4.13 Substation Grounding

4.13.1 Electric Shock and Its Effects on Humans

To properly design a grounding (called equipment grounding) for the high-voltage lines and/or substations, it is important to understand the electrical characteristics of the most important part of the circuit, the human body. In general, shock currents are classified based on the degree of severity of the shock they cause. For example, currents that produce direct physiological harm are called primary shock currents. Whereas currents that cannot produce direct physiological harm but may cause involuntary muscular reactions are called secondary shock currents. These shock currents can be either steady state or transient in nature. In alternating current (ac) power systems, steadystate currents are sustained currents of 60 Hz or its harmonics. The transient currents, on the other hand, are capacitive discharge currents whose magnitudes diminish rapidly with time.

Table 4.6 gives the possible effects of electrical shock currents on humans. Note that threshold value for a normally healthy person to be able to feel a current is about 1 mA. (Experiments have long ago established the well-known fact that electrical shock effects are due to current, not voltage [11].) This is the value of current at which a person is just able to detect a slight tingling sensation on the hands or fingers due to current flow. Currents of approximately 10–30 mA can cause lack of muscular control. In most humans, a current of 100 mA will cause ventricular fibrillation. Currents of higher magnitudes can stop the heart completely or cause severe electrical burns. The ventricular fibrillation is a condition where the heart beats in an abnormal and ineffective manner, with fatal results. Therefore, its threshold is the main concern in grounding design.

Currents of 1 mA or more but less than 6 mA are often defined as the secondary shock currents (let-go currents). The let-go current is the maximum current level at which a human holding an energized conductor can control his muscles enough to release it. The 60 Hz minimum required body current leading to possible fatality through ventricular fibrillation can be expressed as where t is in seconds in the range from approximately 8.3 ms to 5 s.

$\begin{array}{cccc}I\hfill & =\hfill & \frac{0.116}{\sqrt{t}}A\hfill & (4.109)\hfill \end{array}$

The effects of an electric current passing through the vital parts of a human body depend on the duration, magnitude, and frequency of this current. The body resistance considered is usually between two extremities, either from one hand to both feet or from one foot to the other one.

Experiments have shown that the body can tolerate much more current flowing from one leg to the other than it can when current flows from one hand to the legs. Treating the foot as a circular plate electrode gives an approximate resistance of 3ρs, where ρs is the soil resistivity. The resistance of the body itself is usually used as about 2300 Ω hand to hand or 1100 Ω hand to foot [12]. However, IEEE Std. 80-1976 [14] recommends the use of 1000 Ω as a reasonable approximation for body resistance. Therefore, the total branch resistance can be expressed as

$\begin{array}{cccc}R\hfill & =\hfill & 1000+1.5{\rho }_s\mathrm{\Omega }\hfill & (4.110)\hfill \end{array}$

for hand-to-foot currents and

$\begin{array}{cccc}R\hfill & =\hfill & 1000+6{\rho }_s\mathrm{\Omega }\hfill & (4.111)\hfill \end{array}$

for foot-to-foot currents, where ρs is the soil resistivity in ohm meters. If the surface of the soil is covered with a layer of crushed rock or some other high-resistivity material, its resistivity should be used in Equations 4.110 and 4.111.

Since it is much easier to calculate and measure potential than current, the fibrillation threshold, given by Equation 4.109, is usually given in terms of voltage. Therefore, the maximum allowable (or tolerable) touch and step potentials, respectively, can be expressed as

$\begin{array}{cccc}{V}_{\mathrm{touch}}\hfill & =\hfill & \frac{0.116(1000+1.5{\rho }_s)}{\sqrt{t}}\mathrm{V}\hfill & (4.112)\hfill \end{array}$

and

$\begin{array}{cccc}{V}_{\mathrm{step}}\hfill & =\hfill & \frac{0.116(1000+6{\rho }_s)}{\sqrt{t}}\mathrm{V}\hfill & (4.113)\hfill \end{array}$

Table 4.7 gives typical values for various ground types. However, the resistivity of ground also changes as a function of temperature, moisture, and chemical content. Therefore, in practical applications, the only way to determine the resistivity of soil is by measuring it.

Example 4.13

Assume that a human body is part of a 60 Hz electric power circuit for about 0.25 s and that the soil type is average earth. Based on the IEEE Std. 80-1976, determine the following:

1. Tolerable touch potential
2. Tolerable step potential

Solution

1. Using Equation 4.112,

   $V_{\mathrm{touch}}=\frac{0.116(1000+1.5{\rho }_s)}{\sqrt{t}}=\frac{0.116(1000+1.5\times 100)}{\sqrt{0.25}}\cong 267\mathrm{V}$

2. Using Equation 4.113,

   $V_{\mathrm{step}}=\frac{0.116(1000+6{\rho }_s)}{\sqrt{t}}=\frac{0.116(1000+6\times 100)}{\sqrt{0.25}}\cong 371\mathrm{V}$

4.13.2 Ground Resistance

Ground is defined as a conducting connection, either intentional or accidental, by which an electric circuit or equipment becomes grounded. Therefore, grounded means that a given electric system, circuit, or device is connected to the earth serving in the place of the former with the purpose of establishing and maintaining the potential of conductors connected to it approximately at the potential of the earth and allowing for conducting electric currents from and to the earth of its equivalent.

Thus, a safe grounding design should provide the following:

1. A means to carry and dissipate electric currents into ground under normal and fault conditions without exceeding any operating and equipment limits or adversely affecting continuity of service
2. Assurance for such a degree of human safety so that a person working or walking in the vicinity of grounded facilities is not subjected to the danger of critic electrical shock

However, a low ground resistance is not, in itself, a guarantee of safety. For example, about three or four decades ago, a great many people assumed that any object grounded, however crudely, could be safely touched. This misconception probably contributed to many tragic accidents in the past. Since there is no simple relation between the resistance of the ground system as a whole and the maximum shock current to which a person might be exposed, a system or system component (e.g., substation or tower) of relatively low ground resistance may be dangerous under some conditions, whereas another system component with very high ground resistance may still be safe or can be made safe by careful design.

Ground potential rise (GPR) is a function of fault-current magnitude, system voltage, and ground (system) resistance. The current through the ground system multiplied by its resistance measured from a point remote from the substation determines the GPR with respect to remote ground.

The ground resistance can be reduced by using electrodes buried in the ground. For example, metal rods or counterpoise (i.e., buried conductors) are used for the lines, while the grid system made of copper-stranded copper cable and rods is used for the substations.

The grounding resistance of a buried electrode is a function of (1) the resistance of the electrode itself and connections to it, (2) contact resistance between the electrode and the surrounding soil, and (3) resistance of the surrounding soil, from the electrode surface outward. The first two resistances are very small with respect to soil resistance and therefore may be neglected in some applications. However, the third one is usually very large depending on the type of soil, chemical ingredients, moisture level, and temperature of the soil surrounding the electrode.

Table 4.8 gives typical resistivity values for various ground types. However, the resistivity of the ground also changes as a function of temperature, moisture, and chemical content. Therefore, in practical applications, the only way to determine the resistivity of soil is by measuring it.

Table 4.9 presents data indicating the effect of moisture contents on the soil resistivity. The resistance of the soil can be measured by using the three-electrode method or by using self-contained instruments such as the Biddle Megger Ground Resistance Tester.

The human body can tolerate slightly larger currents at 25 Hz and about five times larger at direct current (dc). Similarly, at frequencies of 1,000 or 10,000 Hz, even larger currents can be tolerated.

In the case of lighting surges, the human body appears to be able to tolerate very high currents, perhaps on the order of several hundreds of amperes [17].

When the human body becomes a part of the electric circuit, the current that passes through it can be found by applying Thévenin’s theorem and Kirchhoff’s current law (KCL), as illustrated in Figure 4.40. For dc and ac at 60 Hz, the human body can be substituted by a resistance in the equivalent circuits. The body resistance considered is usually between two extremities, either from one hand to both feet or from one foot to the other one.

Experiments have shown that the body can tolerate much more current flowing from one leg to the other than it can when current flows from one hand to the legs. Figure 4.40a shows a touch contact with current flowing from hand to feet. On the other hand, Figure 4.40b shows a step contact where current flows from one foot to the other. Note that in each case the body current Ib is driven by the potential difference between points A and B.

Currents of 1 mA or more but less than 6 mA are often defined as the secondary shock currents (let-go currents). The let-go current is the maximum current level at which a human holding an energized conductor can control his or her muscles enough to release it. For 99.5% of population, the 60 Hz minimum required body current, IB, leading to possible fatality through ventricular fibrillation can be expressed as

$\begin{array}{ccccc}{I}_b\hfill & =\hfill & \frac{0.116}{\sqrt{t_s}}\mathrm{A}\hfill & \text{for 50 kg body weight}\hfill & (4.114\mathrm{a})\hfill \end{array}$

or

$\begin{array}{ccccc}{I}_b\hfill & =\hfill & \frac{0.157}{\sqrt{t_s}}\mathrm{A}\hfill & \text{for 70 kg body weight}\hfill & (4.114\mathrm{b})\hfill \end{array}$

where t is in seconds in the range from approximately 8.3 ms to 5 s.

The effects of an electric current passing through the vital parts of a human body depend on the duration, magnitude, and frequency of this current. The body resistance considered is usually between two extremities, either from one hand to both feet or from one foot to the other one. Figure 4.41 shows five basic situations involving a person and grounded facilities during fault.

Note that in the figure the mesh voltage is defined by the maximum touch voltage within a mesh of a ground grid. But the metal-to-metal touch voltage defines the difference in potential between metallic objects or structures within the substation site that may be bridged by direct hand-to-hand or hand-to-feet contact. However, the step voltage represents the difference in surface potential experienced by a person bridging a distance of 1 m with the feet without contacting any other grounded object.

On the other hand, the touch voltage represents the potential difference between the GPR and the surface potential at the point where a person is standing while at the same time having a hand in contact with a grounded structure. The transferred voltage is a special case of the touch voltage where a voltage is transferred into or out of the substation from or to a remote point external to the substation site [12].

Finally, GPR is the maximum electrical potential that a substation grounding grid may have relative to a distant grounding point assumed to be at the potential of remote earth. This voltage, GPR, is equal to the maximum grid current times the grid resistance. Under normal conditions, the grounded electrical equipment operates at near-zero ground potential. That is, the potential of a grounded neutral conductor is nearly identical to the potential of remote earth. During a ground fault, the portion of fault current that is conducted by substation grounding grid into the earth causes the rise of the grid potential with respect to remote earth.

Exposure to touch potential normally poses a greater danger than exposure to step potential. The step potentials are usually smaller in magnitude (due to the greater corresponding body resistance), and the allowable body current is higher than the touch contacts. In either case, the value of the body resistance is difficult to establish.

As said before, experiments have shown that the body can tolerate much more current flowing from one leg to the other than it can when current flows from one hand to the legs. Treating the foot as a circular plate electrode gives an approximate resistance of 3ρs, where ρs is the soil resistivity. The resistance of the body itself is usually used as about 2300 Ω hand to hand or 1100 Ω hand to foot.

However, IEEE Std. 80-2000 [12] recommends the use of 1000 Ω as a reasonable approximation for body resistance. Therefore, the total branch resistance, for hand-to-foot currents, can be expressed as

$\begin{array}{ccccc}{R}_b\hfill & =\hfill & 1000+1.5{\rho }_s\mathrm{\Omega }\hfill & \text{for touch voltage}\hfill & (4.115\mathrm{a})\hfill \end{array}$

and, for foot-to-foot currents,

$\begin{array}{ccccc}{R}_b\hfill & =\hfill & 1000+6{\rho }_s\mathrm{\Omega }\hfill & \text{for touch voltage}\hfill & (4.115\mathrm{b})\hfill \end{array}$

where ρs is the soil resistivity, Ω-m. If the surface of the soil has covered with a layer of crushed rock or some other high-resistivity material, its resistivity should be used in Equations 4.56 and 4.57. The touch voltage limit can be determined from

$\begin{array}{cccc}{V}_{\mathrm{touch}}\hfill & =\hfill & \left(R_b+\frac{R_f}{2}\right)I_b\hfill & (4.116)\hfill \end{array}$

and

$\begin{array}{cccc}{V}_{\mathrm{step}}\hfill & =\hfill & \left(R_b+2R_f\right)I_b\hfill & (4.117)\hfill \end{array}$

where

$\begin{array}{cccc}{R}_f\hfill & =\hfill & 3C_s{\rho }_s\hfill & (4.118)\hfill \end{array}$

where

Rb is the resistance of human body, typically 1000 Ω for 50 and 60 Hz
Rf is the ground resistance of one foot
Ib is the rms magnitude of the current going through the body in A, per Equations 4.114a and 4.114b
Cs is the surface layer derating factor based on the thickness of the protective surface layer spread above the earth grade at the substation (per IEEE Std. 80-2000, if no protective layer is used, then Cs = 1)

Since it is much easier to calculate and measure potential than current, the fibrillation threshold, given by Equations 4.114a and 4.114b, is usually given in terms of voltage. Thus, if there is no protective surface layer, for a person with body weight of 50 or 70 kg, the maximum allowable (or tolerable) touch voltages, respectively, can be expressed as

$\begin{array}{ccccc}{V}_{\mathrm{touch}50}\hfill & =\hfill & \frac{0.116(1000+1.5{\rho }_s)}{\sqrt{t_s}}\mathrm{V}\hfill & \text{for 50 kg body weight}\hfill & (4.119\mathrm{a})\hfill \end{array}$

and

$\begin{array}{ccccc}{V}_{\mathrm{touch}70}\hfill & =\hfill & \frac{0.157(1000+1.5{\rho }_s)}{\sqrt{t_s}}\mathrm{V}\hfill & \text{for 70 kg body weight}\hfill & (4.119\mathrm{b})\hfill \end{array}$

If the event of no protective surface layer is used, for the metal-to-metal touch in V, Equations 4.119a and 4.119b become

$\begin{array}{ccccc}{V}_{\mathrm{mm}-\text{touch}50}\hfill & =\hfill & \frac{116}{\sqrt{t_s}}\text{V}\hfill & \text{for 50 kg body weight}\hfill & (4.119\mathrm{c})\hfill \end{array}$

and

$\begin{array}{ccccc}{V}_{\mathrm{mm}-\text{touch}70}\hfill & =\hfill & \frac{157}{\sqrt{t_s}}\text{V}\hfill & \text{for 70 kg body weight}\hfill & (4.119\mathrm{d})\hfill \end{array}$

If a protective layer does exists, then the maximum allowable (or tolerable) step voltages, for a person with body weight of 50 or 70 kg, are given, respectively, as

$\begin{array}{ccccc}{V}_{\mathrm{step}50}\hfill & =\hfill & \frac{0.116(1000+6C_s{\rho }_s)}{\sqrt{t_s}}\mathrm{V}\hfill & \text{for 50 kg body weight}\hfill & (4.120\mathrm{a})\hfill \end{array}$

and

$\begin{array}{ccccc}{V}_{\mathrm{step}70}\hfill & =\hfill & \frac{0.157(1000+6C_s{\rho }_s)}{\sqrt{t_s}}\mathrm{V}\hfill & \text{for 70 kg body weight}\hfill & (4.120\mathrm{b})\hfill \end{array}$

If a protective layer does exists, then the maximum allowable (or tolerable) touch voltages, for a person with body weight of 50 or 70 kg, are given, respectively, as

$\begin{array}{ccccc}{V}_{\mathrm{touch}50}\hfill & =\hfill & \frac{0.116(1000+1.5C_s{\rho }_s)}{\sqrt{t_s}}\mathrm{V}\hfill & \text{for 50 kg body weight}\hfill & (4.120\mathrm{c})\hfill \end{array}$

$\begin{array}{ccccc}{V}_{\mathrm{touch}70}\hfill & =\hfill & \frac{0.157(1000+1.5C_s{\rho }_s)}{\sqrt{t_s}}\mathrm{V}\hfill & \text{for 70 kg body weight}\hfill & (4.120\mathrm{d})\hfill \end{array}$

The earlier equations are applicable only in the event that a protection surface layer is used. For metal-to-metal contacts, use ρs = 0 and Cs = 1. For more detailed applications, see IEEE Std. 2000 [12]. Also, it is important to note that in using the earlier equations, it is assumed that they are applicable to 99.5% of the population. There are always exceptions.

4.13.3 Reduction of Factor Cs

Note that according to IEEE Std. 80-2000, a thin layer of highly resistive protective surface material such as gravel spread across the earth at a substation greatly reduced the possible shock current at a substation. IEEE Std. 80-2000 gives the required equations to determine the ground resistance of one foot on a thin layer of surface material as

$\begin{array}{cc}{C}_s=1+\frac{1,6b}{{\rho }_s}{\displaystyle \sum _{n=1}^{\infty }{K}^n{R}_{m(2n{h}_s)}}\hfill & (4.121)\hfill \end{array}$

and

$\begin{array}{cc}{C}_s=1-\frac{0.09(1-\rho /{\rho }_s))}{2h_s+0.09}\hfill & (4.122)\hfill \end{array}$

where

$\begin{array}{cc}K=\frac{\rho -{\rho }_s}{\rho +{\rho }_s}\hfill & (4.123)\hfill \end{array}$

where

Cs is the surface layer derating factor (it can be considered as a corrective factor to compute the effective foot resistance in the presence of a finite thickness of surface material); see Figure 4.42
ρs is the surface material resistivity, Ω-m
K is the reflection factor between different material resistivity
ρ is the resistivity of earth beneath the substation, Ω-m
hs is the thickness of the surface material, m
b is the radius of circular metallic disk representing the foot, m
Rm(2nhs) is the mutual ground resistance between two similar, parallel, coaxial plates that are separated by a distance of (2nhs), Ω-m

Note that Figure 4.42 gives the exact value of Cs instead of using the empirical equation (4.64) for it. The empirical equation gives approximate values that are within 5% of the values that can be found in the equation.

Example 4.14

Assume that a human body is part of a 60 Hz electric power circuit for about 0.49 s and that the soil type is average earth. Based on the IEEE Std. 80-2000, determine the following:

1. Tolerable touch potential, for 50 kg body weight
2. Tolerable step potential, for 50 kg body weight
3. Tolerable touch voltage limit for metal-to-metal contact, if the person is 50 kg
4. Tolerable touch voltage limit for metal-to-metal contact, if the person is 70 kg

Solution

1. Using Equation 4.61a, for 50 kg body weight,

   $\begin{array}{ccc}{V}_{\mathrm{touch}50}\hfill & =\hfill & \frac{0.116(1000+1.5{\rho }_s)}{\sqrt{t_s}}\hfill \\ \hfill & =\hfill & \frac{0.116(1000+1.5\times 100)}{\sqrt{0.49}}\hfill \\ \hfill & \cong \hfill & 191\mathrm{V}\hfill \end{array}$

2. Using Equation 4.61b,

   $\begin{array}{ccc}{V}_{\mathrm{step}50}\hfill & =\hfill & \frac{0.116(1000+6{\rho }_s)}{\sqrt{t_s}}\hfill \\ \hfill & =\hfill & \frac{0.116(1000+6\times 100)}{\sqrt{0.49}}\hfill \\ \hfill & \cong \hfill & 265\mathrm{V}\hfill \end{array}$

3. Since ρs = 0,

   $\begin{array}{cc}{V}_{\mathrm{mm}-\text{touch}50}=\frac{116}{\sqrt{t_s}}=\frac{116}{\sqrt{0.49}}=165.7\mathrm{V}\hfill & \text{for 50 kg body weight}\hfill \end{array}$

4. Since ρs = 0,

   $\begin{array}{cc}{V}_{\mathrm{mm}-\text{touch}70}=\frac{157}{\sqrt{t_s}}=\frac{157}{\sqrt{0.49}}=224.3\mathrm{V}\hfill & \text{for 70 kg body weight}\hfill \end{array}$

   Figure 4.43 shows a ground rod driven into the soil and conducting current in all directions. Resistance of the soil has been illustrated in terms of successive shells of the soil of equal thickness. With increased distance from the electrode, the soil shells have greater area and therefore lower resistance. Thus, the shell nearest the rod has the smallest cross section of the soil and therefore the highest resistance. Measurements have shown that 90% of the total resistance surrounding an electrode is usually with a radius of 6–10 ft.

   Figure 4.43

   

   Resistance of earth surrounding an electrode.

The assumptions that have been made in deriving these formulas are that the soil is perfectly homogeneous and the resistivity is of the same known value throughout the soil surrounding the electrode. Of course, these assumptions are seldom true. The only way one can be sure of the resistivity of the soil is by actually measuring it at the actual location of the electrode and at the actual depth.

Figure 4.44 shows the variation of soil resistivity with depth for a soil having uniform moisture content at all depths [23]. In reality, however, deeper soils have greater moisture content, and the advantage of depth is more visible. Some nonhomogeneous soils can also be modeled by using the two-layer method [20].

The resistance of the soil can be measured by using the three-electrode method or by using selfcontained instruments such as the Biddle Megger Ground Resistance Tester. Figure 4.45 shows the approximate ground resistivity distribution in the United States.

If the surface of the soil is covered with a layer of crushed rock or some other high-resistivity material, its resistivity should be used in the previous equations. Table 4.9 gives typical values for various ground types. However, the resistivity of ground also changes as a function of temperature, moisture, and chemical content. Thus, in practical applications, the only way to determine the resistivity of soil is by measuring it.

In general, soil resistivity investigations are required to determine the soil structure. Table 4.9 gives only very rough estimates. The soil resistivity can vary substantially with changes in temperature, moisture, and chemical content. To determine the soil resistivity of a specific site, it is required to take soil resistivity measurements. Since soil resistivity can change both horizontally and vertically, it is necessary to take more than one set of measurements. IEEE Std. 80-2000 [12] describes various measuring techniques in detail. There are commercially available computer programs that use the soil data and calculate the soil resistivity and provide a confidence level based on the test. There is also a graphical method that was developed by Sunde [20] to interpret the test results.

4.13.4 Soil Resistivity Measurements

Table 4.9 gives estimates on soil classification that are only an approximation of the actual resistivity of a given site. Actual resistivity tests therefore are crucial. They should be made at a number of places within the site. In general, substation sites where the soil has uniform resistivity throughout the entire area and to a considerable depth are seldom found.

4.13.4.1 Wenner Four-Pin Method

More often than not, there are several layers, each having a different resistivity. Furthermore, lateral changes also take place, however, with respect to the vertical changes; these changes usually are more gradual. Hence, soil resistivity tests should be made to find out if there are any substantial changes in resistivity with depth. If the resistivity varies considerably with depth, it is often desirable to use an increased range of probe spacing in order to get an estimate of the resistivity of deeper layers.

IEEE Std. 81-1983 describes a number of measuring techniques. The Wenner four-pin method is the most commonly used technique. Figure 4.46 illustrates this method. In this method, four probes (or pins) are driven into the earth along a straight line, at equal distances apart, driven to a depth b. The voltage between the two inner (i.e., potential) electrodes is then measured and divided by the current between the two outer (i.e., current) electrodes to give a value of resistance R. The apparent resistivity of soil is determined from where

$\begin{array}{cc}{\rho }_a=\frac{4\pi aR}{1+\left(2a/\sqrt{a^2+4b^2}\right)}-\frac{a}{\sqrt{a^2+b^2}}\hfill & (4.124)\hfill \end{array}$

where

ρa is the apparent resistivity of the soil, Ω-m
R is the measured resistivity, Ω
a is the distance between adjacent electrodes, m
b is the depth of the electrodes, m

In the event that b is small in comparison to a, then

$\begin{array}{cc}{\rho }_a=2\pi aR\hfill & (4.125)\hfill \end{array}$

The current tends to flow near the surface for the small probe spacing, whereas more of the current penetrates deeper soils for large spacing. Because of this fact, the previous two equations can be used to determine the apparent resistivity ρa at a depth a.

The Wenner four-pin method obtains the soil resistivity data for deeper layers without driving the test pins to those layers. No heavy equipment is needed to do the four-pin test. The results are not greatly affected by the resistance of the test pins or the holes created in driving the test pins into the soil. Because of these advantages, the Wenner method is the most popular method.

4.13.4.2 Three-Pin or Driven Ground Rod Method

IEEE Std. 81-1983 describes a second method of measuring soil resistivity. It is illustrated in Figure 4.47. In this method, the depth (Lr) of the driven rod located in the soil to be tested is varied. The other two rods are known as reference rods. They are driven to a shallow depth in a straight line. The location of the voltage rod is varied between the test rod and the current rod.

Alternatively, the voltage rod can be placed on the other side of the driven rod. The apparent resistivity is found from

$\begin{array}{cc}{\rho }_a=\frac{2\pi L_{r}R}{\mathrm{In}\left(8L_r/d\right)-1}\hfill & (4.126)\hfill \end{array}$

where
Lr is the length of the driven rod, m
d is the diameter of the rod, m
R is the measured resistivity, Ω

A plot of the measured resistivity value ρa versus the rod length (Lr) provides a visual aid for finding out earth resistivity variations with depth. An advantage of the driven-rod method, even though not related necessarily to the measurements, is the ability to determine to what depth the ground rods can be driven. This knowledge can save the need to redesign the ground grid. Because of hard layers in the soil such as rock and hard clay, it becomes practically impossible to drive the test rod any further resulting in insufficient data.

A disadvantage of the driven-rod method is that when the test rod is driven deep in the ground, it usually losses contact with the soil due to the vibration and the larger diameter couplers resulting in higher measured resistance values. A ground grid designed with these higher soil resistivity values may be unnecessarily conservative. Thus, this method presents an uncertainty in the resistance value.

4.14 Substation Grounding

Grounding at substation has paramount importance: Again, the purpose of such a grounding system includes the following:

1. To provide the ground connection for the grounded neutral for transformers, reactors, and capacitors
2. To provide the discharge path for lightning rods, arresters, gaps, and similar devices
3. To ensure safety to operating personnel by limiting potential differences that can exist in a substation
4. To provide a means of discharging and de-energizing equipment in order to proceed with the maintenance of the equipment
5. To provide a sufficiently low-resistance path to ground to minimize rise in ground potential with respect to remote ground

A multigrounded, common neutral conductor used for a primary distribution line is always connected to the substation grounding system where the circuit originates and to all grounds along the length of the circuit. If separate primary and secondary neutral conductors are used, the conductors have to be connected together provided the primary neutral conductor is effectively grounded.

The substation grounding system is connected to every individual equipment, structure, and installation so that it can provide the means by which grounding currents are connected to remote areas. It is extremely important that the substation ground has a low ground resistance, adequate current-carrying capacity, and safety features for personnel. It is crucial to have the substation ground resistance very low so that the total rise of the ground system potential will not reach values that are unsafe for human contact.*

The substation grounding system normally is made of buried horizontal conductors and driven ground rods interconnected (by clamping, welding, or brazing) to form a continuous grid (also called mat) network. A continuous cable (usually it is 4/0 bare copper cable buried 12–18 in. below the surface) surrounds the grid perimeter to enclose as much ground as possible and to prevent current concentration and thus high gradients at the ground cable terminals. Inside the grid, cables are buried in parallel lines and with uniform spacing (e.g., about 10 × 20 ft).

All substation equipment and structures are connected to the ground grid with large conductors to minimize the grounding resistance and limit the potential between equipment and the ground surface to a safe value under all conditions. All substation fences are built inside the ground grid and attached to the grid in short intervals to protect the public and personnel. The surface of the substation is usually covered with crushed rock or concrete to reduce the potential gradient when large currents are discharged to ground and to increase the contact resistance to the feet of personnel in the substation.

IEEE Std. 80-1976 [13] provides a formula for a quick simple calculation of the grid resistance to ground after a minimum design has been completed. It is expressed as

$\begin{array}{cc}{R}_{\mathrm{grid}}=\frac{{\rho }_s}{4r}+\frac{{\rho }_s}{L_r}\hfill & (4.127)\hfill \end{array}$

where

ρs is the soil resistivity, Ω-m
L is the total length of grid conductors, m
R is the radius of circle with area equal to that of grid, m

IEEE Std. 80-2000 [19] provides the following equation to determine the grid resistance after a minimum design has been completed:

$\begin{array}{cc}{R}_{\mathrm{grid}}=\frac{{\rho }_s}{4}\sqrt{\frac{\pi }{A}}\hfill & (4.128)\hfill \end{array}$

Also, IEEE Std. 80-2000 provides the following equation to determine the upper limit for grid resistance to ground after a minimum design has been completed:

$\begin{array}{cc}{R}_{\mathrm{grid}}=\frac{{\rho }_s}{4}\sqrt{\frac{\pi }{A}}+\frac{{\rho }_s}{L_T}\hfill & (4.129)\hfill \end{array}$

where
Rgrid is the grid resistance, Ω
ρs is the soil resistance, Ω-m
A is the area of the ground, m2
LT is the total buried length of conductors, m

But, Equation 4.129 requires a uniform soil resistivity. Hence, a substantial engineering judgment is necessary for reviewing the soil resistivity measurements to decide the value of soil resistivity. However, it does provide a guideline for the uniform soil resistivity to be used in the ground grid design. Alternatively, Sverak [19] provides the following formula for the grid resistance:

$\begin{array}{cc}{R}_{\mathrm{grid}}={\rho }_s\left[\frac{1}{L_T}+\frac{1}{\sqrt{20A}}\left(1+\frac{1}{1+h\sqrt{20/A}}\right)\right]\hfill & (4.130)\hfill \end{array}$

where

Rgrid is the substation ground resistance, Ω
ρs is the soil resistivity, Ω-m
A is the area occupied by the ground grid, m2
H is the depth of the grid, m
LT is the total buried length of conductors, m

IEEE Std. 80-1976 also provides formulas to determine the effects of the grid geometry on the step and mesh voltage (which is the worst possible value of the touch voltage) in volts. Mesh voltage is the worst possible value of a touch voltage to be found within a mesh of a ground grid if standing at or near the center of the mesh. They can be expressed as

$\begin{array}{cc}{E}_{\mathrm{step}}=\frac{{\rho }_s\times K_s\times K_i\times I_G}{L_s}\hfill & (4.131)\hfill \end{array}$

and

$\begin{array}{cc}{E}_{\mathrm{mesh}}=\frac{{\rho }_s\times K_m\times K_i\times I_G}{L_m}\hfill & (4.132)\hfill \end{array}$

where

ρs is the average soil resistivity, Ω-m
Ks is the step coefficient
Km is the mesh coefficient
Ki is the irregularity coefficient
IG is the maximum rms current flowing between ground grid and earth, A
Ls is the total length of buried conductors, including cross connections, and (optionally) the total effective length of ground rods, m
Lm is the total length of buried conductors, including cross connections, and (optionally) the combined length of ground rods, m

Many utilities have computer programs for performing grounding grid studies. The number of tedious calculations that must be performed to develop an accurate and sophisticated model of a system is no longer a problem.

In general, in the event of a fault, overhead ground wires, neutral conductors, and directly buried metal pipes and cables conduct a portion of the ground fault current away from the substation ground grid and have to be taken into account when calculating the maximum value of the grid current. Based on the associated equivalent circuit and resultant current division, one can determine what portion of the total current flows into the earth and through other ground paths. It can be used to determine the approximate amount of current that did not use the ground as flow path. The faultcurrent division factor (also known as the split factor) can be expressed as

$\begin{array}{cc}{S}_{\mathrm{split}}=\frac{I_{\mathrm{grid}}}{3I_{ao}}\hfill & (4.133)\hfill \end{array}$

where

Ssplit is the fault-current division factor
Igrid is the rms symmetrical grid current, A
Iao is the zero-sequence fault current, A

The split factor is used to determine the approximate amount of current that did not use the ground flow path. Computer programs can determine the split factor easily, but it is also possible to determine the split factor through graphs. With the Y ordinate representing the split factor and the X axis representing the grid resistance, it is obvious that the grid resistance has to be known to determine the split factor.

As previously said, the split factor determines the approximate amount of current that does use the earth as return path. The amount of current that does enter the earth is found from the following equation. Hence, the design value of the maximum grid current can be found from

$\begin{array}{cc}{L}_G=D_f\times I_{\mathrm{grid}}\hfill & (4.134)\hfill \end{array}$

where

IG is the maximum grid current, A
Df is the decrement factor for the entire fault duration of tf s
Igrid is the rms symmetrical grid current, A

Here, Figure 4.48 illustrates the relationship between asymmetrical fault current, dc decaying component, and symmetrical fault current and the relationship between variables IF, If, and Df for the fault duration tf.

The decrement factor is an adjustment factor that is used in conjunction with the symmetrical ground fault-current parameter in safety-oriented grounding calculations. It determines the rms equivalent of the asymmetrical current wave for a given fault duration, accounting for the effect of initial dc offset and its attenuation during the fault. The decrement factor can be calculated from

$\begin{array}{cc}{D}_f=\sqrt{1+\frac{T_a}{I_f}(1-e^{-(2t_f/T_a)})}\hfill & (4.135)\hfill \end{array}$

where tf is the time duration of fault, s

$T_a=\frac{X}{\omega R}\text{dc offset time constant, s}$

Here, tf should be chosen as the fastest clearing time. (The fastest clearing time includes breaker and relay time for transmission substations.) It is assumed here that the ac components do not decay with time.

The symmetrical grid current is defined as that portion of the symmetrical ground fault current that flows between the grounding grid and surrounding earth. It can be expressed as

$\begin{array}{cc}{I}_{\mathrm{grid}}=S_f\times I_f\hfill & (4.136)\hfill \end{array}$

where
If is the rms value of symmetrical ground fault current, A
Sf is the fault-current division factor

IEEE Std. 80-2000 provides a series of current based on computer simulations for various values of ground grid resistance and system conditions to determine the grid current. Based on those splitcurrent curves, one can determine the maximum grid current.

4.15 Ground Conductor Sizing Factors

The flow of excessive currents will be very dangerous if the right equipments are not used to help dissipate the excessive currents. Ground conductors are means of providing a path for excessive currents from the substation to ground grid. Hence, the ground grid can then spread the current into the ground, creating a zero potential between the substation and the ground. Table 4.10 gives the list of possible conductors that can be used for such conductors.

In the United States, there are only two types of conductors that are used, namely, copper and/or copper-clad steel conductors that are used for this purpose. The copper one is mainly used due to its high conductivity and the high resistance to corrosion. The next step is to determine the size of ground conductor that needs to be buried underground.

Thus, based on the symmetrical conductor current, the required conductor size can be found from

$\begin{array}{cc}{I}_f=A_{{\mathrm{mm}}^2}{\left[\left(\frac{TCAP\times {10}^{-4}}{t_c\times {\alpha }_r\times {\rho }_r}\right)\mathrm{In}\left(\frac{K_0+T_{\max }}{K_0+T_{\mathrm{amb}}}\right)\right]}^{1/2}\hfill & (4.137)\hfill \end{array}$

if the conductor size should be in mm2, it can be found from

$\begin{array}{cc}{A}_{{\mathrm{mm}}^2}=\frac{I_f}{{\left[\left(\frac{TCAP\times {10}^{-4}}{t_c\times {\alpha }_r\times {\rho }_r}\right)\mathrm{In}\left(\frac{K_0+T_{\max }}{K_0+T_{\mathrm{amb}}}\right)\right]}^{1/2}}\hfill & (4.138)\hfill \end{array}$

Alternatively, in the event that it should be in kcmil, since

$\begin{array}{cc}{A}_{\mathrm{kcmil}}=1.974\times A_{{\mathrm{mm}}^2}\hfill & (4.139)\hfill \end{array}$

then Equation 4.130 can be expressed as

$\begin{array}{cc}{I}_f=5.07\times {10}^{-3}{A}_{\mathrm{kcmil}}{\left[\left(\frac{TCAP\times {10}^{-4}}{t_c\times {\alpha }_r\times {\rho }_r}\right)\mathrm{In}\left(\frac{K_0+T_{\max }}{K_0+T_{\mathrm{amb}}}\right)\right]}^{1/2}\hfill & (4.140)\hfill \end{array}$

Note that both ar and ρr can be found at the same reference temperature of Tr, °C. Also, note that Equations 4.137 and 4.140 can also be used to determine the short-time temperature rise in a ground conductor. Thus, taking other required conversions into account, the conductor size in kcmil can be found from

$\begin{array}{cc}{A}_{\mathrm{kcmil}}=\frac{197.4\times I_f}{{\left[\left(\frac{TCAP\times {10}^{-4}}{t_c\times {\alpha }_r\times {\rho }_r}\right)\mathrm{In}\left(\frac{K_0+T_{\max }}{K_0+T_{\mathrm{amb}}}\right)\right]}^{1/2}}\hfill & (4.141)\hfill \end{array}$

where
If is the rms current (without dc offset), kA
$A_{{\text{mm}}^2}$ is the conductor cross section, mm2
Akcmil is the conductor cross section, kcmil
TCAP is the thermal capacity per unit volume, J/(cm3.°C) (it is found from Table 4.10, per IEEE Std. 80-2000)
tc is the duration of current, s
ar is the thermal coefficient of resistivity at reference temperature Tr, 1/°C (it is found from Table 4.10, per IEEE Std. 80-2000 for 20°C)
ρr is the resistivity of the ground conductor at reference temperature Tr, μΩ-cm (it is found from Table 4.10, per IEEE Std. 80-2000 for 20°C)
K0 is 1/α0 or (1/αr) –Tr, °C
Tmax is the maximum allowable temperature, °C
Tamb is the ambient temperature, °C
If is the rms current (without dc offset), kA
$A_{{\text{mm}}^2}$ is the conductor cross section, mm2
Akcmil is the conductor cross section, kcmil

For a given conductor material, once the TCAP is found from Table 4.10 or calculated from where

$\begin{array}{cc}{\text{TCAP[J/(cm}}^3\cdot °\text{C)] =4}\text{.184 (J/cal)}\times \text{ SH[cal/(g }\cdot °\text{C)]}\times {\text{ SW(g/cm}}^3\text{)}\hfill & (4.142)\hfill \end{array}$

SH is the specific heat, in cal/(g × °C), which is related to the thermal capacity per unit volume, J/(cm3 × °C)

SW is the specific weight, in g/cm3, which is related to the thermal capacity per unit volume, J/(cm3 × °C)

Thus, TCAP is defined by

$\begin{array}{cc}{\text{TCAP[J/(cm}}^3\cdot °\text{C)] =4}\text{.184 (J/cal)}\times \text{ SH[cal/(g }\cdot °\text{C)]}\times {\text{ SW(g/cm}}^3\text{)}\hfill & (4.143)\hfill \end{array}$

Asymmetrical fault currents consist of subtransient, transient, and steady-state ac components and the dc offset current component. To find the asymmetrical fault current (i.e., if the effect of the dc offset is needed to be included in the fault current), the equivalent value of the asymmetrical current IF is found from

$\begin{array}{cc}{I}_F=D_f\times I_f\hfill & (4.144)\hfill \end{array}$

where IF is representing the rms value of an asymmetrical current integrated over the entire fault duration, tc, which can be found as a function of X/R by using Df, before using Equation 4.137 or 4.140 and where Df is the decrement factor and is found from

$\begin{array}{cc}{D}_f={\left[1+\frac{T_a}{t_f}\left({1-e}^{(-2t_f/T_a)}\right)\right]}^{1/2}\hfill & (4.145)\hfill \end{array}$

where
tf is the time duration of fault, s
Ta is the dc offset time constant, s

Note that

$\begin{array}{cc}{T}_a=\frac{X}{\omega R}\hfill & (4.146)\hfill \end{array}$

and for 60 Hz,

$\begin{array}{cc}{T}_a=\frac{X}{120\pi R}\hfill & (4.147)\hfill \end{array}$

The resulting IF is always greater than If. However, if the X/R ratio is less than 5 and the fault duration is greater than 1 s, the effects of the dc offset are negligible.

4.16 Mesh Voltage Design Calculations

If the GPR value exceeds the tolerable touch and step voltages, it is necessary to perform the mesh voltage design calculations to determine whether the design of a substation is safe. If the design is again unsafe, conductors in the form of ground rods are added to the design until the design is considered safe. The mesh voltage is the maximum touch voltage and it is found from

$\begin{array}{cc}{E}_{\mathrm{mesh}}=\frac{\rho \times K_m\times K_i\times I_G}{L_M}\hfill & (4.148)\hfill \end{array}$

where

ρ is soil resistivity, Ω-m
Km is mesh coefficient
Ki is correction factor for grid geometry
IG is maximum grid current that flows between ground grid and surrounding earth, A
Lm is length of Lc + LR for mesh voltage, m

The mesh coefficient Km is determined from

$\begin{array}{cc}{K}_m=\frac{1}{2\pi }\left[\mathrm{In}\left(\frac{D^2}{16\times h\times d}+\frac{{(D+2\times h)}^2}{8\times D\times d}-\frac{h}{4\times d}\right)+\frac{K_{ii}}{K_h}\mathrm{In}\left(\frac{8}{\pi (2\times n-1)}\right)\right]\hfill & (4.149)\hfill \end{array}$

where

d is diameter of grid conductors, m
D is spacing between parallel conductors, m
Kii is irregularity factor (corrective weighting factor that adjusts for the effects of inner conductors on the corner mesh)
Kh is corrective weighting factor that highlight for the effects of grid depth
n is geometric factor
h is depth of ground grid conductors, m

As it can be observed from Equation 4.149, the geometric factor Km has the following variables: (Ds) the spacing between the conductors, (ns) the number of conductors, (d) the diameter of the conductors used, and (h) the depth of the grid. The effect of each variable on the Km is different. Figure 4.49 shows the effect of the spacing (D) between conductors on Km. Figure 4.50 shows the effect of the number of conductors (n) on the Km. Figure 4.51 shows the relationship between the diameter of the conductor (d) and the Km. Figure 4.52 shows the relationship between the depth of the conductor (h) and Km [17].

Note that the value of Kü depends on the following circumstances:

1. For the grids with ground rods existing in grid corners as well as perimeter:

   $\begin{array}{cc}\begin{array}{c}{K}_{ii}=1\hfill \end{array}\hfill & (4.150)\hfill \end{array}$

2. For the grids with no or few ground rods with none existing in corners or perimeter:

$\begin{array}{cc}{K}_{ii}=\frac{1}{{(2n)}^{\frac{2}{n}}}\hfill & (4.151)\hfill \end{array}$

and

$\begin{array}{cc}{K}_h=\sqrt{1+\frac{h}{h_0}}\hfill & (4.152)\hfill \end{array}$

where h0 is grid reference depth = 1 m.

The effective number of parallel conductors (n) given in a given grid are found from

$\begin{array}{cc}n=n_a\times n_b\times n_c\times n_d\hfill & (4.153)\hfill \end{array}$

where

$n_a=\frac{2L_c}{L_p}$

nb is 1, for square grids
nc is 1, for square and rectangular grids
nd is 1, for square, rectangular, and L-shaped grids

Otherwise, the following equations are used to determine the nb, nc, and nd so that

$\begin{array}{cc}{n}_b=\left[\frac{L_p}{4\sqrt{A}}\right]\hfill & (4.154)\hfill \end{array}$

$\begin{array}{cc}{n}_c={\left[\frac{L_X\times L_Y}{A}\right]}^{\frac{0.7A}{L_X\times L_Y}}\hfill & (4.155)\hfill \end{array}$

$\begin{array}{cc}{n}_d=\frac{D_m}{\sqrt{L_X^2+L_Y^2}}\hfill & (4.156)\hfill \end{array}$

where

Lp is the peripheral length of the grid, m
A is the area of the grid, m2
LX is the maximum length of the grid in the x direction, m
LY is the maximum length of the grid in the y direction, m
Dm is the maximum distance between any two points on the grid, m

Note that the irregularity factor is determined from

$\begin{array}{cc}{K}_{ii}=0.644+0.148n\hfill & (4.157)\hfill \end{array}$

The effective buried length (LM) for grids is as follows:

1. With little or no ground rods but none located in the corners or along the perimeter of the grid:

   $L_M\begin{array}{cc}=L_C+L_R\hfill & (4.158)\hfill \end{array}$

   where
   LR is the total length of all ground rods, m
   LC is the total length of the conductor in the horizontal grid, m

2. With ground rods in corners and along the perimeter and throughout the grid:

   $L_M\begin{array}{cc}=L_C+\left[1.55+1.22\left(\frac{L_R}{L_X^2+L_Y^2}\right)\right]L_R\hfill & (4.159)\hfill \end{array}$

   where LR is the length of each ground rod, m.

4.17 Step Voltage Design Calculations

According to IEEE Std. 80-2000, in order for the ground system to be safe, step voltage has to be less than the tolerable step voltage. Furthermore, step voltages within the grid system designed for safe mesh voltages will be well within the tolerable limits, the reason for this is that both feet and legs are in series rather than in parallel and the current takes the path from one leg to the other rather than through vital organs. The step voltage is determined from

$E_{\mathrm{step}}\begin{array}{cc}=\frac{\rho \times K_s\times K_i\times I_G}{L_S}\hfill & (4.160)\hfill \end{array}$

where

Ks is the step coefficient
Ls is the buried conductor length, m

Again, for grids with or without ground rods,

$L_S\begin{array}{cc}=0.75L_C+0.85L_R\hfill & (4.161)\hfill \end{array}$

so that the step coefficient can be found from

$K_S\begin{array}{cc}=\frac{1}{\pi }\left[\frac{1}{2h}+\frac{2}{D+h}+\frac{1}{D}(1-{0.5}^{n-2})\right]\hfill & (4.162)\hfill \end{array}$

where h is the depth of ground grid conductors in meters, usually in the range 0.25 m < h < 2.5 m.

As shown in Equation 4.162, the geometric factor Ks is a function of D, n, d, π, and h. Figure 4.53 shows the relationship between the distance (D) between the conductors and the geometric factor Ks. Figure 4.54 shows the relationship between the number of conductors (n) and the geometric factor Ks. Figure 4.55 shows the relationship between the depth of grid conductors (D) in meter and the geometric factor Ks.

4.18 Types of Ground Faults

In general, it is difficult to determine which fault type and location will result in the greatest flow of current between the ground grid and surrounding earth because no simple rule applies. IEEE Std. 80-2000 recommends to not consider multiple simultaneous faults since their probability of occurrence is negligibly small. Instead, it recommends investigating single line-to-ground and lineto-line-to-ground faults.

4.18.1 Line-to-Line-to-Ground Fault

For a line-to-line-to-ground (i.e., double line-to-ground) fault, IEEE Std. 80-2000 gives the following equation to calculate the zero-sequence fault current:

$I_{a0}\begin{array}{cc}=\frac{E(R_2+j{X}_2)}{(R_1+j{X}_1)\left[R_0+R_2+3R_f+j(X_0+X_2)\right]+(R_2+j{X}_2)(R_0+3R_f+j{X}_0)}\hfill & (4.163)\hfill \end{array}$

where
Ia0 is the symmetrical rms value of zero-sequence fault current, A
E is the phase-to-neutral voltage, V
Rf is the estimated resistance of the fault, Ω (normally it is assumed Rf = 0)
R1 is the positive-sequence system resistance, Ω
R2 is the negative-sequence system resistance, Ω
R0 is the zero-sequence system resistance, Ω
X1 is the positive-sequence system reactance (subtransient), Ω
X2 is the negative-sequence system reactance, Ω
X0 is the zero-sequence system reactance, Ω

The values of R0, R1, R2, and X0, X1, X2 are determined by looking into the system from the point of fault. In other words, they are determined from the Thévenin equivalent impedance at the fault point for each sequence.* Often, however, the resistance quantities given in the earlier equation is negligibly small. Hence,

$I_{a0}\begin{array}{cc}=\frac{E\times X_2}{X_1(X_0+X_2)(X_0+X_2)}\hfill & (4.164)\hfill \end{array}$

4.18.2 Single Line-to-Ground Fault

For a single line-to-ground fault, IEEE Std. 80-2000 gives the following equation to calculate the zero-sequence fault current:

$I_{a0}\begin{array}{cc}=\frac{E}{3R_f+R_0+R_1+R_2+j(X_0+X_1+X_2)}\hfill & (4.165)\hfill \end{array}$

Often, however, the resistance quantities in the earlier equation are negligibly small. Hence,

$I_{a0}\begin{array}{cc}=\frac{E}{X_0+X_1+X_2}\hfill & (4.166)\hfill \end{array}$

4.19 Ground Potential Rise

As said before in Section 4.8.2, the GPR is a function of fault-current magnitude, system voltage, and ground system resistance. The GPR with respect to remote ground is determined by multiplying the current flowing through the ground system by its resistance measured from a point remote from the substation. Here, the current flowing through the grid is usually taken as the maximum available line-to-ground fault current.

GPR is a function of fault-current magnitude, system voltage, and ground (system) resistance. The current through the ground system multiplied by its resistance measured from a point remote from the substation determines the GPR with respect to remote ground. Hence, GPR can be found from

$V_{\mathrm{GPR}}\begin{array}{cc}=I_G\times R_g\hfill & (4.167)\hfill \end{array}$

where

VGPR is the GPR, V
Rg is the ground grid resistance, Ω

For example, if a ground fault current of 20,000 A is flowing into a substation ground grid due to a line-to-ground fault and the ground grid system has a 0.5 Ω resistance to the earth, the resultant IR voltage drop would be 10,000 V. It is clear that such 10,000 V IR voltage drop could cause serious problems to communication lines in and around the substation in the event that the communication equipment and facilities are not properly insulated and/or neutralized.

The ground grid resistance can be found from

$\begin{array}{cc}{R}_g=\rho \left[\frac{1}{L_T}+\frac{1}{\sqrt{20A}}\left(1+\frac{1}{1+h\sqrt{20/A}}\right)\right]\hfill & (4.168)\hfill \end{array}$

where

LT is the total buried length of conductors, m
h is the depth of the grid, m
A is the area of substation ground surface, m2

Figure 4.56 shows the effects of the number of grid conductors (n), without ground rods, on the ground grid resistance. It shows that the area (A) has a substantial influence on the grid resistance. Figure 4.57 shows the relationship between the burial depth of the grid (h), in meter, and the grid resistance. Here, the depth is varied from 0.5 to 2.5 m and the number of conductors from 4 to 10 [17].

In order to aid the substation grounding design engineer, the IEEE Standard 80-2000 includes a design procedure that has a 12-step process, as shown in Figure 4.58, in terms of substation grounding design procedure block diagram, based on a preliminary of a somewhat arbitrary area, that is, the standard suggests the grid be approximately the size of the distribution substation. But, some references state a common practice that is to extend the grid three meters beyond the perimeter of the substation fence.

Example 4.15

Let the starting grid be an 84.5 m by 49.6 m ground grid. Design a proper substation grounding to provide safety measures for anyone going near or working on a substation. Hence, use the IEEE 12-step process shown in Figure 4.58, then build a grid large enough to dissipate the ground fault current into the earth. (A large grounding grid extending far beyond the substation fence and made of a single copper plate would have the most desirable effect for dispersing fault currents to remote earth and thereby ensure the safety of personnel at the surface. Unfortunately, a copper plate of such size is not an economically viable option.)

A grounding system is considered for this three-phase 230 kV system that feeds two step-down transformers that step down the voltage from 230 to 69 kV. The two transformers are connected in parallel with respect to each other. One of the transformers feeds a switchyard. The other one is connected to a transformer bank (which has three single-phase 4 MVA transformers) that steps down the 69 to 13.8 kV and feeds an industrial facility.

One alternative is to design a grid by using a series of horizontal conductors and vertical ground rods. Of course, the application of conductors and rods depends on the resistivity of the substation ground. Change the variables as necessary in order to meet specifications for grounding of the substation. The variables include the size of the grid, the size of the conductors used, the amount of conductors used, and the spacing of each grounding rod. Use 17,000 A as the maximum value fault current, a maximum clearing time of 1 s, and a conductor diameter of 210.5 kcmil, based on the given information. The soil resistivity is 50 Ω-m and the crushed rock resistivity on the surface of the substation is 2500 Ω-m. Assume that the incoming transmission line into substation has no shield wires and but there are four distribution neutrals. Design a grounding grid system by using a series of horizontal conductors and vertical ground rods, based on the resistivity of the soil.

Solution

Step 1: Field data

Assume that a uniform average soil resistivity of the substation ground is measured to be 50 Ω-m. The initial design parameters are given in Table 4.11.

Step 2: Conductor size

The analysis of the grounding grid should be based on the most conservative fault conditions. For example, the fault current 3Iao is assumed maximum value, with all current dispersed through the grid (i.e., there is no alternative path for ground other than through the grid to remote earth). As said before, the maximum value of the fault current is given as 17,000 A; thus, the conductor size is selected based on this current and the duration of the fault. Thus, use 17,000 A as the maximum value fault current, a maximum clearing time of 0.5 s, and a conductor diameter of 210.5 kcmil, which is determined from the following calculation:

$\begin{array}{cccc}{A}_{\mathrm{kcmil}}\hfill & =\hfill & I\times K\times \sqrt{t_c}\hfill & \hfill \\ \hfill & =\hfill & 17\times 10.45\times \sqrt{1}\hfill & \hfill \\ \hfill & =\hfill & 210.5\mathrm{kcmil}\hfill & (4.169)\hfill \end{array}$

However, the conductor selected is 500 kcmil. This is based on the given guidelines so that the size is more than enough to handle the fault current. The diameter of the conductor can be found from Table A.1. Based on the selected conductor, the diameter (d) of the conductor is 0.018 m. The crushed rock resistivity is 2500 Ω-m. Surface derating factor is 0.714.

Step 3: Touch and step voltage criteria

In order to move to the third step in the design process, it is first needed to determine the surface layer derating factor Cs as

$\begin{array}{ccc}{C}_s\hfill & =\hfill & 1-\frac{0.09(1-(\rho /{\rho }_s))}{2h_s+0.09}\hfill \\ \hfill & =\hfill & 1-\frac{0.09(1-50/2500))}{2\times 0.1524+0.09}\hfill \\ \hfill & =\hfill & 0.78\hfill \end{array}$

According to the federal law, all known hazards must be eliminated when GPR takes place for the safety of workers at a work site. In order to remove the hazards associated with GPR, a grounding grid is designed to reduce the hazardous potentials at the surface. First, it is necessary to determine what was not hazardous to the body. For two body types, the potential safe touch and step voltages a human could withstand before the fault is cleared need to be determined from Equations 4.120d and 4.120b, respectively, as

$\begin{array}{ccc}{V}_{\mathrm{touch}70}\hfill & =\hfill & (1000+1.5C_s\times {\rho }_s)\frac{0.157}{\sqrt{t_s}}\hfill \\ \hfill & =\hfill & (1000+1.5\times {0.78}_s\times 2500)\frac{0.157}{\sqrt{0.5}}\hfill \\ \hfill & =\hfill & 871.5\mathrm{V}\hfill \end{array}$

and

$\begin{array}{ccc}{V}_{\mathrm{step}70}\hfill & =\hfill & (1000+6C_s\times {\rho }_s)\frac{0.157}{\sqrt{t_s}}\hfill \\ \hfill & =\hfill & (1000+1.6\times 0.78\times 2500)\frac{0.157}{\sqrt{0.5}}\hfill \\ \hfill & =\hfill & 2819.5\mathrm{V}\hfill \end{array}$

Step 4: Initial design

Step 4 deals with the layout of the grounding conductors and the amount of conductors being used for the design. The initial design consists of factors obtained from the general knowledge of the substation. The preliminary size of the grounding grid system is largely based on the size of the substation to include all dimensions within the perimeter of the fence. To establish economic viability, the maximum area is considered and formed the shape of a square with an area of 4204.6 m2. The spacing of conductors (D) is selected as 4.97 m.

The maximum lengths (Lx) of the conductor in the × direction and the y direction (Ly) are determined to be 84.6 m and 49.6 m, respectively. Based on the information given in this section, the total length of the grounding conductor is 1825 m. The length of the ground rods (Lr) is 3.048 m. A total of 25 ground rods are used, which gives the total length of the ground rods (LR) to be approximately 76 m. Thus, the total conductor length that includes the conductor plus the ground rods is 1901 m.

The depth of the ground grid (h) is determined as 1.525 m below the surface. The next step is to take into account the geometry of the ground grid. Given the length of each side of the grounding grid, it is determined that the shape of the grid design will be a rectangle. The geometric factor can be calculated by determining na, nb, nc, and nd as

$\begin{array}{ccc}{n}_a\hfill & =\hfill & \frac{2\times L_c}{L_p}\hfill \\ \hfill & =\hfill & \frac{2\times 1825}{268}\hfill \\ \hfill & =\hfill & 14\hfill \end{array}$

and

$\begin{array}{ccc}{n}_b\hfill & =\hfill & \sqrt{\frac{L_p}{4\times \sqrt{A}}}\hfill \\ \hfill & =\hfill & \sqrt{\frac{268}{4\sqrt{4204.6}}}\hfill \\ \hfill & =\hfill & 1.03\hfill \end{array}$

and

$\begin{array}{ccc}{n}_c\hfill & =\hfill & {\left[\frac{L_X\times L_Y}{A}\right]}^{\frac{0.7\times A}{L_X\times L_Y}}\hfill \\ \hfill & =\hfill & {\left[\frac{84.6\times 49.6}{4204.6}\right]}^{\frac{0.7\times 4204.6}{84.6\times 49.6}}\hfill \\ \hfill & =\hfill & 1.00\hfill \end{array}$

and

$\begin{array}{ccc}{n}_d\hfill & =\hfill & \frac{D_m}{\sqrt{L_X^2+L_Y^2}}\hfill \\ \hfill & =\hfill & \frac{98.1}{\sqrt{{84.6}^2+{49.6}^2}}\hfill \\ \hfill & =\hfill & 1.00\hfill \end{array}$

The geometric factor is then calculated to be

$\begin{array}{ccc}n\hfill & =\hfill & n_a\times n_b\times n_c\times n_d\hfill \\ \hfill & =\hfill & 14\times 1.03\times 1\times 1\hfill \\ \hfill & =\hfill & 14\hfill \end{array}$

Thus, they are approximately equal to 1 due to the shape of the grid.

Step 5: Grid resistance

A good grounding system provides a low resistance to remote earth in order to minimize the GPR. The next step is to evaluate the grid resistance by using Equation 4.103. All design parameters can be found in Table 4.11. Table 4.12 gives the approximate equivalent impedance of transmission line overhead shield wires and distribution feeder neutrals, according to their numbers. From Equation 4.168 for LT = 1901 m, a grid area of A = 4205 m2, ρ = 50 Ω-m, and h = 1.524 m, the grid resistance is

$\begin{array}{ccc}{R}_g\hfill & =\hfill & \rho \left[\frac{1}{L_T}+\frac{1}{\sqrt{20\times A}}\left(1+\frac{1}{1+h\sqrt{20/A}}\right)\right]\hfill \\ \hfill & =\hfill & 50\left[\frac{1}{1901}+\frac{1}{\sqrt{20\times 4205}}\left(1+\frac{1}{1+1.524\sqrt{20/4205}}\right)\right]\hfill \\ \hfill & =\hfill & 0.35\mathrm{\Omega }\hfill \end{array}$

Step 6: Grid current

In step 6 of the logic flow diagram of the IEEE Std. 80-2000, the amount of current that flows within the designed grid (IG) is determined. The GPR is determined as

$V_{\mathrm{GPR}}=I_G\times R_g$

Determining the GPR and comparing it to the tolerable touch voltage is the first step to find out whether the grid design is a safe design for the people in and around the substation. The next step is to find the grid current IG. But, it is first needed to determine the split factor from the following equation:

$\begin{array}{cc}{S}_f=\left|\frac{Z_{\mathit{\text{eq}}}}{Z_{\mathit{\text{eq}}}+R_g}\right|\hfill & (4.170)\hfill \end{array}$

Since the substation has no impedance line shield wires and four distribution neutrals, from Table 4.12, the equivalent impedance can be found as Zeq = 0.322 + j0.242 Ω. Thus, Rg = 1.0043 Ω and a total fault current of 3Ia0 = 17,000 A, a decrement factor of Df = 1.026. Thus, the current division factor (or the split factor) can be found as

$\begin{array}{ccc}{S}_f\hfill & =\hfill & \left|\frac{Z_{\mathit{\text{eq}}}}{Z_{\mathit{\text{eq}}}+R_g}\right|\hfill \\ \hfill & =\hfill & \left|\frac{(0.322+j0.242)}{(0.322+j0.242)+1.0043}\right|\hfill \\ \hfill & \cong \hfill & 0.2548\hfill \end{array}$

since

$\begin{array}{ccc}{I}_g\hfill & =\hfill & S_f\times 3I_{\mathrm{a}0}\hfill \\ \hfill & =\hfill & 0.2548\times 17,000\hfill \\ \hfill & =\hfill & 4,331.6\mathrm{A}\hfill \end{array}$

thus,

$\begin{array}{ccc}{I}_G\hfill & =\hfill & D_f\times I_g\hfill \\ \hfill & =\hfill & 1.026\times 4,331.6\hfill \\ \hfill & =\hfill & 4,444.2\mathrm{A}\hfill \end{array}$

Step 7: Determination of GPR

As said before, the product of IG and Rg is the GPR. It is necessary to compare the GPR to the tolerable touch voltage, Vtouch70. If the GPR is larger than the Vtouch70, further design evaluations are necessary and the tolerable touch and step voltages should be compared to the maximum mesh and step voltages. Hence, first determine the GPR as

$\begin{array}{ccc}\mathrm{GPR}\hfill & =\hfill & I_G\times R_g\hfill \\ \hfill & =\hfill & 4,444.2\times 0.35\hfill \\ \hfill & =\hfill & 1555.48\mathrm{V}\hfill \end{array}$

Check to see whether

$\mathrm{GPR}>V_{\mathrm{touch}70}$

Indeed,

$1555.2\text{V}>871.5V$

As it can be observed from the results, the GPR is much larger than the step voltage. Therefore, further design considerations are necessary and thus the step and mesh voltages must be calculated and compared to the tolerable touch and step voltage as follows.

Step 8: Mesh and step voltage calculations

1. Determination of the mesh voltage
   In order to calculate the mesh voltage by using Equation 4.148, it is necessary first to calculate the variables Kh, Km, and Kii. Here, the correction factor that accounts for the depth of the grid (Kh) can be determined from Equation 4.152 as

   $\begin{array}{ccc}{K}_h\hfill & =\hfill & \sqrt{1+\frac{h}{h_0}}\hfill \\ \hfill & =\hfill & \sqrt{1+\frac{1.524}{1}}\hfill \\ \hfill & =\hfill & 1.59\hfill \end{array}$

   The corrective factor for grid geometry (Kii) can be calculated from Equation 4.157 as

   $\begin{array}{ccc}{K}_{ii}\hfill & =\hfill & 0.644+0.148\times n\hfill \\ \hfill & =\hfill & 0.644+0.148\times 14\hfill \\ \hfill & =\hfill & 2.716\hfill \end{array}$

2. Comparison of mesh voltage and allowable touch voltage
   Using Kh, and Kii the spacing factor for mesh voltage (Km) can be calculated. Here, the corrective weighting factor that can be used to adjust conductors on the corner mesh (Kii) is considered to be 1.0 due to the shape of the grid being rectangular. Hence,

   $\begin{array}{ccc}{K}_m\hfill & =\hfill & \frac{1}{2\pi }\left[\mathrm{In}\left(\frac{D^2}{16\times h\times d}+\frac{{(D+2\times h)}^2}{8\times D\times d}-\frac{h}{4\times d}\right)+\frac{K_{ii}}{K_h}\mathrm{In}\left(\frac{8}{\pi (2\times 14-1)}\right)\right]\hfill \\ \hfill & =\hfill & \frac{1}{2\pi }\left[\mathrm{In}\left(\frac{{4.97}^2}{16\times 1.524\times 0.018}+\frac{{(4.97+2\times 1.524)}^2}{8\times 4.97\times 0.018}-\frac{1.524}{4\times 0.018}\right)+\frac{1}{1.589}\mathrm{In}\left(\frac{8}{\pi (2\times 14-1)}\right)\right]\hfill \\ \hfill & =\hfill & 0.53\hfill \end{array}$

   Thus, the mesh voltage can now be calculated as

   $\begin{array}{ccc}{E}_m\hfill & =\hfill & \frac{\rho \times I_G\times K_m\times K_{ii}}{L_C+\left[1.55+1.22\left(L_r/\sqrt{L_x^2+L_y^2}\right)\right]L_R}\hfill \\ \hfill & =\hfill & \frac{50\times 4,444.2\times 0.53\times 2.716}{1825+\left[1.55+1.22\left(3.048/\sqrt{{84.6}^2+{49.6}^2}\right)\right]76.2}\hfill \\ \hfill & =\hfill & 164.39\mathrm{V}\hfill \end{array}$

3. Determination of the step voltage
   In order for the ground to be safe, step voltage has to be less than the tolerable step voltages. Also, step voltages within a grid system designed for mesh voltages will be well within the tolerable limits. To determine the step voltage (Estep), unknown variables of Ks and Ls are to be calculated. Thus, the spacing factor for step voltage (Ks) can be found from

   $\begin{array}{ccc}{K}_s\hfill & =\hfill & \frac{1}{\pi }\left[\frac{1}{2\times h}+\frac{1}{D+h}+\frac{1}{D}(1-{0.5}^{n-2})\right]\hfill \\ \hfill & =\hfill & \frac{1}{\pi }\left[\frac{1}{2\times 1.524}+\frac{1}{3.97+1.524}+\frac{1}{4.97}(1-{0.5}^{14-2})\right]\hfill \\ \hfill & =\hfill & 0.22\hfill \end{array}$

   The effective length (Ls) for the step voltage is

   $\begin{array}{ccc}{L}_s\hfill & =\hfill & 0.75\times L_c+0.85\times L_R\hfill \\ \hfill & =\hfill & 0.75\times 1825+0.85\times 76\hfill \\ \hfill & =\hfill & 1433.5\mathrm{m}\hfill \end{array}$

   Thus, the step voltage (Estep) determined as

   $\begin{array}{ccc}{E}_{\mathrm{step}}\hfill & =\hfill & \frac{\rho \times I_G\times K_s\times K_{ii}}{L_s}\hfill \\ \hfill & =\hfill & \frac{50\times 4,444.2\times {0.22}_s\times 2.716}{1433.5}\hfill \\ \hfill & =\hfill & 92.62\mathrm{V}\hfill \end{array}$

Step 9: Comparison of Emesh versus Vtouch

Here, the mesh voltage that is calculated in step 8 is compared with the tolerable touch voltages calculated in step 4. If the calculated mesh voltage Emesh is greater than the tolerable Vtouch70, further design evaluations are necessary. If the mesh voltage Emesh is smaller than the Vtouch70, then it can be moved to the next step and compare Estep with Vstep70. Accordingly,

$\begin{array}{c}{E}_{\mathrm{mesh}}<V_{\mathrm{touch}70}(?)\hfill \\ 164.39\mathrm{V}<871.5\mathrm{V}\hfill \end{array}$

Here, the original grid design passes the second critical criteria in step 9. Hence, it can be moved to step 10 to find out whether the final criterion is met.

Step 10: Comparison of Estep versus Vstep70

This is the final step that the design has to meet before the grounding system is considered safe. At this step, Estep is compared with the calculated tolerable step voltage Vstep70. If

$E_{\mathrm{step}}>V_{\mathrm{step}70}(?)$

A refinement of the preliminary design is necessary and can be accomplished by decreasing the total grid resistance, closer grid spacing, adding more ground grid rods, if possible, and/or limiting the total fault current.

On the other hand, if

$E_{\mathrm{step}}<V_{\mathrm{step}70}(?)$

then the designed grounding grid is considerably safe. Since here,

$92.62\mathrm{V}<2819.8\mathrm{V}$

then for the design,

$E_{\mathrm{step}}<V_{\mathrm{step}70}$

In summary, according to the calculations, the calculated mesh and step voltages are smaller than the tolerable touch and step voltages; therefore, in a typical shock situation, humans (weighting 70 kg) that become part of the circuit during a fault will have only what is considered a safe amount of current passing through their bodies.

There are many variables that can be changed in order to meet specifications for grounding a substation. Some variables include the size of the grid, the size of the conductors used, the amount of conductors used, and the spacing of each ground rod. After many processes an engineer has to go through, the project would then be put into construction if it is approved. Designing safe substation grounding is obviously not an easy task, but there are certain procedures that an engineer can follow to make the designing of substation grounding easier.

4.20 Transmission Line Grounds

High-voltage transmission lines are designed and built to withstand the effects of lightning with a minimum damage and interruption of operation. If the lightning strikes an overhead ground wire (also called static wire) on a transmission line, the lightning current is conducted to ground through the ground wire installed along the pole or through the metal tower. The top of the line structure is raised in potential to a value determined by the magnitude of the lightning current and the surge impedance of the ground connection.

In the event that the impulse resistance of the ground connection is large, this potential can be in the magnitude of thousands of volts. If the potential is greater than the insulation level of the apparatus, a flashover will take place, causing an arc. The arc, in turn, will start the operation of protective relays, causing the line to be taken out of service. In the event that the transmission structure is well grounded and there is a sufficient coordination between the conductor insulation and the ground resistance, flashover can generally be avoided.

The transmission line grounds can be in various ways to achieve a low ground resistance. For example, a pole butt grounding plate or butt coil can be employed on wood poles. A butt coil is a spiral coil of bare copper wire installed at the bottom of a pole. The wire of the coil is extended up the pole as the ground wire lead. In practice, usually one or more ground rods are employed instead to achieve the required low ground resistance.

The sizes of the rods used are usually$\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$ or$\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ in. in diameter and 10 ft in length. The thickness of the rod does not play a major role in reducing the ground resistance as does the length of the rod. Multiple rods are usually used to provide the low ground resistance required by the high-capacity structures. But if the rods are moderately close to each other, the overall resistance will be more than if the same number of rods were spaced far apart. In other words, adding a second rod does not provide a total resistance of half that of a single rod unless the two are several rod lengths apart (actually infinite distance). Lewis [30] has shown that at 2 ft apart the resistance of two pipes (used as ground rods) in parallel is about 61% of the resistance of one of them, and at 6 ft apart it is about 55% of the resistance of one pipe.

Where there is bedrock near the surface or where sand is encountered, the soil is usually very dry and therefore has high resistivity. Such situations may require a grounding system known as the counterpoise, made of buried metal (usually galvanized steel wire) strips, wires, or cables. The counterpoise for an overhead transmission line consists of a special grounding terminal that reduces the surge impedance of the ground connection and increases the coupling between the ground wire and the conductors.

The basic types of counterpoises used for transmission lines located in areas with sandy soil or rock close to the surface are the continuous type (also called the parallel type) and the radial (also called the crowfoot type), as shown in Figure 4.59. The continuous counterpoise is made of one or more conductors buried under the transmission line for its entire length.

The counterpoise wires are connected to the overhead ground (or static) wire at all towers or poles. But, the radial-type counterpoise is made of a number of wires and extends radially (in some fashion) from the tower legs. The number and length of the wires are determined by the tower location and the soil conditions. The counterpoise wires are usually installed with a cable plow at a length of 18 in. or more so that they will not be disturbed by cultivation of the land.

A multigrounded, common neutral conductor used for a primary distribution line is always connected to the substation grounding system where the circuit originates and to all grounds along the length of the circuit. If separate primary and secondary neutral conductors are used, the conductors have to be connected together provided that the primary neutral conductor is effectively grounded. The resistance of a single buried horizontal wire, when it is used as radial counterpoise, can be expressed as [16]

$\begin{array}{cc}R=\frac{\rho }{\pi \mathcal{l}}\left(\mathrm{In}\frac{2\mathcal{l}}{2{(ad)}^{1/2}}-1\right)\mathrm{when}d\ll \mathcal{l}\hfill & \left(4.171\right)\hfill \end{array}$

where

ρ is the ground resistivity, Ω-m
ℓ is the length of wire, m
a is the radius of wire, m
d is the burial depth, m

It is assumed that the potential is uniform over the entire length of the wire. This is only true when the wire has ideal conductivity. If the wire is very long, such as with the radial counterpoise, the potential is not uniform over the entire length of the wire. Hence, Equation 4.141 cannot be used. Instead, the resistance of such a continuous counterpoise whenℓ(r/ρ)1/2 is large can be expressed as

$\begin{array}{cc}R={(r\rho )}^{1/2}\coth \left[\mathcal{l}{\left(\frac{r}{\rho }\right)}^{1/2}\right]\hfill & (4.172)\hfill \end{array}$

where r is the resistance of wire, Ω-m.

If the lightning current flows through a counterpoise, the effective resistance is equal to the surge impedance of the wire. The wire resistance decreases as the surge propagates along the wire. For a given length counterpoise, the transient resistance will diminish to the steady-state resistance if the same wire is used in several shorter radial counterpoises rather than as a continuous counterpoise. Thus, the first 250 ft of counterpoise is most effective when it comes to grounding of lightning currents.

4.21 Types of Grounding

In general, transmission and subtransmission systems are solidly grounded. Transmission systems are usually connected to a grounded wye, but subtransmission systems are often connected in delta. Delta systems may also be grounded through grounding transformers. In most high-voltage systems, the neutrals are solidly grounded, that is, connected directly to the ground. The advantages of such grounding are as follows:

1. Voltages to ground are limited to the phase voltage.
2. Intermittent ground faults and high voltages due to arcing faults are eliminated.
3. Sensitive protective relays operated by ground fault currents clear these faults at an early stage.

The grounding transformers used are normally either small distribution transformers (are connected normally in wye–delta, having their secondaries in delta) or small grounding autotransformers with interconnected wye or zigzag windings, as shown in Figure 4.60. The three-phase autotransformer has a single winding. If there is a ground fault on any line, the ground current flows equally in the three legs of the autotransformer. The interconnection offers the minimum impedance to the flow of the single-phase fault current.

The transformers are only used for grounding and carry little current except during a ground fault. Because of that, they can be fairly small. Their ratings are based on the stipulation that they carry current for no more than 5 min since the relays normally operate long before that. The grounding transformers are connected to the substation ground.

All substation equipment and structures are connected to the ground grid with large conductors to minimize the grounding resistance and limit the potential between equipment and the ground surface to a safe value under all conditions. All substation fences are built inside the ground grid and attached to the grid at short intervals to protect the public and personnel. Furthermore, the surface of the substation is usually covered with crushed rock or concrete to reduce the potential gradient when large currents are discharged to ground and to increase the contact resistance to the feet of personnel in the substation.

As said before, the substation grounding system is connected to every individual equipment, structure, and installation in order to provide the means by which grounding currents are conducted to remote areas. Thus, it is extremely important that the substation ground has a low ground resistance, adequate current-carrying capacity, and safety features for personnel.

It is crucial to have the substation ground resistance very low so that the total rise of the grounding system potential will not reach values that are unsafe for human contact. Therefore, the substation grounding system normally is made up of buried horizontal conductors and driven ground rods interconnected (by clamping, welding, or brazing) to form a continuous grid (also called mat) network.

Notice that a continuous cable (usually it is 4/0 bare stranded copper cable buried 12–18 in. below the surface) surrounds the grid perimeter to enclose as much ground as possible and to prevent current concentration and thus high gradients at the ground cable terminals. Inside the grid, cables are buried in parallel lines and with uniform spacing (e.g., about 10 × 20 ft).

Today, many utilities have computer programs for performing grounding grid studies. Thus, the number of tedious calculations that must be performed to develop an accurate and sophisticated model of a system is no longer a problem.

The GPR depends on grid burial depth, diameter, and length of conductors used, spacing between each conductor, fault-current magnitude, system voltage, ground system resistance, soil resistivity, distribution of current throughout the grid, proximity of the fault electrodes, and the system grounding electrodes to the conductors. IEEE Std. 80-1976 [14] provides a formula for a quick simple calculation of the grid resistance to ground after a minimum design has been completed. It is expressed as

$\begin{array}{cc}R=\frac{\rho }{4r}+\frac{\rho }{L}\mathrm{\Omega }\hfill & (4.173)\hfill \end{array}$

where

ρ is the soil resistivity, Ω-m
L is the total length of grid conductors, m
R is the radius of circle with area equal to that of grid, m

IEEE Std. 80-1976 also provides formulas to determine the effects of the grid geometry on the step and mesh voltage (which is the worst possible value of the touch voltage) in volts. They can be expressed as

$\begin{array}{cc}{V}_{\mathrm{step}}=\frac{K_s{K}_i\rho I_G}{L}\hfill & (4.174)\hfill \end{array}$

and

$\begin{array}{cc}{V}_{\mathrm{mesh}}=\frac{K_m{K}_i\rho I_G}{L}\hfill & (4.175)\hfill \end{array}$

where

Ks is the step coefficient
Km is the mesh coefficient
Ki is the irregularity coefficient

Many utilities have computer programs for performing grounding grid studies. The number of tedious calculations that must be performed to develop an accurate and sophisticated model of a system is no longer a problem.

4.22 Transformer Classifications

In power system applications, the single- or three-phase transformers with ratings up to 500 kVA and 34.5 kV are defined as distribution transformers, whereas those transformers with ratings over 500 kVA at voltage levels above 34.5 kV are defined as power transformers. Most distribution and power transformers are immersed in a tank of oil for better insulation and cooling purposes.

Today, various methods are in use in power transformers to get the heat pot of the tank more effectively. Historically, as the transformer sizes increased, the losses outgrew any means of selfcooling that was available at the time; thus, a water-cooling method was put into practice. This was done by placing metal coil tubing in the top oil, around the inside of the tank. Water was pumped through this cooling coil to get rid of the heat from oil.

Another method was circulating the hot oil through an external oil-to-water heat exchanger. This method is called forced-oil-to-water cooling (FOW). Today, the most common of these forced-oilcooled power transformers uses an external bank of oil-to-air heat exchangers through which the oil is continuously pumped. It is known as type FOA.

In present practice, fans are automatically used for the first stage and pumps for the second, in triple-rated transformers that are designated as type OA/FA/FOA. These transformers carry up to about 60% of maximum nameplate rating (i.e., FOA rating) by natural circulation of the oil (OA) and 80% of maximum nameplate rating by forced cooling that consists of fans on the radiators (FA). Finally, at maximum nameplate rating (FOA), not only is oil forced to circulate through external radiators, but fans are also kept on to blow air onto the radiators as well as into the tank itself. In summary, the power transformer classes are as follows:

OA: Oil-immersed, self-cooled
OW: Oil-immersed, water-cooled
OA/FA: Oil-immersed, self-cooled/forced-air-cooled
OA/FA/FOA: Oil-immersed, self-cooled/forced-air-cooled/forced-oil-cooled
FOA: Oil-immersed, forced-oil-cooled with forced-air cooler
FOW: Oil-immersed, forced-oil-cooled with water cooler

In a distribution substation, power transformers are used to provide the conversion from subtransmission circuits to the distribution level. Most are connected in delta–wye grounded to provide ground source for the distribution neutral and to isolate the distribution grounding system from the subtransmission system.

Substation transformers can range from 5 MVA in smaller rural substations to over 80 MVA at urban stations (in terms of base ratings). As said earlier, power transformers have multiple ratings, depending on cooling methods. The base rating is the self-cooled rating, just due to the natural flow to the surrounding air through radiators. The transformer can supply more load with extra cooling turned on, as explained before.

However, the ANSI ratings were revised in the year 2000 to make them more consistent with IEC designations. This system has a four-letter code that indicates the cooling (IEEE C57.12.00-2000):

First letter—Internal cooling medium in contact with the windings:

O: Mineral oil or synthetic insulating liquid with fire point = 300°C
K: Insulating liquid with fire point >300°C
L: Insulating liquid with no measurable fire point

Second letter—Circulation mechanism for internal cooling medium:

N: Natural convection flow through cooling equipment and in windings
F: Forced circulation through cooling equipment (i.e., coolant pumps); natural convection flow in windings (also called nondirected flow)
D: Forced circulation through cooling equipment, directed from the cooling equipment into at least the main windings

Third letter—External cooling medium:

A: Air
W: Water

Fourth letter—Circulation mechanism for external cooling medium:

N: Natural convection
F: Forced circulation: Fans (air cooling), pumps (water cooing)

Therefore, OA/FA/FOA is equivalent to ONAA/ONAF/OFAF. Each cooling level typically provides an extra one-third capability: 21/28/35 MVA. Table 4.13 shows the equivalent cooling classes in old and new naming schemes.

Utilities do not overload substation transformers as much as distribution transformers, but they do not run them hot at times. As with distribution transformers, the tradeoff is loss of life versus the immediate replacement cost of the transformer. Ambient conditions also affect loading. Summer peaks are much worse than winter peaks. IEEE Std. C57.91-1995 provides detailed loading guidelines and also suggests an approximate adjustment of 1% of the maximum nameplate rating for every degree C above or below 30°C.

The hottest-spot conductor temperature is the critical point where insulation degrades. Above the hot-spot conductor temperature of 110°C, life expectancy of a transformer decreases exponentially. The life of a transformer halves for every 8°C increase in operating temperature. Most of the time, the hottest temperatures are nowhere near this. The impedance of substation transformers is normally about 7%–10%. This is the impedance on the base rating, the self-cooled rating (OA or ONAN).

Problems

1. 4.1 Verify Equation 4.17.
2. 4.2 Derive Equation 4.44.
3. 4.3 Prove that doubling feeder voltage level causes the percent voltage drop in the primary-feeder circuit to be reduced to one-fourth of its previous value.
4. 4.4 Repeat Example 4.2, parts (a) and (b), assuming a three-phase 34.5 kV wye-grounded feeder main that has 350 kcmil 19-strand copper conductors with an equivalent spacing of 37 in between phase conductors and a lagging-load power factor of 0.9.
5. 4.5 Repeat part (a) of Problem 4.4, assuming 300 kcmil ACSR conductors.
6. 4.6 Repeat Problem 4.5, assuming a lagging-load power factor of 0.7.
7. 4.7 Repeat Problem 4.6, assuming AWG #4/0 conductors.
8. 4.8 Repeat Example 4.3, assuming ACSR conductors.
9. 4.9 Repeat Example 4.4, assuming ACSR conductors.
10. 4.10 Repeat Example 4.5, assuming ACSR conductors.
11. 4.11 Repeat Example 4.6, assuming ACSR conductors.
12. 4.12 Repeat Example 4.8, assuming a 13.2/22.9 kV voltage level.
13. 4.13 Repeat Example 4.9 for a load density of 1000 kVA/mil.
14. 4.14 Repeat part d of Example 4.11 for a load density of 1000 kVA/mil.
15. 4.15 A three-phase 34.5 kV wye-grounded feeder has 500 kcmil ACSR conductors with an equivalent spacing of 60 in. between phase conductors and a lagging-load power factor of 0.8. Use 25°C and 25 Hz and find the K constant in %VD per kVA per mile.
16. 4.16 Assume a square-shaped distribution substation service area and that it is served by four three-phase 12.47 kV wye-grounded feeders. Feeder mains are of 2/0 copper conductors are made up of three-phase open-wire overhead lines having a geometric mean spacing of 37 in. between phase conductors. The percent voltage drop of the feeder is given as 0.0005 per kVA-mile. If the uniformly distributed load has a 4 MVA per square mile load density and a lagging-load factor of 0.9, and conductor ampacity is 360 A, find the following:
    1. (a) Maximum load per feeder
    2. (b) Substation size
    3. (c) Substation spacing, both ways
    4. (d) Total percent voltage drop from the feed point to the end of the main
17. 4.17 Repeat Problem 4.15 for a load density of 1000 kVA/mi.
18. 4.18 Assume that a 5 mile long feeder is supplying a 2000 kVA load of increasing load density starting at a substation. If the K constant of the feeder is given as 0.00001%VD per kVA·mi, determine the following:
    1. (a) The percent voltage drop in the main.
    2. (b) Repeat part (a) but assume that the load is a lumped-sum load and connected at the end of the feeder.
    3. (c) Repeat part (a) but assume that the load is distributed uniformly along the main.
19. 4.19 Consider the two-transformer bank shown in Figure P3.1 of Problem 3.3. Connect them in open-delta primary and open-delta secondary.
    1. (a) Draw and label the voltage-phasor diagram required for the open-delta primary and open-delta secondary on the given 0° reference line.
    2. (b) Show the connections required for the open-delta primary and open-delta secondary. Show the dot markings.
20. 4.20 A three-phase 12.47 kV wye-grounded feeder main has 250 kcmil with 19-strand, copper conductors with an equivalent spacing of 54 in. between phase conductors, and a lagging-load power factor of 0.85. Use 50°C and 60 Hz, and compute the K constant.
21. 4.21 Suppose that a human being is a part of a 60 Hz electric power circuit for about 0.25 s and that the soil type is dry soil. Based on the IEEE Std. 80-1976, determine the following:
    1. (a) Tolerable touch potential
    2. (b) Tolerable step potential
22. 4.22 Consider the square-shaped distribution substation given in Example 4.10. The dimension of the area is 2 × 2 miles and served by a 12,470 V (line-to-line) feeder main and laterals. The load density is 1200 kVA/mi2 and is uniformly distributed, having a lagging power factor of 0.9. A young distribution engineer is considering selection of 4/0 copper conductors with 19 strands and 1/0 copper conductors, operating 60 Hz and 50°C, for main and laterals, respectively. The geometric mean distances are 53 in. for main and 37 in. for the lateral. If the width of the service area of a lateral is 528 ft, determine the following:
    1. The percent voltage drop at the end of the last lateral, if the laterals are also three-phase four-wire wye grounded.
    2. The percent voltage drop at the end of the last lateral, if the laterals are single-phase twowire wye grounded. Apply Morrison’s approximation. (Explain what is right or wrong in the parameter selection in the problem mentioned earlier.) Any suggestions?
23. 4.23 Resolve Example 3.8 by using MATLAB. Assume that all the quantities remain the same.

References

1. Fink, D. G. and H. W. Beaty: Standard Handbook for Electrical Engineers, 11th edn., McGraw-Hill, New York, 1978.

2. Seely, H. P.: Electrical Distribution Engineering, 1st edn., McGraw-Hill, New York, 1930.

3. Van Wormer, F. C.: Some aspects of distribution load area geometry, AIEE Trans., 73(2), December 1954, 1343–1349.

4. Denton, W. J. and D. N. Reps: Distribution substation and primary feeder planning, AIEE Trans., 74(3), June 1955, 484–499.

5. Westinghouse Electric Corporation: Electric Utility Engineering Reference Book—Distribution Systems, Vol. 3, East Pittsburgh, PA, 1965.

6. Morrison, C.: A linear approach to the problem of planning new feed points into a distribution system, AIEE Trans., pt. III (PAS), December 1963, 819–832.

7. Sciaca, S. C. and W. R. Block: Advanced SCADA concepts, IEEE Comput Appl Power, 8(1), January 1995, 23–28.

8. Gönen, T. et al.: Toward automated distribution system planning, Proceedings of the IEEE Control of Power Systems Conference, Texas A& M University, College Station, TX, March 19–21, 1979, pp. 23–30.

9. Gönen, T.: Power distribution, in The Electrical Engineering Handbook, 1st edn., Academic Press, New York, 2005, pp. 749–759, Chapter 6.

10. Bricker, S., L. Rubin, and T. Gönen: Substation automation techniques and advantages, IEEE Comp. Appl. Power, 14(3), July 2001, 31–37.

11. Ferris, L. P. et al.: Effects of electrical shock on the heart, Trans. Am. Inst. Electric. Eng., 55, 1936, 498–515.

12. Gönen, T.: Modern Power System Analysis, Wiley, New York, 1988.

13. IEEE Standard 399-1980: Recommended Practice for Industrial and Commercial Power System Analysis, 1980, IEEE, New York.

14. IEEE Standard 80-1976: IEEE Guide for Safety in AC Substation Grounding, 1976, IEEE, New York.

15. ABB Power T & D Company, Inc.: Introduction to Integrated Resource T & D Planning, Cary, NC, 1994.

16. McDonald, D. J.: Substation integration and automation, in Electric Power Substation Engineering, 2nd edn., CRC Press, Boca Raton, FL, 2003, Chapter 7, 7-1–7-22.

17. Keil, R. P.: Substation grounding, in J. D. McDonald, ed., Electric Power Substation Engineering, 2nd edn., CRC Press, Boca Raton, FL, 2007, Chapter 11, 11-1–11-23.

18. Farr, H. H.: Transmission Line Design Manual, U.S. Department of Interior, Water and Power Resources Service, Denver, CO, 1980.

19. Institute of Electrical and Electronics Engineers, IEEE Std. 80-2000: IEEE Guide for Safety in AC Substation Grounding, IEEE, Piscataway, NJ, 2000.

20. Sunde, E. D.: Earth Conduction Effects in Transmission Systems, Macmillan, New York, 1968.

21. Gönen, T.: Engineering Economy for Engineering Managers: With Computer Applications, Wiley, New York, 1990.

22. Gonen, T.: Electric Power Transmission System Engineering, 2nd edn., CRC Press, Boca Raton, FL, 2009.

23. National Bureau of Standards Technical Report 108, Department of Commerce, Washington DC, 1978.

* Mesh voltage is the worst possible value of a touch voltage to be found within a mesh of a ground grid if standing at or near the center of the mesh.

* It is often acceptable to use X1 = X2, especially if an appreciable percentage of the positive-sequence reactance to the point of fault is that of static equipment and transmission lines.