Document Sample
                    Department of Technical and Vocational Education
                               Electrical Power Engineering
                           A.G.T.I Year II (Final Examination)
Date:21-10-200                                                         Time: 8:30 am-13:30 am
                       EP 02015 Electrical Wiring And Installation
                           ATTEMPT ANY SIX QUESTIONS
1.    (a) Describe the required knowledge on which fields made good electrical installation.
      (b) State the various range of supply voltage.
      (c) Define the frequency of supply.

2.    (a) What do you understand the single phase installation?
      (b) What do you understand the 3 phase, 4 wire installation?
      (c) Draw the circuit diagram of voltage obtains from 3 phase , 4 wires distribution?
      (d) Draw the wiring diagram of 2-way intermediate switching from four positions?

3.    (a) What are the methods of earthing?
      (b) Explain the method of earthing in electrical installation?

4.    (a) What is light?
      (b) Define regular, scatter and mixed reflection.
      (c) Which method are base on the modern light of light generation?

5     (a) What is a Parallel Circuit? Explain with its circuit diagram.
      (b) Explain the termination of PVC Twin and Tree core with earth and properties of
         the rigid PVC and flexible PVC.

6.    (a) A 3-phase motor with a full load current 180A, and a power factor of 0.85 is to be
         fed by four single-core, pvc-insulated, copper conductor, non-armoured cables,
         clipped direct on a non-metallic surface at 120-meter run as show in Figures.
         Determine the size of conductor if the permissible voltage drop from the MCB to
         the motor terminal is 3%.
         50 mm2 cable has tabulated current rating of 134A; TVDr = 0.81 and TVDx = 0.26
        70 mm2 cable has tabulated current rating of 171A; TVDr = 0.56 and TVDx = 0.25
        95 mm2 cable has tabulated current rating of 207A; TVDr = 0.42 and TVDx = 0.24


                          IFL = 180A       120 m
                                                                         M   0.85 pf
                             4 x 1 C/Cu/PVC/NA
                             enclosed in conduit

6.   (b) A three-phase circuit consisting of 4 x 10 mm2 Cu/PVC/non-armoured single-core
        cable is slected to feed an electrical appliance which has a design current of 50 A
        at 0.8 power factor as shown in Figure. The circuit length is 30 m and the ambient
        temperature is 25ºC. Determine the voltage drop by taking into consideration a
        lower conductor temperature.
        10 mm2 cable has tabulated current rating of 59 A; TVDr = 3.8 and TVDx = 0


                             IB = 50A     30 m 25. C
                                                                    Appliance       0.8 pf
                               4 x 10 C/Cu/PVC/NA
                                   clipped direct

7.   (a) Explain the Rising Mains Cables systems and Busbar system.
     (b) Discuss the values of Earth-lppo Impedance.

8.   (a) Describe the producer for Load Estimation.
     (b) Describe the requirements for panelboards.

      ******************                  END * * * * * * * * * * * * * * * * * *
1. (a) Solution
       To get satisfactory and safety result from installation a person who works in electrical
installation field must be known a good knowledge in the following field;
       1. Supply system
       2. Consumer’s circuit
       3. Wiring accessories
       4. Conductor and cable
       5. Wiring system
       6. Testing
       7. Instrument and measurement
       8. Earthing
       9. Illumination

1. (b) Solution
       The various range of supply voltage are
       -   Low voltage ( not exceeding 250 volt)
       -   Medium voltage ( above the 250 volt but not exceeding 650 volt)
       -   High voltage ( exceeding 650 volt )

1. (c) Solution
       The alternating current(A) supply is used as standard. The supply voltage will pass
through a complete of positive and negative value cycles, in a given number of every second.
This number is known as frequency of supply
       The SI unit of frequency is Hertz. (1 Hz = 1 C/S)

2. (a) Solution
       Such simple two wire installation is known as single-phase installation (or) a circuit
having one phase (red, yellow, blue) and a neutral.

2. (b) Solution
       The differential loads such as heating, lighting and power circuit are then distributed
over the three-phase, which are identified by colour coading red, yellow and blue, in a given
of load balancing are known as 3 phase, 4 wire installation.
2. (c) Solution



                                                   
                       
                                   
                     
                      
                                                                

                                                            



                   Figure. Voltage obtains from 3 phase, 4 wire distribution

2. (d) Solution



                                                A           B

           Figure. Wiringdiagram of 2-way intermediate switching from 4 position

3. (a) Solution
       The mewthods of earthing are:
       -   direct earthing system
       -   multiple earthed-neutral system
       -   earth leakage circuit breaker system

3. (b) Solution

Direct earthing system
       A system of earthing in which part of the installation are so earthed as specified, but
are not connected within installation to earth trip of an earth leakage circuit breaker.
Multiple earthed-neutral system
        A system of earthing in which part of installation, specified so as earthed, are
connected to general mass of earth and in addition are connected within installation to the
neutral system of supply terminal.

Earth leakage circuit breaker system
        In this system, part of installation are connected to an earth electrode through coil of
earth leakage circuit breaker, which controls the supply that of part of installation in addition
to equipment earth.

4. (a) Solution
        Light is defined as visually radiant energy and it has predication. It can travel through
space in form of electromagnetic wave at constant speed.

4. (b) Solution

Regular reflection
        The parallel beam of light is reflected as such mirror reflection area called regular

Scatter reflection
        If, however a beam of light strikes a white wash ceiling, the incident light is scattered
in all direction. This is known as scatter or diffuse reflection.

Mixed reflection
        Between the extremes of these two there are numerous transition-mixed or
semidiffuse reflection.

4. (c) Solution
        The molders light are based one of following method of light generation.

Temperature radiation
        For age this had been method avariable, example of temperature radiation are tough,
candle, the oil lamp and the gas lamp, etc. The filament lamp too is based.
Gas discharge
        Under certain condition it is possible to pass electrical current through a gas (or) metal
vapour. Visible radiation can be generated in this way because free electron knocks the
electron bond in the gas atom out of their normal path. This lost is followed back to the
original path, give of the energy, in form of radiation. This must more economically than
temperature radiation.

        Certain material can, when exposed an ultra violet or visible radiationb, transform
observed energy into radiation with longer wavelength. This is used in fluorescence lamp and
in mercury-covered lamp.

5. (a) Solution
        Circuit can be buit up from simple three phase circuit. For example, if the two light
points are controlled from one-way switch. In loop wire is taken from each terminal of first
light point and connected to the correspond terminal of the second light point. The result is
that when switch is closed the current has two paths. One through each lamp, so that both are
energized. This circuit is said to be connected in parallel.
        When several circuits have to supply from main switch, the switch wire may all be
connected to one side of supply and the return wire from apparatus to the other side. They are
then in parallel.
        The full electrical pressure of the supply is applied to each circuit. The amount of
power taken by any one circuit is independent of power taken to other side.

                                       Fuse         Switch          Current to flow

                         Phase                                          Light 1       Light 2
                     Supply terminal


                                                      Return wire

                                       Figure. Parallel Circuit
5. (b) Solution
The Termination of PVC Twin and Tree core with Earth
       Two methods are used as;
(i)    Using a sharp knife, slice a cable lengthwise, opened a cable up and cut off the
(ii)   With pair of slide cutter, split end of the supply and hole out the end of the earth
       conductor. This may then be help with a pair of piler and pulled back along the fuse
       cable, splitting open the sheathe. Excess sheathe may be cut away.

The Rigid PVC
       The rigid PVC as used in conduit is a thermoplastic polymer and has the following
       -      It has high tensile strength.
       -      It can be bent by hand if warmed up.
       -      It has high electrical resistant.
       -      It has weather resistant.
       -      It does not creale under stress at normal temperature.
       -      It is low flammable.
       -      It is self-extinguished when source of heat is removed.
       -      It is used for special suddle when use of fictuation temperature.

The Flexible PVC (used in cable insulation)
       -      It has high electrical resistant.
       -      It has weather resistant.
       -      It should be leept clear of other plastic to avoid of migration of plasticizer.

6. (a) Solution
       IB         = 180 A
       cos θ = 0.85
       L          = 120 m
       70 mm cable has tabulated current rating of 171 A is initially selected. The line-to –
line voltage.
                           T V D r cos θ  T V D x sinθ x I B xL
        VdropLL       
                           (0.56 x 0.85  0.25 x 0.53) x 180 x 120
                       = 13.144 V or 2.6% of 400V
       So, 95 mm2 cable has tabulated current rating of 207A is selected.

6. (b) Solution
       IB          = 50 A
       cos θ       = 0.8
       L           = 30 m
       Ta          = 25  C
       VD          =?
        t 50  t a    I
                    ( B )2
         tp  tr      It

                              IB 2
       t50         = ta  (      ) (t p  t r )

                               50 2
                   = 25 + (       ) (70  30)
                   = 25 + 28.72
                   = 53.72  C
At TVD at t50 = 53.72  C
                                230  53.72
   TVD53.72        = TVD (                 )
                                 230  70
                   = 3.8 (          )
                   = 3.8 x 0.959
                   = 3.64
                       T V D r cos θ  T V D x sinθ x I B xL
     VdropLL      
                       (3.64 x 0.8  0) 50 x 30
                   = 4.368 V
         The voltage drop 4.368 V is actually 95.9% of voltage drop obtain multiplying TVDr
         The conductor temperature at 29.5 A will be
                            IB 2
         T29.5   = ta  (      ) (t p  t r )

                             29.5 2
                 = 25 + (        ) (70  30)
                 = 25 + 10
                 = 35  C
TVD at 35  C
                              230  35
         TVD35 = TVD (                 )
                              230  70
                 = 0.88 TVD
         The actual voltage is 0.88% of voltage drop obtain multiplying a TVD value.

7. Solution
         (a) Rising Mains Cables system and Busbar system (10 marks)
         Please  see in page 71-72 of textbook

(b) Values of Earth-loop Impedance (6 marks)
         The go/no-go type of tester simply gives an indication of weather the circuit is
suitable for semi-enclosed or cartridge fuse having a fusing factor exceeding 1.5. Since these
require the lowest impedance of all types of earth-leakage protection. If values above this are
obtained and a check reveats no fault on the circuits then the earth electroder resistance must
be tested.
         The line-earth loop tester and the step-down transformer neutral earth loop tester both
relay on a mains supply being available and therefore cannot be used prior to the electric
board being requested to provide a supply. The hand generator type of instrument can
however, be used to check the impedance of the earth electrode and trace high-impedance
section of the installation prior to a supply being connected. It must be remembered however
that this type of instrument normally only injects a very a small test current and consequently
may not detect earth-continuity paths of low current carrying capacity.
8. Solution
(a) Producer for Load Estimation
       A suggested producer for determining the design currents in various sections of a
bulding is given in the following steps.
       • Determine the quantity of each type of load and the power requirements. Estimate
         the design current in each final circuit based on the rating of the connected load.
        Apply an appropriate demand factor to obtain the maximum demand of earth final
       • Determine the type of connection from the final DB to the main DB and continue up
         to the incoming switchboard. Estimate the design current in each distribution circuit
         using an appropriate coincidence factor.
       • Determine spare capacity to be provided for load growth.

(b) The Requirements for panelboards
       Please  see in page 113 of textbook

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