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Faults - Scenario 6

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Below is a modified version of Example D14 (Simplified Fault Current Calculation) in Appendix D of the 2017 Philippine Electrical Code (PEC).

This post is almost identical to the previous scenario, except that this time, an alternative common base value is used. The goal here is to show that regardless of the common base values utilized and the new per-unit values computed, the actual values remain practically the same.

SITUATION

An industrial complex receives 230 V, 60 Hz from a bank of three single-phase distribution transformers interconnected into a three-phase configuration. Each distribution transformer is rated 100 kVA, and the entire bank has an impedance of 5%. The transformer bank taps into a 34.5 kV supply with a 1,000 MVA short-circuit capacity.

Using the per-unit method, what maximum symmetrical fault currents may occur in each of the fault points "a", "b", and "c"? From these fault currents, what are the minimum symmetrical kiloAmpere Interrupting Capacity (kAIC) ratings needed for each molded case circuit breaker (MCCB) A, B, and C?


ANALYSIS

1.) ESTABLISH COMMON BASE VALUES TO EXAMINE ALL AS ONE UNIFIED CIRCUIT.

New base power, 3ph: S,base,new = 1,000 MVA

No need to establish a common base voltage. By retaining the existing base voltages, the per-unit supply voltage for each sub-circuit is pegged at 1 pu.

2.) CONVERT PER-UNIT VALUES FROM THE GIVEN BASE VALUE TO THE COMMON BASE VALUE.

2.1.) SOURCE SUB-CIRCUIT

Actual power, 3ph: S,src,3ph = 1,000 MVA
Actual voltage, LL: V,src,LL = 34.5 kV
Actual impedance, 1ph: Z,src,1ph = (V,src,LL)^2 / S,src,3ph

For the source sub-circuit, the actual values can be used as given base values.

Base power, 3ph: S,base = S,src,3ph
Base voltage, LL: V,base = V,src,LL
Base impedance, 1ph: Z,base = (V,base)^2 / S,base

2.1.1.) Source impedance, given, per-unit:

Z,src,pu = Z,src,1ph / Z,base
Z,src,pu = [ (V,src,LL)^2 / S,src,3ph ] / [ (V,base)^2 / S,base ]
Z,src,pu = [ (V,src,LL)^2 / S,src,3ph ] / [ (V,src,LL)^2 / S,src,3ph ]
Z,src,pu = 1 pu

2.1.2.) Source impedance, new, per-unit:

Z,src,new,pu = Z,src,1ph / Z,base,new
Z,src,new,pu = Z,src,pu * Z,base / Z,base,new
Z,src,new,pu = Z,src,pu * (V,base^2 / S,base) / (V,base^2 / S,base,new)
Z,src,new,pu = Z,src,pu * V,base^2 * S,base,new / (V,base^2 * S,base)
Z,src,new,pu = Z,src,pu * S,base,new / S,src,3ph
Z,src,new,pu = 1 * 1,000 MVA / 1,000 MVA
Z,src,new,pu = 1 pu

2.2.) TRANSFORMER LOAD-SIDE SUB-CIRCUIT

Base power, 3ph: S,base = 300 kVA
Base voltage, LL: V,base = 230 V
Base impedance, 1ph: Z,base = (V,base)^2 / S,base

2.2.1.) Transformer impedance, given, per-unit: Z,xmr,pu = 5% = 0.05 pu

2.2.2.) Transformer impedance, new, per-unit:

Z,xmr,new,pu = Z,xmr,1ph / Z,base,new
Z,xmr,new,pu = Z,xmr,pu * Z,base / Z,base,new
Z,xmr,new,pu = Z,xmr,pu * (V,base^2 / S,base) / (V,base^2 / S,base,new)
Z,xmr,new,pu = Z,xmr,pu * V,base^2 * S,base,new / (V,base^2 * S,base)
Z,xmr,new,pu = Z,xmr,pu * S,base,new / S,base
Z,xmr,new,pu = 0.05 pu * 1,000 MVA / 300 kVA
Z,xmr,new,pu = 166.6667 pu

2.2.3.) Conductor impedance, 1L:

For the cable conductors in the load-side circuit, their specifications are 250 mm^2 THW in steel conduit at 100 ft long (30.4878 m).

According to 2017 PEC Table 10.1.1.9 [ or 2017 NEC Chapter 9 Table 9 ] AC Resistance and Reactance for 600 V Cables, the resistance and reactance for such specification are 0.029 ohms per 305 meters and 0.048 ohms per 305 meters, respectively. [ 305 meters ~= 1,000 feet ]

Z,wire,1L = [ (0.029 + j0.048) ohms / 305 m ] * 30.5 m
Z,wire,1L = (0.0029 + j0.0048) ohms
Z,wire,1L = 0.005608 ohms < 58.86 deg

2.2.4.) Conductor impedance, per-unit:

Z,wire,pu = Z,wire,1L / Z,base,new
Z,wire,pu = Z,wire,1L / (V,base^2 / S,base,new)
Z,wire,pu = 0.005608 ohms / (230V^2 / 1,000 MVA)
Z,wire,pu = 106.0113 pu

2.3.) MOTOR SUB-CIRCUIT

Base power, 3ph: S,base = 100kVA
Base voltage, LL: V,base = 230V
Base impedance, 1ph: Z,base = (V,base)^2 / S,base

2.3.1.) Motor impedance, given, per-unit:

In 2017 PEC Appendix D Example D14, the per-unit impedance of each motor is not directly specified, but it gives us a clue on where the estimate is acquired: "IEEE Std 141".

IEEE Std 141-1993 (IEEE Recommended Practice for Electric Power Distribution for Industrial Plants) Section 4.5.4.1 (First-cycle Duties for Fuses and Circuit Breakers) Paragraph 5 states that:

"The low-voltage circuit breaker application guide, IEEE Std C37.13-1981, recommends subtransient impedances (typically 0.16 to 0.20 per unit) for all motors and allows estimates of typical symmetrical first-cycle contributions from connected low-voltage motors to substation bus short circuits at 4 times rated current (the equivalent of 0.25 per unit impedance)."

In determining circuit breaker fault interrupting capacity, the fault current during the first cycle is usually the biggest concern, and so its equivalent per-unit impedance is used accordingly.

Z,mtr,pu = 0.25 pu

2.3.2.) Motor impedance, new, per-unit:

Z,mtr,new,pu = Z,mtr,1ph / Z,base,new
Z,mtr,new,pu = Z,mtr,pu * Z,base / Z,base,new
Z,mtr,new,pu = Z,mtr,pu * (V,base^2 / S,base) / (V,base^2 / S,base,new)
Z,mtr,new,pu = Z,mtr,pu * V,base^2 * S,base,new / (V,base^2 / S,base)
Z,mtr,new,pu = Z,mtr,pu * S,base,new / S,base
Z,mtr,new,pu = 0.25 pu * 1,000 MVA / 100kVA
Z,mtr,new,pu = 2,500 pu

3.) FAULT CALCULATIONS

3.1.) AT FAULT POINT "a"

3.1.1.) One-line diagram for circuit at fault point "a":
o|---V,pu---Z,a,pu---"a"---|>

o|---V,pu---Z,src,new,pu---Z,xmr,new,pu---"a"---|>

3.1.2.) Short-circuit current at fault point "a", per-unit:

i,a,pu = V,pu / Z,a,pu
i,a,pu = V,pu / (Z,src,new,pu + Z,xmr,new,pu)
i,a,pu = 1 / (1 + 166.6667)
i,a,pu = 0.005964 pu

3.1.3.) Short-circuit current at fault point "a", 1L:

i,a,sc = i,a,pu * i,base,new
i,a,sc = i,a,pu * S,base,new / (sqrt(3) * V,base)
i,a,sc = 0.005964 pu * 1,000 MVA / (sqrt(3) * 230 V)
i,a,sc = 14,970.9435 Amps
i,a,sc = 14.971 kiloAmps

3.1.4.) Minimum interrupting capacity of MCCB A:

3-phase molded case circuit breaker A >= 15 kAIC symmetrical

3.2.) AT FAULT POINT "b"

3.2.1.) One-line diagram for circuit at fault point "b":
o|---V,pu---Z,b,pu---"b"---|>

o|---V,pu---|---Z,src,new,pu---Z,xmr,new,pu---|---"b"---|>
            |----Z,mtr1,new,pu---Z,wire,pu----|
            |----Z,mtr2,new,pu---Z,wire,pu----|

3.2.2.) Short-circuit current at fault point "b", per-unit:

i,b,pu = V,pu / Z,b,pu

Z,b,pu = (Z,src,new,pu + Z,xmr,new,pu) || (Z,mtr1,new,pu + Z,wire,pu) || (Z,mtr2,new,pu + Z,wire,pu)
Z,b,pu = 1 / [ 1/(Z,src,new,pu + Z,xmr,new,pu) + 1/(Z,mtr1,new,pu + Z,wire,pu) +1/(Z,mtr2,new,pu + Z,wire,pu) ]
Z,b,pu = 1 / [ 1/(1 + 166.6667) + 1/(2,500 + 106.0113) + 1/(2,500 + 106.0113) ]
Z,b,pu = 147.766 pu

i,b,pu = 1 / 147.766
i,b,pu = 0.006767 pu

3.2.3.) Short-circuit current at fault point "b", 1L:

i,b,sc = i,b,pu * i,base,new
i,b,sc = i,b,pu * S,base,new / (sqrt(3) * V,base)
i,b,sc = 0.006767 pu * 1,000 MVA / (sqrt(3) * 230 V)
i,b,sc = 16,986.649 Amps
i,b,sc = 16.987 kiloAmps

3.2.4.) Minimum interrupting capacity of MCCB B:

3-phase molded case circuit breaker B >= 17 kAIC symmetrical

3.3.) AT FAULT POINT "c"

3.3.1.) One-line diagram for circuit at fault point "c":
o|---V,pu---Z,c,pu---"c"---|>

o|---V,pu---|---Z,src,new,pu---Z,xmr,new,pu---|---Z,wire,pu---"c"---|>
            |----Z,mtr1,new,pu---Z,wire,pu----|
            |----Z,mtr2,new,pu---Z,wire,pu----|

3.3.2.) Short-circuit current at fault point "c", per-unit:

i,c,pu = V,pu / Z,c,pu

Z,c,pu = [ (Z,src,new,pu + Z,xmr,new,pu) || (Z,mtr1,new,pu + Z,wire,pu) || (Z,mtr2,new,pu + Z,wire,pu) ] + Z,wire,pu
Z,c,pu = [ 1 / [ 1/(Z,src,new,pu + Z,xmr,new,pu) + 1/(Z,mtr1,new,pu + Z,wire,pu) +1/(Z,mtr2,new,pu + Z,wire,pu) ] ] + Z,wire,pu
Z,c,pu = [ 1 / [ 1/(1 + 166.6667) + 1/(2,500 + 106.0113) + 1/(2,500 + 106.0113) ] ] + 106.0113
Z,c,pu = 147.766 + 106.0113
Z,c,pu = 253.7773 pu

i,c,pu = 1 / 253.7773
i,c,pu = 0.00394 pu

3.3.3.) Short-circuit current at fault point "c", 1L:

i,c,sc = i,c,pu * i,base,new
i,c,sc = i,c,pu * S,base,new / (sqrt(3) * V,base)
i,c,sc = 0.00394 pu * 1,000 MVA / (sqrt(3) * 230 V)
i,c,sc = 9,890.2611 Amps
i,c,sc = 9.89 kiloAmps

3.3.4.) Minimum interrupting capacity of MCCB C:

3-phase molded case circuit breaker C >= 10 kAIC symmetrical

CONCLUSION

The maximum symmetrical fault currents are 14.971 kiloAmperes at point "a", 16.987 kiloAmperes at point "b", and 9.89 kiloAmperes at point "c".

The minimum symmetrical kiloAmpere Interrupting Capacity ratings needed then is 15 kAIC for 3ph MCCB A, 17 kAIC for 3ph MCCB B, and 10 kAIC for 3ph MCCB C.

==========

REFERENCES

1.) Philippine Electrical Code (SI Modernized Metric System)

1.1.) 2017 PEC Table 10.1.1.9 Alternating-Current Resistance and Reactance for 600-Volt Cables, 3-Phase, 60Hz, 75degC - Three Single Conductors in Conduit.

1.2.) 2017 PEC Appendix D Example D14 Simplified Fault Current Calculation.

2.) National Electrical Code (US Inch-Pound System)

2.1.) 2017 NEC Chapter 9 Table 9 Alternating-Current Resistance and Reactance for 600-Volt Cables, 3-Phase, 60Hz, 75degC (167degF) - Three Single Conductors in Conduit.

2.2.) [Not found in 2017 NEC Annex D] (Simplified Fault Current Calculation).

3.)  IEEE Std 141-1993: IEEE Recommended Practice for Electric Power Distribution for Industrial Plants

3.1.) IEEE Std 141-1993 Section 4.5.4.1 First-cycle Duties for Fuses and Circuit Breakers.

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