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# soln08 - 10.34 Fall 2006 Homework#8 Solutions Problem 1...

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10.34 – Fall 2006 Homework #8 - Solutions Problem 1 – Conduction/Convection (Beers’ text 6.B.1) This problems involves heat conduction in a flowing Newtonian fluid with a know velocity profile (incompressible flow in a tube). The governing equation for the system is the conduction equation with convection in the z-direction: 2 T λ 1 T 2 T v ⋅∇ ( ρ C T ) = λ T v = r + P z P r r z 2 z ρ C r r 2 T λ 1 2 T 2 T T 2 U 1 = + + R z ρ C r r r 2 z 2 P In this case, U is the average velocity of fluid and can be calculated from the given Reynolds number and fluid properties. The boundary conditions for this problem were given in the radial part of the problem, but were somewhat ambiguous in the z-direction. In the z-direction, a valid boundary condition for upstream would be that T = T 0 at some distance upstream of the temperature change at the wall. A BC for downstream could be that the gradient (dT/dz) far downstream is zero, or that T = T 1 far downstream. The zero derivative boundary condition is probably better since it will be less likely to skew a solution where you do not go far enough in the z-direction. The boundary conditions used in the problem were: T r z = − 0 R Pe , 2 R Pe = T T r z , = 10 = T ( ) ( ) 1 ( = , ) = T T 1 0 z z < 0 0 T r ( r = 0, ) T r R z z = 0 The boundary conditions in z are equivalent to starting 2 characteristic conduction times upstream and going to 10 downstream of the temperature jump. It is often useful to make problem non-dimensional before solving in order to formulate the problem in terms of as few parameters as possible. For this problem, we choose the following non-dimensional variables: η r ξ z θ T T 0 where : Pe = 2 UR and α = λ R R Pe T W T 0 α ρ C P Carefully applying these non-dimensional variables, one will find the following equation: Flow r R T 1 T 0 T wall z = 0 Figure by MIT OCW. Cite as: William Green, Jr., course materials for 10.34 Numerical Methods Applied to Chemical Engineering, Fall 2006. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

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z , 1 1, , 1, , 1 i j i j i j i j i j + + θ 2 θ η 2 2 θ ( 2 ) θ + η + η 1 η = 0 η η 2 Pe ξ 2 ξ ( , = − 2 ) = 0 , = 10 ) = 1 θ η ξ ( θ η ξ θ η 1, 1 0 ξ ξ < 0 0 η θ ( η = 0 ( = ξ ) = = 0, ξ ) You can see that the non-dimensional form is much simpler, and the boundary conditions are extremely simple, which is a nice aspect since the units in many problems can be tedious. We also multiply through by the radius term to avoid any potential problems at r = 0.
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• Fall '04
• ClarkColton
• Chemical reaction, Boundary value problem, MIT OpenCourseWare, Massachusetts Institute of Technology, Numerical Methods Applied

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soln08 - 10.34 Fall 2006 Homework#8 Solutions Problem 1...

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