{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW2Soln - Problem 16.1 1 2T 2T T = = 0 For only radial heat...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 16.1 a. Steady State: 0 = t T , For only radial heat conduction: 0 1 2 2 2 2 2 = = z T T r θ Fourier’s field equation reduces to 0 1 1 2 2 = = + r T r r r r T r r T b. Integrating part a twice ( ) 2 1 ln K r K T + = Applying the two boundary conditions ( ) 2 1 ln K r K T i i + = ( ) 2 1 ln K r K T o o + = Subtracting these two equations: ( ) = o r i r o i T T K ln 1 Using this relationship K 2 is determined as: ( ) ( ) i o r i r o i i r T T T K ln ln 2 = or ( ) ( ) o o r i r o i o r T T T K ln ln 2 = The general temperature profile solution is: ( ) ( ) ( ) ( ) ( ) i i o r i r o i i o r i r o i i o r i r o i T r r T T r T T T r T T T + = + = ln ln ln ln ln ln or ( ) ( ) ( ) ( ) ( ) o o o r i r o i o o r i r o i o o r i r o i T r r T T r T T T r T T T + = +
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}