s1.4-6 - Problem 1.4-6 A current lead must be designed to...

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Problem 1.4-6 A current lead must be designed to carry current to a cryogenic superconducting magnet, as shown in Figure P1.4-6. T H =30 0K T C =300K D =1cm L =20cm Ic =1000amp current lead Figure P1.4-6: 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 ohm-m 17x10 ohm-m 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/m-K. 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 (ohm-m)" rho_e=17e-9 [ohm-m]+(T-300 [K])*5e-11 [ohm-m/K] end "Inputs" L=20 [cm]*convert(cm,m) "length of current lead" T_H=300 [K] "hot end temperature"
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T_C=100 [K]
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s1.4-6 - Problem 1.4-6 A current lead must be designed to...

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