Question

The figure below shows a hollow sphere made of a material with a temperature-dependent conductivity. The conductivity of the material is given by: k=bT where b = 0.01 W/m-K and T is the temperature in K. TH Tc Tout The sphere is used to contain liquid neon; therefore, the inner surface of the sphere at rin = 2 cm is held at Tc = 30 K. The outer surface of the sphere is exposed to ambient temperature; therefore, the temperature at rout = 5 cm is 300 K. a.) Sketch the temperature distribution in the sphere (i.e., sketch T as a function of r). Make sure that you get the qualitative features of your sketch right. Explain why your sketch looks the way it does. b.) Derive the ordinary differential equation that governs this problem. c.) What are the boundary conditions for this problem? d.) Solve the governing differential equation from (b) - you should end up with a solution that involves two unknown constants of integration. e.) Use the boundary conditions from (c) with the solution from (d) in order to obtain two equations in the two unknown constants. f.) Type the inputs for the problem and the equations from (e) into EES in order to evaluate the undetermined constants. g.) Prepare a plot of the temperature as a function of radius using EES. Compare it with your sketch from (a).' h.) Determine the rate of heat transfer to the liquid neon contained in the sphere.

Answer 1

kin=2cm Tc = 30k kout -sum Th= Tout = 300k 2 Tc TH TH distribution Tc sketen temp. rout b) since temperature at surface is higher and its decreasing as we go inside the sphere. It is reoba not linear. Ordinary differential Eq with variable ik' is given as Assuming Steady state and 1-D heat conduction - 0 K e²a7 ele2 de dee d (0.01 T. er at de Integrate w.e.t. le О. 01 т. Эт Dr TDT = 100C, dr M² Integrating both sides Т2 100. e +C2 -211 -2+1

e) Boundary 2см Conditions: at re-rein; T = Tc=30k ③ at r = roud; T = Th=300k ③ som d) put boundary condition in Eq ☺ 302 = 200 la +26₂ 0.02 Am put boundary condition in 300² = – 200 g + 262 0.05 - 200C -(-2003 300²-302 0,05 C = 14.85 put in ② 1 C₂ = 74700 e) Final Eqs after putting Ca & (z ino 2 +2970 = 149400 re no knowledge of EES f) Soury