2The earth fault loop impedance of that part of the system external to the installation. Ze = the external earth fault loop impedance of the supply cable Zs = the earth fault loop impendence of the Internal current carrying circuit. R1+R2 = the sum of the resistance of the line conductor (R1) and the circuit protective conductor (R2) between the point of utilisation and the origin of the installation. Equitation: Zs = Ze + (R1 + R2)
3Ohms Law Resistance of the cable in ohms = R Current being drawn from the circuit = IVoltage being delivered to the circuit = VPower being developed by the circuit = PThis can give us two triangles for formulaResistance triangle Power triangleVPIRVI
4Ohms Law V2 P I P x R Power Voltage P - watts V - volts Voltage I2 X RP x RPower VoltageP - watts V - voltsVoltageV = voltsPowerP = wattsV x II x RVRVICurrentI = amperesResistanceR = ohmsPVPI2PRV2P
5525 – Voltage drop in consumers’ installations Regulation directs readers to a table in Appendix 12. pg 314DescriptionLightingOther uses(i) Low voltage installations supplied directly from a public low voltage distribution system3%5%(ii) Low voltage installation supplied from private LVsupply (*)6%8%(*) The voltage drop within each final circuit should not exceed the values given in (i)Where a distribution circuit (from origin) feeds another distribution board, the total allowable voltage drop should be split between the distribution circuit(s) and final circuit(s).
6Voltage drop calculations Where the voltage drop is in volts and:mV is the millivolts dropped per ampere per metre taken from Tables 4D1A to 4J4A pgIb is the design current of the circuit (the current intended to be carried)Length is the length of the circuit in metres. The division by 1000 gives the answer in volts.
7Volt Drop Cable Calculation A lighting circuit protected by a BS3036 fuse consisting of 10 x 100W luminaires is wired in single copper pvc 700C insulated cable installed with two other circuits in conduit fixed to a wall.The circuit length is 23m and the ambient temperature of the surroundings is 350C.Carry out a cable calculation and check that the circuit complies with BS7671:2008 incorporating Amendment noStep 1. Gather the relevant information from the question.10 x 100W luminaires = 1000W700C PVC single copper cables run in conduit fixed to wall = Reference method B (BS7671 Table 4A2 p.317)BS3036 protective device = Cf = (BS7671 Appendix 4 p.308)Circuit Length = 23mGrouping with 2 other circuits (total 3) = Cg = 0.7 (BS7671 Table 4C1 p.327)Ambient temperature 350C = Ca = 0.94 (BS7671 Table 4B1 p.325)
8Volt DropStep 2. Calculate the circuit design current. (Ib) using ohms law power triangle I = P / V Ib = P = 1000 =4.35A V 230 Step 3. Determine the rating of the protective device (In) In ≥ Ib = 5A (BS7671 Table 41.2 p.55) Step 4. Determine the minimum current carrying capacity of the cable (It) (BS7671 Appendix 4 p.308) It = In = In = 5 = 5 = 10.48A Correction factors (Cg x Ca x Cf ) (0.7 x 0.94 x 0.725) Cable must be capable of carrying 10.48A
9Volt DropStep 5. Identify a suitable cable. = (BS7671 Table 4D1A p.332 reference method B) 1mm2 cable has a rating of 13.5A Step 6. Calculate the volt drop. (Vd) = (BS7671 Table 4D1B page 333 reference method A and B) Voltage drop = (mV/A/m)x Ib x L = 44 x 4.35 x 23 = 4.4V Appendix 4 page 314 This circuit complies with the requirements of BS7671 in that the calculated volt drop 4.4V is less than the maximum permitted 3% for a single phase lighting circuits. Single phase; 3% of 230 = 6.9V Therefore the cable is suitable.
10Resistance of copper and aluminium conductors Uo = the normal voltage to earth Zs = the earth fault loop impedance and : Zs = Ze + (R1 + R2) Ze = the external earth fault loop impedance of the supply cable Zs = the earth fault loop impendence of the Internal current carrying circuit. R1+R2 = the sum of the resistance of the line conductor (R1) and the circuit protective conductor (R2) between the point of utilisation and the origin of the installation.
11Resistance of copper and aluminium conduct Similarly, in order to design circuits for compliance with BS7671 limiting values of earth fault loop impedance given in Tables 41.2 to 41.4, it is necessary to establish the relevant impedance of the circuit conductors concerned at their operating temperatures. For example 1.5mm2 with 1.0mm2 earth c.p.c. is given as mΩ/m therefore if the cable is 20mts in length this will be x 20 = 0.60Ω which is the value of R1+R2 1000
12Electrical systems and earthing arrangements TTTN-C-STN-S
13TN-S system Typical quoted value for Ze is 0.8 W Z means the total opposition to current flow Ze means external Zs means system
14TN-C-S systemTypical quoted value for Ze is 0.35 W
15TT systemThis system affords a relatively high Ze value, which must not exceed 21 W Ra ≤ 200 W
16TN-C system A TN-C system is uncommon in the United Kingdom. Neutral and protective functions are combined in a single conductor (a PEN conductor) throughout the system (although, the term CNE is sometimes used for such a conductor forming part of the distributor’s lines).The exposed-conductive-parts of the installation are connected to the PEN conductor, and hence to the earthed point of the source of energy.Regulation 8(4) of the Electricity Safety, Quality and Continuity Regulations 2002 prohibits the use of PEN conductors in consumers’ installations.
17Nominal Voltages in the UK Single phase (U0) 230 V (Actual 240 V)Three Phase (U) 400 V (Actual 415 V)NOTES:
19DEAD OR LIVE TESTING? ‘DEAD’ TESTING Prescribed tests: ContinuityPolarityInsulation resistance‘LIVE’ TESTINGPrescribed tests:Earth fault loop impedanceFault current measurementRCD operation
20Insure these items are tested External ZeEarth electrode RA (if applicable)Continuity of protective conductorsInsulation between Live conductors and earthInsulation between Live conductorsPolarityEarth Loop-ImpedanceOperation of Residual Current deviceFunctional Testing
21VERIFICATION & CERTIFICATION - CERTIFICATES Three types of Certificate could be used for a domestic electrical installation:-An Periodic Inspection (Electrical Condition Report)An Electrical Installation CertificateA Minor Works Certificate
22TipsLets look at a typical question from the exam and follow it through at various stages.
23Example QuestionBS7671 identifies that the cross-sectional area of a conductor shall be determined byA the admissible maximum temperatureB the nominal voltageC voltage tolerancesD the earthing systemy
24TipsNow if we go straight to the BS7671 index and look up cross-sectional area of conductors and cables page 442. We find a reference to Sec 524, we would now look in part 1 chapter 32 regulation 6 page 19. On reading the regulation we find: The cross-sectional area of conductors shall be determined for both normal operating conditions and where appropriate, for fault conditions according to: ( 1 ) the admissible maximum temperature. Therefore we know that ( A ) must be the answer.
25Example QuestionBS7671 identifies that the cross-sectional area of a conductor shall be determined byA the admissible maximum temperatureB the nominal voltageC voltage tolerancesD the earthing system
26TipsLets try another example and follow that through!
27A residual current B harmonic current C line current D neutral current Example QuestionThe algebraic sum of the currents in the live conductors of a circuit at a point in the electrical installation is known as theA residual currentB harmonic currentC line currentD neutral currenty
28TipsLooking in the BS7671 index the only phrase we can find from the question is Electrical installation definition Part 2 page 444. We then go to Part 2 Definitions we can only find Residual current. Algebraic sum of the currents in the live conductors of a circuit at a point in the electrical installation. Page 33 We now know that ( A ) is the answer.
29A residual current B harmonic current C line current D neutral current Example QuestionThe algebraic sum of the currents in the live conductors of a circuit at a point in the electrical installation is known as theA residual currentB harmonic currentC line currentD neutral current
30TipsAs we can see from the last question it is not always plain sailing in finding out an answer! But keeping calm and looking in the question for the various words or the answers, you can find what you are looking for. Now you try the next one!
31Which one of the following cannot be used as basic protection Example QuestionWhich one of the following cannot be used as basic protectionA insulation of live partsB barriers or enclosuresC protective earthing and bondingD obstaclesy
32Example AnswerWhich one of the following cannot be used as basic protection Options a,b,d, all provide basic protection (see Section 416 and 417) option c provides fault protection therefore C is the answer.A insulation of live partsB barriers or enclosuresC protective earthing and bondingD obstaclesy