Short Circuit Calculations in 60 Seconds

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Electrical Fault Level Calculations

Using the MVA Method

ith modern day personal computers, hand calculations for electrical fault level are becoming a ith modern day personal computers, hand calculations for electrical fault level are becoming a thing of the

thing of the past. past. The classical hand caThe classical hand calculations, either the ohmic lculations, either the ohmic method or the method or the per unitper unit method, will

method, will need many need many formulas formulas and conversions. and conversions. The ohmThe ohmic method ic method is cumberis cumbersome whensome when there are

there are several differeseveral different voltage nt voltage levels. levels. The per The per unit method unit method is not mis not much better uch better because of because of the many the many conversions of data

conversions of data to the choosen to the choosen base values. base values. The complexitThe complexity is significantly increay is significantly increased when symmetrsed when symmetricalical component theory is used to solve single phase to

component theory is used to solve single phase to earth faults, double phase to earth faults, and phase toearth faults, double phase to earth faults, and phase to phase faults. Most electrical engineers will phase faults. Most electrical engineers will blindly memorize these abstract formula and blindly memorize these abstract formula and cumbersome conversions. When these cumbersome conversions. When these engineers are needed to provide on the spot engineers are needed to provide on the spot estimates of fault level which are quick and estimates of fault level which are quick and reasonably accurate, they will often fail to reasonably accurate, they will often fail to deliver.

deliver. When When software software programmes programmes areare used, it is not uncommon to have errors in used, it is not uncommon to have errors in modelling and data entry, which will modelling and data entry, which will produce fault level several order of produce fault level several order of magnitude in error from the correct value. magnitude in error from the correct value. This article describes the MVA method, a This article describes the MVA method, a hand calculation method which is easy to hand calculation method which is easy to use, easy to remember, quick and accurate. use, easy to remember, quick and accurate.

Solution

The The

The The MVA method MVA method is a is a modification of modification of the ohmic the ohmic method. method. The first The first step is step is to convert to convert thethe typical single line diagram to the equivalent MVA single line diagram, and then to reduce the MVA single typical single line diagram to the equivalent MVA single line diagram, and then to reduce the MVA single line diagram into a

line diagram into a single MVA value single MVA value at the point of at the point of fault. fault. The componentThe components of a typical s of a typical single line are thesingle line are the utility

utility source, source, transformers, transformers, motors, motors, cables cables and interand internal nal generators. generators. Figure Figure 5.1 5.1 is is a a typical typical single single lineline diagram. diagram.

W

10MVA, 10MVA, 22/6.6kV 22/6.6kV 9% 9% 3C/300mm 3C/300mm22_{cable of 1km}_{cable of 1km} 6.6kV 6.6kV Generator Generator 3MVA, 3MVA, 6.6kV, 6.6kV, 15% 15% Utility source Utility source 22kV, 25kA fault level 22kV, 25kA fault level

38oh 38oh m m Motor Motor 1MVA, 1MVA, 12% 12% 2MVA, 2MVA, 6.6/0.4k 6.6/0.4k V 6% V 6% 400volts 400volts Motor Motor 0.4MVA, 0.4MVA, 15% 15% 3-phase phase fault fault A A BB CC DD E E FF 10MVA, 10MVA, 22/6.6kV 22/6.6kV 9% 9% 3C/300mm 3C/300mm22_{cable of 1km}_{cable of 1km} 6.6kV 6.6kV Generator Generator 3MVA, 3MVA, 6.6kV, 6.6kV, 15% 15% Utility source Utility source 22kV, 25kA fault level 22kV, 25kA fault level

38oh 38oh m m Motor Motor 1MVA, 1MVA, 12% 12% 2MVA, 2MVA, 6.6/0.4k 6.6/0.4k V 6% V 6% 400volts 400volts Motor Motor 0.4MVA, 0.4MVA, 15% 15% 3-phase phase fault fault A A BB CC DD E E FF FIGURE 5.1

FIGURE 5.1_:: _{Typical Single Line}_{Typical Single Line}

Chapter

5

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22kV Utility Source

The MVA value will be

The MVA value will be 3 _{3 x}_{x 22}_{22 x}_{x 25}_{25 =}_{= 952MVA.}_952MVA. The utility source has a 25kA fault level.

The utility source has a 25kA fault level. 10MVA Transformer

10MVA Transformer

10 10 The MVA value will be

0.09

0.09 = 111 MVA= 111 MVA The transformer has 9% impedance

The transformer has 9% impedance 2MVA Transformer

2MVA Transformer

22 The MVA value will be

0.06

0.06 = 33 MVA= 33 MVA The transformer has 6% impedance

The transformer has 6% impedance 6.6kV Motor

6.6kV Motor

0.12

0.12 = 8.3 MVA= 8.3 MVA

The motor has a sub-transient reactance of 12% and will contribute fault current to the fault. The motor has a sub-transient reactance of 12% and will contribute fault current to the fault. 400 Volts Motor

400 Volts Motor

0.4 0.4 The MVA value will be

0.15

0.15 = 2.7 MVA= 2.7 MVA The motor has sub-transient reactance of

The motor has sub-transient reactance of 15% and will contribute fault current to 15% and will contribute fault current to the fault.the fault. Internal Generator

Internal Generator

0.15

0.15 = 20 MVA= 20 MVA

The generator is synchronized to the utility source and has a sub-transient reactance of 15%. The generator is synchronized to the utility source and has a sub-transient reactance of 15%. 22kV Cable

22kV Cable

V V 22 The MVA value will be

Z

Z ,,

Where : V is the phase to phase voltage in kV. Where : V is the phase to phase voltage in kV.

Z is the per phase impedance in Z is the per phase impedance in ohm.ohm.

22 x 22 22 x 22 The MVA value will be

0.2

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MVA Single Line

Figure 5.2

Figure 5.2 is the equivalent is the equivalent MVA single line of MVA single line of the typical single line the typical single line of Figure 5of Figure 5.1. .1. The next The next step isstep is to reduce

to reduce the MVA the MVA single line to single line to a single a single MVA value MVA value at the point at the point of fault. of fault. The reduThe reduction uses basicction uses basic mathematics, either

mathematics, either add up add up the MVA the MVA values or values or “parallel up” “parallel up” the MVA the MVA values. values. Figure 5.3 Figure 5.3 illustrates theillustrates the steps for

steps for the reduction the reduction of the Mof the MVA single line VA single line to a single to a single MVA value MVA value at the point at the point of fault. of fault. The fault The fault levellevel for a 3 phase fault at 400 volts is 28.7MVA or 41.4kA.

for a 3 phase fault at 400 volts is 28.7MVA or 41.4kA.

Advantages of the MVA Method

 There is no ne There is no need to convert ed to convert impedance from impedance from one voltage to aone voltage to another, a requnother, a requirement in irement in thethe ohmic method .

ohmic method .

 There is no need to select a common MVA base and then to convert the data to the common There is no need to select a common MVA base and then to convert the data to the common MVA

MVA base, base, a a requirement requirement in in the the per per unit unit method. method. The The formulas formulas for for conversion conversion are are complexcomplex and not easy to remember.

and not easy to remember.

 _{Both the ohmic m}_{Both the ohmic method and per u}_{ethod and per unit method usually end}_{nit method usually end up with small decim}_{up with small decimals.}_{als. It is more}_{It is more} prone to make mistakes in the decimal with resulting errors several orders of magnitude from prone to make mistakes in the decimal with resulting errors several orders of magnitude from the correct value.

the correct value.

 The MVA The MVA method uses method uses large whole large whole numbers. numbers. This makes This makes for easier for easier manipulation and manipulation and hencehence less prone to errors.

less prone to errors.

Single Phase to Earth

Fault

So far the calculations were for three So far the calculations were for three phase fault.

phase fault. The MVA method can be used The MVA method can be used to calculate single phase to earth fault, and to calculate single phase to earth fault, and illustrated

illustrated in in Figure Figure 5.4. 5.4. The The positivepositive sequence MVA will be the value calculated sequence MVA will be the value calculated in the previous example, and in most in the previous example, and in most applications the positive sequence MVA will applications the positive sequence MVA will be the same as the negative sequence MVA. be the same as the negative sequence MVA. The zero sequence MVA will usually be The zero sequence MVA will usually be different from the positive sequence MVA. different from the positive sequence MVA. For example in Figure 5.1, only the 2MVA For example in Figure 5.1, only the 2MVA transformer will contribute to the earth fault transformer will contribute to the earth fault at 400 volts through the neutral connected solid to earth. The zero sequence MVA of the 2MVA at 400 volts through the neutral connected solid to earth. The zero sequence MVA of the 2MVA

6.6kV 6.6kV Utility source Utility source A A 400volts 400volts 3-phase phase fault fault 952 952 Cable Cable 2420 2420 10MVA 10MVA transformer transformer 111 111 ₂₀₂₀ Generator Generator B B CC DD 10MVA 10MVA transformer transformer 3 333 88..33 2.7 2.7 E E FF 6.6kV 6.6kV Utility source Utility source A A 400volts 400volts 3-phase phase fault fault 952 952 Cable Cable 2420 2420 10MVA 10MVA transformer transformer 111 111 ₂₀₂₀ Generator Generator B B CC DD 10MVA 10MVA transformer transformer 3 333 88..33 2.7 2.7 E E FF FIGURE 5.2

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transformer is equal to the positive sequence. The zero sequence

transformer is equal to the positive sequence. The zero sequence MVA of the 2MVA transformer is equal toMVA of the 2MVA transformer is equal to the positive sequence.

the positive sequence.

22 MVA value

MVA value of the transfof the transformer of ormer of ( (

0.06

0.06 ) MVA or 33.3 MVA ) MVA or 33.3 MVA

Voltage Drop During Motor Starting

The MVA method can also be used to calculate th

The MVA method can also be used to calculate the voltage drop during large e voltage drop during large motor starting. motor starting. TheThe voltage drop is equal to the motor starting MVA divided by the sum of the motor starting MVA and the voltage drop is equal to the motor starting MVA divided by the sum of the motor starting MVA and the

short-circuit

short-circuit MVA. MVA. Figure Figure 5.5 5.5 is is an an example. example. A A constant constant 1 1 MVA MVA load load is ais assumed ssumed before before the the starting starting of of the

the large large motor. motor. The The MVA MVA value value of of the the transformer transformer is is 50MVA. 50MVA. The The 1MVA 1MVA load load at at 400 400 volts volts will will bebe 1 x 50

1 x 50 seen as a (

seen as a (

1 + 50

1 + 50 ) MVA ) MVA or 0or 0.98 MVA .98 MVA load at load at 22kV. 22kV. The voltage The voltage at 22kat 22kV due V due to the to the load will load will bebe 952

952 0.98 + 952

0.98 + 952 or 99.9%.or 99.9%.

During motor starting, the combined load at 400 volts will be (1 + 4) MVA or 5 MVA. During motor starting, the combined load at 400 volts will be (1 + 4) MVA or 5 MVA.

5 x 50 5 x 50 The 5MVA load at

The 5MVA load at 400 volts will be 400 volts will be seen as seen as ( (

5 + 50

5 + 50 ) MVA or 4.55MVA load ) MVA or 4.55MVA load 952

952 at 22kV.

at 22kV. The voltage The voltage at 22kV at 22kV due to due to the motor the motor starting will bestarting will be

4.55 + 952 4.55 + 952

MVA or 99.5%. MVA or 99.5%.

Hence the voltage drop to the motor starting will be (99.9 – 99.5)% or 0.4% at

Hence the voltage drop to the motor starting will be (99.9 – 99.5)% or 0.4% at 22kV.22kV.

Conclusion

The MVA

The MVA method is emethod is easy to learn, asy to learn, easy to reasy to remember, qemember, quick and uick and accurate. accurate. The author The author has beenhas been using the MVA method for the past 13 years for small and large projects, and has found it most powerful using the MVA method for the past 13 years for small and large projects, and has found it most powerful for on the spot estimates.

for on the spot estimates.

6.6kV 6.6kV A A BB 33 33 E E C C 123.3 123.3 D D F F 2.7 2.7 123.3 123.3==95 + 20 + 8.395 + 20 + 8.3 26 26 E E FF 2.7 2.7 1 1 33 33 1 1 123.3 123.3 + + 26 26= ( = ( ))-1-1 28.7 28.7 E E FF 28.7 = 26 + 2.7 28.7 = 26 + 2.7 400volts 400volts 6.6kV 6.6kV A A 95 95 B B 33 33 E E C C 20 20 D D 8.3 8.3 F F 2.7 2.7 1 1 952 952 1 1 2420 2420 1 1 111 111 + + ++ 95 95= ( = ( ))-1-1 400volts 400volts 400volts

400volts 400volts400volts

6.6kV 6.6kV A A BB 33 33 E E C C 123.3 123.3 D D F F 2.7 2.7 123.3 123.3==95 + 20 + 8.395 + 20 + 8.3 26 26 E E FF 2.7 2.7 1 1 33 33 1 1 123.3 123.3 + + 1 1 33 33 1 1 123.3 123.3 + + 26 26= ( = ( ))-1-1 28.7 28.7 E E FF 28.7 = 26 + 2.7 28.7 = 26 + 2.7 400volts 400volts 6.6kV 6.6kV A A 95 95 B B 33 33 E E C C 20 20 D D 8.3 8.3 F F 2.7 2.7 1 1 952 952 1 1 2420 2420 1 1 111 111 + + ++ 1 1 952 952 1 1 2420 2420 1 1 111 111 + + ++ 95 95= ( = ( ))-1-1 400volts 400volts 400volts

400volts 400volts400volts

FIGURE 5.3

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END END --400volts 400volts Positive Positive sequence sequence MVA MVA Negative Negative sequence sequence MVA MVA Zero Zero sequence sequence MVA MVA 400volts 400volts 28.7 28.7 28.7 28.7 33.3 33.3 400volts 400volts 10 10

Single phase to earth fault = 3 x 10MVA Single phase to earth fault = 3 x 10MVA

= 30MVA = 30MVA

= 43kA at 400volts = 43kA at 400volts FIGURE 5.4

FIGURE 5.4_::_{MVA Diagram for Single Phase to Earth Fault}_{MVA Diagram for Single Phase to Earth Fault}

M M Before motor Before motor starting starting 3MVA, 3MVA, 6% 6% 99.9% 99.9% Source of Source of 952MVA 952MVA fault level fault level 98.0% 98.0% Load of Load of 1MVA 1MVA During motor During motor starting starting 3MVA, 3MVA, 6% 6% 99.5% 99.5% Source of Source of 952MVA 952MVA fault level fault level 90.9% 90.9% Load of Load of 1MVA 1MVA 1MVA 1MVA motor. motor. Starting Starting MVA is MVA is 4 times 4 times 22kV 22kV 400volts 400volts 22kV 22kV 400volts 400volts M M Before motor Before motor starting starting 3MVA, 3MVA, 6% 6% 99.9% 99.9% Source of Source of 952MVA 952MVA fault level fault level 98.0% 98.0% Load of Load of 1MVA 1MVA During m During motor otor

starting starting 3MVA, 3MVA, 6% 6% 99.5% 99.5% Source of Source of 952MVA 952MVA fault level fault level 90.9% 90.9% Load of Load of 1MVA 1MVA 1MVA 1MVA motor. motor. Starting Starting MVA is MVA is 4 times 4 times 22kV 22kV 400volts 400volts 22kV 22kV 400volts 400volts FIGURE 5.5