CODE CALCULATIONS Lesson 11
An 8 AWG stranded conductor is somewhat larger in diameter than a solid conductor of the same size. How much larger is the diameter of an 8 AWG stranded conductor than that of an 8 AWG solid conductor?
0.018 in. Chapter 9, Table 8
A 10 AWG copper stranded conductor is somewhat larger in diameter than a solid conductor of the same material and size. How much greater is the resistance in ohms per 1,000 feet (Ω/kft) of an uncoated 10 AWG stranded copper conductor than that of a 10 AWG solid copper conductor?
0.03 Ω/kft Chapter 9, Table 8
What minimum size aluminum wire is needed for a 440-volt, 3-phase branch circuit supplying a 60-ampere load at a distance of 325 feet and holding the voltage drop within a reasonable limit of efficiency?Use k = 21.2.
210.19(A), Informational Note No. 3 Reasonable voltage drop = 3% Vd = supply voltage × 3% = 440 × 0.03 = 13.2 V cmils = (k × L × I × 1.73) ÷ Vd = (21.2 × 60 × 325 × 1.73) ÷ 13.2 = 715,182 ÷ 13.2 = 54,180 cmils Chapter 9, Table 8, Area, cmils Read (more than) 54,180 = 66,360 cmils 66,360 cmils = 2 AWG aluminum
What minimum size copper wire is needed for a 230-volt, single-phase branch circuit supplying a 65-ampere load at a distance of 250 feet to limit the voltage drop within the recommended Code limit?
210.19(A), Informational Note No. 3 Voltage drop limit = 3% Vd = supply voltage × 3% = 230 × 0.03 = 6.9 V cmils = (k × I × 2L ) ÷ Vd = (12.9 × 65 × 2 × 250) ÷ 6.9 = 419,250 ÷ 6.9 = 60,761 cmils Chapter 9, Table 8, Area, cmils Read (more than) 60,761 = 66,360 cmils 66,360 cmils = 2 AWG copper
Calculate the resistance of 300 feet of 10 AWG solid aluminum conductor. Calculate the answer to two decimal places.
Chapter 9, Table 8, DC resistance 10 AWG solid aluminum = 2.0 Ω/kft R = (DC Res. × L) ÷ 1,000 = (2.0 × 300) ÷ 1,000 = 600 ÷ 1,000 = 0.60 Ω
Calculate the resistance of 500 feet of 250,000 circular mils uncoated copper. Calculate the answer to five decimal places.
Chapter 9, Table 8, DC resistance 250 kcmil uncoated copper = 0.0515 Ω/kft R = (DC Res. × L) ÷ 1,000 = (0.0515 × 500) ÷ 1,000 = 25.75 ÷ 1,000 = 0.02575 Ω
Calculate the resistance of 200 feet of 8 AWG coated solid copper conductor. Calculate the answer to four decimal places.
Chapter 9, Table 8, DC resistance 8 AWG solid coated copper = 0.786 Ω/kft R = (DC Res. × L) ÷ 1,000 = (0.786 × 200) ÷ 1,000 = 157.2 ÷ 1,000 = 0.1572 Ω
What is the voltage at the load with a 240-volt supply for a single-phase branch circuit supplying a 40-ampere load at a distance of 72 feet with 8 AWG THW stranded, uncoated copper conductors at maximum operating temperature? Calculate the answer to two decimal places.
Chapter 9, Table 8, DC resistance Table 310.4(A) 8 AWG stranded uncoated copper = 0.778 Ω/kft Temp. correction unnecessary Vd = (DC Res. × I × 2L) ÷ 1,000 = (0.778 × 40 × 2 × 72) ÷ 1,000 = 4,480 ÷ 1,000 = 4.48 V Vload = Vsupply − Vd = 240 − 4.48 = 235.52 V at load
What is the voltage drop on a single-phase, 240-volt branch circuit supplying a 48-ampere load at a distance of 88 feet using 4 AWG RHW aluminum conductors at maximum operating temperature? Calculate the answer to two decimal places.
Chapter 9, Table 8, DC resistance Table 310.4(A) RHW = 75°C rated Temperature correction unnecessary 4 AWG aluminum = 0.508 Ω/kft Vd = (DC res. × I × 2L) ÷ 1,000 = (0.508 × 48 × 2 × 88) ÷ 1,000 = 4,290 ÷ 1,000 = 4.29 V
What is the voltage drop on a 480-volt branch circuit supplying 96 amperes to a 3-phase load 200 feet from supply with 3 AWG THHN uncoated copper conductors at maximum operating temperature? Calculate the answer to two decimal places.
Chapter 9, Table 8, DC resistance Table 310.4(A) THHN = 90°C Temperature correction necessary Multiply by 1.05 3 AWG uncoated copper = 0.245 Ω/kft Vd = (DC Res. × I × 1.73L × 1.05) ÷ 1,000 = (0.245 × 96 × 1.73 × 200 × 1.05) ÷ 1,000 = 8,540 ÷ 1,000 = 8.54 V
What is the voltage drop of a single-phase branch circuit supplying a 28-ampere load at a distance of 58 feet with 8 AWG THHN stranded aluminum conductors at maximum operating temperature? Calculate the answer to two decimal places.
Chapter 9, Table 8, DC resistance Table 310.4(A) THHN = 90°C Temperature correction necessary Multiply by 1.05 8 AWG stranded aluminum = 1.28 Ω/kft Vd = (DC Res. × I × 2L × 1.05) ÷ 1,000 = (1.28 × 28 × 2 × 58 × 1.05) ÷ 1,000 = 4,365.312 ÷ 1,000 = 4.37 V
Calculate the resistance of 250 feet of 6 AWG THHN aluminum conductor at maximum operating temperature. Calculate the answer to four decimal places.
Chapter 9, Table 8, DC resistance Table 310.4(A) THHN = 90°C rated Temperature correction necessary Multiply by 1.05 6 AWG aluminum = 0.808 Ω/kft R = (DC Res. × L × 1.05) ÷ 1,000 = (0.808 × 250 × 1.05) ÷ 1,000 = 212.1 ÷ 1,000 = 0.2121 Ω
What is the voltage drop on a 3-phase, 480-volt branch circuit supplying a 120-ampere load at a distance of 65 feet with 1 AWG THW uncoated copper conductors at maximum operating temperature? Calculate the answer to two decimal places.
Chapter 9, Table 8, DC resistance Table 310.4(A) THW = 75°C rated Temperature correction unnecessary 1 AWG uncoated copper = 0.154 Ω/kft Vd = (DC Res. × I × 1.73L) ÷ 1,000 = (0.154 × 120 × 1.73 × 65) ÷ 1,000 = 2,080 ÷ 1,000 = 2.08 V
Calculate the resistance of 150 feet of 14 AWG THWN solid uncoated copper conductor at maximum operating temperature. Calculate the answer to four decimal places.
Chapter 9, Table 8, DC resistance Table 310.4(A) THWN = 75°C rated Temperature correction unnecessary 14 AWG solid uncoated copper = 3.07 Ω/kft R = (DC Res. × L) ÷ 1,000 = (3.07 × 150) ÷ 1,000 = 46.05 ÷ 1,000 = 0.4605 Ω
What is the voltage drop on a single-phase, 460-volt branch circuit supplying a 215-ampere load at a distance of 118 feet, with 4/0 AWG THWN copper uncoated conductors at maximum operating temperature? Calculate the answer to three decimal places.
Chapter 9, Table 8, DC resistance Table 310.4(A) THWN= 75°C rated Temperature correction unnecessary 4/0 AWG uncoated copper = 0.0608 Ω/kft Vd = (DC Resist. × I × 2L ) ÷ 1,000 = (0.0608 × 215 × 2 × 118) ÷ 1,000 = 3,084.992 ÷ 1,000 = 3.085 V
What is the voltage drop on a single-phase branch circuit supplying a 42-ampere load at a distance of 125 feet with 6 AWG TW uncoated copper conductors at maximum operating temperature? Calculate the answer to two decimal places.
Chapter 9, Table 8, DC resistance Table 310.4(A) TW = 60°C Temperature correction necessary Divide by 1.05 6 AWG uncoated copper = 0.491 Ω/kft Vd = (DC Res. × I × 2L) ÷ (1,000 × 1.05) = (0.491 × 42 × 2 × 125) ÷ (1,000 × 1.05) = 5,155.5 ÷ 1,050 = 4.91 V
Calculate the resistance of 190 feet of 12 AWG TW solid uncoated copper conductor at maximum operating temperature. Calculate the answer to four decimal places.
Chapter 9, Table 8, DC resistance Table 310.4(A) TW = 60°C rated Temperature correction necessary Divide by 1.05 12 AWG solid uncoated copper = 1.93 Ω/kft R = (DC Res. × L) ÷ (1,000 × 1.05) = (1.93 × 190) ÷ (1,000 × 1.05) = 366.7 ÷ 1,050 = 0.3490 Ω
Which of the following does not specifically change the direct current resistance of a particular conductor?
Conductor insulation Chapter 9, Table 8
Calculate the resistance of 175 feet of 14 AWG stranded uncoated copper conductor. Calculate the answer to four decimal places.
Note: Chapter 9, Table 8, DC resistance 14 AWG stranded uncoated copper = 3.14 Ω/kft R = (DC Res. × L) ÷ 1,000 = (3.14 × 175) ÷ 1,000 = 549.5 ÷ 1,000 = 0.5495 Ω
What is the maximum current for a single-phase 230-volt branch circuit supplying a load at a distance of 140 feet with 4 AWG copper conductors and holding the voltage drop to 2%?Use k = 12.9. Calculate the answer to two decimal places.
Vdmax = supply voltage × 2% = 230 × 0.02 = 4.6 V cmils 4 AWG copper = 41,740 cmils I max = (cmils × Vd) ÷ (k × 2L ) = (41,740 × 4.6) ÷ (12.9 × 2 × 140) =192,004 ÷ 3,612 = 53.16 A
What is the maximum length of a single-phase, 110-volt branch circuit supplying a 40-ampere load with 6 AWG aluminum conductors permitted without exceeding a 3% voltage drop?Use k = 21.2. Calculate the answer to two decimal places.
Vdmax = supply voltage × 3% = 110 × 0.03 = 3.3 V cmils 6 AWG aluminum = 26,240 cmils L = (cmils × Vd) ÷ (2 × k × I ) = (26,240 × 3.3) ÷ (2 × 21.2 × 40) = 86,592 ÷ 1,696 = 51.06 ft. max. length (one way)
What is the maximum length of a single-phase, 120-volt branch circuit supplying a 10-ampere load with 14 AWG solid copper conductors permitted without exceeding a 3% voltage drop?Use k = 12.9. Calculate the answer to two decimal places.
Vdmax = supply voltage × 3% = 120 × 0.03 = 3.6 V cmils 14 AWG copper = 4,110 cmils L = (cmils × Vd) ÷ (2 × k × I ) = (4,110 × 3.6) ÷ (2 × 12.9 × 10) = 14,796 ÷ 258 = 57.35 ft
What size copper conductors are needed to supply a 3-phase, 208-volt, 200-ampere load at a distance of 250 feet and not exceed a 3% voltage drop?
Vdmax = supply voltage × 3% = 208 × 0.03 = 6.24 V cmils = (k × L × I × 1.73) ÷ Vd = (12.9 × 200 × 250 × 1.73) ÷ 6.24 = 1,115,850 ÷ 6.24 = 178,822 cmils Chapter 9, Table 8, Area, cmils Read (more than) 178,822 = 211,600 cmils 211,600 cmils = 4/0 AWG copper
A 3-phase service is 240 volts at the service equipment. Presently, there is a 5-volt drop to the downstream distribution panel. What size copper branch-circuit conductors are needed to supply a 38-ampere, 3-phase branch-circuit load at a distance of 110 feet and not exceed a total 5% voltage drop on the feeder plus the branch circuit?
Vdmax = supply voltage × 5% = 240 × 0.05 = 12 V Vd (BC) = Vdmax − Vd (feeder) = 12 − 5 = 7 V (branch circuit) cmils = (k × I × 1.73 × L ) ÷ Vd (BC) = (12.9 × 38 × 1.73 × 110) ÷ 7 = 93,285.06 ÷ 7 = 13,326 cmils Chapter 9, Table 8, Area, cmils Read (more than) 13,326 = 16,510 cmils 16,510 cmils = 8 AWG copper