3.1 Substation Ground Grid Design & IEEE 80 Touch/Step Voltage Criteria

Substation Ground Grid Design & IEEE 80 Safety Criteria

Substation grounding is one of the most heavily tested subjects on the NCEES PE Electrical: Power examination. An engineered grounding system must accomplish two primary, non-negotiable objectives: providing a safe environment to protect personnel from lethal electric shock during fault conditions, and establishing a low-impedance path to ground for power system fault currents, lightning discharges, and switching surges to ensure reliable operation of protective relays.

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1. Fundamentals of Ground Potential Rise (GPR)

When a single line-to-ground (SLG) or double line-to-ground (DLG) fault occurs inside or near an electrical substation, a large fault current enters the earth through the substation ground grid. Because the earth and the ground grid have finite electrical resistance ($R_g$), this current injection raises the electrical potential of the entire grid and the surrounding soil relative to a remote, undisturbed earth reference.

$\text{GPR} = I_g \times R_g$

Where:

• $\text{GPR}$ = Ground Potential Rise (Volts, $\text{V}$)
• $I_g$ = Maximum symmetrical grid fault current flowing between the grid and earth (Amperes, $\text{A}$)
• $R_g$ = Substation ground grid resistance to remote earth (Ohms, $\Omega$)

Grid Current Division Factor ($S_f$) and Decrement Factor ($D_f$)

The total fault current ($I_f$) at a substation bus does not entirely flow through the ground grid into the earth. A portion of the current returns through overhead ground wires (OHGW / static wires), neutral conductors, and underground cable sheaths. The design current is calculated as:

$I_g = D_f \times S_f \times I_f$

Where:

• $S_f$ = Fault current division factor ($0 < S_f \le 1$), representing the fraction of fault current conducted directly into the earth by the grid.
• $D_f$ = Decrement factor accounting for the DC offset component during subtransient fault conditions (typically $1.0$ to $1.25$ depending on fault clearing time $t_f$ and system $X/R$ ratio).
• $I_f$ = Symmetrical root-mean-square (RMS) phase-to-ground fault current (Amperes, $\text{A}$).

                               +-------------------------+
                               |  Substation Ground Grid |
   Remote Earth (0 V) <--------|       Resistance Rg     |<------- Fault Current (Ig)
          |                    +-------------------------+
          |                                 |
          +================ GPR = Ig x Rg ==+

Exam Tip: If $\text{GPR} \le E_{touch,tol}$, the grounding installation is unconditionally safe because even the maximum theoretical potential of the substation structure cannot exceed the tolerable shock limit. However, in almost all high-voltage substations, $\text{GPR}$ exceeds several kilovolts ($5\text{--}25\text{ kV}$), requiring detailed mesh voltage ($E_m$) and step voltage ($E_s$) verification.

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2. Voltage Definitions Under IEEE Std 80

IEEE Std 80 (IEEE Guide for Safety in AC Substation Grounding) defines four distinct spatial potential differences that can appear across a human body during a fault:

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3. Human Body Impedance & Dalziel Shock Model

Electrocution hazard analysis relies on the extensive ventricular fibrillation research performed by Charles F. Dalziel. The human body is modeled as an internal electrical resistance in series with contact resistances at the points of entry and exit.

Standard Body Resistance ($R_b$)

IEEE Std 80 standardizes the internal resistance of the human body from hand to both feet, or foot to foot, as a constant:

$R_b = 1000\ \Omega$

Foot Contact Resistance ($R_f$)

A human foot is modeled as a flat metallic circular disc of radius $b = 0.08\text{ m}$ ($8\text{ cm}$) resting on the surface of the earth. In uniform soil of resistivity $\rho_s$ ($\Omega\cdot\text{m}$), the ground contact resistance of a single foot is:

$R_f = \frac{\rho_s}{4 b} = \frac{\rho_s}{4(0.08)} = 3.125 \rho_s \approx 3 \rho_s$

Equivalent Resistance for Touch and Step Configurations

1. Step Voltage Configuration (Feet in Series): Current flows into one foot, through the legs and torso, and out of the other foot ($R_{total} = R_b + 2 R_f$): $R_{th,step} = 1000 + 2(3 \rho_s) = 1000 + 6 C_s \rho_s$

2. Touch Voltage Configuration (Feet in Parallel): Current flows into one hand, down through the body, and divides equally across both feet in parallel ($R_{total} = R_b + R_f / 2$): $R_{th,touch} = 1000 + \frac{3 \rho_s}{2} = 1000 + 1.5 C_s \rho_s$

Where $C_s$ is the surface layer derating factor.

  TOUCH VOLTAGE CIRCUIT                    STEP VOLTAGE CIRCUIT
       (Hand-to-Feet)                           (Foot-to-Foot)

         [Hand: 0 V]                             [Foot 1]
              |                                      |
         +----+----+                            +----+----+
         | Rb=1000 | (Body)                     | Rf1=3Csρ| (Foot 1 Contact)
         +----+----+                            +----+----+
              |                                      |
         +----+----+                            +----+----+
         | Rf / 2  | (Parallel Feet)            | Rb=1000 | (Body Resistance)
         | 1.5 Csρ |                            +----+----+
         +----+----+                                 |
              |                                 +----+----+
       [Earth Surface]                          | Rf2=3Csρ| (Foot 2 Contact)
                                                +----+----+
                                                     |
                                                 [Foot 2]
   R_total = 1000 + 1.5 Cs ρ_s             R_total = 1000 + 6.0 Cs ρ_s

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4. Surface Layer Derating Factor ($C_s$)

In substation yards, a layer of high-resistivity material—typically crushed rock or granite gravel with resistivity $\rho_s = 2000\text{--}5000\ \Omega\cdot\text{m}$ and thickness $h_s = 0.08\text{--}0.15\text{ m}$ ($3\text{--}6\text{ inches}$)—is spread over the native soil (resistivity $\rho$).

Because the crushed rock layer is thin relative to the electrical field of a foot, the actual foot resistance is reduced by the underlying lower-resistivity native soil. IEEE Std 80 models this with the surface layer derating factor $C_s$:

$C_s = 1 - \frac{0.09 \left(1 - \frac{\rho}{\rho_s}\right)}{2 h_s + 0.09}$

Where:

• $h_s$ = Surface rock layer thickness in meters ($\text{m}$)
• $\rho$ = Native soil resistivity ($\Omega\cdot\text{m}$)
• $\rho_s$ = Surface crushed rock layer resistivity ($\Omega\cdot\text{m}$)
• If no surface layer is present ($h_s = 0$ or $\rho_s = \rho$), $C_s = 1.0$.

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5. Allowable Body Current & Tolerable Voltage Criteria

Dalziel's empirical formula establishes the $99.5%$ non-fibrillation threshold current as a function of shock duration ($t_s$, in seconds, for $0.03\text{ s} \le t_s \le 3.0\text{ s}$):

Tolerable Voltage Equations (50 kg Criteria)

Combining Ohm's law ($E = I_B \times R_{total}$) yields the governing IEEE Std 80 tolerable voltage equations for a $50\text{ kg}$ human:

$\mathbf{E_{touch,50}} = \left(1000 + 1.5 C_s \rho_s\right) \frac{0.116}{\sqrt{t_s}}$

$\mathbf{E_{step,50}} = \left(1000 + 6.0 C_s \rho_s\right) \frac{0.116}{\sqrt{t_s}}$

For a $70\text{ kg}$ body, replace $0.116$ with $0.157$.

Key Principle: Notice that $E_{step}$ is always substantially higher than $E_{touch}$ because $R_{th,step}$ includes two feet in series ($6.0 C_s \rho_s$) whereas $R_{th,touch}$ includes two feet in parallel ($1.5 C_s \rho_s$). Consequently, touch voltage is almost always the limiting criterion in substation ground grid design.

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6. Ground Grid Resistance Estimation ($R_g$)

For preliminary design and PE exam calculations, simple analytical formulas estimate the total resistance to remote earth of a buried ground grid.

Sverak's Formula (Preferred IEEE 80 Method)

Sverak incorporated the grid burial depth ($h$) and total grid buried conductor length ($L_T$):

$R_g = \rho \left[ \frac{1}{L_T} + \frac{1}{\sqrt{20 A}} \left(1 + \frac{1}{1 + h \sqrt{\frac{20}{A}}}\right) \right]$

Where:

• $\rho$ = Soil resistivity ($\Omega\cdot\text{m}$)
• $L_T$ = Total length of buried grid conductors and ground rods ($\text{m}$)
• $A$ = Area occupied by the ground grid ($\text{m}^2$)
• $h$ = Grid burial depth ($\text{m}$, typically $0.5\text{--}1.5\text{ m}$)

Laurent & Niemann Formula (Upper Bound Approximation)

$R_g \approx \frac{\rho}{4 r} + \frac{\rho}{L_T} = \frac{\rho}{4 \sqrt{A / \pi}} + \frac{\rho}{L_T}$

Where $r = \sqrt{A / \pi}$ is the equivalent circular grid radius.

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7. Conductor Sizing for Thermal Fusing (Onderdonk's Formula)

Ground grid conductors must withstand the maximum expected fault current without exceeding the annealing or melting temperature of the material. IEEE Std 80 uses Onderdonk's equation:

$A_{kcmil} = I \times K_f \sqrt{t_c}$

Or in metric units ($A_{mm^2}$):

$I = A_{mm^2} \sqrt{\left(\frac{\text{TCAP} \times 10^{-4}}{t_c \alpha_r \rho_r}\right) \ln\left(\frac{K_0 + T_m}{K_0 + T_a}\right)}$

Where:

• $I$ = Symmetrical fault current magnitude ($\text{kA}$ RMS)
• $A$ = Conductor cross-sectional area ($\text{kcmil}$ or $\text{mm}^2$)
• $t_c$ = Fault duration / clearing time (seconds)
• $T_m$ = Maximum allowable conductor temperature ($1083^\circ\text{C}$ for copper melting, $450^\circ\text{C}$ for exothermic welded copper connections, $250^\circ\text{C}$ for bolted joints)
• $T_a$ = Ambient temperature ($^\circ\text{C}$, typically $40^\circ\text{C}$)
• $K_f$ = Conductor material constant (for annealed copper with $450^\circ\text{C}$ exothermic weld limit, $K_f \approx 7.01$ for $t_c$ in seconds and $A$ in $\text{kcmil}$).

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8. Step-by-Step Worked Calculation Example

Problem Statement

A $115\text{ kV}$ transmission substation features a ground grid covering an area of $50\text{ m} \times 50\text{ m}$ ($A = 2500\text{ m}^2$) buried at depth $h = 0.5\text{ m}$. The total length of buried grid conductors plus ground rods is $L_T = 1500\text{ m}$.

Given site parameters:

• Native soil resistivity: $\rho = 100\ \Omega\cdot\text{m}$
• Surface crushed rock layer: thickness $h_s = 0.10\text{ m}$ ($10\text{ cm}$), resistivity $\rho_s = 2500\ \Omega\cdot\text{m}$
• Maximum grid fault current: $I_g = 6.0\text{ kA} = 6000\text{ A}$
• Fault duration: $t_s = 0.50\text{ s}$
• Design basis: $50\text{ kg}$ body weight

Calculate:

1. The surface layer derating factor $C_s$.
2. The tolerable touch voltage ($E_{touch,50}$) and tolerable step voltage ($E_{step,50}$).
3. The ground grid resistance $R_g$ using Sverak's formula.
4. The Ground Potential Rise ($\text{GPR}$) and assess whether a full mesh voltage analysis is required.

Solution Walkthrough

Step 1: Calculate Surface Layer Derating Factor ($C_s$)

$C_s = 1 - \frac{0.09 \left(1 - \frac{\rho}{\rho_s}\right)}{2 h_s + 0.09} = 1 - \frac{0.09 \left(1 - \frac{100}{2500}\right)}{2(0.10) + 0.09}$

$C_s = 1 - \frac{0.09 (1 - 0.04)}{0.20 + 0.09} = 1 - \frac{0.09(0.96)}{0.29} = 1 - \frac{0.0864}{0.29} = 1 - 0.2979 = \mathbf{0.7021}$

Step 2: Calculate Tolerable Touch and Step Voltages

First, evaluate the body current factor for $t_s = 0.50\text{ s}$:

$I_{B,50} = \frac{0.116}{\sqrt{0.50}} = \frac{0.116}{0.7071} = 0.16405\ \text{A}$

Now, calculate the equivalent contact resistance for touch and step:

$R_{th,touch} = 1000 + 1.5 C_s \rho_s = 1000 + 1.5(0.7021)(2500) = 1000 + 2632.88 = 3632.88\ \Omega$

$E_{touch,50} = 3632.88\ \Omega \times 0.16405\ \text{A} = \mathbf{596.0\ \text{V}}$

For step voltage:

$R_{th,step} = 1000 + 6.0 C_s \rho_s = 1000 + 6.0(0.7021)(2500) = 1000 + 10531.5 = 11531.5\ \Omega$

$E_{step,50} = 11531.5\ \Omega \times 0.16405\ \text{A} = \mathbf{1891.7\ \text{V}}$

Step 3: Calculate Grid Resistance $R_g$ (Sverak's Formula)

Evaluate intermediate terms for $A = 2500\text{ m}^2$, $h = 0.5\text{ m}$, and $L_T = 1500\text{ m}$:

$\sqrt{20 A} = \sqrt{20 \times 2500} = \sqrt{50000} = 223.61\text{ m}$

$\sqrt{\frac{20}{A}} = \sqrt{\frac{20}{2500}} = \sqrt{0.008} = 0.08944\text{ m}^{-1}$

$1 + h \sqrt{\frac{20}{A}} = 1 + 0.5(0.08944) = 1.04472$

$\frac{1}{1 + h \sqrt{\frac{20}{A}}} = \frac{1}{1.04472} = 0.9572$

$R_g = 100 \left[ \frac{1}{1500} + \frac{1}{223.61} (1 + 0.9572) \right] = 100 \left[ 0.000667 + \frac{1.9572}{223.61} \right]$

$R_g = 100 [0.000667 + 0.008753] = 100 [0.009420] = \mathbf{0.942\ \Omega}$

Step 4: Calculate GPR and Design Compliance

$\text{GPR} = I_g \times R_g = 6000\text{ A} \times 0.942\ \Omega = \mathbf{5652\ \text{V}}$

Compliance Decision: Since $\text{GPR} = 5652\text{ V} > E_{touch,50} = 596.0\text{ V}$, the grid is not unconditionally safe. A detailed calculation of mesh voltage ($E_m$) and step voltage ($E_s$) across each grid quadrant must be conducted to ensure $E_m \le 596.0\text{ V}$ and $E_s \le 1891.7\text{ V}$.

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9. Common NCEES Exam Pitfalls

Pitfall 1: Mixing up 50 kg vs. 70 kg Constants
The $50\text{ kg}$ constant is $0.116$, while the $70\text{ kg}$ constant is $0.157$. Unless a problem specifically instructs you to use the $70\text{ kg}$ criterion, IEEE 80 and standard NCEES exam problems default to $50\text{ kg}$.

Pitfall 2: Forgetting the Parallel vs. Series Foot Coefficients
In touch voltage, the two feet are in parallel, giving $R_f / 2 = 1.5 C_s \rho_s$. In step voltage, the two feet are in series, giving $2 R_f = 6.0 C_s \rho_s$. Never confuse the $1.5$ and $6.0$ multiplier terms!

Pitfall 3: Total Fault Current vs. Grid Current
Do not use the total bus fault current ($I_f$) directly to compute $\text{GPR}$ when a current division factor ($S_f$) is provided. Always apply $I_g = S_f \times I_f$.