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FURTHUR ELECTRICALPOWER (FEP)0.6 - 33KV DG Set Power System (GRADED)

2011

CHARLES EASTLAND

5/2/2011


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FURTHUR ELECTRICAL POWER (FEP)

Overhead Linesand Cables, Symmetrical Faultsand Protection Schemes,

PowerSystem Transientsand Synchronous Machines

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CONTENTS

1.0 SUMMARY.............................................................................................................. 2

2.0 INTRODUCTION..................................................................................................... 4

3.0 REPORT & ANALYSIS........................................................................................... 7

Investigate overhead lines and cables.7

T and models to evaluate performance ...13

Investigate symmetrical faults and protection schemes...17

Investigate and describe the components in a protection scheme .19

Analyse the propagation of surges ..22

Bewley lattice diagram to analyse multiple reflections .25

Investigate the synchronising and control of synchronous machines27

4.0 CONCLUSION....................................................................................................... 30

5.0 DISCUSSION......................................................................................................... 34

6.0 REFERENCES....................................................................... ................................ 38


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1.0 SUMMARY

This report entails information ofOverhead lines and cables, Symmetrical faults and

protection schemes, Power system transients andS ynchronous machines

And my findings are:

Different types of cables used in power systems with a power rating of 20MW to connect to a

range of underground cables for the following voltages: 600/1000V, 11KV, 22KV, and 33KV.

A balanced three-phase supply to a load consisting of 3 impedances, each of value (4 + j3),

connected in a delta arrangement, supplied from a 0.4kV, three-phase balanced supply. The

lines connecting the load to the supply each are designed to have an impedance of (1 + j4)

1) The line current drawn from the supply

2) The current drawn in each impedance of the load

3) The total complex power provided by the supply

A load connected to a distribution transformer, which will introduce balancing currents. Given

that the line supply voltage is 1000V, calculate the current in each load.


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A simulation of a phase-to-earth fault occurring on a 33kV power distribution line 40km long

supplied by a 20MW diesel generator with a fault occurring at a distance of 13km having a nd

a fault path resistance of 10 is:

The components used in a biased differential prote ction scheme and the components used

in transformer protection

Are:

Biased differential relays, current transformers, high spee d differential relays, and high

impedance differential relays

A transient occurs in the power system when the network changes fromone steady state into

another. Such as whenlightning hits the ground in the vicinity of a high -voltage

transmissionline or when lightning strikes a substation directly. The majority of powersyst em

transients are, however, the result of a switching action.

And

Using the attached Bewley lattice diagram I have calculated the voltages at junctions A, B,

and C when a switching surge propagates through the system (answers in p.u. values).

At time t = 1ms

Voltage at junction A= - 0.080 + 0.143 -0.010 = 0.053 pu

Voltage at junction B= 0.199 + 0.004 + 0.032 + 0 = 0.235 pu

Voltage at junction C at t- = 0.040 + 0.127 = 0.167 pu

Voltage at junction C at t+ = 0.040 + 0.127 + 0.005 = 0.172 pu

Finally

Synchronisation and/or paralleling of two or more generators onto the power network grid are

carried out manually using the three -lamp method or a synchro-scope,


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2.0 INTRODUCTION

This report is a study of Overhead lines and cables, Symmetrical faults and protection

schemes, Power system transients and Synchronous machine s

With the following information:

As a power systems engineer I am involved in designing and planning distribution systems

and local generation systems, such as DG sets of up to 20MW. Your tasks and

responsibilities range from line design, protection scheme s, and power quality all the way to

generation monitoring and control.

In this project I will investigate parts of the power system and solve problems related to

power systems performance and surge control.

Tasks 1: Investigateoverhead lines and cables

(Compare different types of cables used in power systems M1.1)

I am planning the installation of a permanent DG set with a power rating of 20MW to connect

to a range of underground cables for the following voltages: 6 00/1000V, 11KV, 22KV, and

33KV. Describe what types of cables are typically used and how these compare?

Giving a brief, technical description of each type and highlight the changes in cable design

as the transmitted voltages go up.

Task B: (use T and models to evaluate performance D1.1)

1.1 I am to design a balanced three -phase supply to a load tha t consists of three

impedances, each of value (4 + j3), connected in a delta arrangement.

The load will be

supplied from a 0.4kV, three-phase balanced supply. The lines connecting the load to t he

supply each are designed to have an impedance of (1 + j4) . In order to make the correct

choices in terms of equipment used a nd electrical power transferred I will need the following

information:




1.2A load is to be connected to a distribution transformer, which wil l introduce balancing

currents. The load is complex and connected in delta (Figure 1), and in order to inves tigate

this load in terms of imbalances it may need to be manipulated.

Given that th e line supplyvoltage is 1000V, calculate the current in eac h load.Note: ZA=15, ZB=10+j10, ZC=12-j15


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Task 2: Investigate symmetrical faults and protection schemes

As part of your power system design you are asked to simulate a phase to earth fault

occurring on the 40km long 33KV line supplied by the 20MW DG, set at a distance of 13km.

The fault path resistance is set to be 10.

Given that the Generator impedances and line

impedances are as follows:

ZS1 = ZS2 = 1+j15, ZS0=0.7+j7 , and Z11 = Z12 = 2+j30, Z10= 8+j50

Draw the supply system with fault as a one -line diagram and calculate the fault current.

Task D: (investigate and describe the components in a protection scheme M2.1)

The power system that is supplied by the 20MW DG sets has a modern protection system

and you are to investigate the components used in differential protection. Briefly describe

the components used in a biased differential protection scheme, orcomponents used in

transformer protection, e.g. how they operate, relay resistance, CT currents, saturated CTs

Relay setting surrents, etc.

Task 3: Analyse power system transients

Task E: (analyse the propagation of surges M3.1)

Appendix BTravelling Waves investigates transients in power systems and the power

network.

For the power system design that uses the 20MW DG sets I am asked to investigate powersystem transients and in particula r the propagation of surges. My work will focus on two

types of surges: due to switching, and due to lightning strikes. With the main task to analyse

how these surges propagate through the power network.

I willuse/draw network maps with equipment identified to use as examples of surge

propagation. My report will also include effects of surges on the power systems and the

effects of surge reflections.

Task F: (use a Bewley lattice diagram to analyse multiple reflections D3.1)

Three 20MW DG sets A, B, and C are connected together over a distance of 40km to form amini grid, where generators C and B are connected together by a cable with a propagation

velocity of 300 x 10^6 m/s. The impedances on the lines are as follows:

500 700 76 700

Using the attached Bewley lattice diagram (Appendix A, page 57) I calculated the voltages

at junctions A, B, and C when a switching surge propagates through the system (answers in

p.u. values). The data in Figure 4.10 applies.


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Using the rest of the information in Appendix A to analyse reflections and explain how these

affect power the power system.

Task 4: Investigate the synchronising and control of a synchronous machine

Task G: (describe and compare two methods of synchronising M4.1)

Two 20MW DG sets shown in figure 2 are used for the local generation system to be

operated in parallel in order to supply power during peak demand times.

Describe the simplest and the most advanced method of synchronising these two

synchronous generators and compare them in terms of reliablitiy, degree of automation,

cost, and monitorability.

Note: You may like to produce a comparison table to deliver information more efficiently.

Figure 2


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3.0 REPORT & ANALYSIS

3.1 Task A: (compare different types of cables used in power systems M1.1)

Recommended current ratings:

The Current rating of power cable is defined by the maximum intensity of current (amperes)

which can flow continuously through the cable, under permanent loading conditions, without

any risk of damaging the cable or deterioration of its electrical properties.

The value given in the tables are valid for one circuit in a three phase system under conditions

specified. For grouping cables rating factors must be used.

The current carrying capacities mentioned in RELEMAC technical data are intende d as a guide,

to assist operating engineers in selecting cables for safety and reliability. [1]

Basic assumptions and conditions of installation :

y Max. conductor Temperature: 90 C.

y Ambient Ground Temperature: 30 C.

y Ambient Air Temperature: 40 C.

y Thermal resistivity of soil: 150C.Cm/W

Depth of laying (to highest point of the cables laid direct in the ground)

y 3.3,6.6 & 11kV Cables : 90 cm

y 22 and 33kV Cables: 105cm

Max. Conductor Temperature

y for Short Circuit : 250 C.

600/1000V Cable

The cable of 600/1000V rating which embracesthe 230/400V standard voltage for domesti c

supplies has four conductors, three for the phase currents and a neutral. Th e typical design

is illustrated in theFigures below. The conductors, except for the smallest sizes, have a

shaped cross-section, so that when the cores (the in sulated conductors) are laid up together

they form a compact circular cable wi th minimum spaces in the centre and at the sides at the

rounded corners of the cores, to be filled with paper strings or jute yarns. It also reduces the

cable diameter and thereforethe amount of lead sheath and armouring.


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Steel Wire Armoured Cable or SWA Cable, is a hard -wearing mains electricity cable

with class 2 stranded copper conductors, XLPE (Cross-Linked Polyethylene) insulation,

PVC (Polyvinyl Chloride) bedding, aluminium or steel wire armour and a black PVC

sheath. SWA Cable is sometimes referred to as Booklet Armoured Cable, Mains or

Power Cable.The voltage rating of an SWA Cable is 600/1000V

.

For higher voltages the cables are generally three -core. The constructionfor voltages up to

11 kV is similar in principle to that of the 600/1000V cable. The insulation thickness is

greater, of course, and the manufacturing processes differ in detail to provide for the higher

operating electrical stresses. For voltages above 11 kV, the belted construction gives place

to the screened cable as the standard. At 11 kV both types are provided for in the

standards, but the belted type has the greater usage.

11kV Cable

A typical three-core screened 11 kV cable is illustrated below. The carbon paper screen

applied to each conductor is a standard feature for cables rated at 11 kV and above in BS

6480. It is to reduce the electrical stress at the conductor surface by smooth ing out the


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profile and to exclude from the field the small spaces between the wires of the outer layer,

which can otherwise be sites for discharge.

A carbon paper screen is also applied over the insulation of single -core 11 kV cables and

over the belt of 11 kV belted cables to eliminate discharge in spaces between the outside of

the insulation and the inside of the lead sheath in places where the latter does not make

close contact.

11kV Cable is a power cable with armouring and a series of screens. Its class 2

stranded plain copper conductors are surrounded by a conductor screen made of semi -

conducting material. This screen is then covered with a layer of XLPE insulation and an

insulation screen also made up of a semi -conducting material.

Between the insulation and armour, 11kV Cable has a copper tape metallic screen, a

PETP (Polyethylene Terephthalate) filler, binding tape separator and PVC bedding. A

layer of aluminium or steel wire armour sits over the bedding covered by a red or

black PVC sheath . The voltage rating of an 11kV Cable is 6350/11000V.

The low smoke zero halogen 11kV Cable has the same number and type of layers

as the PVC version. For the 11kV Cable to be used in public areas it must have LS0H

bedding and an LS0H sheath.

In the UK this type of cable, mainly in single -core form, is favoured for power station cabling,

where lightness and convenience of terminating are major considerations. Three -core

designs are also used for site supplies. For underground distribution at 11 kV, the XLPE

cable does not compete economically with the paper -insulated aluminium-sheathed cable,

but work is in progress on standardizing and assessing XLPE cable design, including trial

installations, in preparation for any change in the situation. Overse as, where circumstances

are different, XLPE cable is the type in major demand. With manufacturing facilities


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increasingly orientated to this market, XLPE -insulated cables constitute a large proportion of

UK production. [2]

33kV Cable

33kV Cable is made up of the same individual layers as a PVC 11kV Cable. The only

difference is that the voltage rating on a 33kV Cable is 19000/33000V.

From 3.8 kV up to 33 kV, XLPE and EPR insulated cables are covered by BS 6622 which

specifies construction, dimensions an d requirements. The polymeric forms of cable

insulation are more susceptible to electrical discharge than impregnated paper and at the

higher voltages, where the electrical stresses are high enough to promote discharge, it is

important to minimize gaseous spaces within the insulation or at its inner and outer surfaces.

SWA Cable LS0H

The low smoke zero halogen or LS0H, version of SWA Cable is constructed in exactly

the same way as the PVC SWA Cable with copper conductors, XLPE insulation and

aluminium or steel wire armour. The only difference is that LS0H SWA Cable has LS0H

bedding and an LS0H sheath. LS0H SWA Cable is used in public areas because the

sheath only emits very low levels of smoke and non -toxic levels of halogen gas when

exposed to fire (usually under 0.5% HCl emission). Find out more about SWA LS0H

Cable.

Armoured cables (BS Standards)

Where SWA Cable is concerned, BS5467 and BS6724 are the British Standards thatspecify requirements for construction, and describe methods of testing for

thermosetting insulated, armoured cables with rated voltages of 600/1000V and

1900/3300V. BS6724 applies to those cables that produce lower levels of smoke and

corrosive products when exposed to fire in specified tests.

Both 11kV Cable (PVC) and 33kV Cable comply with BS6622, whereas 11kV Cable

(LS0H) meets the requirements of BS7835.

The BS6622 and BS7835 standards look at the specifications and test methods for the

construction, dimensions and mechanical and electrical properties of thermosetting

insulated, armoured cables with voltage ratings of 3 800/6600V up to 19000/33000V.

All armoured cables related to these standards are used in fixed installations in

industrial areas and buildings.

Difference between AWA & SWA

A single core armoured cable will always have a layer of aluminium wire armour (AW A)

instead of steel wire armour (SWA). This is because the steel in SWA has a much lower

conductivity and therefore higher resistance than aluminium.


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If it were used in a single core cable the magnetic field generated would induce an

electric current in the armour (eddy current) and combined with the increased

resistance would have a heating effect.

AWA is non-magnetic and has a much better conductivity (lower resistance), so can

conduct these induced currents to earth more efficiently than steel. SWA is used in

multi-core armoured cables because the electromagnetic fields from the neighbouring

cores effectively cancel each other out, meaning less current is induced into the

armour.[3]

The table on page 12 lists all relevant standards relating to the voltages of 0.6 33kV


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Overhead Linesand Cables, Symmetrical Faultsand Protection Schemes, PowerSystem

Voltage Standard Conductor Insulation Sheath Armour Col

600/1000V BS6724

Class 2

stranded plain

annealed

copper

conductor to BS

EN 60228:2005

(previously

BS6360)

PVC or XLPE

(Cross Linked

Polyethylene)

LSZH (Low

Smoke Zero

Halogen)

Single Core:

Aluminium Wire

Armoured

(AWA)

MultiCore: Steel

Wire Armoured

(SWA)

Blac

11KV BS6622

Class 2

stranded plain

copper to BS

EN 60228:2005

(previously

BS6360)

XLPE (Cross-

Linked

Polyethylene)

Type GP8 to

BS7655

PVC

(Polyvinyl

Chloride)

Type TM1 to

BS7655

Single Core:

Aluminium Wire

Armoured

(AWA)

MultiCore: Steel

Wire Armoured

(SWA)

Red

Blac

33KV BS6622

Class 2

stranded plain

copper

conductor to BS

EN 60228:2005(previously

BS6360)

XLPE (Cross-

Linked

Polyethylene)

Type GP8 to

BS7655

PVC

(Polyvinyl

Chloride)

Type TM1 to

BS7655

Single Core:

Aluminium Wire

Armoured

(AWA)

Multi-core: SteelWire Armoured

(SWA)

Red

Blac


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Task B: (use T and models to evaluate performance D1.1)

1.1 I am to design a balanced three-phase supply to a load that consists of three

impedances, each of value (4 + j3), connected in a delta arrangement. The load will be

supplied from a 0.

4kV, three -phase balanced su pply.

The lines connecting the load to thesupply each are designed to have an impedance of (1 + j4).

In order to make the correct choices in terms of equipment used and electrical power

transferred you need the following information:





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1.2 A load is to be connected to a distribution transformer, which wil l introduce balancing

currents. The load is complex and connected in delta (Figure 1), and in order to inves tigate

this load in terms of imbalances it may need to be manipulated. Given that th e line supply

voltage is 1000V, calculate the current in eac h load.Note: ZA=15, ZB=10+j10, ZC=12-j15

In order to investigate the load in Figure 1.2 (a) , it is necessary for the above circuit to be

redrawn as shown in Figure 1.2 (b), where the arrangement is referred to as delta

connected ormesh connected

Figure 1.2 (b)

Finally replacing the delta connection shown in Figure 1.2 (b) by an equivalent star

connection shown below in Figure 1.2 (c)

Figure 1.2 (c)


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Task 2: Investigate symmetrical faults and protection schemes

As part of the power system design I have been asked to simulate a phase to earth fault


The fault path resistance is set to be 10.

Given that the Generator impedances and lineimpedances are as follows:

ZS1 = ZS2 = 1+j15, ZS0 = 0.7+j7, and Z 11 = Z12 = 2+j30, Z10= 8+j50

And

Draw the supply system with fault as a one -line diagram and calculate the fault current


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Then


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Task D: (investigate and describe the components in a protection scheme M2.1)

GROUND FAULT (G/F) PROTECTION SCHEMES

A sensitive protection for ground faults will use one of the following approaches:

y In case the power supply source (such as the transformer) is a part of the system, a

CT and relay can be provided in the ground connection of the neutral of the

transformer.

y By a single current transformer enclosing all phase and neutral conductors (ca lled as

core balance or zero sequence CT). Such a transformer detects the ground fault

currents and can operate a sensitive relay.

y By individual current transformer in phase and neutral conductors and providing a

relay in summation circuit.

Adding special ground fault equipment to the system to sense even low value of earth faultcurrents and trip the circuit faster. Inclusion of the neutral in Figures b and c above is forcancelling any unbalance currents that may flow in the neutral from be ing sensed as groundfaults [3].


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In steady state conditions with the use of perfect current transformers (CTs), relays and

symmetry of connections, stability should be theoretically possible. As a result, no

anomalous tripping should occur due to faults ou tside the zone of protection. Practically,

differences will happen in both the magnitude and phase of the currents entering and leaving

the zone of protection.

Therefore, CT characteristics cannot be perfectly matched and DCcomponents will become involved under fault conditions.

The use of these means that currents, which are fed into the relay from both sides of the

power transformer, may not balance perfectly. Therefore, any imbalance must be

compensated. The application of biased relays and/or interposing CTs allows for this.

Biased differential relays permit low settings and fast operating times, even when a

transformer is fitted with an on -load tap-changer. The biasness progressively increases the

amount of spill current required for operation as the magnitude of the through current

increases.

Interposing CTs balance the currents supplied to a relay where there would otherwise be animbalance due to the rati os of the main CTs. Interposing CTs are installed between the

secondary winding of the main CT and the relay. In addition to this application, interposing

CTs are also utilised in establishing a delta connection in the elimination of zero sequence

currents, where required.

Differential protection is a unit scheme that compares the current on the primary side of a

transformer with that on the secondary side. Where a difference exists (other than that due

to the voltage ratio) it is assumed that the transform er has developed a fault and the plant is

automatically disconnected by tripping the relevant circuit breakers. The principle of

operation is made possible by virtue of the fact that large transformers are very efficient and

hence under normal operation po wer-in equals power-out.

Differential protection detectsfaults on all of the plant and equipment within the protected zone, including inter -turn short

circuits.

Basically, unit protection schemes compare the current entering and leaving the protected

zone. Any difference will indicate the presence of a fault within the zone. By operation of the

appropriate relays the associated circuit breakers can be made to trip thus isolating the

faulty equipment from the power network [4].

The operating principle emp loyed by transformer differential protection is the Merz -Price

circulating current system as shown below. Under normal conditions I 1and I2 are equal and

opposite such that the resultant current through the relay is zero. An internal fault produces

an unbalance or 'spill' current that is detected by the relay, leading to operation [5].

A high impedance relay is defined as a relay or relay ci rcuit whose voltage setting is not less

than the calculated maximum voltage which can appear across its terminals under the

assigned maximum through fault current condition. It can be seen from the figure below that

during an external fault the through fa ult current should circulate between the current

transformers secondary. The only current that can flow through the relay circuit is that due to

any difference in the current transformer outputs for the same primary current. Magnetic


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saturation will reduce the output of a current transformer and the most extreme case for

stability will be if one current transformer is completely saturated and the other unaffected.

To obtain high speed operation for internal faults, the knee point voltage, VK, of the CTs

must be significantly higher than the stability voltage, VS. This is essential so that the

operating current through the relay is a sufficient multiple of the applied current setting.

The knee-point voltage of a current transformer marks the upper limit of the roughly linear

portion of the secondary winding excitation characteristic. This is defined exactly in the IEC

standards as that point on the excitation curve where a 10% increase in exciting voltage

produces a 50% increase in exciting current. The current transformers should be of equal

ratio, of similar magnetizing characteristics and of low reactance construction.

In a biased differential relay, the through current is used to increase the setting of the

differential element. For heavy through faults, it is unlikely that the CT outputs at each zone

end will be identical, due to the effects of CT saturation. In this case a differential current can

be produced. However, the biasing will increase the relay setting, such that the differential

spill current is insufficient to operate the relay.

A biased differential protection function uses the two sets of three -phase current

measurement inputs (IA, IB, IC, IA2, IB2 and IC2), which are connected to measure thephase current at the neutral end and terminals of the machine, as shown above [6].


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Task E: (analyse the propagation of surges M3.1)

For the power system design that uses the 20MW DG sets I am asked to investigate power

system transients and in particula r the propagation of surges. My work will focus on two

types of surges: due to switching, and due to lightning strikes.

With the main task to analysehow these surges propagate through the power network.

The purpose of a power system is to transport and distribute the electricalenergy generated

in the power plants to the consumers in a safeand reliable way. Aluminium and copper

conductors are used to carrythe current, transformers are used to bring the electrical energy

to theappropriate voltage level, and generators are used to take care of theconversion of

mechanical energy into electrical energy. When we speak ofelectricity, we think of current

flowing through the conductors from generatorto load.

A transient occurs in the power system when the network changes fromone steady state into

another.

This can be, for instance, the case whenlightning hits the ground in the vicinity of ahigh-voltage transmissionline or when lightning strikes a substation directly. The majority of

powersystem transients is, however, the result of a switching action. Load -breakswitches

and disconnectors switch off and switch on parts of thenetwork under load and no -load

conditions.

A high-voltage circuit breaker is an indispensable piece of equipment in the power system.

For the analysis of simple switching transients and for carrying out large system studies, it is

often sufficient to model a circuit breaker as an i deal switch.

The voltage transient is a source -side phenomenon. At the load side,where the capacitor

bank or the cables are connected, there is no changein voltage.

The trapped charge causesa constant voltage, but when thesupply voltage has the opposite polarity, nearly two times

the peak valueof the supply voltage is present across the breaker contacts.

A very specific capacitive switching duty is back-to-back switching. Inthis situation, a large

capacitor bank is connected at the load side of the breaker but also at the supply side.

When switching the capacitive current in a back -to-back situation,in the most onerous

situation, a re-ignition occurs when the load -sidecapacitor has a polarity opposite to that

from the supply-side capacitor.

The discharge current then flows from one capacitor bank to the othervia the stray

inductance (caused by the loop of the bus bars and theconnecting wires).

The frequency ofthe current oscillation and the peak ofthe current are therefore much higher than in the case

of single-capacitor bank switching.

Switching unloaded high-voltage transmission lines in and out of serviceis, in principle, the

same as switching a capacitor bank because an unloadedtransmission line has a dominantly

capacitive behaviour. In addition , asmall voltage jump occurs because of the Ferranti -rise

effect. High-voltagetransmission lines also have travel times and must therefore be

representedby distributed elements instead oflumped elements .


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When a transmissionline is energised after closing of a switching device, the resulting

voltage-wave reflects and causes doubling of the voltage at the open end of the line.

Switching unloaded high-voltage transmission lines in and out of serviceis, in principle, the

same as switching a capacitor bank because an unloadedtransmission line has a dominantlycapacitive behaviour. In addition, asmall voltage jump occurs because of the Ferranti -rise

effect.

For the analysis of the li ghtning-induced over-voltages, a difference is made between the

following:

y Lightning strokes in the vicinity of high -voltage transmission lines, which do no t hit

the conductors themselves.

y Direct lightning strokes on the line conductors injecting a current wave on the line.

y Lightning strokes on the transmission towers or on the ground wires.

Cloud-to-ground strokes can hit substations, transmission line towers,and transmission lines

directly, but a considerable number of atmosphericdischarges are between clouds. When the

charged clouds are floatingabove, for instance, a high -voltage transmission line, they induce

chargeaccumulation on the line conductors . When the lightningstroke equalises the charge

difference between the clouds, the ratherslowly accumulated charge on the conductors has

to disappear at once.This results in tran sient currents and over-voltages.


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After a transmission line tower ishit by lightning the instant the stepped leader reaches the

tower, thethundercloud having a potential of EVOLTS has a conducting path to theground. The

charge difference between the thundercloud and the earth isequalised by a travelling wave,

moving upwards at approximately half thevelocity of light. The discharge takes place in

about 50 s, the branchesare brightly illuminated, and the amplitude of the current is in therangeof 20100 kA. The characteristic impedance of the lightning current pathlies in the

order of the characteristic impedance of overhead transmissionlines ZL = 500. The stroke

current ISTROKE will divide into three parts.

One part of the stroke current, ITOWER, will flow through the tower structureto ground and the

remaining part of the stroke current divides equallyand flows in opposite directions through

the ground wire, as shown in Figure below.

ISTROKE = ITOWER + IGROUND WIRE

When lightning strikes directly on a phase conductor of a transmissionline, it can beregarded as a current injection Ion the line, which dividesitself into two equal parts at the

point of strike. The voltages generated bythe divided currents travel in both directions along

the line, away from thepoint of strike. When the characteristic impedance of the phase

conductoris ZPHASE, the voltage is related to the lightning current Iby

V= 0.5IZPHASE

Switching actions, short -circuits, lightning strokes, and disturbancesduring normal operation

often cause temporary over -voltages and highfrequencycurrent oscillations. The power

system must be able to withstandthe over -voltages without damage to the system

components.

[7]


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Task F: (use a Bewley lattice diagram to analyse mu ltiple reflections D3.1)

Three 20MW DG sets A, B, and C are connected together over a distance of 40km to form a

mini grid, where generators C and B are connected together by a cable with a propagation

velocity of 300 x 10^6 m/s.

The impedances on the lines are as follows:

50070076 700

Using the attached Bewley lattice diagram above I have calculated the voltages at junctions

A, B, and C when a switching surge propagates through the system (answers in p.u. values).

Also using the rest of the information in Appendix A to analyse reflections and explain how

these affect power the power system.

The overhead lines beyond A and C on either side are extremely long and reflections need

not be considered from their far ends. Determine using the Bewley lattice diagram the

overvoltages at the 3 substations, at an instant 1_ ms after a voltage surge of magnitude

unity and duration 3/4 reaches the substation A from the outside.

The transmission and reflection coefficients can be calculated as follows.


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Substation A,

Substation B,

Substation C,

The single transit times of the two lines connecting the substations are

The lattice diagram must be set out, such that the intervals AB and BC are in the proportions

to the times 1/4 ms and 3/8 ms respectively. Since the step voltage incident at substation A

is of duration 3/4 ms, only reflections that have occurred after 3/4 ms prior to the present will

be in existence.

At time t = 1_ ms



Since a surge arrives at junction C at the instant of interest, we can define values either just

before or just after the time.


Voltage at junction C at t+ = 0.040 + 0.127 + 0.005 = 0.172 pu


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Task G: (describe and compare two methods of synchronising M4.1)

Two 20MW DG sets shown in figure 2 are used for the local generation system to be

operated in parallel in order to supply power during peak demand times.

This section will explain the reasons for paralleling, the methods by which it is carried out,

the setting up procedures and possible problems that may arise.

Parallel Operation may be necessary for the following reasons:

y To increase the capacity of an existing system.

y Size and weight may preclude the use of one large unit.

y Allows non-interruption of the supply when servicing is required.

In order to parallel AC Generators satisfactorily, certain basic conditions have to be met.

These are as follows:

1) All systems must have the same voltage.

2) All systems must have the same phase rotation.

3) All systems must have the same frequency.

4) All systems must have the same angular phase relationship.

5) Systems must share the load with respect to their ratings.

An important aspect of parallel operation is load sharing. The total load, comprising a kW or

active component and a kVAR or reactive component, must be shar ed by the systems with

respect to their normal ratings. The kW component is adjusted by purely mechanical means

and requires relatively fine speed control of the prime mover (engine). It is advisable to fit a

limited range governor to avoid large adjustmen ts of speed when in parallel. The kVAr

component is a function of the AC generator excitation.

When machines are in parallel, the magnitude of the field excitation will not directly influence

the output voltage (depending upon the relative size of the Generator to the bus -bar system)it does however adjust the internal power factor at which a partic ular machine operates. For

instance, an over-excited AC generator will produce a lagging pf current from that Generator.

If a difference in excitation exists, then circulating currents will flow, limited only by the

internal machine reactance.


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This current will appear as a zero p.f. leading or lagging current, depending on the machine

excitation, and will either subtract or add to the total current that each machine supplies.

Reactive current, either leading or lagging, is by virtue of the 90 -degree phase displacement,

quite commonly described as being quadrature.

The Generators must therefore be provided with equipment to sense this reactive current,and limit it to an acceptable level.

In order that the incoming machine may now take its share of the load, the governor control

should be held in the speed raise position, until the desired load is indicated by the kW meter

and ammeter. Conversely, if too much load is applied holding the gov ernor control in the

speed lower position can reduce it.

It is most important that the total load be shared in respect of their normal ratings and the

meter readings should be compared with the name plate data. In any event, unequal load

sharing requires correction to avoid mechanical problems which occur when diesel engines

are run light for any considerable time.

It is important to differentiate between unbalancedloading caused simply by the operator failing to spread the load equally over the two sets ,

and by circulating currents unbalancing the ammeter readings.

A synchro-scope and/or lights, is required to detect the angular phase displacement. If lights

are used three different connections are possible. For paralleling with the lights dim, they

must be connected across like phases or like lines (single phase), i.e. UU, V -V or L1-L1. For

paralleling with lights bright they should be connected across unlike phases, i.e. U -V etc.

If a three-lamp system is used with the lamps connected across U -W, V-V and W-U the

lamps will 'rotate' and give an indication which machine is running fast. Synchronism is

reached with two lamps bright and one dark and in some respects this connection gives a

closer visual indication of the point of synchronism.

To ensure the sets are in phase and at the same frequency (synchronised) it is necessary to

check before closing the generator onto the supply by one of the following methods: -

1) By the use of a proprietary Synchro -scope

2) By the 2 lamp method so that they are both showing bright when the machines are in

synchronism

3) By the 2 lamp method so connected that they are both showing dark when the

machines are in synchronism

4) By the 3-lamp method so connected that one is dark and the other two are equal

brightness when the machines are in synchronism

The lamps should be rated for at least twice the machine voltage or it will be necessary to

connect two or three in series. A more preferred method is a resistor, in series with each

lamp. The following diagrams illustrate the co nnections:


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NOTE: If the neutral is solidly linked, then only one set of lamp/resistors is required, because

the return path is through the neutral link. This, of course, only applies to the lamps dim or

lamps bright connection, and not on the three lamp connection.

Modern installations prefer to use semi or fully automatic synchronising equipment, which

allows breaker closure only when conditions are correct.

A minimum amount of instrumentation is required to ensure the above information is

satisfactorily monitored, comprising an ammeter, a wattmeter and a reverse power relay. No

voltmeter is specified for each system because it is preferred to use one voltmeter on the

distribution or synchronising panel with a select or switch for each system.

This eliminatesany possible meter inaccuracies. A reverse power relay is essential as any engine shut

down, from low oil pressure or temperature etc. will result in other systems motoring the

failed set, with consequent overload to the remaining systems, and/or damage to the

motored engine. Only one frequency meter is required with the facility of being switched to

the bus-bar, or the incoming system.

When an additional set is being connected to the bus -bars with the synchro-scope/lights

switch in the on position, a point may be reached where the incoming machi ne voltage starts

to fluctuate. This only occurs when the frequency difference is at its greatest. As the

frequencies approach each other, no further instability is noti ced.


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4.0 CONCLUSION

For task 1 of this project I had to research different types of cables, which would be

encounteredin underground low -voltage (LV) and medium-voltage (MV) networks.

I did extensive research into this by accessing Relmac and Nexans website and technical

manuals, as well as research into the various BS and international standards relating to LV

and MV networks

For task 2 (1.1) I had to investigate the load and find the line current drawn from the

supply

I used the formulae below to findby dividing the supply voltage by the load impedance

Then simply converting from j notation to polar form and dividing the whole numbers and

subtracting the polar numbers.

Then I found the line current drawn from the supplyusing the same formulae and since

the system is assumed to be balanced

This gave me the individual line current drawn .

I then found the total complex power provided by the supply using the formulae below

Then taking the square of 97.09 and multiplying by 4.12, the polar numbers cancelling out.

And the total power was found using the formulae below

Finally converting from polar to j notation adding and converting back to polar form

In order to investigate task 3 (1.2) the load in Figure 1.2 (a) , it is necessary for the circuit to

be redrawn as shown in Figure 1.2 (b), where the arrangement is referred to as delta

connected ormesh connected


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Figure 1.2 (a)

Figure 1.2 (b)

Finally replacing the delta connection shown in Figure 1.2 (b) by an equivalent star

connection shown below in Figure 1.2 (c)

Figure 1.2 (c)

And for task 4 in order to investigate this load in terms of imbalances ithad to be

manipulated, and the formulae used below

Then simply convert from j notation to polar form and group them, then multiply the whole

numbers (not polar numbers)


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Finally converting back from polar to j notation

The same process is used for finding the Line current ( regarding changing and addingfrom polar to j noyaion and visa -versa. Only this time dividing the line impedance by the

source voltage.

For task 5 as part of the power system design I had to simulate a phase to earth fault


The fault path resistance is set to be 10. With give generator impedances I used the

formulae below

Again the same principle for using and converting from j notation to polar form applies to the

above problem.

For task 6 of this report researched the components, w hich could be found in a biaseddifferential protection scheme and those that would typical ly be encountered in the modernprotection scheme of a power system that is supplied by the 20MW diesel generator sets.

For task 7 I researched the components, which could be found in a biaseddifferential

protection scheme, and those that would typical ly be encountered in the modern protection

scheme of a power system that is supplied by the 20MW diesel generator sets.


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For task8 of this report I investigated power system transien ts with particular attention to the

propagation of surges and analysis into h ow surges propagate through the power network,

focusing in particular on two main types of surg es

y Switching

y Lightning strikes

My research found that overvoltage surges can be show n to propagate through overhead

lines with the waves partly reflecting and transmitting, as they reach the end of the line or a

junction of transmission lines.

For task 9of this report I used a Bewley lattice diagram to calculate the voltages at three

junctions A, B and C due to switching surges propagating

The transmission and reflection coefficients at substation A was calculated using theformulae below

Substation A,

And the same process as above for,

Substation B,

And again for ,

Substation C,


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The single transit times of the two line s connecting the substations were calculated as

Since a surge arrives at junction C at the instant of interest, we can define values either just

before or just after the time. Therefore only reflections that have occurred after this duration

1ms period prior to the present wo uld be in existence.

At time t = 1ms




Voltage at junction C at t+ = 0.

040 + 0.

127 + 0.

005 = 0.

172 pu

The final task 8 of this report was to investigate the synchronisations of 2 DG sets of

20MWs which are to operate in parallel.And describe the simplest and the mostadvanced

method of synchronising

My research found that the synchronisation and/or paralleling of two or more generators can

be achieved manually using the three -lamp method or asynchro -scope, or automatically

using AVRs.


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5.0 DISCUSSSION

The cross-sectional area of the conductors chosen should be of the optimum size to carry

the specified load current or short circuit short term current without overheating and should

be within the required limits for voltage drop, also t he insulation applied to the cable must beadequate for continuous operation at the specified working voltage with a high degree of

thermal stability, safety and reliability.

All materials used in the construction must be carefully selected in order to ensure a high

level of chemical and physical stability throughout the life of the cab le in the selected

environment, ensuring the cable must be mechanically strong, and sufficiently flexible to

withstand the re-drumming operations in the manufacturers works, handling during transport

or when the cable is installed by direct burial, in trenches, pulled into ducts or laid on cable

racks.

Adequate external mechanical and/or chemical protection must be applied to the insulationand metal or outer sheathing to enable it to withstand the required environmental service

conditions.[1]

Conductors are stranded plain annealed copper or aluminium. Conductors for cables having

a rated voltage of 6350/11000 volts and above are screened with semi -conducting carbon

papers. Insulation is by means of paper tapes applied helically, with the outer layer for twin

and multi-core cables numbered for core ident ification as follows:

Twin Core 0, 1 Three Core 1, 2, 3 Four Core 0, 1, 2, 3

It is intended that the numbers 1 2 and 3 indicate phase conductors and the neutral.

Beltedcables in twin and multi -core construction are laid up and filled and a further la yer of paper

tape is applied forming a compact and circular cable. For belted cables of 6350/11000 volts

a semi-conducting carbon paper tape is applied over the belt insulation.

Screened cables have he cores screened with a perforated metallised paper of a non-

ferrous metal tape. The screened cores are then laid up and bound with fabric tape

incorporating interwoven copper wires.Impregnation is by means of the mass impregnation

method utilising special non -draining insulating compound (M.I.N.D. process).

Lead or lead alloy sheathing in is applied, the choice of pure metal or of a particular alloy

being determined by the nature of the service conditions.

Bedding consists of two compounded paper tapes applied over the lead or lead alloy sheath

with further layers of fibrous material and bitumen compound. Alternatively if required an

extruded PVC bedding is available. Armouring is normally single galvanised steel wire or

double steel tape although double galvanised steel wire can be supplied if speci fically

required. Steel wire or steel tape armoured single core cables should not be used for AC

circuits owing to the induction effect. If armoured single core cable is essential in these

circumstances then non-magnetic armouring can be supplied.


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Overall finish can be wire armour only left bright for installation indoors or in dry protected

situations, or can be finished with compounded layers of fibrous material and bitumen. This

serving is whitewashed to prevent adhesion between turns on the drum. A lternatively an

extruded PVC over-sheath can be provided if required. This may be essential where

particularly onerous conditions are encountered as PVC offers good protection against awide range of chemicals or where a pleasing finish is required . [8]

Transformers are predominantly the most expensive piece of plant on the network and

therefore require many forms of protection; however we are investigating earth/ground fault

protection.

Balanced earth fault devices operate, as the name implies, on balan ced principles. The

current transformers (CTs) are positioned in each phase and during normal running of the

system the distribution of current across the phases summate to zero and a balance is thus

obtained in the Delta winding. If an earth fault occurs on any of the three phases then an

imbalance occurs which initiates the operation of the relay and consequently the opening ofthe breaker, both on the 11kV and the 33kV via an inter -tripping relay or closing of fault

throwers on the 33kV feeder.

Restricted earth fault relays also operate on the balanced principle, CTs are positioned

around each phase connected in parallel and a CT positioned around the earth connection

with a relay connected across these CTs, this types of connection covers a particular zone.

With a fault outside the protected zone the current flow through the CTs positioned around

the phases is equal to the current flow through the CT positioned around the earth

connection, and the is therefore no imbalance and no relay operation.

However when a fault occurs within the protected zone, there is current flow through the

earth CT but no current flow through the phase CTs, there is therefore an imbalance,initiating operation of the relay with the subsequent tripping of the circuit breaker .

Stand by earth fault protection backs up the restricted earth fault and the secondary feeder

protection. This needs to have a longer operating time in order to obtain discrimination

between these protective devices. It will operate if the faulty feeder b reaker has failed to trip,

if the fault is on the 11kV breakers or within the restricted earth fault zone including the

transformer secondary windings.

With this type of protection, a single CT is connected in the earth lead of the star connected

transformer and any current flow in the earth/neutral lead will energise the relay via the CT,

and operate the relay via the d.c supply.

Inter-tripping is used between substances to ensure isolation of faulty sections on the

network. For example, a transformer fau lt will operate the breaker that protects it, but remote

opening of breakers at the other end of the line will required. When a transformer trips, a

signal is sent via pilot cables to operate relevant breakers.

In some circumstances, the cost of installing /maintaining these pilots is prohibited and a fault

thrower is utilised. This basically earths the 33kV feeder on one phase causing the breaker

at the remote end to operate a fault thrower unit.


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With 33/11kV transformers, any relay operation (except windin g temperature) that trips the

11kV transformer will normally also requires the tripping of the source 33kV circuit breaker.

[8]

Lightning is one of the most widely studied and documented natural phenomena.

It is alsoone of the main causes of transient over-voltages in electrical systems. A proper

understanding of lightning is essential for planning protection against lightning strikes so that

no untoward damage is caused to buildings and electrical installations.

A lot of research has been done over a n umber of years worldwide and several publications

as well as national and international standards have evolved which give us a good insight

into this phenomenon.

Some of these are:

y AS 1768: 1991 Australian standard on lightning protection

y ANSI/NFPA 780 National lightning protection code

y IEEE 142: 1991 IEEE green book (Chapter 3)

y IEC 1024:1993 Protection of structures against lightning.

Modern day industries and businesses rely largely on electronic systems for their smooth

functioning, be it industrial driv es, distributed control systems, computer systems and

networking equipment or communication electronics. These electronic devices often work

with very low power and voltage levels for their control and communications and cannot

tolerate even small over -voltages or currents. Induced voltages from nearby power circuits

experiencing harmonic current flow can also cause interference in the systems carrying

communication signals and can result in malfunctions due to erroneous or noisy signal

transmission.

Due to this sensitive nature of electronic and communication equipment, any facility that

houses such equipment needs to have its electrical wiring and grounding systems planned

with utmost care so that there are no unpredictable equipment failures or malfuncti on.

Another problem is that of voltage spikes that occur in the power supply. Some of these may

originate from the external grid but some others may originate from other circuits within the

same premises. The result of such voltage/power surges is invariab ly the failure of the

electronic device itself. A typical example of an external voltage disturbance is a lightning

stroke near an overhead power transmission system.

Such transients can also happen dueto switching on or off large transformers.

The transformers when charged draw a momentary inrush current and this can reflect as a

voltage disturbance. Similarly, switching off an inductive load (say a coil energizing a

contactor) causes a brief voltage spike due to the collapse of the magnetic field i n the

magnetic core. If other equipment is connected in parallel with the inductan ce (after the

switching point), they will experience the surge. [4].


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When synchronising the closing of the breakers if done by hand must still be carefully

controlled as the time for the auxiliary relay and circuit breaker closing time will be a factor in

determining the exact point on the wave at which the two systems close together. Most

persons who have carried out this work on numerous occasions are aware of the delay and

will ensure the close take place just as prior to the Top Dead Centre or about 5 -10 degreesbefore the two are in phase. This allows for the closing time to five th e final Primary close of

the main and standby, or two parallel sets, to hit just at the correct in phase time.

The more powerful set will pull the other set, or in the case of the main electricity supply and

generator the mains will pull the auxiliary set , into true phase very quickly, but the action of

closing way out of phase will put a very severe mechanical loading onto the set and my

cause mechanical failure within the standby set. [8]

The typical lightning strike is comprised of three strokes, with the first stroke surging up to

20,000 amps lasting approximately 30 millionths of a second. Subsequent strokes are about

half the intensity.

Although theoretically possible, it is economically unfeasible to protect acontroller against a direct strike. The protection would cost many times the value of the

equipment it is protecting. Lightning -induced transients are electrical impulses induced in a

conductor simply by being in the vicinity of a lightning strike.

These transients may travel considerable distances from the physical strike location

especially when conductors are present. It is these potentially harmful voltages and currents

that we attempt to suppress. The closer the strike to the equipment, and the longer the

attached conductors (AC line and field valve wiring), the larger the induced transient.

The voltage from a lightning strike rises very fast, typically to its peak in a few millionths of a

second. This energy must be returned t o earth as quickly as possible, through a low

impedance path, otherwise damage to the electronics will occur.

At this speed, the

inductance of the ground system is much more important than its DC resistance. Increasing

the surface area of the conductors an d decreasing their length reduces inductance in the

ground system.

The build-up of electrons on the ground system also must be dissipated to prevent

dangerous voltages in the equipment. The ground system must therefore be of sufficient size

to absorb the energy.


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6.0 REFRENCES

[1] http://www.relemaccables.com

[2]Electrical Pocket BookE.A. ReevesMartin J. Heathcote

[3]http://www.nexans.co.uk

[4]Vijayaraghavan,MarkBrown,Malcolm Barnes : Practical Grounding, Bonding,

Shielding and Surge Protection

[5]Bayliss& Hardy: Transmission and Distribution Electrical Engineering

[6]http://www.transformerworld.co.uk/diffprot.htm

[7] Transients in Power Systems Lou van derSluis

[8] SchneiderElectric HVAP Training (Manual)

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