This book is aimed at the private electrical installation designer although the technical fundamentals are also applicable to the electricity utility transmission and distribution sector.
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Copyright © IDC Technologies 2011. All rights reserved.
First published 2013
ISBN: 978-1-922062-08-6
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1 Overhead Conductor Design
1.1 Inputs required for design
1.2 Design of electrical specifications of OH conductors
1.3 Design of mechanical specifications of OH conductors
2 Cable Sizing
2.1 Cable sizing issues
2.2 Cable selection calculations
3 Design of a HV Switchboard Installation
3.1 Introduction
3.2 Standardisation
3.3 Inputs
3.4 Design step-1: Basic equipment configuration
3.5 Design step-2: Ratings of equipment and devices
3.6 Switchboard
4 Lightning Protection of Structures
4.1 The Rolling Sphere Method (RSM) for analysis of lightning protection adequacy
5 Touch and Step Potential
5.1 Voltage of grounding system during faults
Appendices
Appendix A – Solutions to Calculations
Appendix B – Design of Earthing Systems:Extract from AS2067
Appendix C – Practical Examples
Appendix D – Solutions to Practical Examples
Appendix E – Substation Earthing Design: Worked Example
Appendix F – Substation Earthing: Design Exercises
Appendix G – Case Study: Earthing for a HV Zone Substation
Appendix H – BS EN 61936-1:2010 Power Installations Exceeding 1 kV a.c. (UK standards)
The design of each component of the OH line installation should be done with great care in view of the safety hazards associated with the bare overhead conductors and the necessity to minimize the risks of failure.
The following minimum criteria should be kept in mind while designing OH systems:
The design comprises of the following broad sections:
The sizing of the conductor is based on several considerations as detailed below.
The conductor should be able to carry the maximum full load current under maximum ambient temperature conditions within permissible temperature limits of the conductor. The conductor depending on the material of construction is designed to be able to withstand a maximum temperature without any reduction in its mechanical strength. The sizing of the conductor should also take into account any future loads planned to be connected to the OH line in the reasonably foreseeable future.
Another important input required for determination of the conductor size is the magnitude of fault level in the system. The conductor should be able to carry the short time fault currents in the system without sustaining any damage. Faults in the system can result in large flow of currents of the order of several thousands of amperes through the conductors producing enormous heating and magnetic effects. The conductors should be designed to withstand such large heating effects and magnetic forces. The three phase fault level at any point in an electrical system is given the following formula:
Fault level (MVA) = √3 x System voltage(kV) x Maximum fault current(kA)
The fault level of the system is dependent on the following factors:
It is generally assumed as a standard that the fault can exist for a maximum of three seconds during which time the protective devices will come into action and isolate the faulty section from the rest of the system. Hence, it is required to design the OH components and equipment to withstand the fault currents for a minimum period of three seconds. Flow of large magnitude fault currents, heat up the conductor beyond permissible limits and cause annealing or melting of the conductor which in turn reduce the strength of the conductor.
The selection of the conductor size is also based on the maximum allowable voltage drop (e.g. 5%) in the OH line. The voltage drop is dependent on the current in the conductor and the impedance of the line. The voltage drop is related to the total series impedance of the OH line and the current as follows:
Vdrop = Imax x Z
Where
Z = R + jX
Where
The resistance of the conductor is a function of temperature. The relationship between resistance of the conductor and temperature is given by the formula:
R = ρ x l/A
Where ‘ρ’ is the resistivity of the conductor, ‘l’ is the length of the conductor and ‘A’ is the area of cross section of the conductor. The relationship between the resistances at different temperatures is given by the relation,
R2/R1 = (T0 +T2)/ (T0 +T1)
R2 = Resistance at temperature T2
R1 = Resistance at temperature T1
T0 = Constant and is
= 234.5 for annealed copper of 100% conductivity
= 241 for hard drawn copper of 97.3% conductivity
= 228 for hard drawn aluminum of 61% conductivity
An important phenomenon to be considered in the resistance offered by the conductor is ‘Skin effect’. Skin effect refers to the inclination of current to flow more readily towards the periphery of the conductor rather than in the centre of the conductor. This phenomenon therefore results in increasing the resistance of the conductor.
The resistance data of the conductor is invariably provided by the manufacturer of the conductor in the form of tables. Safety margin is always provided in sizing the conductor, however the cost implications due to conductor size must be taken into consideration in the selection of the conductor.
In the earlier section, the various factors going into the design of the electrical specifications of the OH conductor were discussed. In addition to the electrical specifications, there is a need to look into the mechanical properties to be possessed by the conductor for the specific application. In this section, the various mechanical specification requirements of the OH conductors are discussed.
The decision on the mechanical properties of the conductor at a minimum is based on the following:
In order to decide the mechanical specifications of the conductor, it is necessary to understand the various mechanical terms associated with OH conductors.
Parabola is defined as the shape assumed by a cable suspended between two support points, when the cable is supporting a uniform horizontal load. For example, the cables used for supporting the deck in a suspension type bridge take the form of a parabola.
Catenary is the natural shape assumed by a conductor whose weight is constant per unit of arc length, when the conductor is suspended freely between two support points. The shape assumed by OH conductors therefore is a catenary. However, the formulae related to the calculations of catenary are more involved and complex than that of a parabola. If the Sag (described later in the section) of the conductor is less than 9% of the span length, then the difference between a parabola and a catenary is observed to be less than 1%. Hence, for all practical purposes of calculation, the shape of the OH line is assumed to be that of a parabola.
Span is the horizontal distance between the two supporting points of the overhead conductor. The Span of the section is dependent on:
In general, smaller spans are used in urban areas attributable to reasons of conductor clearances, right of way, pole locations and constraints in providing stays. Longer spans are mostly used in rural areas.
In practical situations all the spans of an OH line are hardly of the same length, hence the use of the Ruling span. Ruling span is defined as the assumed uniform span which would most closely resemble the variety of spans in a particular section of the line. Sag and clearances are calculated based on the Ruling span and used for spotting the structures. The condition that needs to be satisfied is that the spans in the line should not be more than twice or less than half of the Ruling span.
Ruling span is given by the equation:
Wind span is defined as the length of span that determines the transverse load on the support structure due to force of wind on the conductor. Wind span is equal to half of the sum of the adjacent spans. Refer to Figure 1.1.
Weight span is defined as the length of span between the lowest points of the catenaries on either side of a structure. Weight span determines the vertical loading on the structure due to the weight of the conductor. Refer to Figure 1.1.
Sag is defined as the vertical distance between the point where the overhead conductor is attached to the support poles and the lowest point in the conductor as shown in the Figure 1.2.
Factors affecting Sag are:
Sag is given by the formula:
S = WL2/8T
Where:
S is the Sag of the conductor at mid span (m)
W is the weight of the conductor per unit length (N/m) including wind and ice loads.
Conductor weight is normally provided by the manufacturer in terms of Kg/km. This should be converted to N/m using the formula WN = (9.81 X WKg)/1000
L is the horizontal span length between supports
T is the tension of the conductor at the lowest point of conductor (N)
Consider a conductor strung between two support poles having a span of 200 mts. The tension of the conductor is 5368 N and the weight of the conductor per unit length is 1.893 N/m. Calculate the Sag of the conductor.
Adequate amount of Sag should be provided in the OH line during installation due to the following reasons. An excess amount of Sag can result in reduction of safety clearances due to expansion of the conductor during hot weather, ice loading during winter season and any tilting of the poles. A low value of Sag on the other hand will result in increased conductor tension due to contraction of the conductor during cold weather conditions and may result in snapping of the conductor. Safety clearances below the conductor should be observed in calculation of Sag design values.
Stringing tables are used for selecting the minimum amount of Sag of the conductor at a specified temperature based on the type of conductor and the span length. In actual construction, overhead conductors are not installed and sagged as a single dead end span between adjacent rigid supports. The conductor is installed and sagged in one operation in a line section of may be several unequal spans. Freewheeling stringing sheaves are installed in the structures in between the dead end supports which permit the conductor to move freely between the spans.
Slack is defined as the difference in the lengths between the straight line across the conductor supports and the distance along the conductor length. For a level span, the slack is given by:
K = (8 x S2) /(3 x L)
Where
K = Slack (m)
S = Mid span Sag (m)
L = Length of Span (m)
Calculate the Slack for the data provided in the above problem.
The length of the conductor is given by the equation:
L = S x (1 + (W x D)/ (3 x T))
Where
S is the horizontal length of the Span
D is the Sag of the conductor
W is the weight of the conductor per unit length (including wind and ice loads)
T is the tension at the lowest point of the conductor
Calculate the length of the conductor for the case given in the earlier calculations.
The horizontal swing is provided by the following equation:
SW = P X D X L2 /8 X T
Where SW is the horizontal swing at mid span in m
W is the conductor weight in N/m
P is the wind pressure under final conditions in Pa
D is the conductor diameter in m
L is the length of the span in m
T is the final tension in N
Conductor tension is affected by the various following factors;
An increase in the temperature of the conductor results in an increase in length of the conductor. The increase in conductor length is given by the formula:
ΔL = α x T x S
Where α is the coefficient of thermal expansion of the conductor
T is the increase in temperature in deg C
S is the length of the conductor in m
Calculate the increase in length of the conductor for the case given in the earlier calculations.
Effect of increase in temperature of conductor
An increase in temperature therefore results in an increase in conductor Sag and a reduction in tension. The conductor tension is given by the formula
T = (W x L2) /(8 x S)
Where
T is the tension
W is the weight of the conductor per unit length (N/m)
L is the horizontal span length between supports
S is the vertical Sag at the mid span of the conductor
The second significant factor affecting tension is wind. The wind load acting on the conductor results in an increase in tension of the conductor. This increase in tension results in stretching of the conductor defined by the formula;
ΔL = (TO – TI)/ (E x A)
Where
TI is the initial tension in Newtons
TO is the final tension in Newtons
E is the coefficient of elasticity
A is the cross section of the conductor in metres
The wind load on the conductor would cause an inclined Sag on the conductor.
In locations prone for snow fall and ice formation, the buildup of ice on the conductor has a significant bearing on the tension of the conductor. Ice buildup results in an increase in weight of the conductor. The buildup also increases the area of the conductor subjected to wind force. Both of these phenomena result in increased Sag of the conductor.
Sag of the conductor tends to increase over a period of time due to settling of strands and metallurgical creep. Hence a higher tension may be kept during installation of the conductor to compensate for the effects of aging.
Tension of the conductors should be carefully set during the installation since tension values exceeding the strength of the conductor can result in the breaking of the conductor.
The calculations required to predict the Sag and tension behaviour of an overhead line conductor are complex and are usually performed with sophisticated computer programs. These calculations involve the simultaneous applications of equations for Sag-Tension relationships, conductor stress strain characteristics and change in conductor length as a function of conductor temperature. The program calculates the change in tension of a conductor given its initial stringing condition and nominated operating conditions. The horizontal swing and vertical sag at mid span under this tension are also calculated.
The tension under the operating conditions is calculated using the equation given below:
(c1wl/ T)2 – T = ( c1w0l/ T0)2 – T0 + c2t
Where c1 = √ EA/ 24, c2 = αÉA
The following are the inputs required to be fed into the computer program:
The program then determines:
The computed data from the computer is then used for preparing the Sag-Tension and stringing table. By changing one or more input variables, it is possible to obtain alternative conductor designs which in turn can be used to determine the most practical design for a particular distribution line. Sag-Tension table is then used for the installation of the conductor.
The conductor’s calculated Sag and Tension are based on the span, the type and size of the conductor, the loading conditions and specified tension limit.
Design of an electrical itablenstallation involves sizing of cables. The size of the cables selected should be adequate of both normal current and short time fault withstand current. In the case of fuse-protected feeders, the short circuit current is quickly interrupted and thus permits use of lower cable sizes. In the case of breakers, the fault current is not interrupted instantaneously and depending on the protection, a duration of 0.5 to 3 second of short circuit current flow is possible. Thus the method of protection system sometimes becomes the deciding factor for cable size rather than the steady load current.
The cable sizing study selects the conductor size based on the minimum conductor material and cross-sectional area necessary to meet defined feeder current carrying capability, and associated voltage drop criteria.
Feeder sizes are based on the design load value from the Demand Load Study. The Demand Load Study calculates the total connected demand and design load in each branch of the power system. Some loads are defined as continuous loads and as such, the design load value is larger than its demand load value. The branch circuits that serve continuous loads are preferably rated so that not more than 80% of the feeder current rating is used. The feeder current rating is defined as the current in amperes that a conductor may carry continuously under the conditions of use without exceeding its temperature rating. Selecting a design load value of 125% (1/80%) of the demand load value meets the above criteria.
The cable sizing study is based on the calculations of the following criteria:
The sizing study selects the cable that best meets defined current rating values and has the smallest cross-sectional area. The feeder design load value in amperes is first determined and then the feeder design current rating is arrived at. The design load value is the rated size of the load multiplied by specified demand factors and the long continuous load factor (or design factor).
The feeder design rating is the product of the rated current, the derating factors and the number of parallel cables. Derating factors are determined based on the ambient temperature, thermal resistivity of soil and depth of laying of the cable for cables laid underground, cable grouping factor and duct bank design detail criterion.
Once the cable size that meets the current rating conditions is selected, the voltage drop on the cable is calculated based on the design load current and power factor, cable impedance and length. If the voltage drop criterion is exceeded, the next larger cable size is selected and a reiteration of the comparison of cross-sectional area, rated current and voltage drop is performed.
The voltage drop is calculated using the following formula:
Where ‘R’ is the resistance, ‘X’ is the reactance of the cable and VLL is the line to line voltage. If the voltage drop exceeds the specified level, the next largest cable size is chosen and the calculation performed again in order to select a cable or combination of parallel cables which meets the rating criteria of minimum cross-sectional area and acceptable voltage drop.
Thus the following factors should be taken into consideration in cable sizing:
Calculation 5:
In an industrial plant, a new process section requires power that will be supplied by an underground cable. The section will be fed by a 1 MVA 22/3.3kV transformer with fault current of 11.7 kA. The load is 800kVA, 3.3kV, 3-phase mixed load (service transformer, motors, UPS, HVAC, etc.). Expected utilization is 70% with a 0.8 power factor. The maximum permissible voltage drop of the circuit is 5%. The length of the cable is 650 m. The cable is buried in soil at a depth of 1.5 m. The soil thermal resistivity is 1.5 K.m/W.
Soil Thermal Resistivity (K·m/W) | Soil Thermal Resistivity Derating Factor |
---|---|
Buried Direct | |
0.5 | 1.88 |
0.7 | 1.62 |
1 | 1.5 |
1.5 | 1.28 |
2 | 1.12 |
2.5 | 1 |
3 | 0.9 |
Using the cable Table 2.1 provided below, determine the cable size to be used based on cable with copper conductor and give your reasons for that choice
Designing an HV switchgear installation is a complex task and the designer needs to look at various available options and interface with other related engineering disciplines. The switchboard is a part of an industrial plant distribution system and feeds unitized transformer substations and a few essential motor loads. Design of such the installation should be an exercise in selecting the right option among various available options and the selected option should be in line with the overall design philosophy being adopted by the plant in question. The design should, in addition, address the various interfacing requirements with the premises in which it is housed.
Various steps are involved in the design of an indoor MV metal enclosed switchboard. The steps include:
An important aspect that is not to be missed is that while choosing the equipment, due consideration must be given to makes, types and ratings that are already being used in the other existing units of the plant or those under advanced planning stage. The main purpose is to reduce variety, to use existing skills of operation and maintenance personnel and to conserve the inventory of spares and consumables. This consideration naturally will not be applicable if there are any complaints in the existing makes/types of equipment attributable to manufacturing quality.
In order to properly identify the equipment in question, it should be given a designation and should be referred in all documentation. The industry may already use some standard naming convention, which must also be adopted for the new equipment.
The following are the basic inputs required for a MV switchboard with example data that are typical of industrial systems.
Site conditions
Ambient temperature (limits) | Maximum 45 Deg C Minimum 5 Deg C |
Relative humidity | Maximum of 85% |
Dust | Fine dust present in environment |
Presence of flammable dust/gases | None |
Altitude | 100 m above MSL |
Electrical system
System voltage/frequency | 6600V, 50 Hz |
Permissible variations | Voltage +/- 10%, Frequency +/- 3% |
System earthing type | Low resistance earthed |
Earthing impedance value | 15 Ohms |
The Figure 3.1 shows the overall distribution diagram indicating a switchboard, the loads which it feeds and other feeder details.
Existing equipment data to be used as the basis of specifications
Type | |
---|---|
Equipment detail/basic ratings | Metal clad switchboard 6600 V, 800/1600 amperes busbar, 30 kA/40 kA for 1 second short time rating (symmetrical) |
Type of circuit breaker | Vacuum breaker of ratings 400A, 800A, 1600 A, mounted on draw-out carriage with safety interlocks between breaker position, racking mechanism, control plug-socket and earthing switch. Rupturing capacity of 30kA/40kA symmetrical used. |
Circuit breaker control | Switching from remote control panel, draw-out of carriage by manual lever, test switch for local test operations only |
Motor switching panel | Mechanically latched fuse protected Vacuum contactors having a current rating of 400A with under voltage trip, shunt trip and closing supply from in-built control transformer power from the incoming voltage. Contactor mounted on a draw out carriage |
Motor starting | DOL started with control from DCS, local control near the motor for testing only |
Bus and feeder earthing | Integral earth switch for earthing of cables and bus earthing |
Operating experience | Very good |
Control and signaling supply | 110V DC |
Spring charging motor supply | 230V AC single-phase |
Source of control power | Dry type auxiliary transformer connected from the incoming supply to switchboard with fuse protected isolator for control |
Current transformer secondary rating used | 5A |
Enclosure IP category | IP 51 |
Based on the requirements given in earlier sections, a configuration is to be decided and represented in the form of a basic single line diagram of the switchboard MVDS-03 as shown in Figure 3.2. In this scheme the following configuration has been indicated.
Switchgear with two incomers; one from normal source and another from the emergency source with the latter normally in OFF position.
A sectionalizer to segregate the essential loads, which are on a separate section of the bus. Interlocking to ensure that the emergency section incomer can be fed only from either the incomer of that section or the sectionalizer. Normally sectionalizer will be ON.
For better protection against inadvertent paralleling due to interlock failure, bus voltage will be checked and will be used to block the closing if the bus is live.
All panels except the motors will be breaker controlled. Motors will be contactor controlled. Both circuit breakers and contactors will be vacuum type. Panels will be of draw out design. All features will be as in existing switchboards.
Additional transformer and motor feeders required for later expansion are shown in dotted lines. The loads of these feeders are considered for equipment sizing.
In this step, the ratings of feeders and other associated devices are to be decided by calculation using the input data.
For each outgoing/incoming feeder
For the switchboard as a whole
Feeder sizing for the motor feeder shown in Figure 3.2
The current of a motor under rated load condition (full load current or FLC) can be calculated based on the formula below.
Power factor (pf) and efficiency values at rated load are to be considered in the above calculation.
Where pf and efficiency values are not readily available a pf of 0.85 and efficiency of 0.9 can be assumed, being typical values.
An overloading factor of 1.1 can be taken into account since a motor can operate indefinitely at a small overload. (This will also be useful in case the motor has to work at the lowest possible voltage limit of -10% for long periods). Thus the feeder rating can be calculated as:
Feeder rating = FLC * Overload factor
This is the minimum current rating that the feeder and switching device must be designed for. For example a 6600V motor of 320 kW will have FLC of 36 amperes approximately and will need a feeder rating of 40 amperes at a minimum.
The rating of the selected equipment must not be less than the current arrived at as above. There may be recommended derating factors by switchgear manufacturers for operations at ambient conditions other than those for which the equipment is rated. Thus an equipment rated for operation at 40 Deg C in open air may require derating when placed inside an enclosure and at an ambient temperature of 45 Deg C. The equipment rating at specified condition is given by:
Feeder rating < Standard rating * Temperature derating * Enclosure derating
The rating of the switching device (breaker, isolator, contactor etc.) must be selected on the above basis. Starting current of a cage motor can be assumed as 600% of FLC (app 220A). This value taken along with the starting time will be useful to select the motor HV fuse. Fusing characteristic of typical HV fuses are provided by different manufacturers and can be used for selection. (A motor starting calculation may be also performed to assess the voltage drop and starting time for motor DOL starting to verify that the bus voltage does not drop over 10% during starting).
Typical fuse characteristic curves are shown in Figure 3.3 and 3.4. The highest value of normal current (in this case starting current) should not initiate melting curve in Figure 3.3 within the starting time of the motor. Fuse range 2R (70 amps) is the minimum rating of fuse satisfying both conditions. The fuse will melt in a range of 15 to 35 seconds at a value of 100 times the “R’ number, in this case 200 Amps. For starting current of 220 amps, the fuse blows out after 12 seconds as per melt time curve (3.3). If the starting time is likely to be more than this value, a higher fuse rating will be necessary.
The lowest current under fault condition (in this case 5000A as given in the section on inputs) should result in fusing with the delay given by the curve in Figure 3.4. The manufacturer’s data indicates a maximum interrupting capacity of 50 kA (symmetrical) for this fuse and is therefore safe at the maximum fault level given (15 kA) in this case.
CT current must be as close as possible to the rated current and a value of 40/5 will be suitable. Protection CT can also have the same ratio and a 10P10 class is adequate as the main function is to operate overload and other protections but not short circuit faults (which are cleared by fuse). Rated CT burden must be adequate to supply the devices (metering or protection as the case may be). 15VA burden is normally specified and is adequate provided no long wiring is involved (example: for meters mounted remotely from switchgear).
Minimum metering will be provided on the panel in the form of an ammeter with a selector switch to read the current in any of the phases. A meter with compressed scale graduated 0-40-240 Amps of primary current (40 to 240 is the compressed scale range) is required.
Protection against overload (thermal image type), single phasing, earth fault and stall condition will be provided using a comprehensive motor protection relay. Numerical relay with adjustable thermal curves is a preferred choice.
Protection against loss of voltage by an undervoltage relay will be required to ensure that the circuit is disconnected if the supply voltage fails. The protection should be able to distinguish between transient sag in the supply and a power failure and should trip for the latter condition only. Instantaneous undervoltage relay in tandem with a timer relay can achieve this objective.
Feeder sizing for the transformer feeder shown in Figure 3.2
The transformer rating given is 1000 kVA. Full load primary current can be calculated by the formula:
Substituting, we get a current of 87.4 amps. A factor of 1.1 can be considered for short time overloading and the feeder rating will therefore be 96 amps.
The feeder will be switched using a circuit breaker of vacuum type. The minimum current rating of the circuit breaker after applying derating should be more than the feeder rating. The minimum rating of circuit breaker viz., 400A is amply suited for this purpose.
Circuit breaker interrupting capacity of 30 kA is selected as already applied in the system and has a sufficient reserve capacity considering the maximum fault level of 15 kA. The CT selection should be done considering the capability for withstanding the short circuit level of 15 kA for duration of 1 second (this is the period normally considered for breaker-fed feeders). Also, the protection relay setting requirements must be kept in mind.
IDMTL relay in combination with instantaneous high-set element is normally chosen. The instantaneous element should be set so as not to operate for terminal short circuits on the transformer secondary. For this rating of transformer, the impedance is normally about 5% and therefore the secondary short circuit current may be as high as 20 times the rated current (around 1.8 kA). As the instantaneous elements may not provide setting as high as 2000% it will be convenient to consider 200/5 as the CT ratio. A relay range of 400 to 1600% and CT class and 10P15 class for protection will be appropriate.
Earth fault protection will be provided using an instantaneous relay of 20 to 80% range using the CT summation connection of all three phases. The preferred form of relaying is a numerical relay with adjustable characteristic with the required IDMTL and instantaneous current relay functions.
Metering is normally limited to an ammeter with phase sector switch. An energy meter can be added for monitoring process energy consumption if called for. Usually, ammeters may have to be additionally provided near the local control point of the motor, in which case CT burden should consider the leads also into account. Transformer-mounted protection devices such as oil/winding temperature alarm/trip and Buchholz relay will be wired to the switchboard for alarm/tripping.
The various ratings of the switchboard are calculated now.
These ratings will be based on the configuration of the switchboard and the maximum current expected to be supplied by the incomers. In this case, incomer-1 should be capable of handling the entire load of the switchboard and incomer-2 will be sized for supplying the loads of the emergency section only. The following need mention.
While calculating the load, the probable extra loading for future expansion is to be considered as well, to avoid major changes in the switchboard later.
A diversity factor must be applied to the sum of the individual feeder load currents to account for the fact that all of them will not operate simultaneously at their maximum values. Derating must be applied as in the previous cases.
Incomer 1 feeds 4 transformers and three motors including future loads. The sum of the load currents is thus equal to: 4*87.4 + 4*36, which works out to about 494 amps. Dividing this figure by the diversity factor (assumed as 1.2) and then again by a derating factor (assumed as 0.9 for temperature of 45 Deg C), the minimum rating of circuit breaker can be arrived at; in this case it is 457 amperes. The next higher rating of 800 amps can be selected for incomer-1 breaker and for bus bar. Incomer 1 CT will have a primary rating of 500 amps.
Incomer 2 will only feed 4 motors in the worst case. The rating required is 133 amps. A rating of 400 amps for the incomer breaker, sectionalizer and bus bar of this section is considered adequate. CT ratio of 200/5 can be selected. (However, in the interest of standardization both incomers and sectionalizer breaker can be selected to have identical rating of 800 amps if the cost difference is not appreciable).
Sectionalizer will have a breaker and bus-riser compartment with the Potential Transformer (PT) of each section being integrated within this arrangement. PTs will be of star/star connection. Output of each PT will be looped to individual panels in the respective sections.
All the breakers will be designed for fault current of 30 kA and the busbars for short time withstand capacity of 30 kA for 1 second.
Each incoming feeder will be provided with an auxiliary transformer panel. The transformer will be 3-phase 6600V/415V (assuming a standard LV 4-wire supply of 415/240V). The transformer will primarily supply AC power to the switchboard for auxiliaries and for spring charging motor as well as to the charging panel of DC system. Substation lightning can optionally be included. The supplies will be taken from the switchboard to a separate AC distribution panel for feeding to the above loads. A changeover contactor scheme will be provided for automatic transfer of output. Normally the AC feed will be from the transformer of incomer-1 (normal source) but will switch to the standby incomer-fed transformer upon failure of normal source.
The transformer will have a provisional rating of 10 kVA and will be of dry type with fuse protection. The fuse-transformer combination will be mounted on a draw-out truck with suitable interlocks.
In order to reduce the number of grading steps and considering that bus faults are rare in indoor switchboards, no protection relays will be provided in the sectionalizer and incomers. Protection will be provided at the sending end of the incomer feeders only. This will cover faults of both cable and busbars and will act as backup to feeder protection. This is acceptable considering that the probability of faults in metal enclosed switchgear busbars is very low.
The data computed above is incorporated in the final single line diagram as shown in Figure 3.5.
In this section, we mainly discuss the protection of structures against lightning strikes by the placement of air terminals. Lightning protection can be achieved by providing Air terminals and enclosing a building within a mesh of conductors as shown in the Figure 4.1.
The spacing of the mesh will depend on how well the building is to be protected. A closer spacing means better protection. A practical example of a building protected by such a mesh is shown in Figure 4.2.
This figure shows a building with an elevated part in the top (may be a lift house, or a water storage tank) with an antenna placed on top of it. The roof conductors (also known as horizontal air terminals) are placed both on the elevated portion as well as the roof slab itself and are interconnected closely. The antenna itself is bonded with the roof conductors. The down conductors lead the lightning discharge current to the ground rods, which are interconnected at ground level to form a ring. It is also usual to interpose a test link in the down conductor, to isolate the ground electrodes for enabling periodic testing of electrode resistance. Edges and corners of a building are vulnerable and will have to be provided with air terminals
The protection provided to the building will depend largely upon the number and spacing between air terminals on the roof. Air terminals can both be horizontal conductors or vertical rods. The distribution of air terminals can be arrived at by using the Rolling Sphere Method as described in the section below.
The basis for ascertaining the adequacy of mesh type protection is the Rolling Sphere method. In this method, an imaginary sphere is rolled over the protecting structure and the shaded areas, which the sphere cannot touch, are within the protection zone. The radius of the sphere can vary between 20 m to 60 m depending on the degree of protection required.
The standard protection will consider a radius of 45 m and increased degree of protection can be obtained by reduction of the radius. This method is also referred to as the Electro-Geometric method in some literature; since the protection efficiency is determined using an empirically calculated protective sphere radius which is also related to the peak lightning current. (Refer to Figure 4.3 and Table 4.1).
The table also indicates the minimum values of lightning stroke current (peak) which will be intercepted by the protection system. Note that higher the level of protection, smaller is the lightning current value that can be intercepted.
It can also be seen that better the protection, smaller is the sphere radius considered in the protection system design and lower is the minimum peak current of a lightning flash, which the system can protect against. About 80% of the ground flashes tend to exceed a peak current value of 20 kA while 99% of the flashes will exceed 3 kA. It can be seen from the above table that a system design based on a sphere of radius 60 m can only protect against lightning flashes above 15.7 kA (peak current). About 15% of the flashes can be of magnitudes lower than this value and can thus bypass the protection. Compared with this, a design based on a sphere of 20 m radius can offer protection against all flashes exceeding a peak current value of 2.9kA. Such a system can protect a structure from more than 99% of the flashes leaving it unprotected only against 1% of the flashes, which in any case cannot inflict much damage because of their lower energy levels.
When applying the Rolling Sphere method to flat surfaces, a horizontal conductor placed just on the roof (in contact with it) cannot offer protection to the rest of the roof, because the sphere can be rolled all over the horizontal plane of the roof. Earlier it was deemed to be protected by such horizontal air terminals provided that the spacing of these conductors is not greater than 5m, 10m, 15m or 20m for protection levels I, II, III and IV respectively. The corner points and edges however need to be protected with conductors laid along these edges. This is the classical mesh type of protection. However, in some of the other standards (such as AS/NZ 1768), such an approach is not recommended. The protection efficacy in these standards is considered strictly as per the RSM principle, thus making it mandatory to provide air terminals which are at a minimum height of 0.5m above the protected surface.
This is applicable both for vertical and horizontal conductors of the lightning protection system acting as air terminals. The protection range offered by an air terminal can be calculated from the formula:
Where
r is the horizontal distance in metres from the air terminal which will be protected by the terminal
h is the height of the terminal in metres
a is the radius of sphere in metres for the protection level chosen as given in table 4.2
For a vertical air terminal, r is the radius of the circle with center at the air terminal location and any point on the structure lying within the circle is protected.
For a horizontal terminal, r represents the distance on either side up to which the structure is protected from direct strikes.
Also note that the value of h cannot be zero. This means that the air terminals (vertical or horizontal) should be placed at a height from the protected plane. The complete protection of a large structure, thus means an array of air terminals. These terminals are normally arranged in the form of a grid in the case of vertical terminals and parallel conductors in the case of horizontal terminals. The spacing of the grid/parallel conductors is given by the following equations.
For vertical air terminals, the grid spacing should at least be:
For horizontal conductors the distance between parallel conductors should not be less than:
The Rolling Sphere Method as described above has certain shortcomings. In any building there are certain vulnerable points, which are prone to lightning attachment. This is because of field intensification that occurs around pointed features of a structure as well as corners and edges. This needs to be considered and air terminals must be provided to take care of protection of such vulnerable points
The vulnerable points in decreasing orderare as follows:
While planning protection, the most vulnerable features (points and corners) should be first provided with air terminals and rolling sphere method should be used to check coverage of edges. If these are not covered, then more terminals will have to be added. In other words, the sphere should rest on the air terminals without coming into contact with the rest of the structure. The least vulnerable areas such as flat surfaces should be checked if they are covered with the air terminals. If not, additional terminals should be provided to ensure complete coverage of protection. A mix of horizontal air terminals (conductors placed at a slight elevation above the adjacent surfaces) and vertical air terminals can protect a structure completely.
The other problem with the classical values of protection radius is that the design becomes too conservative, particularly when applied to large flat surfaces such as a building roof. More realistic designs without excessive material usage for such surfaces (but without compromising the protection efficiency) are possible with the use of dual sphere radii shown in Table 4.2 below, the purpose of which is explained in this section.
The radius of sphere values shown in brackets are applicable for evaluating protection of flat surfaces and are higher than the sphere radius value applicable for features which are more vulnerable to lightning attachment (listed earlier).
For large flat surfaces, the value of r (in Formula 4.1) is calculated using the larger value shown within brackets. For example, for protection level III, a will have a value of 90m when planning air terminals for a large flat roof. However, for verifying protection of more vulnerable features such as edges, the smaller value should be considered. For protection level III, this value will be 45m.
Naturally for flat surfaces the r value to be used will be the one calculated using the larger sphere radius as explained above.
These principles are illustrated with an example. A building of 70m length, 50m width and 20m height needs to be provided with lightning protection system of level III protection. The first step will be to protect the top edges and corners by 1m high vertical conductors. The protection range offered can be calculated by substituting values in the formula 4.1 as:
The value of r works out to 9.4m. The lower value of a is substituted in this case because, we are protecting the edge of the roof, which is more susceptible to a strike.
The minimum spacing of conductors required along the edge is 2* 9.4 which is 18.8m. Providing 5 equally spaced terminals along the 70m side (spacing 17.5m) and an additional 2 terminals along the width of 50m (spacing 16.7m) will adequately protect the edges. Figure 4.4 below illustrates the protection along the 70m edges of the roof. The vertical flat surfaces of the structure are also protected as the sphere of 45m radius cannot come into contact with the vertical surfaces.
These terminals can however not provide protection for the entire roof because of the large area involved. Further protection is needed by an array of vertical terminals of height say 1.25m.
The range of protection offered by vertical terminals placed on a flat roof can be calculated by substitution of values in formula 4.1 as:
which gives a value of 14.94m (say 15m). The terminals thus have a protection radius of 15m. The use of the value 120 being the higher of the two values of a, since it is the flat roof we are protecting. Four such terminals arranged in the middle of the roof (in combination with the vertical rods already placed along the edges) can give adequate protection. Figure 4.5 illustrates this.
An alternative scheme of protection can also be worked out using the following:
Horizontal conductors along the roof edges placed 1m above the edge and 2 additional horizontal conductors also 1m from the surface or 2 additional 1.5m high vertical rods placed symmetrically along the 70m axis
Figure 4.6 illustrates this arrangement.
For tall buildings with large vertical surfaces, the protection of these surfaces can be assessed using the larger of the two radii values shown in Table 4.2. Figure 4.7 illustrates an example of a level III protection scheme. Note the use of sphere of radius 90m for the larger vertical side and 4m radius value for the smaller surfaces.
The connection of the enclosure of electrical equipment to the grounding electrodes does not automatically ensure that it becomes safe to touch if there is a fault in that equipment. The voltage of the enclosure will go up with respect to remote earth and can attain substantial values. What is of interest is the voltage difference between the enclosure and local earth mass. Refer to the following diagram (Figure 5.1) which shows the voltage rise of the ground grid of an entire HV outdoor substation and the local soil mass of the substation when there is a ground fault within the substation.
It is clear from this figure that there is a considerable rise of voltage in the ground grid with reference to true (remote) earth. However, the voltage at the surface of the soil within the mesh also goes up correspondingly (Ground potential rise or GPR in short). So the voltage, which a person within the substation area can be subjected to, is not as high as the voltage of the mesh above remote earth.
In fact the voltage rise of the mesh (grid) itself can be calculated by the simple formula:
Where
Vg is the ground grid voltage with reference to remote earth
Ig is the current flowing into the earth (soil) through the ground mesh
Rg is the ground resistance of the grid
Ig here may not be the same as the ground fault current at the point of fault, since part of the fault current may be diverted through the other metallic connections that exist between the source and fault point. This can be the overhead shielding wires of a transmission line, the metallic armor of the feeding cables from the source or any other metallic connection. In fact, in industrial distribution systems, ample connections are established between the source and the consumer installations and only a very small component of the actual fault current may flow through the soil, thus limiting the voltage of the grounding system at the fault point with reference to true earth. Refer to Figures 5.2 and 5.3 to illustrate this point.
Figure 5.2 shows a system where the ground fault current has to go through the soil because of the absence of a direct metallic connection between the source and the load. Figure 5.3 shows a system where a metallic path for ground current is present between the load and the source. Needless to say, the system shown in Figure 5.3 will result in only a small fraction of the ground fault current flowing through soil. In such a case, the ground resistance of the grounding system does not greatly influence the safety of the system. This is not true for the system shown in Figure 5.2, where the flow of fault current is through the soil and involves the grounding system at the point of fault as well as the supply source. The voltages of both grounding systems with respect to the soil mass can therefore become high. The objective of good grounding system design is therefore to ensure that such a thing does not happen.
The safety of the grounding system involves not only the voltage rise that can happen during faults but also the time for which such voltage rise can take place, because the human body’s ability to withstand an electric current is a function of the current and the time of exposure. This is where protection systems play a major role. In order to reduce danger, the fault must be correctly sensed and the faulty equipment isolated from the system within the shortest possible time at the same time maintaining necessary discrimination to ensure that electric supply to other healthy parts of the installation is not affected. In circuits provided with autoreclosing of breakers, the total time of exposure (the sum of the duration of the first fault and the duration of fault on auto-reclosing) is to be considered for assessing safety. The design of grounding system for an outdoor HV substation is a complex case and we will discuss this aspect in this text. The principles involved may be extended to cover the design of other types of installations as well.
The output of the design should indicate:
Touch Voltage
Touch voltage is defined as the voltage at any point of contact with uninsulated metal work:
Step Voltage
Step voltage is defined as the difference in surface potential experienced by a person bridging a distance of 1 mtr with his feet apart, without contacting any other earthed object:
The following assumptions are made for calculation of the maximum permissible values of touch and step potential.
The values of safe touch and step potentials are given by the following equations.
where
Cs is a function of the crushed rock layer thickness hS and reflection factor K arrived at based in the graph shown in Figure 5.4.
The value of K is taken from the expression
ρ and ρs being the resistivity of soil and the surface layer respectively. If the resistivity of the surface layer is the same as that of the soil below, then K takes the value of 0 and Cs becomes 1.
The total length of buried conductor is required to be calculated for use in computing the ground grid resistance. Let us assume that the substation is located in a square plot of say 70 m side. Assuming that the mesh will have a dimension of 7m x 7m, the total length of buried mesh forming the electrode is 70 x 22, which works out to 1540m.
Note:
It is assumed here that there are no vertical electrodes attached to the mesh, which is often done in order to ensure contact with deeper soil layers of lower resistivity.
The resistance of the above ground grid Rg can be computed using the formula:
Where:
The above formula is valid for grid depths of 0.25m to 2.5m, which covers most practical cases encountered. In case vertical rods are also used in combination with the horizontal ground grid conductors, the resistance gets further reduced. Standard IEEE 80 indicates the method of calculating the combined resistance of such a system.
The ground potential rise of the ground grid can be computed as the product of Rg arrived at in the previous step and the grid current during a fault. The fault current computed in step 2 for grid conductor sizing may not completely flow into the soil from the grid. Often a part of the current may get diverted through other metallic connections such as the shield wire, buried pipe work etc. The grid current is therefore computed as the product of the fault current and the current division factor Sf. The standard IEEE 80 illustrates various methods used for arriving at the value Sf to the required degree of accuracy. The following equation computes the ground potential rise (also called as grid voltage).
Eg = Sf • If • Rg
If the value of Eg is less than the tolerable touch voltage calculated earlier, the preliminary design can be considered as adequate. Note that the potential rise calculated here is the voltage that the grid assumes with reference to the remote earth and therefore can be expected to be rather high. It is therefore likely that the condition stated above will not be fulfilled.
A further calculation is therefore required to compute the value of mesh voltage and check whether this is lower than the tolerable touch voltage. The mesh voltage is the maximum voltage within a mesh in a ground grid. Refer to Figure 5.5 below which shows the mesh voltage and the voltages to which personnel at different points in a substation can be subjected during a ground fault.
The worst case of mesh voltage in an equally spaced ground grid is to be found in the corner grids and this is the value, which has to be checked for safety. This voltage can be found graphically using Figure 5.6 shown here.
For a 10m x 10m grid with 7 meshes on a side, the corner mesh potential can be seen to be approximately 20% of the grid voltage (Eg).
i.e., Em = 0.2 • Eg
Alternatively, the mesh voltage can also be arrived at by using a formulae provided in the standard.
The earthing system of a high voltage switchyard consists of a grid with conductors buried in the ground in the form of a number of 20 metre square meshes (refer to Figure 5.7). The earth resistance of the grid was measured to be 0.9 ohms. When a fault current of 10 kA flows into the ground through the grid what will be the earth potential rise (EPR) of the substation earth grid with respect to true ground?
The centre of the mesh has a voltage of 5 kV with reference to true ground and the voltage gradient between the centre point and the mesh is assumed to be uniform (linear).
What will be the touch and step voltage that will be impressed on the persons A and B respectively? A is touching the structure S which is connected to the mesh and B stands on the ground within the mesh as shown. All external impedances can be neglected.
Consider a conductor strung between two support poles having a span of 200 mts. The tension of the conductor is 5368 N and the weight of the conductor per unit length is 1.893 N/m. Calculate the Sag of the conductor.
The Sag of the conductor is given by:
Sag = WL2/8T
Where
W = 1.893 N/m
L = 200 m
T = 5368 N
Therefore
Sag of the conductor = (1.893 x 200 x 200) / (8 x 5368) = 1.76 m
Calculate the Slack for the data provided in the above problem.
The slack of the conductor for the above Sag calculation is given by
Slack = 8 x 1.76 x 1.76/(3 x 200)
= 0.04 m
Calculate the length of the conductor for the case given in the earlier calculations.
The length of the conductor is given by the equation:
L = S x (1 + (W x D)/ (3 x T))
Where
S is the horizontal length of the Span
D is the Sag of the conductor
W is the weight of the conductor per unit length (including wind and ice loads)
T is the tension at the lowest point of the conductor
The length of the conductor seen in the earlier example is given by:
L = 200 X (1 + (1.893 X 1.76)/ (3 X 5368))
= 200.04 m
Calculate the increase in length of the conductor for the case given in the earlier calculations.
The increase in conductor length with temperature is given by the formula:
ΔL = α x T x S
Where α is the coefficient of thermal expansion of the conductor
T is the increase in temperature in deg C
S is the length of the conductor in m
Therefore the increase in length of the conductor considered in the earlier example for a 30°C temperature rise is:
ΔL = 13.9 X 10-6 X 30 X 200.04
= 0.08 m
In an industrial plant, a new process section requires power that will be supplied by an underground cable. The section will be fed by a 1 MVA 22/3.3kV transformer with fault current of 11.7 kA. The load is 800kVA, 3.3kV, 3-phase mixed load (service transformer, motors, UPS, HVAC, etc.). Expected utilization is 70% with a 0.8 power factor. The maximum permissible voltage drop of the circuit is 5%. The length of the cable is 650 m. The cable is buried in soil at a depth of 1.5 m. The soil thermal resistivity is 1.5 K.m/W.
Soil Thermal Resistivity (K·m/W) | Soil Thermal Resistivity Derating Factor |
---|---|
Buried Direct | |
0.5 | 1.88 |
0.7 | 1.62 |
1 | 1.5 |
1.5 | 1.28 |
2 | 1.12 |
2.5 | 1 |
3 | 0.9 |
Using the cable Table 2.1 provided below, determine the cable size to be used based on cable with copper conductor and give your reasons for that choice
Power in kVA | = 800 kVA |
Voltage | = 3.3kV |
Full load current | = kVA/√3 x V |
= 800/ (√3 x 3300) | |
= 140 A |
Armoured, XLPE insulated, three core cable is chosen over PILC (Paper insulated lead covered) for its superior insulation properties, higher current ratings and temperature of 90°C (compared to PILC at 70°C), higher tolerance to movement, low charging current and also for its cost effectiveness. Three core cable is chosen as cost effective option and preferred over three numbers of single core cables unless the required cable size is greater than 300mm2 (to be confirmed by calculations below).
The Cable that is selected is 95mm2 XLPE, obtained from the provided Table of the cable manufacturer. The current rating of the 95mm2 cable is 290A as per the Table.
Neglecting cable impedance, the fault current to be carried by the cable = 11.7kA
The derating factors applied are:
De-rating for depth of laying @ 1500mm directly in soil = 0.92 (Current carrying capacity of the cable decreases with increasing depth of burial of the cable)
De-rating for soil thermal resistivity @ 1.5 K.m/W = 1.28 (Increase in thermal resistivity reduces the current carrying capacity of the cable)
De-rating ground temperature @ 25°C = 1.0 (Increase in ground temperature reduces the current carrying capacity of the cable)
Capacity of cable after derating = 290 * 0.92 * (1/1.28) * 1 = 208.4 A
The capacity of the derated cable is well above the required capacity of 140A.
The cable size is required to handle the expected fault current level of 11.7kA. Fault current capacity of a 70mm2 copper XLPE 6.35/11kV cable is 10.0 kA according to the cable catalogue and hence 70mm2 cable size is not suitable.
Smallest cable suitable for phase fault of 11.7 kA is 95 mm2 which as a capacity of 13.6 kA.
Voltage drop at rated current = √3 x Z x I x D/1000
= 1.73 x 0.247 x 140 x 1250/1000.
= 74.78 V
The maximum permissible voltage drop is 5%, 5% of 3.3kV = 165 volts
(Hence the actual voltage drop is within the permissible voltage drop)
The XLPE, three core cable satisfies the conditions of current capacity after derating, the permissible voltage drop and the fault carrying capacity and hence is suitable for the application.
The earthing system of a high voltage switchyard consists of a grid with conductors buried in the ground in the form of a number of 20 metre square meshes (refer to Figure 5.7). The earth resistance of the grid was measured to be 0.9 ohms. When a fault current of 10 kA flows into the ground through the grid what will be the earth potential rise (EPR) of the substation earth grid with respect to true ground?
The centre of the mesh has a voltage of 5 kV with reference to true ground and the voltage gradient between the centre point and the mesh is assumed to be uniform (linear).
What will be the touch and step voltage that will be impressed on the persons A and B respectively? A is touching the structure S which is connected to the mesh and B stands on the ground within the mesh as shown. All external impedances can be neglected.
EPR calculated using a current dispersion factor (Sf) of 1.
Figure 5.7 shows the ground voltage gradient with the calculated EPR. The voltage difference between the grid and the centre point of the mesh (in the soil) is the difference between EPR and the voltage at the centre of the mesh.
Voltage difference | = 9000 – 5000 |
= 4000 Volts |
The distance from the centre point of the mesh and the grid (Point S) is 10 m since the mesh size is 20 x 20m.
The ground voltage gradient between the centre point and the mesh is linear, therefore the voltage will increase at a rate of 400V per meter.
Design of an earthing system can be accomplished by the process outlined in Figure 8.1 and described in the following points:
A series of design tables is also provided in Appendix B to simplify the design process for distribution substations in locations which meet the boundary conditions specified within Appendix B.
NOTES:
Physical implementation of the design needs to consider material selection (sizing, jointing, materials, covering), joint types, protection (i.e. back-up protection clearing time), labelling, uneven current distribution, ability to withstand mechanical and corrosive influences, inspection and testing requirements (i.e. commissioning prior to installation completion). For direct buried or exposed conductors a minimum size of 35 mm2 copper equivalent conductor is considered prudent. Further guidance regarding sizing of conductors to meet thermal requirements is given in the ENA EG1. Aluminium conductor shall not be used for in-ground connections. The design layout shall always be prepared with a view to simplifying the installation process, simplifying ongoing monitoring needs, and minimizing exposure to ‘external risks’ (i.e. damage through vandalism or theft).
A method for calculating the voltage limit is described in Figure A1 below. Figure A2 (Typical touch voltage limits) provides two characteristics (C1 and C2) which may be used for the contact scenarios as described.
NOTES TO FIGURE A1:
C1 curve: | Special locations—High public contact likelihood assuming bare hands and bare feet and thus negligible additional series impedance |
C2 curve: | Normal locations—Normal public contact likelihood with typical mix of footwear and thus additional series impedance |
The voltage limit can be taken as a tolerable touch voltage UT.
The values for the two curves C1 and C2 used in Figure A2 are shown in Table A1.
An 11kV 2MVA generator is protected by a 3-phase high impedance differential relay.
The generator sub-transient reactance is 15% and coverage of greater than 70% is required.
The CT’s are class ‘X’ defined as 0.125PL100R0.4 with 100/5 ratio.
The relay burden is 1VA @ setting.
An overcurrent relay is connected via lead wires to a class ‘P’ CT.
CT ratio: 200/1
CT burdenL 2.5 VA
CT accuracy: 10P20
RLOAD = 1 Ω
Relay burden = 0.25 VA
Relay setting = 1 A
Find the relay setting voltage.
It is required to specify a CT for O/C IDMT application, both the CT secondary and the relay rated current being 1A. The relay burden is 0.25VA at rating and the CT burden is 5VA. The loop resistance is 0.5 Ω. The maximum fault current referred to the CT secondary is 25 times the relay rating. The ranges available for composite error and ALF are:
ALF: 5, 10, 15, 20, 30 Composite error: 5P, 10P
In the following system, the generator and transformer specification are given as below:
G: X = 26% 66.6 MVA 11 kV
T1: X = 12.5% 75 MVA 11/145 kV
What is the magnitude of a three-phase fault that occurs on the 132kV line close to the 132kV busbar?
Calculate the fault level for a 3Ø fault as F
With bus section CB closed
With bus section CB open
A power transformer 132/11kV, 10MVA, Z = 5% is connected to load at the 11kV side. Design the 11kV cable. Assume the cable is consisted of two circuits, directly buried in ground. The cable installation and site conditions are given below:
Soil temperature = 30°C
Soil Thermal resistivity = 1.5 k.m/W
Depth of burial = 1 m
Spacing = 0.3 m
Length = 50 m
Load power factor = 0.8 lagging
An 11kV 2MVA generator is protected by a 3-phase high impedance differential relay.
The generator sub-transient reactance is 15% and coverage of greater than 70% is required.
The CT’s are class ‘X’ defined as 0.125PL100R0.4 with 100/5 ratio.
The relay burden is 1VA @ setting.
Solution:
Ifault = 2MVA / (√3 x 11kV x 0.15) = 100 x 0.66 ≈ 700A (prim) = 35A (sec)
Minimum requirement for knee point: VK/VS = 2
VS = VK / 2 = 100/2 = 50 V (from the CT class given)
VS = Ifault secondary side * (RLead + RCT) → 50 = 35 (RLead + 0.4) → RLead = 1.03 Ω
Selecting the right current setting for the relay will affect the POC:
Let’s assume: PSM = 15% → Pickup = Ir = 0.15 x 5 = 0.75 A
RRelay = VA / (ISetting)2 = 1 / 0.752 = 1.77 Ω
VS = ISetting (RRelay + RST) → 50 = 0.75 (1.77 + RST) → RST = 65 Ω
The primary operating current (POC) should be less than 30% of the minimum fault current. This ensures sufficient relay current during internal fault conditions for high speed operation.
POC should also be in excess of the rated current, so that if one of the CT’s gets saturated, the relay does not show mal-operation.
Since in each protection zone, there are several CT’s in parallel with the relay and each other, the combined CT magnetising currents will increase the primary operating current:
POC = CTR (Ir + nIe)
POC = (100/5)(0.75 + 6 x 0.125) = 30 A
IFull load = 2MVA / (√3 x 11kV) = 105 A
POC / IFull load = 30/105 = 0.19 = 29%
Protection coverage = 100% – 29% = 71%(above 70%, thus acceptable)
An overcurrent relay is connected via lead wires to a class ‘P’ CT.
CT ratio: 200/1
CT burdenL 2.5 VA
CT accuracy: 10P20
RLOAD = 1 Ω
Relay burden = 0.25 VA
Relay setting = 1 A
Find the relay setting voltage.
Solution:
Relay burden = 0.25 VA
Relay setting = 1 A
Therefore: RRelay = 0.25/12 = 0.25 Ω
Voltage delivered to the CT burden:
Vburden = 2.5VA / 1A = 2.5 V
Voltage supplied by the CT @ ALF = 20
VS = 20 x 2.5 V = 50 V
Min stability voltage required:
VS = ALF In (2RLead + RRelay)
VS = 20 x 1 (2 x 1 + 0.25) = 45 V (CT acceptable due to 50>40)
It is required to specify a CT for O/C IDMT application, both the CT secondary and the relay rated current being 1A. The relay burden is 0.25VA at rating and the CT burden is 5VA. The loop resistance is 0.5 Ω. The maximum fault current referred to the CT secondary is 25 times the relay rating. The ranges available for composite error and ALF are:
ALF: 5, 10, 15, 20, 30 Composite error: 5P, 10P
Solution:
CT burden = RLoop + RRelay = 0.5 Ω + (0.25VA / 12) = 0.75 Ω
CT rated voltage = 5VA / 1A = 5 V
CT output voltage @ ALF = 0.75 Ω x 25 A = 18.75 V
ALF = 18.75 / 5 = 3.75 Select: ALF = 5
For O/C applications, 10% composite error is suitable. Select: 10P
CT specification: 10P5 5VA
In the following system, the generator and transformer specification are given as below:
G: | X = 26% | 66.6 MVA | 11 kV |
T1: | X = 12.5% | 75 MVA | 11/145 kV |
What is the magnitude of a three-phase fault that occurs on the 132kV line close to the 132kV busbar?
Solution:
Sbase = 100 MVA , Vbase = 132kV
(Xfault)Gen = 26 x (100/66.6) x (11/132)2 = 0.27%
(Xfault)TX = 12.5 x (100/75) x (145/132)2 = 20.1%
(Xfault)Total = 20.1 + 0.27 = 20.37% = 0.2037 p.u
Sfault = [Sbase/(Xfault)Total] = 100 MVA / 0.2037 = 491 MVA
(Ifault)3Ø = [Sfault/√3 (Vbase)] = 491 MVA / (√3 x 132kV) = 2.15 kA
Calculate the fault level for a 3Ø fault as F
With bus section CB closed
With bus section CB open
Solution:
Sbase = 0.5 MVA
(Xfault)Source = [Sbase/(Sfault)] = 0.5/250 = 0.002 p.u
Zbase = [(Vbase)2/(Sbase)] = [(11kV)2/(0.5 MVA)] = 242 Ω
Rcable = [(Rcable)Ω/(Zbase)] = 0.252/242 = 0.001 p.u
Xcable = [(Xcable)Ω/(Zbase)] = 0.0817/242 = 0.00034 p.u
With bus section closed:
(Xfault)Total = 0.001 + j0.0273 → (Xfault)Total = 0.0273 p.u → Sfault = 0.5 MVA / 0.0273 p.u = 18.3 MVA
With bus section open:
(Xfault)Total = 0.001 + j0.0523 → (Xfault)Total = 0.0523 p.u → Sfault = 0.5 MVA / 0.0523 p.u = 9.56 MVA
A power transformer 132/11kV, 10MVA, Z = 5% is connected to load at the 11kV side. Design the 11kV cable. Assume the cable is consisted of two circuits, directly buried in ground. The cable installation and site conditions are given below:
Soil temperature = 30°C
Soil Thermal resistivity = 1.5 k.m/W
Depth of burial = 1 m
Spacing = 0.3 m
Length = 50 m
Load power factor = 0.8 lagging
Solution:
Full load current at LV = 525 A
2 circuits: Each circuit = 525/2 = 262A
Derating factors:
Soil thermal resistivity = 0.92
Soil temperature = 0.96
Depth of laying = 0.98
Grouping = 0.89
Total derating factor = 0.77
Select: 3-core 95 sqmm: Rated current = 292A ; Derated current = 292 x 0.77 = 225 A < 262 A
This cable is not acceptable.
Select: 3-core 120 sqmm: Rated current = 330A ; Derated current = 330 x 0.77 = 254 A < 262 A
This cable is not acceptable.
Select: 3-core 150 sqmm: Rated current = 367A ; Derated current = 367 x 0.77 = 283 A > 262 A
This cable is acceptable.
From the tables:
Conductor fault current = 21.4 kA for 1 sec
Screen fault current = 10.1 kA for 1 sec
AC resistance = 0.16 ohm/km
AC reactance = 0.099 ohm/km
Voltage drop test
Volt drop = I x L (Rac cosϕ + Xac sinϕ) = 262 x 0.05 (0.16 x 0.8 + 0.099 x 0.6) = 2.455 V
Volt drop percentage = (2.455/11000) x 100% = 0.022% (acceptable)
Short circuit test
IK = (K x A) / √t
(K) PVC = 96.4 ; A = 150 sqmm ; t = 1 sec
IK = [(96.4 x 150) / √1] = 14460 A = 14.46 kA
Calculation of fault at the LV bus:
S fault = 10 MVA / 0.05 pu = 200 MVA
I fault = [200 MVA / (√3 x 11kV)] = 10.5 kA
14.46 KA > 10.5 kA ; This cable ia acceptable.
The earthing system in a plant / facility is very important for a few reasons, all of which are related to either the protection of people and equipment and/or the optimal operation of the electrical system. These include:
This calculation is based primarily on the guidelines provided by IEEE Std 80 (2000), “Guide for safety in AC substation grounding”.
The earthing calculation aids in the proper design of the earthing system. Using the results of this calculation, you can:
This calculation should be performed when the earthing system is being designed. It could also be done after the preliminary design has been completed to confirm that the earthing system is adequate, or highlight the need for improvement / redesign. Ideally, soil resistivity test results from the site will be available for use in touch and step potential calculations (if necessary).
The sizing of earthing conductors should always be performed, but touch and step potential calculations (per IEEE Std 80 for earth faults with a return path through remote earth) are not always necessary.
For example, when all electricity is generated on-site and the HV/MV/LV earthing systems are interconnected, then there is no need to do a touch and step potential calculation. In such a case, all earth faults would return to the source via the earthing system (notwithstanding some small leakage through earth).
However, where there are decoupled networks (e.g. long transmission lines to remote areas of the plant), then touch and step potential calculations should be performed for the remote area only.
This calculation is based on IEEE Std 80 (2000), “Guide for safety in AC substation grounding”. There are two main parts to this calculation:
IEEE Std 80 is quite descriptive, detailed and easy to follow, so only an overview will be presented here and IEEE Std 80 should be consulted for further details (although references will be given herein).
The following information is required / desirable before starting the calculation:
Determining the minimum size of the earthing grid conductors is necessary to ensure that the earthing grid will be able to withstand the maximum earth fault current. Like a normal power cable under fault, the earthing grid conductors experience an adiabatic short circuit temperature rise. However unlike a fault on a normal cable, where the limiting temperature is that which would cause permanent damage to the cable’s insulation, the temperature limit for earthing grid conductors is the melting point of the conductor. In other words, during the worst case earth fault, we don’t want the earthing grid conductors to start melting!
The minimum conductor size capable of withstanding the adiabatic temperature rise associated with an earth fault is given by re-arranging IEEE Std 80 Equation 37:
Where A is the minimum cross-sectional area of the earthing grid conductor (mm2)
ρt is the energy of the maximum earth fault (A2s)
Tm is the maximum allowable (fusing) temperature (°C)
Ta is the ambient temperature (°C)
αr is the thermal coefficient of resistivity (°C – 1)
ρr is the resistivity of the earthing conductor (μΩ.cm)
TCAP is the thermal capacity of the conductor per unit volume(Jcm – 3°C – 1)
The material constants Tm, αr, ρr and TCAP for common conductor materials can be found in IEEE Std 80 Table 1. For example. commercial hard-drawn copper has material constants:
Tm = 1084 °C
αr = 0.00381 °C – 1
ρr = 1.78 μΩ.cm
TCAP = 3.42 Jcm – 3°C – 1.
As described in IEEE Std 80 Section 11.3.1.1, there are alternative methods to formulate this equation, all of which can also be derived from first principles).
There are also additional factors that should be considered (e.g. taking into account future growth in fault levels), as discussed in IEEE Std 80 Section 11.3.3.
When electricity is generated remotely and there are no return paths for earth faults other than the earth itself, then there is a risk that earth faults can cause dangerous voltage gradients in the earth around the site of the fault (called ground potential rises). This means that someone standing near the fault can receive a dangerous electrical shock due to:
The earthing grid can be used to dissipate fault currents to remote earth and reduce the voltage gradients in the earth. The touch and step potential calculations are performed in order to assess whether the earthing grid can dissipate the fault currents so that dangerous touch and step voltages cannot exist.
The resistivity properties of the soil where the earthing grid will be laid is an important factor in determining the earthing grid’s resistance with respect to remote earth. Soils with lower resistivity lead to lower overall grid resistances and potentially smaller earthing grid configurations can be designed (i.e. that comply with safe step and touch potentials).
It is good practice to perform soil resistivity tests on the site. There are a few standard methods for measuring soil resistivity (e.g. Wenner four-pin method). A good discussion on the interpretation of soil resistivity test measurements is found in IEEE Std 80 Section 13.4.
Sometimes it isn’t possible to conduct soil resistivity tests and an estimate must suffice.
When estimating soil resistivity, it goes without saying that one should err on the side of caution and select a higher resistivity. IEEE Std 80 Table 8 gives some guidance on range of soil resistivities based on the general characteristics of the soil (i.e. wet organic soil = 10 Ω.m, moist soil = 100 Ω.m, dry soil = 1,000 Ω.m and bedrock = 10,000 Ω.m).
Applying a thin layer (0.08m – 0.15m) of high resistivity material (such as gravel, blue metal, crushed rock, etc) over the surface of the ground is commonly used to help protect against dangerous touch and step voltages. This is because the surface layer material increases the contact resistance between the soil (i.e. earth) and the feet of a person standing on it, thereby lowering the current flowing through the person in the event of a fault.
IEEE Std 80 Table 7 gives typical values for surface layer material resistivity in dry and wet conditions (e.g. 40mm crushed granite = 4,000 Ω.m (dry) and 1,200 Ω.m (wet)).
The effective resistance of a person’s feet (with respect to earth) when standing on a surface layer is not the same as the surface layer resistance because the layer is not thick enough to have uniform resistivity in all directions. A surface layer derating factor needs to be applied in order to compute the effective foot resistance (with respect to earth) in the presence of a finite thickness of surface layer material. This derating factor can be approximated by an empirical formula as per IEEE Std 80 Equation 27:
Where Cs is the surface layer derating factor
ρ is the soil resistivity (Ω.m)
ρa is the resistivity of the surface layer material (Ω.m)
hs is the thickness of the surface layer (m)
This derating factor will be used later in Step 5 when calculating the maximum allowable touch and step voltages.
A good earthing grid has low resistance (with respect to remote earth) to minimise ground potential rise (GPR) and consequently avoid dangerous touch and step voltages. Calculating the earthing grid resistance usually goes hand in hand with earthing grid design – that is, you design the earthing grid to minimise grid resistance. The earthing grid resistance mainly depends on the area taken up by the earthing grid, the total length of buried earthing conductors and the number of earthing rods / electrodes.
IEEE Std 80 offers two alternative options for calculating the earthing grid resistance (with respect to remote earth) – 1) the simplified method (Section 14.2) and 2) the Schwarz equations (Section 14.3), both of which are outlined briefly below. IEEE Std 80 also includes methods for reducing soil resistivity (in Section 14.5) and a treatment for concrete-encased earthing electrodes (in Section 14.6).
IEEE Std 80 Equation 52 gives the simplified method as modified by Sverak to include the effect of earthing grid depth:
Where Rg is the earthing grid resistance with respect to remote earth (Ω)
ρ is the soil resistivitiy (Ω.m)
Lr is the total length of buried conductors (m)
A is the total area occupied by the earthing grid (m2)
The Schwarz equations are a series of equations that are more accurate in modelling the effect of earthing rods / electrodes. The equations are found in IEEE Std 80 Equations 53, 54, 55 and 56, as follows:
Where Rg is the earthing grid resistance with respect to remote earth (Ω)
R: is the earth resistance of the grid conductors (Ω)
R2 is the earth resistance of the earthing electrodes (Ω)
Rm is the mutual earth resistance between the grid conductors and earthing electrodes (Ω)
And the grid, earthing electrode and mutual earth resistances are:
Where ρ is the soil resistivity (Ω.m)
Lc is the total length of buried grid conductors (m)
α΄ is for conductors buried at depth h metres and with cross-sectional radius r metres, or simply r for grid conductors on the surface
A is the total area covered by the grid conductors (m2)
Lr is the length of each earthing electrode (m)
nr is number of earthing electrodes in area
b is the cross-sectional radius of an earthing electrode (m)
K1 and K2 are constant coefficients depending on the geometry of the grid
The coefficient K1 can be approximated by the following:
The coefficient K2 can be approximated by the following:
Where in both cases, L/R is the length-to-width ratio of the earthing grid.
The maximum grid current is the worst case earth fault current that would flow via the earthing grid back to remote earth. To calculate the maximum grid current, you firstly need to calculate the worst case symmetrical earth fault current at the facility that would have a return path through remote earth (call this Ik,e). This can be found from the power systems studies or from manual calculation. Generally speaking, the highest relevant earth fault level will be on the primary side of the largest distribution transformer (i.e. either the terminals or the delta windings).
Not all of the earth fault current will flow back through remote earth. A portion of the earth fault current may have local return paths (e.g. local generation) or there could be alternative return paths other than remote earth (e.g. overhead earth return cables, buried pipes and cables, etc). Therefore a current division factor Sf must be applied to account for the proportion of the fault current flowing back through remote earth.
Computing the current division factor is a task that is specific to each project and the fault location and it may incorporate some subjectivity (i.e. “engineeing judgement”). In any case, IEEE Std 80 Section 15.9 has a good discussion on calculating the current division factor. In the most conservative case, a current division factor of Sf = 1 can be applied, meaning that 100% of earth fault current flows back through remote earth.
The symmetrical grid current Ig is calculated by:
The symmetrical grid current is not the maximum grid current because of asymmetry in short circuits, namely a dc current offset. This is captured by the decrement factor, which can be calculated from IEEE Std 80 Equation 79:
Where Df is the decrement factor
tf is the duration of the fault (s)
TA is the dc time offset constant (see below)
The dc time offset constant is derived from IEEE Std 80 Equation 74:
Where is the X/R ratio at the fault location
f is the system frequency (Hz)
The maximum grid current IG is lastly calculated by:
One of the goals of a safe earthing grid is to protect people against lethal electric shocks in the event of an earth fault. The magnitude of ac electric current (at 50Hz or 60Hz) that a human body can withstand is typically in the range of 60 to 100mA, when ventricular fibrillation and heart stoppage can occur. The duration of an electric shock also contributes to the risk of mortality, so the speed at which faults are cleared is also vital. Given this, we need to prescribe maximum tolerable limits for touch and step voltages that do not lead to lethal shocks.
The maximum tolerable voltages for step and touch scenarios can be calculated empirically from IEEE Std Section 8.3 for body weights of 50kg and 70kg:
Touch voltage limit – the maximum potential difference between the surface potential and the potential of an earthed conducting structure during a fault (due to ground potential rise):
Step voltage limit – is the maximum difference in surface potential experience by a person bridging a distance of 1m with the feet without contact to any earthed object:
Where Etouch is the touch voltage limit (V)
Estep is the step voltage limit (V)
Cs is the surface layer derating factor (as calculated in Step 2)
ρs is the soil resistivity (Ω.m)
ts is the maximum fault clearing time (s)
The choice of body weight (50kg or 70kg) depends on the expected weight of the personnel at the site. Typically, where women are expected to be on site, the conservative option is to choose 50kg.
Normally, the potential difference between the local earth around the site and remote earth is considered to be zero (i.e. they are at the same potential). However an earth fault (where the fault current flows back through remote earth), the flow of current through the earth causes local potential gradients in and around the site. The maximum potential difference between the site and remote earth is known as the ground potential rise (GPR). It is important to note that this is a maximum potential difference and that earth potentials around the site will vary relative to the point of fault.
The maximum GPR is calculated by:
Where GPR is the maximum ground potential rise (V)
IG is the maximum grid current found earlier in Step 4 (A)
Rg is the earthing grid resistance found earlier in Step 3 (Ω)
Now we just need to verify that the earthing grid design is safe for touch and step potential. If the maximum GPR calculated above does not exceed either of the touch and step voltage limits (from Step 5), then the grid design is safe.
However if it does exceed the touch and step voltage limits, then some further analysis is required to verify the design, namely the calculation of the maximum mesh and step voltages as per IEEE Std 80 Section 16.5.
The mesh voltage is the maximum touch voltage within a mesh of an earthing grid and is derived from IEEE Std 80 Equation 80:
Where :: ρs is the soil resistivity (Ω.m)
IG is the maximum grid current found earlier in Step 4 (A)
Km is the geometric spacing factor (see below)
Ki is the irregularity factor (see below)
LM is the effective buried length of the grid (see below)
Geometric Spacing Factor Km
The geometric spacing factor Km is calculated from IEEE Std 80 Equation 81:
Where D is the spacing between parallel grid conductors (m)
h is the depth of buried grid conductors (m)
d is the cross-sectional diameter of a grid conductor (m)
Kh is a weighting factor for depth of burial =
Kii is a weighting factor for earth electrodes /rods on the corner mesh
Kii = 1 for grids with earth electrodes along the grid perimeter or corners
for grids with no earth electrodes on the corners or on the perimeter n is a geometric factor (see below)
Geometric Factor n
The geometric factor n is calculated from IEEE Std 80 Equation 85:
With
nb = 1 for square grids, or otherwise
ne = 1 for square and rectangular grids, or otherwise
nc = 1 for square, rectangular and L-shaped grids, or otherwise
Where Lc is the total length of horizontal grid conductors (m)
Lp is the length of grid conductors on the perimeter (m)
A is the total area of the grid (m2)
Lx and Ly are the maximum length of the grids in the x and y directions (m)
Dm is the maximum distance between any two points on the grid (m)
Irregularity Factor Ki
The irregularity factor Ki is calculated from IEEE Std 80 Equation 89:
Where n is the geometric factor derived above
Effective Buried Length LM
The effective buried length LM is found as follows:
Where Lc is the total length of horizontal grid conductors (m)
LR is the total length of earthing electrodes / rods (m)
Where Lc is the total length of horizontal grid conductors (m)
LR is the total length of earthing electrodes / rods (m)
Lr is the length of each earthing electrode / rod (m)
Lx and Ly are the maximum length of the grids in the x and y directions (m)
Step Voltage Calculation
The maximum allowable step voltage is calculated from IEEE Std 80 Equation 92:
Where :: ρs is the soil resistivity (Ω.m)
IG is the maximum grid current found earlier in Step 4 (A)
Ks is the geometric spacing factor (see below)
Ki is the irregularity factor (as derived above in the mesh voltage calculation)
Ls is the effective buried length of the grid (see below)
Geometric Spacing Factor Ks
The geometric spacing factor Ks based on IEEE Std 80 Equation 81 is applicable for burial depths between 0.25m and 2.5m:
Where D is the spacing between parallel grid conductors (m)
h is the depth of buried grid conductors (m)
n is a geometric factor (as derived above in the mesh voltage calculation)
Effective Buried Length LS
The effective buried length Ls for all cases can be calculated by IEEE Std 80 Equation 93:
Where Lc is the total length of horizontal grid conductors (m)
LR is the total length of earthing electrodes / rods (m)
What Now?
Now that the mesh and step voltages are calculated, compare them to the maximum tolerable touch and step voltages respectively. If:
then the earthing grid design is safe.
If not, however, then further work needs to be done. Some of the things that can be done to make the earthing grid design safe:
In this example, the touch and step potential calculations for an earthing grid design will be performed. The proposed site is a small industrial facility with a network connection via a transmission line and a delta-wye connected transformer.
The soil resistivity around the site was measured with a Wenner four-pin probe and found to be approximately 300 Ω.m.
A thin 100mm layer of blue metal (3,000 Ω.m) is proposed to be installed on the site. The surface layer derating factor is:
Step 3: Earthing Grid Resistance
A rectangular earthing grid (see the figure right) with the following parameters is proposed:
Using the simplified equation, the resistance of the earthing grid with respect to remote earth is:
Suppose that the maximum single phase to earth fault at the HV winding of the transformer is 3.1kA and that the current division factor is 1 (all the fault current flows back to remote earth).
The X/R ratio at the fault is approximately 15, the maximum fault duration 150ms and the system nominal frequency is 50Hz. The DC time offset is therefore:
The decrement factor is then:
Finally, the maximum grid current is:
Based on the average weight of the workers on the site, a body weight of 70kg is assumed for the maximum touch and step potential. A maximum fault clearing time of 150ms is also assumed.
The maximum allowable touch potential is:
The maximum allowable step potential is:
The maximum ground potential rise is:
The GPR far exceeds the maximum allowable touch and step potentials, and further analysis of mesh and step voltages need to be performed.
Mesh Voltage Calculation
The components of the geometric factor na, nb, nc and nd for the rectangular grid are:
Therefore the geometric factor n is:
The average spacing between parallel grid conductors D is:
where We and Lg are the width and length of the grid respectively (e.g. 50m and 90m) nr and nc is the number of parallel rows and columns respectively (e.g. 6 and 7)
The geometric spacing factor Km is:
The irregularity factor Ki is:
The effective buried length LM is:
Finally, the maximum mesh voltage is:
The maximum allowable touch potential is 1,720V, which exceeds the mesh voltage calculated above and the earthing system passes the touch potential criteria (although it is quite marginal).
The geometric spacing factor Ks is:
The effective buried length Ls is:
Finally, the maximum allowable step voltage is:
The maximum allowable step potential is 5,664V, which exceeds the step voltage calculated above and the earthing system passes the step potential criteria. Having passed both touch and step potential criteria, we can conclude that the earthing system is safe.
As can be seen from above, touch and step potential calculations can be quite a tedious and laborious task, and one that could conceivably be done much quicker by a computer. Even IEEE Std 80 recommends the use of computer software to calculate grid resistances, and mesh and step voltages, and also to create potential gradient visualisations of the site.
Computer software packages can be used to assist in earthing grid design by modeling and simulation of different earthing grid configurations. The tools either come as standalone packages or plug-in modules to power system analysis software (such as PTW’s GroundMat or ETAP’s Ground Grid Design Assessment. Examples of standalone packages include SES Autogrid and SafeGrid.
A ground rod of 40 mm diameter is driven into soil to a depth of 250 cm. Calculate the resistance of the soil surrounding this rod at distances of 10 cm, 15 cm, 20 cm, 40 cm, 60 cm, 100 cm, 200 cm, 250 cm, 300 cm, 500cm and 800 cm from the rod. Soil resistivity can be taken as 100 ohm meter. Express the values as % of the resistance at 800 cm.
Hint: Assume soil shells (slices) of thickness 5 cm up to 20 cm distance; 10 cm between 20 and 100 cm distance and 50 cm after 100 cm distance. Area may be calculated using the inner face of the shell. Make a tabular form for recording the resistance of successive shells and calculate cumulative resistance values as you go along. Calculate the resistance of the shells of soil around an earth electrode. Solve using the principle illustrated in Figure 1.1 below and the formula
Where
R is the Resistance of the shell in Ohms
L is the thickness of the shell in meters
A is the inner surface area of the shell in sq. meters
And is the soil resistivity in ohm meters
Given that:
Ground rod length is 2.5 m
Soil resistivity is 100 Ohm m
Diameter of electrode is 40 mm (4 cm)
Step 1 Divide the soil around the electrode into a number of slices.
The first slice is from the periphery of rod to 5 cm from the center
The next slice is 5 to 10 cm and so on
Step 2 The inner surface area
This can be calculated as the sum of the concentric cylindrical surface and the hemispherical bottom surface.
Cylindrical surface area for the first slice can be calculated using the formula
Area = 2 * π * Radius of the cylinder * Length
Substituting
Hemispherical area can be calculated using the formula
Area = 2 * π * Radius * Radius
Total surface area is the sum of A1 and A2
A = 0.317 sq. m
Resistance of the shell can be calculated using the formula
Where
R is the Resistance of the shell in Ohms
L is the thickness of the shell in meters
A is the inner surface area of the shell in sq. meters
And ρ is the soil resistivity in ohm meters (100 in this case)
Substituting these values
Step 3
Repeat this for subsequent slices and tabulate the results.
Step 4
The slice thickness given in the hint under the problem may be used for calculation.
Step 5
Calculate cumulative resistance using this table.
Step 6
Express the cumulative resistance as % of resistance calculated for soil mass up to a radius of 8 m.
Note
It would be of interest to compare the values obtained with the values given in Table below.
The values calculated using a spreadsheet program are shown below and can be compared with manually obtained values.
Radial distance cm | Surface area sq.m | Resistance of slice Ohm | Cum. Resistance Ohm | % of Cum. Res to value at 8 m |
---|---|---|---|---|
2 | ||||
5 | 0.32 | 9.47 | 9.47 | 24.80 |
10 | 0.80 | 6.24 | 15.71 | 41.15 |
15 | 1.63 | 3.06 | 18.77 | 49.16 |
20 | 2.50 | 2.00 | 20.77 | 54.40 |
30 | 3.39 | 2.95 | 23.72 | 62.12 |
40 | 5.28 | 1.89 | 25.61 | 67.08 |
50 | 7.29 | 1.37 | 26.98 | 70.67 |
60 | 9.43 | 1.06 | 28.04 | 73.45 |
70 | 11.69 | 0.86 | 28.90 | 75.69 |
80 | 14.08 | 0.71 | 29.61 | 77.55 |
90 | 16.59 | 0.60 | 30.21 | 79.13 |
100 | 19.23 | 0.52 | 30.73 | 80.49 |
150 | 22.00 | 2.27 | 33.00 | 86.44 |
200 | 37.71 | 1.33 | 34.33 | 89.92 |
250 | 56.57 | 0.88 | 35.21 | 92.23 |
300 | 78.57 | 0.64 | 35.85 | 93.90 |
350 | 103.71 | 0.48 | 36.33 | 95.16 |
400 | 132.00 | 0.38 | 36.71 | 96.15 |
450 | 163.43 | 0.31 | 37.01 | 96.95 |
500 | 198.00 | 0.25 | 37.27 | 97.62 |
550 | 235.71 | 0.21 | 37.48 | 98.17 |
600 | 276.57 | 0.18 | 37.66 | 98.64 |
650 | 320.57 | 0.16 | 37.82 | 99.05 |
700 | 367.71 | 0.14 | 37.95 | 99.41 |
750 | 418.00 | 0.12 | 38.07 | 99.72 |
800 | 471.43 | 0.11 | 38.18 | 100.00 |
A driven rod electrode of made of 40 mm diameter GI pipe is buried to a depth of 3m in a soil of resistivity 50 ohm m. Find the resistance of the ground rod. In order to improve the resistance, 12 such rods are used in a rectangular array and are bonded together. Calculate the overall ground resistance of this array.
The resistance of a ground electrode can be calculated once the soil resistivity is known. For a rod driven vertically into ground the electrode resistance is given by the following formula.
where
R is the resistance of the Electrode in Ohms
ρ is the soil resistivity in Ohm meters
L is the length of the buried part of the electrode in meters and
D is the outer diameter of the rod in meters
A simplified formula for an electrode of 5/8” (16mm) diameter driven 10’ (3m) into the ground is:
where
R is the resistance of the Electrode in Ohms and
ρ is the soil resistivity in Ohm meters
Step 1 Arrive at Resistance of a single electrode
Given:
Soil resistivity (ρ) is 50 Ohm m
Electrode dia (d) is 40 mm (0.04 m)
Length L is 3 m
Substituting in the formula
where
R is the resistance of the Electrode in Ohms
ρ is the soil resistivity in Ohm meters
L is the length of the buried part of the electrode in meters and
D is the outer diameter of the rod in meters
R can be calculated as 14.31Ohms
Step 2 Resistance of parallel combination
Given that there are 12 electrodes connected together. The resistance of parallel combination is R*F/N where N = 12 and F can be arrived at from the Table given in Fig 2.1. The table in Table 2.1 shows the value of the factor F in the formula.
No. of Rods | F |
---|---|
2 | 1.16 |
3 | 1.29 |
4 | 1.36 |
8 | 1.68 |
12 | 1.80 |
16 | 1.92 |
20 | 2.00 |
24 | 2.16 |
For N=12 F is 1.8
Substituting
Resistance of the combination is 14.31*1.8/12
Result = 2.147 Ohms
Calculate for the above example the ground fault current that can be safely dissipated into the ground for the following fault clearance times.
To calculate the current carrying capacity for individual electrodes use the formula:
where
I is the maximum permissible current in Amperes
d is the outer diameter of the rod in meters
L is the length of the buried part of the electrode in meters and
ρ is the soil resistivity in Ohm meters and
t is the time of the fault current flow in seconds
Given that
Soil resistivity (ρ) is 50 Ohm m
Electrode dia (d) is 40 mm (0.04 m)
Length L is 3 m
Time t has different values 0.1. Sec. 0.5 sec. 1 sec and 5 sec.
Substituting in the formula, safe current for a single electrode is:
Total current for 12 electrodes can be taken as 12 * I
The values can be calculated as
I for 0.1 Sec. | = | 22410 amps |
I for 0.5 Sec. | = | 10022 Amps |
I for 1 Sec. | = | 7086 Amps |
I for 5 Sec. | = | 3169 Amps |
A test using the 4-pin method is conducted to determine the soil resistivity at a substation site. The test data are as follows.
Spike length | 150 cm |
Spike spacing | 40 m |
Soil temperature during test | 30 deg. C |
Soil moisture during test | 20% |
Resistance value | 2 .6 ohm |
Calculate the soil resistivity under test conditions. Also calculate the resistivity at soil moisture content of 12% and soil temperature of 10 Deg. C.
Hint: Consider that the soil type as typical top soil. Use tables 4.1 and 4.2 given below.
Moisture content % | Resistivity in Ohm M | ||
---|---|---|---|
Top soil | Sandy loam | Red Clay | |
2 | *** | 1850 | *** |
4 | *** | 600 | *** |
6 | 1350 | 380 | *** |
8 | 900 | 280 | *** |
10 | 600 | 220 | *** |
12 | 360 | 170 | 1800 |
14 | 250 | 140 | 550 |
16 | 200 | 120 | 200 |
18 | 150 | 100 | 140 |
20 | 120 | 90 | 100 |
22 | 100 | 80 | 90 |
24 | 100 | 70 | 80 |
Temperature Deg. C | Resistivity Ohm M |
---|---|
-5 | 700 |
0 | 300 |
0 | 100 |
10 | 80 |
20 | 70 |
30 | 60 |
40 | 50 |
50 | 40 |
Answer:
Soil resistivity can be measured using a ground resistance tester or other similar instruments using Wenner’s 4-pin method as shown in the Figure 4.1 below. The two outer pins are used to inject current into the ground (called current electrodes) and the potential developed as a result of this current flow is measured by the two inner pins (potential electrodes).
Soil resistivity is given by the equation:
Where:
= Soil resistivity
S = Spacing between test spikes
R = Measured value of soil resistance
Calculating the soil resistivity under the test conditions of 30°C & 20% moisture.
Calculating the soil resistivity with a moisture content of 12%.
Assuming the soil type is top soil, the data in the Table 4.1 indicates that with all other factors remaining constant, when the moisture decreases from 20% to 12%, the soil resistivity will increase by a facter of three. Applying this to the calculated resistivity above, the soil resistivity at 12% moisture is:
Calculating the soil resistivity with a moisture content of 12% and temperature 10 deg C. Referring to Table 4.2 it is seen that the resistivity increases with decrease in temperature. Using the data in the Table the soil resistivity with moisture content of 12% and temperature of 10 deg C,
= (1960.354) x 80/60
= 2613.80 Ώm
This case study refers to a project of the construction of a 132/33kv transmission zone substation in an arid desert region for the design of a suitable earthing system for the substation. Soil testing in laboratory is conducted and soil resistivity measurements are conducted at site in order to assess the quality of the soil and take necessary steps towards a suitable design for the earthing system of the substation. The following data is the data obtained from the testing activities.
Soil resistance measurement data at various locations at the site and resistivity calculation
Note – With 1 mtr spacing, we are not getting any values
It is observed from the earth resistance measurements at site that the soil resistivity is high of the order of several hundred ohms. It is also observed from the lab test report that the major proportion of the soil is made up of coarse particles and consisting of cobbles and gravel. The moisture content of the soil is also found to be very low of the order of 1.9%. As a result the, electrical resistance offered by the soil is high necessitating soil treatment. As such the soil condition is not ideally suited for earthing purposes in the proposed substation, without taking recourse to some corrective actions to improve the soil electrical resistivity conditions.
The soil is treated with Bentonite clay in order to reduce the electrical resistivity of the soil and improve electrical earthing conditions. Bentonite clay is a natural clay of volcanic origin and has the desirable properties of absorbing and retaining moisture for a very long time. In the event of higher temperatures, only the outer portion of the clay dries out retaining the internal moisture. In addition, the clay also does not cause any corrosion problems to the earth electrodes unlike salts, that are some times used for improving earth electrical resistance. The rocky soil was excaved and replaced by bentonite clay. After the bentonite clay treatment and initial watering, the soil resistivity dropped down to less than 10 ohm.metre. The number of vertical electrodes of 3 metres length was also installed in adequate numbers to reduce the electrode resistance and also to ensure that the bottom of the electrodes made contact with the underground water table.
The UK implementation of Cenelec Standard EN 61936-1:2010 incorporating CENELEC corrigenda March 2011 from IEC 61936-1:2010, incorporating IEC corrigendum March 2010 is BS EN 61936-1:2010 .
BS EN 50522:2010. Earthing of power installations exceeding 1 kV a.c. is the complementary standard specifically with regard to the earthing of these installations.
BS EN 61936-1:2010 together with BS EN 50522:2010 replace BS 7354:1990 which has been withdrawn.
The differences in the earthing design in the UK approach as compared with IEC/CENELEC Standards are as follows:
“Part 1: Common rules” of the Standard provides common rules for the design and the erection of electrical power installations in systems with nominal voltages above 1 kV a.c. and nominal frequency up to and including 60 Hz, so as to provide safety and proper functioning for the intended use.
Section 1 of the Standard defines the scope of the Standard.
For the purpose of interpreting the standard, an electrical power installation is considered to be one of the following:
The electrical power installation includes, among others, the following equipment:
The Standard does not apply to the design and erection of any of the following:
The standard does not apply to the design of factory-built, type-tested switchgear for which separate IEC standards exist. This standard does not apply to the requirements for carrying out live working on electrical installations. If not otherwise required in this standard, for low-voltage electrical installations the standard series IEC 60364 applies.
Section 2 of the BS EN 61936-1 2010 Standard lists the normative reference Standards for the application of the Standard.
Section 3 of the Standard lists the terms and definitions as applied to the Standard.
Section 4 of the Standard specifies the fundamental requirements. Section 4.1 specifies the general requirements. Following are the general requirements:
Preferences for specific maintenance features and identification of the safety requirements to be met are defined by the user. Where, specific design criteria is required, these may be agreed between the user and the manufacturer. The design of the installation shall take into account the working procedures of the user. Additional agreements between the manufacturer/ contractor/ planner and user/ orderer/ owner shall be followed for the design and erection of power installations.
Section 4.2 deals with the following electrical requirements for the installation:
Section 4.3 specifies the following mechanical requirements.
The equipment and supporting structures, foundations shall withstand the anticipated mechanical stresses under normal and exceptional load conditions. Normal loads consist of dead load, tension load, erection load, ice load and wind load. Exceptional loads relate to the simultaneous acting of the dead load and tension load with the largest of occasional loads such as switching forces, short-circuit forces, vibration and loss of conductor tension.
Section 4.4 specifies requirements of the installation for climatic and environmental conditions. Installations, including the devices and equipment in the installations shall be designed for operation under the climatic and environmental conditions and the selection based on specification of factors such as the presence of:
For the selection of a suitable indoor or outdoor equipment for the installation, the following factors are considered:
Special conditions related to altitude, pollution levels, temperature and humidity, vibration shall also be factored into the selection process.
Section 4.5 deals with any requirement of special requirements related to the following:
Section 5 deals with insulation requirements of the equipment and installation and the selection of insulation level. These requirements include minimum clearances to be maintained between live parts and earth and clearances between live parts of phases to avoid flashovers. The section also discusses about the selection of insulation level according to the established highest voltage for installation Um and/or impulse withstand voltage. The factors of methods of neutral earthing and rated withstand values are taken into consideration. The minimum clearances in air are provided in Table 1, Table 2 and Annex A of the Standard and apply for altitudes up to 1 000 m above sea level.
Section 6 specifies the requirements of the equipment in the installation. The general requirements of equipment are as follows:
Section 7 deals with installations comprising the following:
Section 8 relates to safety measures. The topics covered under this section are:
Section 9 provides details of protection, control and auxiliary systems. This section comprises of the following sections:
Section 10 pertains to earthing systems.
General – This clause provides the criteria for design, installation, testing and maintenance of an earthing system for ensuring the safety of human life and that the integrity of equipment connected and in proximity to the earthing system is maintained.
Safety criteria – Flow of current into the body of a human in the region of the heart can result in ventricular fibrillation. The curve for obtaining the current limit is provided in IEC/TS 60479-1:2005. This body current limit when converted to voltage limits is used for comparing with the calculated touch and step voltages in the design of the earthing system taking into consideration the following factors:
The fault occurrence, fault current magnitude, fault duration and presence of human beings are probabilistic in nature. Generally, meeting the requirements of touch voltage satisfies the step voltage requirements, since the tolerable step voltage limits are much higher than touch voltage limits due to the different current path through the body.
For installations where high-voltage equipment is not located in closed electrical operating areas, a global earthing system should be used to prevent touch voltages resulting from HV faults exceeding the low voltage limit.
Following are the various functional requirements of the earthing system.
The earthing system:
Where high- and low-voltage earthing systems exist in proximity to each other and do not form a global earthing system, part of the EPR from the HV system can be applied on the LV system either by interconnection of all HV with LV earthing systems or by separation of HV from LV earthing systems. However, compliance shall be met with reference to the requirements of step, touch and transfer potentials within a substation and at LV installation supplied from that substation.
In locations where the LV system is totally confined within the area covered by the HV earthing system, both earthing systems shall be interconnected even if there is no global earthing system. In locations where the earthing system of the HV installation is part of a global earthing system or connected to a multi-earthed HV neutral conductor in a balanced system, then full compliance is ensured. In locations not having a global earthing system, the minimum requirements for interconnection of low-voltage and high-voltage earthing systems based on EPR limits specified in the Standard shall be used to identify those situations where interconnection of earthing systems with low-voltage supply outside the high-voltage installation is feasible.
If high-voltage and low-voltage earthing systems are separate, the method of separating earth electrodes shall be chosen such that no danger to persons or equipment can occur in the low voltage installation. This means that step, touch and transfer potentials and stress voltage in the LV installation caused by a high-voltage fault are within the appropriate limits.
In locations where the LV systems are located in the zone of influence of the HV substation earthing, special consideration should be given. Common earthing system can be used for industrial and commercial installations since it is not possible to separate earthing systems due to the close proximity of equipment.
Section 10.3 discusses about the design of earthing systems.
The steps in the designing of an earthing system are as follows:
After the above criteria are satisfied, the design can be refined, if necessary, by repeating the above steps. Detailed design is necessary to ensure that all exposed conductive parts, are earthed. Bonding should be made between the earthing system and the structural earth electrodes. Metallic structures with cathodic protection may be separated from the earthing system.
The various following types of earth faults at the different voltage levels, within and outside the installation should be analysed to determine the worst case fault situation. Power system faults:
High voltage and current surges occur due to lightning and switching operations on long cable sections, operating GIS disconnectors or carrying out back-to-back capacitor switching. The earthing system should also include the earthing for the lightning protection system.
Where more than one building or location are involved, interconnection should be made for the earthing system of each. Protection measures must be taken to prevent damages to sensitive equipment connected between different buildings or locations during lightning strokes. Non-metallic media, such as fibre optic cable, should be used where possible, for the exchange of low-level signals between such locations.
Section 10.4 requires that protective measures shall be taken to ensure the safety of persons during fault conditions where construction work involves an existing earthing system.
Section 10.5 is on measurements on earthing systems. The adequacy of the earthing design can be verified through measurements after the construction. These measurements may include earthing system impedance, prospective touch and step voltages at relevant locations and transferred potential, if appropriate.
Section 10.6 is about maintainability of earthing systems. Maintainability of earthing systems comprise of the following:
Section 11 deals with inspection and testing. This section of the Standard discusses about the need and requirement for inspection and testing of, equipment with technical specifications and of installations for compliance with the Standard. The points of inspection and testing should be agreed between the user and the supplier of equipment. The inspections and testing comprise of:
Section 12 deals with operation and maintenance manual.
Each installation should have:
Manufacturers of major components of installation should provide operation and maintenance manuals and test and in-service reports. Emergency routes to the nearest hospital and emergency phone numbers should be displayed in a visible location in the installation.
Annexure-A of BS EN 61936-1 2010 provides values of rated insulation levels and minimum clearances for various levels of voltages, based on current practice in some countries.
Annexure-B of BS EN 61936-1 2010 provides the method of calculating permissible touch voltages.
Annexure-C of BS EN 61936-1 2010 provides details on the Permissible touch voltage according IEEE 80.
Annexure-D of BS EN 61936-1 2010 provides a Earthing system design flow chart.
Annexure-D of BS EN 61936-1 2010 provides the Protection measures against direct lightning strokes. Following methods have been listed as providing sufficient protection level for avoiding direct lightning strokes with a high degree of certainty but without detailed studies of insulation coordination.
Annexure-ZA of BS EN 61936-1 2010 provides special national conditions (National characteristic or practice that cannot be changed even over a long period, e.g. climatic conditions, electrical earthing conditions). For the countries in which the relevant special national conditions apply these provisions are normative, for other countries they are informative.
Annexure-ZB of BS EN 61936-1 2010 provides A-deviations which are National deviation due to regulations, the alteration of which is for the time being outside the competence of the CENELEC national member. This European Standard does not fall under any Directive of the EC. In the relevant CENELEC countries these A-deviations are valid instead of the provisions of the European Standard until they have been removed.
Annexure-ZC of BS EN 61936-1 2010 provides Normative references to international publication-96 with their corresponding European publications. The listed referenced documents are indispensable for the application of BS EN 61936-1 2010.
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