Infinite Power source

[Strahan]

I'm in the process of calculating let through power (fault capacity) of transformers and I'm a little confused as to what is meant by infinite power source. I understand the formula KVA/%Z/100 but this is only true when connected to an infinite power source. If it is connected to a 480v supply with ample protection would this be considered an infinite supply?

 

[jim dungar]

In the context of your equation, an infinite power source would be able to supply an unlimited amount of current into a short circuit. In short circuit analyzes the primary current limiting items are the impedance of transformers and conductors.

 

[ramsy]

I believe there are several fault calc methods. I have notes on the Point-to-Point method, which assumes an infinite power source. Perhaps the assumption is the massively-paralleled services operating in larger urban areas.

Here's the formula I played with. You'll need to check my use of Z. Forgot if it was a %Z or a service/feeder impedance, but apparantly it matches IEEE Std. 241-1990, with an assumed 85% power factor.

Xfm SCA = Vph/(ZxLN*SQRT(?))

Vph = Phase Voltage
Z = Impedance ?
LN = 1 for 3?, 2 for 1?

Here are my notes, if you can make sense of them.

Note: This Point-to-Point implement of 1/Z per foot varies with Pwr Factor and conduit types, based on resistance and reactance, per NEC Tbl.9 formulas.

Setting power factor to 85% matches "C" table results, published by IEEE Std. 241-1990, IEEE Recommended Practice for Commercial Building Power Systems.

Note: For the special case of single-phase/split-phase (center tapped) circuits, such as 240/120 the Line-to-Neutral fault current at the transformer is considerably larger than the Line-to-Line fault current.

As the distance along the feeder from the split-phase transformer?s secondary increases, the L-L fault current will eventually become larger than the L-N fault current. The higher of these two values should always be used in determining the minimum AIR rating of the downstream single-phase equipment.

 

[Strahan]

In the context of your equation, an infinite power source would be able to supply an unlimited amount of current into a short circuit. In short circuit analyzes the primary current limiting items are the impedance of transformers and conductors.

So with providing protection on the primary side of the transformer this "infinite" supply would not exist. Is this correct? Basically with this formula its telling you what fault current the transformer is capable of handling before self destructing am I correct?

 

[wirenut1980]

Basically with this formula its telling you what fault current the transformer is capable of handling before self destructing am I correct?

That is an interesting way of putting it.:grin:

The words "infinite supply to the transformer primary" are normally used in this context to give a conservative/maximum amount of fault current that is available at the transformer secondary terminals. I like using this method rather than using actual fault current available because it allows for utility future conductor upsizing, or changes in transformer impedance if the transformer is changed out, which might increase available fault current to the customer.

 

[gar]

080729-0749 EST

ramsy:

Please explain the logic of your statement:

Note: For the special case of single-phase/split-phase (center tapped) circuits, such as 240/120 the Line-to-Neutral fault current at the transformer is considerably larger than the Line-to-Line fault current.

.

 

[bob]

So with providing protection on the primary side of the transformer this "infinite" supply would not exist. Is this correct? Basically with this formula its telling you what fault current the transformer is capable of handling before self destructing am I correct?

This calculation is giving you the max fault at the secondary terminals. This fault has nothing to due with the destruction of the transformer.
It is sometimes use to spec the fault requirements of the panels down stream.
The actual fault is less because you actually have the utility transmission and distribution system impedance included

 

[jim dungar]

So with providing protection on the primary side of the transformer this "infinite" supply would not exist. Is this correct? Basically with this formula its telling you what fault current the transformer is capable of handling before self destructing am I correct?

Short circuit currents calculated by use of an infinite bus are peak values. Withstand (SCCR) ratings determine how long the component can tolerate/supply the peak values

In general primary side protective devices are not viewed as reducing the infinite bus current as they have an Ipeak value that flows for an amount of time before they operate. An infinite source will usually exist reduced only by the primary system impedance (usually from transformers and conductors). We assume that the power grid has enough paralleled sources that it can easily supply any amount of current while maintaining the nominal voltage.

 

[LarryFine]

. . . I'm a little confused as to what is meant by infinite power source.

Here's my understanding of that: when doing fault-current calcs, the term "infinite bus" means they want you to presume that the utility transformers' primaries have access to an unlimited amount of current, so the transformers' impedances are the first numbers you use in the calcs.

In other words, you don't get to include any of the primary distribution equipment's impedances in your calculations. The transformers' secondaries' voltages, ampacities, and impedances are where you start.

Now, using the term "infinite source" may mean they want you exclude the secondary impedances as well, meaning that the service drop or lateral are the first impedance numbers you get to use. Check to be sure.

 

[gar]

080729-0952 EST

Strahan:

When electrical equivalent circuits are drawn these usually have voltage and/or current sources as part of the circuit. A voltage source has no internal impedance, its voltage is invariant relative to load current. A current source has no internal impedance, its current is invariant relative to load.

In the real world any source will have some internal impedance. The real world source may be represented by the ideal voltage or current source and some internal impedance.

The real world source may be represented by a voltage source and a series internal impedance or identically by a current source with that impedance in parallel with the current source.

An ideal voltage source is your infinite bus because the voltage is invariant relative to load.

In the real world if your voltage source does not change much with load then you can consider it an ideal voltage source. In the electronics field over a limited range you can build a very good voltage source.

In a power distribution system a 25 KVA transformer can be considered, as a first approximation, a very light load on the supply lines, and therefore the supply lines are an infinite bus. Suppose the transformer secondary output is 240 V open circuit and the transformer impedance is 5%. Full load current would be 25,000/240 = 104.2 A. Since the transformer internal impedance is 5% this means that the secondary short circuit current would be about 20 * 104 = 2000 A. Change the internal impedance to 2.5% and the short circuit current is about 4000 A. This is because we are assuming that the primary voltage to the transformer is invariant.

This transformer internal impedance is largely a result of two components, winding resistance and leakage inductance. The values of the internal impedance equivalent circuit may be somewhat different between light load, meaning normal full load, and short circuit conditions.

.

 

[jghrist]

Everyone is making this much more complicated than it really is.

Current is Voltage divided by Impedance. The impedance includes the transformer impedance plus the impedance of the entire system ahead of the transformer. The impedance of the system is normally a lot lower than the transformer impedance, and if it is neglected (assumed to be zero), then the source is considered to be of "infinite" capacity. This gives a fault current that is higher than actual, but usually not by a whole lot.

 

[ramsy]

Please explain the logic of your statement:

As it applies to fault calcs, the logic for greater L-N fault current is given as follows:

Note: "When you have a line to neutral fault you have the full primary winding involved but only half of the secondary winding. The impedance of the half winding condition is different from the full winding. Therefore you must make an adjustment to %R and %X. The multiplier usually is 1.5 x full winding %R and 1.2 time full winding %X."

In my speadsheet application when 75% of the core leg, phase set, is shown as 4/3, it matched the LICE point-to-point results shown by the application these notes came from.

 

[Strahan]

Of course this fault current will be found at the secondary terminals of the transformer. Thanks guys I wasn't thinking clearly. After reading these responses I get the understanding "infinite" supply is a term used in order to calculate maximum let through current (or fault current) that the transformer will deliver. As long as I size my protection devices correctly hopefully this will never be encountered. Thanks Guys!

 

[benaround]

Of course this fault current will be found at the secondary terminals of the transformer. Thanks guys I wasn't thinking clearly. After reading these responses I get the understanding "infinite" supply is a term used in order to calculate maximum let through current (or fault current) that the transformer will deliver. As long as I size my protection devices correctly hopefully this will never be encountered. Thanks Guys!

It really has nothing to do with how you size the protection devices, it has to

do with their capacity to interupt the available short circuit current.