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Technical Brief

Homogeneous–Heterogeneous Reactions in Boundary-Layer Flow of a Nanofluid Near the Forward Stagnation Point of a Cylinder

[+] Author and Article Information
Qingkai Zhao

Collaborative Innovation Center for Advanced Ship and
Deep-Sea Exploration (CISSE),
State Key Laboratory of Ocean Engineering,
School of Naval Architecture,
Ocean and Civil Engineering,
Shanghai Jiao Tong University,
Shanghai 200240, China
e-mail: qkzhao@sjtu.edu.cn

Hang Xu

Collaborative Innovation Center for Advanced Ship and
Deep-Sea Exploration (CISSE),
State Key Laboratory of Ocean Engineering,
School of Naval Architecture,
Ocean and Civil Engineering,
Shanghai Jiao Tong University,
Shanghai 200240, China
e-mail: hangxu@sjtu.edu.cn

Longbin Tao

School of Marine Science and Technology,
Newcastle University,
Newcastle NE1 7RU, UK
e-mail: longbin.tao@newcastle.ac.uk

1Corresponding author.

Contributed by the Heat Transfer Division of ASME for publication in the JOURNAL OF HEAT TRANSFER. Manuscript received July 3, 2016; final manuscript received September 19, 2016; published online November 16, 2016. Assoc. Editor: Alan McGaughey.

J. Heat Transfer 139(3), 034502 (Nov 16, 2016) (4 pages) Paper No: HT-16-1438; doi: 10.1115/1.4034902 History: Received July 03, 2016; Revised September 19, 2016

A mathematical model describing the homogeneous–heterogeneous reactions in the vicinity of the forward stagnation point of a cylinder immerged in a nanofluid is established. We assume that the homogeneous reaction is given by isothermal cubic autocatalator kinetics, while the heterogeneous reaction is chosen as first-order kinetics. The existence of multiple solutions through hysteresis bifurcations is discussed in detail for the various diffusion coefficients of reactant and autocatalyst.

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Figures

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Fig. 1

Physical model and coordinate system

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Fig. 2

ξ(0) against K with Pr = 1, Nb = 1, Nt = 0.05, Sc = 1, and ε = 1.5 for some values of Ks

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Fig. 3

ξ(0) against K with Pr = 1, Nb = 1, Ks = 0.025, Sc = 1, and ε = 1 for different values of Nt

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Fig. 4

ξ(0) against Ks with Pr = 1, Nb = 1, Nt = 0.05, Sc = 1, and ε = 1.5 for different values of K

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Fig. 5

ξ(0) against ε with Pr = 1, Nb = 1, Nt = 0.05, Sc = 1, and Ks = 0.025 for different values of K

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Fig. 6

ξ(0) against Nt with Pr = 1, Sc = 1, Nb = 1, ε = 1.5, and Ks = 0.025 for different values of K

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