1、1 -. - -. 7 fabricated of any metal, solid or hitranded. If stranded, the conductor should form a compact circular unit. 1.3.2 Close-fitting insulation, concentric with the conductor, consisting of one or more layers of different materials. 2. GENERAL METHOD The method is based on the fact that unde
2、r conditions of thermal equilibrium the rate at which heat is produced in the conductor (current squared x conductor resistance) is equal to the rate at which the heat is conducted through the insulation. The latter also equals the rate at which heat is trans- ferred from the outer insulation bounda
3、ry to the surrounding air and walls. Currents are found by solving simultaneously the equations for the rate of heat flow (1) through the insulation and .(2) to the surroundings from the insulation surf ace. A graphical method is employed. A more detailed description with supporting experimental dat
4、a is found in the following references: Continuous Current and Temperature Riss in Aircraft Cables, M. Schach, AIEE Transactbm, Part II, volume 71, 1952, pages 197-209. Attachment to Meeting Report No. 8 EIA Subcommittee SQ-12.4, on Current Rating of Hook- Up Wire, December 8,1955. e 3. SINGLE WIRES
5、 3.1 The calculation of current rating for sin h is a known function of b and T2 - Ta, and r is a known function of T2 - Ta. Specific values of h and r are chosen as indicated in the previous section. Step 1. Flot P versus T2 from the equation 2rK (Ti-Tz) P= log, b/a (3) P=rb (h+r) (Ta-Ta) (4) I = (
6、P/RTi) va (6) Step 1. Plot on the same graph aa step (1) P versus T2 from the equation The intersection of the curves of step8 (1) and (2) gives the desired value of P. Step 3. Solve for I from using P determined in previous steps. 3.1.4 Illustradon of proceure for c&ing current rating Given : No. 1
7、6 hook-up wire a = 0.0666 in. b = 0.0826 in. K = 0.0037 watt in.- OC- Ti = O6C Ta = 20C &I = 0.476 X 10-8 ohm in.- . EIA 214 58 m 3234600 00b3279 T m Ta T2-T8 h r P RS-2 1 4 Page 3 30 60 70 90 100 10 30 60 70 80 ,0110 ,0126 .O137 ,0146 ,0150 .O036 .O0396 ,00435 .O048 .O061 ,0378 .128 ,238 ,852 .414
8、Substituting in equation (3) 2rK (Ti-Te) loge b/a P= = 0.0618 ( 105-T2) watts 0 Since the equation is linear in T2, we need only two points P = O watts for T2 = 105C P = 0.618 watts for TP = 96C P = 7 b (h + r) (T2 - T8) = 0.269 (h +r) (T2 - 20) watts Substituting in equation (4) The values of h and
9、 r are taken from Figures 1 and 2 P versus Ta is plotted in Figure 3, and the intersection of the two curves gives a value of 0.403 wattrr for P. = 29.1 amps .476 X P I= 1” = EIA 214 58 m 3234b00 00b3280 b m &- RS-2 14 Page 4 o O f u 9 c Overall Diameter b, Inches Flgure 1 e 6 e - EIA 214 58 m 323qb00 0063283 8 m i I RS-2 14 Page 5 Figure 2 RS-2 14 1 Page6 EIA 214 58 m 3234b00 0063282 T m 20 40 60 80 T:! - OC 100 Figure 3
