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1-9eses01 (Q2, Q3)

The document discusses settings for protective relays on generators. It analyzes fault currents and selects pickup values, time dial multipliers, and curves for the protective relays. Despite adjusting some settings to be as fast as possible, it does not provide backup overcurrent protection to the generators due to the thermal damage curve of (9ESES01, 9ESES01). Recommendations include using other types of curves besides Extremely Inverse Time.

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mahmoud12122012
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27 views4 pages

1-9eses01 (Q2, Q3)

The document discusses settings for protective relays on generators. It analyzes fault currents and selects pickup values, time dial multipliers, and curves for the protective relays. Despite adjusting some settings to be as fast as possible, it does not provide backup overcurrent protection to the generators due to the thermal damage curve of (9ESES01, 9ESES01). Recommendations include using other types of curves besides Extremely Inverse Time.

Uploaded by

mahmoud12122012
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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1- 9ESES01 (Q2,Q3)

1-1 I fl = S/(sqrt(3)*V) = 2500/(sqrt(3)*6.6) =219 A

1-2 I max = 219 * 1.25 = 273.75 A

1-3 CT : 250 / 1 A 15 VA 5P20

1-4 I max.sec = 273.75 / 250 = 1.095

1-5 Relay :

1-5-1 Manufacturer : Square D

1-5-2 Model : Sepam Series 40

1-5-3 Function :

1-5-3-1 51P :

1-5-3-1-1 Curve : Extremely Inverse Time

1-5-3-1-2 PU : 1.1 ( Pick up value)

1-5-3-1-3 TM : 2.69 (Time Dial Multiplier)

2- 9ESES01 (Q5)

2-1 I lf = I fl (9ESES01-Q3) – I lf (9ESES01-Q9) = 219 – 38.4 = 180.6 A

Or

I lf = I fl (9ESES01-Q2) – I lf (9ESES01-Q7) = 219 – 12.4 = 206.6 A

So , We take

I lf = 206.6 A

2-2 I max = 206.6 * 1.25 = 258.25 A

2-3 CT : 1000 / 1 A 30 VA 5P20

2-4 I max.sec = 258.25 / 1000 = 0.25825 A

2-5 Relay :

2-5-1 Manufacturer : Square D

2-5-2 Model : Sepam Series 40

;ϭͿ



2-5-3 Function :

2-5-3-1 51P :

2-5-3-1-1 Curve : Extremely Inverse Time

2-5-3-1-2 PU : 0.3 ( Pick up value)

1-5-3-1-3 TM : 1.59 (Time Dial Multiplier)

3- 9ESES01 (Q1,Q4)

3-1 I lf = I lf (motor) + I (starting) + 3 * I (Safe Shutdown)

= 98.4 + 1023/(sqrt(3)*6.6) + 3 * 523/(sqrt(3)*6.6)

= 325 A

3-2 I max = 325 * 1.25 = 406.25 A

3-3 CT : 1000 / 1 A 30 VA 5P20

3-4 I max.sec = 406.25 / 1000 = 0.40625 A

3-5 Relay :

3-5-1 Manufacturer : Square D

3-5-2 Model : Sepam Series 40

3-5-3 Function :

3-5-3-1 51P :

3-5-3-1-1 Curve : Extremely Inverse Time

3-5-3-1-2 PU : 0.4 ( Pick up value)

3-5-3-1-3 TM : 0.44 (Time Dial Multiplier)

3-5-3-2 50P : (If Applicable)

I fmax (Maximum Fault Current at Far End) = 5.16 KA

I (instantaneous.) = 1.75* I fmax = 1.75*5160 = 9030 A (To avoid overreach due to DC


component)

I (inst,sec.) = 9030/1000 = 9.03 A

;ϮͿ



So we set (50P) as follows

3-5-3-2-1 I set = 9.1

3-5-3-2-2 With Zero Time Delay

4- 1ESES01 (Q4)

4-1 I lf = I lf (motor) + I (starting) + 3 * I (Safe Shutdown)

= 98.4 + 1023/(sqrt(3)*6.6) + 3 * 523/(sqrt(3)*6.6)

= 325 A

4-2 I max = 325 * 1.25 = 406.25 A

4-3 CT : 1000 / 1 A 30 VA 5P20

4-4 I max.sec = 406.25 / 1000 = 0.40625 A

4-5 Relay :

4-5-1 Manufacturer : Square D

4-5-2 Model : Sepam Series 40

4-5-3 Function :

4-5-3-1 51P :

4-5-3-1-1 Curve : Extremely Inverse Time

4-5-3-1-2 PU : 0.4 ( Pick up value)

4-5-3-1-3 TM : 0.1 (Time Dial Multiplier)

5- 2ESES01 (Q12)

5-1 I lf = I lf (motor) + I (starting) + 3 * I (Safe Shutdown)

= 98.4 + 1023/(sqrt(3)*6.6) + 3 * 523/(sqrt(3)*6.6)

= 325 A

5-2 I max = 325 * 1.25 = 406.25 A

5-3 CT : 1000 / 1 A 30 VA 5P20

5-4 I max.sec = 406.25 / 1000 = 0.40625 A

;ϯͿ



5-5 Relay :

5-5-1 Manufacturer : Square D

5-5-2 Model : Sepam Series 40

5-5-3 Function :

5-5-3-1 51P :

5-5-3-1-1 Curve : Extremely Inverse Time

5-5-3-1-2 PU : 0.4 ( Pick up value)

5-5-3-1-3 TM : 0.1 (Time Dial Multiplier)

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