Design Guide for Short Circuit Calculation Generation System

TATA CONSULTING ENGINEERS

SECTION: TITLE

TCE.M6-EL-700-6621

DESIGN GUIDE FOR GENERATOR SYSTEM FAULT CALCULATION

SHEET i OF iii


TATA CONSULTING ENGINEERS 414, Veer Savarkar Marg Prabhadevi MUMBAI 400 025

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1992.03.30 FORM NO. 020R2

TATA CONSULTING ENGINEERS TCE.M6-EL-700-6621


SECTION:REV SHT

SHEET ii OF iii

REVISION

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ISSUE R0 FORM NO. 120 R1

TATA CONSULTING ENGINEERS

SECTION:CONTENTS

TCE.M6-EL-700-6621

SHEET iii OF iii


CONTENT

SR.NO.

TITLE

SHEET NO.

1.0

SCOPE

1

2.0

ABBREVIATIONS USED

1

3.0

REFERENCE DIAGRAM

1

4.0

DATA

1

5.0

PROCEDURE

2

6.0

DISCUSSION

3

7.0

REFERENCE

3

8.0

ANNEXURE-1 ONE LINE DIAGRAM

4

9.0

ANNEXURE-2 DATA REQUIRED FROM THE

5

GENERATOR MANUFACTURER 10.0

ANNEXURE-3 SAMPLE CALCULATION

6




SECTION:WRITE-UP

SHEET 1 OF 19

1.0

SCOPE

1.1

The scope of this design guide is to provide a detailed method to calculate the fault currents when a 3 phase fault occurs at any point in the following sections of the power plant electrical network. a)

The generator main power evacuation circuit consisting of the generator, the main bus duct till generator transformer and the section upto the grid.

b)

The unit auxiliary transformer section consisting of tap off section of the busduct upto the unit auxiliary transformer.

1.2

The fault currents calculated would be required to assign the rating of the equipment operating in the electrical network indicated above such as Extra High Voltage Circuit Breaker, Bus Duct and Generator Circuit Breaker (if provided)

2.0

ABBREVIATIONS USED GT : UAT : EHVCB: GCB : SEB : pu :

Generator Transformer Unit Auxiliary Transformer Extra High Voltage Circuit Breaker Generator Circuit Breaker State Electricity Board Per Unit

3.0

REFERENCE DIAGRAM

3.1

A schematic diagram indicating the equipment such as Generator, GT, UAT, EHVCB and GCB is given in Annexure-1.

4.0

DATA To calculate the fault currents, the ratings/data of the following equipment/system are required.

4.1

GRID : Following details/data of the grid to which the generator is connected are required. i)

The voltage level in kV.

ii)

The maximum rms symmetrical fault level in kA.

iii)

The X/R Ratio ISSUE R0 FORM NO. 120 R1



SECTION:WRITE-UP

SHEET 2 OF 19

4.2

Generator

4.2.1

The data required from the generator manufacturer to carryout the fault calculations are listed in Annexure-2.

4.2.2

Tolerances on the machine parameters shall be specifically confirmed with the manufacturer, in the absence of this confirmation, tolerances as per IEC-34 or relevant standards shall be applied.

4.3

Generator Transformer

4.3.1

The following parameters of the GT are required for fault calculation: a) b) c)

Rated primary and secondary voltages. Rated maximum 3 phase MVA of the transformer Reactance and resistance values of the transformer.

The resistance of the transformer shall be found using the full load copper losses in case the values of the resistances are not directly available. 4.3.2

Tolerances as confirmed by the transformer manufacturer shall be applied. If the tolerances are not furnished by the manufacturer, the tolerances as per relevant standards shall be applied.

4.4

Bus Ducts

4.4.1

The values of resistance and reactance of the busduct are very small quantities of the order of micro ohms. Hence these parameters need not be considered for calculation.

5.0

PROCEDURE

5.1

All calculations shall be made in p.u. system.

5.2

The base MVA and base kV shall preferably be taken as the rated MVA and rated kV of the generator.

5.3

(Base kV)2 L-L Base Impedance = ------------------------ OHMS Base MVA (3 ph)

5.4

Actual Impedance in ohms P.U. Impedance = -----------------------------------Base Impedance in ohms ISSUE R0 FORM NO. 120 R1



5.5

SECTION:WRITE-UP

SHEET 3 OF 19

P.U. Impedance on any other MVA and kV can be converted to the P.U. Impedance on base MVA & base kV as follows: Zpu (New) Zpu (Old)

Zpu (new)

= =

=

P.U. Impedance on base MVA and base kV P.U. Impedance on any other MVA and kV

Zpu (old) x

MVA (new) kV2 (old) -------------- x ----------MV (old) kV2 (new)

6.0

DISCUSSION

6.1

A sample calculation to calculate the symmetrical fault current, the DC component, the asymmetrical fault current and the momentary current has been given in Annexure-3 for a typical 500 MW plant having a generator circuit breaker. The calculation steps have been explained in detail and the same procedure shall be adopted for units of other ratings also.

6.2

While specifying the rating of the equipment operating in the electrical network considered in this design guide such as GCB and EHVCB, etc., the results obtained from the calculations shall be indicated in the data sheet-A of the relevant specification.

7.0

REFERENCES

7.1

The general theory of electrical machines- Adkins B. Chapman & Hall Ltd. 1962.

7.2

ANSI/IEEE Std. C.37.5-1979 IEEE Guide for calculation of fault currents for application of AC High Voltage Circuit Breakers rated on a total current basis.

7.3

Elements of Power System Analysis – W.D.Stevenson Jr.

7.4

IEEE C37.013-1989 Standard for AC High Voltage Generator Circuit Breakers rated on a symmetrical current basis.

7.5

IEC-56 High Voltage AC Breakers-Publication 1987

7.6

IEC-34 Rotating Electrical Machines

7.7

Control and Stability of Electric Power Systems – Anderson and Fouad.


SECTION: ANNEXURE-2


SHEET 5 OF 19


ANNEXURE-2 DATA REQUIRED FROM THE GENERATOR MANUFACTURER Sl.No. Machine Parameters

Symbol

Unit

1.

Rated Voltage (L-L) rms

V

kV

2.

Rated MVA

MVA

MVA

3.

Rated Power Factor

cos φ

-

4.

Rated Current (rms)

I

Amps

5.

Frequency

f

Hz

6.

Armature resistance at 750 C

Ra

ohms

7.

Saturated Direct Axis Steady State Reactance

Xd

p.u.

8.

Saturated Direct Axis Transient Reactance

Xd’

p.u.

9.

Saturated Direct Axis Subtransient Reactance

Xd”

p.u.

10.

Saturated Quadrature Axis Steady State Reactance Xq

p.u.

11.

Saturated Quadrature Axis Subtransient Reactance Xq”

p.u.

12.

Direct Axis Short Circuit Transient Time Constant Td’

Sec

13.

Direct Axis Short Circuit Sub-transient Time Constant

Td”

Sec

14.

Armature Short Circuit Time Constant

Ta

Sec

15.

Quadrature Axis Short Circuit Sub-transient Time Constant

Tq”

Sec




SECTION: ANNEXURE-3

SHEET 6 OF 19

ANNEXURE-3 SAMPLE CALCULATION 1.0

This section outlines the detailed procedure for calculation of fault current in the generator-generator transformer, UAT tap off section. Reference to the figure in Annexure-1 shall be made. The calculations are split into 5 parts.

1.1

Part-1 : Calculates 3 phase fault currents at any point between the generator and generator transformer when generator alone is contributing to the fault.

1.2

Part-2 : Calculates 3 phase fault current at any point between GT and EHVCB when grid/generator is contributing to the fault when the value of 3 phase symmetrical fault current is known.

1.3

Part-3 : Calculates the 3 phase fault currents at the point considered in part 1 when grid alone is feeding the fault.

1.4

Part-4 : Calculates the 3 phase fault currents on the UAT tap off part of the busduct when both the grid and generator are feeding the fault.

1.5

Part-5 : Gives a summary of sizing the equipment performing in this section of the electrical network considered.

2.0

ASSUMPTIONS MADE IN THE CALCULATIONS

2.1

For conservation fault estimation, the saturated value of the reactances are considered.

2.2

The speed of the machine is assumed to be constant for about 60 to 80 milliseconds (3 to 4 cycles) even after the instant of short circuit.

2.3

The switching angle A defines the point in the AC cycle at which the short circuit occurs. The fault current at λ = 00 for unloaded generator case gives the maximum DC component & λ = 900 gives the zero DC component. Hence λ = 00 is assumed. The power angle σ is the angle between E the generated voltage and Vt the terminal voltage of the generator is equal to zero since the value of E and Vt are same for unloaded generator.




SECTION: ANNEXURE-3

SHEET 7 OF 19

2.4

The fault current fed from generator will be maximum when the value of λ and σ are nearly equal. This is applicable for loaded as well as unloaded cases prior to occurrence of the fault. For engineering applications the value of fault current under generator unloaded conditions is adequate.

2.5

The calculations are done using general theory of electrical machines which is based on a general machine concept. This is required to incorporate the finite effect of saliency present in the turbo-generators which takes into consideration the differences between the values of direct and quadrature axis reactances and time constants and thus give accurate results.

2.6

The generator circuit breaker shall be sized such that the maximum values of the symmetrical short circuit current, DC component shall be taken from part 1 and part 3 of the calculation.

2.7

Equivalent system (grid) X/R ratio at typical locations (for quick approximations) are given below. Type of Circuit

X/R

1.

Synchronous machines connected through transformers rated 100 MVA and larger

40-60

2.

Synchronous machines connected through transformers rated 25 to 100 MVA for each three-phase bank.

30-50

3.

Remote synchronous machines connected through transformers rated 100 MVA or larger for each threephase bank, where the transformers provide 90 percent or more of the total equivalent impedance to the fault point.

30-50

4.

Remote synchronous machines connected through transformers rated 10 MVA to 100 MVA for each threephase bank, where the transformers provide 90 percent or more of the total equivalent impedance to the fault point.

15-40

5.

Remote synchronous machines connected through other types of circuits, such as; transformers rated 10 MVA or smaller for each three-phase bank, transmission lines, distribution feeders, etc.

15 or less


SECTION: ANNEXURE-3


SHEET 8 OF 19


General Calculations Base kV = 21 kV Base MVA = 659 MVA

Base Impedance =

(Base kV)2 (21)2 ------------- = ------- = 0.669 Ω Base MVA 659

Following table furnish the parameters of generator, generator transformer and grid. Refer Annexure-2 for details required for generator. a)

GENERATOR

Sl. No.

Parameter

Actual Value

P.U. Value

P.U Value with tolerance

1.

Vt = Terminal Voltage

21.0 kV

1.0

1.05

2.

MVA

659

1.0

1.0

3.

Rated Current

18,118A

1.0

1.0

4.

Power factor

0.85

-

-

5.

Rated frequency

50 Hz

-

-

6.

Xd

-

2.16

1.728

7.

Xq

-

2.16

1.728

8.

Xd’

-

0.271

0.2168

9.

Xd”

-

0.193

0.1544

10.

Xq”

-

0.212

0.1696

11.

Ra at 750 C

0.00176 ohms

0.00263

0.00263

12

Td’

0.98 sec

-

-

13.

Td”

0.03 sec

-

-


SECTION: ANNEXURE-3


SHEET 9 OF 19


14.

Tq”

0.075 sec

-

-

15.

Ta

0.2 sec

-

-

SR. No.

Parameter

b)

GENERATOR TRANSFORMER

1.

Actual Value

P.U. Value

P.U Value with tolerance

HV Volts (LL)

400 kV

-

-

2.

LV Volts

21 kV

-

-

3.

MVA

630 MVA

-

-

4.

Impedance, Zt

15%

0.1569

0.1412

5.

Resistance, Rt

0.005 ohms

0.0075

0.0075

6.

Inductance, Xt

-

-

0.141

c)

GRID

1.

Voltage level

400 kV

-

-

2.

Fault current

40 kA

-

-

3.

Fault MVA

27712 MVA

-

-

4.

Grid X/R

40

-

-

Notes: a)

20% negative tolerance on the value of generator reactance, 5% positive tolerance on generator terminal voltage and 10% negative tolerance on transformer reactance are considered in this particular example. However the tolerances shall be taken as per relevant standards when the manufacturer does not indicate the tolerances in his data sheets.

b)

The transformer resistance value Rt shall be obtained from copper losses or from transformer winding resistance test.

c)

Xt the transformer reactance = (Zt2 – Rt2)1/2

d)

400 kV grid fault level = 40 kA = 27712 MVA ISSUE R0 FORM NO. 120 R1

SECTION: ANNEXURE-3


SHEET 10 OF 19


659 Grid fault impedance in pu = ------------ = 0.02378 pu 27712 PART-1

Vt peak

=

(2)1/2 x Vt ∠ 00

=

(2)1/2 x 1.05 ∠ 00

= 1.4849 pu

2 Xd”Xq” Factor, Xm = --------------- = (Xd” + Xq”)

2 x 0.1544 x 0.1696 ------------------------- = (0.1544 + 0.1696)

0.1616 pu

2 Xd”Xq” = --------------- = (Xq” - Xd”)

2 x 0.1544 x 0.1696 ------------------------- = (0.1696 - 0.1544)

3.4456 pu

Factor, Xn

t

=

Breaker opening time = 60 milliseconds for GCB & EHVCB

ω

=

2πf radians = 314.1593 radians = 18,000 degrees

ωt

=

2πft

λ = 00

= 18000 x 60 x 10-3 = 1080 degrees

& σ = 00

The steady state value of the voltage before short circuit can be analysed to be made up of 2 components namely Direct and Quadrature axes. The direct axis component Vd

= (2)1/2 Vt sin σ

The quadrature axis component Vq = - (2)1/2 Vt cos σ The direct axis component Vd = 0 since the value of σ = 00 and Vq = - 1.4849 pu. The complete expression for the fault current in each phase consists of 2 components Ia1 & Ia2 whose equations are given below for phase ‘a’. The value of fault currents in the other two phases ‘b’ & ‘c’ can be calculated substituting λ by (λ- 2π/3) and (λ - 4π/3) respectively.




SECTION: ANNEXURE-3

SHEET 11 OF 19

Equation 1 : Ia1 = [ Vq/xd +(Vq/Xd’ – Vq/Xd”) e- t/Td” + (Vq/Xd” – Vq/Xd’) e –t/Td”] cos (ωt +λ) - Vq/Xm e-t/Ta cos λ - Vq/Xn e-t/Ta cos (2ωt + λ) Equation 2 : Ia2 = - [ Vd/xq +(Vd/Xq” – Vd/Xq) e- t/Tq” ] sin (2ωt +λ) + Vd/Xm e-t/Ta sin λ - Vd/Xn e-t/Ta sin (2ωt + λ) The total fault current in phase a : Ia = Ia1 + I a2 The equation 2 reduces to zero since Vd = 0 Equation 1 consists of 3 parts Part – a : Iac1 = [ Vq/xd +(Vq/Xd’ – Vq/Xd”) e- t/Td” + (Vq/Xd” – Vq/Xd’) e –t/Td”] cos (wt +λ) is called the AC fundamental Part-b : Iac2 = (- Vq/Xn) e-t/Ta cos (2wt + λ) is called the second harmonic component Part –c : Idc = (- Vq/Xm) e-t/Ta cos λ is called the DC component of the fault current. Total AC component of the fault current = (Part a + Part b) pu Total symmetrical Iac (rms) = (Iac1 + Iac2) / √2 pu Iac peak = Iac x √ 2 pu % DC component = (Idc /Iac peak) x 100 Total Iac asymmetrical

= (Idc2 + Iac2 rms)1/2 ISSUE R0 FORM NO. 120 R1



SECTION: ANNEXURE-3

SHEET 12 OF 19

Substituting the values of the parameters in equation 1 we get by simplifying: Iac1 Iac2 Idc

= = =

- 6.8669 pu 0.3192 pu 6.80699 pu

Iac1 + Iac2 Symm Iac rms = --------------------- = 4.6299 pu √2 (Iacrms2 + Idc2)1/2

Asymm Iac rms

=

Iactual

I base x I pu

=

=

8.23 pu

Symm Iac rms

=

18118 x 4.6299

=

83.88 kA

Asymm Iac rms

=

18118 x 8.23

=

149.11 kA

Idc

=

18118 x 6.80699

=

123.33 kA

% DC

123.33 x 100 = -----------------------√ 2 x 83.88

=

103.97%

A computer package SMS is available which calculates the fault currents which have been computed above manually.


SECTION: ANNEXURE-3



SHEET 13 OF 19

PART-2 a)

Grid Feed The symmetrical fault current fed from the 400 kV grid Iacgrms =

40 kA

X/R ratio of the grid =

40

Iacg (peak)

40 x √2

X/R

=

=

56.57 kA

=

0.127 sec

2πFL / R

L 40 ∴ Time constant , τ = ------ = ----------R 2π x 50 t = EHVCB opening time Idc

=

= 60 milliseconds

= DC current due to X/R of 40 = Idc

=

Iac g (peak) e-t/τ

56.57 e –60X.001 / 0.127

Idc

= 35.27 kA Idc x 100 % DC component = -------------- = Iacg (peak)

35.27 -------- x 100 = 62.35% 56.57

Total Asymmetrical Current = (Iac2grms +Idc2 ) b)

½

=

53.33 kA

Generator Feed The generator feed to the fault current when a fault occurs after EHVCB can be computed by using equation 1 referred in part A of the calculations. However the value of reactances such as Xd, Xd’, Xd”, Xq, Xq” etc. get modified as Xd + Xt, Xd’ + Xt, Xd”+Xt, Xq+Xt, Xq”+Xt etc. respectively where Xt is the GT reactance. The armature time constant Ta which is approximately equal to Ld”/Ra becomes (Ld” + Lt)/(Ra + Rt) where Ld” = Xd”/ 2πf and Lt = Xt / 2πf Substituting the values in equation 1 we get, ISSUE R0 FORM NO. 120 R1

SECTION: ANNEXURE-3



Iac rms=

2.833 x 18118 =

51.33 kA

Idc

2.569 x 18118 =

46.545 kA

=

SHEET 14 OF 19

Iac asymm = (51.332 + 46.5452)1/2 = 69.3 kA These values are on 21 kV side of the GT I The values on 400 kV side of the GT = -----------(400 + 21) ∴ The value of Iac rms on 400 kV side = 51.33 x 1 / 19.05 = 2.69 kA Similarly on 400 kV side Iac asymm = 3.64 kA Idc = 2.44 kA

% DC component

4.2325 x 100 = --------------------√ 2 x 4.67

=

64.08%

Comparing grid feed and generator feed for the fault considered, we find that the grid feed is much larger than the generator feed because of the impedance of the GT. Hence EHVCB should be sized to perform for grid feed.




SECTION: ANNEXURE-3

SHEET 15 OF 19

PART – 3

The fault current fed from grid alone when a 3 phase fault occurs anywhere between the generator and generator transformer LV bushing is calculated in this section. 400 kV The grid impedance = ---------------= 5.774 ohms √ 3 x 40 ∴ Zg =

5.744 ohms

X/R of the grid =

40

∴ Rg = Xg/40 ohms & Xg = 40 Rg But Z2 = R2 ∴ Rg2 = Rg2 =

+ Zg2

X2 - Xg2

Zg2 - (40 Rg)2

Rg2 (1 + 1600) = Zg2 Rg2

(5.774)2 = ----------------(1600 + 1)

=

0.020824 ohms

∴Xg = √ Zg2 - Rg2 = √ 5.7742 - 0.020824 = 5.7722 ohms Assuming 630 MVA and 400 kV the rating of transformer as base MVA and base kV on EHV side of the transformer, the base impedance. 4002 = ----- = 253.97 ohms 630 The fault impedance of the grid in p.u. 5.774 = ---------= 0.02273 p.u. 253.97


SECTION: ANNEXURE-3



SHEET 16 OF 19

The fault reactance of the grid in p.u. 5.7722 = ---------= 0.02273 p.u.= Xg 253.97 The fault resistance of the grid in p.u. 0.020824 = -------------- = 0.000082 p.u. 253.97 The p.u. impedance of the GT

= 0.1412 p.u.

For a fault between the generator and generator transformer, The total fault impedance = 0.1412 + 0.02273 = 0.16393 p.u.

Fault MVA

Base MVA = -------------------Fault impedance

630 = ---------- = 3847.79 MVA 0.16393

∴ Symmetrical fault current Igrms on 21 kV side Ig peak = 105.79 x √ 2 0.02273 Lg = ------------- pu 2π x 50

Similarly Lt

=

3847 .79 = ------------ = √ 3 x 21

105.79 kA

149.605 kA

= 0.00007235 pu & Rg = 0.000082 pu

0.1408 = ------------2 x π x 50

= 0.00044 pu & Rt = 0.0075 pu

The effective time constant to (Lg + Lt) calculate DC component T = ---------(Rg + Rt)

0.00051 = ------------ = 0.06726 secs 0.00758

Assuming 60 milliseconds is required for EHVCB to break fault currents t

=

60 x 10-3 sec

Idc

=

(Igpeak) x e-t/C




Idc

=

149.605 x e-60 x 10 –3/0.06726

Idcg

=

61.3 kA on 21 kV side

Asumm Iacg = √ Idcg2 + Igrms2

% DC component =

SECTION: ANNEXURE-3

SHEET 17 OF 19

= √ 105.792 + 61.32 = 122.27 kA

61.3 ---------- x 100 149.605

=

41%




SECTION: ANNEXURE-3

SHEET 18 OF 19

PART-4

This part calculates the fault currents on the UAT tap off portion of the busduct. The total fault current will be equal to the sum of the fault currents fed from the generator and the grid. Hence the total fault current will be equal to the sum of the fault currents calculated in Part 1 and Part 3. ∴ Total Iac symm

=

83.88 kA + 105.79 kA = 189.67 kA

Total Iac asymm

=

149.11 + 122.27

Total Idc

=

123.33 kA + 61.3 kA = 184.63 kA

= 271.36 kA




SECTION: ANNEXURE-3

SHEET 19 OF 19

PART-5 SUMMARY OF SIZING EQUIPMENT PERFORMING IN THE SECTION OF THE NETWORK UNDER CONSIDERATION 1. GCB : Symmetrical Breaking Capacity

= 106 kA (grid feed)

DC current = 123.33 kA (generator feed) 123.33 % DC component = ------------------ x 100 = 82.3% 106 x √ 2 Asymmetrical breaking capcity

=

Peak Fault Making Current

106 x 2.55

=

162.62 kA = 271 kA

GCB shall be sized for the above values of the short circuit current. The generator circuit breaker as per IEEE C 37.013 – 1989 meeting the above rating shall be specified. 2. EHVCB:

Symmetrical Breaking Capacity

=

40 kA

% DC component

=

63%

Asymmetrical Breaking Capacity

=

54 kA

Peak fault making current

= 40 x 2.55

= 102 kA

3. BUSDUCT a)

Main busduct between generator and generator transformer (Higher of the generator & grid feed shall be taken) Short time current Momentary rating

b)

= =

106 kA for 1 second 271 kA (Peak)

Portion of the busduct from UAT tapoff point till UAT HV terminals.

This portion of the bus duct is susceptible to fault currents fed from both generator and EHV grid. Hence short time current = 84 kA + 106 kA = 190 kA for 1 sec. Momentary rating = 190 x 2.55 = 485 kA (Peak) ISSUE R0 FORM NO. 120 R1

Design Guide for Short Circuit Calculation Generation System

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