Installation Rules-past papers.

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  • AndyD
    Diamond Member

    • Jan 2010
    • 4946

    #16
    You can find free ftp hosting but it comes at the price of pestering and advertising so it's not ideal.

    I can help you make them into a torrent and I'll carry them for a while on my seedbox that way you don't need to fork out all the hosting bandwidth.

    Alternatively you can mail then a few at a time to my gmail address which will receive large files and I'll make the torrent for you and upload it. Send me a pm if you're interested.
    _______________________________________________

    _______________________________________________

    Comment

    • mikilianis
      Bronze Member

      • Dec 2008
      • 125

      #17
      I am open to correction but I can recall that when I was doing a course at night tech to prepare for the exams, there was some discussion as to how one qualifies as a 3 phase installation electrician, and the main requirement was that one had to have passed a trade test as an electrician or millwright, and that if you had passed T3 you did not have to have a trade test pass, is it still or was that a requirement
      Mike

      Comment

      • shuudeni
        Junior Member
        • Sep 2010
        • 14

        #18
        Thanks in advance. You can try my email: mosesshuudeni@gmail.com

        Comment

        • Maccie
          Email problem
          • Aug 2010
          • 14

          #19
          I have sent the papers that In was able to print legibly and to a size my email could handle to shuudini and Esajoda.

          I cannot scan the 2010 paper as my scanner is kaput.

          The papers I sent are what I recieved from the lecturer, and theres none from 2010 on the disc.

          Comment

          • shuudeni
            Junior Member
            • Sep 2010
            • 14

            #20
            Ou, I had recieved the question papers, thanks very much and good luck for the exam, because I believe you will need it.

            Comment

            • Fikani2010
              Junior Member
              • Nov 2010
              • 21

              #21
              Paper 1 and Paper 2 Study Guide and Exam Generator

              Hello All,
              I have compiled a comprehensive study guide for both papers with worked out examples of how to answer exam questions with the emphasis on calculations, also I have developed what I call an exam generator, a software which generate 10 sample exam questions, you can generate as much exam samples as you wish. My package is guaranteed to give anyone attempting these exams a pass and with a lot of hard work manage to grab for themselves distinctions on both papers. I passed my exams with distinction...

              If you like more info, please drop an email.

              Thanks

              Comment

              • jvorster
                New Member
                • Nov 2010
                • 1

                #22
                Hey guys, I'm writing my 2nd paper Installation Rules in November, but I have no clue on what to study. I have some old question papers, but to learn and figure out all of them in two weeks time is going to be impossible, and there's no tutors or classes here. Can someone tell me what I must know for the test?

                Comment

                • shuudeni
                  Junior Member
                  • Sep 2010
                  • 14

                  #23
                  That sounds exciting. Try me at mosesshuudeni@gmail.com

                  Comment

                  • Fikani2010
                    Junior Member
                    • Nov 2010
                    • 21

                    #24
                    Hello all, for those who are interested in receiving a study guide and an Exam Generator are welcome to contact me at: fikani2010@hotmail.com

                    To your success
                    Fikani

                    Comment

                    • Fikani2010
                      Junior Member
                      • Nov 2010
                      • 21

                      #25
                      Hi jvorster, two weeks! Is this your first attempt? please contact me at fikani2010@hotmail.com I will send you something to work on.

                      To your success
                      Fikani

                      Comment

                      • Fikani2010
                        Junior Member
                        • Nov 2010
                        • 21

                        #26
                        SAMPLE EXAMPLE:
                        Name FOUR different methods of how the voltage drop of a cable can be determined and give examples for each method.

                        ANSWER:
                        EXAMPLE: Suppose a balanced three phase load of 100A is to be supplied using a 35mm2 copper cable over a distance of 100m. Calculate the Voltage Drop (USING FOUR DIFFERENT METHODS).

                        Let Vd stand for Voltage Drop to be calculated.

                        We are given: I = 100A and L = 100m (35mm2 of a copper (Cu) cable)
                        From Table E1 page 307 R = 0.63 Ω/Km for a 35mm2 Cu. NB: We must express resistance in Ohms, to do this use:
                        R = Value of R in Ω/Km (for a given cable size from page 307 of the Regulation Book) x L (in meters). We also have to convert Km to meters: 1Km = 1000m. Therefore, R in Ohms = Value of R in Ω/1000m x L (in meters)

                        METHOD 1
                        Use this formula taken from page 308 of the Regulation Book:
                        Vd = (Fv x I x R x L) /1000
                        Where,
                        Vd is the voltage drop, in volts
                        Fv is the multiplication factor determined from Table E3 on page 311 of the Regulation Book and Fv = 1 for a balanced three-phase load
                        I is the current in amperes (100A)
                        R is the resistance in ohms per kilometer (Ω/Km) from Table E1 and it is 0.63 Ω/Km for a 35mm2 copper cable
                        L is the length in meters (100m)

                        Vd = (Fv x I x R x L) /1000
                        So substituting values we have
                        Vd = (1 x 100A x 0.63 Ω/Km x 100m)/1000 = 6.30 V

                        METHOD 2
                        Use: Vd = I x R
                        Where,
                        Vd is the voltage drop, in volts
                        I is the current in amperes
                        The value of R in ohms/Km from Table E1 is 0.63 Ω/Km for a 35mm2 copper cable
                        Now determine R in ohms using:
                        R = Value of R in Ω/Km (for a given cable size from page 307 of the Regulation Book) x L (in meters). We also have to convert Km to meters: 1Km = 1000m. Therefore, R in Ohms = Value of R in Ω/1000m x L (in meters)
                        R = (0.63 Ω/1000m) x 100m = 0.063 Ω
                        Vd = I x R
                        So substituting values we have
                        Vd = I x R = 100 A x 0.063 Ω = 6.30 V

                        METHOD 3
                        Use this formula Vd = (mV/A/m) x A x m [Volts]
                        Where,
                        mV is the unit of voltage in milli-volts. NB 1V = 1000mV
                        A is the unit of current in amps [100A]
                        m is the unit of length of the cable in meters [100m]
                        (mV/A/m) = 1.10 for a 35mm2 copper (Cu) determined from Table 6.3(b) on page 105 or Table 6.4(b) on page 107 of the Regulation Book – The difference between the two tables is that Table 6.3(b) makes provision for a 1mm2 whereas Table 6.4(b) does not. In the exam you will most probably be given Table 6.3(b) [Please familiarize yourself with the use of these Tables and also Table 6.4(a) – Current Carrying Capacity]. NB: From the Tables 6.3(b) or 6.4(b) take the value of z (the impedance) corresponding to the given cable size and not the value of r (resistance) or x (the reactance).
                        Vd = (mV/A/m) x A x m [Volts]
                        So substituting values we have
                        Vd = (mV/A/m) x A x m = 1.10mV/A/m x 100A x 100m = 11000mV = 11V
                        For phase to neutral Vd = 11V/√3 = 6.35V *[NB: See section 6.2.7 page 120]

                        METHOD 4
                        Use Vd = (I x pL)/A
                        Where,
                        A is area of cable in this case 35mm2
                        p = 0.0223, the resistivity of copper conductors
                        Vd = (I x pL)/A
                        So substituting values we have
                        Vd = (I x pL)/A = (100A x 0.0223 x100m) / 35 mm2 = 6.37 V
                        ADDITIONAL METHODS
                        In the case of a load with impedance, the voltage drop can be calculated from the following
                        Vd = (Fv x I x (R cosθ + X cosθ) x L)/1000
                        The phase angle θ of the load is determined by power factor = cos θ
                        The reactance X may be obtained from cable manufactures
                        OR
                        Voltage drop = (1.72 x I x R) / √3
                        OR Alternatively the value of the Voltage Drop may be derived from a monogram
                        END OF SAMPLE EXAMPLE:

                        APRIL 2010 EXAM QUESTION
                        QUESTION 1: SANS 10142 PART 1 OF 2008: ANNEXURE E: VOLT DROP (COMPULSORY)
                        1.1 With reference to FIGURE 1, on the DIAGRAM SHEET (attached), calculate the following:
                        1.1.1 The total volt drop of the supply (4)
                        1.1.2 The total resistance of the supply cable (1)
                        1.1.3 Test the answer in QUESTION 1.1.1 above by making use of any other method. (3)
                        1.2 By making use of TABLE E1 (attached) only, calculate the actual length of a 10 mm2 copper ECC conductor if the total resistance of the conductor is 0,22 ohm. (2)
                        [10]

                        ANSWER TO QUESTION 1 (PRIL 2010 PAPER II:
                        1.1. With reference to FIGURE 1, on the DIAGRAM SHEET (attached), calculate the following:
                        1.1.1. The total volt drop of the supply (4 Marks)
                        Assumption: The installation uses copper cables
                        NB: We know that the voltage drop shall not exceed 11.5V [5% of 230V] according to the code.
                        Given:
                        Point of control (DB): Supply voltage is 230V
                        The resistive load of 50A is 65m away from the point of control [I=50A, L=65m]
                        From the Point of control to the cable JOINT we have 40m of 16 mm2 copper cable [R = 1.4 Ω/Km, given this from Table E1]
                        From the cable JOINT to the load we have 25m of 10 mm2 copper cable [R = 2.2 Ω/Km, given this from Table E1]
                        Use this formula from page 308 Vd = (Fv x I x R x L) /1000
                        Where,
                        Vd is the voltage drop, in volts
                        Fv is the multiplication factor determined from Table E3 on page 311 we have Fv = 2 for a single-phase load
                        I is the current in amperes
                        R is the resistance in ohms per kilometer (Ω/Km) from Table E1
                        L is the length in meters [40m and 25m lengths of different cables]
                        Vd = (Fv x I x R x L) /1000
                        So substituting values we have:
                        Vd = (2 x 50A x 1.4 x 40m)/1000 + (2 x 50A x 2.2 x 25m)/1000 = 5.6V + 5.5V = 11.10 V

                        1.1.2 The total resistance of the supply cable (1 Mark)
                        R = Value of R in Ω/Km (for a given cable size from page 307 of the Regulation Book) x L (in meters). We also have to convert Km to meters: 1Km = 1000m. Therefore, R in Ohms = Value of R in Ω/1000m x L (in meters)
                        Resistance of a 16 mm2 = 1.4 Ω/Km x 40m = 1.4 Ω/1000m x 40m = 0.056 Ω
                        Resistance of a 10 mm2 = 2.2 Ω/Km x 25m = 2.2 Ω/1000m x 25m = 0.055 Ω
                        Total Resistance of the supply cable = Resistance of a 16 mm2 + Resistance of a 10 mm2=0.056 Ω + 0.055 Ω = 0.111 Ω

                        1.1.3 Test the answer in QUESTION 1.1.1 above by making use of any other method. (3 Marks)
                        Alternative method 1:
                        Use: Vd = I x R
                        Where,
                        Vd is the voltage drop, in volts
                        I is the current in amperes [which is 50A]
                        The value of R in ohms/Km from Table E1 is 1.4 Ω/Km for a 16mm2 and 2.2 Ω/Km for a 10mm2 copper cable
                        We have already determined R in ohms from the previous question [0.111 Ω]
                        Vd = I x R
                        So substituting values we have
                        Vd = I x R = 50A x 0.111 Ω = 5.55 V
                        NB: Remember that we have to account for the voltage drop on neutral as well, which is 5.55V in this case.
                        Therefore, Total Voltage Drop = 5.55V + 5.55V = 11.10V

                        Alternative method 2:
                        Use Vd = (I x pL)/A
                        Where,
                        A is area of cable [16mm2 and 10mm2]
                        p = 0.0223, the resistivity of copper conductors
                        Vd = (I x pL)/A
                        So substituting values we have
                        Vd on a 16mm2 cable = (I x pL)/A = (50A x 0.0223 x40m) / 16mm2 = 2.7875V
                        Vd on a 10mm2 cable = (I x pL)/A = (50A x 0.0223 x25m) / 10mm2 = 2.7875V
                        Vd of the supply = Vd on a 16mm2 + Vd on a 10mm2 = 2.7875V + 2.7875V = 5.575V
                        Again remember that we have to account for the voltage drop on neutral as well, which is 5.575V in this case.
                        Therefore, Total Voltage Drop = 5.575V + 5.575V = 11.15V

                        Alternative method 3:
                        *** Use this formula Vd = (mV/A/m) x A x m [Volts]
                        Determine the mV/A/m either from Table 6.3(b) or 6.4(b)
                        mV/A/m for a 16mm2 is 2.8
                        mV/A/m for a 10mm2 is 4.4
                        Vd = (mV/A/m) x A x m [Volts]
                        So substituting values we have:
                        Vd on a 16mm2 cable = mV/A/m x A x m = 2.8 mV/A/m x 50A x 40m = 5600 mV = 5.60V
                        Vd on a 10mm2 cable = mV/A/m x A x m = 4.4 mV/A/m x 50A x 25m = 5500 mV = 5.50V
                        Vd of the supply = Vd on a 16mm2 + Vd on a 10mm2 = 5.60V + 5.50V = 11.10V

                        *** Please note that neither Table 6.3(b) nor Table 6.4(b) was given in the exam, so we would not be able to use this method to substantiate our answer, but it is given here to prove that our answer is indeed right.

                        1.2. By making use of TABLE E1 (attached) only, calculate the actual length of a 10 mm2 copper ECC conductor if the total resistance of the conductor is 0,22 ohm. (2 Marks)

                        Recall that resistance in Ohms can also be obtained by using the following method:
                        R = Value of R in Ω/Km (for a given cable size from page 307 of the Regulation Book) x L (in meters). We also have to convert Km to meters: 1Km = 1000m. Therefore, R in Ohms = Value of R in Ω/1000m x L (in meters)
                        We are given 0.22 ohms and we have to calculate the actual length of a 10 mm2 ECC, we can apply the above formula:
                        We are given R = 0.22 ohm
                        From Table E1, R in Ω/Km is 2.2 for a 10 mm2
                        The only unknown now is the length (L) of the cable
                        R = Value of R in Ω/1000m x L (in meters)
                        So substituting values we have
                        0.22 ohm = 2.2 ohm/1000m x L
                        0.22 ohm = 0.00022 ohm/m x L
                        L = 0.22 ohm / (0.00022 ohm/m)
                        L = 100m
                        Therefore, the actual length of a 10 mm2 ECC is 100m for a given resistance value


                        Refer to figure 4 and calculate the following:
                        1. Calculate the minimum cable size for the loads.
                        2. Touch voltage at L3.
                        3. PSCC at L1
                        NOTE: Use a three core cable and ignore suppliers transformer and mains supply characteristics.
                        FIGURE 4
                        CIRCUIT DIAGRAM NOT ACCORDING TO SCALE

                        TO GET THE COMPLETE ANSWER TO THE ABOVE QUESTION AND MANY MORE EXAMPLES, ORDER THE COMPLETE STUDY GUIDE AND AN EXAM QUESTIONS GENERATOR SOFTWARE FROM – FIKANI AT: fikani2010@hotmail.com or give me a call at 072 634 8547.
                        I have Installation Rules Paper1 study guide as well. The study guides consists of a selection of questions and answers similar to the ones you will find on the official exam. All you need to do is use your Regulation Book to answer the questions in each study guide and attempt as many questions as you can that will be generated by the Exam Generator software. Then, once that is done you will be ready to take the exam. Remember, I guarantee that you will pass provided that you follow the instructions correctly! Should you use these study materials and still fail the exams, then send me the original copy of your exam and a copy of your results, I will gladly refund the cost of the study material. However, you are not going to need this guarantee if you follow the instructions as mentioned in the study guide.
                        Why Use This Study Material?
                        The study material is a no-nonsense guide for the Installation Rules Paper 1 exam and Installation Rules Paper 2 exam. It focuses on the areas most likely to be on the exam (9 out of 10 Questions generated by the Exam Generator Software will be in the actual exam), plus it provides background information to help you understand some of the more complex concepts and detailed, step by step calculation examples. The focus is on preparing you for the exams and to shorten your preparation time for your exams. The study material offers a self-paced method of preparing yourself for the exams. You do not have to guess what to study; every Module guides you an in depth questions, detailed coverage, and review questions (NO GARBAGE). This step-by-step structure identifies what you need to study, gives you all the facts, and rechecks what you know. If you work very hard, you can pass both papers with distinctions.
                        GOOD LUCK!

                        Comment

                        • cvmostert
                          Junior Member
                          • Feb 2008
                          • 19

                          #27
                          Hi guys/gals, I have papers and answers for paper 1 and 2, I just passed the first paper and registered for the second. I can make them available. I need a couple of papers/memos to make my set complete.

                          does anyone have the following:

                          paper 1:
                          2006 april - memo
                          2006 nov - memo
                          2009 apr - memo
                          2010 aug - memo

                          paper 2:
                          2007 nov, apr - papers
                          2009 apr, nov - memos

                          I have 2010 apr aug nov papers and memos.

                          Regards,

                          CVM

                          Comment

                          • naidoos7
                            Email problem
                            • Jul 2011
                            • 1

                            #28
                            Hi

                            I used the info you provided thqanks alot, I have some new papers how can I upload them or mail it to someone to do that. Thanks again for the info provided.

                            Comment

                            • cvmostert
                              Junior Member
                              • Feb 2008
                              • 19

                              #29
                              Hi, If you are not able to upload them, email them to me at. chris ( at ) allerliefste (dot) co (dot) za - just replace the words in brackets with the actual characters.
                              Once my list is complete or anyone asks, I will upload my complete set of papers!
                              This helps allot! Thanks!
                              CVM

                              Comment

                              • SPARKIE
                                New Member
                                • Sep 2011
                                • 2

                                #30
                                Installation Rules

                                Baie dankie vir Paper 2 vraestelle- Kan iemand dalk dieselfde doen vir Paper 1

                                Dankie

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