Question

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.

Answer 1

a) For radial heat transfer to be zero along the perfectly insulated adiabatic surface;
`(dT_(y) )/(dr)`
`|_(r-0)` = 0

b) For constant temperature;
`T_(y)`(
`r_(i)`) =
`T_(C)`(
`r_(i)`)

c) The heat transfer in the conducting rod and the cladding material is the same, i.e;
`k_(r)`
`(dT_(y) )/(dr)`
`|_(ri)` =
`k_(c)`
`(dT_(c) )/(dr)`
`|_(ri)`

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

`k_(c)`
`(dT_(c) )/(dr)`
`|_(r0)` = h(
`T_(c)`(
`r_(0)` ) -
`T_(\infty)` )

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
`(d)/(dr)`( kr
`(dT)/(dr)` ) + 1/r²
`(d)/(d\beta )`( ( k
`(dT)/(dr)` ) +
`(d)/(dz)`( k
`(dT)/(dr)`) + 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.

`k_(y)`/r
`(d)/(dr)`( r
`(dT_(y) )/(dr)` ) + q = 0

Apply the conditions for cladding by substituting 0 for q

`(d)/(dr)`( r
`(dT_(r) )/(dr)` ) = 0

Apply the following boundary conditions;

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

`(dT_(y) )/(dr)`
`|_(r-0)` = 0

b) For constant temperature

`T_(y)`(
`r_(i)`) =
`T_(C)`(
`r_(i)`)

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

`k_(r)`
`(dT_(y) )/(dr)`
`|_(ri)` =
`k_(c)`
`(dT_(c) )/(dr)`
`|_(ri)`

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

`k_(c)`
`(dT_(c) )/(dr)`
`|_(r0)` = h(
`T_(c)`(
`r_(0)` ) -
`T_(\infty)` )