ECNG 3015Industrial and Commercial Electrical Systems Lecturer Prof Chandrabhan Sharma #7 System Earthing
SYSTEM EARTHINGReason for earthing: - To provide means to carry electric currents to earth under normal and fault conditions without exceeding any operating and equipment limits or adversely affecting continuity of service. - To ensure that a person in the vicinity of the grounded facilities is not exposed to the danger of critical electric shock.
Earth resistivity: - measured either Ω.m - affected by moisture content of the soil → since conduction in soil is electrolytic in natureSoil resistivity falls sharply when moisture content falls below 22 % byweight. Electrodes should be buried deep enough - to ensure that the moisture content of the surrounding soil does not vary from season to season. - and have permanent contact with moist soil
The Human factor - normal, healthy person can feel 1 mA - 10-25 mA can cause loss of muscular control - 100 mA can result in ventricular fibrillation (death), burns etc.Note:For practical grounding studies, the threshold of ventricular fibrillation isthe major concern, i.e. most grounding mats are designed to limit the currentthrough the average human body to below threshold.
Studies have shown that 99.5 % of healthy persons can tolerate acurrent through the heart without going into V.F. given by 0.116 Ib ....................................(1) T where I b maximumbody currentin A T - is thedurat ionof current in secondsNote:Large currents can be accommodated once ‘T’ is kept below the value given byequation 1.
STEP AND TOUCH VOLTAGESThe body can tolerate more current flowing from one leg to theother than it can from one hand to the legs.TOUCH VOLTAGE: (refer to the equivalent circuit below.) Ib Rg Ig earth rod resistivity of the soil must be very low to ensure that most of the current flows to ground.
The circuit below shows the hazards due to touch voltages.The body current Ib is driven by the potential between A and B.The worst possible touch potential (mesh potential) occurs at ornear to the centre of a grid mesh.
STEP VOLTAGE: (refer to the equivalent circuit below.) Here, current flows from one foot to another due to the potential difference between A and B. Hence the step voltage is the voltage which a man would receive across the body by taking a 1m step in a radial direction from the earth electrode.
In both cases; Rb = 1000 Ω (IEEE Std 80 -2000) Rf = 3 s where s – soil resistivity and treating the foot as a circular plate electrodeActually, the body itself has a total measured resistance of 2300 Ω hand tohand and 1100 Ω hand to foot.From the equivalent circuit for touch voltage, the effectiveresistance under consideration for hand to foot would be: Rf but R b 1000 and Rf 3 R effective,Touch Rb s 2 R effective,Touch 1000 1.5 s
And for foot to foot currents: R effective,Step R b 2R f again, R b 1000 and Rf 3 s R effective,Step 1000 6 s Applying the above two equations to equation (1), we have: (0.116)(10 1.5 s ) 116 0.17 00 s E Touch max I b (1000 1.5 s ) T T (0.116)(10 6 s ) 00 116 0.7 s and EStep max I b (1000 6 s ) T TThe grid voltage depends upon three basic factors: 1. Ground resistivity 2. Available fault current 3. Grid geometry
1. Ground Resistivity: Type of ground Resistivity (Ωm) Wet organic soil 10 Moist soil 102 Dry soil 103 Bed rock 104 Eelctricfield (V/l ) RA Volume Resistivity m Current Density(I/A) l Where R resistance A cross- sectionarea and l length
2. Available fault current:The maximum ground fault current (S.L.G) is given by : 3V I " 3Rg (R1 R 2 R 0 ) j(X1 X 2 X0 ) Where V phase voltage R g grid resistanceto earth R1 , R 2 , R 0 positive,negativeand zero seq. resistance " X1 positiveseq. subtransient reactance and X 2 , X 0 negativeand zero seq. reactance
To account for the initial d.c. offset of the fault current, I ismultiplied by an appropriate correction factor known as thedecrement factor. Shock Cycles (60 Hz) Decrement factor Duration (s) Fault duration 0.008 1/2 1.65 0.1 6 1.25 0.25 15 1.10 5 30 1.00 Shock and Fault Duration
A more accurate value of the decrement factor (D) is given by -2 T 1 1 X X R D T . 1 e T R Where T durationof fault (s) (obtained from relay setting) system frequency in radians/s X totalsystem reactance R totalsystem reactance
3. Grid Geometry:The potential rise of points protected by a grounding mat dependson such factors as: a. grid burial depth b. length and diameter of conductor c. space between each conductor d. distribution of current throughout the grid e. proximity of the fault electrode and the system grounding electrodes to the grid conductors
EARTH ELECTRODES:1. Hemisphere electrode at the earth surfaceFor a hemisphere of radius ‘r’ from which a total current ‘I’ spreadsout in a radial direction into the earth: I Current density (i ) at a distance ( x) 2πx 2
The voltage drop at distance x across a hemispherical shell of thickness dx is given by: I Voltage drop (dv) i dx 2 dx (V) 2πx dv I Potential gradient e x (V/m) dx 2πx 2Vx = voltage drop from ‘r’ to ‘x’ x I I 1 1i.e. Vx 2 dx Vx r 2πx 2π r xVoltage at t heelectrodesurface w.r.t remoteearthpot ential I (x ) E0 2πr
I and at distance x Ex 2π x E0 The resistance R at the electrode surface is R I 2π rQuestion:For a hemispherical electrode, r = 0.5 m, = 10 .m andI = 100 A. Find the resistance at the electrode surface and thepotential gradient at distance 0.6m from the electrode surface. Soln : a. T heresistanceat t heelect rodesurface is R 2π r 10 3.18 2π (0.5)
Ib. P otentialgradient,e x 2 , 2πx(where x is thehorizontal distancefrom thecentreof theh/sphere) 10 x 100 131.5 V/m 2π(0.5 0.6) 2 If 50 m th e x en 657.6V/m E step is fatal
2. Spherical electrodeHemisphere electrode at the earth surface with insulated lead wireburied at depth, ‘h’ having radius ‘r’. Case 1. ‘h’ very large: The area of the sphere is twice that of the hemisphere so all the previous equations would be halved. I Potential gradient, e x 4πx 2 and potential thesurface of thesphereand at distance x is at I I E0 and E x respectively 4πr 4πx
E0AND the resistance R at the electrode surface is R I 4π rCase 2. ‘h’ finite – h>> rThe non-uniformity due to the earths surface may be eliminated byconsidering identical and equal current flowing from an imageelectrode a distance ‘h’ above the earth surface.See diagram on next page.
T hepot ent ial point P1 is due t o bot h elect rodes: at I 1 1 i.e. E P1 4π x1 xFor a point on the surface of the electrode: x1 = r , x’ = 2h I 1 1 potential thesurface E 0 at 4π r 2h I r E0 1 4π r 2h E0 r But R 1 I 4π r 2h
For a point P2 on t heEart hSurface : I 2 i.e. E P2 4π x2 h2 Potential gradient at thesurface: dE P2 I 2x e P2 3 dx 4π (x2 2 2 h )and when x 0 thepotential theearthsurface verticall aboveis : at y I 2 EV 4π h
thepotential between th surface of thesphereand e theearthsurface verticall aboveit is (T ouchVoltage): y I 1 3 E0 - EV 4π r 2h Touch voltage increases with increase in ‘h’Similarly,it can be shown thatwhen x 0.6 themaximum voltageis : I 2 e max 2 i.e. (Step Voltage) 4π (1.6h) Step voltage decreases with depth of electrode.
3. Rod electrodeA rod or pipe in which the length of the rod (l ) >> the diameterof the rod (d) can be broken up into ‘n’ spherical electrodes. For therod theearthresistanceis : 4l R ln 2π l d d If theradius of thesphere then the 2 l diameteris d and thenumber of spheresn d
I also ex 2π x(l 2 x2 ) 2xand E 0 E x 0.366 i log (Step Voltage) dwhere i current/le ngth x horizontal distanceon theearth surface from therod s
EARTH MAT DESIGNKnowing the tolerable step and touch voltage limits, we can nowdesign and construct the ground system for a substation using astep by step procedure as follows:1. Determine the soil characteristics (resistivity).2. Determine the maximum possible fault current (SLG).3. Prepare preliminary design of grounding system.4. Calculate the resistance of the designed grounding system in (3).5.Calculate the maximum grid potential rise.6.Calculate the step voltages at periphery of the mat.
7. Calculate the internal step and touch voltages.8. Refine the design so as to satisfy 5, 6 and 7 for safety.9. Construct the mat as per design (make sure measured values compares favourably with the calculated values)10. Review 5, 6, 7, 8 based on actual measurement and modify ( additional electrodes, screens, barriers etc.).
Step 3 – Preliminary grounding designHaving obtained (1) and (2), we can now do a grounding design.Assumption – the grounding system will take the form of a grid of horizontally buried conductors3.1 Determination of conductor size on the grid:This can be estimated by the equation below; Tm Ta log10 1 234 Ta I A 33s
whereI currentin AmpsA coppercross- sectionarea (circular mills or mm2 )s timein secondsduring which I is appliedTm maximumallowable temp. C inTa ambient te in C mp.1974circular mills 1 mm2
3.2 Determination of conductor length required for gradientcontrolIn this part of the analysis, Touch voltages are used for thebasis of analysisAlso, Touch voltages from a grounded structure to the centerof a rectangle of grid mesh are used instead of touch voltagesat a horizontal distance of 1 m.Laurent developed the following approximate equations Estep = 0.1 0.5 i Etouch = 0.6 0.8 i Emesh = i
Where Estep = step voltage over a horizontal distance of 1 m Etouch = touch voltage over a horizontal distance of 1 m from the grid conductor Emesh = potential difference in volts, from grid conductor to ground surface at center of a grid mesh = soil resistivity (Ωm) i = current in amp (per m of buried conductor), flowing into the ground
In order to allow for the non-uniformity in flow of ground current per unit length of buried conductor we write: IE mesh k mki Lwhere km coefficien which ta intoaccount th physicalparameters t kes e of thegrid D, d, h ki an irregularity correctionfactor averageresistivit ( m) y I max.totalr.m.scurrentflowing between ground grid and earth [adjust for future growth] L totallengthof buried conductor(m)
For safe touch vol tages we must have: 116 0.17 s I E mesh k mki t L k mki I t From which L 116 0.17 s where s resistivit immediatel beneath th foot y y e t max.durationof shock (s)
Step 4 – Calculation of grounding system resistance R 4r L where R resist anceof t hegrounding syst em r t heradius in m of a circular plat ehavingt hesame area as t hatoccupiedby t hegrid derived from t hecircular plat eelect rodeformula 4 eliminat eerrorsdue t o t hefact t hatt hegrid is not a L t ue circular plat e r
Step 5 – Maximum Grid Potential Rise : E E IR where I max.s/c currentin thegrid R resistanceof thegrid 116 0.17 s if E E touch then the preliminar design is o.k. y t Step 6 – Step voltage at the periphery for homogeneous soil = s I 116 0.7 s E stepmax kski L t similar toE mesh but k m ks
Step 7 –If the length of the conductor calculated before is used (L), thenthe internal step and touch voltages should be within the limits.The values can be decreased by increasing the length ofconductor.Aside: Actually: if d = diameter of conductor D = spacing n = no. of parallel conductors h = depth of burial 1 D2 1 3 5 7 2n 1 km ln ln .......... 2π 16hd π 4 6 8 2n 2 where t henumber of t ermsin ln is (n - 2)
1 1 1 1 1 and k s ......... π 2h D h 2D 3D where thenumber of termsin thebrackets nStep 8 –If calculations based on preliminary design indicate dangerouspotential can exist, the following remedies can be taken:a. Decrease R by either - Increasing area of grid, or - Increase L by the use of earth rods - Spacing the grid closer which increases Lb. Addition of a relatively high resistance surface layer (crushed stone) to increase the resistance in series with the body. Use approximately 4’ thickness of ½’ washed gravel.
Example:An electric utility plans to install a new 132 kV substation to feed anindustrial load. The proposed area of the substation is a rectangle ofdimension 42 m x 100 m. The soil in the area is non-homogenous with s =3000 Ωm and = 1316 Ωm. The system has the following impedances X” =40.5, X2 = 41 and X0 = 41 Ω respectively. The protection is so designed thatthe clearing time is 0.4 s. Given km = 0.568 and ks = 0.814 and an irregularityfactor ki = 2.0. Estimate, taking account of a decrement factor of D of 1.3and future growth of fault level of 125%:a. The minimum length of copper conductor required to keep touch and step voltages within safe limits.b. The resistance of the earth mat using the estimated copper in (a).c. Give a proposed layout of the grid calculate the actual length of conductor used (Maximum grid spacing = 6 m and length of earth rods = 3 m).d. For the layout in (c), check whether the maximum step voltage developed is within tolerable limits.
Solution:Estimated fault current for existing system is: 3VLN 3 132 x 103Ifault (SLG fault) " x 1260A j(X1 X 2 X0 ) 181.5 3 Design fault current 1260 x D x growth factor 1260 x 1.3 x 1.25 2047.5A k mki I t(a) Recall, t heconduct orlengt h L 116 0.17 s 0.568x 2.0 x 1316 x 2047.5x 0.4 116 (0.17 x 3000) 3092.5m
(b) Resistanceof earthmat is given by R 4r L 1 1 1316 4r L Now πr 2 42 x 100 r 36.6m 1 1 R 1316 9.4 4(36.6) 3092.5(c) Layout Let the spacingon the m side be 4 m and on the42 m side be 6m 100 100 42 T hen the lengthof conductorfor grid 1 42 1 100 1892m 4 6
3092 5 1892 . Min.no.of 3 m earthrods required 400 3Eitherreduce thespacingor bury 440rods (10%)Let the of earthrods be 440 no. totallengthof conductoris now 440 x 3 1320m Hence totallengthof copper to used 1892 1320 be 3212m I 0.814x 2.0 x 1316 x 2047.5(d) Max.E step kski L 3212 1365.7V
116 0.7 s 116 (0.7 x 3000) T olerableE step t 0.4 3503.8V Since Estepmax Esteptolerable then the design is good Aside T hemax.grid potential rise E IR 2047.5x 9.4 19.276kVTo avoid this being transferred, it is necessary to isolate thesubstation circuits from the supply circuits i.e. substation groundnot the common ground, therefore all m/c and other equipmentneutrals should be grounded at their site.