1. A school classroom with general dimension of 24 x 30 feet is lighted with 10 fluorescent lamp 4F 40 T12 WW rapid start lamp. Calculate the initial and maintaned illumination in foot candle and lux, assuming that the coefficient of utilization is 0.35 and the maintenance factor is 0.70.
2. An office room with general dimension of 8 x 20 meters is to be lighted at an average maintaned foot candle of 50. How many 3 lamp fixtures of 120 centimeterslong F40 T12 WW rapid start fluorescent lamps are required assuming 0.38 cu and 0.75 mf?
Sunday, February 28, 2010
Sunday, February 21, 2010
COST ESTIMATION
ESTIMATION GUIDE
- Prepare paper, pencils, scale and rulers. Mark papers indicating the panel no., circuit number and the location of the circuit run.
- Study plans, drawing and specifications.
2.1 Coordinate with Civil, Mechanical and Architectural Estimators about the following
2.1.1 Height between floors
2.1.2 Drop ceilings and ceiling supports
2.1.3 Height between finish floor and ceilings
2.1.4 Major beams and columns thru which conduits may not pass
2.1.5 Other architectural/civil/mechanical drawings indicating positions of the lights, special outlets or aircon unit equipment.
2.2 Check and make a physical count of the following
2.2.1 Lighting fixtures – number of each type of fixtures
2.2.2 Convenience outlets – duplex
2.2.3 Special outlet
2.2.4 Panel boards – make a complete description of each panel board.
The description should include:
a. main breaker rating or lugs only
b. no. of branches per ampere trip
c. kAIC
2.2.5 Other electrical equipment to be supplied by contractor
2.3 Study carefully the circuit runs and the riser diagram together with the schedule of load.
- Determine the approximate length of wire and conduit per circuit.
3.1 For the conduit (each circuit) - measure the length from the last outlet to the panel using the scale. The trace of the route must be followed as per drawing.
3.2 For the wire – measure the length between outlets and the length shall be multiplied by the no. of wires. The sum of the products (lengths x the no of wires) shall be the approximate length of wire.
3.3 Sum up the total length of conduit per size and divide by 3. Round off and add 10%.
3.4 Sum up the total length of wire for each size and divide by 150 to get the no of rolls. Round off and add 10%.
3.5 Set aside the papers and data temporarily.
- Determine the approximate length of wire and conduct for the panel homerun to the main panel or main distribution panel.
4.1 Conduits – measure the length of the run. Check the shortest possible route and avoid obstructions. Total length divide 3 and add 5%
4.2 Wire – multiply the length of conduct by the following constants
a. 2- for two-wire single phase
b. 3- for two-wire single phase with neutral
c. 3- for three-wire, 3-phase
d. 4- for 3-phase, 4 wire
- Boxes
5.1 Octagonal boxes – provide one box for each lighting fixtures
5.2 Utility box 4”x 2” – provide one box for each switch, duplex outlet or special outlet (small)
5.3 Square box 4” – provide one box if the conduits terminating exceed 4 conduits or special big outlets.
5.4 Square
5.5 Pull boxes – provide one box for every 18 meters of conduits length depending upon the length of run. Other pull boxes may be designated by plans. Check with the designer/consultant about the sizes.
6. Fittings
6.1 For PVC pipes
6.1.1 Couplings – provide 1 coupling for every length plus 1 coupling for every termination.
6.1.2 Elbows – provide 1-900 elbow for every quarter turn for sizes of 32 mm (1 1/4”) and above.
6.1.3 Cement – provide 1 can for every 10 length of conduit.
6.1.4 End bells – provide 1 for every termination.
6.2 For RSC conduits
6.2.1 Coupling – provide 1 additional coupling for every 5 lengths
6.2.2 Elbows- provide one 90-degree elbow every 90-degree turn for sizes of 25 mm (1”) diameter above.
6.2.3 Locknut and bushing – provide one pair for every termination.
6.3 For electrical metallic tubing
6.3.1 Couplings- one set for every length
6.3.2 Elbows- use on site bended EMT
6.3.3 Adapters w/ locknut and bushing- for every termination are 1 adapter and 1 pair of locknut and bushings.
6.4 Straps – two (2) straps for every length of conduit. In sizes of 25 mm diameter and above use clamps especially for RSC or EMT conduits.
6.5 Other fittings
6.5.1 Wire trays/cable trays – check with drawings and consultant/designer
6.5.2 Cable trough / duets – check w/ drawings especially that the drawings may have specific sizes.
7. Other Considerations
7.1 for lighting fixtures – add 1 m of wire for every termination or lighting fixtures
7.2 for convenience outlets
a. Add 0.8 m. for every C.0. to the length of pipe and 1m of wire for every termination
b. Add 0.4 m above the height of counters if the C.O. is above the counter in addition to the height of the counter. Add also 1 m of wire for every termination.
7.3 for homeruns terminating a panel boards add 2 meters of wires for every circuit.
7.4 provide an empty conduit for every spare circuit per panel
7.5 normally the electronic and communication circuits will be in separate sheets and have an ECE estimate
7.6 provide 1 connector for every termination # 6 up.
8. Summarize the lists of materials as follows
8.1 conduits – total of each size
8.2 fittings - total of each type/size
8.3 boxes - total of each type/size
8.4 panels – per panel and lowest canvassed price
8.5 Wires - total length of each wire size
8.6 Connectors – (solderless less) total termination of each size for wire #6 wire and above.
8.7 Tape – 1 roll PVC tape for every 100 m of wire plus 1 roll of rubber tape for every 200m of wire.
8.8 other materials must be itemized.
9. Costing
9.1 get the unit cost for each item and deduct all discounts.
9.2 from the total cost add 5% to 10% mark up
9.3 For all others materials like duets, panels, transfer switches, safety switches, and etc.- get the price from the fabricator net (less discounts) and add 5% markup.
10. Preparation of Bid or Asking Price
10.1 Material cost
Conduits Fitting Boxes__________________________
Wires and wiring Devices________________________
Lighting fixtures _______________________________
Safety Devices_________________________________
Service Entrance and Mains_______________________
Others________________________________________
_______________________
Subtotal A
10.2 Labor Cost
10.1 If materials are imported
a. Labor Cost is 20 % of subtotal A
b. Supervision is 3% of subtotal A
c. Mark-up is 1.25 % of subtotal A
10.2 If the conduits and most materials are locally available
a. Labor cost is 25 % to 30% of subtotal A
b. Supervision cost is 4% to 5% of subtotal A
c. Mark up cost is 2 % of subtotal A
10.3 Contingencies – an allowance of 5% to 7% of the total cost of materials and labor
10.4 Overhead – this include the cost of transportation, office staff tools and equipment depreciation, papers and office supplies to representation, and cost of money.
- Normally 7% to 10% of the cost of materials is the cost of overhead.
10.5 Permits – show the plans to the municipal electrical engineer or his assistant and request for an estimate. Add 5% to cover the exingencies.
10.6 a. the sum cost as computed in 10.1 and 10.5 is to be multiplied by 0.03 to get the contractor’s tax.
b. Add the contractor’s tax to the sum of sections 10.1 to 10.5 and round off. This will be your bid price.
ELECTRIC MOTOR AND OVERCURRENT PROTECTION
Parts of a Motor Branch Circuit
CIRCUIT FOR MOTOR LOAD
just a few of the situations that can be protected against with the careful choice of protective devices. If left unprotected, motors will continue to operate even under abnormal conditions. The excessive current causes the motor to overheat, which in turn causes the motor winding insulation to deteriorate and ultimately fail. Good motor overload protection can greatly extend the useful life of a motor. Because of a motor’s characteristics, many common overcurrent devices actually offer limited or no protection.
Motor Starting Currents
When an AC motor is energized, a high inrush current occurs. Typically, during the initial half cycle, the inrush current is often higher than 20 times the normal full load current. After the first halfcycle the motor begins to rotate and the starting current subsides to 4 to 8 times the normal current for several seconds. As a motor reaches running speed, the current subsides to its normal running level. Typical motor starting characteristics are shown in Curve 1.
Because of this inrush, motors require special overload protective devices that can withstand the temporary overloads associated with starting currents and yet protect the motor from sustained overloads. There are four major types. Each offers varying degrees of protection.
Fast Acting Fuses
To offer overload protection, a protective device, depending on its application and the motor’s service factor (S.F.), should be sized at 115% or less of motor F.L.A. for 1.0 S.F. or 125% or less of motor F.L.A. for 1.15 or greater S.F. However, as shown in Curve 2, when fast-acting, non-time-delay fuses are sized to the recommended level the motors inrush will cause nuisance openings.
A fast-acting, non-time-delay fuse sized at 300% will allow the motor to start but sacrifices the overload protection of the motor. As shown by Curve 3 below, a sustained overload will damage the motor before the fuse can open.
A - Motor circuit conductor 125 % of current full load
B' - Motor disconnecting means
B - Motor Branch Circuit; over current protection usually 300% of full load current for safety switch (SS) or 250% of full load current for air circuit breaker (ACB)
NOTE: These values are for general purposes motor only, values of B will depend on type and class of motor. The range is from 150% to 300% of the full load current, and shall in no case exceed 400% of full load current.
C - Motor Controller - rated according to motor horsepower
D - Motor Running Over current and overload protection: setting is from 115% to 125% of full load current. Value of 1.15 nd 1.25 is called service factor (SF)
1. Circuit Conductor - for the size of conductor having a maximum ampacity greater than the calculated value to protect the conductor from burning due to overheating. Faulty wiring is caused by the undersized conductor wire.
2. Branch Circuit Over current Protection - for protection against short circuit ground fault which cause over current flow. It should be capable of carrying the starting current of the motor.
3. Motor - controller ( magnetic contactor ) - used to start and stop the motor. It includes any switch or device capable of interrupting the stalled rotor current of motor.
4. Running Over current and Overload Protection - used to protect motor controller and motor against excessive heating due to motor overload and failure to start.
B' - Motor disconnecting means
B - Motor Branch Circuit; over current protection usually 300% of full load current for safety switch (SS) or 250% of full load current for air circuit breaker (ACB)
NOTE: These values are for general purposes motor only, values of B will depend on type and class of motor. The range is from 150% to 300% of the full load current, and shall in no case exceed 400% of full load current.
C - Motor Controller - rated according to motor horsepower
D - Motor Running Over current and overload protection: setting is from 115% to 125% of full load current. Value of 1.15 nd 1.25 is called service factor (SF)
1. Circuit Conductor - for the size of conductor having a maximum ampacity greater than the calculated value to protect the conductor from burning due to overheating. Faulty wiring is caused by the undersized conductor wire.
2. Branch Circuit Over current Protection - for protection against short circuit ground fault which cause over current flow. It should be capable of carrying the starting current of the motor.
3. Motor - controller ( magnetic contactor ) - used to start and stop the motor. It includes any switch or device capable of interrupting the stalled rotor current of motor.
4. Running Over current and Overload Protection - used to protect motor controller and motor against excessive heating due to motor overload and failure to start.
CIRCUIT FOR MOTOR LOAD
Name plate of the motor
25 hp,220 volts, 3 phase; 3 wires
60 Hz, 0.84 power factor 90.5% efficiency
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.
25 hp,220 volts, 3 phase; 3 wires
60 Hz, 0.84 power factor 90.5% efficiency
SOLUTION:
1. Solve for the current load:
1 horsepower = 746 watts
I = (load in hp x 746 w)/ (k x E x pf x n)
where:
k - 1.0 for 2 wire single phase DC
1.73 for 3 wire, 3 phase AC
2.0 for 3 wire single phase AC or DC
3.0 for 4 wires, 3 phase AC
E - voltage between the neutral and live wire or between two live wires if no neutral line exists
I - Current in any live wire except Neutral Line
pf - power factor
N - efficiency
2. Applying the formula
I = (25 hp x 746)/(1.73 x 220 volts x .84 x 90.5%)
= 64.45 A
3. Determine the Motor feeder. The current load of a motor multiplied by 125% (Code requirements)
4. Find the size of the conductor wire. Refer to Table, for 80.56 A, use any of the following:
3 - 38 mm2 THW or RHW copper wire
3 - 50 mm2 TW copper wire
3 - 80 mm2 TW aluminum or copper clad alum
3 - 50 mm2 THW or RHW aluminum or copper clad aluminum
The allowable ampacities of the above wires in Table was derated by 80% to carry the 80.56 A current load.
5. Solve for the size of conduit pipe. Refer to Table, use 50 mm diameter pipe.
6. Determine the size or rating of the over current protection. The Code provides "The maximum over current protection for a single motor or a combination of motors should be, 250% of the ampacity of the largest motor plus the sum of the full load current of the other motors.
Therefore:
64.45 A x 250% = 161.15 A minimum
7. Refer to Table. Use 150 A fuse or trip breaker. It is the nearest standard rating which does not exceed the 161.12 A current load as computed.
MATERIALS FOR MOTOR INSTALLATION:
1. 25 hp Induction motor 230 volts, 3 phase, 1800 rpm, 60 Hz at 40 degree temperature rise
2. Magnetic thermal overload control with contractors.
3. Service entrance cap 50 mm with locknut
4. 38 mm2 THW or RHW copper wire.
5. 50 mm2 diameter IMT or RSC conduit pipe
6. Conduit clamp with screw, 50 mm conduit pipe
7. TPST safety switch or circuit breaker 150 or 250 volts
The quantity of materials depends upon the area and choice of the designing Engineer
CALCULATION PROCEDURES IN FINDING THE SIZE OF FEEDER AND THE OVERLOAD CURRENT PROTECTION FOR A GROUP OF MOTORS
Four 3 - phase motor 220 volts squirrel cage induction motor designed for 40 degree Celsius temperature rise at 1800 rpm, 60 Hz
SOLUTION:
1. Determine the main feeder of the motors. Apply 25% of the biggest motor current load plus the sum of the other motors.
(45 x 1.25) + 39 + 29 +21
= 145.25 A
2. Refer to Table. For the 145.25 A current load use any of the following conductor wires:
3 - 80 mm2 THW or RHW copper wire
3 - 100 mm2 TW copper wire
3- 125 mm2 THW or RHW aluminum or copper clad aluminum
3 - 150 mm2 TW clad aluminum
1. Solve for the current load:
1 horsepower = 746 watts
I = (load in hp x 746 w)/ (k x E x pf x n)
where:
k - 1.0 for 2 wire single phase DC
1.73 for 3 wire, 3 phase AC
2.0 for 3 wire single phase AC or DC
3.0 for 4 wires, 3 phase AC
E - voltage between the neutral and live wire or between two live wires if no neutral line exists
I - Current in any live wire except Neutral Line
pf - power factor
N - efficiency
2. Applying the formula
I = (25 hp x 746)/(1.73 x 220 volts x .84 x 90.5%)
= 64.45 A
3. Determine the Motor feeder. The current load of a motor multiplied by 125% (Code requirements)
4. Find the size of the conductor wire. Refer to Table, for 80.56 A, use any of the following:
3 - 38 mm2 THW or RHW copper wire
3 - 50 mm2 TW copper wire
3 - 80 mm2 TW aluminum or copper clad alum
3 - 50 mm2 THW or RHW aluminum or copper clad aluminum
The allowable ampacities of the above wires in Table was derated by 80% to carry the 80.56 A current load.
5. Solve for the size of conduit pipe. Refer to Table, use 50 mm diameter pipe.
6. Determine the size or rating of the over current protection. The Code provides "The maximum over current protection for a single motor or a combination of motors should be, 250% of the ampacity of the largest motor plus the sum of the full load current of the other motors.
Therefore:
64.45 A x 250% = 161.15 A minimum
7. Refer to Table. Use 150 A fuse or trip breaker. It is the nearest standard rating which does not exceed the 161.12 A current load as computed.
MATERIALS FOR MOTOR INSTALLATION:
1. 25 hp Induction motor 230 volts, 3 phase, 1800 rpm, 60 Hz at 40 degree temperature rise
2. Magnetic thermal overload control with contractors.
3. Service entrance cap 50 mm with locknut
4. 38 mm2 THW or RHW copper wire.
5. 50 mm2 diameter IMT or RSC conduit pipe
6. Conduit clamp with screw, 50 mm conduit pipe
7. TPST safety switch or circuit breaker 150 or 250 volts
The quantity of materials depends upon the area and choice of the designing Engineer
CALCULATION PROCEDURES IN FINDING THE SIZE OF FEEDER AND THE OVERLOAD CURRENT PROTECTION FOR A GROUP OF MOTORS
Four 3 - phase motor 220 volts squirrel cage induction motor designed for 40 degree Celsius temperature rise at 1800 rpm, 60 Hz
SOLUTION:
1. Determine the main feeder of the motors. Apply 25% of the biggest motor current load plus the sum of the other motors.
(45 x 1.25) + 39 + 29 +21
= 145.25 A
2. Refer to Table. For the 145.25 A current load use any of the following conductor wires:
3 - 80 mm2 THW or RHW copper wire
3 - 100 mm2 TW copper wire
3- 125 mm2 THW or RHW aluminum or copper clad aluminum
3 - 150 mm2 TW clad aluminum
3. Determine the main over current protection. The National Electrical Code provides that:
" The protection rating or setting of a motor shall be 250 % percent (maximum) of the full load current of the biggest motor being served plus the sum of the full load current of the other motors."
(45 x 125%) x (250% +39 + 29 + 21)
140.625 + 89 = 229.625 A (maximum)
4. Refer to Table. Select a fuse or trip breaker that is nearest to standard rating that will not exceed 229.62 A. Use 200 A.
OVERLOAD PROTECTION
Overcurrents
An overcurrent exists when the normal load current for a circuit is exceeded. It can be in the form of an overload or short-circuit. When applied to motor circuits an overload is any current, flowing within the normal circuit path, that is higher than the motor’s normal full load amperes (F.L.A.). A short-circuit is an overcurrent which greatly exceeds the normal full load current of the circuit. Also, as its name infers, a short-circuit leaves the normal current carrying path of the circuit and takes a “short cut” around the load and back to the power source. Motors can be damaged by both types of currents.Single-phasing, overworking and locked rotor conditions areOvercurrents
just a few of the situations that can be protected against with the careful choice of protective devices. If left unprotected, motors will continue to operate even under abnormal conditions. The excessive current causes the motor to overheat, which in turn causes the motor winding insulation to deteriorate and ultimately fail. Good motor overload protection can greatly extend the useful life of a motor. Because of a motor’s characteristics, many common overcurrent devices actually offer limited or no protection.
Motor Starting Currents
When an AC motor is energized, a high inrush current occurs. Typically, during the initial half cycle, the inrush current is often higher than 20 times the normal full load current. After the first halfcycle the motor begins to rotate and the starting current subsides to 4 to 8 times the normal current for several seconds. As a motor reaches running speed, the current subsides to its normal running level. Typical motor starting characteristics are shown in Curve 1.
Because of this inrush, motors require special overload protective devices that can withstand the temporary overloads associated with starting currents and yet protect the motor from sustained overloads. There are four major types. Each offers varying degrees of protection.
Fast Acting Fuses
To offer overload protection, a protective device, depending on its application and the motor’s service factor (S.F.), should be sized at 115% or less of motor F.L.A. for 1.0 S.F. or 125% or less of motor F.L.A. for 1.15 or greater S.F. However, as shown in Curve 2, when fast-acting, non-time-delay fuses are sized to the recommended level the motors inrush will cause nuisance openings.
A fast-acting, non-time-delay fuse sized at 300% will allow the motor to start but sacrifices the overload protection of the motor. As shown by Curve 3 below, a sustained overload will damage the motor before the fuse can open.