Electrical Design 101.pdf

ELECTRICAL DESIGN 1 THIS COURSE DEALS WITH THE STUDY OF ELECTRICAL SYSTEM DESIGN, INSTALLATION AND COST ESTIMATION FOR SINGLE AND MULTIFAMILY DWELLING UNITS GUIDED BY THE PROVISIONS OF THE PHILIPPINE ELECTRICAL CODE (PEC) AND OTHER RELEVANT LAWS AND STNDARDS.

BASIC CONCEPT IN ELECTRICAL DESIGN

RECEPTACLES AND WIRING DEVICES ELECTRICAL CONDUCTORS AND INSULATORS

Electrical Conductors are substances that offer a very low resistance to current flow. Insulators are substances that offer a very high resistance to current flow. List of some good electrical conductors: Silver Zinc Copper Platinum Aluminum Iron Nickel Tin Brass Lead

List of some insulating materials: Rubber Asbestos Porcelain Thermoplastics Varnish Paper Slate Oils Glass Wax Mica

Dry air Latex

WIRES AND CABLES Wires are those electrical conductors which are 8 mm2 (AWG no. 8) or smaller, while cables are those larger than the wires. They are either solid or stranded. Stranded wire - consists of a group of wires twisted to form metallic string. The total circular-mil area of a stranded wire is found by multiplying the circular mil area of each strand by the total number of strand. Cord is the term given to an insulated stranded wire. CIRCULAR MIL. This is the unit of cross section in the American wire gauge. The term “mil” means one-thousandth of an inch (0.001 in.). It is the area of a circular wire having a diameter of one mil. To find the number of circular mils in a circle of a given diameter, we have to square the number of mils in the diameter. Area in circular mil = ( diameter in mils )2 1 inch = 1,000 mils MCM = 1,000 circular mils

SQUARE MIL. It is the area of a square having its side equal to 1 mil. Square mil = ( sides )2 = ( 1 mil )2 = ( 0.001 in.)2 = 1 x 10-6in.2 Square mil = 0.7854 x circular mils CONDUCTOR AREAS:

CONVERSION FACTOR Square inch = square mil x 0.000001 Square mil = square inch x 1,000,000

Square mil = circular mils x 0.7854 Circular mil = square mils x 1.273 Millimeter = inches x 25.4 Square mm = circular mils x 0.0005067

DIFFERENT TYPES OF CABLES 1. Armored Cable. This type of cable, the type AC is a fabricated assembly of insulated conductors enclosed in flexible metalsheath. Armored cable is used in both exposed and concealed work. 2. Metal Clad Cable. Cable of the type MC is a factory assembled cable of one or more conductors, each individually insulated and enclosed in a metallic sheath of interlocking tape, or a smooth or corrugated tube. This type is used specifically for services, feeders, branch circuits, either exposed or concealed and for indoor or outdoor work. 3. Mineral Insulated Cable. This type of cable, type MI, is a factory assembly of one or more conductors insulated with a highly compressed refractory mineral insulation and enclosed in liquid-tight and gas-tight continuous copper sheath. The type MI is used in dry, wet or continuously moist location as service, feeders or branch circuit. 4. Nonmetallic Sheathed Cable. Types NM and NMC are factory assembled two or more insulated conductors having a moisture-resistant outer sheath, flame-retardant and non-metallic material. These types are used specifically for one or two dwelling not exceeding 3 storey buildings. 5. Shielded Nonmetallic Sheathed Cable. This type of cable, the type SNM, is a factory assembly of two or more insulated conductors in an extruded core or moisture-resistant and flameretardant material, covered with an overlapping spiral metal tape. This type is used in hazardous locations and in cable trays or in raceways. 6. Service Entrance Cable. This is a single conductor or multiconductor assembly provided with or without an over-all covering, primarily used for services and of the types SE and USE. 7. Underground Feeder and Brach Circuit Cables. This type of cable, the type UF cable is a moisture-resistant cable used for underground, including direct burial in the ground, as feeder or branch circuit. 8. Power and Control Tray Cable. Type TC cable is a factory assembly of two or more insulated conductors with or without associated bare or covered grounding under a metallic sheath. This is used for installation in cable trays, raceways or where supported by a messenger wire. 9. Flat Cable Assemblies. This is an assembly of parallel conductors formed integrally with an insulating material web designed specifically for field installation in metal surface raceway. Cables of this type are the types FC. 10. Flat Conductor Cable. This type of cable, type FCC consists of three or more flat conductors placed edge to edge, separated and enclosed within an insulating assembly. This used for general purpose, appliance branch circuits and for individual branch circuits specifically on hard, smooth, continuous floor surfaces, etc.

12.Medium Voltage Cables. MV cable is a single or multiconductor solid dielectric insulated cable rated 2,001 volts or higher and is used for power systems up to 35,000 volts. The MV cables are of different types and characteristics.

RACEWAYS Raceways are channels designed for holding wires, cables or bus-bars, which are either made of metal or insulating materials. The common types of raceways in household wiring are the a) conduits, b) connectors, and c) others. a) Conduits Conduits, pipes or tubings are the most common electrical raceway. According to the type of materials used, conduit maybe classified as either metallic such as steel pipes or nonmetallic such as PVC, and the like. According to its make, conduits maybe classified as: rigid metal, flexible metal, rigid nonmetal and flexible nonmetal. b) Connectors A connector is a metal sleeve usually made of copper that is slipped over and secured to the butted ends of conductors in making joint. A connector is also called a splicing sleeve. c) Other Raceways Aside from the conduits and connectors there are still numerous types and kinds of raceways, among these are the a) conduit couplings, elbows and other fittings; b) conduit supports, such as clamps, hangers,etc; c) cable trays, cablebus; d) metal raceways;e) nonmetal raceways.

OUTLETS, RECEPTACLES and other WIRING DEVICES OUTLETS. An outlet is a point in the wiring system at which current is taken to supply utilization equipment. The kinds of outlets are: convenience outlet or attachment cap, lighting outlet, and receptacle outlet. A convenience outlet or attachment cap is a device which by insertion in a receptacle, establishes connection between the conductor of the flexible cord and the conductors connected permanently to the receptacle. A lighting outlet is an outlet intended for direct connection of a lampholder, a lighting fixture, or a pendant cord terminating in a lampholder.

A receptacle outlet is an outlet where one or more receptacles are installed.

TYPES OF WIRES A. TYPES T, TW, THW The most ordinary type of plastic insulated wire is the “type T”. It may be used only in dry locations. Some manufactures no longer make the ordinary Type T, instead produce Type TW, which is identical in appearance, but may be used in wet or dry locations. Also available is Type THW, is similar to Type TW but withstand a greater degree of heat, and consequently has a higher ampacity rating in the larger sizes. B. TYPES THHN, THWN These are comparatively new types of wire, consisting of the basic Type THH and THW but with less thermoplastic insulation, and with a final extruded jacket of nylon. Nylon has exceptional insulating qualities and great mechanical strength, all of which results in a wire which is smaller in diameter than ordinary Types T, TW, TW of corresponding size. C. TYPE XHHW In appearance, it resembles Types T, TW, THW but because of somewhat thinner layer of insulation, the over-all diameter is smaller. The insulation is “cross-linked synthetic polymer,” which has an extraordinary properties as to insulating value, heat resistance, and moisture resistance. It may be used in dry or wet locations. While at present, it is an expensive wire, it would be no surprise if in due course of time, this one single type will replace all the many types and subtypes of Type T or R now recognized by the Code. D. RUBBER-COVERED WIRE It consists of copper conductor, tinned to make it easier to remove the insulation, and for easy soldering. Over the copper is a layer of rubber, the thickness of which depends on the size of the wire. Then follows an outer fabric braid which is saturated with moisture-and-fire-resistant compounds; if it is set on fire with a blowtorch, the flame dies out when the torch is removed. E. OTHER TYPES Other types such as the basic Type R, which is suitable for only in dry locations, is no longer being made. The most ordinary kind is Type RHW, which may be used for dry or wet locations. Types RH and RHH have insulation which withstands more heat and therefore have a higher ampacity in the larger size. They may be used only in dry locations.

KINDS OF LOCATIONS DAMP LOCATION Partially protected locations under canopies, marquees, roofed open porches, and like locations, and interior locations subjected to moderate degree of moisture, such as some basements, some barns, and some cold-storage warehouses. DRY LOCATION A location not normally subject to dampness or wetness. A location classified as dry may be temporarily subject to dampness or wetness, as in the case of a building under construction. WET LOCATION Installations underground or in concrete slabs or masonry in direct contact with the earth, and location subject to saturation with water or other liquids, such as vehicle washing areas, and locations exposed to weather and unprotected. HAZARDOUS (CLASSIFIED) LOCATIONS Locations where fire or explosion hazards may exist due to flammable gases or vapors, flammable liquids, combustible dust, or ignitible fibers or flyings. 1. Class I Locations. Class I locations are those in which flammable gases or vapors are or may be present in the air in quantities sufficient to produce explosive or ignitible mixtures. a) Class I, Division 1. A Class I, Division 1 location is a location: I) in which igntible concentrations of flammable gases or vapors can exist under normal operating conditions; or ii) in which ignitible concentrations of such gas vapors may exist frequently because of repair or maintenance operations or because of leakage; or iii) in which breakdown or faulty operation of equipment or processes might release ignitible concentrations of flammable gases or vapors, and might also cause simultaneous failure of electric equipment. b) Class I, Division 2. A Class I, Division 2 location is a location: I) in which volatile flammable liquids or flammable gases are handled, processes, or used, but in which the liquids, vapors, or gases will normally be confines within closed containers or closed systems from which they can escape only in case of accidental rupture or breakdown of such containers or systems, or in case of abnormal operation of equipment; or ii) in which ignitible concentrations of gases or vapors are normally prevented by positive mechanical ventilation, and which might become hazardous through failure or abnormal operation of the ventilating equipment; iii) that is adjacent to Class I, Division 1 location, and to which ignitible concentrations of gases or vapors might occasionally be communicated unless such communication is prevented by adequate positive ventilation from a source of clean air, and effective safeguards against ventilation failure are provided.

Class II Locations. Class II locations are those that are hazardous because of the presence of combustible dust. a) Class II, Division 1. A class II, Division 1 location is a location: I) in which combustible dust is in the air normal operating conditions in quantities sufficient to produce explosive or ignitible mixtures; or ii) where mechanical failure or abnormal operation of machinery or equipment might cause such explosive or ignitible mixtures to be produced, and might also provide a source of ignition through simultaneous failure of electric equipment, operation devices, or from other causes; or iii) in which combustible dusts of an electrically conductive nature may be present in hazardous quantities. b) Class II, Division 2. A Class II, Division 2 location is a location where combustible dust is not normally in the air in quantities sufficient to produce explosive or ignitible mixtures, and dust accumulations are normally insufficient to interfere with the normal operation of electrical equipment or other apparatus, but combustible dust may be in suspension in the air as a result of infrequent malfunctioning of handling or processing equipment and where combustible dust accumulations on, in, or in the vicinity of the electrical equipment may be sufficient to interfere with the safe dissipation of heat from electrical equipment or may be ignitible by abnormal operation or failure of electrical equipment.

3. Class III Locations. Class III locations are those that are hazardous because of the presence of easily combustible fibers or flyings, but in which such fibers or flyings are not likely to be in suspension in the air in quantities sufficient to produce ignitible mixtures. a) Class III, Division 1. A Class III, Divisions 1 location is a location in which easily ignitible fibers or materials producing combustible flyings are handled, manufactured, or used.

b) Class III, Division 2. A Class III, Division 2 location is a location in which easily ignitible fibers are stored or handled.

ELECTRIC CIRCUITS IN BUILDING * SERVICES * No. of Service: A building or other structure served shall be supplied by only one service. *EXCEPTIONS* 1. For fire pump where a separate service is required. 2. For emergency electrical system where a separate service is required.

3. Multiple-Occupancy building 4. Capacity Requirements. Two or more services shall be permitted: a) Where the capacity requirements are in excess of 2,000 amperes at a supply voltage of 600 volts or less; or b) Where the load requirements of a single-phase installation are greater than the serving agency normally supplies through one service; or 5. Building of Large Area ( 10,000 m2 or more Total Area ). 6. For different voltage characteristics, such as for different voltage, frequencies, or phases, or for different uses, such as for different rate schedules.

THE OVERHEAD SERVICE-DROP CONDUCTOR This is the overhead service conductor from the last pole or other aerial support to and including the splices if any, connecting the service entrance conductors at the building or other structure. SIZE AND RATING: a) General. Service drop shall have sufficient ampacity to carry the load without a temperature rise detrimental to the covering or insulation of the conductors and shall have adequate mechanical strength. b) Minimum Size. The conductors shall not be smaller than 8 mm2 copper, 14 mm2 aluminum or copper-clad aluminum. CLEARANCES:

a) Above Roofs. Conductors shall have a vertical clearance of not less than 2,500 mm from the roof surface. b) Vertical Clearance from Ground. 3,100 mm - at the electric service entrance to buildings, or at the drip loop of the building electric entrance, or above areas or sidewalks 3,700 mm - for those areas listed in the 4,600 mm classification when the voltage is limited to 600 volts to ground. 4,600 mm - over residential property and driveways, and those commercial areas not subject to truck traffic.

5,500 mm - over public streets, alleys, roads, parking areas subject to truck traffic, driveways on other than residential property, and other land transversed by vehicles such as cultivated, grazing, forest, and orchard.

UNDERGROUND SERVICE-LATERAL CONDUCTOR This is the underground service conductor between the street main, including any risers at a pole or other structure or from transformers, and the first point of any connection to the serviceentrance conductors in a terminal box or meter or other enclosure with adequate space, inside or outside the building wall. INSULATION. Service-lateral conductor shall withstand exposure to atmospheric and other conditions of use without detrimental leakage of current. *EXCEPTIONS* A grounded conductor shall be permitted to be uninsulated as follows: a) Bare copper used in a raceway. b) Bare copper for direct burial where bare copper is judged to be suitable for the soil conditions. c) Bare copper for direct burial without regard to soil conditions where part of cable assembly identified for underground use. d) Aluminum or copper-clad aluminum without insulation or covering where part of a cable assembly identified for underground use in a raceway or for direct burial. SIZE AND RATING a) General. Service lateral conductors shall have sufficient ampacity to carry the current for the load and shall have adequate mechanical strength. b) Minimum Size. The conductors shall not be smaller than 5.5 mm2 copper or 8.0 mm2 aluminum or copper-clad aluminum. Where two to six service disconnecting means in separate enclosures supply separate loads from one service drop or lateral, one set of service entrance conductors shall be permitted to supply each or several such service equipment enclosures. EXCEPTION: For installations to supply only limited loads of a single branch circuit such as small polyphase power, controlled water heaters and the like, they shall not be smaller than 3.5 mm2 copper or 5.0 mm2 aluminum or copper-clad aluminum.

SERVICE ENTRANCE Service is defined as the portion of the supply which extends from the street main duct or transformer to the service switch or switchboard of the building supply. -it is the conductor and equipment for delivering energy from the electricity supply system to the wiring system of the premises served. TYPES: 1. Overhead Service Entrance The most common type of service entrance employed by the power companies supplying electricity which is either a 2, 3 or 4-wire connection. Generally, the overhead service cable between the building property line and the supply point is supplied by electric company to a limit of 30 meters. 2. The Underground Service Entrance The underground service entrance consists of a raceway conduit extending from the building to the property line where it is tapped to the main. The type of cable recommended is the underground service entrance cable commonly referred to as USE.

SERVICE - ENTRANCE CONDUCTORS No. of Service-Entrance Conductor Sets Each service drop or lateral shall supply only one set of service-entrance conductors. *EXCEPTIONS: 1. Buildings with more than one occupancy. 2. Where two to six service disconnecting means in a separate enclosures are grouped at one location and supply separate loads from one service drop or lateral. SIZE AND RATING: Service entrance conductors shall be of sufficient size to carry the computed loads. Ungrounded conductors shall not be smaller than: 1. 100 A ---- For one family dwelling with six or more 2-wire branch circuits.

2. 60 A ---- For one family dwelling with an initial computed load of 10 kVA above. 3. 40 A ---- For other loads. EXCEPTIONS: 1. For loads consisting of not more than 2 - wire branch circuits, 5.5 mm2 copper or 8.0 mm2 aluminum or copper-clad aluminum. 2. By special permission, for loads limited by demand or by the source of supply, 5.5 mm2 copper or 8.0 mm2 aluminum or copper-clad aluminum. 3. For limited loads of single branch circuit, 3.5 mm2 copper or 5.5 mm2 aluminum or copperclad aluminum. INSTALLATION OF SERVICE CONDUCTORS Service entrance conductors shall be installed in accordance with the applicable requirements of this Code covering the type of wiring method used and limited to the following methods: 1. Open-wiring on insulators 2. Rigid Metal Conduit (RMC) 3. Intermediate Metallic Tubing (IMT) 4. Electrical Metallic Tubing (EMT) 5. Service-Entrance Cables 6. Wireways 7. Busways 8. Auxiliary gutters 9. Rigid Non-Metallic Conduit (RNMC) 10. Cable Bus 11. Mineral-Insulated Metal-Sheated Cable 12. Type MC Cables

PROTECTION: Service entrance conductors subjected to physical damage shall be protected in any of the following ways or methods: 1. By RMC 2. By IMC 3. By RNMC suitable for the location 4. By EMT 5. Type MC cable or other approved means

THE SERVICE EQUIPMENT-DISCONNECTING MEANS GENERAL: The service-disconnecting means shall be provided to disconnect all conductors in a building or other structures from the service-entrance conductor. NUMBER OF DISCONNECTING MEANS: The service disconnecting means for each set or each subset of service entrance conductor shall consist of not more than six switches or six circuit breakers mounted in a single enclosure, or in a switchboard. LOCATION: The service disconnecting means shall be installed either inside or outside the building or other structure at a readily accessible location nearest the point of entrance of the service entrance conductor RATING: The service disconnecting means shall have a rating of not less than the load to be carried. In no case shall the rating be lower than specified through: 1. One circuit installation -- The service disconnecting means shall have a rating of not less than 15 amperes.

2. Two circuit installation -- The service disconnecting means shall have a rating of not less than 30 amperes. 3. One family dwelling -- The service disconnecting means shall have a rating of: 60 A -- where the initial computed loads is 10 kVA or more 100 A -- where the initial installations consist of six or more 2-wire branch circuit. 4. Others -- For all other installations, the service disconnecting means shall have a rating of not less than 40 amperes. NOTES: The service disconnecting means shall simultaneously disconnect all ungrounded conductors and shall be capable of being closed on a fault equal to or greater than the maximum available shortcircuit current. Service entrance conductor shall have a short-circuit protective device in each ungrounded conductors. Fuses shall have an Interrupting Rating no less than the maximum available short circuit current in the circuit at their supply terminals. Circuit breakers shall be free to open in case the circuit is closed on an overload. Circuit breakers shall have an interrupting rating not less than the maximum available short-circuit current at its supply terminals.

THE CIRCUIT BREAKER AND THE FUSE A circuit breaker is an overcurrent protective device also designed to function as a switch. It is equipped with an automatic tripping device to protect the branch circuit from overload and ground fault. A fuse is also an overcurrent protective device with a circuit opening fusible element which opens when there is an overcurrent in the circuit. It is considered as the simplest and the most common circuit protective device used into the house wiring connection. Advantages of circuit breaker over a fuse 1. The circuit breaker acts as a switch aside from its being an overcurrent device.

2. When there is an overcurrent, the circuit breaker trips automatically and after correcting the fault, it is ready to be switched on again, unlike the fuse which has to be discarded and replaced after it is busted. Advantages of fuse over a circuit breaker 1. One of its major advantage is its reliability and stability. It can stay on its position for years and act when called on to act as designed, unlike the circuit breaker which requires proper maintenance and periodic testing to keep it into a tip-top condition. 2. The cost of a fuse is less than that of a circuit breaker. Standard Ampere Ratings of Fuses and Inverse time circuit breakers 15, 20, 25, 30, 40, 45, 60, 70, 80, 90, 100, 110, 125, 150, 200, 225, 250, 300, 350, 400, 450, 500, 600, 700, 800, 1000, 1200, 1600, 2000, 2500, 3000, 4000, 5000 and 6000 Fuses, circuit breakers or combinations shall not be connected in parallel. Exception: Circuit breakers or fuses, factory assembled in parallel, and approved as a unit.

Position of Knife Switches a) Single-throw Knife Switches. Single-throw knife switches shall be so placed that gravity will not tend to close them. Single-throw knife switches, approved for use in the inverted position, shall be provided with a locking device that will ensure that the blades remain in the open position when so set. b) Double-throw Knife Switches. Double-throw knife switches shall be permitted to be mounted so that the throw will be either vertical or horizontal. Where the throw is vertical, a locking device shall be provided to hold the blades in the open position when so set.

FEEDERS AND MAIN Essential considerations being adapted or followed. 1. On large installation, one feeder is provided for each floor. 2. In small installations, one or two feeders is satisfactory. 3. Feeder for motor must be separate and independent from the light circuits. 4. Feeders requiring more than 50 mm diameter conduit should not be used.

5.Feeders should be subdivided if there are several bends or offsets because a 50 mm conduit is the largest that could be economically used. 6. Feeders radiating from the distributing panel should be provided each with a properly rated switch and circuit breaker. 7. Good practice dictates that feeders and main shall be installed inside a conduit pipe as it carries high voltage that should be well protected.

GROUNDING PROTECTION A ground is an electrical connection which may either be intentional or accidental between an electric circuit or equipment and the earth, or to some conducting body that serves in place of the earth. The purpose of grounding a circuit is to fix permanently a zero voltage point in the system. The grounded line of a circuit should not be broken nor fused to maintain a solid and uninterrupted connection to the ground. Grounding could be accomplished in the following manner: 1. Connection to a buried cold water main. 2. Connection to a rod or group of rods. 3. Connection to a buried ground plate.

THE PANELBOARD A panelboard is a single panel or group of panel units designed for assembly in the form of a single panel. This includes buses, automatic overcurrent protective devices, and with or without switches for the control of light, heat or power circuit. It is designed to be placed in a cabinet or cutout box placed in or against a wall or partition and accessible only from the front. Principles applied in installing panel board 1. The approach should be accessible and convenient. 2. The panelboard must be centrally located to shorten the home wiring runs. 3. It must be installed near the load center. As in most cases, panelboard is installed near the kitchen and the laundry where heavy loads are expected. MAIN- is the feeder interior wiring extending from service switch, generator bus, or converter bus to the main distribution.

BRANCH CIRCUIT- is defined as the circuit conductors between the final overcurrent device protecting the circuit and the outlets. This means that the branch circuit is only the wiring between the circuit overcurrent protection device such as fuses or circuit breaker and the outlets. However, it is a common knowledge and practice that the branch circuit comprises the entire circuit including the outlet receptacles and other wiring devices. PROTECTION OF THE BRANCH CIRCUIT Any current in excess of the rated current capacity of the equipment or the rated ampacity of the conductor is calledovercurrent. The causes of overcurrent are: 1. Overload in the equipment conductors. 2. Short circuit or ground fault As per PEC requirement, conductors shall be protected against overcurrent in accordance with their ampacities (Art. 4.5.1.3) Ampacity - is the current-carrying capacity of an electric conductor. CIRCUITRY DESIGN

Circuitry design varies according to the number of designers. However, good circuitry design is based on the following considerations: flexibility of the circuit  It means that the installation can accommodate all probable pattern arrangements and location of the loads for expansion, or future development  · reliability and efficiency of service  It means to have a continuous service and supply of power that are all dependent on the wiring system.  Reliability of electric power in a facility is determined by two factors:  o utility service  o building electric system  · safety of the circuitry  SAFETY means that independent service can be used in lieu of emergency equipment as backup for normal services. For reliability of the circuitry, the following principles should be considered: 

o to provide double emergency power equipment at selected weak points in the system  o that the electrical service and the building distribution system must act together so that the power can reach the desired point of service  o critical loads within the best way to serve them by providing a reliable power either from the outside source, or by standby power package for them  o the system design must readily detect any equipment failure and to be corrected automatically,  · economy as to cost  ECONOMY refers to the initial cost as well as the operating costs. These two cost-factors stand in inverse relationship to one another. OVER DESIGN is as bad as under design. It is wasteful both on initial and operating costs.  The effect of acquiring low cost equipment:  o high energy cost  o higher maintenance cost  o shorter life  · energy consideration  It is a complex one considering the following factors:  o energy laws and codes  o budget  o energy conservation technique  o energy control  · space allocation  It must consider the following:  o easy maintenance  o ventilation  o expandability  o centrality 

o limitation of access 

BRANCH CIRCUIT The branch circuit is classified into: General purpose branch circuit

It supplies outlets for lighting and appliances, including convenience receptacles   

Appliance branch circuit It supplies outlets intended for feeding appliances. 

Individual branch circuit It is designed to supply a single specific item. 

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CIRCUITING GUIDELINES: There are many ways of doing circuitry but there is no optimum or perfect way of doing it. However, there are certain rules and guidelines promulgated by the NEC for flexibility, economical, and convenient way of installing a circuitry. 1. The Code requires sufficient circuitry to supply residential load of 30 watts per square meter in buildings excluding porches, garages, and basements. 2. The requirement of 30 watts per square meter is up to 80 sq. m for a 20 amperes circuit (2400 watts) or 60 sq. m for 15 ampere circuit (1800 circuit). 3. Good practice suggests that the load should not exceed 1600 watts for a 20 amperes circuit and 1200 watts for a 15 amperes circuit. a. Observe a minimum load of 1200 watts on a 15 amperes circuit with a maximum area of 40 sq. m b. A maximum load of 1600 watts on a 20 amperes circuit with a maximum area of 53 sq. m

4. The Code requires a minimum of 20 amperes application branch circuit to feed all small appliance outlets in the kitchen, pantry, dining, and family room 5. The general purpose branch circuit shall be rated at 20 amperes circuit, wired with No. 12 AWG being the minimum size of conductor wire required for all convenience outlets. 6. Plug outlets or convenience receptacles shall be counted in computing the load if it is not included in the load for general lighting circuit. To find the number of outlets for 9 and 12 A loading on a 20 A circuit respectively, we have:

a. For 15 A ckt : 91.5 = 6 outlets b. For 20 A ckt : 121.5 = 8 outlets 7. Convenience receptacles should be planned properly, so that in case of failure by any one of the circuitry, the entire area will not be deprived of power supply. 8. All kitchen outlets should be fed from at least two of these circuits 9. The Code further stipulated that: “all receptacles are potential appliance outlet and at least two circuits shall be supplied to serve them.” 10. Certain outlets in the room should be designed as appliance outlet like: a. All kitchen receptacles b. Dining room receptacles c. One in the living room 11. The Code requires that, “at least one 20 A ckt. Supply the laundry outlets.” 12. If air conditioner is anticipated, provide a separate circuit for this particular appliance

OTHER GOOD PRACTICES IN CIRCUITING 1. Lighting and receptacles should not be combined in a single circuit. 2. Avoid connecting all building lights on a single circuit 3. Lighting and receptacles should be supplied with current from at least 2 ckt. so that if a single line is out, the entire area is not deprived of power. 4. Do not allow combination switch and receptacle outlets 5. Provide at least one receptacle in the bathroom, and one outside the house. Both must be Ground Fault Circuit Interrupter (GFCI) type 6. Provide switch control for closet lights. Pull chain switch is nuisance 7. Convenience outlet though counted as part of the general lighting load shall be limited to 6 convenience outlets on a 15 A ckt and 8 convenience outlet on a 20 A ckt. 8. The Code requires that, at least one 20 A ckt supply shall be installed to the laundry outlets 9. Convenience outlet shall be laid out in such a manner that no point on a wall is more than 2 meters from an outlet. Use a grounding type receptacle only

PEC REQUIREMENTS FOR ADEQUATE WIRING IN SINGLE AND MULTIFAMILY DWELLING UNIT

GENERAL LIGHTING LOADS BY OCCUPANCIES (Table 1.1)

* All receptacle outlets of 20-ampere or less in one-family, two-family and multifamily dwellings and in guest rooms of hotels and motels shall be considered as outlets for general illumination, and no additional load calculations shall be required for such outlets. ** In addition a unit load of 8 volt-amperes per square meter shall be included for general purpose receptacle outlets when the actual number of general purpose receptacle outlets is unknown.

FEEDER DEMAND FACTORS FOR GENERAL LIGHTING LOAD AND SMALL APPLIANCE LOAD (Table 2.1)

*The demand factors of this table shall not apply to the computed load of feeders to areas in hospitals, hotels, and motels where the entire lighting is likely to be used at one time, as in operating rooms, ballrooms, or dining rooms.

DEMAND FACTOR FOR HOUSEHOLD ELECTRIC CLOTHES DRYER (Table 3.1)

DEMAND LOADS FOR HOUSEHOLD ELECTRICRANGES, WALL-MOUNTED OVENS, COUNTER-MOUNTED COOKING UNITS, AND OTHER HOUSEHOLD COOKING APPLIANCES OVER 1.75 KW RATING. COLUMN A TO BE USED IN ALL CASES EXCEPT AS OTHERWISE PERMITTED ON NOTE 3 BELOW (Table 4.1)

Note: Over 12 kW through 27 kW ranges all of same rating. For ranges individually rated more than 12 kW but not more than 27 kW, the maximum demand in Column A shall be increased 5 % for each additional kW of rating of major fraction thereof by which the rating of individual ranges exceeds 12 kW. Note 2: Over 8.75 kW through 27 kW ranges of unequal ratings. For ranges individually rated more than 8.75 kW and of different ratings but no exceeding 27 kW, an average of value of rating shall be computed by adding together the ratings of all ranges to obtain the total connected load (using 12 kW for any range rated less than 12 kW) and dividing by the total number of

ranges; and then the maximum demand in column A shall be increased 5 percent for each kW or major fraction thereof by which this average value exceeds 12 kW. Note 3: Over 1.75 kW through 8.75 kW. In lieu of the method provided in column A, it shall be permissible to add the nameplate ratings of all ranges rated more than 1.75 kW but not more than 8.75 kW and multiply the sum by the demand factors specified in column B or C for the given numbers of appliances. Note 4: Branch circuit load. It shall be permissible to compute the branch-circuit load for one range In accordance with Table 3.3.2.10. the branch-circuit load for one wall-mounted oven or one counter-mounted cooking unit shall be the nameplate rating of the appliance. The branchcircuit load for a counter-mounted cooking unit and not more than two wall-mounted ovens, all supplied from a single branch circuit and located in the same room, shall be computed by adding the nameplate ratings of the individual appliances and treating this total as equivalent to one range. Note 5: This table also applies to household cooking appliances rated over 1.75 kW and used in instructional programs.

FULL LOAD CURRENT IN AMPERES SINGLE PHASE ALTERNATING - CURRENT MOTORS (Table 4.1)

FULL-LOAD CURRENT TWO-PHASE ALTERNATING CURRENT MOTORS (4-WIRE) (Table 5.1)

FULL-LOAD CURRENT THREE-PHASE ALTERNATING-CURRENT MOTORS (Table 6.1)

CONVERSION TABLE OF LOCKED-ROTOR CURRENTS FOR SELECTION OF DISCONNECTING MEANS AND CONTROLLERS AS DETERMINED FROM HORSEPOWER AND VOLTAGE RATING (Table 7.1)

MAXIMUM RATING OR SETTING OF MOTOR BRANCH-CIRCUIT SHORT-CIRCUIT AND GROUND-FAULT PROTECTIVE DEVICES (Table 8.1)

ALLOWABLE AMPACITIES OF INSULATED CONDUCTORS RATED 0-2000 VOLTS, 60º C TO 90ºC NOT MORE THAN THREE CONDUCTORS IN RACEWAY OR CABLE OR EARTH (DIRECTLY BURIED), BASED ON AMBIENT TEMPERATURE OF 30ºC (Table 9.1)

AMPACITY CORRECTION FACTORS (Table 10.1)

ALLOWABLE AMPACITIES OF SINGLE INSULATED CONDUCTORS, RATED 0-2000VOLTS, IN FREE AIR

BASED ON AMBIENT AIR TEMPERATURE OF 30ºC (Table 11.1)

AMPACITY CORRECTION FACTORS (Table 12.1)

+ Unless otherwise specifically permitted elsewhere in this Code, the over current protection for conductor types marked with an obelisk (+) shall not exceeds 15 amperes for 2.0 mm2,20 amperes for 3.5 mm2, and 30 amperes for 5.5 mm2 copper; or 15 amperes for 3.5 mm and 25 amperes for 5.5 mm2aluminum and copper clad aluminum. ALLOWABLE AMPACITIES OF THREE SINGLE INSULATED CONDUCTORS, RATED 0-2000VOLTS, 150º TO 250ºC, IN RACEWAY OR CABLE BASED ON AMBIENT AIR TEMPERATURE OF 40ºC (Table 13.1)

AMPACITY CORRECTION FACTORS (Table 14.1)

ALLOWABLE AMPACITIES FOR SINGLE INSULATED CONDUCTORS, RATED 0-2000VOLTS, 150º TO 250ºC, IN FREE AIR BASED ON AMBIENT AIR TEMPERATURE OF 40ºC (Table 15.1)

AMPACITY CORRECTION FACTORS (Table 16.1)

EXCEPTIONS (based on PEC requirements) 1. The small appliance appliance branch circuit required in a dwelling unit shall supply only the receptacle outlets specified in that section. (b.) 25- and 30-Ampere Branch Circuits. A 25- or 30-ampere branch circuit shall be permitted to supply fixed lightning units with heavy-duty lamp holders in other dwelling unit(s) or appliances shall not exceed 80 percent of the branch-circuit ampere rating. (c.) 40- and 50-Ampere Branch Circuits. A 40- and 50-ampere branch circuit shall be permitted to supply fixed lighting units with heavy-duty lamp holders or infrared heating units in other than dwelling units or cooking appliances that are fastened in place in any occupancy. Receptacle Outlets required (a) General. where flexible cords are used. 2. where flexible cords are specifically permitted to be permanently connected, and are so connected in boxes or fittings approved for the purpose, it shall be acceptable to omit receptacles on such equipment. (b) Dwelling units. In every kitchen, family room, dining room, breakfast room, living room, parlor, library, den, sun room, bedroom, recreation room, or similar rooms, receptacle outlets shall be installed so that no point along the floor kine in any wall space is more than 1800 mm, measured horizontally, from an outlet in that space, including any wall space 600 mm or more in width and the wall space occupied by sliding panels in exterior walls. The wall space afforded by fixed room dividers, such a free-standing bar type counters, shall be included in the 2 meter measurement. In kitchen and dining areas, a receptacle outlet shall be installed at each counter space wider than 300 mm. Counter top spaces separated by range tops, refrigerators, or sinks shall be considered as separate counter top spaces. Receptacles rendered inaccessible by appliances fastened in place or appliances occpying dedicated space shall not be considered as these requires outlets. Receptacles outlets shall, in so far as practicable, be spaced equal distances apart. Receptacle outlets in floors shall not be counted as part of the required number of receptacle outlet unless located close to the wall.

At least one wall receptacle shall be installed in the bathroom adjacent in the basin location. For a one-family dwelling, at least one receptacle outlet shall be installed outdoors. For a one-family dwelling, at least one receptacle outlet in addition to any provided for laundry equipment, shall be installed in each basement and in each attached garage. Outlets in other section of the dwelling unit for special appliances, such as laundry equipment, shall be placed within 1800 mm of the intended location of the appliance. At least one receptacle outlet shall be installed for the laundry. 3. In a dwelling unit that is an apartment or living area in a multifamily dwelling where laundry facilities are provided on the premises that are available to all building occupancies, a laundry receptacle shall not be required. 4.In other than one-family dwellings where laundry facilities are not be installed or permitted, a laundry receptacle shall not be required. As used in this section, a "wall space" shall be considered a wall unbroken along the floor line by doorways, fireplaces, and similar openings. Each wall space 600 or more mm wide shall be treated individually and separately from the other wall spaces within the room. A wall space shall be permitted to include two or more walls or a room(around corners) where unbroken at the floor line. The receptacle outlets required by this section shall be in addition to any receptacle that is part of any lighting fixture or appliances, located within cabinet or cupboard, or located over 1600 mm above the floor. 5. Permanently installed electric baseboard heaters equipped with factory installed receptacle outlets, or outlets provided as a separate assembly by the manufacture, shall be permitted as the required outlet or outlets for the wall space utilized by such permanently installed heaters. Such receptacle outlets shall not be connected to the heater circuits. Lighting Outlets Required: (a) Dwelling units: At least one wall switch controlled lighting outlet shall be installed in every habitable room, in bathrooms, hallways, stairways and attached garage: and at outdoor entrances At least one lighting outlet shall be installed in an attic, underfloor space, utility room and basement only wher these spaces are used for storage or containing equipment requiring servicing. 6. In habitable rooms, other than the kitchen, one or more receptacles controlled by a wall switch shall be permitted in lieu of light outlets 7. In hallways, stairways, and at outdoor entrances remote, central, or automatic control of lighting shall be permitted. FEEDERS: Minimum size or rating. Feeder conductors shall have an ampacity not lower than required to supply the load . The minimum sizes shall be as specified in (a) and (b) below under the conditions stipulated. Feeder conductors for a one family dwelling or amobile home need not be larger than service entrance conductors. (a) For specified circuits. The feeder conductors shall not be smaller than 5.5 square mm where the load

supplied consists of the following number and types of circuits: (1) Two or more 2-wire branch circuits supplied by a 2-wire feeder. (2) MOre than two 2-wire branch circuits supplied by a 3-wire feeder (3) Two or more 3-wire branch circuits supplied by a 3-wire feeder (b) ampacity relative to service entrance conductors. The feeder conductor ampacity shall not be lower than that of the service-entrance conductors 14 square mm or smaller. (c) Overload feeders. Where at any time feeder conductors are or will be overloaded the feeder conductors shall be increased in ampacity to accommodate the ctual load served. OPTIONAL CALCULATION - MULTIFAMILY DWELLING (a) It shall be permissible to compute the feeder or service load of a multifamily dwelling where all the following conditions are met: (1) No dwelling unit is supplied by more than one feeder (2) Each dwelling unit is equipped with electric cooking equipment. EXCEPTION: When the computed load for multifamily dwelling under the section without cooking load exceeds that computed under this section for the identical load plus electric cooking exceeds that computed for the identical load plus the electric cooking, the lesser of the loads may be used. (3) Each dwelling unit is equipped with either electric space heating or air conditioning or both Feeders and service-entrance conductors whose demand load is determined by this optional calculation shall be permitted to have the neutral load determined by this section the connected load to which the demand factors apply shall include the following: (1)1500 watts for each 2-wire, 20 A small appliance branch circuit and each laundry branch circuit (2) 24 watts per square meter for general lighting and general use receptacles (3) The nameplate rating of all appliances that are fastened in place, permanently connected or located to be on a specific circuit, ranges, wall mounted ovens, counter-mounted cooking units, clothes dryer, water heaters and space heaters If water heater elements are so interlocked that all elements cannot be used at the same time, the maximum possible load shall be considered the nameplate load. (4)The nameplate A or kVA rating of all motors and of all low-power factor load. (5) The larger of the air conditioning load or the space heating load.

WIRING CALCULATIONS FOR SINGLE FAMILY DWELLING UNIT

SINGLE FAMILY DWELLING Type of Service - 230 V Single Phase 2 wire 60 Hz Line to ground Current system

PROBLEM: A single family dwelling is to be circuited with the following requirements as shown on the figure above. Determine the: size of the branch circuit wire for lighting outlets size of the conduit pipes size or rating of the fuse protective device

SOLUTION A. Circuit - 1 for lighting load 1. From the Figure above, determine the number of lighting outlets. By direct counting, there are 8 light outlets. The National Electrical Code provides that: "100 watts shall be the maximum load for each household lighting outlet." Adopting the 100 watts per lighting outlet we have: 8 outlets x 100 = 800 watts 2. Determine the Total Current load 800 watts/230 volts = 3.48 amperes 3. Determine the size of conductor wire for circuit - 1 . Refer to Table 9.1 or 11.1. use 2 pieces 2.0 mm2 or No. 14 TW copper wirehaving an ampacity of 15 amperes that is much largr than the 3.48 amperes computed maximum load. 4. Determine the size of the Conduit Pipe. Refer to Table . The smallest diameter of a conduit pipe that could accommodate up to 3 pieces of No.14 TW conductor wire is 13 mm diameter. therefore, specify13 mm diameter conduit pipe. 5. determine the size or rating of the fuse protective device. Refer to Table . Use 15 amperes fuse The National Electrical Code provides that: "Ampacity of the connected load shall not exceed 80% of the amperage capacity of the conductor and the fuse." In Table , the maximum ampacity load of a or 2.0 mm2 No.14 AWG copper wire is 15 amperes. 80 % of 15 is 12, the maximum allowable load of the circuit sufficient enough to carry the 3.48 amperes computed load for a maximum 100 watts per light outlet. therefore the use of 2.0mm2 or No.14 TW is safe.

B, Circuit - 2 For Small Appliance load

SOLUTION The National Electrical Code provides that: "For each single receptacle shall be considered at no less than 180 watts rating." It simply mean that, each convenience outlet, is considered to have a maximum load of not less than 180 watts per plug or gang. thus: 1. From Figure above, there are 6 convenience receptacles for small appliance load. considering that there are two plug outlet, the total number of plug will be: 6 outlets x 2 plug = 12 pieces 2. Solve for the Total Estimated Load 12 x 180 watts per outlet = 2,160 watts 3. Determine the Maximum Expected Current Load: 2,160 watts / 230 volts = 9.39 amperes 4. Determine the Size of the Conductor wire . Refer to Table 9.1 or 11.1. For 9.39 amperes, use 2 pieces 3.5 mm2 or No. 12 TW copper wire for Circuit no.2 5. Determine the Size of the conduit pipe. for the 2 - No. 12 TW wire, refer to Table . Use 13 mm conduit pipe. 6. Determine the Over Current fuse protection. Refer to Table . under the column of fuse and breaker rating, the 20 amperes fuse can safely carry a maximum load of 16 amperes the 80 % of 20 amperes fuse can safely carry a maximum load of 16 amperes the 80% of 20 amperes load permitted by the National Electrical Code on No. 12 circuit wire. COMMENT: 1. On convenience outlet receptacle, the National Electrical Code provides that, " Each single receptacle shall be considered at no less than 180 watts rating." 2. Examining the values given on Table , the 2.0 mm2 or No. 14 AWG TW copper wire has an allowable ampacity rating of 15 amperes. Granting that only 80 % of this 15 amperes is considered the derated value, still 12 amperes is very much larger than the 9.36 amperes computed as maximum load for 6-convenience outlet. Why specify a bigger 3.5 mm2or No. 12 AWG conductor wire? 3. Although the 2.0 mm2or No. 14 AWG wire conductor could safely carry the 9.36 amperes computed load, considering its 15 amperes ampacity rating,yet, we cannot do so because the Code specifically mandated the use of 3.5 mm2or No. 12 AWG copper wire as the minimum size for all types of convenience outlet wiringexcept, for an appliance with limited load wherein a 2.0 mm2or No.14 AWG wire is permitted. C. Circuit - 3 for other loads SOLUTION

1. Examining Figure above, other loads are: 1 - unit electric stove at 1.1 kw = 1,1oo watts 1 - unit water heater at 2.5 kw = 2,500 watts Total Load..........3,600 watts 2. Compute for the current load. Divide: 3,600 watts / 230 volts = 15.65 amperes 3. Determine the size of the service conductor wire. Refer to Table 9.1 or 11.1. 4. For the 15.65 amperes load, use 2 pieces 3.5mm2or No.12 AWG TW copper wire 5. Determine the Size of the Conduit Pipe (if reqd) to Table. Two pieces No. 12 AWG wire can be accommodated comfortably in a 13 mm diameter conduit pipe. Specify 13 mm diameter conduit pipe. 6. Determine the size or rating of the Over-current Protection. Refer to Table . For the 15.65 amperes load use 20 amperes fuse rating. COMMENTS: The fuse rating is 20 amperes. Granting that it will be derated at 80 % x 20, the 16 amperes derated value is still higher than the computed load of 15.65 amperes. Therefore, the 20 amperes fuse over current protection is accepted. FINDING THE SIZE OF SERVICE ENTRANCE The size of service entrance being the supply conductor and equipment for delivering energy from the electricity supply to the wiring system of the building, is also computed based on the total load supplied by the branch circuit. Continuing the solution, we have the following: 1. Solve for the total load of circuit 1 to circuit 3. Total current load = Total connected load / voltage rating = (800 W + 2,160 W + 3,600 W) / 230 V = 28.52 amperes 2. Apply 80% demand factor as permitted by the National Electrical Code. 28.52 x .80 = 22.8 amperes 3. Find the Size of the Service Wire. Refer 22.8 amperes to Table . Use 2- 8.0 mm2or No.8 TW copper wire 4. Determine the size of conduit pipe for the service wire. Refer Table , for No.8 TW copper wire, use 20 mm diameter conduit pipe. COMMENT: 1. A demand factor of 80% was applied considering that not all receptacles and outlets are being used simultaneously. 2. These type of loads are classified as non-continuous load. From Table 9.1, the 5.5 mm2or No.10 AWG copper wire conductor has 30 amperes ampacity which is bigger than 22.8 amperes as

computed. However, we do not specify the use of No.10 AWG wire because the code limits the use of 8.0mm2or No.8 AWG, conductor as minimum size for Service Entrance. 3. The National Electrical Code on Service Entrance provides that: " Service entrance shall have sufficient ampacity to carry the building load. They shall have the adequate mechanical strength and shall not be smaller than 8.0 mm2or 3.2 mm diameter except for installation to supply limited load of a single branch circuit such as small poly-phase power, controlled water heaters and the like and they shall not be smaller than 3.5 mm2or 2.0 mm diameter copper or equivalent. THE MAIN DISCONNECTING MEANS OR SAFETY SWITCH Find the total computed load Circuit - 1 ........ 3.48 amperes Circuit - 2 ........9.39 amperes Circuit - 3 ........15.65 amperes TOTAL........28.52 amperes 2. Use 2 pieces 30 amperes fuse parallel connection 60 amperes 2 pole single throw (PST) 250 volts safety switch 3. Provide 2-double branch circuit cut out with two 15 and 2-20 amperes fuse respectively.

MULTI-ground system and line to line service The protection of branch circuit is tapped to the hot line of live wire. The grounded line being in neutral zero voltage is not protected with fuse. this is one advantage of the MULTI-GROUND SYSTEM being adopted by the electric cooperative implemented by the RURAL ELECTRIFICATION PROGRAM of the government. The branch circuit and cutout should be doubled because the engaged voltage in the line is only 230 V while the other is zero

being grounded ( see figure)

Other electric service system on the other hand, are classified as LINE TO LINE SERVICE wherein the engaged voltage is 115/230 volts which requires FUSE PROTECTIO FOR BOTH LINES. NOTE: The quantity of materials is subject to change depending upon the area and the choice of the designing engineers. For open onstallation, conduit pipe can be changed to split knobs or PDX wires.

SINGLE FAMILY DWELLING Type of Service - 115/230 V Single Phase 3 wire 60 Hz Line to Line system

SOLUTION: Examining the lighting plan of the above figure, there are 19 lighting outlets. Split the 19 outlets into two circuits A and B. A. Circuit - 1 Lighting Load (10 light outlets) 1. The PEC provides that 100 watts be the maximum load per light outlet. thus, for 10 light outlets at 100 watts, multiply: 10 outlets x 100 watts = 1000 watts 2. Compute The Current Load 1000 watts/230 volts = 4.35 amperes 3. Find the size of Branch circuit wire. Refer to Table 9.1 or 11.1. For 4.35 amperes, use 2.0 mm2 TW copper wire. 4. Find the rating of overcurrent protection. Refer to Table. for 4.35 A, use 15 amperes trip breaker. 5. Determine the size of conduit pipe. Refer to Table , for No. 14 TW copper wire, use 13 mm conduit pipe. B. Circuit - 2 Lighting Load (9 light outlets) 1. For 9 outlets, find the Total load in watts. 9 outlets x 100 watts per outlet = 900 watts Divide : 900 watts/230 volts = 3.91 amperes 2. Determine the Size of the Branch circuit Wire. Refer to Table 9.1 or 11.1. For 3.91 A load, use 2.0 mm2or No.14 TW copper wire. 3. Determine the size of the conduit pipe. Refer to Table. For 2 pieces No. 14 TW copper wire, use the 13 mm minimum size of conduit pipe. 4. Determine the size or rating of the overcurrent Protection. Refer to Table. For 3.91 A load, use 15 A load fuse of trip breaker. C. Circuit - 3 For small appliance load Section 3.3.1.2 of the the PEC specif 180 watts load limit per convenience outlet. Thus, 1. Find the number of appliance load outlet and the current load. 6 outlets x 2 gang per outlet x 180 watts 12 x 180 = 2,160 watts Divide: 2160 watts/230 volts = 9.39 A 2. Determine the Size of the Service Wire conductor. Refer to Table 9.1 or 11.1. For the 9.39 A load, specify

the minimum wire gauge for convenience outlet. 2 pieces 3.5 mm2or No.12 TW copper wire 3. Determine the Size of the conduit pipe. Refer to Table. For 2 pieces No.12 TW copper wire. Use 13 mm diameter conduit pipe. 4. Solve for the Size or Rating of the Over current Protection. Refer to Table. For 9.39 A on No. 12 TW copper wire specify: 20 A fuse or trip breaker. D. Circuit- 4 for Small Appliance Load 1. The load of circuit 4 is identical with circuit . Use the same size of wire, conduit, and wire protection rating. E. Circuit - 5 for Range Load 1. Range load (appliance rating) at 8.0 kW = 8000 watts 2. Solve for the Line current. 8000 watts / 230 V= 34.78 A 3. Refer to Table , apply 80% demand load factor . 34.78 x .80df = 27.82 A 4. Determine the Size of the Branch Circuit wire. Refer to Table 9.1 or 11.1. For the 27.82 A, use 8.0 mm2 or No. 8 TW copper wire. 5. Determine the Size of Conduit pipe. Refer to Table. For 2 pieces No.8 wire use 200 mm diameter pipe. 6. Find the Size or Rating of the Fuse or Trip Breaker. Refer to Table. For appliance load, use 40 A fuse or trip breaker. F. Circuit - 6 for Water heater Load 1. one unit water heater at 2.5 kW = 2500 watts 2. The current load will be: 2500 watts/230 volts = 10.86 A 3. Solve for the Size of the Branch circuit wire. Refer to Table 9.1 or 11.1 . For 10.86 A convenience outlet use 2 pcs 3.5 mm2 or No. 12 TW copper wire. 4. Determine the Size of the conduit pipe. Refer to Table. For 2 - No.12 Tw copper wire, use 13 mm conduit pipe. 5. Find the Size or Rating of the Over Current Protection. For the 10.86 A load, use 20 A fuse fuse or trip breaker. G. Circuit 7 and 8 with 1 - unit ACU each 1. One unit ACU at 1.5 Hp is 1.5 Hp x 746 = 1119 watts

Article 6.7 of the Philippine Electrical Code provides that: "BRANCH CIRCUIT CONDUCTOR SUPPLYING A MOTOR SHALL HAVE AN AMPACITY NOT LESS THAN 125% OF THE FULL LOAD CURRENT." 2. Current Load: 1119 watts/230 V = 4.86 A 4.86 A x 125% = 6.07 A 3. Find the Size of the Branch circuit service wire. Refer to Table. The 6.7 A can be served by a 2.0 No.14 TW copper wire, but the Code limits the size of convenience outlet to No. 12 AWG copper wire. SpecifyNo.12 THW copper wire for circuit 7 and 8. 4. Find the Size of the conduit pipe. Refer to Table. For two No.12 wire, use 13 mm conduit pipe. 5. Find the Size and Rating of the Branch Circuit Protection. The Code on branch circuit protection for a single motor provides that" It shall be increased by 250% of the full load current of the motor." thus, 4.86 x 250% = 12.15 A. From Table for a continuous load use 2-30 AT breaker.

CALCULATING THE AMPACITY OF THE SERVICE ENTRANCE CONDUCTOR AND THE MAIN DISCONNECTING MEANS

1. Find The total current load of circuit 1 to circuit 8: lighting load Ct. 1 and Ct.2 ....................1900 watts small appliance load Ct. 3 and Ct. 4.......4320 watts other loads Ct.5 and Ct 6.,......................10500 watts TOTAL LOAD(except the ACU)....16720 watts 2. From Table , OPTIONAL CALCULATION for dwelling Unit, apply demand factor. for the first 10000 w at 100%(df)...........10000 watts subtract: 16720 - 10000 = 6720 watts for other load, multiply by 40% 6720 x .4 .........................................2688 watts Aircon unit at 100% demand factor 2-units at 1119 watts........................2238 watts TOTAL .....................14, 926 watts TOTAL CONNECTED LOAD PLUS 25% OF THE LARGEST MOTOR 1. Ampere I = 14.926w + (25% of 1.119w)/230 V = 63.37 A 2. Find the Size of Main feeder and the Neutral line. - Use 2 - 3.8 mm2 TW copper wire 3. The Neutral Conductor of a 3 - wire line to line supply system shall have an ampacity of not less than

70% of the ungrounded(live wire) conductor or TWO TRADE SIZE SMALLER SIZE THAN THE UNGROUNDED CONDUCTOR. (PEC specs) Therefore use1- 22 mm2 Tw copper wire for the Neutral line. 4. Determine the Size of the Conduit pipe. Refer to Table , use 32 mm diameter pipe. 5. For main Breaker, refer to Table . Use 2 - 100 A 2 - wires 250 V, 2 pole molded air circuity breaker. COMMENT: The total computed load is 63.37 A. The 30 mm2 copper wire could be used considering its 90 A capacity. However, The NEC provides that: " If the computed load exceeds 10000 watts, the conductor and overcurrent protection shall be rated not less than 100 A. THEREFORE USE 2 - 38 mm2 TW WIRE FOR THE MAIN FEEDER AND 2 - 100 A FOR THE MAIN BREAKER.

Small Family Dwelling Type of Service - 230 volts; two wire Line to Ground system

A single family dwelling with a floor area of 80 square meters has the following receptacles and outlets load. LIGHTING: 7 pcs. - 40 watts fluorescent lamps 2 pcs. - 20 watts Incandescent lamps CONVENIENCE OUTLET: 1 - Electric Iron .............................................. 1000 watts 1 - Electric stove.............................................. 1100 watts 2 - Electric fan ................................................ 500 watts 1 - 7 cu. ft Refrigerator .................................... 175 watts 1 - Portable stereo ........................................... 100 watts 1 - 20" TV set ................................................... 300 watts

SOLUTION: A. Circuit 1 - Lighting Load by the Area method 1. Determine the wattage required per square meter area. From, the wattage required per square meter for dwelling units is 24 watts. Multiply: 80 sq.m x 24 watts = 1920 watts

2. Determine the current load. Divide: 1920 watts/ 230 volts = 8.35 A 3. Compute the actual lighting load. Multiply: 7 - Fluorescent lamps x 40 watts = 280 watts 2 - Incandescent bulb x 60 watts = 120 watts TOTAL............ 400 watts 4. Solve the actual current load. Divide: 400 watts/230 volts = 1.74 A 5. Determine th Size of the Branch Circuit wire. From , the 1.74 A is very small load to be carried by 2.0 mm2 or No. 14 TW copper wire. Therefore, the No. 14 wire is safe. 6. Determine the Size of the conduit pipe. Refer to table, for 2- No.14 wire, use 13 mm conduit pipe. 7. Determine the size or rating of the branch circuit protection. Refer to table. For 2.0 mm2 or No.14 copper wire conductor, use 15 A fuse or trip breaker. B. Circuit - 2 For small appliance load 1. Solve for the total appliance current load. LOAD CURRENT = ( 1000 + 1100 + 500 + 175 + 300 + 100) / 230 volts = 3175 watts/230 volts = 13.81 A 2. Determine the size of the Branch circuit wire conductor. Refer to Table. For a convenience load of 13.81 A specify 3.5 mm2 or No. 12 TW copper wire, the minimum size required for convenience outlet. 3. Find the size of the conduit pipe. Refer to Table, for 2 pieces No.12 TW copper wire, use 13 mm diameter pipe. 4. Find the Size or rating of the Protection device. See Table, for 13.81 A, use 1 - 20 A fuse. COMMENT: It is interesting to note that only one 20 A fuse protection was used because the current is a LINE TO GROUND OR MULTI-GROUND SYSTEM where one line is zero voltage being grounded. Unlike the LINE TO LINE SYSTEM of current supply, it is necessary to provide 2 fuses to protect the two line branch circuit. FINDING THE SIZE OF THE SERVICE ENTRANCE OR FEEDER 1. Get the sum total of connected load. Add: Lighting Load.................................... 1920 watts Small appliance load ......................... 3175 watts TOTAL.................................. 5095 watts 2. Solve for the total connected load current. Divide:

5095 watts/230 volts = 22.15 A 3. Find the size of Service Entrance. Refer to Table. For 22.15 A, useNo. 8 TW copper wire, the minimum size for service entrance. 4. For Main Protection, use 1-safety switch, 2 pole, 2 wires, 250 volts. Under the preceding set-up, one safety switch could supply both lighting and convenience outlet at different branch circuit without the use of fuse cutout. This is only applicable to the line to ground or multi-ground system being used by the electric cooperative.

ILLUMINATION CALCULATION AND DESIGN FOR SINGLE FAMILY DWELLING

PRINCIPLES OF ILLUMINATION 6 – 1 INTRODUCTION

Illumination is defined as the intensity of light per unit area. When we talk of illumination, or simply lighting, we are referring to man made lighting. Daylight being excellent is not included, thus, we assume a night time condition. Electric Illumination is the production of light by means of electricity and its applications to provide efficient, comfortable and safe vision. Specifically, when one speaks of lighting design, he refers to only two things: 1. The quantity of light 2. The quality of light Quantity of Light – refers to the amount of illumination or luminous flux per unit area. Quantity of light can be measured and easily handled because it deals with the number of light fixtures required for a certain area. Quality of Light – refers to the distribution of brightness in the lighting installation. It deals with the essential nature or characteristics of light. In short, quality of light is the mixture of all the items related to illumination other than the quantity of light which includes several elements such as:

1. Brightness 2. Brightness ratio or contrast

3. Glare 4. Diffuseness 5. Color 6. Aesthetics 7. Psychological reaction to color and fixtures 8. Economics There are four factors that affect illumination, namely: 1. Brightness 2. Glare 3. Contrast 4. Diffuseness Brightness is the light that seems to radiate from an object being viewed. Brightness or luminance is the luminous flux (light) emitted, transmitted or reflected from a surface. Contrast is the difference in brightness or the brightness ratio between an object and its background. The recommended brightness ratio between an object being viewed and its background is normally 3:1. If a print on a white paper can be clearly seen on a light background, it is due to the effect called contrast. Likewise, if a light object is placed on a dark background, the light object reflects more light and looks brighter although bought have equal illumination. It is for this reason that office furniture are generally light colored, tan or light green for eye comfort. Glare is a strong, steady, dazzling light or reflection. There are two types of glare: 1. Direct Glare is an annoying brightness of light in a person’s normal field of vision. 2. Indirect or Reflected Glare is much more serious and difficult to control. Technically, reflected glare is a glossy object. Diffuseness refers to the control of shadows cast by light. Diffuseness is the degree to which light is shadowless and is therefore a function of the number of directions to which light collides with a particular point and the comparative intensities.

Perfect Diffusion is an equal intensities of light clashing from all directions producing no shadows. A single lamp will cast sharp and deep shadows. A luminous ceiling provides a satisfactory diffuse illumination and less shadows. The color of lighting and the corresponding color of the object within a space is an important consideration in producing a quality of light. There are three characteristics that define a particular coloration. They are: a. Hue – is the quality attribute by which we recognize and describe colors as red, blue, yellow, green, violet and so on. b. Brilliance or Value – is the difference between the resultant colors of the same hue, such as: white is the most brilliant of the neutral colors while black is the last. c. Saturation or Chroma – is the difference from the purity of the colors. Colors of high saturation must be used in well lit spaces. 6 – 2 ESTIMATING ILLUMINATION AND BRIGHTNESS In many respect, it is more important to know luminance measurements than illumination because the eye is more sensitive to brightness than simple illumination. However, it is more difficult to measure luminance than illumination. There are three types of luminance meter, namely: 1. The Comparator type which requires the operator to make a brightness equivalence judgment between the target and the background. 2. The Direct Reading type is basically an illumination meter equipped with a hooded cell arranged to block oblique light. 3. The Accurate Laboratory Instrument which unsuitable for field work.

The quantity of light level of illumination can be easily measured or calculated with the aid of portable foot candle meter. Footcandle (fc) is the amount of light flux density. It is the unit of measure used when describing the amount of light in a room and is expressed in lumens per square foot. Footlambert (fl) is defined as “the luminance of a surface reflecting. Transmitting or emitting one lumen (lm) of illumination per square foot of area in the direction being viewed or the conventional unit of brightness or luminance. In the same manner, the lumens (lm) is the light output generated continuously by a standard wax candle

In our study of light, we are interested in the amount of light that fall on the areas that we want to illuminate. We also want to know the lumens per square foot or square meter in a space. This quantity called Light Flux Density is the common term Foot-candle (fc) represented by the formula: Footcandle = Lumens Area ILLUSTRATION 6 – 1 A 40 – watt fluorescent lamp 120 centimeters long produces 3,200 lumens of light in a room having a general dimension of 10 x 20ft. Find the illumination on the floor. SOLUTION Footcandle = Lumens Area fc = 3,200 lm. = 16 footcandle 10 x 20 ft. The footcandle is an important unit of measure in calculating the desired illumination and layout of fixtures. In the absence of Tables of equivalent footcandles for a particular fixture, a rule of thumb of 10-30-50 illumination level is here presented. 10 – footcandle is adequate for halls and corridors 30 – footcande is sufficient for areas between work stations such as in offices other than desk areas. 50 – footcandle is satisfactory on spaces where office work is done. However, providing an adequate quantity of light alone is not a guarantee for an efficient and comfortable vision. In fact, the quality of light is very important especially where difficult visual needs are required. The luminance or brightness of a diffusely reflecting surface is equal to the product of the illumination and the reflectance. Thus; Luminance = Illumination x Reflectance factor or Footlambert = Footcandle x Reflectance factor

ILLUSTRATION 6 – 2 From illustration 6 – 1, find the luminance if the reflectance factor of the wall is 40%.

SOLUTION 1. Footlambert = Footcandle x Reflectance factor = 16 x 40% = 6.4 Metric Lighting Units In English System of measure, the distance is expressed in feet and the area is in square feet. Under the Metric System (SI) the distance and area are expressed in meters and square meters respectively. Meanwhile, Lumens flux remains in Lumens but illuminationor light flux is expressed in Lux. Thus: Lux = Lumens Area (sq. m.) Table 6 -1 APPROXIMATE REFLECTANCE FACTOR

In the metric system, Luminance or Brightness is expressed in Lambert which is defined as “the luminance or brightness of a surface reflecting, transmitting or emitting one lumen per square centimeter. Millilambert is more conveniently used than the lambert because the value of lambert is greater than what is usually encountered. Table 6 – 2 TABLE OF COMPARISON

ILLUSTRATION 6 – 3 A 40 – watts x 120 centimeters long fluorescent lamp produces 3,200 lumens of light in a room having a general dimension of 10ft. x 20ft. Compute the illumination on the floor comparing the English and the Metric units. SOLUTION BY COMPARISON English Metric (SI) Light Flux = 3,200 lm. …………………. 3,200 lm Area = 10’ x 20’ …………………. 10 x 20 10.76 = 200 sq. ft. ………………… 18.59 sq. m. Illumination = 16 fc ……………………… 172.16 lux Another SOLUTION Convert: 10 feet to meter = 3.048 m. 20……………. = 6.097 m. Lux = 3,200 = 172.19 Lux 3.048 x 6.097 ILLUSTRATION 6 – 4 Compute for the brightness of a fixture with a 1’x 4’ plastic diffuser having a transmittance of .6 and illuminated by 2 pieces 3,200 lm. lamp assuming 100% use of light flux.

SOLUTION 1. Luminance = Total lumens x transmission factor Area of diffuser = 2pcs. x 3,200 x .6

1’ x 4’ = 960 footlambert 1. To obtain the metric equivalent, multiply: Millilambert = Footlambert x 1.076 = 960 x 1.076 = 1032.96 millilambert

The Watts per Square Meter Another method used in determining the illumination is the watts per square meter wherein the floor area is computed from the outside dimensions of the building excluding open porches. Depending upon the size of the room, color of wall and ceiling, types of lighting units and methods of lighting used, the watts per square meter method is may produce 50 to 100 lux which is approximately 5 to 10 footcandles.

1. Twenty watts (20) per square meter will provide an illumination of 100 to 150 lux which is approximately 10 to 15 fc in industrial areas. 2. For commercial areas, two (2) watts per square foot or 22 watts per square meter is will provide from 80 to 120 lux when used with standard quality equipment. 3. Forty (40) watts per square meter will provide about 200 lux which is approximately 20 fc wherein greater illumination is required 4. Sixty (60) watts per square meter will provide about 300 lux or approximately 30 fc which is recommended for many conventional, industrial and commercial requirements. 5. Eighty (80) watts per square meter will provide from 300 to 350 lux and in excess of supplementary lighting is necessary. 6 – 3 COEFFICIENT OF UTILIZATION AND MAINTENANCE FACTOR

The usable Initial footcandle or lux is equal to the footcandle produced by the coefficient of utilization (cu). Initial was emphasized because the output is of a light fixture is reduced with time as the lamp fixture is becomes old and dirty. Lamp output normally drops and is termed asMaintenance factor (mf). And to find the average maintained illumination, we reduce the initial illumination by the maintenance factor. The efficiency of a light fixture is equals the ratio of fixture output lumens to lamp output lumens. What we need is to determine a number indicating the efficiency of the fixture room combination, or how a particular light fixture lights a particular room. This number is normally expressed in decimal value called coefficient of utilization represented by letter (cu). The usable initial footcandle is equal to the footcandle produced by the coefficient of utilization (cu). a.) Initial Footcandle = footcandle x cu. Area b.) Maintenance illumination = Lamp lumens x cu x mf Area * Lamp lumen therefore is simply the rated output of the lamp. TABLE 6 – 3 COEFFICIENT OF UTILIZATION

TABLE 6 – 4 MAINTENANCE FACTOR

ILLUSTRATION 6 – 5

A school classroom with a general dimension of 24 x 30 ft. is lighted with 10 fluorescent of 4F 40 T12 WW rapid start lamp. Calculate the initial and maintained illumination in footcandles (English) and Lux (Metric) assuming that (cu) is 0.35 and (mf) is 0.70.

SOLUTION – 1 (English Measure) 1. Refer to Table 5 – 3. An F 40 T12 WW watts fluorescent lamp has 3,200 lm. output. Multiply: Lamp lumens = 10 fixturesX 4 lamps per fixture X3,200 lumens per lamp = 128,000 lumens Initial footcandle = 128,000 x 0.35 24 x 30 ft. = 62.22 fc x 0.70 mf = 43.55 footcandle SOLUTION – 2 By Metric Measure (SI) Convert feet to meter: 24 ft. = 7.32 m. 30 ft = 9.14 m. Lux = Lumens x cu x mf Area = 10 x 4 x 3,200 x 0.35 x 0.70 7.32 m. x 9.14 m. = 468.75 lux Check the answer: One lux = .09294 468.75 x .09294 = 43.56 fc. Sometimes when the size of the room and the footcandle are given, the problem is how to find the number of lamps required in each fixture. The following example is offered.

ILLUSTRATION 6 – 6 An office room having a general dimension of 8 x 20 meters is to be lghted at an averaged maintained footcandle of 50 fc, How many 3-lamp fixtures of 120 centimeters long F40 T12 WW rapid start fluorescent lamps are required assuming the cu is 0.38 and the mf is 0.75?

SOLUTION 1. Lamp lumens = maintained footcandle x area cu x mf = 50 fc x (8m. x 20 m.) 0.38 x 0.75 = 28,070 lumens 2. Each 40 watt fluorescent lamp has an output of 3,200 lumens, the number of lamps will be: Number of lamps = 28,070 3,200 = 8.77 lamps 3. Since there are 3 lamps for each fixture, divide: 8.77 = 2.93 say 3 lamps in ach fixture 3 Calculation involving a wide area is more confusing than by computing the number of lamp fixtures per bay or per row which is more meaningful and interesting. This could be done easily by using the following formula: Number of mixtures = Illumination x area Lamp per fixture x lumens x cu x mf

This means that the area lighted by a single area is: Area per fixture = lamp per fixture x lumens per lamp x cu x mf Illumination TABLE 6 – 6 EFFICACY OF VARIOUS LAMPS

ILLUSTRATION 6 – 7 An entire office floor is lighted at an averaged maintained 538 lux or 50 fc. The floor measures 20 meters by 50 meters and is divided into bays measuring 4 m. x 5 m. Using 2-lamp of F40 T12 CW rapid start preheat lamp, find the number of fixtures required. Assume an economy grade fixture with a lo cu of 0.35 and mf of 0.70 SOLUTION – 1 Solve for the number of fixtures per bay. Refer to Table 5-3 for F40 T12 CW = 3,150 lm.

No. of Fixtures = Illumination x area Lamp per fixture x lumens x cu x mf Fixtures = 538 lux x (4 m. x 5m. ) 2-lamps x 3,150 lm. x 0.35 x 0.70 = 10,760 = 6.9 fixtures 1,543 Accept 6 pieces per bay to make it symmetrical SOLUTION – 2 1. From the following Formula, substitute the values: Area per Fixtures = Lamp per fixture x lumen/ lamp x cu x mf Illumination Fixtures = 2-lamps per fixture x 3,150 lm. x 0.35 x 0.70 538 lux = 1,543.5 = 2.87 sq.m. per fixture 538 2. Therefore, the number of fixture per bay is: 4 m x 5m = 6.9 say 6 pcs. Per bay for symmetry 2.87 6 – 4 MEASURING FOOTCANDLE The unit measure of illumination is the footcandle or lux in the Metric System which is frequently used when describing the amount of light in a room. It is not enough to know how to calculate the illumination level but it is also equally important to know how to measure them in enclosed space. In measuring illumination level, the footcandle meter is held horizontally with its sensitive surface at least 30 centimeters from the body of the person holding the meter, The meter could be placed on a table and read from a distance to avoid obstruction of the light.

In conducting a general illumination check inside a room, the meter is held at least 80 centimeters above the floor. Reading is undertaken throughout the room and the results are recorded on the plan of the room.

a. All dimensions in meters  b. These spacing apply where desks and benches are next to wall, otherwise, one third the spacing between units is satisfactory.  c. The actual spacing of luminaries is usually less the maximum spacing to suit bay or room dimensions.  d. For mounting height of general diffusing and direct-indirect fixtures.   

6-5 Uniformity of Light The purpose of lighting calculations, by the foot-candle or lux, is to determine the average illumination in a room or lux, is to determine the average illumination in a room to a working level condition. This working level condition refers to the height of 75 centimeters above the floor being the approximate height of the table. The average illumination at the working level is directly related to the maximum spacing of the light to the mounting height ratio represented by the formula.

S/mh where: S = Spacing of light fixtures  mh = mounting height 

Normally, the manufacture of light provides data with respect to spacing and mounting ratio. However, in the event that the manufacturer failed to provide these data, Table 6-8 was presented shoeing the spacing and mounting height ratio for particular lighting conditions.  Table 6-8 SPACING AND MOUNTING HEIGHT RATIO 

ILLUSTRATION 6-8  A room with a ceiling height of 3 meters is to be lighted with direct concentrating fluorescent light. What is the maximum fixtures spacing?  SOLUTION:  1. For spacing and mounting ratio, refer to 6-8. The mounting height ratio od direct concentrating light is 0.40. 

Therefore:

 

2. Substituting the given values, wherein mh is the ceiling height,  S = 0.40 x 3.00  Spacing: S = 1.20 meters maximum side to side of the fixtures.  ILLUSTRATION 6-9  A warehouse will install pendant dome incandescent lamps at a mounting ratio of 1.50 meters. The lamp will be mounted on a grid measuring 5.00 x 5.00 meters. What is the minimum mounting height of the lamps? 

Solution:  Mounting height is; mh = Spacing  Ratio  mh = 5.00 m. = 3.30 meters  1.50 6-6 Classification of Lighting System Lighting system is classified into four types, namely: 1. Direct lighting 2. Semi-direct lighting  3. Semi-indirect-lighting 4. Indirect Lighting Direct Lighting. When the light on an illuminated area is focused downward coming directly from the lighting fixtures.

Semi-Direct Lighting. When the predominant light on the illuminated area is fed directly from the lighting units wherein the greater amount of light is obtained from the ceiling through the reflection.

Semi-Indirect Lighting. A lighting arrangement wherein 5% to 25% of the light is directed downward with more than half of the light focused upward and reflected from the ceiling.

Indirect Lighting. When the light is diffused and reflected from a wide ceiling area. This kind of lighting produces a soft and subdued effect due to low brightness and absence of sharp shadows.

FIGURE 6-6 CHART FOR ESTIMATING LIGHTING LOAD AND ILLUMINATION LEVEL CALCULATED FOR FAIRLY LARGE ROOM

6-7 Street Lighting

The Institute of Integrated Electrical Engineers instituted guidelines for adequate and acceptable illumination of the streets in order to promote safety. This concept was brought about by the continuously increasing speed of motor vehicles using the road. The Philippine Electrical Code Committee prepared the guidelines for a standard practice on design of street lighting installation recommending the proper quantity and quality of light for traffic routes. Definition of Terms Lighting Installation – is defined as the whole of the equipment provided for lighting the roadway comprising the lamps luminaries, means of support and electrical installations including other auxiliaries. Lighting System – refers to an array of luminaires having a characteristic of light distribution. Luminaire – is a housing for one or more lamps comprising a body and any refractor, diffuser or enclosure associated with the lamps. Road Width – is the distance between the edges of the road curbs measured at right angles to the length of the roadway. Outreach – is the distance measured horizontally between the outer of the column or wall face or lamp post and the center of the luminaries. Overhang – is the horizontal distance between the center of luminaires and the adjacent edge of the road. Mounting Height – refers to the vertical distance between the center of the luminaire and the surface of the roadway. Spacing – is the distance between the successive luminaries in an installation.

Maximum Light Utilization – In order to attain the maximum utilization of light from the fixtures, the luminaires should be mounted under the following specifications.

Working Voltage Luminance are properly selected and mounted on a location most feasible and effective with minimum cost. For a 230 volt system, a voltage drop of 5% is allowed although in extreme cases 15% voltage drop is sometimes tolerated. For street illumination, the following formula is adopted.

Where: E=The illumination in lux Al=Average lumens with a typical value of: 20 500 lumens for 40 watts 11 500 lumens for 250 watts 5 400 lumens for 125 watts The value of Al varies depending upon the type of lamp specified. mf- is the maintenance factor which depends on the following: a). Maintenance practice of the company b). Operation of light sources at rate current and voltage

c). Regular replacement of depreciated lamp d). Periodic cleaning of the luminaires either 0.8-0.9 w = Width of the roadway d = Distance between luminaires cu = Coefficient of utilization which is dependent on the type of fixture, mounting height, width of roadway and the lenght of mast arm or outreach.

The values given are based on the favorable reflectances for asphalt road, the recommended illumination should be increased by 50%. For concrete road, the recommended value could be decreased by 25%. In decreasing street illumination, consider the modern lighting today that will be obsolete tomorrow when the minimum light levels are raised. The increasing motor vehicle speed and the incerasing congestiin in the street requires higher level of highway lighting. Therefore, future needs for light should be considered in the design. ILLUSTRATION 6-10 Considering the data as presented on Figure 6-7 when the night pedestrian traffic is estimated oto be light and the night vehicular traffic is to be medium, determine the required lumens if theroad concrete is a pavement. SOLUTION

Refering to Table 6- 9, E= 6.46 for light pedestrian medium traffic classifications. For concrete road, the reflectance will be higher but let us accept the value of 6.46 lumens. Determine the average pole distance. E= 6.46 lumens per sq.m. w= 7.00 meters d= 50 meters mf= 0.9 cu= 0.29 (type A fixture)

Under the Working Voltage, the mean lamp lumens of a 250 watts lamp is 11, 500 lumens, this is the nearest value to 8,662.83 average lumens. Therfore, a 250 watts lamp is acceptable. Computing for the new actual illumination E

This is higher than the 6.46 recommended in table 6- 11. Therefore, the road is considered as adequeately lighted.

POSTED BY ELECTRICAL DESIGN1 AT 4:08 PM LABELS: D. WEEK 6 AND 7

WIRING CALCULATION FOR MULTI-FAMILY DWELLING

Multi - Family Dwelling 4 - door apartment Types of Service: 230 V 2 wire, Line to Ground System Floor Area per unit: 80 sq. m Total Floor Area: 320 sq. m

Determine the branch circuit protection, size of conductor wires and the main header.

SOLUTION: Assume that the dwelling unit is equipped with one 5.1 kW cooking unit; one unit laundry ckt. at 1.5 kW A. Circuit - 1 For Lighting Load per unit (see plan) 1. By the area method, refer to Table, General Lighting Load by occupancy for dwelling units. 80 sq. m x 24 watts per sq. m = 1920 watts 2. Compute for the Lighting Load. Divide: 1920 watts/230 volts = 8.35 A 3. Determine the size of the Branch Circuit conductor wire. Refer to Table. For 8.35 A load, use 2 pieces 2.0 mm2 or No. 14 TW AWG copper wire 4. Determine the size of the conduit pipe. For number 14 AWG , TW wire use 13 mm minimum size of conduit pipe. 5. Determine the size or rating of the branch circuit protection. Refer again to Table. For 8.35 A load on a 2.0 mm2 wire conductor size, use 15 A fuse or trip breaker.

B. Circuit - 2 For Convenience Outlet Load 1. Solve for the total current load. 8 receptacles x 2 gang per outlet x 180 watts = 2880 watts 2. Solve for the appliance current load. Divide. I = 2880 watts/230 volts = 12.52 A 3. Determine the size of the Branch Circuit conductor. Refer to Table 9.1 or 11.1. For a 12.52 A load, a 2.0 mm2 or No.14 TW AWG wire would be sufficient considering its 15 A ampacity that is bigger than 12.52 A as computed 4. But the National Electrical Code limits the size of convenience outlet wire to minimum of 3.5 mm2 or No. 12 AWG copper wire. The code must prevail. Use No. 12 TW. 5. Determine the size of the conduit pipe. Refer to Table . For No. 12 TW wire, use 13 mm diameter pipe. 6. Find the Size of the Branch Circuit fuse protection. Refer to table. For 12.53 A non continuous load on convenience outlet, use 20 AT breaker. C. Circuit - 3 Other Load 1. Laundry Circuit at 1500 watts per circuit (PEC provision)

1500 watts/230 volys = 6.52 A 2. Find the size of the branch circuit conductor. From Table, use 3.5mm2 or No. 12 TW copper wire, the minimum size for convenience outlet. 3. Find the size of the conduit pipe. From Table, use 13 mm diameter pipe. 4. Find the size of the branch circuit fuse protection. From Table. The 6.52 A load on convenience outlet requires 20 A fuse or trip breaker. D. Circuit - 4 Cooking Unit 1. Total Load is 5.1 kW = 5100 watts 2. Refer to Table Demand load for household. For electric range, apply 80% demand factor. Total load x demand factor (Df) 5100 watts x .80 = 4080 watts 3. Compute for the line current load. Divide: 4080 watts/230 volts = 17.74 A 4. Find the size of the branch Circuit wire. Refer to Table 0.1 or 11.1. For 17.74 A line current, use 5.5 mm2 or No.10 TW copper wire. 5. Determine the size of the conduit pipe. From Table, for No.10 TW wire, use 20 mm diameter pipe. 6. Find the size of the branch circuit fuse protection. Refer to Table, for 17.74 A current load, use 30 A fuse or trip breaker. E. Determine the sub-feeder per dwelling 1. Solve for the total connected load per dwelling. Lighting load ........................................ 1920 watts Convenience Load ,............................... 2880 watts Other loads 5.1 + 1.5 kW........................ 6600 watts TOTAL....................................11400 watts 2. Apply 80% demand factor (see Table) TOTAL LINE CURRENT = (11400 watts x.80 df)/230 volts = 39.65 A 3. Determine the Size of the sub-feeder and protection per dwelling for 39.65 A. For Table 9.1 or 11.1, use 8.0 mm2 or No. 8 wire THW copper wire. 4. Find the size of the conduit pipe. For 8.0 mm2 wire, specify 25 mm diameter pipe. 5. Determine the size or rating of the fuse protection. From Table, use 60 A molded Circuit breaker 2- wire 250 volts with solid bus.

F. Determine the Size of the Main Feeder 1. Solve for the Total connected Load on 4 dwelling units at 11400 watts each. Multiply: 11400 watts x 4 = 45600 watts 2. Refer to Table. For 4 dwelling units apply 45% demand factor. Multiply: 45600 watts x .45 = 20520 watts 3. Solve for the line current: I = 20520 watts/230 volts = 89.22 A 4. Determine the Size of the Conductor wire. Refer to Table. For 89.22 A, use 2- 50 mm2 TW copper wire or 2- 38 mm2 THW copper wire. COMMENT: IT will be noted in Table, that the 89.22 A as computed does not exceed 80 % of the 120 allowable ampacity of 50 mm2 TW copper wire or 125 ampacity of 38 mm2 THW copper wire. Therefore, any one of these two types of wire could be used for main feeder. 5. Find the size of conduit pipe. Refer to Table. Use 38 nn diameter RSC or IMT pipe 6. Find the size or rating of the over-current protection. Refer to Table. Use 125 A safety switch,250 volts, 2 pole.

ILLUMINATION CALCULATION AND DESIGN FOR MULTI-FAMILY DWELLING UNIT Interior lighting. (1) Within all buildings of three or fewer storeys in building height, having a building area not exceeding 600 square metres and used for residential occupancies, business and personal services occupancies, mercantile occupancies or medium and low industrial occupancies. (1) Every exit (except those serving not more than one dwelling unit), public corridor or corridor providing access to exit for the public shall be equipped to provide illumination to an average level of not less than 50 lux at floor or tread level and at all points such as angles and intersections at changes of level where there are stairs or ramps. (2) Emergency lighting shall be provided in: (a) Exits; (b) Principal routes providing access to exit in an open floor area; (c) Corridors used by the public;

(d) Underground walkways; and (e) Public corridors. (3) Emergency lighting required in Subsection B(1)(b) shall be provided from a source of energy separate from the electrical supply for the building. (4) Lighting required in Subsection B(2)(b) shall be designed to be automatically actuated for a period of not less than 30 minutes when the electric lighting in the affected area is interrupted. (5) Illumination from lighting required in Subsection B(2)(b) shall be provided to average levels of not less than 10 lx at floor or tread level. (6) Where incandescent lighting is provided, lighting equal to one watt per square metre of floor area shall be considered to meet the requirement in Subsection B(5)(e). (7) Where self-contained emergency lighting units are used, they shall conform to CSA C22.2 No. 141-M, “Unit Equipment for Emergency Lighting”. (8) Every public or service area in buildings, including a recreational camp and a camp for housing of workers, shall have lighting outlets with fixtures controlled by a wall switch or panel. (9) When provided by incandescent lighting, illumination required in Sentence (1) shall conform to Table § 629-36B(1). (j) When other types of lighting are used, illumination equivalent to that shown in Table 36.B.(1) shall be provided.

(2) Within all buildings exceeding three storeys in building height or having a building area exceeding 600 square metres or used for other occupancies not described in Subsection B(1). (a) An exit, a public corridor, a corridor providing access to exit for the public, a corridor serving patients or residents in a Care and Treatment occupancy or Care occupancy, a corridor serving classrooms, an electrical equipment room, a transformer vault and a hoistway pit shall be equipped to provide illumination to an average level not less than 50 lux at floor or tread level and at angles and intersections at changes of level where there are stairs or ramps. (b) Rooms and spaces used by the public shall be illuminated as described in Subsection B(1)(h),(i) and (j). (c) Elevator machine rooms shall be equipped to provide illumination to an average level of not less than 100 lux at floor level.

(d) Every place of assembly intended for the viewing of motion pictures or the performing arts, shall be equipped to provide an average level of illumination at floor level in the aisles of not less than two lux during the viewing. (e) Every area where food is intended to be processed, prepared or manufactured and where equipment or utensils are intended to be (2) Within all buildings exceeding three storeys in building height or having a building area exceeding 600 square metres or used for other occupancies not described in Subsection B(1). (a) An exit, a public corridor, a corridor providing access to exit for the public, a corridor serving patients or residents in a Care and Treatment occupancy or Care occupancy, a corridor serving classrooms, an electrical equipment room, a transformer vault and a hoistway pit shall be equipped to provide illumination to an average level not less than 50 lux at floor or tread level and at angles and intersections at changes of level where there are stairs or ramps. (b) Rooms and spaces used by the public shall be illuminated asdescribed in Subsection B(1)(h),(i) and (j). (c) Elevator machine rooms shall be equipped to provide illumination to an average level of not less than 100 lux at floor level. (d) Every place of assembly intended for the viewing of motion pictures or the performing arts, shall be equipped to provide an average levelof illumination at floor level in the aisles of not less than two lux during the viewing. (e) Every area where food is intended to be processed, prepared or manufactured and where equipment or utensils are intended to be In a service space in which facilites are included to permit a person to enter and to undertake maintenance and other operations; and On a shelf and rack storage system, which includes walkways, platforms, unenclosed egress stairs and exits providing means of egress. (j) The minimum value of the illumination required by Subsections B(2) (h) and (i) shall be not less than one lux. (k) In addition to the requirements of Subsections B(2)(h) to (j), the installation of battery-operated emergency lighting in health care facilities shall conform to the appropriate requirements of CSA Z32, “Electrical Safety and Essential Electrical Systems in Health Care Facilities”. C. For parking lots, walkways, stairs, porches, verandas, loading docks, ramps or other similar areas, a minimum level of illumination of ten lux (0.90 foot-candle) at ground or tread level and at angles and intersections at changes of level where there are stairs or ramps.

Introduction to the Lumen Method The lumen method is applicable to design of a uniform (general) lighting scheme in a space where flexibility of working locations or other activities is required.

The lumen method is applied only to square or rectangular rooms with a regular array luminaires as shown in Figure 2.

2. Lumen Method Calculations The lumen method is based on fundamental lighting calculations. The lumen method formula is easiest to appreciate in the following form.

(1)

where E = average illuminance over the horizontal working plane n = number of lamps in each luminaire N = number of luminaire F = lighting design lumens per lamp, i.e. initial bare lamp luminous

flux UF = utilisation factor for the horizontal working plane LLF = light loss factor A = area of the horizontal working plane 2.1 Light Loss Factor Light loss factor (LLF) is the ratio of the illuminance produced by the lighting installation at the some specified time to the illuminance produced by the same installation when new. It allows for effects such as decrease in light output caused by (a) the fall in lamp luminous flux with hours of use, (b) the deposition of dirt on luminaire, and (c) reflectances of room surfaces over time. In fact, light loss factor is the product of three other factors: (2)

where LLMF = lamp lumen maintenance factor LMF = luminaire maintenance factor RSMF = room surface maintenance factor 2.1.1 Lamp Lumen Maintenance Factor Lamp lumen maintenance factor (LLMF) is the proportion of the initial light output of a lamp produced after a set time to those produced when new. It allows for the decline in lumen output from a lamp with age. Its value can be determined in two ways: (a) by consulting a lamp manufacturer's catalog for a lumen depreciation chart, and (b) by dividing the maintained lumens by the initial lamps.

2.1.2 Luminaire Maintenance Factor Luminaire maintenance factor (LMF) is the proportion of the initial light output from a luminaire after a set time to the initial light output from a lamp after a set time. It constitutes the greatest loss in light output and is mainly due to the accumulation of atmospheric dirt on luminaire. Three factors must be considered in its determination: (a) the type of luminaire, (b) atmospheric conditions, and (c) maintenance interval. 2.1.3 Room Surface Maintenance Factor Room surface maintenance factor (RSMF) is the proportion of the illuminance provided by a lighting installation in a room after a set time compared with that occurred when the room was clean. It takes into account that dirt accumulates on room surfaces and reduces surface reflectance. Figure 4 shows the typical changes in the illuminance from an installation that occur with time due to dirt deposition on the room surfaces. 2.2 Utilisation Factor Utilisation factor (UF) is the proportion of the luminous flux emitted by the lamps which reaches the working plane. It is a measure of the effectiveness of the lighting scheme. Factors that affect the value of UF are as follows: (a) light output ratio of luminaire (b) flux distribution of luminaire (c) room proportions (d) room reflectances (e) spacing/mounting height ratio

2.2.1 Light Output Ratio of Luminaire

Light output ratio of luminaire (LOR) takes into account for the loss of light energy both inside and by transmission through light fittings. It is given by the following expression.

(3) Example 1 The total, upward and downward lamp output from a lamp are 1000 lm, 300 lm and 500 lm respectively. Calculate upward light output ratio (ULOR), downward light output ratio(DLOR), light output ratio (LOR) of luminaire and percentage of light energy absorbed in luminaire.

Amount of light energy absorbed in luminaire = 100 - 80 = 20 % A greater DLOR usually means a higher UF. A simple classification of luminaires according to their distribution is based on flux fractions, as shown in Figure 5. Upward flux fraction (UFF) and downward flux fraction (DFF) are used as a basis of comparison. Example 2 For data given in Example 1 determine upward flux fraction (UFF), downward flux fraction (DFF) and flux fraction ratio(FRR).

Figure 5 Flux Fraction of Various Luminaires

2. Lumen Method Calculations The lumen method is based on fundamental lighting calculations. The lumen method formula is easiest to appreciate in the following form.

(1)

where E = average illuminance over the horizontal working plane

n = number of lamps in each luminaire N = number of luminaire F = lighting design lumens per lamp, i.e. initial bare lamp luminous flux UF = utilisation factor for the horizontal working plane LLF = light loss factor A = area of the horizontal working plane 2.1 Light Loss Factor Light loss factor (LLF) is the ratio of the illuminance produced by the lighting installation at the some specified time to the illuminance produced by the same installation when new. It allows for effects such as decrease in light output caused by (a) the fall in lamp luminous flux with hours of use, (b) the deposition of dirt on luminaire, and (c) reflectances of room surfaces over time. In fact, light loss factor is the product of three other factors: (2) where LLMF = lamp lumen maintenance factor LMF = luminaire maintenance factor RSMF = room surface maintenance factor 2.1.1 Lamp Lumen Maintenance Factor Lamp lumen maintenance factor (LLMF) is the proportion of the initial light output of a lamp produced after a set time to those produced when new. It allows for the decline in lumen output from a lamp with age. Its value can be determined in two ways:

(a) by consulting a lamp manufacturer's catalog for a lumen depreciation chart, and (b) by dividing the maintained lumens by the initial lamps. 2.1.2 Luminaire Maintenance Factor Luminaire maintenance factor (LMF) is the proportion of the initial light output from a luminaire after a set time to the initial light output from a lamp after a set time. It constitutes the greatest loss in light output and is mainly due to the accumulation of atmospheric dirt on luminaire. Three factors must be considered in its determination: (a) the type of luminaire, (b) atmospheric conditions, and (c) maintenance interval. 2.1.3 Room Surface Maintenance Factor Room surface maintenance factor (RSMF) is the proportion of the illuminance provided by a lighting installation in a room after a set time compared with that occurred when the room was clean. It takes into account that dirt accumulates on room surfaces and reduces surface reflectance. Figure 4 shows the typical changes in the illuminance from an installation that occur with time due to dirt deposition on the room surfaces. 2.2 Utilisation Factor Utilisation factor (UF) is the proportion of the luminous flux emitted by the lamps which reaches the working plane. It is a measure of the effectiveness of the lighting scheme. Factors that affect the value of UF are as follows: (a) light output ratio of luminaire (b) flux distribution of luminaire (c) room proportions (d) room reflectances (e) spacing/mounting height ratio

2.2.1 Light Output Ratio of Luminaire Light output ratio of luminaire (LOR) takes into account for the loss of light energy both inside and by transmission through light fittings. It is given by the following expression.

(3)

Example 1 The total, upward and downward lamp output from a lamp are 1000 lm, 300 lm and 500 lm respectively. Calculate upward light output ratio (ULOR), downward light output ratio (DLOR), light output ratio (LOR) of luminaire and percentage of light energy absorbed in luminaire.

Amount of light energy absorbed in luminaire = 100 - 80 = 20 % A greater DLOR usually means a higher UF. A simple classification of luminaires according to their distribution is based on flux fractions, as shown in Figure 5. Upward flux fraction(UFF) and downward flux fraction (DFF) are used as a basis of comparison.

Example 2

For data given in Example 1 determine upward flux fraction (UFF), downward flux fraction (DFF) and flux fraction ratio (FRR).

Figure 5 Flux Fraction of Various Luminaires 2.2.2 Flux Distribution of Luminaire Direct ratio is the proportion of the total downward luminous flux from a conventional installation of luminaires which his directly incident on the working plane. It is used to assess the flux distribution of luminaire. Since the intensity distribution pattern of the light radiated from a luminaire in the lower hemisphere will affect: (a) the quantity of the downward flux falls directly on the working plane and (b) the quantity of flux available for reflection from the walls in a given room,

Direct ratio depends on both the room proportions and the luminaires.Direct ratio has a low value with a narrow room (small room index) and a luminaire which emits most of its light sideways (BZ 10), and on the contrary, a high value with a wide room (large room index) and a luminaire which emits most of its light downwards (BZ 1). 2.2.3 Room Proportion Room index (RI) is the ratio of room plan area to half the wall area between the working and luminaire planes.

(4) where L = length of room W = width of room Hm = mounting height, i.e. the vertical distance between the working plane and the luminaire. 2.2.4 Room Reflectances The room is considered to consist of three main surfaces: (a) the ceiling cavity, (b) the walls, and (c) the floor cavity (or the horizontal working plane). The effective reflectances of the above three surfaces affect the quantity of reflected light received by the working plane.

2.2.5 Spacing to Height Ratio Spacing to Height ratio (SHR or S/Hm) is defined as the ratio of the distance between adjacent luminaires (centre to centre), to their height above the working plane. For a rectangular arrangement of luminaires and by approximation,

(5) where A = total floor area N = number of luminaires Hm = mounting height Under a regular array of luminaires the illuminance on the working plane is not uniform. The closer spaced the luminaires for a given mounting height, the higher the uniformity; or the greater the mounting height for a given spacing, the greater the uniformity. If uniformity of illuminance is to be acceptable for general lighting, (a) SHR should not exceed maximum spacing to height ratio (SHR MAX) of the given luminaire as quoted by the manufacturer, and (b) geometric mean spacing to height ratio of the luminaire layout should be within the range of nominal spacing to height ratio(SHR NOM) of the given luminaire as quoted by the manufacturer, i.e.

(6)

3. Summary of Procedures for Lumen Design Method (a) Calculate the room index. (b) Determine the effective reflectances of the ceiling cavity, walls and floor cavity. (c) Determine the utilisation factor from the manufacturer's data sheet, using the room index and effective surface reflectances as found in (a) and (b) above. (d) Determine the light loss factor. (e) Inert the appropriate variables into the lumen method formula to obtain the number of luminaires required.

(f) Determine a suitable layout. (g) Check that the geometric mean spacing to height ratio of the layout is within the SHR NOM range:

(h) Check that the proposed layout does not exceed the maximum spacing to height ratios (SHR MAX). (i) Calculate the illuminance that will be achieved by the final layout and check against the standard. Example 3 Design a lighting installation for a college seminar room so that the average illuminance is 500 lux on the horizontal working plane, using the data listed below. Suggest the layout and check appropriate spacing to mounting height. Room dimensions: 12 m long x 8 m wide x 3.2 m high Working plane at 0.7 m above floor Reflection factors: Ceiling 70 % Walls 50 % Working plane 20 % Light Loss factor: 0.779 Luminaires: 1800 mm twin tube with opal diffuser Ceiling mounted Downward light output ratio 36 % SHR MAX 1.60 : 1 SHR NOM 1.50 : 1 Dimensions : 1800 mm long x 200 mm wide

Lamps: 1800 mm 75 W plus white 5800 average initial lumens per lamp 2 lamps per luminaire

Solution (a) Initial calculation

From manufacturer's photometric data sheet (Table 3), utilisation factor (UF) is 0.5336 by interpolation.

Therefore, the number of luminairs is 10. Initial check on S/Hm ratio gives:

From the manufacture's photometric data, maximum S/Hm is 1.6 : 1. Therefore, it should be possible to use 10 luminaires. (b) Proposed layout A 5 x 2 array is proposed fro the lighting installation. (A 10 x 1 array is an alternative.) (c) Checking the proposed layout Since 2 x 1.8 m = 3.6 m < 8 m (width of room), the proposed layout will fit. (Usually checking only the linear dimension of the fitting for space is enough as the other dimension (i.e. 200 mm in this case) is much smaller.) For long axis,

For short axis,

Note that if the checks had worked out to be unsatisfactory, the number of luminaires should be reconsidered and the calculations on the illuminance should be repeated. For example, a 3x3 array for lower lux level or a 4x4 array for higher lux level.

Distribution One of the primary functions of a luminaire is to direct the light to where it is needed. The light distribution produced by luminaires is characterized by the Illuminating Engineering Society as follows: Direct ( 90 to 100 percent of the light is directed downward for maximum use. Indirect ( 90 to 100 percent of the light is directed to the ceilings and upper walls and is reflected to all parts of a room. Semi-Direct ( 60 to 90 percent of the light is directed downward with the remainder directed upward. General Diffuse or Direct-Indirect ( equal portions of the light are directed upward and downward. Highlighting ( the beam projection distance and focusing ability characterize this luminaire. The lighting distribution that is characteristic of a given luminaire is described using the candela distribution provided by the luminaire manufacturer (see diagram on next page). The candela distribution is represented by a curve on a polar graph showing the relative luminous intensity 360 around the fixture ( looking at a cross-section of the fixture. This information is useful because it shows how much light is emitted in each direction and the relative proportions of downlighting and uplighting. The cut-off angle is the angle, measured from straight down, where the fixture begins to shield the light source and no direct light from the source is visible. The shielding angle is the angle, measured from horizontal, through which the fixture provides shielding to prevent direct viewing of the light source. The shielding and cut-off angles add up to 90 degrees.

The lighting upgrade products mentioned in this document are described in more detail in Lighting Upgrade Technologies.

ELECTRIC MOTOR AND OVERCURRENT PROTECTION

Parts of a Motor Branch Circuit

A - Motor circuit conductor 125 % of current full load B' - Motor disconnecting means B - Motor Branch Circuit; over current protection usually 300% of full load current for safety switch (SS) or 250% of full load current for air circuit breaker (ACB) NOTE: These values are for general purposes motor only, values of B will depend on type and class of motor. The range is from 150% to 300% of the full load current, and shall in no case exceed 400% of full load current. C - Motor Controller - rated according to motor horsepower D - Motor Running Over current and overload protection: setting is from 115% to 125% of full load current. Value of 1.15 nd 1.25 is called service factor (SF) 1. Circuit Conductor - for the size of conductor having a maximum ampacity greater than the calculated value to protect the conductor from burning due to overheating. Faulty wiring is caused by the undersized conductor wire. 2. Branch Circuit Over current Protection - for protection against short circuit ground fault which cause over current flow. It should be capable of carrying the starting current of the motor. 3. Motor - controller ( magnetic contactor ) - used to start and stop the motor. It includes any switch or device capable of interrupting the stalled rotor current of motor. 4. Running Over current and Overload Protection - used to protect motor controller and motor against excessive heating due to motor overload and failure to start.

CIRCUIT FOR MOTOR LOAD Name plate of the motor 25 hp,220 volts, 3 phase; 3 wires 60 Hz, 0.84 power factor 90.5% efficiency

SOLUTION: 1. Solve for the current load: 1 horsepower = 746 watts

I = (load in hp x 746 w)/ (k x E x pf x n) where: k - 1.0 for 2 wire single phase DC 1.73 for 3 wire, 3 phase AC 2.0 for 3 wire single phase AC or DC 3.0 for 4 wires, 3 phase AC E - voltage between the neutral and live wire or between two live wires if no neutral line exists I - Current in any live wire except Neutral Line pf - power factor N - efficiency 2. Applying the formula I = (25 hp x 746)/(1.73 x 220 volts x .84 x 90.5%) = 64.45 A 3. Determine the Motor feeder. The current load of a motor multiplied by 125% (Code requirements) 4. Find the size of the conductor wire. Refer to Table, for 80.56 A, use any of the following: 3 - 38 mm2 THW or RHW copper wire 3 - 50 mm2 TW copper wire 3 - 80 mm2 TW aluminum or copper clad alum 3 - 50 mm2 THW or RHW aluminum or copper clad aluminum The allowable ampacities of the above wires in Table was derated by 80% to carry the 80.56 A current load. 5. Solve for the size of conduit pipe. Refer to Table, use 50 mm diameter pipe. 6. Determine the size or rating of the over current protection. The Code provides "The maximum over current protection for a single motor or a combination of motors should be, 250% of the ampacity of the largest motor plus the sum of the full load current of the other motors. Therefore: 64.45 A x 250% = 161.15 A minimum 7. Refer to Table. Use 150 A fuse or trip breaker. It is the nearest standard rating which does not exceed the 161.12 A current load as computed.

MATERIALS FOR MOTOR INSTALLATION: 1. 25 hp Induction motor 230 volts, 3 phase, 1800 rpm, 60 Hz at 40 degree temperature rise 2. Magnetic thermal overload control with contractors.

3. Service entrance cap 50 mm with locknut 4. 38 mm2 THW or RHW copper wire. 5. 50 mm2 diameter IMT or RSC conduit pipe 6. Conduit clamp with screw, 50 mm conduit pipe 7. TPST safety switch or circuit breaker 150 or 250 volts The quantity of materials depends upon the area and choice of the designing Engineer CALCULATION PROCEDURES IN FINDING THE SIZE OF FEEDER AND THE OVERLOAD CURRENT PROTECTION FOR A GROUP OF MOTORS Four 3 - phase motor 220 volts squirrel cage induction motor designed for 40 degree Celsius temperature rise at 1800 rpm, 60 Hz SOLUTION: 1. Determine the main feeder of the motors. Apply 25% of the biggest motor current load plus the sum of the other motors. (45 x 1.25) + 39 + 29 +21 = 145.25 A 2. Refer to Table. For the 145.25 A current load use any of the following conductor wires: 3 - 80 mm2 THW or RHW copper wire 3 - 100 mm2 TW copper wire 3- 125 mm2 THW or RHW aluminum or copper clad aluminum 3 - 150 mm2 TW clad aluminum

3. Determine the main over current protection. The National Electrical Code provides that: " The protection rating or setting of a motor shall be 250 % percent (maximum) of the full load current of the biggest motor being served plus the sum of the full load current of the other motors." (45 x 125%) x (250% +39 + 29 + 21) 140.625 + 89 = 229.625 A (maximum) 4. Refer to Table. Select a fuse or trip breaker that is nearest to standard rating that will not exceed 229.62 A. Use 200 A.

OVERLOAD PROTECTION Overcurrents An overcurrent exists when the normal load current for a circuit is exceeded. It can be in the form of an overload or short-circuit. When applied to motor circuits an overload is any current, flowing within the

normal circuit path, that is higher than the motor’s normal full load amperes (F.L.A.). A short-circuit is an overcurrent which greatly exceeds the normal full load current of the circuit. Also, as its name infers, a short-circuit leaves the normal current carrying path of the circuit and takes a “short cut” around the load and back to the power source. Motors can be damaged by both types of currents.Single-phasing, overworking and locked rotor conditions are just a few of the situations that can be protected against with the careful choice of protective devices. If left unprotected, motors will continue to operate even under abnormal conditions. The excessive current causes the motor to overheat, which in turn causes the motor winding insulation to deteriorate and ultimately fail. Good motor overload protection can greatly extend the useful life of a motor. Because of a motor’s characteristics, many common overcurrent devices actually offer limited or no protection. Motor Starting Currents When an AC motor is energized, a high inrush current occurs. Typically, during the initial half cycle, the inrush current is often higher than 20 times the normal full load current. After the first halfcycle the motor begins to rotate and the starting current subsides to 4 to 8 times the normal current for several seconds. As a motor reaches running speed, the current subsides to its normal running level. Typical motor starting characteristics are shown in Curve 1.

Because of this inrush, motors require special overload protective devices that can withstand the temporary overloads associated with starting currents and yet protect the motor from sustained overloads. There are four major types. Each offers varying degrees of protection. Fast Acting Fuses To offer overload protection, a protective device, depending on its application and the motor’s service factor (S.F.), should be sized at 115% or less of motor F.L.A. for 1.0 S.F. or 125% or less of motor F.L.A. for

1.15 or greater S.F. However, as shown in Curve 2, when fast-acting, non-time-delay fuses are sized to the recommended level the motors inrush will cause nuisance openings.

A fast-acting, non-time-delay fuse sized at 300% will allow the motor to start but sacrifices the overload protection of the motor. As shown by Curve 3 below, a sustained overload will damage the motor before the fuse can open.

COST ESTIMATION

ESTIMATION GUIDE Prepare paper, pencils, scale and rulers. Mark papers indicating the panel no., circuit number  and the location of the circuit run. Study plans, drawing and specifications. 2.1 Coordinate with Civil, Mechanical and Architectural Estimators about the following 2.1.1 Height between floors 2.1.2 Drop ceilings and ceiling supports 2.1.3 Height between finish floor and ceilings 2.1.4 Major beams and columns thru which conduits may not pass 2.1.5 Other architectural/civil/mechanical drawings indicating positions of the lights, special  outlets or aircon unit equipment. 2.2 Check and make a physical count of the following 2.2.1 Lighting fixtures – number of each type of fixtures 2.2.2 Convenience outlets – duplex 2.2.3 Special outlet 2.2.4 Panel boards – make a complete description of each panel board. The description should include: a. main breaker rating or lugs only b. no. of branches per ampere trip c. kAIC 2.2.5 Other electrical equipment to be supplied by contractor 2.3 Study carefully the circuit runs and the riser diagram together with the schedule of load.

Determine the approximate length of wire and conduit per circuit. 3.1 For the conduit (each circuit) ‐ measure the length from the last outlet to the panel using  the scale. The trace of the route must be followed as per drawing. 3.2 For the wire – measure the length between outlets and the length shall be multiplied by  the no. of wires. The sum of the products (lengths x the no of wires) shall be the  approximate length of wire. 3.3 Sum up the total length of conduit per size and divide by 3. Round off and add 10%. 3.4 Sum up the total length of wire for each size and divide by 150 to get the no of rolls.  Round off and add 10%. 3.5 Set aside the papers and data temporarily. Determine the approximate length of wire and conduct for the panel homerun to the main  panel or main distribution panel. 4.1 Conduits – measure the length of the run. Check the shortest possible route and avoid  obstructions. Total length divide 3 and add 5% 4.2 Wire – multiply the length of conduct by the following constants a. 2‐ for two‐wire single phase b. 3‐ for two‐wire single phase with neutral c. 3‐ for three‐wire, 3‐phase d. 4‐ for 3‐phase, 4 wire Boxes 5.1 Octagonal boxes – provide one box for each lighting fixtures 5.2 Utility box 4”x 2” – provide one box for each switch, duplex outlet or special outlet  (small) 5.3 Square box 4” – provide one box if the conduits terminating exceed 4 conduits or special  big outlets. 5.4 Square box 4 11/16 – provide one box for one‐inch diameter conduits or for special  purpose outlets. Also provide one box for multiple (6 or more) terminations.

5.5 Pull boxes – provide one box for every 18 meters of conduits length depending upon the  length of run. Other pull boxes may be designated by plans. Check with the  designer/consultant about the sizes. 6. Fittings 6.1 For PVC pipes 6.1.1 Couplings – provide 1 coupling for every length plus 1 coupling for every termination. 6.1.2 Elbows – provide 1‐900 elbow for every quarter turn for sizes of 32 mm (1 1/4”) and  above. 6.1.3 Cement – provide 1 can for every 10 length of conduit. 6.1.4 End bells – provide 1 for every termination. 6.2 For RSC conduits 6.2.1 Coupling – provide 1 additional coupling for every 5 lengths 6.2.2 Elbows‐ provide one 90‐degree elbow every 90‐degree turn for sizes of 25 mm (1”)  diameter above. 6.2.3 Locknut and bushing – provide one pair for every termination. 6.3 For electrical metallic tubing 6.3.1 Couplings‐ one set for every length 6.3.2 Elbows‐ use on site bended EMT 6.3.3 Adapters w/ locknut and bushing‐ for every termination are 1 adapter and 1 pair of  locknut and bushings. 6.4 Straps – two (2) straps for every length of conduit. In sizes of 25 mm diameter and above  use clamps especially for RSC or EMT conduits. 6.5 Other fittings 6.5.1 Wire trays/cable trays – check with drawings and consultant/designer 6.5.2 Cable trough / duets – check w/ drawings especially that the drawings may have  specific sizes.

7. Other Considerations 7.1 for lighting fixtures – add 1 m of wire for every termination or lighting fixtures 7.2 for convenience outlets a. Add 0.8 m. for every C.0. to the length of pipe and 1m of wire for every termination b. Add 0.4 m above the height of counters if the C.O. is above the counter in addition to the  height of the counter. Add also 1 m of wire for every termination. 7.3 for homeruns terminating a panel boards add 2 meters of wires for every circuit. 7.4 provide an empty conduit for every spare circuit per panel 7.5 normally the electronic and communication circuits will be in separate sheets and have  an ECE estimate 7.6 provide 1 connector for every termination # 6 up. 8. Summarize the lists of materials as follows 8.1 conduits – total of each size 8.2 fittings ‐ total of each type/size 8.3 boxes ‐ total of each type/size 8.4 panels – per panel and lowest canvassed price 8.5 Wires ‐ total length of each wire size 8.6 Connectors – (solderless less) total termination of each size for wire #6 wire and above. 8.7 Tape – 1 roll PVC tape for every 100 m of wire plus 1 roll of rubber tape for every 200m of  wire. 8.8 other materials must be itemized. 9. Costing 9.1 get the unit cost for each item and deduct all discounts. 9.2 from the total cost add 5% to 10% mark up

9.3 For all others materials like duets, panels, transfer switches, safety switches, and etc.‐  get the price from the fabricator net (less discounts) and add 5% markup. 10. Preparation of Bid or Asking Price 10.1 Material cost Conduits Fitting Boxes__________________________ Wires and wiring Devices________________________ Lighting fixtures _______________________________ Safety Devices_________________________________ Service Entrance and Mains_______________________ Others________________________________________ _______________________ Subtotal A 10.2 Labor Cost 10.1 If materials are imported a. Labor Cost is 20 % of subtotal A b. Supervision is 3% of subtotal A c. Mark‐up is 1.25 % of subtotal A 10.2 If the conduits and most materials are locally available a. Labor cost is 25 % to 30% of subtotal A b. Supervision cost is 4% to 5% of subtotal A c. Mark up cost is 2 % of subtotal A 10.3 Contingencies – an allowance of 5% to 7% of the total cost of materials and labor 10.4 Overhead – this include the cost of transportation, office staff tools and equipment  depreciation, papers and office supplies to representation, and cost of money.

‐ Normally 7% to 10% of the cost of materials is the cost of overhead. 10.5 Permits – show the plans to the municipal electrical engineer or his assistant and  request for an estimate. Add 5% to cover the exingencies. 10.6 a. the sum cost as computed in 10.1 and 10.5 is to be multiplied by 0.03 to get the  contractor’s tax. b. Add the contractor’s tax to the sum of sections 10.1 to 10.5 and round off. This will be your  bid price.

TERM TEST 1. A school classroom with general dimension of 24 x 30 feet is lighted with 10 fluorescent lamp 4F 40 T12 WW rapid start lamp. Calculate the initial and maintaned illumination in foot candle and lux, assuming that the coefficient of utilization is 0.35 and the maintenance factor is 0.70. 2. An office room with general dimension of 8 x 20 meters is to be lighted at an average maintaned foot candle of 50. How many 3 lamp fixtures of 120 centimeterslong F40 T12 WW rapid start fluorescent lamps are required assuming 0.38 cu and 0.75 mf?