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BS7671 Formula and Tips

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The 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)

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Ohms Law Resistance of the cable in ohms = R Current being drawn from the circuit = I Voltage being delivered to the circuit = V Power being developed by the circuit = P This can give us two triangles for formula Resistance triangle Power triangle V P IR VI

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Ohms Law Power Voltage P - watts V - volts Power P = watts Voltage V = volts Current I = amperes Resistance R = ohms V2RV2R I 2 X R V x I PIPI P x R I x R VRVR PVPV VIVI PI2PI2 V2PV2P PRPR

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DescriptionLightingOther uses (i) Low voltage installations supplied directly from a public low voltage distribution system 3%5% (ii) Low voltage installation supplied from private LV supply (*) 6%8% (*) The voltage drop within each final circuit should not exceed the values given in (i) 525 – Voltage drop in consumers’ installations Regulation directs readers to a table in Appendix 12. pg 314 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).

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Voltage 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 pg I b is the design current of the circuit (the current intended to be carried) Lengthis the length of the circuit in metres. The division by 1000 gives the answer in volts.

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Volt Drop Cable Calculation A lighting circuit protected by a BS3036 fuse consisting of 10 x 100W luminaires is wired in single copper pvc 70 0 C 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 35 0 C. Carry out a cable calculation and check that the circuit complies with BS7671:2008 incorporating Amendment no Step 1. Gather the relevant information from the question. 10 x 100W luminaires = 1000W 70 0 C 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 = 23m Grouping with 2 other circuits (total 3) = C g = 0.7 (BS7671 Table 4C1 p.327) Ambient temperature 35 0 C = C a = 0.94 (BS7671 Table 4B1 p.325)

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Volt Drop Step 2. Calculate the circuit design current. (I b ) using ohms law power triangle I = P / V I b = P = 1000 =4.35A V 230 Step 3. Determine the rating of the protective device (I n ) I n ≥ I b = 5A (BS7671 Table 41.2 p.55) Step 4. Determine the minimum current carrying capacity of the cable (I t ) (BS7671 Appendix 4 p.308) I t = I n = I n = 5 = 5 = 10.48A Correction factors (C g x C a x Cf ) (0.7 x 0.94 x 0.725) Cable must be capable of carrying 10.48A

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Volt Drop Step 5. Identify a suitable cable. = (BS7671 Table 4D1A p.332 reference method B) 1mm 2 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 I b 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.

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Resistance 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.

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Resistance 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 1.5mm 2 with 1.0mm 2 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

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Electrical systems and earthing arrangements TTTN-C-STN-S

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TN-S system Typical quoted value for Z e is 0.8

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TN-C-S system Typical quoted value for Z e is 0.35

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TT system This system affords a relatively high Z e value, which must not exceed 21 R a ≤ 200

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TN-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.

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Nominal Voltages in the UK Single phase (U 0 ) 230 V(Actual 240 V) Three Phase (U) 400 V (Actual 415 V)

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IP code lettering/numbering systempg381 Discuss: IP2X, IP4X IPXXB, and IPXXD

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‘DEAD’ TESTING Prescribed tests: Continuity Polarity Insulation resistance DEAD OR LIVE TESTING? ‘ LIVE’ TESTING Prescribed tests: Earth fault loop impedance Fault current measurement RCD operation

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Insure these items are tested External Ze Earth electrode R A (if applicable) Continuity of protective conductors Insulation between Live conductors and earth Insulation between Live conductors Polarity Earth Loop-Impedance Operation of Residual Current device Functional Testing

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VERIFICATION & CERTIFICATION - CERTIFICATES Three types of Certificate could be used for a domestic electrical installation:- ► An Periodic Inspection (Electrical Condition Report) ► An Electrical Installation Certificate ► A Minor Works Certificate

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Tips Lets look at a typical question from the exam and follow it through at various stages.

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BS7671 identifies that the cross-sectional area of a conductor shall be determined by A the admissible maximum temperature B the nominal voltage C voltage tolerances D the earthing system y Example Question

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Tips Now 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.

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BS7671 identifies that the cross-sectional area of a conductor shall be determined by A the admissible maximum temperature B the nominal voltage C voltage tolerances D the earthing system Example Question

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Tips Lets try another example and follow that through!

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The algebraic sum of the currents in the live conductors of a circuit at a point in the electrical installation is known as the A residual current B harmonic current C line current D neutral current y Example Question

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Tips Looking 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.

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The algebraic sum of the currents in the live conductors of a circuit at a point in the electrical installation is known as the A residual current B harmonic current C line current D neutral current Example Question

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Tips As 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!

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Which one of the following cannot be used as basic protection A insulation of live parts B barriers or enclosures C protective earthing and bonding D obstacles y Example Question

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Which 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 parts B barriers or enclosures C protective earthing and bonding D obstacles y Example Answer

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