https://www.vultr.com

The Everywhere Cloud

Deploy an instance.
Accelerate your application.


Faults - Scenario 8

By -

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 a single-phase version of the three-phase example in the previous scenario without any motors present in the adjacent circuits. The goal here is to examine what magnitudes are to be expected in single-phase circuits compared to three-phase circuits, and how these magnitudes factor into the selection of protective device ratings.

SITUATION

An industrial complex receives 230 V, 60 Hz from a single-phase distribution transformer rated 300 kVA and an impedance of 5%. The transformer taps into a single-phase 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.

New base power, 1ph: S,base,new = 300 kVA

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

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

Base power, 1ph: S,base = S,src,1ph
Base voltage, 1ph: V,base = V,src,1ph
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,1ph)^2 / S,src,1ph ] / [ (V,base)^2 / S,base ]
Z,src,pu = [ (V,src,1ph)^2 / S,src,1ph ] / [ (V,src,1ph)^2 / S,src,1ph ]
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,1ph
Z,src,new,pu = 1 * 300 kVA / 1,000 MVA
Z,src,new,pu = 0.0003 pu

2.2.) TRANSFORMER LOAD-SIDE SUB-CIRCUIT

Base power, 1ph: S,base = 300 kVA
Base voltage, 1ph: 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 * 300 kVA / 300 kVA
Z,xmr,new,pu = 0.05 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 / 300 kVA)
Z,wire,pu = 0.0318 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 / (0.0003 + 0.05)
i,a,pu = 19.8807 pu

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

i,a,sc = i,a,pu * i,base,new
i,a,sc = i,a,pu * S,base,new / V,base
i,a,sc = 19.8807pu * 300kVA / 230V
i,a,sc = 25,931.3478 Amps
i,a,sc = 25.931 kiloAmps

3.1.4.) Minimum interrupting capacity of MCCB A:

1-phase molded case circuit breaker A >= 26 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"---|>

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,b,pu = 0.0003 + 0.05
Z,b,pu = 0.0503 pu

i,b,pu = 1 / 0.0503
i,b,pu = 19.8807 pu

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

i,b,sc = i,b,pu * i,base,new
i,b,sc = i,b,pu * S,base,new / V,base
i,b,sc = 19.8807pu * 300kVA / 230V
i,b,sc = 25,931.3478 Amps
i,b,sc = 25.931 kiloAmps

3.2.4.) Minimum interrupting capacity of MCCB B:

1-phase molded case circuit breaker B >= 26 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"---|>

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,wire,pu
Z,c,pu = (0.0003 + 0.05) + 0.0318
Z,c,pu = 0.0503 + 0.0318
Z,c,pu = 0.0821 pu

i,c,pu = 1 / 0.0821
i,c,pu = 12.1803 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 / V,base
i,c,sc = 12.1803pu * 300kVA / 230V
i,c,sc = 15,887.3478 Amps
i,c,sc = 15.887 kiloAmps

3.3.4.) Minimum interrupting capacity of MCCB C:

1-phase molded case circuit breaker C >= 16 kAIC symmetrical

CONCLUSION

All fault points in this single-phase system experience significantly larger fault currents compared to the previous three-phase scenario. Although the per-unit currents are the same, the actual currents in the single-phase system are concentrated on one circuit instead of being distributed over three circuits "firing" their sine waves in succession "120 degrees" apart in the time domain [ | tan (120 deg) | = square root of 3 ].

The maximum symmetrical fault currents are 25.931 kiloAmperes at point "a", 25.931 kiloAmperes at point "b", and 15.887 kiloAmperes at point "c".

The minimum symmetrical kiloAmpere Interrupting Capacity ratings needed then is 26 kAIC for 1ph MCCB A, 26 kAIC for 1ph MCCB B, and 16 kAIC for 1ph 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).

Comments

Post a Comment

Popular posts from this blog

Standards - Note 3

By -
Image
WIRE AMPACITIES, INSULATORS, AND IMPEDANCES Below are some useful tables pertaining to ampacities, insulators, and impedances of various wires. These are derived from the 2017 Philippine Electrical Code [ or the 2017 National Electrical Code ], but are rearranged in a manner that is easier to interpret and understand. 1.) Allowable Ampacities of Insulated Conductors in Raceway. The areas, ampacities, and insulation types of the following tables are based on 2017 PEC Table 3.10.2.6(B)(16) [ or 2017 NEC Table 310.15(B)(16) ]. 1.1.) Copper Wire Ampacities --------------------------------------- |  Copper  |  Insulation Max Op. Temp | --------------------------------------- |   Area   | 60degC | 75degC | 90degC | |   mm^2   |  Amps  |  Amps  |  Amps  | --------------------------------------- |    2.0   |    15  |    20  |    25  | |    3.5   |    20  |    25  |    30  | |    5.5   |    30  |    35  |    40  | |    8.0   |    40  |    50  |    55  | |   14.0   |    55  |    65
Read more

Standards - Note 4

By -
Image
RACEWAY SIZES (CONDUIT AND TUBING) 1.) Raceway Sizing There are three options for sizing raceways according to 2017 PEC Table 10.1.1.1 [ or 2017 NEC Chapter 9 Table 1 ], based on cross-sectional areas: A.) 53% raceway area for 1 conductor (a multiconductor cable is treated as a single conductor [Note 9]) B.) 31% raceway area for 2 conductors C.) 40% raceway area for over 2 conductors 1.1.) Choosing Between Alternatives For future expansion purposes, option (A) is far from practical. This may have use in very specific installations that are assumed to be not needing any expansions, but upgrades will definitely happen and option (A) will eventually become a bottleneck. Option (B) for 2 conductors having a smaller area than either option (A) for 1 conductor or option (C) for over 2 conductors may seem nonsensical, but it actually does make sense. Having 2 conductors usually means 2 live wires supplying loads that may require an equipment grounding conductor (EGC) later on
Read more

Faults - Scenario 4

By -
Image
This post demonstrates how to estimate the kiloAmpere Interrupting Capacity (kAIC) rating of the first circuit breaker at the secondary side of a transformer using the Infinite Bus Method. The method assumes infinite supply from the primary side with negligible line impedances. Below is a modified version of Example D13 (Available Short Circuit Current)  in Appendix D of the 2017 Philippine Electrical Code (PEC). SITUATION Two adjacent buildings are supplied by two different distribution transformers. One transformer is single-phase rated 100kVA, 230V with 2.5% impedance. The other is three-phase rated 100kVA, 230V line-to-line and also has 2.5% impedance. What minimum kAIC ratings are needed for the first circuit breakers down the line of each transformer to safely interrupt the maximum symmetrical fault currents? ANALYSIS 1.) SINGLE-PHASE TRANSFORMER LOAD-SIDE SUB-CIRCUIT Base Power, 1ph: S,base = 100kVA Base Voltage, 1ph: V,base = 230V Impedance, given: Z
Read more