Direction of flow of capacitance current.

[beyond86]

Hello. I have a question. Can you explain please. I have a long HV line (for example 220 kV). I close breaker with the one end and the other end remains open. Where will go capacitance current? From busbar to HV line or from HV line to busbar?
May be you will suggest some books about this topic. Thanks.

 

[LionelHutz]

AC current flow is still current flow, no matter what the component at the end. It doesn't really have a direction.

You're probably thinking power flow, which has a direction convention to make it easier to describe what is happening in the power system. Capacitance power is generally considered flowing from the capacitors to the power system.

 

[beyond86]

If capacitance power flow to the power system that current also flow to the system, because current and power have one direction.
May be you can explain this image.

 

[crshears]

It depends how you define "capacitance current"...

As an operator/controller with 30+ years of experience and a bit of knowledge, I and my compatriots typically define reactive power, or Volt-Amperes-Reactive, or VARs, as lagging, such that inductive devices like transformers and squirrel-cage induction motors draw or absorb [lagging] VARs from the system, and capacitive devices, like capacitors or radially connected circuits, supply [lagging] VARs.

As a consequence, a 230 kV line on potential from one end only would act as a capacitive device and supply [lagging] VARs from the line into the energizing station's bus.

At least that's the way I have learned to view it.

Hope this helps.

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[Mbrooke]

Could we then say transformers and motors supplying leading VARs?

 

[crshears]

I would say we don't, but yes, we could...

Addendum: had another look at the linked diagram, and in my view what you're suggesting comports well with the terminology and conventions contained therein.

 

[LionelHutz]

I wouldn't use that curve. The curve has power on both axis and then calls the vectors currents, which I would call wrong. If the units on the axis are power, then the vectors should be also power.

In AC systems, current doesn't have a direction. You can not figure out what quadrant the power is in by only measuring the current.

 

[crshears]

True; current is just current, hence my elaboration. Maybe the OP will respond...

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[marks1080]

The direction of current is back and forth and back and forth really fast.... about 60 times a second i think.

The direction of the Watts and Vars I think is what you're really asking. The answer depends on your system conditions.

Generally, when you turn a cap 'On' power flow will flow into the cap until it reaches a steady state(charged), and then it will export VARs. When you turn the cap off it will discharge (quickly, or slowly depending on your set up) with power flowing out - usually to ground.

 

[crshears]

Hey marks, some places in the world it only goes back and forth about 50 times a second, you know...

That being said, I hear where you're coming from; we try to have the mindset of engineers, not salesman, and hence favor technically precise wording over the employment of linguistic elasticity...but we don't always succeed.

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[LionelHutz]

crshears - It seems the claim is that when you're exporting VARs you are also exporting current and I'd agree that there is nothing stopping anyone from saying that if they so chose to. But, expect odd responses when saying that you're exporting current.

 

[crshears]

Hi, Lionel,

Sigh...at myself; or maybe not...

Had a third and then a fourth look at the diagram...I'm no math whiz, but with machine to left and system to right,

[a] pure real power export [generating] shows current as positive and perfectly in phase with [positive] voltage, and therefore pointing to the right of origin; near as I can tell, this is correct

pure real power import [motoring] shows current as negative but still perfectly in phase with [positive] voltage, and therefore pointing to the left of origin; near as I can tell, this is also correct

[c] pure reactive power export [producing "lagging" VARs] shows current as positive but in quadrature upward with [positive] voltage, yielding a positive imaginary jQ product and therefore pointing straight up from origin; near as I can tell, this is correct

[d] pure reactive power import [absorbing "lagging" VARs] shows current as negative but in quadrature downward with [positive] voltage, yielding a negative imaginary jQ product and therefore pointing straight down from origin; near as I can tell, this is also correct

As to the "vectors" with I's on them, as near as I can tell those are the intermediate currents which are a product of real and reactive power and therefore fall neither perfectly in phase nor in quadrature but at some intermediate angle to the [positive] voltage. This accords well with the way the synchronous condensers and generators I've operated over the years have behaved. It is this behaviour which, in my humble opinion, render's the OPr's statement that

If capacitance power flow to the power system that current also flow to the system, because current and power have one direction

problematic.

Please recall, beyond86, that current is a directly measurable [primary?] quantity, whereas power is a derived quantity [product of voltage times current times power factor/angle, or however you compute it]. Do not make the mistake of saying the current flows to or from anything in an AC network, since its direction is reversing numerous times per second. That is only [generally] true when considering DC.

It is also for this reason that real power can flow one way in a circuit while reactive power can simultaneously flow in the other [look at the lower right quadrant of the diagram]; as a matter of fact as an operator I was routinely asked during the overnight hours to reduce the excitation on running generators so as to absorb maximum reactive power for voltage control purposes [but without falling out-of-step / losing synchronism]...all while the unit was at the prime mover's maximum efficiency point of real power production on its capability curve.

Anyone, please correct me if I'm wrong.

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[waross]

Think of a small capacitor connected to 120 Volts.
At first the current will flow into the capacitor to charge it. Once the alternating voltage drops lower than the capacitor voltage the capacitor will discharge, after the first cycle, the charge discharge cycle will become regular.
As the alternating voltage passes the zero point and starts to increase in the opposite direction, the capacitor will still be discharging. At this point in the cycle the current will be in the opposite direction to what the current would be if the load was resistive rather than capacitive.
When you connect an open ended transmission line to a voltage source, you are connecting a capacitor to a voltage source.
Go back to basics and consider a simple capacitor on an AC circuit.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[HamburgerHelper]

Anything that produces vars will see vars flow towards items that consume vars. Things that consume vars are transmission line reactances, transformers, shunt reactors, and loads. In a power flow program, it will look like vars flow from high PU voltages to low PU voltages. An open ended line might have high voltage at the end and produce vars and these vars will flow into the system, feeding anything that is consuming vars and raising the system voltage in the process.

Looking at it as var flow instead of currents I think is easier. There are 4 quadrants that currents can lag or lead the voltage and those produce every combination of real power and reactive power flow. If you want to look at it as currents, you are going to be looking at it as current that leads or lags a certain angle to tell you the capacitive current flow or more specifically real and imaginary/reactive/capacitive current flows. If you look it as var flow, you are not looking at it as voltage to current angles. They both give you the same answer but I think the concept of real and imaginary power flows is easier to grasp than trying to remember what the angles the currents should be to the voltages. If I needed the angles, I think I would first figure out the mw and var flows and just convert the answer to angles.

 

[crshears]

An open ended line might have high voltage at the end and produce vars and these vars will flow into the system, feeding anything that is consuming vars and raising the system voltage in the process.

Ferranti effect...sometimes referred to as LEO [line end open].

We see the same thing on lightly loaded 230 kV and 500 kV lines that export VARS at both ends; when heavily loaded, the line eats all those VARS and then some [surge impedance limit]. Result is that when transfer drops we either take shunt caps o/s or place shunt reactors in service at stations along the path to control voltage; when transfer increases, vice versa.

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[beyond86]

Thanks for your answer and sorry for delay. Let's deal with this question.
The direction is flow current durring the positive half-cycle of the sine wave (1 image). I was mean it when I spoke about direction current.
Next. I completelly agree with waross. In this case, the flow of capacitance current will go from busbar to HV line and it will be really direction of charged particles. But if we'll look at image with quadrature powers, in this case the direction of power (VARs) will be oposite direction of charged particles. Why? Is it just an assumption?

 

[HamburgerHelper]

You are losing the forest from the trees. Current created by VAR flow is +90 or -90 degrees out of phase with the voltage. Your current vector will be I = imaginary(current)*i + real(current). It flows back and forth. The whole point of talking about VARs is to get away from talking about angles the current is in respect to the voltage. Var flow isn't a direction of charged particles but just a representation of the angle the current is in respect to the voltage or you can also look it is as to what portion of the apparent power (power triangle) is dedicated to creating magnetic fields that grow and collapse twice a cycle, not doing any real work.

Talking about power factor is the same way, watts/(apparent power) sounds a lot better than talking about angles and you have an idea of how much you can reduce the current if you raise the power factor to 1.0 through correction.

If you are asking why apparent power S = V*conjugate(i) and not S = V*I, which changes what positive and negative vars represents, that is just an arbitrary convention as far as I know. I don't understand that one. It feels like the arbitrary convention forces you to use the conjugate of the current.