Arrange circuits, control and protection for general ...

Arrange circuits, control

and

protection

for

general electrical installations

Week 3: Prospective Fault Current and Short Circuit Temperature Rise

All writing in BLUE is examinable

All writing in RED is

NOT examinable.

Slide 2 of 41

Prospective Fault Current

To cover:

1) What is it?

2) How to calculate at terminals of Tx:

3) How to calculate at MSB:

4) How to calculate at DB:

Slide 3 of 41

FLI worst case, not enough I.PFC worst case, too much I. Both Ω’s law

Prospective Fault Currents (PFC) (also

known as prospective short-circuit current) is a calculation or measurement ofhow high thecurrent could be inthe event of ashort circuit.

Reference: 2.5.4AS/NZS 3000

Slide 4 of 41

When an arc is initiated, the live conductors are in close enough proximity to create sparking. The arc itself is caused by uncontrolled conduction of electrical current from phase to phase, or from phase to earth/neutral, and this ionises the surrounding air.

When conductive metal is vaporized, a pressure wave develops. A phase to phase, or phase to earth/neutral arc fault can escalate into a three phase arc within a millisecond. The heat energy and intense light produced at this stage is known as the arc flash.

Reference NHP Technical newsletter

Slide 5 of 41

The energy released as a transformer dissipates it’s magnetic field energy can cause enough heat to instantly vaporise copper.

(Heated so hot it bypassesthe liquid phase and goesstraight to a gas causinga fatal explosion)

Slide 6 of 41

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H= I2 R t

The heat generated (in Joules) is to the square of the current times the resistance times the time it is there.

Double the current, quadruple the heating effect

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If we put in adequately rated Circuit Breakers and/or Fault Current Limiters (HRC fuses) we can interrupt fault currents before the switchboards explodes

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Common size CBs1.5 kA3 kA4.5 kA6 kA9 kA

A 6 kA breaker can interrupt faults up to 6000A

Cheap

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The fault current must be interruptedhere before it is reaches it’s maximum.

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3 Feb 2015 fatality in Western Australia

EnergySafety (WA) has imposed new safety precautions for the type of high-voltage (HV) switches involved in the 3 February Morley Galleria Shopping Centre explosion.

The HV oil-insulated combined-fuse switches produced by Long & Crawford Manchester must be completely disconnected from the electricity supply before any person may open the switch’s lid.

Announcing this today the Director of EnergySafety, Ken Bowron said that the Morley Galleria accident wasstark evidence of the inherent danger associated with these switches.”

Slide 12 of 41

Transformers (Tx) have a high PFC

As do batteries,however Solarpanels don’thave high PFCs

A commonIndustrial1 MVA Tx willlikely have a PFC close to 30 kA

Slide 13 of 41

A fault like 3000 10A welders

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0.01Ω

400V0.02Ω

=20kA230V0.01Ω

=23kA

PFC is highest phase to Earth (when at close to the terminals of the transformer)

230V, 0.01Ω0.01Ω

0.01Ω400V, 0.02Ω

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0.01Ω

400V0.04Ω

=10kA230V0.03Ω

=7.67kA

PFC is highest phase to phase (when far from the transformer)

230V, 0.03Ω0.01Ω

0.01Ω

400V,0.04Ω

0.01Ω

0.01Ω

0.01Ω

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The PFC at at the terminals of a TX (I

PFCTX)

Can be calculated

(or measured live using a Fault Loop Impedance tester)

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To calculate IPFCTX

The kVA rating, voltage and % impedance of the TX must be know.

Line current in a three phase system can be calculated using the formulaP = √3 V I (Using line V and I: P = √3 V

L I

L)

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PT – Transformer kVA rating

VL – Line voltage (400 volts)

IL – Line current

PT=√3VLILPT

√3×VL=√3×VL×IL√3×VL

IL=PT

√3×VL

IL=250,000

√3×400V(250KVATX)

=360.85A

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Transformer name plates are stamped with the % Impedance which was measured using the following test...

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At 1%of 6.35 kV=63.5V=90.2 AAt 2%of 6.35kV=127.04V=180.4 AAt 3%of 6.35kV=190.5V=270.6 AAt 4%V6.35kV=254V=360.8A

This is a 4% impedance TX

Full line current

V ADead short in secondary with Ammeter

VariableVoltageSupply

Tx360.8 A per phase at 6.35KV−11kV phase−phase

VL=VP×√3

VP=VL√3

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If a fault causes 360A to flow with only 4% of the voltage, what will 100% of the supply voltage cause?

360A×100%4%

=9kA

( 3600.04

=9kA)

Will also give you the same answer

IPFCTX=IL×100%% impTx

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TXCM

PEN

MENP/E

MSBSM DB

Conductive pipes

Building Materials

Ground

The return path could be any number of parallel paths to earth

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TXCM

0Ω

MSBSM DB

Therefore we take a conservative approach and say that the return path = 0Ω (Estimating a low resistance means we allow for more current and put in better circuit protection devices)

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In the ACT:30,000A in commercial 10,000A in domestic 3Φ 6,000A in domestic 1Φ

SM – Sub mainCM – Consumer mainTX – transformerMSB – Main switch board

(Yet another application of ohms law)

ZTX=Volts (230V)

Perspective fault Current

IPFCMSB=V

ZTX+ZCM

IPFCDB=V

ZTX+ZCM+ZSM

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CM MSB SM1 DB1

ZTX=230V30,000A IPFCMSB=

VZTX+ZCM

SM2

DB2

IPFCDB1=V

ZTX+ZCM+ZSM1

IPFCDB2=V

ZTX+ZCM+ZSM2

TX

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TXMSB DB1

DB2

0.002 0.02

0.04

Example

ZTX=23030,000

=0.00767

IPFCMSB=230

(0.00767+0.002)=23.78KA

IPFCDB2=230

(0.00767+0.002+0.04)=4.63kA

IPFCDB1=230

(0.00767+0.002+0.02)=7.75kA

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To find the values of Zcm, Zsm etc.Look up the value in Tables 34 to 39 AS/NZS 3008.

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Use worst case assuming cold copper

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ZCM=Table 34 0.00703Ω/kM÷1000×60m=0.0042Ω

Example:

What is the resistance of 60m of 300mm2 single core cable?

1) Look up value on Table 352) Divide by 1000 to change Ωs per km to Ωs per m.3) Times value by length

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Select appropriately rated CB

If the calculatedPFC is 4.8kA, Select the next

size up.

A CB or Fuse that is rated to interrupt the maximum (prospective) fault current that could be present at the SWB.

1.5 kA3 kA4.5 kA6 kA9 kA

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Cascading of CB’s may be used if the same brand of CB is installed upstream (at the MSB)

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Example: With a 25 kA CB upstream 6kA breakers can have a rupturing capacity of 14kA

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This is because both circuit breakers start to open together increasing the overall breaking capacity of the pair.

MSB DB

Short Circuit Temperature Rise (SCTR)

To cover:

1) What is it?

2) How to Select cables

Slide 35 of 41

SCTR is “Prospective Fault Current for cables”

All currents caused by a short-circuit shall be interrupted before the temperature of the conductors reaches the permissible limit. (Cables must withstand fault current)

(111 PVC) (143 XLPE)

I2 t<K2S2

Factor dependant on the material

Time CB to trip

CSA

“Let throughenergy”

“Energy the cable can withstand”

<

PFC

I2t is found off manufactures specifications

Reference 5.3 AS/NZS 3008

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Look up I2t

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CSA

40kA2S

10kA

20A CB

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Example X-90

Start 90° (table 1)Finish 250° (table 53)

K rating = 143 (table 52)

Look up k value using Tables 1, 53 and 52 AS/NZS 3008

Example V-90

Start 75° (table 1)Finish 160° (table 53)

K rating = 111 (table 52)

I2 t<K2S2

Will a 2.5mm2 cable withstand temperature rise when subjected to a 10kA fault and protected by a 20A CB?

40kA2S(specs)

111(V-90)

2.5mm2

(CSA)

40k<111x 2.5

40k<77k

YES

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