Problem 1.46
A current lead must be designed to carry current to a cryogenic superconducting magnet, as
shown in Figure P1.46.
T
H
=30
0K
T
C
=300K
D
=1cm
L
=20cm
Ic
=1000amp
current lead
Figure P1.46: Current lead.
The current lead carries
Ic
= 1000 amp and therefore experiences substantial generation of
thermal energy due to ohmic dissipation.
The electrical resistivity of the lead material depends
on temperature according to:
[]
[
]
()
91
1
ohmm
17x10 ohmm
300 K 5x10
K
e
T
ρ
−−
⎡
⎤
=+
−
⎢
⎥
⎣
⎦
(1)
The length of the current lead is
L
= 20 cm and the diameter is
D
= 1 cm.
The hot end of the
lead (at
x
= 0) is maintained at
T
x
=0
=
T
H
= 300 K and the cold end (at
x
=
L
) is maintained at
T
x
=
L
=
T
C
= 100 K.
The conductivity of the lead material is
k
= 400 W/mK.
The lead is installed in a
vacuum chamber and therefore you may assume that the external surfaces of the lead (the outer
surface of the cylinder) are adiabatic.
a.) Develop a numerical model in EES that can predict the temperature distribution within the
current lead.
Plot the temperature as a function of position.
The input information is entered in EES and a function is used to define the electrical resistivity
according to Eq. (1):
$UnitSystem SI MASS RAD PA
K J
$TABSTOPS
0.2 0.4 0.6 0.8 3.5 in
function rho_e(T)
"Input:
T  temperature (K)
Output:
rho_e  electrical resistivity (ohmm)"
rho_e=17e9 [ohmm]+(T300 [K])*5e11 [ohmm/K]
end
"Inputs"
L=20 [cm]*convert(cm,m)
"length of current lead"
T_H=300 [K]
"hot end temperature"
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 Summer '08
 Rothamer
 Thermodynamics, Heat, Heat Transfer, cold end

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