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Thread: Installation Rules-past papers.

  1. #21
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    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

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    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?

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

  4. #24
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    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

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    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

  6. #26
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    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!

  7. #27
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    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

  8. #28
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    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.

  9. #29
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    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

  10. #30
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    Installation Rules

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

    Dankie

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