Answered

Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.