|Home | Sitemap/Articles
One of the main factors that determines the amount of current a conductor is permitted to carry is the ambient, or surrounding, air temperature.
Table 310.15(B)(16), for example, lists the ampacity of not more than three conductors in a raceway in free air. These ampacities are based on an ambient air temperature of 30 degree C, or 868F. If these conductors are to be used in a location that has a higher ambient temperature, the ampacity of the conductor must be reduced because the resistance of copper or aluminum increases with an increase of temperature. Temperature correction factors can be found in Table 310.15(B)(2)(a) and 310.15(B)(2)(b). Table 310.15(B)(2)(a) is for conductors rated at 30 degree C or 868F. The ampacity of conductors in Table 310.15(B)(16) and Table 310.15(B)(17) are based on an ambient air temperature of 30 degree C. The correction factors for conductors rated at 40 degree C are found in Table 310.15(B)(2)(b). The ampacity of conductors in Table 310.15(B)(18) and Table 310.15(B)(19) are based on an ambient temperature of 40 degree C. The correction factors found in Table 310.15(B)(2)(a) are shown below.
More Than Three Conductors in a Raceway:
Table 310.15(B)(16) and Table 310.15(B)(18) list three conductors in a race way. If a raceway is to contain more than three conductors, the ampacity of the conductors must be derated because the heat from each conductor combines with the heat dissipated by the other conductors to produce a higher temperature inside the raceway. NEC Table 310.15(B)(3)(a) lists these correction factors. If the raceway is used in an area with a greater ambient temperature than that listed in the appropriate wire table, the temperature correction factor must also be applied.
EXAMPLE: What is the maximum ampacity of a 4 AWG copper conductor with Type THWN insulation used in an area with an ambient temperature of 438C?
Solution: Determine the ampacity of a 4 AWG copper conductor with Type THWN insulation from the wire table. Type THWN insulation is located in the second column of Table 310.15(B)(16). The table lists an ampacity of 85A for this conductor.
Locate 438C in the far left-hand column of the correction factor chart; 438C falls between 418C and 458C. Follow across to the 758C column. The chart lists a correction factor of 0.82. The ampacity of the conductor in the above wire table is to be multiplied by the correction factor:
85A x 0.82 = 69.7A
What is the maximum ampacity of a 1/0 AWG copper-clad aluminum conductor with Type RHH insulation if the conductor is to be used in an area with an ambient air temperature of 100 degree F? Solution:
Locate the column that contains Type RHH insulation in the copper-clad aluminum section of Table 310.15(B)(16). The table indicates a maximum ampacity of 135 A for this conductor. The chart is used to determine the correction factor for this temperature. Fahrenheit degrees are located in the far right-hand column of the chart; 100 degree F falls between 978F and 1048F. The correction factor for this temperature is 0.91. Multiply the ampacity of the conductor by this factor:
135 A x 0.91 = 122.85 A
EXAMPLE: Twelve 14 AWG copper conductors with Type RHW insulation are to be run in a conduit. The conduit is used in an area that has an ambient temperature of 110 degree F. What is the maximum ampacity of these conductors? Solution
Find the ampacity of a 14 AWG copper conductor with Type RHW insulation. Type RHW insulation is located in the second column of Table 310.15(B)(16). A 14 AWG copper conductor has an ampacity of 20 A. Next, use the correction factor for an ambient temperature of 110 degree F in Table 310.15(B)(2)(a) shown below.
A correction factor of 0.82 will be used:
20 A 3 0.82 = 16.4 A The correction factor located in Table 310.15(B)(3)(a) must now be used. The table indicates a correction factor of 50% when 10 through 20 conductors are run in a raceway:
16.4A x 0.50 = 8.2 A 16.4A x 0.50 = 8.2A
Each 14 AWG conductor has a maximum current rating of 8.2A.
Determining Conductor Size Using the NEC:
Using the NEC to determine the amount of current a conductor is permitted to carry and the proper size conductor to use for a particular application are not the same thing. The several factors that must be considered when selecting a conductor for a specific job are of no consequence when determining the ampacity of a conductor. One of these factors is whether the load is a continuous or non-continuous load. The NEC defines a continuous load as one where the maxi mum current is expected to continue for three hours or more. Most industrial motor and lighting loads, for example, would be considered continuous loads.
NEC 210.19(A)(1) states that conductors must have an ampacity not less than the non-continuous load plus 125% of the continuous load. Basically, the ampacity of a conductor must be 125% greater than the current rating of a continuous load.
Another factor that affects the selection of a conductor is termination temperature limitations. NEC 110.14(C) states that the temperature rating of a conductor must be selected so as not to exceed the lowest temperature rating of any connected conductor, termination, or device. Because the termination temperature rating of most devices is not generally known, NEC 110.14(C)(1)(a) states that conductors for circuits rated at 100 amperes or less are to be selected from the 60 degree C column. This does not mean that conductors with a higher temperature rating cannot be used, but their size must be selected from the 60 degree C column. The only exception to this is motors that are marked with NEMA (National Electrical Manufacturers Association) code letters B, C, or D. NEC 110.14(C)(1)(a)(4) permits conductors for motors with these code letters to be selected from the 758C column.
EXAMPLE: Assume that a motor with a full-load current rating of 28 A is to be connected with copper conductors that have Type THW insulation. The motor is located in an area with an ambient temperature of 30 degree C. The motor is not marked with NEMA code letters, and the termination temperature is not known. What size conductors should be used? Solution Because a motor load is continuous, multiply the full-load current rating by 125%:
28A x 1.25 = 35 A
Refer to NEC Table 310.15(B)(16). Type THW insulation is located in the 758C column. Although this conductor is located in the 758C column, the wire size must be chosen from the 60 degree C column because the termination temperature is not known and the motor does not contain NEMA code letters. The nearest wire size without going under 35 A is an 8 AWG.
Assume that a bank of heating resistors is rated at 28 kW and is connected to 240 V. The resistors operate for more than three hours at a time and are located in an area with an ambient temperature of 868F. Aluminum conductors with Type THWN-2 insulation are to be used to connect the heaters. What size conductors should be used?
Determine the amperage of the load using Ohm's law:
I = P /E
I = 28,000W / 240 V
I = 116.667A
Because the load is continuous, the conductors must have an ampacity 125% greater than the load current.
I = 116.667 A x 1.25
I = 145.834A or 146A
Refer to NEC Table 310.15(B)(16). Type THWN-2 insulation is located in the 90 degree C column. The conductor size, however, must be chosen from the 758C column.
The nearest size aluminum conductor without going less than 146 A is 3/0 AWG.
EXAMPLE: An electric annealing oven is located in an area with an ambient temperature of 125°F. The oven contains a 50-kW electric heating element and is connected to 480 volts. The conductors are to be copper with type THHN insulation. The termination temperature is not known. The furnace is expected to operate more than three hours continuously. What size conductor should be employed to make this connection? Solution Determine the amount of current needed to operate the furnace.
I = P/E
I = 50,000 / 480
I = 104.167 amperes
Because the load is continuous, it must be increased by 125%:
104.167 x 1.25 = 130.2 amperes
The next step is to apply the correction factor for temperature. Type THHN insulation is located in the 90°C column. The correction factor for a temperature of 125°F is 0.76, as shown below. To determine the current rating of the conductor at 125°F, divide the current by the correction factor:
I = 130.2 / 0.76
I = 171.3 amperes
Because the termination temperature is not known and the current is over 100 amperes, the conductor size will be selected from the 75°C column of NEC Table 310.15(B)(16). A 2/0 AWG conductor will be used.
+++++ Electric duct banks.
The requirements of NEC 110.14(C) also require that conductors be selected from Table 310.15(B)(16) as a general rule. NEC Table 310.15(B)(17), for example, lists the current-carrying capacity of conductors located in free air. Because the conductors are generally terminated inside an enclosure, however, they must be chosen from a table that lists the ampacity of conductors inside an enclosure.
For circuits with a current of 100A or greater, NEC 110.14(C)(1)(b) permits conductors to be selected from the 758C column.
Another factor to be considered when selecting a conductor for a particular application is the ambient temperature. Table 310.15(B)(2)(a) states: "For ambient temperatures other than 30°C (86°F), multiply the allowable ampacities shown above by the appropriate factor shown below." The wire tables are used to determine the maximum current-carrying capacity of conductors based on the type of material from which they are made, the type of insulation, and the surrounding air temperature (ambient temperature). The correction factors listed have a value less than 1 for any temperature greater than 30°C (86°F) because conductors become more resistive as temperature increases. This reduces the amount of current they can safely carry.
When determining the conductor size needed for a particular application, you are not determining the maximum current a conductor can carry. You are determining the size conductor needed to carry the amount of current for the particular application. Therefore, when determining conductor size in an area where ambient temperature is a concern, you must divide the needed current by the correction factor instead of multiplying by it. Dividing by the correction factor will result in an increase of the amperage used in selecting the conductor size. This increase is needed to offset the effects of higher temperature on the conductor.
Duct banks are often used when it becomes necessary to bury cables in the ground. An electric duct can be a single metallic or nonmetallic conduit. An electric duct bank is a group of electric ducts buried together. When a duct bank is used, the center points of individual ducts should be separated by a distance of not less than 7.5 inches.
Calculating Conductor Sizes and Resistance
Although the wire tables in the NEC are used to determine the proper size wire for most installations, there are instances in which these tables are not used. The formula in 310.15(c) of the NEC is used for ampacities not listed in the wire tables:
I = v
TC 2 (TA 1 DTD) / RDC(1 + YC)RCA where
TC = conductor temperature in 8C
TA = ambient temperature in 8C
DTD = dielectric loss temperature rise
RDC = DC resistance of conductor at temperature TC
YC = component AC resistance resulting from skin effect and proximity effect
RCA = effective thermal resistance between conductor and surrounding ambient
Although this formula is seldom used by electricians, the NEC does permit its use under the supervision of an electrical engineer.
+++++ Long wire runs have the effect of adding resistance in series with the load. 3000 Feet Load Power source
+++++ A large pipe has less resistance to the flow of water than a small pipe.
+++++ A mil-foot is equal to a piece of wire one foot long and one thousandth of an inch in diameter. 1 Foot -- Mil-foot 0.001''
+++++ Resistivity of materials.
Aluminum Carbon Constantan Copper Gold Iron Lead Manganin Mercury Nichrome Nickel Platinum Silver Tungsten - = CM ft at 20°C--Material--Temp. coeff.( per °C)--Resistivity (K) of Materials
Long Wire Lengths:
Another situation in which it becomes necessary to calculate wire sizes instead of using the tables in the Code is when the conductor becomes excessively long. The listed ampacities in the Code tables assume that the length of the conductor won’t increase the resistance of the circuit by a significant amount.
When the wire becomes extremely long, however, it’s necessary to calculate the size of wire needed.
All wire contains resistance. As wire is added to a circuit, it has the effect of adding resistance in series with the load. Four factors determine the resistance of a length of wire:
1. The type material from which the wire is made. Different types of material have different wire resistances. A copper conductor will have less resistance than an aluminum conductor of the same size and length. An aluminum conductor will have less resistance than a piece of iron wire the same size and length.
2. The diameter of the conductor. The larger the diameter, the less resistance it will have. A large-diameter pipe, for example, will have less resistance to the flow of water than will a small-diameter pipe. The cross-sectional area of round wire is measured in circular mils (CM). One mil equals 0.001 inch. A circular mil is the cross-sectional area of the wire in mils squared. For example, assume a wire has a diameter of 0.064 inch. Sixty-four thousandths should be written as a whole number, not as a decimal or a fraction (642 [64 mil x 64 mil] = 4096 CM).
3. The length of the conductor. The longer the conductor, the more resistance it will have. Adding length to a conductor has the same effect as connecting resistors in series.
4. The temperature of the conductor. As a general rule, most conductive materials will increase their resistance with an increase of temperature.
Some exceptions to this rule are carbon, silicon, and germanium. If the coefficient of temperature for a particular material is known, its resistance at different temperatures can be calculated. Materials that in crease their resistance with an increase of temperature have a positive coefficient of temperature. Materials that decrease their resistance with an increase of temperature have a negative coefficient of temperature.
In the English system of measure, a standard value of resistance called the mil-foot is used to determine the resistance of different lengths and sizes of wire. A mil-foot is a piece of wire 1 foot long and 1 mil in diameter. A chart showing the resistance of a mil-foot of wire at 20 degree C is shown above. Notice the wide range of resistances for different materials. The temperature coefficient of the different types of conductors is listed also.
Calculating Resistance Now that a standard measure of resistance for different types of materials is known, the resistance of different lengths and sizes of these materials can be calculated. The formula for calculating resistance of a certain length, size, and type of wire is R = K x L
______ CM where
R = resistance of the wire
K = ohms-CM per foot
L = length of wire in feet CM = circular mil (area of the wire) This formula can be converted to calculate other values in the formula such as, to find the SIZE of wire to use CM = K x L
______ R to find the LENGTH of wire to use L = R x CM
________ K to find the TYPE of wire to use K = R x CM
EXAMPLE: Find the resistance of a piece of 6 AWG copper wire 550 ft long. Assume a temperature of 20 degree C. The formula to be used is R = K x L
______ CM Solution
The value for K can be found in the table -- it indicates a value of 10.4 V-CM per foot for a copper conductor. The length, L, was given as 550 ft, and the circular mil area of 6 AWG wire is listed as 26,250 in the table.
R = 10.4 V-CM/ft x 550ft
_____ 26,250 CM
R = 5720 V-CM
_____ 26,250 CM
R = 0.218 V
EXAMPLE: An aluminum wire 2250 ft long cannot have a resistance greater than 0.2 V. What size aluminum wire must be used? Solution
To find the size of wire, use
CM = K x L
CM = 17 V-CM/ft x 2250 ft
CM = 38,250 V-CM
CM = 191,250
The nearest standard size conductor for this installation can be found in the American Wire Gauge table. Because the resistance cannot be greater than 0.2V, the conductor cannot be smaller than 191,250 CM. The nearest standard conductor size is 0000 AWG.
Good examples of when it becomes necessary to calculate the wire size for a particular installation can be seen in the following problems.
EXAMPLE: A manufacturing plant has a cooling pond located 4000 ft. from the plant. Six pumps are used to circulate water between the pond and the plant. In cold weather, however, the pumps can freeze and fail to supply water. The plant owner decides to connect electric-resistance heaters to each pump to prevent this problem. The six heaters are connected to a two-conductor cable from the plant at a junction box. The heaters operate on 480 V and have a total current draw of 50 A. What size copper conductors should be used to supply power to the heaters if the voltage drop at the junction box is to be kept to 3% of the applied voltage? Assume an average ambient temperature of 20 degree C.
Solution The first step in the solution of this problem is to determine the maximum amount of resistance the conductors can have without producing a voltage drop greater than 3% of the applied voltage. The maximum amount of voltage drop can be calculated by multiplying the applied voltage by 3%: 480 V x 0.03 = 14.4 V
The maximum amount of resistance can now be calculated using Ohm's law:
R = E / I
R = 14.4V / 50A
R = 0.288 V
The distance to the pond is 4000 ft. Because two conductors are used, the resistance of both conductors must be considered. Two conductors 4000 ft long will have the same resistance as one conductor 8000 ft long. For this reason, a Plant Two-conductor cable Junction box Pumps Cooling pond.
For this installation 300-kcmil (thousand circular mils) cable will be used.
+++++Calculating long wire lengths.
EXAMPLE: The next problem concerns conductors used in a three-phase system. Assume that a motor is located 2500 ft from its power source and operates on 560 V. When the motor starts, it has a current draw of 168 A. The voltage drop at the motor terminals cannot be permitted to be greater than 5% of the source volt age during starting. What size aluminum conductors should be used for this installation? Solution The solution to this problem is very similar to the solution in the previous ex ample. First, find the maximum voltage drop that can be permitted at the load by multiplying the source voltage by 5%:
E = 560 V x 0.05
E = 28 V
+++++The line currents in a three-phase system are 120° out of phase with each other.
+++++ Current flows from lines 2 and 3 to line 1.
+++++ Current flows from line 3 to line 2.
The second step is to determine the maximum amount of resistance of the conductors. To calculate this value, the maximum voltage drop is divided by the starting current of the motor:
The next step is to calculate the length of the conductors. In the previous ex ample, the lengths of the two conductors were added to find the total amount of wire resistance. In a single-phase system, each conductor must carry the same amount of current. During any period of time, one conductor is supplying cur rent from the source to the load, and the other conductor completes the circuit by permitting the same amount of current to flow from the load to the source.
In a balanced three-phase circuit, three currents are 120° out of phase with each other. These three conductors share the flow of current between source and load. ---- two lines labeled A and B have been drawn through the three current waveforms. Notice that at position A the cur rent flow in phase 1 is maximum and in a positive direction. The current flow in phases 2 and 3 is less than maximum and in a negative direction. This condition corresponds to the example shown. Notice that maximum cur rent is flowing in only one conductor. Less than maximum current is flowing in the other two conductors.
Observe the line marking position B. The current flow in phase 1 is zero, and the currents flowing in phases 2 and 3 are in opposite directions and less than maximum. This condition of current flow is illustrated. Notice that only two of the three phase lines are conducting current and that the current in each line is less than maximum.
Because the currents flowing in a three-phase system are never maximum at the same time, and at other times the current is divided between two phases, the total conductor resistance won’t be the sum of two conductors. To calculate the resistance of conductors in a three-phase system, a demand factor of 0.866 is used.
In this problem, the motor is located 2500 ft from the source. The conductor length is calculated by doubling the length of one conductor and then multiplying by 0.866:
Now that all the factors are known, the size of the conductor can be calculated using the formula
Three 500-kcmil conductors will be used.
EXAMPLE: A single-phase motor is located 250 ft from its power source. The conductors supplying power to the motor are 10 AWG copper. The motor has a full-load current draw of 24 A. What is the voltage drop across the conductors when the motor is in operation?
A slightly different formula can be used to calculate the voltage drop on a three-phase system. Instead of multiplying KIL by 2, multiply KIL by the square root of 3:
EXAMPLE: A three-phase motor is located 175 ft. from its source of power. The conductors supplying power to the motor are 1/0 AWG aluminum. The motor has a full-load current draw of 88 A. What is the voltage drop across the conductors when the motor is operating at full load? Solution
Saturday, April 18, 2015 11:28