9/27/01
AC 43.13-1B CHG 1
current at an elevated conductor temperature
using the following information:
(1) The wire run is 15.5 feet long, in
cluding the ground wire.
(2) Circuit current (I2) is 20 amps,
continuous.
(3) Follow the 150 °C corresponding
horizontal line to intersect with #14 wire size,
drop vertically and read 47 Amps at bottom of
chart (current amperes).
(4) Use figure 11-5, left side of chart
reads 0.91 for 20,000 feet, multiple
0.91 x 47 Amps = 42.77 Amps.
(3) The voltage source is 28 volts.
(4) The wire type used has a 200 °C
conductor rating and it is intended to use this
thermal rating to minimize the wire gauge.
Assume that the method described in para
graph 11-66d(6) was used and the minimum
wire size to carry the required current is #14.
(5) Ambient temperature is 50 °C under
hottest operating conditions.
(5) Use figure 11-6, find the derate
factor for 8 wires in a bundle at 60 percent.
First find the number of wires in the bundle (8)
at bottom of graph and intersect with the
60 percent curve meet. Read derating factor,
(left side of graph) which is 0.6. Multiply
0.6 x 42.77 Amps = 26 Amps.
Imax = 26 amps (this is the maximum
current the #14 gauge wire could carry at 50°C
ambient
f. Procedures in example No. 2.
STEP 1: Assuming that the recommended
load bank testing described in para
graph 11-66d(5) is unable to be conducted,
then the estimated calculation methods out
lined in paragraph 11-66d(6) may be used to
determine the estimated maximum current
(Imax). The #14 gauge wire mentioned above
can carry the required current at 50 °C ambient
(allowing for altitude and bundle derating).
(1) Use figure 11-4a to calculate the
Imax a #14 gauge wire can carry.
Where:
T2 = estimated conductor temperature
T1 = 50 °C ambient temperature
TR = 200 °C maximum conductor rated
temperature
L1=15.5 feet maximum run length for size
#14 wire carrying 20 amps from figure 11-2
STEP 2: From paragraph 11-66d (5) and (6),
determine the T2 and the resultant maximum
wire length when the increased resistance of
the higher temperature conductor is taken into
account.
T2 = T1 + (TR − T1 )( I2 / I max )
T2 = 50 °C + (200 C − 50 C)( 20A / 26 A
= 50 °C+(150 °C)(.877)
T2 = 182 °C
L2 = (254.5 °C)(L1) =
(234.5 °C) + (T2)
L2 = (254.5 °C)(15.5Ft)
(234.5 °C) + (182 °C)
L2 = 9.5 ft
(2) Find the temperature differences
(TR-T1) = (200 °C-50 °C) = 150 °C.
Par 11-67
Page 11-27