handout ch3

__________ ___________________ _____________ ________ Electrical Installation

__________________________________________________________________________ Electrical Installation Circuit Design

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CHAPTER 4

Electrical Installation Circuit Design

► Terminology and definitions:

Ampacity: current carrying capacity of electric conductors expressed in amperes. Appliance: utilization equipment. Branch circuit: the circuit conductor between the final over current device protecting the circuit

and the outlet(s). Demand factor: the ratio of the maximum demand of a system, or part of a system, to the total

connected load of a system or the part of the system under consideration. Feeder: all circuit conductors between the service equipment, or the generator switchboard of an

isolated plant, and the final branch circuit over current device. Ground: a conductor connection, whether intentional or accidental, between an electric circuit or

equipment and the earth, or to some conducting body that serves in place of the earth. Lighting outlet: an outlet intended for direct connection of a lamp holder, a light fixture, or a

pendant cord terminating in a lamp holder. Outlet: a point on the wiring system at which current is taken to the utilization equipment Receptacle: a contact device installed at the outlet for the connection of a single attachment plug. Service: the conductor and equipment for delivering energy from electric supply system to the

wiring system of the premises served. Switch board: a large panel, frame or assembly of panels on which are mounted, on the face or

back or both, switches, over current and other protective devices, buses, any usual instruments.

► Wiring Design Criteria

Flexibility: every wiring system should incorporate sufficient flexibility of design in branch circuitry, feeders, and panels to accommodate all portable, patterns, arrangements and locations of electric loads. The degree of flexibility to be incorporated depends in large measure on the type of facility. As part of the design for flexibility, provision for expansion must be provided. It must, however, be emphasized that over design is as bad as under design.

Reliability: the reliability of electrical power within a facility is determined by two factors: the utility’s service and the building’s electrical system.

Safety: the designer must be constantly alert to an initial safe electrical installation and such factors as electrical hazards caused by misuse of equipment or by equipment failure after installation.

Energy consideration: includes limiting voltage drops, power factor correction, use of switches for control, etc

Economic cost: includes initial cost and operating cost Space allocation: concerned with maintenance ease, ventilation, expandability, centrality,

limitation of access, and noise, in addition to the basic item of space adequacy.



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► Design procedure

The steps involved in the electrical wiring design of any facility are outlined below. These may in some instances be performed in different order, or two or more steps may be combined, but the procedure normally used is that listed below. a) Determine with the client the usage of all areas, and type and rating of all client furnished

equipments including their specific electric ratings. b) If the designer could not get the exact electrical rating of all the equipment that are going

to be installed in the building such as plumbing, elevators, kitchen, motors etc, determine their ratings from other consultants.

c) Make an electrical load estimate based on the above collected data, areas involved, previously installed similar installation data and any other pertinent data.

Load Estimation: when initiating the wiring design of a building, it is important to be able to estimate the total building load in order to plan such spaces as transformer rooms, chases, and closet. This information is also required by the local power company well in advance of the start of construction. Of course, an exact load total can be made after completing the design. But such estimation can be made from the knowledge of the loads the building uses.

The electrical loads in any facility can be categorized as: (i) Lighting. (ii) Miscellaneous power, which includes convenience outlets and small motors. (iii) Heating, ventilating, and air conditioning. (iv) Plumbing or sanitary equipment: house water pump, air compressors, and vacuum

pumps etc. (v) Vertical transportation equipment: elevators, moving stairs, and dumbwaiters. (vi) Kitchen equipments. (vii) Special equipments.

d) In cooperation with the local electric utility, decide upon the point of service entrance, type of service run, service voltage, metering location, and building utilization voltage.

The above considerations and general rules affecting service equipment are listed below:

i. A building may be supplied at one point by either a single set or parallel sets of service conductors.

ii. All equipment used for service including cable, switches, meters, and so on, shall be approved for that purpose.

iii. It is recommended that a minimum of 100-amp, 3-wire, 220/380V service be provided for all individual residences.

iv. No service switch smaller than 60 amp or circuit breaker frame smaller than 50 amp shall be used.

v. In multiple occupancy buildings tenants must have access to their own disconnect means.

vi. All building equipment shall be connected on the load side of the service equipment except that service fuses, metering, fire alarm, and signal equipment and equipment serving emergency systems may be connected ahead of the main disconnect.

In computing the size of the service equipment bus, a total is taken of the various feeder loads. Although application of a Diversity Factor to this total is permissible, good



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practice dictates the use of a unity Diversity Factor in order to provide a measure of spare capacity in the service equipment.

(e) Determine the location and estimate the size of all required electric equipment spaces including switchboard rooms, emergency equipment spaces, electric closets, and so forth. NOTE: - Panel boards are normally located in closets but may be located in corridor

walls or elsewhere. This work is necessary at this point to enable the architect to reserve these spaces for the electrical equipment. Once the design is accomplished in detail, the estimated space requirements can be checked and necessary adjustments made.

(f) Design the lighting for the facility. This step is complex and involves a continued

interaction between the architect and the lighting designer. (g) On the same plan, or on a separate plan, as decided, locate all electrical apparatus

including receptacles, switches, motors, and other power consuming apparatus. Under floor duct and ceiling track systems would be shown at this stage. If extensive, a separate plan is made.

(h) On the plan, locate signal apparatus such as phone outlets, speakers, microphones, TV

outlets, fire and smoke detectors, and so on. (i) Make drawing showing all lightings, devices, and power equipments circuit connection to

the appropriate panel board. (j) Prepare the panel schedule (table). This table shows the load distribution over the three

phases and the type of load which is connected on each circuit. At this step, include the separate circuitry for emergency equipments and for spare circuit.

(k) From the panel schedule (table) compute panel loads, And make connection

rearrangement so that you will be able to an optimum power balance over the three phases R, S and T.

(l) Prepare the riser diagram. This includes design of distribution panels, switchboards, and

service equipment. (m) Compute feeder sizes and all protective equipment ratings. (n) Cheek the preceding work.

► Branch Circuit Design ● Guidelines for Residential (a) The NEC requires for residences sufficient circuitry to supply a load of 3w/sq ft in the

building, excluding unfinished spaces such as porches, garages, and basements.

(b) The NEC requires a minimum of two 20-amp appliance branch circuits to feed all the small appliance outlets in the kitchen, pantry, dining room, family room etc. Furthermore, all kitchen outlets must be fed from at least two of these circuits (Avoid placing all the lighting in a building on a single circuit). Also receptacles should be circuited with preferably two, but not more than four on a 20-amp circuit.



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(c) The NEC requires that at least one 20-amp circuit supply to be set for laundry outlets.

This requirement satisfies good practice. If electric clothes dryer is anticipated an individual branch circuit should be supplied to serve this load, via a heavy-duty receptacle.

(d) Do not combine receptacles and switches into a single outlet except where convenience

of use dictates high mounting of receptacles.

(e) Circuit the lighting and receptacles so that each room has parts of at least two circuits. This includes basements and garages.

(f) Supply at least one receptacle in the bathroom and one outside the house (g) Provide switch control for closet lights. (o) In bedrooms supply two duplex outlets at each side of the bed location to accommodate

electric blanket, clocks, radios, lamps, and other such appliances.

(i) Since receptacles are counted as part of general lighting and no additional load is included for them, no limit is placed on the number of receptacle outlets that may be wired to a circuit. But for good practice they should be limited to 6 on a 15-amp circuit and 8 on a 20-amp circuit.

(l) Kitchens should have a duplex appliance outlet every 36 in. of counter space, but no less

than two in addition to the normal wall outlets. (m) A disconnecting means, readily accessible, must be provided for electric ranges, cook

tops, and ovens It is better practice to utilize a small kitchen panel recessed into a corner wall to control the large kitchen appliances and to provide completely safe, accessible disconnecting means. Such an arrangement can also be cheaper if the length of run between the main panel and the kitchen is appreciable.

● Guidelines for Non-residential (a) Schools. Since schools comprise an assembly of varied use spaces, including lecture hall,

laboratory, shop, assembly, office, gymnasium, plus special areas such as swimming pools, photographic labs, and so on, it is not possible to generalize on branch circuit design considerations except for the following:

i. To accommodate the opaque and film projectors frequently used in the classroom,

20-amp outlets wired two receptacles on a circuit are placed at the front and back of each such room. A similar receptacles, wired 6 or 8 to a circuit is placed on each remaining wall.

ii. Light switching should provide: 1) High-low levels for energy conservation and to permit low-level lighting

for film viewing. With fluorescent lighting this can be accomplished by alternate ballast wiring and switching, thus avoiding the high cost of dimming equipment.

2) Separate switching of the lights on the window side of the room, which is often lighted sufficiently by daylight.

iii) Provide appropriate outlets for all special equipment in labs, shops, cooking rooms, and the like.



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iv) Use heavy-duty devices and key operated switches for public area lighting (corridors, etc.), plastic instead of glass in fixtures, and vandal-proof equipment wherever possible. All panels must be locked and should be in locked closets.

v) The NEC requires sufficient branch circuitry to provide a minimum of 3 w/sq ft for general lighting in schools. Refer to the NEC Article No. 220. Unlike residential occupancy this figure does not include receptacles. Receptacles are calculated separately at 180 w each for ordinary convenience outlets.

vi) Keep lighting and receptacles completely separate when circuiting. (b) Office Space

i. In small office spaces (less than 400 sq ft) provide either one outlet for every 40 sq ft, or one outlet for every 10 linear ft of wall space, whichever is greater. In larger office spaces, provide one outlet every 100 to 125 sq ft beyond the initial 400 sq ft (10 outlets). These should comprise wall outlets spaced as above plus floor outlets sufficient to make up the required total. In view of the increasingly heavy loads of office machines, these receptacles should be circuited at no more than 6 to a 20-amp branch circuit, and less if the equipment to be fed so dictates.

ii. Corridors should have a 20-amp, 220-v outlet every 50 ft, to supply cleaning and waxing machines.

iii. As with all non-residential buildings, convenience receptacles are figured at 180 w each.

(c) Stores. In stores, good practice requires at least one convenience outlet receptacle for

every 300 sq ft in addition to outlets required for loads such as lamps, show windows, and demonstration appliances.

► Load Tabulation

While circuiting the loads, a panel schedule is drawn up which lists:

The circuit numbers Load description (the type of the load) Wattage (actually in volt-amperes) The current ratings Number of poles of the circuit-protective device feeding each circuit and

the like. Spare circuits are included to the extent that the designer considers them necessary and consonant with economy, but normally no less than 20% of the number of active circuits. Finally, spaces are left for future circuit breakers, in approximately the same quantity as the number of spare circuits, but always to round off the total number of circuits. A typical panel schedule is shown on the next page as an example.

In calculating panel loads, the following rules apply:

(a) Each specific appliance, device, lighting fixture, or other load is taken at its nameplate rating, except certain kitchen and laundry appliances for which the NEC allows a demand factor. (See NEC Article 220.)

(b) Each convenience outlet, in other than residential spaces, is counted as1.5 amp (180 W). (c) Spare circuits are figured at approximately the same load as the average active circuits. (d) Free spaces are not added into the load.



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(e) Loads for special areas and devices such as show window lighting, heavy-duty lamp holders, and multi outlet assemblies, are taken at the figures given in NEC Article 220.

Note: 1) In calculating total panel load, no demand factors may be applied except

specifically stated in the NEC. This is because feeders are calculated for maximum load to be carried, i.e. 100% demand factor is used.

2) The phase loads have to be approximately equally distributed over the three phases

(if a three-phase supply is utilized in an installation). It is the responsibility of the

(Example)



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designer (or contractor) to circuit the loads so that the phases are as closely balanced in load as possible. If this is not done, one phase will carry considerably more current than the others. Since the panel feeder must be sized for the

maximum phase current, this may lead to an over sized feeder and therefore a waste of money.

Having tabulated and balanced the loads and totaled them by phase, the maximum current is calculated. A portion of the spare capacity available in branch circuit is added to the above total, as the basis for the calculation of the feeder load.

► Feeder Capacity The electric line (cable) that is running from the main distribution line to each sub distribution board is known as Feeder. To achieve economy, the panel feeder must accommodate the initial load plus some portion of the future load. One or more of the following procedures provides spare capacity in feeders:

(a) Provide feeder for initial plus spare, with properly sized conduit. This method is generally most economical.

(b) Provide feeder for initial plus spare, with conduit oversized by one size. Some additional cost is entailed here. This is only used where large load expansion is anticipated.

(c) Provide for initial load plus spare, with an empty conduit for future. This method is expensive because of high conduit cost, and it is infrequently advisable.

EXAMPLE Assume a single floor of an office building 100 ft X 200 ft. Assume also 15% of the area is corridor and storage. Calculate the required number of panels, circuits, and feeder size.

SOLUTION

Office space = 85% of 20,000 sq ft = 17,000 sq ft

Corridor and storage =15% of 20,000 sq ft =3000 sq ft

With respect to minimum loads, NEC specifies that the power supply can be increased by 25% if loads are continuous (3 or more hours). This requirement allows for breakers to heat up in panels while carrying continuous load, and is waived for circuit breaker which are ambient compensated, that is, are rated to carry 100% load. Since we have established 80% of the breaker rating as maximum load, we have already accounted for this factor in circuitry, but must keep it in mind in feeder calculation. Office load

17,000 sq ft @ 5 w/sq ft = 85kw Storage

3000 sq ft @ 0.5 w/sq ft = 1.5kw TOTAL LOAD = 86.5kw

Minimum feeder capacity 1.25 X 86.5 = 108kw. The 25% additional capacity is for continuous load.



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Since this building is of good grade office construction, let us anticipate 40 to 80% expansion, and utilize an initial branch circuit loading of 1300 W per ckt, Assuming that each lightening branch circuit is 1300 W rated.

Number of branch circuit for lightening will be = 86.5 *1000 = 66.54 ~ 67 ckt. 1300

Because of the size of the building, three panels are required to keep branch circuits below 100 ft in length. Also we need to have additional circuits that are provided for receptacles and spares. Receptacles: For the first 400 sq ft 10 receptacles, and for the next 16,600ft2, taking 1 receptacle per 110sqft, there will be 151 receptacles. This will give a total of 161 receptacles in the building. If each branch circuit carries 20A, supplying for 6 receptacles,

No. of receptacle circuits will be = 161 = 27 ckts 6

The total no of circuits for lighting plus receptacles is 67 + 27 = 94 ckts Spares is = 20% of total circuits

94*20% = 94*0.2 = 18.8 ~ 19 Total 0f 113 ckts.

Each panel would then have 113/3 or 38 circuits plus 4 free spaces, for a maximum of 42 poles. Thus, with initially three panel locations we proceed to circuit the lighting and receptacles according to the actual tenant requirements.

NOTE In calculating the panel load for feeder sizing, the actual load as determined by adding the lighting and other loads on the panel. This load is compared to the load by square foot calculation. And the larger figure is used in determining the required panel feeder size. Thus in the above case, the actual load would be compared to 108 kW and the larger used. If the actual number of circuits is less than 67, then 67 ckt must be provided; but if greater, the actual number required must be used. In either case, 20% spare should be used. Assuming even distribution of load, and actual load greater than the minimum 67 ckt, panel load would be 38 ckts each rated @ 1300 W = 49.5 KW, and if 25% future expansion is anticipated that is 12.5 KW (note: it is for this power that the four free spaces are left).

So feeder load would become 62 KW The feeder current is calculated in terms of the panel 3-phase kva thus: I = Kilo Volt Ampere √3 *Power Factor *Mains Voltage * Efficiency Or I = Kilo Watt √3 *Mains Voltage * Efficiency If mains voltage is 380, PF 0.8, 100% efficiency

I = 62 KW = 94.2 amp √3 *380*1

Thus, the above feeder current is 94.2 amp.



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► Riser Diagrams

When all devices are circuited and panels are located and scheduled, we are ready to

prepare a riser diagram. A typical diagram, shown in Figure below, represents a block version of a single-line diagram, as the name implies, vertical relationships are shown. All panels, feeders, switches, switchboards, and major components are shown up to, but not including, branch circuiting. This diagram is an electrical version of a vertical section taken through the building.

The main switchboard shown in the above figure constitutes a combination of service

equipment and feeder switchboard. The service equipment portion of the board comprises the metering and the 4 main switches feeding risers, motor control center (MCC), roof, machine room, and elevators



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*NEC: - National Electrification Code. *EBCS: - Ethiopian Building Code Standard.

► Choosing Cable Size

● Cable Size Design procedure The correct choice of cable size for any installation is dependent upon fundamental aspects of

(a) Environmental conditions and characteristics of protection, (b) Current-carrying capacity of the cable and (c) Voltage drop of the cable.

When current flows through a conductor, the resistance offered by the conductor produces heat. The increase in heat is proportional to the cable resistance, which in turn depends upon the cross-sectional area of the cable. Since overheating damages the insulation, the conductor size must be of adequate size to prevent this from occurring. The requirements of IEE Regulations make it clear that circuits must be designed and the design data made readily available. How then can we begin to design? Clearly, plunging into calculations of cable size is of little value unless the type of cable and its method of installation is known. This in turn will depend on the installation’s environment. At the same time, we would need to know whether the supply was single or three phases, the type of earthing arrangements, and so on. Here then is our starring point. Having ascertained all the necessary details, we can decide on an installation method, the type of cable, and how we will protect against electric shock and over currents. We would now be ready to begin the calculation part of the design procedure. Basically, there are eight stages in such a procedure. These are the same whatever the type of installation, be it a lightening circuit, cooker circuit or a sub main cable feeding a distribution board in a factory. Here then are the eight basic steps in a simplified form:

1. Determine the design current Ib. 2. Select the rating of the protection In 3. Select the relevant correction factors (CFs). 4. Divide In by the relevant CFs to give cable current-carrying capacity 5. Choose a cable size to suit Iz 6. Check the voltage drop 7. Cheek for shock risk constraints 8. Cheek for thermal constraints.

Let us now examine each stage in detail. Design current

In many instances the design current Ib is quoted by the manufacturer, but there are times when it has been calculated. In this case there are two formulae involved, one for single phase and one for three phase:

Single phase:

Ib=P/V



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Three phase: Ib=P/(√3 V) If an item of equipment has a power factor and/or has an efficiency (eff) will have been taken into account. Hence: Single phase:

Ib = (Px1000)/(V*PF*eff) Three phase:

Ib = (Px1000)/ (√3*VL*PF*eff) Nominal setting of protection

Having determined Ib we must now select the nominal setting of the protection In such that In>Ib. this value may be taken from IEE regulations.

Correction factors

When a cable carries its full load current, it can become warm. This is not problem unless its temperature rises further due to other influences, in which case the insulation could be damaged by over heating. These other influences are:

high ambient temperature cable grouped together closely uncleared over currents and contact with thermal insulation.

For each of these conditions there is a correction factor (CF) which will respectively called Ca, Cg, Cf and Ci, and which derates cable current carrying capacity or conversely increases cable size. Ambient temperature Ca

The cable rating in the IEE regulations are on an ambient temperature of 300C, and hence it is only above this temperature that an adverse correction improvement is needed. Grouping Cg

When cables are grouped together they impart heat to each other. Therefore the more cables there are the more heat they will generate, thus increasing the temperature of each cable. IEE regulation also gives factors for such groupings of the same cable sizes. Protection by BS 3036 fuse Cf

Because of the high fusing factor of BS 3036 fuses, the rating of the fuse In, should be less than or equal to 0.725Iz Hence 0.725 is the correction factor to be used when BS 3036 fuses are used. Thermal Insulation Ci

With the modern trend, towards energy saving and the installation of thermal insulation, there may be a need to derate cables to account for heat retention. IEE Regulation gives these factors for situations when thermal insulation touches one side of a cable. However, if a cable is totally surrounded by thermal insulation for more than 0.5 m, a factor of 0.5 must be applied to the tabulated clipped direct ratings. For less than 0.5 m, derating factors Table __ should be applied. Refer to the table on pages __. Application of correction factors



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Some or all of the onerous conditions just outlined may affect a cable along its whole length or parts of it, but not all may affect it at the same time. If all conditions are to appear at the same time consider all correction factors, otherwise take the worst.

Having chosen the relevant correction factors, we now apply them to the nominal rating

of the protection In as divisors in order to calculate the current carrying capacity Iz of the cable. Current carrying capacity The required formula for current carrying capacity Iz is Iz= In/(relevant CFs) Choice of cable size

Having established the current carrying capacity Iz of the cable to be used, it now remains to choose a cable to suit that value. The IEE regulation also lists all the cable sizes, current carrying capacity and voltage drops of varies types of cables.(These data is given from page __ to page __ ).

Voltage drop

The resistance of a conductor increases as the length increases and/or the cross-sectional area decreases. Associated with an increased resistance is a drop in voltage, which means that a load at the end of a long thin cable will not have the full supply voltage available. The IEE regulation requires that the voltage drop Vd should not be so excessive that equipment does not function safely. They further indicate that a drop of no more than 4% of the nominal voltage at the origin of the circuit will satisfy. The voltage drop will be calculated using a formula (adopted by IEE regulation):

Vd = mV * Ib * L where mV- voltage drop in mV obtained from IEE table L- total length of the cable in consideration.

● Fundamental 3-phase Voltage-drop Calculations

These are all based on the basic formula R =ρl/A where ρ (rho) stands for resistivity. ρ = 1.72x10-8 – for copper conductor ρ = 2.83x10-8 – for Aluminum conductor If resistance of a conductor at any temperature different from room temperature is

required it can be calculated using RT= RO (1+ αΔT) Where RT- resistance at the required temperature RO =ρl/A resistance at room temperature α-Expansion coefficient = 0.00393 - for copper = 0.0039 - for Aluminum ΔT- Change in temperature Resistivity is defined as the resistance between two opposite faces of a unit cube of the

conductor material. Many voltage-drop problems involve the determination of resistance by this means and then multiplying by the current to obtain the IR drop. The weakness of this method, as against that adopted by use of the I.E.E. Tables, is that the Tables are much more realistic since they take into account the actual type of cable and conditions of service. ● Diversity Factor



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The diversity factor has an important place in the design of an installation and its final costing. IEE regulation 311-01 deals with this subject. Diversity factor is a factor which is applied to sub main and main cables and their associated gears to reduce:

a) the cross sectional area if the cable conductor, and b) the capacity of the switch gears.

The factor is based on the assumption that the whole of the connected load will not be turned on at the same time. For example, the total lighting load in a dwelling house is rarely switched on at a time. Thus, it can be taken that if the total lighting load is 1000W during the life of the installation, only 66% of the load (660W) will be switched on at any one time. The factor in this instance is 0.66. A factor for diversity shall not be allowed for calculating the size of circuit conductor and switchgears of final sub circuits, other than specified circuits such as cooker circuits. It is noted that the provision of an allowance for diversity is a matter of calling for a special knowledge and experience. Indeed, the application of the diversity should be decided by the engineer responsible for designing each particular installation. The amount by which they are increased or decreased for each installation is a matter for the installation engineer to decide. There are ten types of final circuit fed from wiring to which diversity applies:

lighting heating cooking appliances which are permanently connected motors (other than lifting motors) instantaneous-type water heater thermostatically controlled water heater floor-warming installation thermal-storage space-heating installation 13A fused socket outlets and appliance fed there from and Other socket outlets such as 15A sockets.

The general groups of installation premises are also recognized:

1) Individual domestic installation, including individual flats of a block. 2) Hotels, boarding house, lodging houses etc. 3) Shops, stores, offices and business premises

In the case of lighting for each type of installation, it will be noticed that the more the total lighting load is likely to switched on over definite periods, the smaller is the allowance made for diversity. In a domestic installation, it is estimated that some two- thirds (0.66) of the lighting load will be on at any one time. In a hotel, the figure is 75%(0.75), and in a shop, where virtually all the lights are on for most of the time when the shop is open, the figure is 90% (0.90). It should be noted that no diversity is allowable in the relevant wiring supplying certain types of load. Example 1 From EBCS-10. Table B.1, select cables of suitable current-carrying capacity for the following loads and conditions (p.v.c. cables to BS 6004 into screwed conduit).



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(a) 240 V single-phase sub-mains of lighting load totaling 10.5 kW. Length of run 10 m. Average ambient temperature 25'C, diversity 66% (b) 400 V balanced 3-phase power circuit. Load 18.65 kW, efficiency, 80%, power factor 0.69. Average temperature 30'C. Length of runs 100 m.

Solution of (a) Current taken by load = Power

Voltage = 10.5 * 1000 = 43.75 A

240 Allowing for diversity, maximum current through cables = 43.75 * 66 = 28.88 A

100 If BS 88 32-A circuit breaker is chosen for protection, 32 A rated circuit breaker can be

selected from table 9.1. → In = 32 Amp

The correction factor for ambient temperature from Table A.4 for250C is 1.06. Therefore the required cable rating: → Iz = 32 = 30.2 A 1.06 From Table B.1, choose a 4 mm2 conductor which carries 32A.

Testing for Voltage drop = (mV/Am) * I * l → From table B.2 voltage drop for 4mm2 conductor size = 11mv/Am

→ Voltage drop on cable = 11 mv/Am * 28.8 A * 10 m = 3.168 V → Maximum allowable voltage drop = 2.5% of 240 V = 6V.

Since the actual voltage drop is less than from the allowable maximum voltage drop, selected size is 4 mm2

If BS 3036 fuse is chosen for protection, this fuse type requires a correction factor of

0.725. → In = 32 Amp

Therefore the load current will be : → Iz = In/CF CF- Correction Factors. Ca = 1.06, Cf = 0.725 Required cable rating Iz = 32 Amp

1.06 * 0.725 = 41.64 Amp.

From Table B.1, a 6mm2 conductor carries 41 A. And a 10mm2 conductor carries 57 A. Take 10mm2 diameter conductor.

Testing for Voltage drop = (mV/Am) * I * l → From table B.2 voltage drop for 6mm2 conductor size = 7.3mv/Am

→ Voltage drop on cable = 4.4 mv/Am * 28.8 A * 10 m = 1.27 V → Maximum allowable voltage drop = 2.5% of 240 V = 6V.

Since the actual voltage drop is less than from the allowable maximum voltage

drop, selected size is 10 mm2. Comment: you can easily observe that the conductor size deference in using Circuit

breaker and fuses. Solution of (b)



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efficiency = Output Input

= kW * 1000 √3VI cosφ

80 = 18.65 * 1000 100 √3 * 400 * I * 0.69

I = 18.65 * 1 000 * 100 = 48.77 A.

√3 * 400 * 80 * 0.69

From table 9.1, 50 A circuit-breaker of type BS 3871 can used for protection. → In = 50 Amp

Load current will be : → Iz = In/CF CF- Correction Factors. CF = 1 because Ca = 1. → Iz = 50 A Choose 16 mm2 cable which is capable of carrying 52 A.

Testing for Voltage drop:

Maximum voltage drop = 2.5% of 400 V = 10V. Voltage drop on the cable = (mV/Am) * I * l

= 2.3 * 48.77 * 100 = 11.22 V this is beyond the allowable voltage drop. So, choose the next cable size, which is 25mm2. Voltage drop for 25 mm2 = 1.7 * 43.77 * 100 = 8.29 V

Therefore selected size is 25 mm2. Comment. This is one of the situations where the voltage drop becomes the main determining

Factor of the conductor size.

From table 9.1, 50 A BS 3036 fuse can used for protection. → In = 50 A

→ Correction factor for the fuse is Cf = 0.725 → Load current Iz = In / CF = In / Cf → Iz = 50 A / 0.725 = 68.966 A

From table B.3 select 25mm2 cable which carries 97 A

Testing for Voltage drop:

Voltage drop on the cable = (mV/Am) * I * l = 2.3 * 48.77 * 100

= 13.17 V this is beyond the allowable voltage drop. So, choose the next cable size, which is 25mm2. Voltage drop for 25 mm2 = 1.7 x 48.77 x 100 = 8.29 V

Therefore selected size is 25 mm2. Example 2

A 30 m run of twin and earth p.v.c. non-armored four touching copper cables are situated in an ambient temperature of 350C. Determine the minimum size of cable to supply a 220-V 10-kW load. Protection given by:

(a) Miniature circuit-breaker (m.c.b.)



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(b) Rewirable fuse.

Solution of (a) Ib = p v Ib = 10000 = 41.67 A. 240 Ib = 41.67 A. 50-Amp m.c.b. is adequate for protection of 45.67 Amp. In = 50 A. And from Table B.1 correction factor for 350C = 0.94 Ca = 0.94.

From table A.1,Correction factor for cables group together is 0.75 → Cg = 0.75

Required cable current rating Iz = 50 = 70.9 A 0.94 x 0.75

→ Iz = 70.9 A From Table B.3 (for multi core cable) 16mm2 cable carries 69A.

Testing for Voltage drop: Maximum voltage drop = 2.5% of 220 V = 5.5V Voltage drop on the cable = (mV/Am) * I * l

= (2.8 mV/A m) * 41.67 A * 30m = 3.50 V this is with in the allowable voltage drop.

So, choose cable size of 16mm2.

Solution of (b) Ib = 41.67 A.

Assume that a rewirable fuse type that requires a correction factor of 0.725 is used. So Cf = 0.725 → In = 50 A. Required cable rating Iz = 50 = 97.8 A

0.94 x 0.75 x 0.725 → Iz = 97.8 A

From Table B.3 (for multi core cable) 35 mm2 cable carries 111 Amp. Testing for Voltage drop:

Voltage drop on the cable = (mV/Am) * I * l = ( mV/A m) * 41.67 A * 30m

= V < 5.5V So, choose cable size of 35mm2.

Comment. The example exhibits once again the considerable economic savings which can be

gained by fitting an m.c.b. or correct cartridge fuse in place of the rewirable type. Example 3 A load of 300 kW at 0.78 power-factor is to be supplied at 415 V, 3-phase, through a 3-core copper cable 260 m long. The cross-sectional area of each cable core is 400 mm2.



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Calculate the voltage drop in the cable. Ignore voltage drop due to reactance. (Resistivity of copper may be taken as 0.017 μΩ-m). Three-phase power, P= √3VI cos φ By transposition I= P_____

√3VI cos φ = _ 300 W * 1000 = 535.2 A √3* 415V * 0.78

Resistance per core, R = ρl/A = 0.017Ω-m * 260 m * 10-6 = 0.01105Ω 10-6 * 400 m2

Therefore Voltage drop in the cable = √3*I* R = √3 * 535.2 * 0.01105 = 10.24 V

Comments. The important point to note is the voltage drop in a 3-core cable, when carrying the current in to a balanced 3-phase load, is given by √3*I* R. The cable conductors are presumed to act as a pure resistance without any reactive effects. Problems An apartment having 15 individual rooms each 3.5mX2.5m (see figure below) is to be installed with the following loads: Ventilator (500W), Fridge (1kW), Stove (2kW), six 60W



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lamps, and five socket outlets. a) Locate the approximate area of location of each lamps and sockets. b) Calculate the sub-feeder cable size for this room if the average ambient temperature

of the local area is taken as 35oC and five groups of circuits are running together in a conduit with it. The length of run is 15m and protection is by MCB.

c) Choose the rating of the MCB and the branch circuits breakers.

Fig xxxx Floor Plan for problem no. 6

Data for Electrical Installation



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handout ch3

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