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Passage of an electric current through a long conducting rod of radiusriand thermalconductivitykrresults in uniform volumetric heating at a rate ofq. The conducting rodis wrapped in an electrically nonconducting cladding material of outer radiusroandthermal conductivitykc, and convection cooling is provided by an adjoining fluid. Forsteady-state conditions, write appropriate forms of the heat equations for the rod andcladding. Express appropriate boundary conditions for the solution of these equations.

Sagot :

nuhulawal20

Answer:  

a) For radial heat transfer to be zero along the perfectly insulated adiabatic surface; [tex]\frac{dT_{y} }{dr}[/tex][tex]|_{r-0}[/tex] = 0

b) For constant temperature; [tex]T_{y}[/tex]([tex]r_{i}[/tex]) = [tex]T_{C}[/tex]([tex]r_{i}[/tex])

c) The heat transfer in the conducting rod and the cladding material is the same, i.e; [tex]k_{r}[/tex][tex]\frac{dT_{y} }{dr}[/tex] [tex]|_{ri}[/tex] =  [tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{ri}[/tex]

d) The convection surface conduction by cooling fluid will be;

[tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{r0}[/tex] = h( [tex]T_{c}[/tex]( [tex]r_{0}[/tex] ) - [tex]T_{\infty}[/tex] )

 

Explanation:  

Given the data in question;

we write the general form of the heat conduction equation equation in cylindrical coordinates with internal heat generation.

1/r[tex]\frac{d}{dr}[/tex]( kr[tex]\frac{dT}{dr}[/tex] ) + 1/r² [tex]\frac{d}{d\beta }[/tex](  ( k[tex]\frac{dT}{dr}[/tex] ) + [tex]\frac{d}{dz}[/tex]( k[tex]\frac{dT}{dr}[/tex]) + q = 0

where radius of cylinder is r, thermal conductivity of the cylinder is k, and q is heat generated in cylinder.

Now, Assume one dimensional heat conduction

lets substitute the condition for conducting rod with steady state condition.

[tex]k_{y}[/tex]/r [tex]\frac{d}{dr}[/tex]( r[tex]\frac{dT_{y} }{dr}[/tex] ) + q = 0

Apply the conditions for cladding by substituting 0 for q

[tex]\frac{d}{dr}[/tex]( r[tex]\frac{dT_{r} }{dr}[/tex] ) = 0

Apply the following boundary conditions;  

a) For radial heat transfer to be zero along the perfectly insulated adiabatic surface;

[tex]\frac{dT_{y} }{dr}[/tex][tex]|_{r-0}[/tex] = 0

b) For constant temperature

[tex]T_{y}[/tex]([tex]r_{i}[/tex]) = [tex]T_{C}[/tex]([tex]r_{i}[/tex])

c) The heat transfer in the conducting rod and the cladding material is the same, i.e

[tex]k_{r}[/tex][tex]\frac{dT_{y} }{dr}[/tex] [tex]|_{ri}[/tex] =  [tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{ri}[/tex]  

d) The convection surface conduction by cooling fluid will be;

[tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{r0}[/tex] = h( [tex]T_{c}[/tex]( [tex]r_{0}[/tex] ) - [tex]T_{\infty}[/tex] )

View image nuhulawal20
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