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# IEEE 1584-2018 – In Depth Arc Flash Calculations Using Mathcad Express & Etap 19.0 Last updated: Sep 28, 2020

Previously, we have written a detailed blog on the major changes in the newly approved IEEE 1584-2018 standard. If you haven't checked it out, please click here!

Now, in this blog we will explain the arc flash calculations which are used in the official standard of IEEE 1584-2018 in a very simple and intuitive manner.

Many Arc flash softwares including:

Performs Arc flash calculations in the back end and gives only the final result in the form of Tables, report and Arc flash labels. In this way, user will not be able to grasp the idea of manual calculations, but as a Power Systems Engineer or Protection engineer, you must know about the manual calculations, on which the whole arc flash study calculations rely or based upon.

For this blog, we will limit our discussion only for HV Arc flash calculations which are widely used in High voltage applications. We will perform the hand calculations for Arc flash study as defined in the latest standard of IEEE 1584-2018.

#### Particular Scenario or Case:

Let's consider a Model One line for 15 kV Switchgear (HV system) with HCB electrode configuration which we have made on the Etap 19.0 version.

The reason why we have selected a 15 kV Switchgear is because we want to analyze the results for High voltage arc flash.

#### Assumptions:

Hello there! On a related topic, we previously wrote a blog about IEEE 1584-2018 Published – 2018 IEEE Guide for Performing Arc Flash Hazard Calculations. If this peaks your interest, check it out and let us know what you think

## Arc Flash Hazard Calculations in 5 Steps Using IEEE 1584-2018 Latest Standard:

IEEE 1584-2018 has provided a summary of the steps required to apply to this model which are as follows:

1. Determine the arcing current (Iarc)
2. Determine the arc duration (T) or fault clearing time (Tfault) using the same arcing current determined in "Step no.1"
3. Determine the Box Size Correction Factor (CF)
4. To determine the incident energy (I.E)
5. To determine the arc-flash boundary (AFB)

We will follow the exact steps as described above and perform the calculations manually on Mathcad software, then find out the results.

Before we begin, we first need to configure our bus or electrode configuration according to the standards of Electrode configurations based on voltage levels as defined in new IEEE 1584 2018.

However, IEEE 1584 2018 has included 2 more electrode configurations in its latest standard.

• VCB (Vertical Electrodes in a Closed Box)
• VCBB (Vertical Electrodes terminated in an insulating barrier, Metal "Box" Enclosure)
• HCB (Horizontal Electrodes in a Closed Box)
• VOA (Vertical Electrodes in Open Air)
• HOA (horizontal Electrodes in Open Air)

"If you are curious about the electrode configurations and their differences, then please comment below, we will make a new blog post specifically targeting on the 5 Electrode configurations and their effect on the Incident energy level and Arc flash calculations."

or this particular case, we have chosen the HCB (Horizontal Electrodes inside a Metal box/Enclosure) electrode configuration, in which the Incident energy is more compare to other configurations. We are using it for the worst case scenario.

Without wasting time, let's move on to the calculations step by step.

### First Step: Calculation for Intermediate Average Arcing Current (Iarc) – MV Switchgear

According to the latest standard of IEEE 1584-2018, we need to first determine the coefficients for our first equation based on the HCB (Horizontal Electrodes in a Cubic Box) configuration and voltage levels. According to the given table, we need to calculate the arcing currents at three different open-circuit voltage (Voc). Equation 1 is given by:

Where,

Ibf = Bolted fault current for three phase faults (kA)

G = Conductor Gap between the conductors or electrodes (mm)

Iarc_600 = Average RMS arcing current at the voltage level (Voc) of 600v (kA)

Iarc_2700 = Average RMS arcing current at the voltage level (Voc) of 2700v (kA)

Iarc_14300 = Average RMS arcing current at the voltage level (Voc) of 14300v (kA)

k1 to k10 = Coefficients provided in Table 1

lg = log10

Now, start plugging the variables from above table to obtain the value of arcing currents at the voltage levels of 600, 2700 and 14300 volts.

We will follow the same procedures for voltage levels "2700v" and "14300v". Just change the Constants (k1, k2, k3, …, k10) as given in Table 1.

Now, we have Arcing currents for 3 different voltage levels, but we need to find also the reduced arcing current (Iarc_min) in order to determine if the arcing current variation has an effect on the operating time of protective devices.

Because we already know that if the arcing fault current is less then Protective devices will take longer time to interrupt the fault current, result in a high Incident energy.

In addition to that, when arcing fault current is high, then protective devices will take less amount of time to interrupt the fault current, result in a low value of incident energy level.

That's why we are performing an additional thing in order to determine the reduced and minimum amount of fault current.

### 2nd Step: Arcing current Variation Factor - Calculation

#### Arcing Current Variation Correction Factor:

For this reason, IEEE 1584-2018 has introduced a new correction factor, which will be used to calculate a second set of arc duration, using the reduced arcing current (Iarc_min). Equation for arcing current variation correction factor:

Equation 2 is given by:

#### VarCf:= k1.Voc6 + k2.Voc5 + k3.Voc4 + k4.Voc3 + k5.Voc2 + k6.Voc+ k7 = 0.023

Where,

VarCf = Arcing current variation correction factor(kA)

Iarc = Final or intermediate RMS arcing current (kA)

Voc = Open-circuit voltage between 0.208 kV and 15.0 kV

K1 to k7 = Coefficients provided in Table 2

To determine a lower bound of the average rms arcing current, we will use the above equation (Equation no 2) and coefficients as provided in Table 2.

As we are using the HCB electrode configuration, that's why we will take the values of k1 to k10 from the 3rd row. By using the Mathcad tool, we¬ can easily get our equation 2 solved by plugging all the variables.

Important Note:

The variables and constant value used in the correction factor {1-(0.5 x VarCf)} has a great significance.

• The "0.5" coefficient indicates that variation is applied to the average arcing current to obtain a lower-bound value arcing current.
• 208 volts ≤ Voc ≤ 600 Volts: Iarc (Final current Only)
• • 600 volts ≤ Voc ≤ 15000 volts: Iarc_600 , Iarc_2700 , Iarc_14300 (intermediate average arcing currents). The final Iarcvalue inherits the correction factor.

To determine the final arcing current, incident energy, and arc-flash boundary at a specific voltage, first, we need to calculate the intermediate values for the three voltage levels of 600 V, 2700 V, and 14,300 V. Then use the interpolation Equation (16) to Equation (24) throughout to determine the final estimated values.

Interpolation equations for Arcing Current (Iarc) – Equation 16 to 18

Where,

Iarc_1 = first Iarc interpolation term between 600 V and 2700 V (kA)

Iarc_2 = second Iarc interpolation term used when Voc is greater than 2700 V (kA)

Iarc_3 = third Iarc interpolation term used when Voc is less than 2700 V (kA)

Voc = The open-circuit voltage (system voltage) (kV)

The final arcing current is found from Iarc_2 since the Voltage (Voc) > 2700 volts.

Iarc = Iarc_2

Iarc = 16.506 kA

### 3rd Step: Determine the Box Size Correction Factor (CF)

In order to determine the box size correction factor, use the equations provided in section 4.8. of the official document of IEEE 1584-2018.

But, first we need to determine if the enclosure should be classified as "Typical" or "Shallow".

Shallow Enclosure: Depth less than or equal to 203.2 mm (8 in)

Typical Enclosure: Depth greater than 203.2 mm (8 in)

The effect of depth is only considered if the system voltage is less than 600 V.

Since, our switchgear voltage is higher than 600 volts and the depth is higher than 203.2 mm. That's why we will consider the "Typical enclosure" in our case.

Next, the equivalent width and height can be determined based on the reasoning provided in Table 6. However, we cannot cover this thing in this blog. If you want to read in detail, you can refer to section 4.8.3 of IEEE 1584-2018 standard.

IEEE 1584-2018 has provided us 2 equation w.r.t that. Using equations (11) and (12). Since the width and height are higher than 1244.6 mm both equations are solved with a value 1244.6 (maximum width or height).

Where,

Height1 = The equivalent enclosure height

Width1 = The equivalent enclosure width

Width = The actual enclosure width (mm)

Height = The actual enclosure height (mm)

Voc = The Open Circuit System voltage

A = Constant, equal to 4 for VCB and 10 for VCBB and HCB

B = Constant, equal to 20 for VCB, 24 for VCBB and 22 for HCB

#### In our scenario:

The equivalent enclosure size can be determined using equation (13) given in the document.

The enclosure size correction factor is determined from equation (14) since the enclosure is "Typical". The coefficients b1, b2, b3 come from Table 7 for HCB electrode configuration.

Equation 13 and 14 is given below.

The enclosure size correction factor is determined from equation 14 on the right since the enclosure is "Typical".

The coefficients b1, b2, b3 come from Table no 7 given above for each electrode configurations.

Where,

b1 to b3 = Coefficients based on Electrode configurations

CF = Enclosure size correction Factor

EES = Equivalent enclosure size used to find the correction factor. For typical box, enclosures the minimum value of EES is 20

### 4th Step: Determine the Incident Energy (I.E)

Use Equation (3) to Equation (6) as follows and Table 3, Table 4, and Table 5 to determine the intermediate incident energy values.

We have already present a general formula which will be used afterwards to calculate the incident energy for 600v, 2700v and 14300v.

Where,

E600, E2700, and E14300 are the incident energy at 600 volts, 2700 volts and 14300 volts respectively, measured in J/cm2

T = Arc duration (ms)

G = Gap distance between the electrodes (mm)

Iarc_600 = Arcing current for 600 volts (kA)

Iarc_2700 = Arcing current for 600 volts (kA)

Iarc_14300 = Arcing current for 600 volts (kA)

Ibf = Bolted fault current for three phase faults (kA)

D = The distance between the electrodes and calorimeters (mm)

CF = Correction factor for enclosure size

Lg = Log10

k1 to k13 = Coefficients provided in Table 3, table 4 and table 5.

Now, we will calculate Incident energy for each voltage level step by step and in an intuitive manner.

#### 1. Determine the Incident Energy (I.E) for 600 volts and HCB configuration

Using the coefficients from Table 3 an equation (3) to find the intermediate incident energy for 600 volts.

E600 = 20.468 J/cm2

#### 2. Determine the Incident Energy (I.E) for 2700 volts and HCB configuration

E2700 = 32.115 J/cm2

#### 3. Determine the Incident Energy (I.E) for 14300 volts and HCB configuration

E14300 = 37.358 J/cm2

Now,

Interpolation equations for Incident Energy (I.E) – Equation 19 to 21

Where,

E1 = The first E interpolation term between 600 volts and 2700 volts

E2 = The second E interpolation term when voltage is greater than 2700 volts

E3 = The third E interpolation term when voltage is less than 2700 volts

And,

The final incident energy is found from E2 since the Voltage (Voc) > 2700 volts

E = E2 = 37.132 J/cm2

E = 37.132 J/cm2

### 5th Step: Determine the Arc Flash Boundary (AFB)

The final and last step is to find the arc flash boundary (AFB), IEEE 1584-2018 has provided two intermediate equations i.e. (22) and (23) to determine the intermediate arc-flash boundary values.

In the same manner, we will calculate the incident energy for each voltage level step by step and in an intuitive manner.

#### 1. Determine the Arc-flash Boundary for 600 volts and HCB configuration

Using the coefficients from Table 3 an equation (7) to find the intermediate arc flash boundary for 600 volts.

AFB600 = 1.826 * 103 mm

#### 2. Determine the Arc-flash Boundary for 2700 volts

Using the coefficients from Table 4 an equation (8) to find the intermediate arc flash boundary for 2700 volts.

AFB2700 = 2.683 * 103 mm

#### 3. Determine the Arc-flash Boundary for 14300 volts

Using the coefficients from Table 4 an equation (8) to find the intermediate arc flash boundary for 14300 volts

AFB14300 = 3.073 * 103 mm

Interpolation equations for The Final Arc Flash Boundary (AFB) – Equation 16 to 18

As, we have determined the intermediate Arc flash boundary for the three voltage levels of 600 volts, 2700 volts and 14300 volts. Now, we will use the interpolation equations (22) to (24) to determine the final Arc flash boundary.

Where,

AFB1 = first AFB interpolation term between 600 V and 2700 V (mm)

AFB2 = second AFB interpolation term used when Voc is greater than 2700 V (mm)

AFB3 = third AFB interpolation term used when Voc is less than 2700 V (mm)

The final arc-flash boundary is found from AFB2 (AFB2700) since the voltage, Voc>2700 volts

AFB := AFB2700 = 2.683 x 103 mm

Finally, we have completed all the steps which are required to find the Arc flash boundary. 