Consider the diagram below
R1=RA=kAL=2×0.12×10.01=0.040C/W
R2=R4=RC=kAL=20×0.04×10.05=0.060C/W
R3=RB=kAL=8×0.04×10.05=0.160C/W
R5=RD=kAL=15×0.06×10.01=0.110C/W
R6=RE=kAL=35×0.06×10.01=0.050C/W
R7=RF=kAL=2×0.12×10.06=0.250C/W
R8=0.120.00012=0.0010C/W
Rmid,11=R21+R31+R41=0.061+0.161+0.061
Rmid,1=0.0250C/W
Rmid,21=R51+R61=0.111+0.051
Rmid,2=0.0340C/W
RTotal=0.04+0.025+0.034+0.25+0.001=0.350C/W
Q=0.35300−100=571.43W
QTotal=571.430.12×15×8=190476.67W
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