CENTRAL ELECTRICITY AUTHORITY (MEASURES RELATING TO SAFETY AND - - PowerPoint PPT Presentation

central electricity authority measures relating to safety
SMART_READER_LITE
LIVE PREVIEW

CENTRAL ELECTRICITY AUTHORITY (MEASURES RELATING TO SAFETY AND - - PowerPoint PPT Presentation

CENTRAL ELECTRICITY AUTHORITY (MEASURES RELATING TO SAFETY AND ELECTRIC SUPPLY) REGULATIONS, 2010, AND EARTHING PRACTICES IN MINES. AJAY SINGH, DIRECTOR OF MINES SAFETY(ELECTRICAL), CENTRAL ZONE, DGMS, DHANBAD REG 100: PROTECTIVE EQUIPMENT.-


slide-1
SLIDE 1

CENTRAL ELECTRICITY AUTHORITY (MEASURES RELATING TO SAFETY AND ELECTRIC SUPPLY) REGULATIONS, 2010, AND EARTHING PRACTICES IN MINES.

AJAY SINGH, DIRECTOR OF MINES SAFETY(ELECTRICAL), CENTRAL ZONE, DGMS, DHANBAD

slide-2
SLIDE 2

REG 100: PROTECTIVE EQUIPMENT.- (1) appropriate equipment shall be suitably placed in the mines for automatically disconnecting supply to any part of the system, where a fault, including an earth fault, occurs and fault current shall not be more than 750 milliampere in installations of voltage exceeding 250 V and up to 1100 V for below ground mines and oil fields and 50 ampere in installations of voltage exceeding 1100 V and up to 11 kV in open cast installations of voltage exceeding 1100 V and up to 11 kV in open cast mines and the magnitude of the earth fault current shall be limited to these specified values by employing suitably designed, RESTRICTED NEUTRAL SYSTEM OF POWER SUPPLY.

slide-3
SLIDE 3
  • Body resistance= 1000 ohm
  • SHOCK VOLTAGE = 317V
  • Flow of current through body ≃

≃ ≃ ≃317mA

  • Death most likely.
  • Body resistance= 1000 ohm
  • Extra resistance introduced in the path of

earth fault current = 423Ω

  • SHOCK VOLTAGE = 30.15V
  • Flow of current through body ≃

≃ ≃ ≃30mA

  • CHANCES OF SURVIVAL.
slide-4
SLIDE 4

EFFECT OF RESISTANCE OF EARTH RETURN PATH ON SHOCK VOLTAGE/CURRENT

For 45 OHM plus NGR, SHOCK VOLTAGE = 30.15V / 30mA For 100 OHM plus NGR, SHOCK VOLTAGE = 60.61V / 60mA For 10 OHM plus NGR, SHOCK VOLTAGE = 7.32V / 7mA

slide-5
SLIDE 5

Concept of touch potential

slide-6
SLIDE 6
slide-7
SLIDE 7

EQUIVALENT CIRCUIT:

slide-8
SLIDE 8
slide-9
SLIDE 9

For a 550V system, per phase voltage is 317V and the value of NGR to restrict earth fault current within 750mA will be 423 Ω. The magnitude of shock voltage ( or touch potential) will be: = (550/√3) x 45 / (423 + 45) = 30.53 V which is well within safe limits. Now, the characteristics between touch potential and time shall be referred so as to appropriately set the timing of earth leakage relay (source: AS-4871), failing which the protective scheme won’t be fool- proof and may not offer desired protection. proof and may not offer desired protection.

slide-10
SLIDE 10
slide-11
SLIDE 11

CONVENTIONAL TOUCH POTENTIAL ANALYSIS

slide-12
SLIDE 12
  • 1. Touch potentials resulting from ‘compliant’ fault

limits and loop impedances are not necessarily ‘safe’.

  • 2. AS/NZS4871.1:2012

relates the maximum duration of human exposure to prospective touch voltages that do not usually result in touch voltages that do not usually result in harmful physiological effects on any person subjected to that touch voltage (i.e. ‘safe’).

slide-13
SLIDE 13

MAXIMUM DURATION OF 50Hz TOUCH VOLTAGE AS PER AS4871;

slide-14
SLIDE 14
slide-15
SLIDE 15

DURATION OF 50Hz TOUCH VOLTAGE

slide-16
SLIDE 16

At 162V, total protection clearance time (relay setting plus contactor actuation time) must be ~100msec. If the operating time of our protection exceeds 100msec we must increase the NER impedance or reduce the return earth continuity limit accordingly.

  • 1. The presence of an NER does not make the system safe.
  • 2. Safety is reliant on active earth leakage protection.
  • 3. If our active protection fails to operate we must restrict the

continuous touch potential to less than 25V to remain safe.

slide-17
SLIDE 17

5 Amp earth fault current limit, 35+40 = 75 Ohms earth return impedance. Prospective touch voltage is 75/(115+75)*577 = 228V. Clearance time: earth leakage relay 50 msec interposing relay delay 20 msec circuit breaker delay 130 msec. the total clearance time is around 200 msec.

slide-18
SLIDE 18

200msec clearance time for the Lp curve requires 110V touch potential limit. Can adjust the NER, E/C limit or both

slide-19
SLIDE 19

Resultant touch voltage is 134V

  • Clearance time

400msec (E/L relay 250msec, interposing relay 20msec, CB delay 130msec). At 400msec clearance time, touch voltage req’d ~57V Back calculate current limitation for 35 Ohm return impedance as 1.8A

slide-20
SLIDE 20

PROPOSED AMENDMENT OF REG 100(1):

In the interest of safety, the earth fault current shall not be more than 750 milliampere in installations of voltage exceeding 250V and up to 1100V for below ground mines and oil fields, and 10 A in installations of voltage exceeding 1100V and up to 11KV for belowground mines, open cast mines and oil mines or oil fields, and the magnitude of the earth fault current shall be limited to these specified values by employing suitably designed, restricted neutral system of power supply including neutral monitoring protection. Appropriate equipment shall be suitably placed for automatically disconnecting supply to any part of the system in the event of earth fault or earth leakage. Provided that the settings of protective devices thereof shall not exceed 100 mA for individual apparatus of voltage Provided that the settings of protective devices thereof shall not exceed 100 mA for individual apparatus of voltage up to 1100V. Provided further that in installations where magnitude of aggregate capacitive current is more than 10A, the

  • wner of installations or user shall limit the earth fault current to appropriate value, with the approval of Electrical

Inspector. Appropriate PPE shall be used by the personnel for handling such apparatus to protect against electric shock due to touch potential.

slide-21
SLIDE 21

UNGROUNDED SYSTEM:

IA= IB= IC = 1A UNFAULTED

IA= 0A IB= IC = √3*1 = 1.73A IF= - (IB + IC ) = 3A PHASE-A FAULTED

N

slide-22
SLIDE 22

RESTRICTED NEUTRAL SYSTEM

  • 1. Touch

potential for a typical 3.3KV network with solidly neutral system will be around 1905.25V.

  • 2. Touch

potential for a typical 3.3KV network with 50A restricted neutral system will be around 1905.25÷(38.10+45)Ax 45Ω=1031.72V.

  • 3. Touch

potential for a typical 3.3KV network with 5A restricted neutral system will be around 1905.25÷(381.05+45)Ax 45Ω=201.23V.

slide-23
SLIDE 23

Vs > VR

In case when the proportion of capacitive current is lesser than resistive current: desairable

VR > VS

In case when the proportion of capacitive current is more than resistive current: not desairable

slide-24
SLIDE 24
slide-25
SLIDE 25

What about 10:1 tripping ratio? Under 4871:2012 test current for E/L relay is 120% or 1.2:1 Typical tripping ratios in UK are 3:1, US are 2.5:1

  • Tripping ratio should be maintained high as possible, but at 350mA trip (5:1) system would be

safe

  • Increased trip current avoids sympathetic trips on unfaulted outlets and relays still

proven/tested to trip

slide-26
SLIDE 26

JAMADOBA REGIONAL SUB-STATION, M/S TATA: 10MVA, 33/11KV TRANSFORMER

CASE STUDIES:

JEETPUR COLLIEY, M/S SAIL: 2MVA, 11/3.3KV TRANSFORMER

1. 2.

slide-27
SLIDE 27

MONIDIH PROJECT SUB-STATION, M/S BCCL: 10MVA, 33/6.6KV TRANSFORMER

3.

  • 1. With 4 nos. of feeders in circuit = 5.67A
  • 2. With 3 nos. of feeders in circuit = 4.32A
  • 3. With 1 nos. of feeders in circuit = 1.07A

Magnitude of capacitive currents for various schemes:

  • 3. With 1 nos. of feeders in circuit = 1.07A
slide-28
SLIDE 28

√(62+152) = 16.15A √(62+102) = 11.66A √(4.52+152) = 15.66A √(4.52+102) = 10.96A

slide-29
SLIDE 29

NGR5

Icapacitive

NGR10

NGR15

VNGR10 VNGR15

current through human body of 1000Ω resistance

INGR10

INGR15 TOUCH POTENTIAL or SHOCK VOLTAGE

VNGR5 INGR5

netwok voltage

201V 1270Ω 762Ω 381Ω 217mA 212mA

6A

635Ω

423Ω

420V

610V

217V

254Ω

402V

573V

402mA

573mA

212V 201mA 364mA

498mA 11KV 6.6KV 3.3KV 6A

190Ω

127Ω

364V

498V

420mA

610mA 6A

380Ω

slide-30
SLIDE 30
  • Capacitance between

A and B in the delta system = Cc + 0.5CC = 1.5Cc .

  • C1

1.5 Cc = 0.5C1, C1 = 3 Cc 1.5Cc .

C0 = C1 + CS = CS + 3CC = A

slide-31
SLIDE 31
slide-32
SLIDE 32

Cb CS

C0 = C1 + CS = CS + 3CC

CA-BC CC+ CS

slide-33
SLIDE 33
slide-34
SLIDE 34

Thank you all.

For any doubt feel free to contact me at 9771499979, 9434070734 ajaysingh.ddms@gmail.com