**Short-Circuit Calculations**

**A Handbook to Accompany the**

**Short-Circuit Calculation Program**

**From MSHA's Approval and Certification Center**

**By Wayne L. Carey**

**January 4, 2006**

**Short-Circuit Calculations**

[Introduction]{.underline}

One of the jobs of the Mine Safety and Health Administration's (MSHA's)
Approval and Certification Center (A&CC) is to approve electrical
equipment for underground mines. An important part of this job is to
determine if circuit protective devices are sized and set correctly.
Sizing of circuit protective devices can only be done when maximum
available short-circuit current is known. Setting of the devices can
only be done when minimum available short-circuit current is known. The
A&CC has a computer program available through the MSHA website
([www.msha.gov](http://www.msha.gov/)) to calculate available
short-circuit currents in mine power systems. This program, other
programs, or hand calculation can be used to perform short-circuit
calculations. Any method used requires specific data on the mine power
system. All this data is converted to resistance and reactance values
for each component in the system. Impedance can then be calculated and
then minimum and maximum available currents. This paper describes the
data needed for each type of circuit component and how to obtain the
data. It also describes methods used by the computer program to
calculate available short-circuit currents.

[Utility]{.underline}

The first step in analyzing a power system is to get the data for the
power available at the site, the utility data. This data can be obtained
from the power company. When calling the power company, explain the type
of information you need and ask for the engineering department. It may
be helpful to explain why you need the information.

The power company will be able to supply this information for the point
in the power system where their responsibility for the power system ends
and the customer's responsibility starts. A common location for this
point is the secondary of a pole or pad mounted transformer. If the
customer is responsible for the transformer, the transition point would
be the primary of the transformer. Sometimes a pole mounted disconnect
will be the transition point. The power company will specify where in
the system their responsibility ends.

The data needed is the line to line voltage (V~LL~), short-circuit kVA
(kVA~SC~), and X/R. Obtaining the voltage is simple enough.
Short-circuit kVA is the power available at a bolted three phase fault.
Bolted means all three phases connected together with no added
impedance. X/R is the ratio of reactance to resistance in the supply.
Short-circuit kVA and X/R may need to be derived from other data.

Short-circuit current (I~SC~) is sometimes supplied by the power company
rather than short-circuit kVA. This current is the current in one phase
of a three phase bolted fault. The short-circuit kVA can be calculated
from the short-circuit current using the following equation.

$\text{kVA}_{\text{SC}} = \frac{\sqrt{3} \ast I_{\text{SC}} \ast V_{\text{LL}}}{\text{1000}}$
(Eq. 1)

If you're not doing these calculations every day, it is sometimes hard
to remember when to include the square root of three factor. Power in a
three phase circuit is three times the single phase power or three times
the current in one phase times the line to neutral voltage (V~LN~). The
line to neutral voltage is the line to line voltage divided by the
square root of three.

$\text{kVA}_{\text{SC}} = \frac{3 \ast I_{\text{SC}} \ast V_{\text{LN}}}{\text{1000}} = \frac{3 \ast I_{\text{SC}}}{\text{1000}} \ast \frac{V_{\text{LL}}}{\sqrt{3}} = \frac{\sqrt{3} \ast I_{\text{SC}} \ast V_{\text{LL}}}{\text{1000}}$
(Eq. 2)

Power factor (PF) is sometimes specified instead of X/R. This must be
the short-circuit power factor. Power factor is defined as the cosine of
the angle between voltage and current. X/R is the tangent of this same
angle. X/R can be found from power factor by taking the tangent of the
inverse cosine of the power factor.

$\frac{X}{R} = \text{tan}\left( \text{cos}^{- 1}\text{PF} \right)$ (Eq.
3)

When neither X/R nor power factor are specified for the utility, it is
usually safe to assume the impedance of the utility is all reactance and
X/R is infinite. Unless there are many miles of transmission line, the
impedance of the utility will be mainly reactance in the generator. This
is all the data needed for the utility. From this data, the
short-circuit calculation program can be used to calculate impedance
(Z), resistance (R), and reactance (X). This data can also be calculated
manually.

Impedance is calculated from V~LL~ and short-circuit kVA.

$Z = \frac{V_{\text{LL}}^{2}}{\text{kVA}_{\text{SC}} \ast \text{1000}}$
(Eq. 4)

Resistance and reactance are then calculated from the impedance using
X/R.

Since: $Z = \sqrt{X^{2} + R^{2}}$ (Eq. 5)

$R = \frac{Z}{\sqrt{1 + \left( \frac{X}{R} \right)^{2}}}$ (Eq. 5a) And
$X = R \ast \left( \frac{X}{R} \right)$ (Eq. 5b)

Resistance and reactance are calculated at the voltages for the points
in the circuit where the short-circuit currents are calculated. For
example, even though the utility voltage may be 69 kilovolts (kV), if
the short-circuit currents are being calculated further down the circuit
where the voltage is 2400 volts, the resistance and reactance will be
calculated at 2400 volts. The computer program initially calculates
resistance and reactance for the utility at the utility voltage. As the
program works down through a circuit, encountering transformers, it
converts the resistance and reactance to the new voltage by multiplying
by the ratio of the voltages squared.

$R_{2} = R_{1} \ast \frac{V_{2}^{2}}{V_{1}^{2}}$ (Eq. 6)

Where: R~2~ = resistance at secondary voltage

R~1~ = resistance at primary voltage

V~2~ = secondary voltage

V~1~ = primary voltage

This formula is the result of conservation of energy. Energy into the
transformer (V~1~^2^/R~1~) equals energy out of the transformer
(V~2~^2^/R~2~).

[Transformers]{.underline}

Transformers are specified by output voltage (V), kVA rating, percent
impedance (%Z), and X/R ratio. The X/R ratio is the ratio of reactance
to resistance. This information, with the exception of X/R, is usually
on the transformer nameplate. If X/R is not specified on the nameplate,
the transformer manufacturer may be able to supply this. When contacting
the transformer manufacturer, it may be helpful to have the transformer
serial number. If X/R cannot be obtained, a value of 4.9 is typical and
can be used in calculations. Impedance (Z) is calculated from V, kVA,
and %Z.

$Z = \frac{\% Z \ast V^{2}}{\text{kVA} \ast \text{100},\text{000}}$ Or
$Z = \frac{\% Z}{\text{100}} \ast \frac{V^{2}}{\text{VA}}$ (Eq. 7)

Resistance and reactance are then calculated from Z and X/R as they were
for the utility. Again, these resistances and reactances are for a
short-circuit at the secondary of the transformer. If the short-circuit
is at a point further down the circuit and after another transformer,
the voltage at the short-circuit should be used in equation 7.
Alternatively, the calculated resistances and reactances can be
converted to the new voltage by multiplying by the ratio of the voltages
squared.

[Three Winding Transformers]{.underline}

The most common three winding transformers used in mining are located in
underground power centers. They are usually specified like two separate
transformers with no interwinding impedance. They can be modeled as two
separate transformers. When three winding transformers have separate
interwinding impedance specified, it is usually specified as reactance
(X) and resistance (R) in ohms. Three sets of X and R must be supplied;
primary-secondary (ps), primary-tertiary (pt), and secondary-tertiary
(st). X~ps~ and R~ps~ are measured in the primary with the secondary
short-circuited and the tertiary open. X~pt~ and R~pt~ are measured in
the primary with the tertiary short-circuited and the secondary open.
X~st~ and R~st~ are measured in the secondary with the tertiary
short-circuited and the primary open. The three winding transformer is
modeled as follows.

![](media/image1.png){width="2.1770833333333335in"
height="1.0208333333333333in"}

The impedances are related by the following formulas.

$Z_{\text{ps}} = Z_{p} + Z_{s}$ (Eq. 8a)

$Z_{\text{pt}} = Z_{p} + Z_{t}$ (Eq. 8b)

$Z_{\text{st}} = Z_{s} + Z_{t}$ (Eq. 8c)

$Z_{p} = 0\text{.}5 \ast \left( Z_{\text{ps}} + Z_{\text{pt}} - Z_{\text{st}} \right)$
(Eq. 8d)

$Z_{s} = 0\text{.}5 \ast \left( Z_{\text{ps}} + Z_{\text{st}} - Z_{\text{pt}} \right)$
(Eq. 8e)

$Z_{t} = 0\text{.}5 \ast \left( Z_{\text{pt}} + Z_{\text{st}} - Z_{\text{ps}} \right)$
(Eq. 8f)

In the equations, all impedances must be referred to a common voltage.
Note that X~st~ and R~st~ are measured at the secondary voltage. If all
impedances are to be at the primary voltage, X~st~ and R~st~ must be
transferred.

[Cables]{.underline}

The minimum data needed for cables is size, length, voltage rating, and
type. Size, voltage rating, and type of cables is embossed into the
jacket. Size and type of aerial cable may be shown on a drawing or
listed on an invoice. Lengths may be measured, determined from mine
maps, or paced off and estimated. Some cables such as types SHD and MPF
have different constructions for different voltage ratings and the
voltage rating will need to be specified. It will be embossed in the
cable jacket. It should be noted if more than one cable is used in
parallel. The spacing of conductors of aerial cable will affect
inductance. The computer program uses a default spacing of three feet.
If the spacing is different than three feet, it should be noted. The
computer program has stored values for resistance and reactance per
thousand feet for most commonly used cable types. This data is also
available in cable standards and manufacturer's specifications.
Resistance is specified at ambient temperature.

Cable operating temperature has an effect on the resistance of a cable.
Most cables have a rated operating temperature of 90 °C. Aerial cable is
rated 75 °C. If different operating temperature ratings are specified,
they should be noted. The computer program uses fully loaded cables at
their rated operating temperature to calculate minimum available
currents. The cables have higher resistances at their rated operating
temperature ratings than at ambient temperature. The resistance at rated
operating temperature can be calculated from the resistance at ambient
temperature using the following formula.

$R_{2} = R_{1}\left\lbrack 1 + \alpha\left( T_{2} - T_{1} \right) \right\rbrack$
(Eq. 9)

Where: R~2~ = resistance at operating temperature

R~1~ = resistance at ambient temperature

T~2~ = rated operating temperature

T~1~ = ambient temperature

α = temperature coefficient of resistivity corresponding to temperature
T~1~ (0.00393 for copper at 20 °C)

When calculating maximum available current by hand, the resistance of
the cable at ambient temperature should be used. However, the current
that causes the cable to reach its highest temperature may not be the
maximum available current. If the cable is initially at a temperature
between ambient and its rated operating temperature and a short-circuit
occurs, it is possible that the cable will reach a higher final
temperature at a current lower than the maximum available current. The
current depends on the cable's resistance, but the resistance depends on
the current. Therefore, the computer program calculates current that
causes maximum temperature with an initial temperature of ambient and
with an initial temperature at one degree increments up to the cable's
rated operating temperature. This is one place where the power of the
computer really comes in handy. The computer program also uses the
subtransient reactance (discussed below) to calculate temperature after
one cycle of short-circuit. This temperature is used as the initial
temperature along with transient reactance in an additional calculation
to calculate temperature at the end of the short-circuit (when the
circuit protective device opens). The maximum available current
calculated in the computer program is actually the current that causes
maximum temperature.

Reactance of most cables is published by the manufacturer. The reactance
in ohms per 1000 feet of aerial cables with one foot spacing can be
found with the following formula.

$X_{L} = 0\text{.}\text{02298} \ast \text{ln}\left( \frac{1}{\text{GMR}} \right)$
(Eq. 10)

GMR is the geometric mean radius in feet. It can be calculated by
multiplying the wire O.D. in inches by .03245.

$\text{GMR} \simeq R \ast e^{- 0\text{.}\text{25}} \simeq 0\text{.}\text{3894} \ast D \simeq 0\text{.}\text{03245} \ast d$
(Eq. 11)

Where: R = wire radius in feet

D = wire diameter in feet

d = wire diameter in inches

The reactance of aerial cable depends on the spacing between wires.
Reactance at spacings other than one foot can be calculated with the
following formula.

$X_{\text{new}} = X_{\text{old}}\left\lbrack 1 + \frac{\text{ln}\left( \text{spacing} \right)}{\text{ln}\left( \frac{1}{\text{GMR}} \right)} \right\rbrack$
(Eq. 12)

[Motors]{.underline}

Motors are a source of short-circuit current. They will act like
generators when a short-circuit occurs. At a minimum, motors must be
specified by horsepower which will be listed on the motor nameplate.
Their reactance may also be specified. Motors have two values of
reactance, subtransient and transient. Subtransient reactance is the
reactance of the motor during the first cycle of the short-circuit.
Transient reactance is the reactance of the motor during the remainder
of the short-circuit. The computer program supplies typical values of
subtransient and transient reactance for motors. Resistance is
considered to be negligible. AC motors may be further identified as
induction or synchronous. Synchronous motors can be 6-pole or 8- to
14-pole. The subtransient and transient reactance will vary depending on
whether the motor is induction, 6-pole synchronous, or 8- to 14-pole
synchronous. It also varies between induction motors rated less than or
equal to 600 volts and induction motors rated greater than 600 volts.
Induction motors will not contribute to a short-circuit after the first
cycle. Typical values of reactance are listed in Table 1^1^.

+---------------------------+----------------------+------------------+
| **Type of Motor**         | **Subtransient       | **Transient      |
|                           | Reactance (per       | Reactance**      |
|                           | unit)**              |                  |
|                           |                      | **(per unit)**   |
+---------------------------+----------------------+------------------+
| DC                        | 0.15                 | 0.30             |
+---------------------------+----------------------+------------------+
| 6-pole Synchronous        | 0.15                 | 0.23             |
+---------------------------+----------------------+------------------+
| 8- to 14-pole Synchronous | 0.20                 | 0.30             |
+---------------------------+----------------------+------------------+
| Induction ≤600 Volts      | 0.25                 | \-               |
+---------------------------+----------------------+------------------+
| Induction \>600 Volts     | 0.17                 | \-               |
+---------------------------+----------------------+------------------+

Table 1

Subtransient and transient reactances are usually given in per unit
values. The actual value of reactance can be found with the following
formula.

$X = \frac{X_{\text{PU}} \ast V^{2}}{\text{746} \ast \text{HP}}$ (Eq.
13)

Where: X~PU~ = per unit reactance

V = voltage

HP = horsepower

Subtransient reactance should be used to calculate maximum available
short-circuit current. Minimum available short-circuit current does not
usually include contributions from motors, but when it does, transient
reactance should be used in the calculation. A&CC does not include motor
contribution in minimum available short-circuit current. When adding
impedances working down through a circuit, the impedance of a motor
should be added in parallel with the total impedance up to the point in
the circuit where the motor contributes. This reduces the impedance at
that point. Adding impedances in parallel is most easily done by first
converting resistance and reactance to conductance and susceptance,
adding the conductances and susceptances, and converting the conductance
and susceptance back to resistance and reactance. Conductance,
susceptance, resistance, and reactance are related by the following
formulas.

$G = \frac{R}{R^{2} + X^{2}}$ (Eq. 14a) $- B = \frac{X}{R^{2} + X^{2}}$
(Eq. 14b)

$R = \frac{G}{G^{2} + B^{2}}$ (Eq. 14c) $X = \frac{- B}{G^{2} + B^{2}}$
(Eq.14d)

Where: R = resistance

X = reactance

G = conductance

B = susceptance

[Generators]{.underline}

Generators are treated just like motors. They are specified by kVA,
subtransient reactance, and transient reactance. The kVA will be listed
on the generator nameplate. Table 2 lists typical values of reactance
for dc and four types of ac generators^2^. The four types are two-pole
turbine, four-pole turbine, salient pole with dampers and salient pole
without dampers.

  ---------------------------- --------------------- ---------------------
  **Type of Generator**        **Subtransient        **Transient Reactance
                               Reactance (per        (per unit)**
                               unit)**               

  DC                           0.15                  0.30

  2-Pole Turbine               0.09                  0.15

  4-Pole Turbine               0.14                  0.23

  Salient Pole with Dampers    0.20                  0.30

  Salient Pole without Dampers 0.30                  0.30
  ---------------------------- --------------------- ---------------------

Table 2

Reactance is calculated with the following formula.

$X = \frac{X_{\text{PU}} \ast V^{2}}{\text{1000} \ast \text{kVA}}$ (Eq.
15)

Where: X~PU~ = per unit reactance

V = voltage

kVA = kilovolt-ampere rating

[Capacitors]{.underline}

Capacitors are specified by kVAR on their nameplates. This is kVA
reactive. If no tolerance is specified, 15% may be used. Capacitors will
feed a short-circuit just like motors and generators. The following
formula calculates reactance for a capacitor using a tolerance of 15%.

$X = \frac{V^{2} \ast 1\text{.}\text{15}}{\text{kVAR} \ast \text{1000}}$
(Eq. 16)

Where: V = voltage

kVAR = kilovolt-amperes reactive

[Rectifiers]{.underline}

Efficiency is the only rating needed for rectifiers. If no efficiency
rating is specified, 99% can be assumed. Current through the rectifier
is reduced by the factor Efficiency (%) / 100.

[Minimum Available Current]{.underline}

Minimum available current is calculated for a line to line arcing fault
using the following formula.

$I_{\text{MIN}} = \frac{0\text{.}\text{95} \ast V_{\text{LL}}}{2 \ast Z_{\text{MAX}}}$
(Eq. 17)

Where: V~LL~ = line to line voltage

Z~MAX~ ^=^ maximum impedance

This formula is used for AC and the DC output from a three phase
rectifier. The factor of 0.95 accounts for voltage fluctuations. The
maximum impedance is calculated from the maximum resistances and
reactances for all the elements in the circuit. For AC circuits, the
current is further reduced by multiplying by an arcing fault factor,
K~A~. This factor is listed in Table 3 for various voltages^3^.

  ----------------------------------------- -----------------------------
  **Voltage (V)**                           **K~A~**

  V ≤ 480                                   0.85

  480 \< V ≤ 600                            0.90

  600 \< V ≤ 1040                           0.95

  1040 \< V                                 1.0
  ----------------------------------------- -----------------------------

Table 3

For DC circuits the arc voltage depends on the current. First, the
current is calculated with Equation 17. If the calculated current is
greater than or equal to 600 amperes, the arc voltage is 60 volts. If
the calculated current is less than 600 amperes, the arc voltage is
calculated with the following formula^4^.

$V_{\text{ARC}} = e^{\left\lbrack \frac{\left( \text{1842} - I \right)}{\text{303}} \right\rbrack}$
(Eq. 18)

Where: I = initially calculated current

The voltage is then reduced by the arc voltage and the available current
is recalculated.

[Maximum Available Current]{.underline}

Maximum available current for AC circuits is calculated for a
three-phase bolted fault using the following equation.

$I_{\text{MAX} - \text{AC}} = \frac{V_{\text{LL}}}{\sqrt{3} \ast Z_{\text{MIN}}}$$$
(Eq. 19)

Where: V~LL~ = line to line voltage

Z~MIN~ ^=^ minimum impedance

Maximum available current for DC circuits is calculated for a line to
line bolted fault using the following equation.

$I_{\text{MAX} - \text{DC}} = \frac{V_{\text{LL}}}{2 \ast Z_{\text{MIN}}}$
(Eq. 20)

Where: V~LL~ = line to line voltage

Z~MIN~ ^=^ minimum impedance

[Example]{.underline}

A mine is supplied 95 MVA of power at 34.5 kV with an X/R ratio of 5.23.
The power travels 1200 feet through number 2, ACSR aerial cable with
three-foot spacing to a substation. The substation has a secondary
voltage of 12.47 kV and is rated 10 MVA with 6.08% impedance. The power
then travels 6000 feet through number 2/0, 15 kV, mine power feeder
cable to an underground power center. The power center has a secondary
voltage of 1040 volts and is rated 1350 kVA with 5.0% impedance. Power
then travels 850 feet through number 2/0, 2 kV, shielded trailing cable
to a continuous miner. If a short-circuit occurs on the continuous miner
at the point where the trailing cable ends, what are the minimum and
maximum available short-circuit currents?

![](media/image2.png){width="2.9784722222222224in"
height="5.228472222222222in"}

Figure 1

1\. Starting at the utility, calculate the impedance using equation 4.
Remember to use the voltage at the short-circuit, 1040 volts.

$$$Z = \frac{V_{\text{LL}}^{2}}{\text{kVA} \ast \text{1000}} = \frac{\text{1040}^{2}}{\text{95},\text{000} \ast \text{1000}} = 0\text{.}\text{0114}\ \Omega$

From the impedance and X/R ratio, find X and R.

$$R = \frac{Z}{\sqrt{1 + \left( \frac{X}{R} \right)^{2}}} = \frac{\text{.}\text{0114}}{\sqrt{1 + 5\text{.}\text{23}^{2}}} = 0\text{.}\text{0021}\ \Omega$$

$$X = R \ast \left( \frac{X}{R} \right) = 0\text{.}\text{0021} \ast 5\text{.}\text{23} = 0\text{.}\text{0112}\ \Omega$$

2\. Now, find the reactance and minimum and maximum resistance for the
aerial cable. The GMR must first be found with Equation 11. From
manufacturer's specifications, the O.D. of number 2, ACSR wire is 0.316
inch.

$$$\text{GMR} \simeq 0\text{.}\text{03245} \ast d = 0\text{.}\text{03245} \ast 0\text{.}\text{316}\text{=0 }.\text{ }0103\text{'\} \{}$

The reactance is then found with Equation 10.

$$X_{L} = 0\text{.}\text{02298} \ast \text{ln}\left( \frac{1}{\text{GMR}} \right) = 0\text{.}\text{02298} \ast \text{ln}\left( \frac{1}{0\text{.}\text{0103}} \right) = 0\text{.}\text{105}\ \Omega$$

> Since this is the impedance for one-foot spacing, you must correct for
> three-foot spacing using Equation 12.

$$X_{\text{new}} = X_{\text{old}}\left\lbrack 1 + \frac{\text{ln}\left( \text{spacing} \right)}{\text{ln}\left( \frac{1}{\text{GMR}} \right)} \right\rbrack = 0\text{.}\text{105} \ast \left\lbrack 1 + \frac{\text{ln}(3)}{\text{ln}\left( \frac{1}{0\text{.}\text{0103}} \right)} \right\rbrack = 0\text{.}\text{1304}\ \Omega$$

> The reactance of 1000 feet of number 2, ACSR aerial cable is 0.1304Ω.
> The reactance for 1200 feet is 1.2 \* 0.1304 = 0.1565Ω. This needs to
> be converted from the overhead line voltage to the voltage at the
> short-circuit.

$$X = 0\text{.}\text{1565} \ast \frac{1\text{.}\text{04}^{2}}{\text{34}\text{.}5^{2}} = 0\text{.}\text{0001}\ \Omega$$

> From the manufacturer's specifications, the resistance of number 2,
> ACSR aerial cable is 1.753 Ω per mile at 75 °C. The resistance of 1200
> feet is 0.3984 Ω at 75 °C. Multiplying by the ratio of the voltages
> squared gives a maximum resistance of 0.0004 Ω. To get the minimum
> resistance, the resistance at 75°C must be converted to the resistance
> at 20 °C using Equation 9.

$$R_{\text{20}} = \frac{R_{\text{75}}}{1 + \alpha\left( \text{75} - \text{20} \right)} = \frac{0\text{.}\text{0004}}{1 + 0\text{.}\text{00393} \ast \text{55}} = 0\text{.}\text{0003}\ \Omega$$

3\. The next component in the system is the substation. First, find the
impedance using Equation 7.

$$Z = \frac{\% Z \ast V^{2}}{\text{kVA} \ast \text{100},\text{000}} = \frac{6\text{.}\text{08} \ast \text{1040}^{2}}{\text{10},\text{000} \ast \text{100},\text{000}} = 0\text{.}\text{0066}\ \Omega$$

> The X/R ratio is not specified for the substation. A value of 4.9 can
> be assumed. Using this value and the impedance, find X and R.

$$R = \frac{Z}{\sqrt{1 + \left( \frac{X}{R} \right)^{2}}} = \frac{\text{.}\text{0066}}{\sqrt{1 + 4\text{.}9^{2}}} = 0\text{.}\text{0013}\ \Omega$$

$$X = R \ast \left( \frac{X}{R} \right) = 0\text{.}\text{0013} \ast 4\text{.}9 = 0\text{.}\text{0065}\ \Omega$$

4\. Next is the 6000 feet of number 2/0, 15 kV, mine power feeder cable.
From manufacturer's specifications, reactance of 1000 feet of this cable
is 0.038 Ω. The reactance of 6000 feet at 1040 volts is calculated as
follows:

$$X = 6 \ast \text{.}\text{038} \ast \frac{\text{1040}^{2}}{\text{12},\text{470}^{2}} = 0\text{.}\text{0016}\ \Omega$$

> From manufacturer's specifications, resistance of this cable at 20°C
> is 0.0792 Ω per 1000 feet. Resistance of 6000 feet at 1040 volts would
> be:

$$R_{\text{MIN}} = 6 \ast 0\text{.}\text{0792} \ast \frac{\text{1040}^{2}}{\text{12},\text{470}^{2}} = 0\text{.}\text{0033}\ \Omega$$

> The maximum resistance will be at 90 °C, the rated operating
> temperature of mine power feeder cable.

$$R_{\text{MAX}} = R_{\text{MIN}} \ast \left\lbrack 1 + \alpha\left( \text{90} - \text{20} \right) \right\rbrack = 0\text{.}\text{0033} \ast \left( 1 + 0\text{.}\text{00393} \ast \text{70} \right) = 0\text{.}\text{0042}\ \Omega$$

5\. The power center is the next component in the circuit. First find
the impedance.

$$Z = \frac{\% Z \ast V^{2}}{\text{kVA} \ast \text{100},\text{000}} = \frac{5 \ast \text{1040}^{2}}{\text{1350} \ast \text{100},\text{000}} = 0\text{.}\text{0401}\ \Omega$$

> The X/R ratio is not specified for the substation. A value of 4.9 can
> be assumed. Using this value and the impedance, find X and R.

$$R = \frac{Z}{\sqrt{1 + \left( \frac{X}{R} \right)^{2}}} = \frac{0\text{.}\text{0401}}{\sqrt{1 + 4\text{.}9^{2}}} = 0\text{.}\text{0080}\ \Omega$$

$$X = R \ast \left( \frac{X}{R} \right) = 0\text{.}\text{0080} \ast 4\text{.}9 = 0\text{.}\text{0393}\ \Omega$$

6\. Last is the 850 feet of number 2/0, 2 kV, shielded cable. From
manufacturer's specifications, reactance of 1000 feet of this cable is
0.031 Ω. The reactance of 850 feet is 0.85 \* 0.031 = 0.0264 Ω. From
manufacturer's specifications, resistance of this cable at 20°C is
0.0839 Ω per 1000 feet. The minimum resistance (R~MIN~) of 850 feet is
0.85 \* 0.0839 = 0.0713 Ω. The maximum resistance is the resistance at
the 90 °C rated operating temperature of mine power feeder cable.

$$R_{\text{MAX}} = R_{\text{MIN}} \ast \left\lbrack 1 + \alpha\left( \text{90} - \text{20} \right) \right\rbrack = 0\text{.}\text{0713} \ast \left( 1 + 0\text{.}\text{00393} \ast \text{70} \right) = 0\text{.}\text{0909}\ \Omega$$

7\. Now that all the resistances and reactances are calculated, we total
them and calculate impedances and available currents.

  ----------------------- --------------- --------------- ---------------
  Component               Minimum         Maximum         Reactance (Ω)
                          Resistance (Ω)  Resistance (Ω)  

  Utility                 0.0021          0.0021          0.0112

  Aerial Cable            0.0003          0.0004          0.0001

  Substation              0.0013          0.0013          0.0065

  Mine Power Feeder Cable 0.0033          0.0042          0.0016

  Power Center            0.0080          0.0080          0.0393

  Trailing Cable          0.0713          0.0909          0.0264

  Total                   0.0863          0.1069          0.0851
  ----------------------- --------------- --------------- ---------------

Table 4

$$Z_{\text{MIN}} = \sqrt{R_{\text{MIN}}^{2} + X^{2}} = \sqrt{0\text{.}\text{0863}^{2} + 0\text{.}\text{0851}^{2}} = 0\text{.}\text{1212}\ \Omega$$

$$Z_{\text{MAX}} = \sqrt{R_{\text{MAX}}^{2} + X^{2}} = \sqrt{0\text{.}\text{1069}^{2} + 0\text{.}\text{0851}^{2}} = 0\text{.}\text{1366}\ \Omega$$

$I_{\text{MIN}} = \frac{0\text{.}\text{95} \ast K_{A}V_{\text{LL}}}{2 \ast Z_{\text{MAX}}} = \frac{0\text{.}\text{95} \ast 0\text{.}\text{95} \ast \text{1040}}{2 \ast 0\text{.}\text{1366}} = \text{3436}$
Amperes

$I_{\text{MAX}} = \frac{V_{\text{LL}}}{\sqrt{3} \ast Z_{\text{MIN}}} = \frac{\text{1040}}{1\text{.}\text{732} \ast 0\text{.}\text{1212}} = \text{4954}$
Amperes

[Conclusion]{.underline}

Accurate calculations of available short-circuit currents can be made
without a power systems specialization in electrical engineering.
Symmetrical components and the per-unit system can be left for more
advanced analyses. Hand calculations can still be tedious though. A
computer program can make short work of the calculations.

[References]{.underline}

1.  [Recommended Practice for Electric Power Distribution for Industrial
    Plants, Std. 141-1976]{.underline}, IEEE, 1976.

2.  [Electrical Transmission and Distribution Reference
    Book]{.underline}, East Pittsburgh, PA, Westinghouse Electric
    Corporation, 1964.

3.  William S. Vilcheck, George Fesak, and William J. Helfrich,
    "Instantaneous Circuit Breaker Settings for the Short-Circuit
    Protection of Three-Phase 480-, 600-, and 1040-V Trailing Cables,"
    [IEEE Transactions on Industry Applications Volume IE-17 No. 4, pp.
    362-368]{.underline}, IEEE, July/August, 1981.

4.  George Fesak, William S. Vilcheck, William J. Helfrich, and David C.
    Deutsch, "Instantaneous Circuit Breaker Settings for the
    Short-Circuit Protection of Direct Current 300- and 600-V Trailing
    Cables," [IEEE Transactions on Industry Applications Volume IE-17
    No. 4, pp. 369-375]{.underline}, IEEE, July/August, 1981.