Over Current Grading For Different Relays

Published Date: 02 Nov 2017

ABSTRACT:

In this assignment, it is required to perform an over-current grading for different relays with different characteristics according to the given schematic respectively.

When a system has two or more relays, it is necessary to grade them with co-ordination of the relays with each other otherwise the relays might be working on a same time causing healthy sections disturbance and might not be working at a desired time which damages the whole system.

The characteristic of each relay has to be plotted in such a way that the operating relay time should be decreased with the increasing of fault current and for grading the relay time is being added to operating time to provide the grading between different relays to achieve a good protection scheme for the given entire system. However, a differential protection scheme provides the protection on the basis of comparison of current transformers. Hence, the main aim of the grading is to provide the proper grading with complete protection.

INTRODUCTION:

By using the given data firstly fuse A, fuse C, relay B and relay J characteristics are plotted by taking 3.3kv as the base voltage then the grading is done with the upstream relays.

The following specifications were provided with which all the calculations should be done:

Commence fault levels from source fault level of 500MVA

Use of 0.4s discrimination interval between the relays

Fault currents on 3.3kv base voltage

Origin of vertical time axis is 0.01second

Origin of horizontal time axis is 10A

The relay setting current is (P121)from 0.1xIn to 25xIn

The relay high-set setting has the following range of 0.5 to 40xIn

The relay TMS(P121) is from 0.01 to 1.5

Minimum relay operating time is 0.02s

Motor rating is 200kW, 0.8p.f., 3300v, 50HZ

For over-current relays, C.T. winding resistance=0.05Ω and one way lead resistance=0.1Ω.

For differential relays, charging current=45A, single lead resistance=0.15Ω, knee-point voltage=38v, winding resistance=0.214Ω and X/R ratio=20.

The characteristics of the fuse A and fuse C are given as follows:

Fuse A characteristics:

Fault Current (A)

430

650

1150

2000

3506

Arc Extinction Time (s)

100

10

1.0

0.1

0.01

Fuse C characteristics:

Fault Current (A)

300

400

500

600

700

800

Arc Extinction Time (s)

120

25

7.0

2.2

0.9

0.45

P121 (RELAY B): PLOT

Given data for the relay B is as follows

Type : extremely inverse

C.T ratio : 100/5A

TMS setting : 0.1

IDMT setting : 1.0 In

Instantaneous current setting : No

But for an extremely inverse type relay

t = x TMS

From IDMT settings, setting current, I set = 1.0 In = 1*100=100A

Whereas the actual current is the twice of the setting current that is

I actual = 2*100=200A

Therefore

t =

t = 2.67 seconds.

The relay fastest operating time is 0.02sec but however for some fault levels , the operating time may be smaller than 0.02sec.

The maximum fault current at location B by using the 3 phase fault level of 41.3MVA is

If max =

If max = = 7226A

Multiple of setting

operating current at 3300v in amps

Operating time for TMS=0.1 in sec

1.1

110

38.10

1.2

120

18.18

1.5

150

6.40

2

200

2.67

4

400

0.53

6

600

0.23

8

800

0.13

10

1000

0.08

20

2000

0.02

72.26

7226 (If max)

0.02

Key point:

The minimum operating time, t= 0.02 sec, however it is not sure that operating time is less than 2 times setting.

1.1 P121 (RELAY J): PLOT

The relay J can be plotted same as the relay B and the position of relay J is at the source at 11kv and hence it is important to provide the grading for rest of the system.

Given data for the relay J is as follows

Type : standard inverse characteristics

C.T ratio : 3000/5A

TMS setting : 0.45

IDMT setting : 1.21 In

Instantaneous current setting : No

But for a standard inverse characteristics type relay

t = x TMS

here from the C.T ratio (primary) In = 3000A

also given IDMT setting, I set = 1.21 In = 1.21 x 3000=3630A

Whereas the actual current is the twice of the setting current that is

I actual = 2*3630 = 7260A

Therefore now

t =

t = 4.51 seconds.

By using the multiples of setting current, the operating times for the relay J is plotted.

Maximum fault current calculations:

The maximum fault current at location J by using the 3 phase fault level of 500MVA at the source is given by

If max =

If max = = 26.243kA

If max =26.243KA at 11kv

Key point:

The voltage base of this relay J has to be changed from 11kv to 3.3kv, therefore with respect to 3.3kv

If max =

= 26.243x x

= 87.477 x A

If max = 87.477kA at 3.3 Kv

Multiple of setting

Operating current at 11kv in amps

Operating current at 3.3kv in kA

Operating time for TMS=0.45

1.1

3.99

13.3

33.01

1.2

4.35

14.5

17.24

1.4

5.08

16.93

20.73

2

7.26

24.2

4.51

2.8

10.164

33.86

3.02

4

14.52

48.4

2.24

4.5

16.33

54.43

2.06

5

18.15

60.5

1.92

6

21.78

72.6

1.72

0.007

26.243

87.477

1.56

Key point:

1) It is not sure that operating time is less than 2 times setting.

2) Definite time is achieved for SI characteristics only after 31x Is

1.1) Introduction to relay settings:

To plot the characteristics of all the relays present in the entire system it is important to represent them on a common base voltage.

Therefore, let the voltage base be 3.3 KV base.

After plotting and studying the characteristics of fuse A, fuse C , relay B and relay J the next stage is to plot the characteristic and consider the over current relay settings at 3.3KV.

1.2) P121: Relay D: (plot):

To maintain an adequate time margin at maximum fault level with relay B, the relay D must also have a current and time setting for discrimination

Hence, extremely inverse characteristic must be adopted for relay D in order to allow adequate discrimination with the 160A fuse at small currents.

The settings should ensure that the relay will not pick up if the 200Kw motor is started when the transformer is at full load current.

Therefore, the relay setting is

(Transformer F.L.C.) + ((n-1) Motor F.L.C.)

Here n= 5 which is the multiple of the motor at full load current.

Transformer full load current =

=

=875A

Motor full load current =

=

= 43.74A

Therefore the relay setting current is

=(Transformer Full Load Current.) + ((n-1) Motor Full Load Current)

= 875 + ((5-1) x 43.74)

= 1049.96 A

Allowing 95% reset and 10% safety margin we get

IS = = 1216A

For the graph 1216A is rounded to nearest value that is 1220 A .

In terms of C.T ratio the setting current is

IS = = 1.22 In

( since 1000ª is the primary of C.T relay D )

Therefore current should be set greater than 1.22 In also 0.4 sec discrimination must be taken between relay D and relay B at the maximum.

3 phase fault on the 3.3KV at 41.3 MVA that is

The fault current =

= = 7226A (IACTUAL)

Therefore fault current is 7226A at 3.3kV base voltage.

Now, operating time at TMS=1 for maximum fault current is given as

t = x TMS

t =

t= 2.35 sec.

now, for discrimination between relay D and relay B

required total operating time = (relay B operating time )+( grading margin)

= 0.02+0.4

= 0.42 sec

This time is very slow, and hence a new TMS will be required that is

TMS =

TMS = =0.178

TMS = 0.18 as it is set to the nearest value.

Therefore the relay D for TMS = 0.18 can be plotted by using the following table

Multiple of setting

operating current at 3300v in amps

Operating time for TMS4=0.18

1.1

1342

68.57

1.2

1464

32.72

1.4

1708

15.0

1.6

1952

9.23

2.0

2440

4.80

4.0

4880

0.96

5.92

7226(max fault)

0.42

1.3 Relay E (plot):

When grading is done through a delta/star transformer due to a phase to phase fault condition on the low voltage side of star which produces the same value of fault current in one of the phase on the high voltage side of delta.

Hence, discrimination for relay D at 0.866 x 7226 A and 7226 A at relay E must be ensured failure of which can result in inadequate discrimination.

Extremely inverse chat be characteristics must be chosen for relay E for proper discrimination with relay D and other downstream equipments.

Relay current and time setting should be maintained for time margin with transformer on full load current.

Moreover, the relay should not pick up if the 200kw motor is started.

On high voltage side of transformer instantaneous high set elements can be applied.

Therefore, the relay E setting is

=(Transformer Full Load Current.) + ((n-1) Motor Full Load Current)

Where n=5 is the multiple of the motor at full load current.

Transformer full load current =

=

=262.4 A

Motor full load current =

=

= 13.12 A

Therefore the relay setting current is

=(Transformer Full Load Current.) + ((n-1) Motor Full Load Current)

= 262.4 + ((5-1) x 13.12)

= 314.88 A

Allowing 95% reset and 10% safety margin we get

IS =

IS = 364.6 A

With respect to 3.3KV side

Now IS = 364.6 x

= 364.6 x

IS = 1215 A

Therefore, the setting current of relay E is same as relay E and hence there is a chance that the relay E could pick up before relay D.

Hence to avoid this a setting of 1.1 times the relay D is considered for relay E settings

Therefore, I set = 1.1 x I set (for relay D)

= 1.1 x 1220 A

= 1342 A

= 1350 A as the nearest value.

That is I set = 1.35 In

In terms of C. T ratio

I s = 1.35 x 300 = 405 A (on 11Kv base)

I s = 1.35 x 1000 1350 A (on 3.3 KV base)

0.4 sec discrimination is considered between relay D and relay E.

Also 3 phase fault on the 3.3Kv at 237MVA is given as

Max fault current =

= = 41464.2 A

Therefore, operating time for relay E at TMS =1 is

t = x TMS

where IACTUAL = maximum fault current at 3.3KV busbar of 41.3 MVA

t =

t = 2.89 sec

Fault conditions must be considered for both sides to grade across the transformer.

Therefore relay D fault current = 0.866 x If max

= 0.866 x 7226 A

= 6258 A

Relay D operating time for above fault level is

t = x TMS

t =

t = 0.57 sec

Required TMS setting =

= 0.34

Therefore TMS of relay E = 0.34

High set for relay E:

Given:

Setting range of relay E for high set = 0.5 to 40 In in step sizes of 0.05 In

The hi set can be set to 120% to 130% of the maximum LV fault level. This will give fast clearing times for faults on the transformer HV bushings and part of the way into the transformer.

The hi set is set in multiples of the IDMT setting current

Therefore, hi set current = 1.3 x 7226 = 9393.8 A

With respect to 11 KV

Hi set current = 9393.8 x 3.3KV/11KV = 2818.14 A

In terms of C.T ratio that is 300 A primary

Select current = 9.39 In =9.4 In

At 11 KV base Is = 9.4 x 300 = 2820 A

3.3 KV base Is = 9.4 x 1000 = 9400 A

For TMS = 0.34 and 1.35 In relay curve can be plotted by using the following table:

Multiple of setting

operating current at 3300v in amps

Operating time for TMS=0.34

1.1

1485

129.52

1.2

1620

61.81

1.4

1890

28.33

1.6

2160

17.43

1.8

2430

12.14

2.0

2700

9.07

3.0

4050

3.4

4.0

5400

1.81

6.0

8100

0.78

6.96(hi set value)

9400

0.02

30.7(max fault current at 237 MVA)

41464.2

0.02

P521: Relay F (plot):

The basic operating principle of this type of relay protection is to see the difference between the currents leaving and entering a protected zone. The relay operates when this difference of two currents exceeds the threshold value.

Relay F given is an differential current relay with p521 pilot wire protection.

The relay adopts the biasing techniques to provide good stability which is helpful to avoid mal-operation of the relay.

Due to C.T saturation during external faults differential currents may be generated.

The bias current is the average of the measured current where as the differential current is the vector sum of the currents leaving and entering the protected zone

Therefore, I bias =

I differential = I1 +I2

The operating characteristic of p521 is shown in figure 1 below:

Figure: relay bias characteristics

Relay F settings are as follows:

Relay F characteristic settings :

The four protection setting characteristics are defined as follows:

Is1 : the basic differential current setting which gives the minimum pick up level of the relay

K1 : the smaller percentage bias used when the bias current is below Is2

Is2 : A higher percentage bias k2 is used for above the bias current threshold setting.

Bias

K2 : to improve relay stability by using the higher percentage bias setting.

The default settings are used for the relay which is given as follows

K1 = 30% = 0.3

K2 = 150% = 1.5

Is2 = 2.0 p.u

Capacitive charging current :

The setting on the most sensitive phase should be always greater than the steady state line charging current to avoid the mal-operation on line charging currents.

Now,

Is1 is given as

Is1 > 2.5 x

Is1 > 2.5 x 45/450 = 0.25 A

Therefore, Is1 = 0.25 In

The next setting available is 0.26 A

Current differential operating time setting: (td)

Td is set to zero initially to trip the relay instantaneously.

The IEC knee point voltage is given by

Vk ≥ KS . KT . IN (RCT + 2RL)

Here

kT = operating time factor

Ks = dimensioning factor

RL = one way lead impedance from CT to relay = 0.15 Ω

RCT = CT dc resistance = 0.214Ω

IN = CT nominal secondary current = 1 A

Vk ≥ KS . KT (0.214 + 2(0.15))

Also the dimensioning factor Ks for x/R < 40 is given as

KS = 0.023 x If (X/R + 55) + 0.9 (X/R + 26)

The fault current, If = =

= 10.392kA

Now,

Ifault =

=

= 10392/450

= 23.09 A

Now, the dimensioning factor is given as follows

KS = 0.023 x (20 + 55) + 0.9 (20 + 26)

KS = 0.023 x (75) + 0.9 (46)

KS = 81.236

KT value for td =0 should be always taken as 1

Now, the knee point voltage is

VK ≥ 81.236 x 1 x 1 (0.214 + 2(0.15))

VK ≥ 81.236 x (0.214 + 0.3)

VK ≥ 41.75V

Given knee point voltage should be 38 V to get this new value of KT is needed that is

KT =

KT = 0.910

Again substituting new value of KT in we get

VK ≥ 81.236 x 0.91 x 1 (0.214 + 2(0.15))

VK ≥ 81.236 x 0.91 x 1 (0.214 + 0.3)

VK ≥ 81.236 x 0.91 x 0.51

VK ≥ 37.99V

Also value of KT is to be adjusted accordingly as it is a constant which depends on differential operating time setting (td)

Therefore

KT = 1 – (6.2 x tIdiff)

Td =

= = 0.0145 sec = 14.5 msec

Therefore as KT = 0.91 , set td =15msec the next setting available will be 20msec.

Relay G plot :

For overcurrent relay G an extremely inverse time characteristic is chosen to provide the proper discrimination with the upstream protection . At its high set setting relay G should discriminate with relay E.

The protection at location G can also be provided by p521 instead of p121.

The settings must be same as at the extreme end of the feeder that is as follows:

The default settings are used for the relay which is given as follows

K1 = 30% = 0.3

K2 = 150% = 1.5

Is2 = 2.0 p.u

Given data for relay G

Maximum full load current per feeder = 420A

C.T ratio = 450/5A

Allowing 95% reset and 10% safety margin we get

Is= 420 x

= 486.31 A

= 486 A at 11kv

110% of setting should be considered that is

In terms of C.T ratio (450 A primary)

Is = 1.1 x In

= 1.1 x 450 = 495 A at 11 kv base.

With respect to 3.3kv base

Is =

The high set value is 9400 A which should be set with relay G that is I actual = high set =9400 A

Also the operating time at TMS=1 is

t = x TMS

t =

t = 2.54sec

now for relay E we have

I actual = high set =9400 A

Is = 1350 A

TMS= 0.34

Therefore operating time is

t = x TMS

t =

t= 0.57sec

required operating time = 0.57 +0.4 = 0.97 sec

Therefore required TMS is given as

TMS =

=

= 0.381

TMS = 0.39 (next value)

Relay G can be plotted at TMS =0.39 and IS = 1650A at 3.3 kv base.

Multiple of setting

Primary current at 3300v in amps

Operating time for TMS=0.1

1.1

1815

148.57

1.2

1980

70.90

1.4

2310

32.5

1.8

2970

13.92

2.0

3300

10.4

3.0

4950

3.9

4.0

6600

2.08

5.0

8250

1.30

5.7

9400

0.99

Fault calculation: there should be proper discrimination between relay E and relay G at 198 MVA

If max =

If max = = 34.641kA

The operating time of relay E is = 0.02 sec

But for 34.64kA it should be = 0.02+ 0.4 = 0.42 sec

Now, the actual operating time for relay G at 34.64kA level is

t = x TMS

t =

t= 0.071sec

Proper discrimination must be ensured where as the multiple of current setting is given as

I =

=

= 13.83

Hence, the primary setting = = 2510A at 3.3kV base

Primary setting at 11kv base =

In terms of C.T ratio (450A primary)

Is = 753/450

= 1.67 In = 1.68 In

Therefore relay G settings should be increased to 1.68 In that is 168% to ensure good discrimination.

At 11kv base

Is = 1.68 x 450 = 756 A

At 3.3 kv base

Is = 1.68 x 1500= 2520 A

Fault calculation: maximum fault current at 294 MVA on 3.3 kv base is

If max =

If max = = 51436.67A = 51.437 KA

Therefore the operating time of the relay can be ensured as

t = = 0.42sec

Now set TMS =1

The relay G can be plotted by using the following table at TMS=1 and Is = 1.68 x In

Multiple of setting

operating current at 3300v in amps

Operating time for TMS=1

1.1

2772

380.95

1.2

3024

181.81

1.4

3528

83.33

1.6

4032

51.28

2.0

5040

26.67

4.0

10080

5.34

6.0

15120

2.28

10

25200

0.81

20.41( If max)

51437

0.19

The curve is plotted upto the maximum fault level of 51.437 KA at 294 MVA

P121 relay H (plot):

Supply authority relay J and relay H should have proper discrimination for this reason standard inverse characteristics must be selected for relay H.

Given data for relay H

C.T ratio = 3000/5A

Maximum full load current per feeder = 2775 A

Allowing 95% reset and 10% safety margin we get current setting as

IS = = 3213.15A = 3213 A

In terms of C.T ratio (3000A primary ) the setting current is

IS = = 1.071 In

For 11kv base

Is = 1.07 x 3000 = 3213 A

Also for 3.3 KV base

Is = = 10710 A

For parallel sharing feeder relay H will see full fault current that is 41464A where as relay G will see only half of it that is 20732

Therefore operating time is

t = x TMS

t =

t= 1.2sec

with discrimination

required operating time , t= 1.2 + 0.4 = 1.6 sec

now, actual operating time for relay H at 41464A is

t = x TMS

t =

t= 5.10sec

Therefore, required TMS =

= 1.6 / 5.10 = 0.313 =0.31

Therefore, the curve for relay H can be plotted at TMS = 0.31 and IS = 1.071 In

Multiple of setting

operating current at 3300v in amps

Operating time for TMS=0.31

1.1

11781

22.75

1.2

12852

11.88

1.4

14994

6.43

1.6

17136

4.60

2.0

21420

3.11

3.0

32130

1.95

4

42840

1.54

4.802

51437

1.36

Overview:

Relay Type

Relay

location

C.T.

Ratio

Relay settings

Time

Current

P121

B

100/5

TMS = 0.1

1.0In

P121

D

1000/5

TMS = 0.18

1.22In

P121

E

300/5

TMS = 0.34

1.35In

P521

F

450/1

td = 0.02

IS1 = 0.26In

P521

G

450/1

td = 0.02

IS1 = 0.26In

P121

450/5

TMS = 1.0

1.68In

P121

H

3000/5

TMS = 0.31

1.07In

P121

J

3000/5

TMS = 0.45

1.21In

Figure : over current grading characteristics

CONCLUSION:

The coordination of different relays has been achieved successfully and the corresponding characteristics and graphs has been drawn to ensure a good protection scheme from high fault currents such that the relays will operate in much shorter time(few Milli seconds) to provide complete protection for the given schematic system. Hence the continuity of supply can be maintained.