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Unformatted text preview: the distance from the center.
Note on Insulation:
By definition, heat cannot pass through (ideal) insulation, therefore the heat flow lines
can have no component along insulation., In other words, heat flow must be parallel to
insulation. Or, put another way, the heat flow lines § = constant are the same as the
insulation lines.
(C) Mixed Boundary value Problem
Solve for temperature in the following situation:
insulation
20º
50º 1 This is a classical situation with T independent of r. Referring again to Topic 9, we find
F(z) = iALn(z) + B
Looking at the real part,
T(z) = A Arg(z) + B
and we get
B = 50 and A = 60/π,
giving
60
T(z) = 50 ø
π
Notice that the heat flow lines are
lnr = constant
which is consistent with the above drawing (insulation = semicircles)
(D) Using Conformal Mappings
Find the temperature distribution in the following situation: 37 0º
–1 Insulation 1 20º
It looks best to map this thing onto something like this: 0º 20º –π/2 Insulation π/2
which we can do with the inverse sine function (see the discussion of what the sine
function does muuuch earlier). On the target strip, the temperature is given by
10...
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This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.
 Fall '03
 StefanWaner
 Math, Algebra, Geometry, Complex Numbers

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