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AC 43.13-1B CHG 1 9/27/01 The size #14 wire selected using the methods outlined in paragraph 11-66d is too small to meet the voltage drop limits from figure 11-2 for a 15.5 feet long wire run. STEP 3: Select the next larger wire (size #12) and repeat the calculations as follows: L1=24 feet maximum run length for (7/35 = 20 percent) will be carrying power cur rents nearly at or up to capacity. STEP 1: Refer to the “single wire in free air” curves in figure 11-4a. Determine the change of temperature of the wire to determine free air ratings. Since the wire will be in an ambient of 60 oC and rated at 200° C, the change of to temperature is 200 °C - 60 °C = 140 °C. Fol low the 140 °C temperature difference hori zontally until it intersects with wire size line on figure11-4a. The free air rating for size #20 is 21.5 amps, and the free air rating for size #22 is 16.2 amps. STEP 2: Refer to the “bundle derating curves” in figure 11-5, the 20 percent curve is selected since circuit analysis indicate that 20 percent or less of the wire in the harness would be car rying power currents and less than 20 percent of the bundle capacity would be used. Find 35 (on the abscissa) since there are 35 wires in the bundle and determine a derating factor of 0.52 (on the ordinate) from the 20 percent curve. STEP 3: Derate the size #22 free air rating by multiplying 16.2 by 0.52 to get 8.4 amps in- harness rating. Derate the size #20 free air- rating by multiplying 21.5 by 0.52 to get 11.2 amps in-harness rating. STEP 4: Refer to the “altitude derating curve” of figure 11-6, look for 60,000 feet (on the ab scissa) since that is the altitude at which the vehicle will be operating. Note that the wire must be derated by a factor of 0.79 (found on the ordinate). Derate the size 12 gauge wire ure 11-2. carrying 20 amps from fig Imax = 37 amps (this is the maximum current the size #12 wire can carry at 50 °C ambient. Use calculation methods outlined in para graph 11-69 and figure 11-4a. T2 =50oC+(200oC-50oC)( 20A/37A= 50 oC+(150 oC)(-540)=131oC L2 = 254.5 oC(L ) 1 234.5 o C + (T2 ) (254.5 oC)(24ft) (234.5 oC)+(131 oC) L = 6108 366 = L2 = 2 (254.5 oC)(24ft) 366 =16.7ft The resultant maximum wire length, after ad justing downward for the added resistance as sociated with running the wire at a higher tem perature, is 15.4feet, which will meet the original 15.5 foot wire run length requirement without exceeding the voltage drop limit ex pressed in figure 11-2. 11-69. COMPUTING CURRENT CARRY- ING CAPACITY. a. Example 1. Assume a harness (open or braided), consisting of 10 wires, size #20, 200 °C rated copper and 25 wires, size #22, 200 °C rated copper, will be installed in an area where the ambient temperature is 60 °C and the vehicle is capable of operating at a 60,000-foot altitude. Circuit analysis reveals #22 harness rating by 8.4 amps by 0.79 to get 6.6 amps. size #20 harness rating by 11.2 amps by 0.79 to get 8.8 amps. multiplying Derate the multiplying that 7 of the 35 wires in the bundle STEP 5: To find the total harness capacity, multiply the total number of size #22 wires by the derated capacity (25 x 6.6 = 165.0 amps) and add to that the number of size #20 wires Page 11-28 Par 11-68PDF Image | AFS-640
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