901.12 Electrical Components - Revision history

Hoskir: /* 901.12.1 Voltage Drop and Wire Sizes */ updated per RR3881

2024-05-29T13:27:09Z

901.12.1 Voltage Drop and Wire Sizes: updated per RR3881

← Older revision		Revision as of 08:27, 29 May 2024
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	Since the resistance of a wire conductor is inversely proportional to the cross-sectional area of the wire, increased voltage drops are developed as the wire diameters decrease.		Since the resistance of a wire conductor is inversely proportional to the cross-sectional area of the wire, increased voltage drops are developed as the wire diameters decrease.

	The areas of wires in circular mils (mm<sup>2</sup>) and the resistance in ohms per 1000 ft. (1000 m) to be used in calculations are shown in ~~the~~ [[~~901~~.~~10 Preparation of Plans~~#~~901~~.10.~~3 Cable and Conduit~~\|~~tables of EPG 901~~.10.3 Cable ~~and Conduit~~]]. The voltage drop in any line is calculated from the formula:		The areas of wires in circular mils (mm<sup>2</sup>) and the resistance in ohms per 1000 ft. (1000 m) to be used in calculations are shown in [[902.15_Designing_a_Traffic_Signal#902.15.6.2_Power_Cable\|902.15.6.2 Power Cable]]. The voltage drop in any line is calculated from the formula:

	: ENGLISH		: ENGLISH
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	::: Total Area = 6.22 mm<sup>2</sup>.		::: Total Area = 6.22 mm<sup>2</sup>.

	The wire size is that with the area next above, as shown in Table ~~1 (8-01~~.~~1) of Figure 901~~.~~10 (8-01~~.~~11)~~ or #8 AWG (10 mm<sup>2</sup>) (The minimum size used in cable-conduit is #8 AWG (10 mm<sup>2</sup>)). When several circuits are carried from one control station, determine the wire size for each circuit separately. #10 AWG (6 mm<sup>2</sup>) pole and bracket cable is used in all installations regardless of the size of the conductor cable. To check the preceding result, use the basic formula:		The wire size is that with the area next above, as shown in [[902.15_Designing_a_Traffic_Signal#Table_902.15.6.2.1_Properties_for_Conductors\|Table 902.15.6.2.1 Properties for Conductors]] or #8 AWG (10 mm<sup>2</sup>) (The minimum size used in cable-conduit is #8 AWG (10 mm<sup>2</sup>)). When several circuits are carried from one control station, determine the wire size for each circuit separately. #10 AWG (6 mm<sup>2</sup>) pole and bracket cable is used in all installations regardless of the size of the conductor cable. To check the preceding result, use the basic formula:

	Voltage Drop = Current x Resistance		Voltage Drop = Current x Resistance

	To determine the current in amperes, use length of wires (2 x distance) x number of luminaires x line amperes. Table ~~8-01~~.1 ~~of Figure 8-01.11~~ shows the resistance of #8 AWG (10 mm<sup>2</sup>) wire in ohms/1000 ft = 0.78 (ohms/1000 m = 2.2).		To determine the current in amperes, use length of wires (2 x distance) x number of luminaires x line amperes. [[902.15_Designing_a_Traffic_Signal#Table_902.15.6.2.1_Properties_for_Conductors\|Table 902.15.6.2.1 Properties for Conductors]] shows the resistance of #8 AWG (10 mm<sup>2</sup>) wire in ohms/1000 ft = 0.78 (ohms/1000 m = 2.2).

	Therefore:		Therefore:

Smithk: /* 901.12.1 Voltage Drop and Wire Sizes */

2020-08-14T17:13:46Z

901.12.1 Voltage Drop and Wire Sizes

← Older revision		Revision as of 12:13, 14 August 2020
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	* High Mast Light Poles - #2 AWG (35 mm<sup>2</sup>) MAXIMUM		* High Mast Light Poles - #2 AWG (35 mm<sup>2</sup>) MAXIMUM

	If circuit loading or voltage drop requires a larger wire size than the maximum then one or more of the following can be considered:		If circuit loading or voltage drop requires a larger wire size than the maximum, then one or more of the following can be considered:

	* Increase the system voltage from 240 volts to 480 volts.		* Increase the system voltage from 240 volts to 480 volts.

EPGsysop: Per TS, replaced LED wattage references with A,B and C nomenclature

2018-02-13T16:53:57Z

Per TS, replaced LED wattage references with A,B and C nomenclature

← Older revision		Revision as of 11:53, 13 February 2018
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	'''901.12.1.2 Example to Determine Necessary Wire Cable Size'''		'''901.12.1.2 Example to Determine Necessary Wire Cable Size'''

	Given: Five ~~170 watt~~ LED luminaires on a 480-volt circuit at 1200, 1700, 3700, 4300 and 4600 ft. (350, 500, 1100, 1300 and 1400 m), respectively, from the control station.		Given: Five LED-B luminaires on a 480-volt circuit at 1200, 1700, 3700, 4300 and 4600 ft. (350, 500, 1100, 1300 and 1400 m), respectively, from the control station.

	There must be sufficient current to operate the first luminaries as well as the next four luminaires, etc. Operating current is obtained from the manufacturer of the luminaire. For our example we will assume 0.71 amps.		There must be sufficient current to operate the first luminaries as well as the next four luminaires, etc. Operating current is obtained from the manufacturer of the luminaire. For our example we will assume 0.71 amps.
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	Circuit breakers are protective devices for over-current conditions. When the current passing through the circuit breaker coils exceeds a predetermined amount, the magnetic field developed is sufficiently large in magnetomotive force to trip the contacts and open the faulty circuit. The amount of current required to operate the trip mechanism is referred to as the "trip rating". Proper protection of the circuit requires a breaker with the correct "tripping" current value. This value, "trip rating", can readily be computed by totaling the number of luminaires for each circuit breaker and thereby obtaining the total current being used in the circuit.		Circuit breakers are protective devices for over-current conditions. When the current passing through the circuit breaker coils exceeds a predetermined amount, the magnetic field developed is sufficiently large in magnetomotive force to trip the contacts and open the faulty circuit. The amount of current required to operate the trip mechanism is referred to as the "trip rating". Proper protection of the circuit requires a breaker with the correct "tripping" current value. This value, "trip rating", can readily be computed by totaling the number of luminaires for each circuit breaker and thereby obtaining the total current being used in the circuit.

	To prevent unnecessary tripping of the breaker during surges and in-rush currents, the total current is usually multiplied by a factor of 1.3. Using the previous example: 5 each - ~~170-watt LED~~ luminaires require a normal operating current of 5 x (0.71 ampere) = 3.67 amperes.		To prevent unnecessary tripping of the breaker during surges and in-rush currents, the total current is usually multiplied by a factor of 1.3. Using the previous example: 5 each LED-B luminaires require a normal operating current of 5 x (0.71 ampere) = 3.67 amperes.

	The line loss load in amperes is added as follows:		The line loss load in amperes is added as follows:

EPGsysop: Per TS, provided guidance to allow the use of LED luminaires for roadway lighting.

2018-02-09T23:29:35Z

Per TS, provided guidance to allow the use of LED luminaires for roadway lighting.

Show changes

Smithk: minor clarification

2010-03-30T16:26:04Z

minor clarification

← Older revision		Revision as of 11:26, 30 March 2010
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	Since the resistance of a wire conductor is inversely proportional to the cross-sectional area of the wire, increased voltage drops are developed as the wire diameters decrease.		Since the resistance of a wire conductor is inversely proportional to the cross-sectional area of the wire, increased voltage drops are developed as the wire diameters decrease.

	The areas of wires in circular mils (mm<sup>2</sup>) and the resistance in ohms per 1000 ft. (1000 m) to be used in calculations are shown in the [[901.10 Preparation of Plans#901.10.3 Cable and Conduit\|tables of 901.10.3, Cable and Conduit]]. The voltage drop in any line is calculated from the formula:		The areas of wires in circular mils (mm<sup>2</sup>) and the resistance in ohms per 1000 ft. (1000 m) to be used in calculations are shown in the [[901.10 Preparation of Plans#901.10.3 Cable and Conduit\|tables of EPG 901.10.3 Cable and Conduit]]. The voltage drop in any line is calculated from the formula:

	: ENGLISH		: ENGLISH

Smithk: /* 901.12.1 Voltage Drop and Wire Sizes */ appearance of equations enhanced

2008-08-13T14:12:31Z

901.12.1 Voltage Drop and Wire Sizes: appearance of equations enhanced

← Older revision		Revision as of 09:12, 13 August 2008
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	The areas of wires in circular mils (mm<sup>2</sup>) and the resistance in ohms per 1000 ft. (1000 m) to be used in calculations are shown in the [[901.10 Preparation of Plans#901.10.3 Cable and Conduit\|tables of 901.10.3, Cable and Conduit]]. The voltage drop in any line is calculated from the formula:		The areas of wires in circular mils (mm<sup>2</sup>) and the resistance in ohms per 1000 ft. (1000 m) to be used in calculations are shown in the [[901.10 Preparation of Plans#901.10.3 Cable and Conduit\|tables of 901.10.3, Cable and Conduit]]. The voltage drop in any line is calculated from the formula:

	ENGLISH		: ENGLISH
	~~Volts drop = Distance (ft) x Amperes x 25/(Area, in circular mils)~~

	~~METRIC~~		: <math>Volts\; drop=\frac{Distance\;(ft) \times \ Amperes \times \ 25}{Area\;(circular\;mils)}</math>
	Volts drop = Distance (m) x Amperes ~~x 0.04125/~~(~~Area, in mm<sup>2~~</~~sup~~>)

	~~Transposing this formula, the required area in circular mils (mm<sup>2</sup>) can be computed~~:		: METRIC

	~~ENGLISH~~		: <math>Volts\; drop=\frac{Distance\;(m) \times \ Amperes \times \ 0.04125}{Area\;(mm^2)}</math>

	~~Area~~ in circular mils ~~= Distance x Amperes x 25~~/~~(Permissible voltage drop~~)		Transposing this formula, the required area in circular mils (mm<sup>2</sup>) can be computed:

	~~METRIC~~		: ENGLISH

	Area in mm<~~sup~~>2<~~/sup~~> = Distance x Amperes x 0.04125/(Permissible voltage drop)		: <math>Area\; in\; circular\; mils=\frac{Distance \times \ Amperes \times \ 25}{Permissible\;voltage\; drop}</math>

			: METRIC

			: <math>Area\; in\; mm^2=\frac{Distance \times \ Amperes \times \ 0.04125}{Permissible\;voltage\; drop}</math>

	With the constant wattage ballast, the permissible voltage drop is five percent of the system voltage. A factor of 0.95 is to be applied to the system voltage to allow for line fluctuations. This results in an allowable drop of 23 volts for a 480-volt circuit, and 11 volts for a 240-volt circuit.		With the constant wattage ballast, the permissible voltage drop is five percent of the system voltage. A factor of 0.95 is to be applied to the system voltage to allow for line fluctuations. This results in an allowable drop of 23 volts for a 480-volt circuit, and 11 volts for a 240-volt circuit.

Smithk: correction (denominators corrected in voltage drop equations)

2008-08-13T13:02:10Z

correction (denominators corrected in voltage drop equations)

← Older revision		Revision as of 08:02, 13 August 2008
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	ENGLISH		ENGLISH
	Volts drop = Distance (ft) x Amperes x 25		Volts drop = Distance (ft) x Amperes x 25/(Area, in circular mils)
	circular mils

	METRIC		METRIC
	Volts drop = Distance (m) x Amperes x 0.04125		Volts drop = Distance (m) x Amperes x 0.04125/(Area, in mm<sup>2</sup>)
	~~mm2~~

	Transposing this formula, the required area in circular mils (mm<sup>2</sup>) can be computed:		Transposing this formula, the required area in circular mils (mm<sup>2</sup>) can be computed:
Line 31:		Line 29:
	ENGLISH		ENGLISH

	Area in circular mils = Distance x Amperes x 25		Area in circular mils = Distance x Amperes x 25/(Permissible voltage drop)
	Permissible voltage drop

	METRIC		METRIC

	Area in mm<sup>2</sup> = Distance x Amperes x 0.04125		Area in mm<sup>2</sup> = Distance x Amperes x 0.04125/(Permissible voltage drop)
	Permissible voltage drop

	With the constant wattage ballast, the permissible voltage drop is five percent of the system voltage. A factor of 0.95 is to be applied to the system voltage to allow for line fluctuations. This results in an allowable drop of 23 volts for a 480-volt circuit, and 11 volts for a 240-volt circuit.		With the constant wattage ballast, the permissible voltage drop is five percent of the system voltage. A factor of 0.95 is to be applied to the system voltage to allow for line fluctuations. This results in an allowable drop of 23 volts for a 480-volt circuit, and 11 volts for a 240-volt circuit.

Smithk: article construction

2008-02-01T19:17:54Z

article construction

New page

[[image:901.12 Electrical Components.jpg|right|300px]]

==901.12.1 Voltage Drop and Wire Sizes==

Voltage drop is an important factor in determining wire sizes. The voltage drop in any electrical circuit is directly dependent upon current and wire resistance. According to Ohm's Law the voltage drop in a line is equal to the current in amperes multiplied by the resistance of the line in ohms:

E = I x R

Where:

E = voltage

I = Current (in amperes)

R = Resistance (in ohms)

Since the resistance of a wire conductor is inversely proportional to the cross-sectional area of the wire, increased voltage drops are developed as the wire diameters decrease.

The areas of wires in circular mils (mm2) and the resistance in ohms per 1000 ft. (1000 m) to be used in calculations are shown in the [[901.10 Preparation of Plans#901.10.3 Cable and Conduit|tables of 901.10.3, Cable and Conduit]]. The voltage drop in any line is calculated from the formula:

ENGLISH
Volts drop = Distance (ft) x Amperes x 25
circular mils

METRIC
Volts drop = Distance (m) x Amperes x 0.04125
mm2

Transposing this formula, the required area in circular mils (mm2) can be computed:

ENGLISH

Area in circular mils = Distance x Amperes x 25
Permissible voltage drop

METRIC

Area in mm2 = Distance x Amperes x 0.04125
Permissible voltage drop

With the constant wattage ballast, the permissible voltage drop is five percent of the system voltage. A factor of 0.95 is to be applied to the system voltage to allow for line fluctuations. This results in an allowable drop of 23 volts for a 480-volt circuit, and 11 volts for a 240-volt circuit.

'''901.12.1.1 Maximum Wire Sizes.''' Maximum wire sizes for lighting branch circuits (cables between control station and light poles) are as follows:

* Standard Light Poles (30 and 45 ft. mounting height) and Underpass Luminaires - #6 AWG (16 mm2) MAXIMUM

* High Mast Light Poles - #2 AWG (35 mm2) MAXIMUM

If circuit loading or voltage drop requires a larger wire size than the maximum then one or more of the following can be considered:

* Increase the system voltage from 240 volts to 480 volts.

* Split the circuits into additional branch circuits. This may require parallel cable conduits in some parts of the system.

* Locate the control station closer to the lighting system.

* Install additional control stations.

Availability and proximity of power sources and system voltages is considered in obtaining the best solution. Where alternate configurations can achieve the same results, a rough cost comparison is made to determine the most economical configuration.

Wire sizes larger than #0 AWG (70 mm2) are not recommended for power supply cables (cables between power supply and base-mounted control station) since larger wire sizes are difficult to handle in the control stations and power supplies. If a larger wire size is required, a terminal strip is provided in the control cabinet to reduce the wire size.

'''901.12.1.2 Example to Determine Necessary Wire Cable Size'''

Given: Five 250-watt lamps on a 480-volt circuit at 1200, 1700, 3700, 4300 and 4600 ft. (350, 500, 1100, 1300 and 1400 m), respectively, from the control station.

There must be sufficient current to operate the first lamp as well as the next four lamps, etc.

Operating current is determined using 95 percent of the system voltage to allow for line fluctuations. In addition, the lamp wattage is increased by a factor of 1.3 to allow for ballast load. The formula for power is P = V x I and the formula for current is I = P / V.

(250 watts x 1.3) / (480 volts x 0.95) = 0.71 amps

It is also necessary to add 5% to the wire length to allow for snaking.

ENGLISH

1200 ft x 1.05 x 5 lamps x 0.71 amps x 25/23 (Permissible Drop) = 4862 cir. mils to 1st lamp

500 ft x 1.05 x 4 lamps x 0.71 x 25/23 = 1621 cir. mils to 2nd lamp

2000 ft x 1.05 x 3 lamps x 0.71 x 25/23 = 4862 cir. mils to 3rd lamp

600 ft x 1.05 x 2 lamps x 0.71 x 25/23 = 972 cir. mils to 4th lamp

300 ft x 1.05 x 1 lamp x 0.71 x 25/23 = 243 cir. mils to 5th lamp

Total Area = 12,560 cir. mils.

METRIC

350 m x 1.05 x 5 lamps x 0.71 amps x 0.04125/23 (Permissible Drop) = 2.34 mm2 to 1st lamp

150 m x 1.05 x 4 lamps x 0.71 x 0.04125/23 = 0.80 mm2 to 2nd lamp

600 m x 1.05 x 3 lamps x 0.71 x 0.04125/23 = 2.41 mm2 to 3rd lamp

200 m x 1.05 x 2 lamps x 0.71 x 0.04125/23 = 0.54 mm2 to 4th lamp

100 m x 1.05 x 1 lamp x 0.71 x 0.04125/23 = 0.13 mm2 to 5th lamp

Total Area = 6.22 mm2.

The wire size is that with the area next above, as shown in Table 1 (8-01.1) of Figure 901.10 (8-01.11) or #8 AWG (10 mm2) (The minimum size used in cable-conduit is #8 AWG (10 mm2)). When several circuits are carried from one control station, determine the wire size for each circuit separately. #10 AWG (6 mm2) pole and bracket cable is used in all installations regardless of the size of the conductor cable. To check the preceding result, use the basic formula:

Voltage Drop = Current x Resistance

To determine the current in amperes, use length of wires (2 x distance) x number of lamps x line amperes. Table 8-01.1 of Figure 8-01.11 shows the resistance of #8 AWG (10 mm2) wire in ohms/1000 ft = 0.78 (ohms/1000 m = 2.2).

Therefore:

ENGLISH

2 wires x 1200 ft x 1.05 x 5 lamps x 0.71 x 0.78 ohms/1000 ft = 6.98 volt drop to 1st lamp

2 wires x 500 ft x 1.05 x 4 lamps x 0.71 x 0.78 ohms/1000 ft = 2.33 volt drop to 2nd lamp

2 wires x 2000 ft x 1.05 x 3 lamps x 0.71 x 0.78 ohms/1000 ft = 6.98 volt drop to 3rd lamp

2 wires x 600 ft x 1.05 x 2 lamps x 0.71 x 0.78 ohms/1000 ft = 1.40 volt drop to 4th lamp

2 wires x 300 ft x 1.05 x 1 lamp x 0.71 x 0.78 ohms/1000 ft = 0.35 volt drop to 5th lamp

Total Drop = 18.03 volts and 18.03 volts/(480 volts x .95) = 3.96%

METRIC

2 wires x 350 m x 1.05 x 5 lamps x 0.71 x 2.2 ohms/1000 m = 5.74 volt drop to 1st lamp

2 wires x 150 m x 1.05 x 4 lamps x 0.71 x 2.2 ohms/1000 m = 1.97 volt drop to 2nd lamp

2 wires x 600 m x 1.05 x 3 lamps x 0.71 x 2.2 ohms/1000 m = 5.90 volt drop to 3rd lamp

2 wires x 200 m x 1.05 x 2 lamps x 0.71 x 2.2 ohms/1000 m = 1.31 volt drop to 4th lamp

2 wires x 100 m x 1.05 x 1 lamp x 0.71 x 2.2 ohms/1000 m = 0.33 volt drop to 5th lamp

Total Drop = 15.25 volts and 15.25 volts/(480 volts x .95) = 3.34%

Circuit loading due to line losses is also to be calculated for the purpose of sizing circuit breakers. The formula for power, P = I x V is used. Total current on each segment and the voltage drop as calculated above is used:

ENGLISH

5 lamps x 0.71 amps x 6.98 volts = 25 watts consumed to the 1st lamp

4 lamps x 0.71 amps x 2.33 volts = 7 watts consumed to the 2nd lamp

3 lamps x 0.71 amps x 6.98 volts = 15 watts consumed to the 3rd lamp

2 lamps x 0.71 amps x 1.40 volts = 2 watts consumed to the 4th lamp

1 lamp x 0.71 amps x 0.35 volts = 0 watts consumed to the 5th lamp

Total line loss load = 49 watts.

METRIC

5 lamps x 0.71 amps x 5.74 volts = 21 watts consumed to the 1st lamp

4 lamps x 0.71 amps x 1.97 volts = 6 watts consumed to the 2nd lamp

3 lamps x 0.71 amps x 5.90 volts = 13 watts consumed to the 3rd lamp

2 lamps x 0.71 amps x 1.31 volts = 2 watts consumed to the 4th lamp.

1 lamp x 0.71 amps x 0.33 volts = 0 watts consumed to the 5th lamp

Total line loss load = 42 watts.

==901.12.2 Circuit Breakers==

Circuit breakers are protective devices for over-current conditions. When the current passing through the circuit breaker coils exceeds a predetermined amount, the magnetic field developed is sufficiently large in magnetomotive force to trip the contacts and open the faulty circuit. The amount of current required to operate the trip mechanism is referred to as the "trip rating". Proper protection of the circuit requires a breaker with the correct "tripping" current value. This value, "trip rating", can readily be computed by totaling the number of luminaires for each circuit breaker and thereby obtaining the total current being used in the circuit.

To prevent unnecessary tripping of the breaker during surges and in-rush currents, the total current is usually multiplied by a factor of 1.3. Using the previous example: 5 each - 250-watt luminaires require a normal operating current of 5 x (0.71 ampere) = 3.67 amperes.

The line loss load in amperes is added as follows:

42 watts/(480 volts x 0.95) + 3.6 amps = 3.7 amps

Then: Trip rating = 3.7 amps x (1.3) = 4.8 amps.

Conventional circuit breakers for lighting systems are used in the following trip ratings: 15, 20, 25, 30, 35, 40, 50, 70, 90 and 100 amperes. For the above example, a 15-amp breaker would be used.

[[Category:901 Lighting]]