By Günter Brenn
This publication presents analytical suggestions to a couple of classical difficulties in delivery tactics, i.e. in fluid mechanics, warmth and mass move. increasing computing energy and extra effective numerical equipment have elevated the significance of computational instruments. despite the fact that, the translation of those effects is usually tricky and the computational effects must be verified opposed to the analytical effects, making analytical ideas a precious commodity. additionally, analytical options for delivery methods supply a far deeper realizing of the actual phenomena curious about a given strategy than do corresponding numerical ideas. even though this booklet basically addresses the desires of researchers and practitioners, it might even be precious for graduate scholars simply coming into the sphere.
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Additional info for Analytical Solutions for Transport Processes: Fluid Mechanics, Heat and Mass Transfer
45) In solving this equation, we use the approach of Tomotika developed for the axisymmetric cylindrical case for the polar cylindrical problem as well . 47) respectively. The two Eqs. 47) are solved by separation of variables, which reveals the functions ψcz,1 and ψcz,2 as products of eigenfunctions of the two operators in the two cylindrical coordinates and in time. 48) ψcz,1,m = C1r m + C2 r −m eimθ−αt , where m is a mode number and plays the role of a wave number in the direction of the polar angle θ .
Corresponding to the flow field, this stream function does not depend on the coordinate x in the flow direction.
60) 2 reads for the stream function ψcθ . 61) and the operator to the fourth power is 4 =r E cθ ∂ ∂r 1 ∂ ∂ r r ∂r ∂r 1 ∂ r ∂r + 2r ∂ ∂r 1 ∂ r ∂r ∂2 ∂z 2 + ∂4 . 4 Axisymmetric, Linear, Unsteady Flow Analytical solutions of Eq. 60) are found for linear flow fields, where the Jacobian 2 ψcθ with the z derivative is either negligible or vanishes exactly and the product of E cθ of ψcθ is small of the same order. The linearisation for cylindrical flows in the r, z plane leads to the equation − 1 ∂ 2 + E cθ ν ∂t 2 ψcθ = 0 .