ELECTRICAL DESIGN 1
THIS COURSE DEALS WITH THE STUDY OF ELECTRICAL SYSTEM
DESIGN, INSTALLATION AND COST ESTIMATION FOR SINGLE AND
MULTI-FAMILY DWELLING UNITS GUIDED BY THE PROVISIONS OF
THE PHILIPPINE ELECTRICAL CODE (PEC) AND OTHER RELEVANT
LAWS AND STNDARDS.
COST ESTIMATION
ESTIMATION GUIDE
1. Prepare paper, pencils, scale and rulers. Mark
papers indicating the panel no., circuit number
and the location of the circuit run.
2. 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.
3. 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.
4. 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
5. 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 90degree 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.
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. Determine the Motor feeder. The current load of a
motor multiplied by 125% (Code requirements)
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. Find the size of the conductor wire. Refer to Table,
for 80.56 A, use any of the following:
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.
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
CIRCUIT FOR MOTOR LOAD
The allowable ampacities of the above wires in Table
was derated by 80% to carry the 80.56 A current load.
Name plate of the motor
25 hp,220 volts, 3 phase; 3 wires
60 Hz, 0.84 power factor 90.5% efficiency
SOLUTION:
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.
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
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.
I - Current in any live wire except Neutral Line
3. Service entrance cap 50 mm with locknut
pf - power factor
4. 38 mm2 THW or RHW copper wire.
N - efficiency
5. 50 mm2 diameter IMT or RSC conduit pipe
2. Applying the formula
6. Conduit clamp with screw, 50 mm conduit pipe
I = (25 hp x 746)/(1.73 x 220 volts x .84 x 90.5%)
7. TPST safety switch or circuit breaker 150 or 250 volts
= 64.45 A
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
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.
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.
VERLOAD 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-
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.
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 footcandle) 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
(a) the type of luminaire,
(b) atmospheric conditions, and
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:
(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)
where LLMF = lamp lumen maintenance factor
2.2 Utilisation Factor
LMF = luminaire maintenance 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:
RSMF = room surface maintenance factor
2.1.1 Lamp Lumen Maintenance Factor
(a) light output ratio of luminaire
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:
(b) flux distribution of luminaire
(c) room proportions
(d) room reflectances
(a) by consulting a lamp manufacturer's catalog for a
lumen depreciation chart, and
(e) spacing/mounting height ratio
(b) by dividing the maintained lumens by the initial
lamps.
2.2.1 Light Output Ratio of Luminaire
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:
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 %
2.2.3 Room Proportion
A greater DLOR usually means a higher UF.
Room index (RI) is the ratio of room plan area to half
the wall area between the working and luminaire
planes.
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.
(4)
where L = length of room
Example 2
W = width of room
For data given in Example 1 determine upward flux
fraction (UFF), downward flux fraction (DFF) and flux
fraction ratio (FRR).
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
Figure 5 Flux Fraction of Various Luminaires
(c) the floor cavity (or the horizontal working plane).
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,
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
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).
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
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
(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)
Working plane at 0.7 m above floor
Reflection factors: Ceiling 70 %
Walls 50 %
Working plane 20 %
3. Summary of Procedures for Lumen Design Method
Light Loss factor: 0.779
(a) Calculate the room index.
Luminaires: 1800 mm twin tube with opal diffuser
(b) Determine the effective reflectances of the ceiling
cavity, walls and floor cavity.
Ceiling mounted
Downward light output ratio 36 %
(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.
SHR MAX 1.60 : 1
SHR NOM 1.50 : 1
(d) Determine the light loss factor.
Dimensions : 1800 mm long x 200 mm wide
(e) Inert the appropriate variables into the lumen
method formula to obtain the number of luminaires
required.
Lamps: 1800 mm 75 W plus white
(f) Determine a suitable layout.
2 lamps per luminaire
5800 average initial lumens per lamp
(g) Check that the geometric mean spacing to height
ratio of the layout is within the SHR NOM range:
Solution
(a) Initial calculation
(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.
From manufacturer's photometric data sheet (Table 3),
utilisation factor (UF) is 0.5336 by interpolation.
Therefore, the number of luminairs is 10.
Example 3
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.
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
2. Refer to Table Demand load for household. For
electric range, apply 80% demand factor.
1. Solve for the total current load.
8 receptacles x 2 gang per outlet x 180 watts = 2880
watts
Total load x demand factor (Df)
5100 watts x .80 = 4080 watts
3. Compute for the line current load. Divide:
2. Solve for the appliance current load. Divide.
4080 watts/230 volts = 17.74 A
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. 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.
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.
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.
5. Determine the size of the conduit pipe. Refer to Table
. For No. 12 TW wire, use 13 mm diameter pipe.
E. Determine the sub-feeder per dwelling
1. Solve for the total connected load per dwelling.
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.
Lighting load ........................................ 1920 watts
Convenience Load ,............................... 2880 watts
Other loads 5.1 + 1.5 kW........................ 6600 watts
C. Circuit - 3 Other Load
TOTAL....................................11400 watts
1. Laundry Circuit at 1500 watts per circuit (PEC
provision)
2. Apply 80% demand factor (see Table)
TOTAL LINE CURRENT = (11400 watts x.80 df)/230 volts
1500 watts/230 volys = 6.52 A
= 39.65 A
2. Find the size of the branch circuit conductor. From
Table, use 3.5 mm2 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.
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.
D. Circuit - 4 Cooking Unit
1. Total Load is 5.1 kW = 5100 watts
F. Determine the Size of the Main Feeder
2. The quality of light
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:
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.
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 238 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.
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
ILLUMINATION CALCULATION AND DESIGN FOR SINGLE
FAMILY DWELLING
PRINCIPLES OF ILLUMINATION
6 – 1 INTRODUCTION
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.
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.
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.
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
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.
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.
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.
The quantity of light level of illumination can be easily
measured or calculated with the aid of portable foot
candle meter.
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:
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
illumination or 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 as Maintenance 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 X 3,200
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
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.
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
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 directindirect 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 611. Therefore, the road is considered as adequeately
lighted.
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
maximum 100 watts per light outlet. therefore the use
of 2.0 mm2 or No.14 TW is safe.
B, Circuit - 2 For Small Appliance load
SOLUTION
SOLUTION
The National Electrical Code provides that:
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."
"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:
8 outlets x 100 = 800 watts
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:
2. Determine the Total Current load
6 outlets x 2 plug = 12 pieces
800 watts/230 volts = 3.48 amperes
2. Solve for the Total Estimated Load
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 wire having an ampacity of 15 amperes
that is much largr than the 3.48 amperes computed
maximum load.
12 x 180 watts per outlet = 2,160 watts
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, specify
13 mm diameter conduit pipe.
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
Adopting the 100 watts per lighting outlet we have:
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
3. Determine the Maximum Expected Current Load:
2,160 watts / 230 volts = 9.39 amperes
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 wiring
except, for an appliance with limited load wherein a 2.0
mm2or No.14 AWG wire is permitted.
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
C. Circuit - 3 for other loads
SOLUTION
2. Apply 80% demand factor as permitted by the
National Electrical Code.
1. Examining Figure above, other loads are:
28.52 x .80 = 22.8 amperes
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:
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 8.0 mm2or
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.
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
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.
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
MULTI-ground system and line to line service
1000 watts/230 volts = 4.35 amperes
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)
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)
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.
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
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.
Section 3.3.1.2 of the the PEC specif 180 watts load limit
per convenience outlet. Thus,
F. Circuit - 6 for Water heater Load
1. Find the number of appliance load outlet and the
current load.
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.
1. one unit water heater at 2.5 kW = 2500 watts
2. The current load will be:
2500 watts/230 volts = 10.86 A
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.
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
E. Circuit - 5 for Range Load
4.86 A x 125% = 6.07 A
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
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. Specify No.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
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:
1. Find The total current load of circuit 1 to circuit 8:
" If the computed load exceeds 10000 watts, the
conductor and overcurrent protection shall be rated not
less than 100 A.
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
THEREFORE USE 2 - 38 mm2 TW WIRE FOR THE MAIN
FEEDER AND 2 - 100 A FOR THE MAIN BREAKER.
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
Small Family Dwelling
Type of Service - 230 volts; two wire
Line to Ground system
subtract: 16720 - 10000 = 6720 watts
for other load, multiply by 40%
6720 x .4 .........................................2688 watts
A single family dwelling with a floor area of 80 square
meters has the following receptacles and outlets load.
Aircon unit at 100% demand factor
2-units at 1119 watts........................2238 watts
LIGHTING:
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 use 1- 22 mm2 Tw
copper wire for the Neutral line.
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:
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.
1920 watts/ 230 volts = 8.35 A
3. Compute the actual lighting load. Multiply:
FINDING THE SIZE OF THE SERVICE ENTRANCE OR
FEEDER
7 - Fluorescent lamps x 40 watts = 280 watts
2 - Incandescent bulb x 60 watts = 120 watts
TOTAL............ 400 watts
1. Get the sum total of connected load. Add:
4. Solve the actual current load. Divide:
Lighting Load.................................... 1920 watts
Small appliance load ......................... 3175 watts
TOTAL.................................. 5095 watts
400 watts/230 volts = 1.74 A
2. Solve for the total connected load current. Divide:
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.
5095 watts/230 volts = 22.15 A
3. Find the size of Service Entrance. Refer to Table. For
22.15 A, use No. 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.
PEC REQUIREMENTS FOR ADEQUATE WIRING IN SINGLE
AND MULTI-FAMILY DWELLING UNIT
GENERAL LIGHTING LOADS BY OCCUPANCIES
(Table 1.1)
* All receptacle outlets of 20-ampere or less in onefamily, 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)
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 branch-circuit load for a countermounted cooking unit and not more than two wallmounted 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.
DEMAND LOADS FOR HOUSEHOLD ELECTRIC RANGES,
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.
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 BRANCHCIRCUIT
SHORT-CIRCUIT AND GROUND-FAULT PROTECTIVE
DEVICES
(Table 8.1)
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 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)
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 10.1)
AMPACITY CORRECTION FACTORS
(Table 16.1)
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 mm2 aluminum and copper
clad aluminum.
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 30ampere 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 50ampere 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 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:
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.
(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
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.
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.
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.
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.
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:
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 3wire 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
(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.
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
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 106 in.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
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 onethousandth 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
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.
web designed specifically for field installation in metal
surface raceway. Cables of this type are the types FC.
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.
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.
4. Nonmetallic Sheathed Cable. Types NM and NMC are
factory assembled two or more insulated conductors
having a moisture-resistant outer sheath, flameretardant and non-metallic material. These types are
used specifically for one or two dwelling not exceeding
3 storey buildings.
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.
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 flame-retardant material,
covered with an overlapping spiral metal tape. This type
is used in hazardous locations and in cable trays or in
raceways.
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
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.
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.
OUTLETS, RECEPTACLES and other WIRING DEVICES
C. TYPE XHHW
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.
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 “crosslinked 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
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.
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-fireresistant compounds; if it is set on fire with a blowtorch,
the flame dies out when the torch is removed.
E. OTHER TYPES
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.
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
B. TYPES THHN, THWN
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 coldstorage 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.
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.
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.
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.
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 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.
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
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.
THE OVERHEAD SERVICE-DROP CONDUCTOR
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.
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.
b) Class III, Division 2. A Class III, Division 2 location is a
location in which easily ignitible fibers are stored or
handled.
SIZE AND RATING:
ELECTRIC CIRCUITS IN BUILDING
* SERVICES *
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.
No. of Service:
A building or other structure served shall be supplied by
only one service.
b) Minimum Size. The conductors shall not be smaller
than 8 mm2 copper, 14 mm2 aluminum or copper-clad
aluminum.
*EXCEPTIONS*
CLEARANCES:
1. For fire pump where a separate service is required.
a) Above Roofs. Conductors shall have a vertical
clearance of not less than 2,500 mm from the roof
surface.
2. For emergency electrical system where a separate
service is required.
b) Vertical Clearance from Ground.
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.
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 service-entrance 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.
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.
*EXCEPTIONS*
A grounded conductor shall be permitted to be
uninsulated as follows:
TYPES:
a) Bare copper used in a raceway.
1. Overhead Service Entrance
b) Bare copper for direct burial where bare copper is
judged to be suitable for the soil conditions.
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.
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
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.
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 copperclad aluminum.
SERVICE - ENTRANCE CONDUCTORS
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.
No. of Service-Entrance Conductor Sets
Each service drop or lateral shall supply only one set of
service-entrance conductors.
*EXCEPTIONS:
EXCEPTION: For installations to supply only limited
loads of a single branch circuit such as small polyphase
1. Buildings with more than one occupancy.
PROTECTION:
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 2wire 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 copper-clad 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
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 short-circuit 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.
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
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) 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.
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.
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.
Advantages of circuit breaker over a fuse
FEEDERS AND MAIN
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
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.
Principles applied in installing panel board
1. The approach should be accessible and convenient.
4. Feeders requiring more than 50 mm diameter
conduit should not be used.
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.
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
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.
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.
PROTECTION OF THE BRANCH CIRCUIT
Grounding could be accomplished in the following
manner:
1. Connection to a buried cold water main.
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)
Any current in excess of the rated current capacity of
the equipment or the rated ampacity of the conductor
is called overcurrent.
The causes of overcurrent are:
1. Overload in the equipment conductors.
2. Connection to a rod or group of rods.
3. Connection to a buried ground plate.
Ampacity - is the current-carrying capacity of an electric
conductor.
THE PANELBOARD
CIRCUITRY DESIGN
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.
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.
·
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