protection system

Hi gents!

We are recently working on a substation having 132/11kV system with 36MVA transformer (Y-Y with treasery winding). There are differential, O/C, U/V, O/V, E/F and all other transformer protections is there with individual protection for different feeders (S/C, E/F, O/C).

I have following quires:

1. What is the difference b/w unrestricted and restricted earth fault protection, how they differ and what are there selecting criteria according to application, which one provided more accurate and complete protection?
2. What is the setting procedure for differential protection?
3. Selection criteria of protection relays setting i.e. O/V, U/V, O/C and S/C.
4. What is directional O/C relay, how they differ from no directional relay, is there any advantage for using directional relay?
5. Plz info me any web site given me details about protection relays (cables feeder protection, …)?
6. Is any one having info about E-Tap software?


Best regards
ASD

slip ring motor resistance

Hi sir

I have seen your effective calculations for selection of slip ring motor resistance
I have still some confusion in my mind as I have following relation to calculate the resistance
Rotor coefficient is
K = U (lock rotor voltage) /I (rated current of rotor at rated torque). Root 3

in (ohm /phase)

And rotor resistance is then R = S. K (ohm /phase)

Slip = S =(Ns – N / Ns)

Therefore for motor rotor current = 224 A
Rotor V = 579v with rmp = 1477, 4 pole

The value is

K = 1.5

S = (1500-1477/1500) = 0.0153

R = 0.0153 * 1.5 = 0.022 ohm per phase

Plz help me in this regard


An advance thanks to all senior who will guided me.

filtre ckt for harmonics

hi friends

We have a filter circuit in order to reduce the harmonic current and improve the pf
Rating is: 600 MVA

Specification of air-cooled inductor:
Rated v= 11kv, rated L = 8.65 mH, rated I = 334A, rated f = 50HZ, harmonic A = 40A harmonic f = 150Hz,

Specification of capacitor:
Rated O/p = 500KVAR, rated V = 8.28KV, rated C = 23.2 micro F, rated I = 60A, rated f = 50Hz

The connection is such that with one coil (inductor) we have 6-cap bank (i.e. with 3 inductor we have 18 cap, 6 cap / inductor) in double star configuration.

I have find out the tuned frequency as

f filter = 1 / 2 *pi * root LC = ½ *3.14 * root (8.56 m H * (6*23.2 micro F)) = 145 Hz

1. We have fundamental frequency of 50 Hz, and hence the 5th harmonic frequency is 150 HZ. Why the harmonic frequency I have calculated differ from the 5ht harmonic i.e 150HZ, is there some correction is required?

2. Also the said filter ckt is connect to 30 MVA transformer, 132/11KV, %Z= 10

The inductance of transform is therefore
L Transformer = L total – L filter

L total = (1/2* pi *f)^2 * 1/ c
L total = (1/2* pi *50)^2 * 1/ 6*23.2 micro F = 72.7 mH

L transformer = 72.7 mH – 8.65mH = 64.05 mH (Is this correct?)

3 Is the 5th harmonic current is higher than the other harmonic current i.e. 7ht and other higher harmonics.


4. Why should we connect the capacitor in double star instead of delta connection? Is relation? (C delta = 3 C star)

muzamil

earth wire size fro motors

how to calculate the SWG for earth wire , it also depend on load ?
some body tell me the general rule i.e earth wire size half the cables used for equipment.plz tell me in detail

p.f calculation for transformer

We have system of two single-phase transformers in one unit (Varivolt system) i.e. secondary of transformer T1 is connected to the primary of second transformer T2. By varying the secondary of T1 we get variation in primary and secondary of T2.The specification of transformers:

T1: 1 phase 1302KVA, 50HZ, primary 11kv, 118A, secondary 330-1210v, 1207A with Z%= 11
T2: 1 phase 1500KVA, 50HZ, primary 1119kv, 1250A, secondary 2*93.3 v, 2*7500A with Z%= 6

B/w two connection i.e. secondary of T1 and Primary of T2 we have 3 individual cap banks, each rating is

Qn = 300KVAr, Un= 1200V, Cn=663microF, Ic = 250A, Hz = 50

The load at secondary of T2 is an arc furnace with max load of 850KW.

As I know through the seimens hand book that the cap rating for transform is near to the 80-90% of the reactive full load power of the transform. With formula
Qtr = Q0 + Uz/100% (S/Sn) ^2 * S

Where Qtr = transformer reactive power, Q0 is the transformer no load power, S = apparent power that a transformer is supplied, Sn = rated power for transformer, Uz is the % impedance of transformer.

Here in our case
Q0 from seimens graph = 27 Kvar
S= 1302 for full load
Thus Qtr = 27 + 11/100 (1302/1302)^2 * 1302 = 170KVAR

Which is not fulfill in our case, can you suggest me what is the exact relation to calculate the cap bank KVAr and step required to control the p.f in case of transformer, is the bank rating equal to kva rating of transformer?

Is there any effect on the secondary voltage of the transform if the cap bank is over rated i.e. if load on transform is reduced?