XLPE cable

XLPE(Cross Linked Poly Ethylene) Cables have advantages over conventional paper insulated/ Oil filled cables

í          Lower dielectric losses

í          Lower weight

í          Higher permissible continuous and short circuit current withstand capacity

í          Environment friendly - no risk of oil leakage in to environment

í          Easy installation of accessories

í          Maintenance free

í          No problem with vertical  installation

í          Improved flame retardant  property with PVC sheath, due to absence of any oil etc.

Manufacturing range

í          Cable sizes                                          : 95 sq mm - 2000 sq mm

í          Voltage grade                                    : 66 kV - 220 kV (extendable up to 400 kV)

í          The basic material for XLPE is low density polyethylene(LDPE)- which is thermoplastic. For EHV, the cross linking takes place in an electrically heated tube of extrusion line , in an inert gas atmosphere.The gas pressure in tube is kept around  5-10 bar to avoid formation of voids due to peroxide decomposition.

Purpose of cross linking

î          To improve thermal  rating - continuous , overload & short circuit temperature ratings

î          To improve mechanical properties

î          To improve deformation resistance

Property                                                                     PE                           XLPE

Max  continuous operating temperature ( °C)                            70                           90

Max  short circuit temperature                ( °C)                         150                         250

Dielectric constant                                                                    2.3                          2.3

Dielectric strength(kV/mm)                                                     > 22                        > 22

Tan d                                                                                   0.0005                   0.0005 

Series compensated lines - Distance protection

Why series compensation?

1. Increased power transfer capability
2. Better system stability
3. Reduced transmission system losses

Benefits

1.    Optimization of power capabilities

2.    Power sharing

3.    Technical & environmental benefits

Criteria

1.    Strong transmission links

2.    Maximum power transfer within the steady state stability limits

Possible measures

1.    Increase no of parallel feeders

2.    Increase transmission voltage

3.    Decrease the transfer impedance using series compensation

• Spark gap – to bypass C during high current faults
• Circuit breaker – to close during high current faults and also to discharge TRV (Transient Recovery Voltage)
• To control the voltage across the series capacitor

 Problems:

1. Voltage inversion
2. Current inversion
3. Distance estimation

Current Inversion

·         I_R inductive as impedance to fault has X_L which is inductive

·         I_S capacitive as impedance to fault has X_C which is capacitive

·         For an internal fault, I_R  & I_S are in opposite directions

·         Probability of non- operation of relay for a high impedance fault

·         Current inversion affects directional, distance, phase comparison scheme and differential protection

Voltage Inversion

·         If XC > m XL, voltages V & V’ are out of phase

·         Line side VTs detect a forward fault correctly

·         Bus side VTs detect a reverse fault correctly

Distance Estimation

·         Errors in distance estimates

·         Correct estimation when capacitor out of service – High current fault

·         Impedance reduces because of C ( ZL – j XC )

·         The set distance characteristic over reaches for a high impedance fault

·         Mho element reach can be reduced to 90% when C is in service

·         Reach reduces to 50% when C is out of service

·         For a high impedance fault close to the capacitor, relay of line section AB sees it as a reverse fault

·         Relay of adjacent line section may mal operate depending on the location of the fault

·         Hence, series capacitor can cause nuisance tripping

·         Memory polarization for voltage inversion à polarizing memory should be long enough

·         Blocking of zone-1 element for high impedance faults to prevent over-reach when capacitor is in circuit              

·         Reduce the reach of zone-1 when capacitor is in circuit.

·         Using line differential protection à excellent choice for a series compensated line 

Parallel Transformer Protection

If a transformer is connected in parallel with another transformer which is already energized, magnetizing inrush will occur in both transformers.  The dc component of the inrush associated with the switched transformer creates a voltage drop across the line resistance between the source and the transformer.  This voltage causes an inrush in the opposite direction in the transformer that was already connected.  After a time the two currents become substantially equal and since they flow in opposite directions in the transmission line they cancel and produce no more voltage drop in the line resistance.  The two currents then become a single circulating current flowing around the loop circuit made up of the two transformers in series - the rate of decay being determined by the R/L ratio of the transformer.

Parallel transformers are typically protected by directional overcurrent and earthfault protection on the LV side set to look back into the transformers. Where an LV bus section exists the directional relays can be replaced by non-directional relays, with the addition of a non-directional overcurrent and earthfault relay at the bus-section.

Protection are:

1. Directional Over current

2. Directional Earth fault

Busbar Protection Calculation

A typical 132kV double bus generating station is made up of two 100MVA generators and associated step-up transformers, providing power to the high voltage system, by means of four overhead transmission lines. If the main bus is sectionalised with a bus section circuit breaker and the reserve bus is split by bus section isolators, calculate for a high impedance circulating current scheme having 4 zones and an overall check feature, as follows :-

1.    The stability level of the busbar protection.

2.    The discriminating and check features primary fault settings.

3.    The value of stabilising resistor required with the MCAG14 relay whose current setting range is 0.2 – 0.8A and has a burden of 1VA at the current setting.

4.    The supervision relay primary setting, using a VTX31 with a voltage setting range of 2 – 14 volts. Assume the relay burden at the setting is 2 milli VA.

5.    Determine the CT knee point voltage requirements.

Given that the switchgear rating is 3500MVA, the system voltage is 132kV solidly earthed, the maximum pilot resistance is 4 ohms and that the non-linear resistors to be connected across the buswires to limit the peak volts follow the expression V=900I^0.25. The current transformers are of ratio 500/1 amp, have a secondary internal resistance of 0.7 ohms and have a magnetising impedance of 2000 ohms.

The layout of the station is a shown in the diagram.

Solution:

1.    The stability level of the busbar protection is governed by the switchgear rating which in this instance is
 = 15300 Amps

2.    The primary operating current of busbar protection is normally between 2% and 10% of the stability level i.e. between 306 amps and 1530 amps. Where 30% of the minimum fault current is more than the full load current of the largest circuit, the primary operating current should be made more than the full load current i.e. 130% of full load current.

Full load current =

                                  = 438 Amps

\ Primary operating current    = 1.3 x 438

                                                      = 569 Amps

Discriminating Feature

Relay voltage setting     =                                           =
                                          = 144 volts

Magnetising current taken by each C.T. at 144 volts
                                          =
                                          = 0.072 Amps

Metrosil r.m.s. current at 144V is given by:

 = = 1.36mA
In practice the metrosil current is considered to be negligible and is ignored.

To obtain a primary operating current of 569 amps the relay current setting must be calculated as follows :-

Primary operating current        = CT ratio (I_RELAY + n.I_CT)

Where n = maximum number of CT’s per zone (i.e. 5)

\ I_RELAY    === 0.778A

The nearest available setting is 0.8A.

This gives an actual primary operating current of :-

Primary operating current        = 500 (0.8 + 5 x 0.072)

                                                      = 500 x 1.16

                                                      = 580 Amps

Check Feature

Relay setting voltage                 = 144 volts

Relay setting current (I_s)           = 0.8 Amps

Maximum number of circuits   = 6

Primary operating current        = 500 (0.8 + 6 x 0.072)

                                                      = 500 x 1.232

                                                      = 616 Amps

3.    Ohmic value of stabilising resistor (R_ST)
      RST    =                   =                   = 178.44 ohms

4.    Voltage setting of the supervision relay Vsp = I (R // Zm)

R = Rst + Rr = resistance of stabilising resistor and relay

Zm = parallel impedance of CT magnetising impedances assuming 1 CT is open circuit

I = supervision operating current

Whenever possible the supervision primary operating current should not be more than 25 amps or 10% of the smallest circuit, whichever is the greater.

For the discriminating zone there are 5 CT’s. Therefore assuming one CT is open circuit :-

      Zm      =

                        = 500 ohms
     R         = Rst + Rr = 178.44 +                = 180 ohms

Assuming I = 25 amps
      V         =

                  = 6.62V

For the check zone there are 6 CT’s. Therefore assuming one CT is open circuit
      Zm      =                  = 400 ohms
      V         =                  = 6.2V

5.    Knee point voltage of the current transformers should not be less than twice the relay setting voltage.
      V_K        = 2 x 144
                  = 288 volts    

Directional Earth fault

Step 1: Operating signal & polarizing signal

Requirements are similar to directional overcurrent 

i.e.       need operating signal

            and polarising signal

Operating Signal

obtained from residual connection of line CT's

i.e.       I_op  =  3I_o

Polarising Signal

The use of either phase-neutral or phase-phase voltage as the reference becomes inappropriate for the comparison with residual current.

Most appropriate polarising signal is the residual voltage.

Residual Voltage

Notes :

1.         VT primary must be earthed.

2.         VT must be of the '5 limb' construction (or 3 x single phase units)

Step 2: Relay Characteristic Angle

0  -  Resistance earthed systems

45 (I lags V)  -  Distribution systems (solidly earthed)

60 (I lags V)  -  Transmission systems (solidly earthed)

Current Polarising

A solidly earthed, high fault level (low source impedance) system may result in a small value of residual voltage at the relaying point.  If residual voltage is too low to provide a reliable polarising signal then a current polarising signal may be used as an alternative.

The current polarising signal may be derived from a CT located in a suitable system neutral to earth connection.

e.g.

Directional overcurrent

Step 1: Establishing Direction

Operating region

S₁  =  Reference Direction  =  Polarising Signal  =  V_POL

S₂  =  Current Signal  =  I

OPERATION when S₂ is within ±90° of S₁  :-

Restrain region

RESTRAINT when S₂ lags S₁ by between 90° and 270° :-

Step 2 : Polarizing Voltage selection

OPERATE SIGNAL =          I_A

POLARISING SIGNAL :-     Which voltage to use ?

                                                            Selectable from

                                                                           V_A

                                                                           V_B

                                                                           V_C

                                                                           V_A-B

                                                                           V_B-C

                                                                           V_C-A

Case 1:

Applied Voltage        :           V_A

Applied Current        :           I_A

is this connection suitable for a typical power system ?

           No, the fault current at near to zero sensitive point and operation of relay will be insensitive. Polarity voltage will also change.

Case 2:

Applied Voltage   :    V_BC

Applied Current   :    I_A

is this connection suitable for a typical power system ?

         Yes, Because Polarising voltage remains healthy & Fault current in centre of characteristic

Step 3: Relay connection angle

          The angle between the current applied to the relay and the voltage applied to the relay at system unity power factor

e.g.  90° (Quadrature) Connection  :    I_A  and V_BC

Step 4: Relay Characteristic Angle (RCA)

     The angle by which the current applied to the relay must be displaced from the voltage applied to the relay to produce maximum operational sensitivity

(i) 90° Connection - 45° R.C.A.

Application :

1.    Plain feeder, zero sequence source behind relay

(ii) 90° Connection - 30° R.C.A

Application

1.    Plain or Transformer Feeder :-  Zero Sequence Source in Front of Relay

2.    Transformer Feeder :-  Delta/Star Transformer in Front of Relay