Cahier technique no 195 - Schneider Electric

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2.4 And if the CT cannot be manufactured?When a CT manufacturer says that he is unableto manufacture the requested CT, nine times outof ten this is because the CT has beenincorrectly specified. To eliminate all thecumulated safety margins taken by all the peopleinvolved, the CT must be redefined on the basisof real needs:c real currents in the installation,c types of protection, required power,c discrimination study and protection plan(settings).This approach must be adopted whenever thespecification leads to a non standard CT. Costs,lead times and safety are the factors at stake.Let us take an example:We have calculated the class X of a 1000/5 CTfor a generator differential protection, assumingthat X’’ = 15 %.Not knowing the exact characteristics of thegenerator, we have assumed that the generatorI n equals the CT I n .This results in: If= 100 In of the CT15i.e. V k u 2 x 6.7 x 5 (R ct + 2 R L )We have rounded off 6.7 as 7.We assumed:2 R L = 300 m of 2.5 mm 2 , i.e. 2.4 Ωhence: V k u 70 R ct + 168.Since this CT requires two other windings, thisvalue could not be achieved using the standardmould.The solution was found by using 4 mm 2connections and by requesting the generatorcharacteristics. Then:2 R L = 1.5 ΩGenerator I n = 830 A.X’’ = 25 %, hence:V k u 2 x830 x 10025x51000(R +1.5) ctV k = 33.2 R ct + 50The difference is marked and shows theimportance of obtaining the right information andof knowing the safety margins.If the CT is declared impossible to manufacture,a solution, i.e. a compromise, must be foundbetween all those involved. There is always away out, which can be found with the help ofspecialists.As an example, here are a few leads:c play on the equivalences between CTs (seenext section),c reduce the safety coefficient (for instance2 to 1.5 for an overcurrent protection),c change the secondary from 5 to 1 A (see fig. 11 ),c increase wiring cross-section,c overrate the CTs (primary I n ),c move the relay with respect to the CT,c use matching CTs with low consumption,c and so on.The overrating of a CT can solve amanufacturing problemLet us take two examples:c A 100/1 CT with a load of 2.5 VA requires anALF of 25 for an overcurrent protection.The standard CTs proposed are 2.5 VA-5P20. Ifa CT with a ratio of 150/1 - 2.5 VA-5P20 isproposed, the ALF need will be reduced in theCT primary ratio, i.e. necessaryALF = 25 x (100/150) = 16.7. An ALF of 20 isthus sufficient!c If the class X requested for a CT isproportional to a through current or a primary I sc ,these values are multiplied by the CT ratio; thus,the required knee point voltage will be less for anoverrated CT, unless its increasing resistanceR ct starts to neutralise the ratio benefit.In all cases, it will be possible to create a higherknee point voltage than with a CT of lower ratio,as it is proportional to the number of secondaryturns.Globally, the chance of obtaining workablecharacteristics will be greater.The same reasoning can be made for a 1 Asecondary CT compared with a 5 A CT.However, the factor gain of 5 obtained on theformula by the CT ratio is often completelyerased, if not reversed, by a far greater increaseof secondary winding resistance.Length (m) 5 10 20 50 100 200 400Wiring losses (VA) for:I n= 1 A 0.04 0.08 0.16 0.4 0.8 1.6 3.2I n= 5 A 1 2 4 10 20 40 80Fig. 11 : losses in wiring for a 2.5 mm 2 cross-section (8 Ω/km at 20° C). With 1 A, losses are 25 times less.Cahier Technique Schneider Electric no. 195 / p.12
Indeed, the space required for the number ofturns x 5 results in reduction in wiring crosssection,thus naturally increasing its linearresistance. The new resistance can thus beamply multiplied by 10 with respect to the 5 A CT.c If you are tempted to impose a CT overrating,you must check the repercussions of the changein ratio.For example:v If the CT supplies a pilot wire differentialprotection, you must ensure that thecorresponding CT at the other end of the line hasalso the same ratio change.v In the case of a restricted earth protection, youmust ensure that:- the CT installed on the neutral point is alsomodified,- the earth fault detection is not compromised bythe overrating.v For all protection types, you must check thatsetting of the protection is still possible.Optimisation of the differential protection CTsLet us take the example of a transformerdifferential protection (see fig. 12 ).c Calculating the through current.The transformer impedance limits the through5 x 100current to: (P sct = = 62.5 MVA) .8v Short-circuit power becomes:600 62.5P sc = x56.6 MVA600+62.5=.v The through current at the secondary is:- 11 kV side:656.6 x 10I f1 3x 10 x 5= = 49.5 A ,11 3 300- 3.3 kV side:656.6 x 10If2 3. x 10 x 5= = 49.5 A ,33 3 1000c Formulas to be applied for V k (standardprotection):v Calculating the matching CTs5with a ratio of:5/3V ka mini[ ]4I= f1 R sr + 3 ( R L3 + Rp)3v Calculating main CTs- 11 kV side: 300/5V k p1 min = 4I f1 (R ct + R L1 + R sp ) + V ka mini- 3.3 kV side: 1000/54 I (R R R )V k p2 min = f2 ct + L2 + r55 3Distance500 mI sc = 31.5 kA (P sc 600 MVA)300/51000/511 kV3.3 kV5 MVAZ sc = 8 %R L1R rR L2Fig. 12 : transformer differential protection.87TR L3c Optimisation approach.Let us examine the case of the 300/5 CT placedin the 11 kV switchboard.v First hypothesis5The matching CT is the one proposed as5/3standard by the relay manufacturer. It is locatedwith the relay on the 3.3 kV side. Wiring is2.5 mm 2 throughout.R L1 = 4 ΩR L2 = 0.08 ΩR L3 = 0.024 ΩR sr = 0.25 Ω, secondary winding resistance ofthe matching CT,R sp = 0.15 Ω, primary winding resistance of thematching CT,R p = 0.02 Ω, relay resistance.We find:- V ka mini = 43.7 V (standard V ka = 58 V),- V kp1 mini = 198 R ct + 847Cahier Technique Schneider Electric no. 195 / p.13
Page 1 and 2: Collection Technique ..............
Page 3 and 4: no. 195Current transformers:specifi
Page 5 and 6: Current transformers:specification
Page 7 and 8: A 15 VA-5P20 CT has a guaranteed er
Page 9 and 10: 2 Examples of specification errors2
Page 11 and 12: v The second possibility concerned
Page 13: account instead of the maximum thro
Page 17 and 18: 3 Equivalence of the various possib
Page 19 and 20: 4 ConclusionThis Cahier Technique c
Page 21: Schneider ElectricDirection Scienti

Cahier technique no 195 - Schneider Electric

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