Effect of the Spacing Design of Two Alternate Baffles on the Performance of Heat Exchangers

The present study investigates numerically the heat transfer process based forced convective flow of an incompressible fluid in a two-dimensional rectangular channel. Two baffles are imposed periodically on the lower and upper walls. The study mainly focused on the influence of the arrangement and s...

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Published inDiffusion and defect data. Solid state data. Pt. A, Defect and diffusion forum Vol. 415; pp. 53 - 72
Main Authors Filali, Abdelkader, Salhi, Najim, Salhi, Jamal-Eddine, Amghar, Kamal
Format Journal Article
LanguageEnglish
Published Zurich Trans Tech Publications Ltd 27.04.2022
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ISSN1012-0386
1662-9507
1662-9507
DOI10.4028/p-826179

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Summary:The present study investigates numerically the heat transfer process based forced convective flow of an incompressible fluid in a two-dimensional rectangular channel. Two baffles are imposed periodically on the lower and upper walls. The study mainly focused on the influence of the arrangement and spacing separating the baffles on the heat transfer's intensification. The values of the Reynolds number for the present turbulent flow regime were chosen in the range of 104 to 8.73 × 104. The equations resulting from the three conservation laws, namely continuity, Navier-Stokes, and energy equations, are solved numerically based on the finite volume method. SIMPLE algorithm is used to overcome the pressure-velocity coupling, and k-ε model is used for the computation of turbulent patterns. Numerical simulations are carried out to study the dynamic and thermal behavior influenced by the control parameters. The physical quantities calculated are the axial velocity, the local, mean Nusselt numbers and the friction coefficient. The obtained results show that the friction coefficient decreases proportionally with the increase of Re number, and the local Nusselt number increases with the Reynolds number. As the spacing between the baffles decreases, the NR ratio increases, and as the Reynolds number increases, NR decreases NR = 6.13, 5.31, 4.62, and 4.30 for case P1, NR = 5.1, 4.5, 3.89, and 3.64, for case P2, NR = 5.00, 4.45, 8.83, and 3.51, for case P3, for equal Reynolds number, 104, 2×104, 4×104, 8.73×104, respectively.
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ISSN:1012-0386
1662-9507
1662-9507
DOI:10.4028/p-826179