Shunt Capacitor Bank Modeling: short-circuit/breaker-rating studies

[siberhusky]

Do any of you regularly include shunt capacitor banks in your overall system models when performing short-circuit calculations/Breaker rating studies, or do you consider this a dynamic situation that is distinct from this process?

- john

 

[cuky2000]

Shunt capacitor do not contribute to short circuit. However, switching capacitors produce high Inrush current, high transient frequency and peak overvoltage as high as 2 pu, imposing on the circuit breaker additional stresses that should be considered during the selection and rating of the device. Consult ANSI/IEEE C37.012. for additional information on general or definite purpose circuit breakers rating.

 

[davidbeach]

cuky2000, I disagree. The shunt capacitors will offset some of the system inductance, reducing the system impedance. The reduced system impedance will cause higher fault currents. Whether this is a significant change or not depends on a number of factors and will vary from case to case.

 

[cuky2000]

Hello Davidbeach,

You have good point regarding the reduction of impedance. If we could quantify that during SC this could help to understand better the contribution of capacitor banks to the overall SC current.

Let's take a break for a minute and check what IEEE std 141 discusses about this topic.

 

[davidbeach]

cuky2000, I was thinking of the capacitors as sources of fault current, certainly not in terms of currents to be interrupted because the transient will die off in a portion of the first cycle. My point was that they would reduce the impedance between the fault and the source, thereby allowing higher fault currents than if they were not there, kinda like bigger wire. Whether that reduction in circuit impedance will be of any significance is another matter and would have to be evaluated case by case.

 

[jghrist]

Quick example:

Base 30000 kVA, 13.2 kV, 1312A
system impedance j·0.20 pu
line impedance j·0.1 pu

Without shunt capacitor:
If = 1/(j·0.2 + j·0.1) = -j·3.3333 pu = 4373A

With 10000 kvar shunt capacitor at 1/2 point on line:

XC = -j·3 pu
Solving a 2 loop equation,
If = -j·3.3803 pu = 4435A 1.4% higher

 

[siberhusky]

Thank you everyone for your input.... At the moment, I am reviewing IEEE C37.012.

thx,

- john

 

[waross]

I think there is a difference between source impedance and system impedance. There is also a difference between series impedance and parallel impedance.
Series capacitors to counteract line impedance will reduce system impedance and source impedance and allow higher fault currents.
To ilustrate my point, please visualise a distribution system.
There is a 13kv distribution line. It has a given impedance and available fault current.
There is a 13kv:480v transformer. It feeds a main switch.
The main switch feeds a bus to which are connected:
Load "A" through breaker "A",
A motor through breaker "B",
And a capacitor bank through breaker "C".
The capacitor bank will have no effect on the fault current available from the system.
Imagine a fault at the bus bar.
The current from the supply system will be the same.
The current in the fault will be the sum of the current from the system and the regenerated current from the motor. The current from the motor will be considerable. With the capacitors in the circuit to provide magnetizing current the motor contribution would probably be greater. The effect of the capacitors will be influenced by the capacity of the capacitor bank in relation to the connected inductive load.
Depending on the point in the cycle at which the fault occurs, and the speed at which it becomes "Bolted" there will probably be a capacitive discharge into the fault one at least one phase. I would suggest that although the current duty cycle may not be much increased, the magnetic forces trying to remove the bus bars from their mounts may be much increased.
Although the current in the fault is greater than the source fault current, the main breaker is only seeing the source fault current, not the total fault current.
Now imagine a fault on the feeder to load "A".
The main breaker still sees only the system fault current, but breaker "A" sees the total fault current.
I have seen a coupling explode and both the driven shaft and the motor shaft bent by an arcing short on the motor feeder. The decelleration torque which caused the damage was a reflection of the motors contribution to the fault current. I imagine that with a bolted fault and/or capacitors the motors contribution to the fault current would have been higher.
These comments apply to a plant installation, not a transmission line installation.
I have understood from the posts that this may be a industrial plant installation or it may be a transmission line installation.
Please be gentle with me if I have misunderstood the type of system under discussion.
Comments gentlemen?
Respectfully

 

[davidbeach]

My responses have been based on capacitors connected between the source and the point at which the fault current is being calculated. waross is correct that capacitors between the fault and point at which fault current is being calculated will not raise the current that needs to be interrupted at that point.

 

[dpc]

Capacitors are generally ignored when doing short-circuit calculations - this is allowed by ANSI C37 AFAIK. But this is a simplication of the issue.

Conrad St. Pierre has a good discussion of the impact of capacitors during short circuits in his "A Practical Guide to Short Circuit Calculations".

BTW, I hear rumors that the IEEE Violet(?) Book on SC calcs may finally be coming out??

 

[jghrist]

I initially thought that only series capacitors would reduce the impedance to the fault. That's why I made the simple fault calculation of my previous post. Even though the capacitor is a shunt capacitor, it effectively reduces the source impedance.

In that example, the current before the capacitor would be 4361A, a 0.23% decrease from the case without the capacitor.

 

[satnam123]

Waross,
I have already a question here on site but could not help asking one more here. Please excuse my inquisitiveness.
In your post, "The main breaker still sees only the system fault current, but breaker "A" sees the total fault current."
I beleive what you are saying is that if there is a 3 Ph fault and One part of current is coming from system and the second from Motor and both currents will add at the breaker connected to motor.
But the main breaker feeding the switchboard will only see the system contribution. Am I correct?
Thanks

 

[KJvR]

The only influence shunt capacitance have on standard load flow / fault level software is the initial voltage used before fault inception. This is for both IEEE/ANSI and IEC methods. If you use system voltage or use IEC min/max voltage to calculate min/max fault levels, capacitance will have no influence. This can also be seen when compiling impedance matrix for numerical method load flow calcs - a shunt capacitor is treated as a load.

 

[waross]

satnam123
Let me restate my statement. I think you're correct but I am not totally sure of your wording.
The main breaker will see the supply system contribution to the fault.
The motor breaker will see the motor contribution to the fault.
The breaker feeding the fault will see the sum of the supply system contribution and the motor contribution to the fault.
The currents will combine on the bus system which is common to all three breakers.
Any contribution to the fault by the capacitors would be similar to the motor contribution. However, the magnitude and existence of capacitor contribution is being debated.

I have three observations.
1> The total energy supplied to a fault by a motor is limited by the rotational energy of the motor.
2> The rate at which the motor supplies it's stored energy to a fault may well be significantlly greater when a capacitor bank is supplying magnetizing current to the motor. The result may be a greater curent for a shorter time.
3> Cucky2000's post of IEEE Std. 141 indicates that the capacitors may contribute an extremely high transient short curcuit discharge current. However the duration is usually short enough that calculated power frequency short curcuit duty current is not significantly increased by adding the capacitor discharge.
Please note, IEEE Std 141 recognises extremely high transient short-circuit discharge currents. Although the standard does not mandate the consideration of these transients in normal calculations, the use of the word "Usually" and the word "Significantly" imply the existence of special cases where the contribution to fault currents by capacitors may be significant.
I feel that a discussion of these contributions is in order and should not be cut off by reference to a standard which does recognise these currents.
Respectfully

 

[KJvR]

This discussion is about circuit breaker rating / short circuit studies. My thoughts and questions on the influence of shunt capacitors on circuit breaker ratings are:
(a) Symm breaking current: As already stated, it will keep up motor excitation and the motor contribution decay will be slower. How should we claculate this decay? IEC provides certain empirical equations and it is also in the draft copy of the "Violet" book. It does not state under what conditions it was derived.
(b) TRV: The capacitor will reduce the TRV.
(c) Peak withstand current: The higher the current contribution at intial fault inception, the higher the resonant frequency (smaller inrush current limiting reactor). The circuit breaker's rated peak inrush current limit is specified for mechanical forces at power frequency. The cap bank's high frequency discharge will therefore not influence the initial peak current rating of the circuit breaker.
(d) DC-offset: How will the cap influence the X/R ratio (and initial 1/2 peak current and 3 cycle DC off-set)? It seems as existing methods (Hand calcs and software based on IEEE and IEC) ignores this.

I will appreciate any comments.

 

[waross]

Two observations re the decay of the motor contribution.
The motor contribution is greatly influenced by the decelleration of the motor. If the initial rate of decay of the motor contribution is slower, then the initial rate of decelleration will be greater, As a result I expect that the decay curve will be different. Initially less decay, but then dropping faster and terminating sooner. The mechanical inertia of the motor system will determine if the difference in decay curves is significant. eg; is there sufficient inertial energy to hold up the motor contribution until the breaker clears?
In many cases in the field, the duration of the motor contribution may be limited by mechanical failure, most likely a coupling.
respectfully