghrist]

Yes, Ohm's law, but not necessarily as simple as V = I * Z. If the values of the capacitor and the network components are such that the net frequency is near the resonance frequency of the network, then the change in the voltage may be much larger than that.

Resonance at the fundamental frequency is unlikely. The capacitor bank would have to have the same Mvar rating as the system short circuit MVA. See IEEE Std 519-1992, Eq 8.1:

h_r = sqrt(MVAS_sc/Mvar_cap)

 

[LionelHutz]

150kVAR doesn't seem like a capacitor size that would cause much of a line voltage change on a 12kV system.

I did notice that your source doesn't look correct at 0MVAsc.

 

[yyoungengineers]

Thanks for the comments.
I have also came to that conclusion, the 150 KVAR bank will only raise the voltage with 0....%

I always thought that overcompensation leads to overvoltages, they seemed like a synonym to me.

please provide me an article where the difference is elaborated.

 

[jghrist]

I always thought that overcompensation leads to overvoltages, they seemed like a synonym to me.

It can lead to overvoltage. See Pwrtran's 16 Jul post. If Qc > Q, then the voltage drop can be negative (voltage rise), but not necessarily. Voltage rise is not necessarily an overvoltage either.