abb_manual_10e_03_print_07.pdf

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3.4 Examples of calculation

More complex phase fault calculations are made with computer programs (Calpos ).See Section 6.1.5 for examples.

When calculating short-circuit currents in high-voltage installations, it is often sufficientto work with reactances because the reactances are generally much greater inmagnitude than the effective resistances. Also, if one works only with reactances in thefollowing examples, the calculation is on the safe side. Corrections to the reactancesare disregarded.

The ratios of the nominal system voltages are taken as the transformer ratios. Insteadof the operating voltages of the faulty network one works with the nominal system

Example:

Three-phase busbars 40 m long, each conductor comprising three copper bars80 mm 10 mm ( A = 2400 mm 2), distance D = 30 cm, f = 50 Hz. According to the curve,L = 3.7 10 7 H/m; and so

X = 3.7 10 7 H /m 314 s 1 40 m = 4.65 m .

The busbar arrangement has a considerable influence on the inductive resistance.

The inductance per unit length of a three-phase line with its conductors mounted onedge and grouped in phases (Fig. 3-20 and Fig. 13-2a) is relatively high and can beusefully included in calculating the short-circuit current.

Small inductances can be achieved by connecting two or more three-phase systems inparallel. But also conductors in a split phase arrangement (as in Fig. 13-2b) yield verysmall inductances per unit length of less than 20 % of the values obtained with themethod described. With the conductors laid flat side by side (as in the MNS system) theinductances per unit length are about 50 % of the values according to the method ofcalculation described.

Fig. 3-20

Inductance L of busbars of rectangular cross section


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voltage. It is assumed that the nominal voltages of the various network components arethe same as the nominal system voltage at their respective locations. Calculation isdone with the aid of the %/ MVA system.

Example 1

To calculate the short-circuit power S" k, the peak short-circuit current i p and thesymmetrical short-circuit breaking current I a in a branch of a power plant station servicebusbar. This example concerns a fault with more than one infeed and partly commoncurrent paths. Fig. 3-21 shows the equivalent circuit diagram.

For the reactances of the equivalent circuit the formulae of Table 3-4 give:

1.1 100 110Network reactance x Q = = = 0.0138 %/MVA,S kQ 8000u K 13Transformer 1 x T1 = = = 0.1300 %/MVA,S rT1 100x d 11.5Generator x G = = = 0.1227 %/MVA,S rG 93.7u K 7Transformer 2 x T2 = = = 0.8750 %/MVA,S rT2 8

I rM / I start 1Induction motor x M1 = 100 = 100 = 7.4349 %/ MVA,S rM 5 2.69I rM / I start 1Induction- x M2 = 100 = 100 = 5.4348 %/ MVA.

motor group S rM 5 8 0.46

For the location of the fault, one must determine the total reactance of the network. Thisis done by step-by-step system transformation until there is only one reactance at theterminals of the equivalent voltage source: this is then the short-circuit reactance.

Calculation can be made easier by using Table 3-20, which is particularly suitable forcalculating short circuits in unmeshed networks. The Table has 9 columns, the first ofwhich shows the numbers of the lines. The second column is for identifying the parts andcomponents of the network. Columns 3 and 4 are for entering the calculated values.The reactances entered in column 3 are added in the case of series circuits, while thesusceptances in column 4 are added for parallel configurations.Columns 6 to 9 are for calculating the maximum short-circuit current and thesymmetrical breaking current.To determine the total reactance of the network at the fault location, one first adds thereactances of the 220 kV network and of transformer 1. The sum 0.1438 % /MVA is incolumn 3, line 3.The reactance of the generator is then connected in parallel to this total. This is doneby forming the susceptance relating to each reactance and adding the susceptances(column 4, lines 3 and 4).The sum of the susceptances 15.1041 %/ MVA is in column 4, line 5. Taking thereciprocal gives the corresponding reactance 0.0662 %/ MVA, entered in column 3, line5. To this is added the reactance of transformer 2. The sum of 0.9412 % /MVA is in

column 3, line 7.The reactances of the induction motor and of the induction motor group must then beconnected in parallel to this total reactance. Again this is done by finding thesusceptances and adding them together.


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The resultant reactance of the whole network at the site of the fault, 0.7225%/MVA, isshown in column 3, line 10. This value gives

1.1 100 % 1.1 100 %S k = = 152 MVA, (column 5, line 10).x k 0.7225 % /MVA

To calculate the breaking capacity one must determine the contributions of theindividual infeeds to the short-circuit power S k.

The proportions of the short-circuit power supplied via transformer 2 and by the motorgroup and the single motor are related to the total short-circuit power in the same wayas the susceptances of these branches are related to their total susceptance.

Contributions of individual infeeds to the short-circuit power:

0.1345Contribution of single motor S kM1 = 152 = 14.8 MVA,1.381

0.184Contribution of motor group S kM2 = 152 = 20.3 MVA,1.381

1.0625Contribution via transformer 2 S kT2 = 152 = 116.9 MVA.1.381

The proportions contributed by the 220 kV network and the generator are foundaccordingly.

8.150Contribution of generator S kG = 116.9 = 63.1 MVA,15.104

6.954Contribution of 220 kV network S kQ = 116.9 = 53.8 MVA.15.104

The calculated values are entered in column 5. They are also shown in Fig. 3-21b.

To find the factors and q

When the contributions made to the short-circuit power S k by the 220 kV network, thegenerator and the motors are known, the ratios of S k / S r are found (column 6). Thecorresponding values of for t v = 0.1 s (column 7) are taken from Fig. 3-5.Values of q (column 8) are obtained from the ratio motor rating / number of pole pairs(Fig. 3-6), again for t v = 0.1 s.

Single motorS kM1 14.8 motor rating 2.3 = = 5.50 = 0.74 = = 1.15 q = 0.59S rM1 2.69 no. pole pairs 2

Motor groupS kM2 20.3 motor rating 0.36 = = 5.52 = 0.74 = = 1.12 q = 0.32S rM2 8 0.46 no. pole pairs 3

S kG 63.1Generator = = 0.67 = 1S rG 93.7

For the contribution to the short-circuit power provided by the 220 kV network, = 1,see Fig. 3-5, since in relation to generator G 3 it is a far-from-generator fault.


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Contributions of individual infeeds to the breaking capacity

The proportions of the short-circuit power represented by the 220 kV network, thegenerator and the motors, when multiplied by their respective factors and q, yield thecontribution of each to the breaking capacity, column 9 of Table 3-20.

Single motor S aM1 = q S kM1 = 0.74 0.59 14.8 MVA = 6.5 MVA

Motor group S aM2 = q S kM2 = 0.74 0.32 20.3 MVA = 4.8 MVA

Generator S aG = S kG = 1 63.1 MVA = 63.1 MVA

220 kV network S aQ = S kQ = 1 53.8 MVA = 53.8 MVA

The total breaking capacity is obtained as an approximation by adding the individualbreaking capacities. The result S a = 128.2 MVA is shown in column 9, line 10.

Table 3-20

Example 1, calculation of short-circuit current

1 2 3 4 5 6 7 8 9Component x 1 S k S k / S r q S a

x % /MVA MVA/% MVA (0.1s) (0.1s) MVA

1 220 kV network 0.0138 53.8 1 53.82 transformer 1 0.1300 3 1 and 2 in series 0.1438 6.9541 4 93.7 MVA generator 0.1227 8.1500 63.1 0.67 1 63.15 3 and 4 in parallel 0.0662 15.1041 6 transformer 2 0.8750 7 5 and 6 in series 0.9412 1.0625 116.9 8 induction motor

2.3 MW /2.69 MVA 7.4349 0.1345 14.8 5.50 0.74 0.59 6.59 motor group = 3.68 MVA 5.4348 0.1840 20.3 5.52 0.74 0.32 4.8

10 fault location7, 8 and 9 in paral le l 0 .7225 1.3810 152.0 128.2

At the fault location:

S k 152.0 MVAI k = = = 14.63 kA, 3 U n 3 6.0 kV

I p = 2 I k = 2.0 2 14.63 kA = 41.4 kA (for = 2.0),

S a 128.2 MVAI a = = = 12.3 kA. 3 U n 3 6.0 kV

Example 2

Calculation of the phase-to-earth fault current I k1.

Find I k1 at the 220 kV busbar of the power station represented by Fig. 3-22.

Calculation is made using the method of symmetrical components. First find thepositive-, negative- and zero-sequence reactances X 1, X 2 and X 0 from the networkdata given in the figure.


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Fig. 3-21

a) Circuit diagram, b) Equivalent circuit diagram in positive phase sequence with equivalent voltage source at fault location, reactances in %/MVA: 1 transformer 1,2 transformer 2, 3 generator, 4 motor, 5 motor group, 6 220 kV network, 7 equivalent voltage at the point of fault.

Zero-sequence reactances (index 0)

A zero-sequence system exists only between earthed points of the network and thefault location. Generators G1 and G 2 and also transformer T1 do not thereforecontribute to the reactances of the zero-sequence system.

Positive-sequence reactances (index 1)

1Overhead line X 1L = 50 0.32 = 8 21.1 (220 kV) 2220 kV network X = 0.995 = 6.622

8000 MVA(21 kV) 2Power plant unit X G = 0.14 = 0.494 125 MVA

(220 kV) 2X T = 0.13 = 48.4 130 MVA

X KW = KKW ( 2r X G + X T)

1.1K KW = 1 + (0.14 0.13) 0.6

220X KW = 1.093 [()2 0.494 + 48.4 ] = 112.151 21At the first instant of the short circuit, x 1 = x 2. The negative-sequence reactances arethus the same as the positive-sequence values. For the generator voltage: U rG = 21 kVwith sin rG = 0.6, the rated voltages of the transformers are the same as the systemnominal voltages.


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Overhead lineX 0L = 3.5 X 1L = 28 2 circuits in parallel

220 kV network X 0Q = 2.5 X 1Q = 16.555

Transformer T 2 X 0T2 = 0.8 X 1T 1.093 = 42.321

With the reactances obtained in this way, we can draw the single-phase equivalentdiagram to calculate I k1 (Fig. 3-22b).

Since the total positive-sequence reactance at the first instant of the short circuit is thesame as the negative-sequence value, it is sufficient to find the total positive and zerosequence reactance.

Calculation of positive-sequence reactance:1 1 1 = + x 1 = 11.598 x 1 56.076 14.622

Calculation of zero-sequence reactance:1 1 1

= + x 0 = 21.705 x 0 42.321 44.556

Fig. 3-22

a) Circuit diagram, b) Equivalent circuit diagram in positive phase sequence, negative phase sequence and zero phase sequence with connections and equivalent voltage

source at fault location F for I k1.With the total positive-, negative- and zero-sequence reactances, we have

1.1 3 U n 1.1 3 220I k1 = = = 9.34 kA.x 1 + x 2 + x 0 44.901


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The contributions to I k1 represented by the 220 kV network (Q) or power station (KW)are obtained on the basis of the relationship

I k1 = I 1 + I 2 + I 0 = 3 I 1 with I 0 = I 1 = I 2 = 3.11 kA

to right and left of the fault location from the equations:

I k1Q = I 1Q + I 2Q + I 0Q , and I k1KW = I 1KW + I 2KW + I 0KW.

The partial component currents are obtained from the ratios of the respectiveimpedances.

56.08I 1Q = I 2Q = 3.11 kA = 2.47 kA70.70

42.32I 0Q = 3.11 kA = 1.51 kA86.88I 1KW = 0.64 kA

I 0KW = 1.60 kA

I k1Q = (2.47 + 2.47 + 1.51) kA = 6.45 kA

I k1KW = (0.641 + 0.64 + 1.60) kA = 2.88 kA

Example 3

The short-circuit currents are calculated with the aid of Table 3-2.

1.1 (0.4) 220 kV network: x 1Q = 0.995 = 0.0007 250

r 1Q 0.1 x 1Q = 0.00007

(0.4) 2Transformer: x 1T = 0.058 = 0.0147 0.63

(0.4)2

r 1T = 0.015 = 0.0038 0.63x 0T = 0.95 x 1T = 0.014 r 0T r 1T = 0.0038

Cable: x 1L = 0.08 0.074 = 0.0059 r 1L20 = 0.08 0.271 = 0.0217 r 1L80 = 1.24 r 1L20 = 0.0269 x 0L 7.36 x 1L = 0.0434 r 0L20 3.97 r 1L20 = 0.0861 r 0L80 = 1.24 r 0L20 = 0.1068

Maximum and minimum short-circuit currents at fault location F 1

a. Maximum short-circuit currents

Z 1 = Z 2 = (0.0039 + j 0.0154) ; Z 0 = (0.0038 + j 0.0140)

1.0 0.4I k3 = kA = 14.5 kA 3 0.0159 3I k2 = I k3 = 12.6 kA

2 3 1.0 0.4I k1 = kA = 15.0 kA.0.0463


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Table 3-21

Summary of results

Fault Max. short-circuit currents Min. short-circuit currentslocation 3p 2p 1p 3p 2p 1p

kA kA kA kA kA kA

Fault location F 1 14.5 12.6 15.0 13.8 12.0 14.3

Fault location F 2 6.9 6.0 4.0 6.4 5.5 3.4

The breaking capacity of the circuit-breakers must be at least 15.0 kA or 6.9 kA.Protective devices must be sure to respond at 12 kA or 3.4 kA. These figures relate tofault location F1 or F2.

b. Minimum short-circuit currents

The miminum short-circuit currents are calculated with c = 0.95.

Maximum and minimum short-circuit currents at fault location F 2

a. Maximum short-circuit currents

Z 1 = Z 2 = (0.0265 + j 0.0213) ; Z 0 = (0.0899 + j 0.0574)

1.0 0.4I k3 = kA = 6.9 kA 3 0.0333 3I k2 = I k3 = 6.0 kA

2 3 1.0 0.4I k1 = kA = 4.0 kA.0.1729

b. Minimum short-circuit currents

The minimum short-circuit currents are calculated with c = 0.95 and a temperature of80 C.

Fig. 3-23 a) Circuit diagram of low-voltage network,b) Equivalent diagram in component systems and connection for single- phase fault


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Examples of use Networks of limited Overhead-line Cable networks High-voltage networksextent, power plant networks 10230 kV system (123 kV) to 400 kVauxiliaries 10123 kV e. g. in towns (protective multiple

earthing in I. v. network)

Between system Capacitances, (inst. trans- Capacitances, Capacitances, (Capacitances),and earth are: former inductances) Suppression coils Neutral reactor Earth conductor

Z 0 / Z 1 1/j C E very high resistance inductive: 4 to 60 2 to 4 resistive: 30 to 60Z 1

Current at fault site with Ground-fault Residual ground- Ground-fault current I k1single-phase fault current fault currentCalculation (approximate) I E (capacitive) I R

I E j 3 C E E 1 I R 3 C E ( + j ) E 1 = loss angle = interference

3.5 Effect of neutral point arrangement on fault behaviour in three-phase high-voltage networks above 1 kV

Table 3-22

Arrangement of neutral isolated with arc current-limiting low-resistance earthpoint suppression coil R or X

c U nE 1 = = E 3

3 E 1I k1 = I R j (X 1 + X 2 + X 0)

I k1 3 X 1 3 = = I k3 2 X 1 + X 0 2 + X 0 / X 1 (continued)


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Table 3-22 (continued)

Arrangement of neutral isolated with arc current-limiting low-resistance earthpoint suppression coil R or X

I k2 / I k3 I CE / I k3 I R / I k3 inductive : 0.05 to 0.5 0.5 to 0.75resistive : 0.1 to 0.05

U LEmax / U n 1 1 to (1.1) inductive: 0.8 to 0.95 0.75 to 0.80resistive: 0.1 to 0.05

U 0max / U n 0.6 0.6 to 0.66 inductive: 0.42 to 0.56 0.3 to 0.42resistive: 0.58 to 0.60

Voltage rise in yes yes no nowhole network

Duration of fault 10 to 60 min 10 to 60 min < 1 s < 1 sPossible short-time earthing with subsequent selectivedisconnection by neutral current (< 1 s)

Ground-fault arc Self-quenching Self-quenching Partly self-quenching Sustainedup to several A usually sustained

Detection Location by disconnection, ground-fault wiping-contact Selective disconnection by Short-circuitrelay, wattmeter relay. (With short-time earthing: dis- neutral current (or short- protectionconnection by neutral current) circuit protection)

Risk of double earth fault yes yes slight no

Means of earthing Earth electrode voltage U E < 125 V Earth electrode voltage U E > 125 V permissibleDIN VDE 0141 Touch voltage 65 V Touch voltages 65V

Measures against interfer- Generally not Not necessary Overhead lines: possibly required if approaching overence with communication necessary a considerable distancecircuits Cables: generally not necessaryDIN VDE 0228 needed only with railway block lines

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