Substation Current Limiting Reactor Sizing Calculation

The current limiting reactor sizing calculation is very straightforward. But there are a couple of important things you need to watch out for.

I’m going to go over the calculation for sizing current limiting reactors for substations. It’s a common calculation I do.

For this example, we’ll do the sizing for a typical substation located in a large city.

We’ll assume the incoming voltage into our substation is 115,000-volts.

Also, keep the following in mind for this discussion:

  • I’m going to refer to current limiting reactors as “reactors” from time to time.
  • Each phase of our circuit will have a current limiting reactor installed. So our calculated reactor size will be used for each phase of our 3\theta AC power system.

Important Note: circuits have a source and a load. Thus, there’s a predefined path for electrons to follow in a circuit to perform certain functions. 

When a path in your electrical circuit forms that shouldn’t be there, we call that a short circuit.

Now, the job of current limiting reactors is to reduce short circuit currents.

What is a current limiting reactor?

A simple large coil of wire made to produce a magnetic field. It’s basically an inductive coil.

Electric energy turns into magnetic energy.

So as current flows through the reactor, the reactor’s magnetic field expands. Then eventually stabilizes and the magnetic field stores the energy.

The flowing current through the reactor is proportional to the produced magnetic flux.

This magnetic field then “opposes” the current flow. In other words, a back EMF (counter Electromotive Force) forms from the induced voltage. It then acts against the applied voltage.

The changing current in the inductor produces this back emf. The key to understand is that the reactor limits the current but doesn’t eliminate it.

If the reactor eliminated the current, then we wouldn’t have a back emf, to begin with. So don’t think of the voltage sources opposing each other, but rather balancing each other out.

So there’s nothing too fancy going on here. The reactor is a simple piece of equipment.

What’s important is the size of the wound-up wire and the number of turns in the reactor coil. You want the wire to allow the flow of the designed full current while keeping true power losses to a minimum.

And of course, to limit the short circuit current. That’s the entire point of the reactor here.

Let’s dig a little deeper now.

Understanding equations related to current limiting reactors

To get the ball rolling, Impedance (Z) is Z = R + jX, where:

  • R = resistance
  • X = reactance (more specifically, inductive reactance in our case)

Now when Current (I) flows through an impedance, there’s a voltage drop of V = IZ. Also, there are the following effects on power:

  • Watt losses of: P = I^{2}R
  • Var “losses” of Q = I^{2}X

Important Note: var losses aren’t a watt loss. Where watt losses are energy lost to heat. Rather, the vars aren’t lost, they’re used by the reactor. 

I’m using the term “losses” to balance the reactive power in my discussion. Similar to how we balance real power with real losses.

Now, to keep our watt losses as low as possible, we want a reactor with low resistance.

But, we need the inductive reactance to be high. The high inductive reactance rating will limit the short circuit current flow.

The reactance rating dictates the magnetic field energy density. The greater this energy density, the greater the back emf produced.

We see this in the Q = I^{2}X equation. The var “losses” increase with the square of the current flow. So, as the current increases from a short circuit, the stronger the magnetic field becomes.

Thus, the reactor creates a larger back emf, as represented by: V = L\dfrac{di}{dt}. This also explains why the reactor can’t eliminate the current.

Without a change in current over time, we wouldn’t have a back emf.

In short, we choose the reactor’s inductive reactance to be low enough for an acceptable voltage drop during normal operation. But, high enough to restrict the short circuit current through the created back emf.

When to use a current limiting reactor?

If the short circuit current in your power system exceeds the interrupting rating of your electrical equipment. This includes certain electrical devices as well.

By electrical equipment and devices, I’m talking about the following as examples:

  • Switchboards
  • Switchgear
  • Motor control centers
  • Circuit breakers
  • Fuses

For sake of simplicity, I’m going to reference the above items all as “equipment”.

Each of these pieces of equipment has a rating to withstand a maximum short circuit. This is your equipment’s “interrupting rating.”

So you always choose your equipment’s interrupting rating to be greater than the maximum calculated short circuit. You do this before you buy your equipment.

I’ll discuss this a bit more later.

That said, the calculated short circuit value hinges on your power source. Your power source can be any one of the following:

  • Utility power
  • Generator
  • Renewable energy such as a solar PV system that back feeds into the power grid

Important Note: the available short circuit current differs at every point in a power system. It’s driven by various variables such as voltage and impedance. 

When you apply for a new electric service, you’re given the available short circuit current by your utility. This value is typically provided at the terminals of the utility transformer that feeds you.  

This value decreases from the utility transformer downstream to your electrical equipment. This is because of the impedance of the conductors from the utility transformer to you. 

What causes the short circuit current to increase?

Now, imagine the following happens:

  • Your facility power requirements increase
  • New power sources added to your facility
  • Your utility source expands

All these scenarios will affect the short circuit current. For example, imagine if the number of substations from your utility source grows year after year.

Let’s say your utility’s substation capacity doubles from 25,000 MVA to 50,000 MVA.

As a result, the short circuit currents in the power grid will increase too.

BUT,  your existing electrical equipment will still have the same interrupting ratings. The interrupting ratings are fixed values dictated by the factory that manufactures your equipment.

So, you’d be screwed!

Current limiting reactor schematic

To illustrate, below is an overly simplified one-line diagram with a current limiting reactor.  It shows the short circuit current traveling from the utility power source to the load.

You can see how the circuit breaker interrupting rating is 65,000 amps. But, the short circuit current rating is 70,000 amps.

Thus, we installed the current limiting reactor to drop the short circuit current below 65,000 amps. Because it’s not practical to replace all your existing electrical equipment.

current limiting reactor schematic

Important Note: every piece of equipment has an interrupting rating. It’s the highest current at a given voltage a piece of equipment can safely interrupt. 

Imagine a 480-volt rated electrical equipment has an interrupting capacity of 65,000 amps. This means it can safely interrupt any short circuit that’s below 65,000 amps. 

It’s a code violation to have equipment that can’t withstand the maximum available short circuit rating in your power system. Because someone can be badly hurt, or even die. 

Current limiting reactor sizing calculation

Lafayette Public Power Authority (Photo Credit: publicpowerorg)

Finally, let’s do our current limiting reactor sizing calculation.

Again, we’re doing this calculation for a fairly large substation in a large size city.

Let’s assume the power transformer rating is: 24/32/40 MVA (OA/FA/FA)

Also, the transformer has the following voltage ratings:

  • Primary voltage: 115,000-volts
  • Secondary voltage: 12,470-volts

We’ll use the largest transformer rating for our calculation. Thus, 40 MVA.

I_{Full\:load} = \frac{40MVA}{115kV\times\surd(3)} = 200.8 Amps

Important Note: OA/FA/FA are liquid transformer cooling classes.

OA: self-cooling through natural ventilation 

FA: forced air-cooling with the help of fans

FA: another set of fans for additional forced air-cooling 

I’m going to use the per-unit system to do my calculation. I’ll assume you’re familiar.

The system impedance to our station is:

I_{Fault}\: at\: 115kV = 20,000 Amps

I_{Base} = \frac{24MVA}{115kV\times\surd(3)} = 120.5

I_{pu} = \frac{20,000}{120.5} = 166.0\:Amps

Z_{pu} = \frac{V_{pu}}{I_{pu}}} = \frac{1.0}{166.0} = 0.0060\:pu\:\Omega

Now, the existing station fault impedance at the 12.47kV rating and 24MVA base is:

I_{Base} = \frac{24MVA}{12.47kV\times\surd(3)} = 1,111.2 \:Amps

I_{Fault}\:at\:the\: 12.47kV\: network = 14,719\: Amps

I_{pu\: Fault} = \frac{14,719 Amps}{1,111.2 Amps} = 13.246\:pu\:Amps

Z_{pu}\: Fault = \frac{V_{pu}}{I_{pu}} = \frac{1.0}{13.246\:pu} = 0.0755\:pu\:\Omega

Current limiting reactors for transformers connected in parallel

Next, at a large substation, you’ll commonly find transformers connected in parallel. You do this when you need to supply larger loads.

In our case, we have 3 transformers and their associated current limiting reactors designed in parallel. The desired total fault impedance including the system impedance of Z = 0.0060\: pu\: \Omega is 0.0755\: pu\: \Omega.

So, our three transformers and their reactors connected in parallel is 0.0755\: pu\: \Omega - 0.0060\: pu\: \Omega = 0.0695\: pu\: \Omega.

And each transformer and its reactor alone would be 3 \times 0.0695\: pu\: \Omega = 0.2085\: pu\: \Omega

Now, we’ll assume a maximum transformer design impedance of 10.5% (or 0.105\: pu\: \Omega) on a 24 MVA base.

Thus, the current limiting reactor impedance would need to be:

0.2085 = 0.105 = 0.1035 at 24 MVA.

Now, we’ll reflect everything on our 40 MVA base:

0.1035 \times \frac{40 MVA}{24 MVA}= 0.1725 = 17.25% at 40 MVA

The transformer equivalent impedance at 40 MVA would be:

0.105 \times \frac{40 MVA}{24 MVA}= 0.175 = 17.5% at 40 MVA

The current limiting reactor rating 

The reactor rating would need to be: 115kV_{L-L}/66.4kV_{L-N}, 40 MVA, 3\theta, 201\: Amps, Z = 17.5% on a 40 MVA base.

Also, the reactor will need bracing for 20kA available fault current.

Next, the transformer rating would be 24/32/40 MVA (OA/FA/FA) with Z = 10.5% on a 24 MVA base.

Important Note: the Z value for a current limiting reactor and transformer is critical. The higher these impedances are, the less short circuit current we’ll have. 

But, you can’t blindly choose a high impedance current limiting reactor and transformer. Not only is it expensive, but it can increase your circuit’s reactance to unacceptable levels.

Calculating the short circuit current without the current limiting reactor

This calculation will show us how the short circuit rating will be greater than 20,000 Amps. Because we don’t have the current limiting reactors installed to limit the short circuit current.

Z_{Fault\:pu} = Z_{System\:pu} + \frac{1}{3}(Z_{Transit\:pu})

= 0.0060\:pu\: \Omega + \frac{1}{3}(0.105\:pu\: \Omega})

= 0.0410\: pu\: \Omega

I_{pu} = \frac{V_{pu}}{Z_{pu}} = \frac{1.0\: pu\:Volts}{0.0410\: pu\: \Omega}

= 24.38\: pu\: Amps

I_{Fault} = 24.38\: pu\: Amps \times 1,111.2\:Amps

= 27,085\:Amps\:at\:12.47kV

It’s clear as day we need the current limiting reactors. But, it’s important to know reactors have the following drawbacks too:

  • They increase the total reactance of a circuit. As a result, this causes a reactive voltage drop and a lagging power factor. Thus, the power system’s voltage regulation becomes poor.
  • The requirement for physical space to provide clearance from the potential magnetic flux exposure.
  • The magnetic field produced by the coil may induce currents in large nearby objects.
  • Cost to purchase and install the current limiting reactors.

The drawbacks differ though depending on the type of reactors you choose. Also, where you install the reactors in your power system.

It’s always best to install reactors where they’ll be most effective. Sounds obvious, but in practice, this isn’t unfortunately always the case.

Important Note: current limiting reactors are good options to reduce short circuit current. But, you need to carefully consider their usage. 

You should always complete a short circuit and load flow study to better understand a reactor’s impact on your power system.

Current limiting reactor sizing calculation wrap up

Hopefully, you learned a thing or two about current limiting reactors. Also, about the impact of short circuit currents in power systems.

The current limiting reactor sizing calculation is fairly straightforward too. You just need to have information on your power system to do the calculation.

But, you’ll need to do additional calculations as well. This way, you can weigh the negatives and positives of installing reactors.

As with anything in engineering, installing reactors isn’t a quick decision you can make.

What are your thoughts on current limiting reactors? Have you used or seen current limiting reactors in your line of work before? Have you done a current limiting reactor sizing calculation before?


Featured Image Photo Credit: publicpowerorg (image cropped)

SUBSCRIBE TO ENGINEER CALCS NEWSLETTER

Get daily articles and news delivered to your email inbox

Leave a Comment