Unsteady Flow and Heat Transfer of a Dusty Fluid through a Rectangular Channel

Abstract

The present discussion deals with the study of an unsteady flow and heat transfer of a dusty fluid through a rectangular channel under the influence of pulsatile pressure gradient along with the effect of a uniform magnetic field. The analytical solutions of the problem are obtained using variable separable and Fourier transform techniques. The graphs are drawn for the velocity fields of both fluid and dust phases under the effect of Reynolds number. Further, changes in the Nusselt number are shown graphically, and, on the basis of these, the conclusions and discussions are given.

1. Introduction

The concept of an unsteady flow and heat transfer of a dusty fluid has a wide range of applications in refrigeration, air conditioning, space heating, power generation, chemical processing, pumps, accelerators, nuclear reactors, filtration and geothermal systems, and so forth. One common example of heat transfer is the radiator in a car, in which the hot radiator fluid is cooled by the flow of air over the radiator surface. On this basis many mathematicians were attracted by this field.

Saffman [1] has formulated the governing equations for the flow of dusty fluid and has discussed the stability of the laminar flow of a dusty gas in which the dust particles are uniformly distributed. Datta et al. [2] have obtained the solution of unsteady heat transfer to pulsatile flow of a dusty viscous incompressible fluid in a channel. Heat transfer in unsteady laminar flow through a channel was analyzed by Ariel [3]. Ghosh et al. [4] have made the solution for hall effects on MHD flow in a rotating system with heat transfer characteristics. Ezzat et al. [5] analyzed a space approach to the hydromagnetic flow of a dusty fluid through a porous medium.

Some researchers like Anjali Devi and Jothimani [6] have discussed the heat transfer in unsteady MHD oscillatory flow. Further, Malashetty et al. [7] have investigated the convective magnetohydrodynamic two phase flow and heat transfer of a fluid in an inclined channel. Palani and Ganesan [8] have discussed the heat transfer effects on dusty gas flow past a semi-infinite inclined plate. Attia [9] has investigated an unsteady MHD Couette flow and heat transfer of dusty fluid with variable physical properties.

Unsteady hydromagnetic flow and heat transfer from a nonisothermal stretching sheet immersed in a porous medium was discussed by Chamkha [10]. Mishra et al. [11] have studied the two-dimensional transient conduction and radiation heat transfer with temperature-dependent thermal conductivity. MHD flow and heat transfer of a dusty visco-elastic stratified fluid down an inclined channel in porous medium under variable viscosity was analyzed by Chakraborty [12]. Shawky [13] has investigated the solution for pulsatile flow with heat transfer of dusty magnetohydrodynamic Ree-Eyring fluid through a channel.

Gireesha et al. [14] have obtained the analytical solutions for velocity fields using variable separable method for an unsteady flow of dusty fluid through a rectangular channel under the influence of pulsatile pressure gradients and in the absence of a magnetic field. In continuation of this paper and with the help of the above cited papers we have studied an unsteady flow and heat transport in a dusty fluid through a rectangular channel under the influence of a pulsatile pressure gradient in the presence of uniform magnetic field and viscous dissipation term. Further, heat transfer analysis and the effect of Reynolds number, Prandtl number, and Nusselt number have been considered. This paper presents three methods of solution, namely, perturbation technique, Fourier decomposition, and finite Fourier transform, to obtain useful results on the problem. Finally, the graphical representation of velocity fields of both fluid and dust phases and changes in the Nusselt number are drawn for different values of Reynolds number and Prandtl number.

2. Equations of Motion

The governing equations of motion and energy for two phases are given by [1] the following.

For fluid phase,𝜕𝑢∇⋅𝑢=0,+()1𝜕𝑡𝑢⋅∇𝑢=−𝜌∇𝑝+𝜈∇²𝑓𝑢+𝜏_𝑣(𝑣−𝑢)−𝜎𝐵²₀𝜌𝜌𝑢,𝜕𝐸+((𝑢)()𝜕𝑡𝑢⋅∇𝐸)=𝑄+𝑣−⋅𝐹+𝑘∇⋅∇𝑇+Φ_𝑓. (2.1)

For dust phase,𝜕𝑣∇⋅𝑣=0,+()1𝜕𝑡𝑣⋅∇𝑣=𝜏_𝑣(𝑣),𝑁𝑢−𝜕𝐸_𝑝+𝜕𝑡𝑣⋅∇𝐸_𝑝=−𝑄−Φ_𝑑. (2.2) We have following nomenclature. 𝐸=𝑐_𝑝𝑇, 𝐸_𝑝=𝑐_𝑚𝑇_𝑝, 𝑄=𝑁𝑐_𝑝(𝑇_𝑝−𝑇)/𝜏_𝑇 is the thermal interaction between fluid and dust particle phases, 𝐹=𝑁(𝑣−𝑢)/𝜏_𝑣 is the velocity interaction force between the fluid and dust particle phase, 𝜏_𝑣=𝑚/6𝜋𝑎𝜇=𝑚/𝐾 is the velocity relaxation time of the dust particles, 𝜏_𝑇=𝑚𝑐_𝑝/4𝜋𝑎𝑘 is the thermal relaxation time of the dust particles, 𝑘∇.(∇𝑇) is the rate of heat added to the fluid by conduction in unit volume, Φ_𝑓andΦ_𝑑 are the viscous dissipation of fluid and dust particles. 𝑢, 𝜌, 𝑝, 𝜈, 𝑇, 𝑐_𝑝, and 𝑘 are, respectively, the velocity vector, density, pressure, kinematic viscosity, temperature, specific heat, and thermal conductivity of the fluid, 𝑣, 𝑁, 𝑇_𝑝, 𝑐_𝑚, and 𝑚 are, respectively, the velocity vector, number density, temperature, specific heat, mass concentration of dust particles, 𝐾=6𝜋𝑎𝜇 is the Stoke's resistance coefficient, and 𝑡 is the time.

3. Formulation of the Problem

Consider an unsteady flow of an incompressible, viscous, electrically conducting fluid with uniform distribution of dust particles through a rectangular channel. It is assumed that the flow is due to the time-dependent pressure gradient and applied uniform magnetic field. Both the fluid and the dust particle clouds are supposed to be static at the beginning. The dust particles are assumed to be spherical in shape and uniform in size. The number density of the dust particles is taken as a constant throughout the flow. The flow is taken along 𝑧-axis, and it is as shown in Figure 1. For the above described flow the velocities of both fluid and dust particles are given by 𝑢=𝑢𝑥,𝑦,𝑡𝑘,𝑣=𝑣𝑥,𝑦,𝑡𝑘. (3.1)

Figure 1 

Schematic diagram of dusty fluid flow in a rectangular channel.

4. Solution of the Problem

The governing equations from (2.1) and (2.2) can be decomposed as follows.

For fluid phase,𝜕𝑢1𝜕𝑡=−𝜌𝜕𝑝𝜕𝜕𝑧+𝜈²𝑢𝜕𝑥²+𝜕²𝑢𝜕𝑦²+𝑓𝜏_𝑣(𝑣−𝑢)−𝜎𝐵²₀𝜌𝑢,𝜌𝑐_𝑝𝜕𝑇=𝜕𝑡𝑁𝑐_𝑝𝜏_𝑇𝑇_𝑝+𝑁−𝑇𝜏_𝑣(𝑣−𝑢)²𝜕+𝑘²𝑇𝜕𝑥²+𝜕²𝑇𝜕𝑦²4+𝜇3𝜕𝑢𝜕𝑥²+𝜕𝑢𝜕𝑦². (4.1)

For dust phase,𝜕𝑣=1𝜕𝑡𝜏_𝑣(),𝑐𝑢−𝑣_𝑚𝜕𝑇_𝑝=𝑐𝜕𝑡𝜏^𝑝_𝑇𝑇−𝑇_𝑝−𝜇_𝑝𝑁43𝜕𝑣𝜕𝑥²+𝜕𝑣𝜕𝑦², (4.2) where 𝑢 and 𝑣 denote the velocity of the fluid and the dust phases, respectively.

The boundary conditions of the given problem are taken as𝑢=0,𝑣=0,𝑇=𝑇₀at𝑦=0,𝑢=0,𝑣=0,𝑇=𝑇₁at𝑦=ℎ,𝑢=0,𝑣=0,𝑇=𝑇₂at𝑥=−𝑑,𝑢=0,𝑣=0,𝑇=𝑇₃at𝑥=𝑑. (4.3) Since we have assumed that the pulsatile pressure gradient has influence on the flow, we have that−1𝜌𝜕𝑝𝜕𝑧=𝐴1+𝜖𝑒^𝑖𝜔𝑡, (4.4) where 𝜖 is a small quantity and 𝐴 and 𝜔 are constants.

To make the above system dimensionless, introduce the following nondimensional variables:𝑢=𝑢𝜔𝐴,𝑣=𝑣𝜔𝐴,𝑡=𝑡𝜔,𝜃=𝑇−𝑇₀𝑇₁−𝑇₀,𝑥𝜉=ℎ𝑦,𝜂=ℎ𝑧,𝜓=ℎ,𝑝𝑝=𝐴𝜌ℎ,𝜃_𝑝=𝑇_𝑝−𝑇₀𝑇₁−𝑇₀,𝑃_𝑟=𝜇𝑐_𝑝𝑘,Re=𝜔ℎ²𝜈,Re_𝑝=𝜔ℎ²𝜈_𝑝𝐴,Ec=²𝜔²𝑐_𝑝𝑇₁−𝑇₀, (4.5) where ℎ is the distance between plates, Ec the Eckert number, 𝑃_𝑟 the Prandtl number, Re and Re_𝑝 the Reynolds numbers of fluid and dust phases, 𝑥 and 𝑦 the space coordinates along and perpendicular to the plates, 𝜃 and 𝜃_𝑝 the dimensionless fluid and dust phase temperatures, 𝜇 the viscosity of fluid, and 𝜈_𝑝 and 𝜌_𝑝 the kinematic viscosity and density of the dust particles.

Using the above nondimensional variables in (4.1) and (4.2) and dropping the bars, one can get that𝜕𝑢𝜕𝑡=−𝜕𝑝+1𝜕𝜓𝜕Re²𝑢𝜕𝜉²+𝜕²𝑢𝜕𝜂²+𝑓𝛼(𝑣−𝑢)−𝛽𝑢,𝜕𝜃=1𝜕𝑡Re𝑃_𝑟𝜕²𝜃𝜕𝜉²+𝜕²𝜃𝜕𝜂²+𝛼₁𝛽₁𝜃_𝑝−𝜃+𝛼𝛽₁Ec(𝑣−𝑢)²+Ec4Re3𝜕𝑢𝜕𝜉²+𝜕𝑢𝜕𝜂²,𝜕𝑣(),𝜕𝑡=𝛼𝑢−𝑣𝜕𝜃_𝑝𝜕𝑡=𝛾𝛼₁𝜃−𝜃_𝑝−𝛾EcRe_𝑝𝛽₂43𝜕𝑣𝜕𝜉²+𝜕𝑣𝜕𝜂², (4.6) where 𝛼=1/𝜔𝜏_𝑣, 𝛽=𝜎𝐵²₀/𝜌𝜔, 𝛼₁=1/𝜔𝜏_𝑇, 𝛽₁=𝑁/𝜌, 𝛽₂=𝑁/𝜌_𝑝, and 𝛾=𝑐_𝑝/𝑐_𝑚.

The dimensionless boundary conditions are𝑢=0,𝑣=0,𝜃=0at𝜂=0,𝑢=0,𝑣=0,𝜃=1at𝜂=1,𝑢=0,𝑣=0,𝜃=𝑇_𝑎at𝜉=−𝑟,𝑢=0,𝑣=0,𝜃=𝑇_𝑏at𝜉=𝑟, (4.7) where 𝑇_𝑎=(𝑇₂−𝑇₀)/(𝑇₁−𝑇₀), 𝑇_𝑏=(𝑇₃−𝑇₀)/(𝑇₁−𝑇₀), and 𝑟=𝑑/ℎ.

The nondimensional form of pressure gradient is given by−𝜕𝑝=𝜕𝜓1+𝜖𝑒^𝑖𝑡. (4.8) Now assume the solutions of the velocities and temperature of both fluid and dust phases as𝑢𝜉,𝜂,𝑡=𝑢₀𝜉,𝜂+𝜖𝑢₁𝜉,𝜂𝑒^𝑖𝑡,𝑣𝜉,𝜂,𝑡=𝑣₀𝜉,𝜂+𝜖𝑣₁𝜉,𝜂𝑒^𝑖𝑡,𝜃𝜉,𝜂,𝑡=𝜃₀𝜉,𝜂+𝜖𝜃₁𝜉,𝜂𝑒^𝑖𝑡+𝜖²𝜃₂𝜉,𝜂𝑒^2𝑖𝑡,𝜃_𝑝𝜉,𝜂,𝑡=𝜃_𝑝0𝜉,𝜂+𝜖𝜃_𝑝1𝜉,𝜂𝑒^𝑖𝑡+𝜖²𝜃_𝑝2𝜉,𝜂𝑒^2𝑖𝑡. (4.9)

Substituting (4.8) and (4.9) in (4.6) and equating the coefficient of the similar powers of 𝜖 on both sides, then we obtain the following set of equations.

Steady part (coefficient of 𝜖⁰):1𝜕Re²𝑢₀𝜕𝜉²+𝜕²𝑢₀𝜕𝜂²+𝑓𝛼𝑣₀−𝑢₀−𝛽𝑢₀𝛼𝑢=−1,₀−𝑣₀1=0,Re𝑃_𝑟𝜕²𝜃₀𝜕𝜉²+𝜕²𝜃₀𝜕𝜂²+𝛼₁𝛽₁𝜃_𝑝0−𝜃₀+𝛼𝛽₁Ec𝑣₀−𝑢₀²+Ec4Re3𝜕𝑢₀𝜕𝜉²+𝜕𝑢₀𝜕𝜂²=0,𝛾𝛼₁𝜃₀−𝜃_𝑝0−𝛾EcRe_𝑝𝛽₂43𝜕𝑣₀𝜕𝜉²+𝜕𝑣₀𝜕𝜂²=0. (4.10)(4.11)(4.12)

Unsteady part (coefficient of 𝜖):1𝜕Re²𝑢₁𝜕𝜉²+𝜕²𝑢₁𝜕𝜂²+𝑓𝛼𝑣₁−𝑢₁−𝛽+𝑖𝑢₁𝛼𝑢=−1,₁−𝑣₁−𝑖𝑣₁1=0,Re𝑃_𝑟𝜕²𝜃₁𝜕𝜉²+𝜕²𝜃₁𝜕𝜂²+𝛼₁𝛽₁𝜃_𝑝1−𝜃₁+2𝛼𝛽₁Ec𝑣₀−𝑢₀𝑣₁−𝑢₁+2Ec4Re3𝜕𝑢₀𝜕𝜉𝜕𝑢₁+𝜕𝜉𝜕𝑢₀𝜕𝜂𝜕𝑢₁𝜕𝜂−𝑖𝜃₁=0,𝛾𝛼₁𝜃₁−𝜃_𝑝1−2𝛾EcRe_𝑝𝛽₂43𝜕𝑣₀𝜕𝜉𝜕𝑣₁+𝜕𝜉𝜕𝑣₀𝜕𝜂𝜕𝑣₁𝜕𝜂−𝑖𝜃_𝑝1=0. (4.13)(4.14)(4.15)

Unsteady part (coefficient of 𝜖²):1Re𝑃_𝑟𝜕²𝜃₂𝜕𝜉²+𝜕²𝜃₂𝜕𝜂²+𝛼₁𝛽₁𝜃_𝑝2−𝜃₂+𝛼𝛽₁Ec𝑣₁−𝑢₁²+Ec4Re3𝜕𝑢₁𝜕𝜉²+𝜕𝑢₁𝜕𝜂²−2𝑖𝜃₂=0,𝛾𝛼₁𝜃₂−𝜃_𝑝2−𝛾EcRe_𝑝𝛽₂43𝜕𝑣₁𝜕𝜉²+𝜕𝑣₁𝜕𝜂²−2𝑖𝜃_𝑝2=0. (4.16)(4.17) Now, the corresponding dimensionless boundary conditions are as follows:𝑢₀=0,𝑣₀=0,𝜃₀=𝑇_𝑎𝑢at𝜉=−𝑟,₀=0,𝑣₀=0,𝜃₀=𝑇_𝑏𝑢at𝜉=𝑟,₀=0,𝑣₀=0,𝜃₀𝑢=0at𝜂=0,₀=0,𝑣₀=0,𝜃₀𝑢=1at𝜂=1;₁=0,𝑣₁=0,𝜃₁𝑢=0at𝜉=−𝑟,₁=0,𝑣₁=0,𝜃₁𝑢=0at𝜉=𝑟,₁=0,𝑣₁=0,𝜃₁𝑢=0at𝜂=0,₁=0,𝑣₁=0,𝜃₁𝜃=0at𝜂=1;₂𝜃=0at𝜉=−𝑟,₂𝜃=0at𝜉=𝑟,₂𝜃=0at𝜂=0,₂=0at𝜂=1. (4.18)(4.19)(4.20) By substituting (4.11) in (4.10), one can get𝜕²𝑢₀𝜕𝜉²+𝜕²𝑢₀𝜕𝜂²−𝛽Re𝑢₀=−Re. (4.21) To solve (4.21), we assume that the solution is in the form𝑢₀𝜉,𝜂=𝑋𝜉,𝜂+𝑌(𝜉). (4.22) Substituting 𝑢₀(𝜉,𝜂) in (4.21), then we obtain that𝜕²𝑋𝜕𝜉²+𝜕²𝑌𝜕𝜂²+𝜕²𝑋𝜕𝜂²−𝛽Re(𝑋+𝑌)=−Re. (4.23) so that𝜕²𝑌𝜕𝜂²𝜕−𝛽Re𝑌+Re=0,₂𝑋𝜕𝜉²+𝜕²𝑋𝜕𝜂²−𝛽Re𝑋=0. (4.24)(4.25) The corresponding boundary conditions will become𝑢₀−𝑟,𝜂=𝑋−𝑟,𝜂+𝑌(−𝑟)=0,𝑢₀(𝑟)𝑢𝑟,𝜂=𝑋𝑟,𝜂+𝑌=0,₀(𝜉,0)=𝑋(𝜉,0)+𝑌(𝜉)=0,𝑢₀(𝜉,1)=𝑋(𝜉,0)+𝑌(𝜉)=0. (4.26) By solving (4.24) and using the method of separation of variables to (4.25), we obtain the solution in the form𝑢₀=1𝜉,𝜂𝛽1−cosh𝛽Re𝜉cosh+2𝛽Re𝑟𝛽^∞_𝑛=1sin𝑛𝜋𝑟𝜉sinh𝐴₁𝜂−1−sinh𝐴₁𝜂sinh𝐴₁×1−𝑛𝜋𝑛𝜋𝑟²𝐴²₁cosh−𝛽Re𝑟𝛽Re𝑛𝜋𝐴²₁(−1)^𝑛,𝑣₀𝜉,𝜂=𝑢₀𝜉,𝜂, (4.27) where 𝐴₁=(𝛽Re𝑟²+𝑛²𝜋²)/𝑟².

Here one can observe that the velocity of steady part of the fluid and the dust phases is the same.

In a similar manner, by the method of separation of variables, the solution of (4.13), and using the boundary conditions (4.19), one can obtain that𝑢₁𝜉,𝜂=Re𝑄²()1−cosh𝑄𝜉()+cosh𝑄𝑟2Re𝑄^2∞_𝑛=1sin𝑛𝜋𝑟𝜉sinh𝐵𝜂−1−sinh𝐵𝜂×1sinh𝐵−𝑛𝜋𝑛𝜋𝑟²𝐵²()−𝑄cosh𝑄𝑟𝑛𝜋𝐵²²(−1)^𝑛,𝑣₁=𝛼𝑢𝛼+𝑖¹, (4.28) where 𝑄²=Re[𝑓𝛼𝑖/(𝛼+𝑖)+(𝛽+𝑖)] and 𝐵=(𝑄²𝑟²+𝑛²𝜋²)/𝑟².

We define the finite Fourier sine transform of 𝜃 and 𝜃_𝑝 as𝐹_𝑠(𝜃)=^𝑟_−𝑟𝜃𝜉,𝜂sin𝑛𝜋𝑟𝜉,𝐹_𝑠𝜃_𝑝=^𝑟_−𝑟𝜃_𝑝𝜉,𝜂sin𝑛𝜋𝑟𝜉. (4.29) Eliminating 𝜃_𝑝0 from (4.12), we get that𝜕²𝜃₀𝜕𝜉²+𝜕²𝜃₀𝜕𝜂²=𝐻₁43𝜕𝑢₀𝜕𝜉²+𝜕𝑢₀𝜕𝜂²,𝜃_𝑝0𝜉,𝜂=𝜃₀−EcRe_𝑝𝛼₁𝛽₂43𝜕𝑣₀𝜕𝜉²+𝜕𝑣₀𝜕𝜂², (4.30)(4.31) where 𝐻₁=EcRe𝑃_𝑟(𝛽₁/Re_𝑝𝛽₂−1/Re).

Applying the finite fourier sine transform to (4.30) with respect to the variable 𝜉 and to boundary conditions, one obtains that𝑑²𝐹_𝑠𝜃₀𝑑𝜂²−𝑛²𝜋²𝑟²𝐹_𝑠𝜃₀=^∞_𝑛=1𝑛𝜋𝑟𝑇_𝑏−𝑇_𝑎(−1)^𝑛+4𝐻₁38𝜋²𝛽Re𝛽²×^∞_𝑛=1𝑛²𝛽Re𝑟²+4𝑛²𝜋²sinh𝛽Re𝑟cosh×1𝛽Re𝑟−𝑛𝜋𝑛𝜋𝑟²𝐴²₁cosh−()𝛽Re𝑟𝛽Re−1^𝑛𝑛𝜋𝐴²₁×sinh𝐴₁𝜂−1−sinh𝐴₁𝜂sinh𝐴₁,𝐹_𝑠𝜃₀(𝜉,0)=0,𝐹_𝑠𝜃₀(𝜉,1)=0. (4.32)(4.33) The temperature of fluid is obtained by solving (4.32) with the help of boundary conditions (4.33) as𝐹_𝑠𝜃₀=^∞_𝑛=1𝑟𝑇𝑛𝜋^𝑏−𝑇_𝑎(−1)^𝑛+32𝜋²𝛽Re𝐻₁3𝛽^2∞_𝑛=1𝑛²𝛽Re𝑟²+4𝑛²𝜋²×𝑟²𝐴²₁𝑟²−𝑛²𝜋²sinh𝛽Re𝑟cosh1𝛽Re𝑟−𝑛𝜋𝑛𝜋𝑟²𝐴²₁cosh−()𝛽Re𝑟𝛽Re−1^𝑛𝑛𝜋𝐴²₁×cosh𝑛𝜋𝑟𝜂+1−cosh(𝑛𝜋/𝑟)𝜂sinh(𝑛𝜋/𝑟)sinh𝑛𝜋r𝜂−^∞_𝑛=1𝑟𝑇𝑛𝜋^𝑏−𝑇_𝑎(−1)^𝑛+32𝜋²𝛽Re𝐻₁3𝛽^2∞_𝑛=1𝑛²𝛽Re𝑟²+4𝑛²𝜋²×𝑟²𝐴²₁𝑟²−𝑛²𝜋²sinh𝛽Re𝑟cosh1𝛽Re𝑟−𝑛𝜋𝑛𝜋𝑟²𝐴²₁cosh−()𝛽Re𝑟𝛽Re−1^𝑛𝑛𝜋𝐴²₁×sinh𝐴₁𝜂−1−sinh𝐴₁𝜂sinh𝐴₁. (4.34) Now taking the inverse finite Fourier sine transform to (4.34), one can obtain that𝜃₀𝜉,𝜂=^∞_𝑟=12𝑟^∞_𝑛=1𝑟𝑇𝑛𝜋^𝑏−𝑇_𝑎(−1)^𝑛+32𝜋²𝛽Re𝐻₁3𝛽²×^∞_𝑛=1𝑛²𝛽Re𝑟²+4𝑛²𝜋²𝑟²𝐴²₁𝑟²−𝑛²𝜋²sinh𝛽Re𝑟cosh×1𝛽Re𝑟−𝑛𝜋𝑛𝜋𝑟²𝐴²₁cosh−()𝛽Re𝑟𝛽Re−1^𝑛𝑛𝜋𝐴²₁×cosh𝑛𝜋𝑟𝜂+1−cosh(𝑛𝜋/𝑟)𝜂sinh(𝑛𝜋/𝑟)sinh𝑛𝜋𝑟𝜂−^∞_𝑛=1𝑟𝑇𝑛𝜋^𝑏−𝑇_𝑎(−1)^𝑛+32𝜋²𝛽Re𝐻₁3𝛽^2∞_𝑛=1𝑛²𝛽Re𝑟²+4𝑛²𝜋²×𝑟²𝐴²₁𝑟²−𝑛²𝜋²sinh𝛽Re𝑟cosh𝛽Re𝑟sinh𝐴₁𝜂−1−sinh𝐴₁𝜂sinh𝐴₁×1−𝑛𝜋𝑛𝜋𝑟²𝐴²₁cosh−()𝛽Re𝑟𝛽Re−1^𝑛𝑛𝜋𝐴²₁sin𝑛𝜋𝑟𝜉. (4.35) The temperature of dust 𝜃_𝑝0 is obtained by substituting 𝜃₀ in (4.31).

Using (4.15) in (4.14) and boundary conditions (4.19), with the help of finite fourier sine transform technique, one can get the solution for 𝜃₁ as 𝜃₁=𝜉,𝜂_∞𝑟=12𝑟16Re𝜋²𝐻₃13𝛽𝑄^∞_𝑛=1𝑛²𝐴²₁−𝑞²₁1𝑄²𝑟²+4𝑛²𝜋²sinh(𝑄𝑟)()×1cosh𝑄𝑟−𝑛𝜋𝑛𝜋𝑟²𝐴²₁cosh−()𝛽Re𝑟𝛽Re−1^𝑛𝑛𝜋𝐴²₁+𝛽Re𝑄^2∞_𝑛=1𝑛²𝐵²−𝑞²₁1𝛽Re𝑟²+4𝑛²𝜋²×1−𝑛𝜋𝑛𝜋𝑄²𝑟²+𝑛²𝜋²()−𝑄cosh𝑄𝑟²(−1)^𝑛𝑛𝜋𝐵²sinh𝛽Re𝑟cosh×𝑞𝛽Re𝑟cosh₁𝜂+1−cosℎ𝑞₁sinh𝑞₁sinh𝑞₁𝜂+16Re𝜋²𝐻₃13𝛽𝑄^∞_𝑛=1𝑛²𝐴²₁−𝑞²₁1𝑄²𝑟²+4𝑛²𝜋²×1−𝑛𝜋𝑛𝜋𝑟²𝐴²₁cosh−()𝛽Re𝑟𝛽Re−1^𝑛𝑛𝜋𝐴²₁sinh(𝑄𝑟)()×cosh𝑄𝑟sinh𝐴₁𝜂−1−sinh𝐴₁𝜂sinh𝐴₁+𝛽Re𝑄^2∞_𝑛=1𝑛²𝐵²−𝑞²₁1𝛽Re𝑟²+4𝑛²𝜋²×1−𝑛𝜋𝑛𝜋𝑄²𝑟²+𝑛²𝜋²()−𝑄cosh𝑄𝑟²(−1)^𝑛𝑛𝜋𝐵²sinh𝛽Re𝑟cosh×𝛽Re𝑟sinh𝐵𝜂−1−sinh𝐵𝜂sinh𝐵sin𝑛𝜋𝑟𝜉. (4.36) Using 𝜃₁, we get the expression for 𝜃_𝑝1 as 𝜃_𝑝1𝜉,𝜂=𝛾𝛼₁𝛾𝛼₁𝜃+𝑖¹−2𝛾EcRe_𝑝𝛽₂𝛾𝛼₁4+𝑖3𝜕𝑣₀𝜕𝜉𝜕𝑣₁+𝜕𝜉𝜕𝑣₀𝜕𝜂𝜕𝑣₁,𝜕𝜂 (4.37) where 𝑞₁=𝐻₂+𝑛²𝜋²𝑟²,𝐻₂=Re𝑃_𝑟𝛼₁𝛽₁𝑖𝛾𝛼₁,𝐻+𝑖+𝑖₃=2EcRe𝑃_𝑟𝛼₁𝛽₁𝛾Re_𝑝𝛽₂𝛾𝛼₁𝛼+𝑖()−1𝛼+𝑖.Re (4.38)

Similarly, the solutions of (4.16) and (4.17) using boundary conditions (4.20) are obtained as 𝜃₂𝜉,𝜂=^∞_𝑟=12𝑟8Re²𝑄³𝐻₅𝑄^∞_𝑛=11𝑟−𝑄+4𝜋²𝐻₆3^∞_𝑛=1𝑛²𝑄²𝑟²+4𝑛²𝜋²1𝐵²−𝑞²₂×sinh(𝑄𝑟)()1cosh𝑄𝑟−𝑛𝜋𝑛𝜋𝑄²𝑟²+𝑛²𝜋²1()−𝑄cosh𝑄𝑟²(−1)^𝑛𝑛𝜋𝐵²+cosh𝑞₂𝜂1−cosh𝑞₂sinh𝑞₂sinh𝑞₂𝜂+sinh𝐵𝜂−1−sinh𝐵𝜂sinh𝐵×sin𝑛𝜋𝑟𝜉. (4.39) From (4.17), one can get that 𝜃_𝑝2𝜉,𝜂=𝛾𝛼₁𝛾𝛼₁𝜃+2𝑖²−𝛾EcRe_𝑝𝛽₂𝛾𝛼₁4+2𝑖3𝜕𝑣₁𝜕𝜉²+𝜕𝑣₁𝜕𝜂², (4.40) where 𝑞₂=𝐻₄+𝑛²𝜋²𝑟²,𝐻₄=Re𝑃_𝑟2𝑖𝛼₁𝛽₁𝛾𝛼₁,𝐻+2𝑖+2𝑖₅=−EcRe𝑃_𝑟𝛼𝛽₁(𝛼+𝑖)²,𝐻₆=EcRe𝑃_𝑟𝛼₁𝛽₁𝛾Re_𝑝𝛽₂𝛾𝛼₁𝛼+2𝑖𝛼+𝑖²−1.Re (4.41)

5. Results and Discussion

Figures 2, 3, 4, 5, 6, 7, and 8  represent the velocity and temperature fields, respectively, for the fluid and dust particles, which are parabolic in nature. Here we can see that the path of fluid particles is much steeper than that of dust particles. Further, one can see that if the dust is very fine, that is, the mass of the dust particles is negligibly small, then the relaxation time of dust particle decreases and ultimately as 𝜏_𝑣→0 the velocities of fluid and dust particles will be the same. Also we see that the fluid particles will reach the steady state earlier than the dust particles. Further, one can observe the impressive effect of Reynolds number on the velocity fields. It means that the Reynolds number is favorable to the velocity fields, that is, the velocity profiles for both fluid and dust particles increases as the Reynolds number increases.

(a)

(b)

Figure 2 

Variation of fluid velocity 𝑢0(𝜉,𝜂) with 𝜉 and 𝜂 (for Re=1 and Re=10).

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(b)

Figure 3 

Variation of dust velocity 𝑣0(𝜉,𝜂) with 𝜉 and 𝜂 (for Re=1 and Re=10).

(a)

(b)

Figure 4 

Variation of fluid velocity 𝑢1(𝜉,𝜂) with 𝜉 and 𝜂 (for Re=1 and Re=10).

(a)

(b)

Figure 5 

Variation of dust velocity 𝑣1(𝜉,𝜂) with 𝜉 and 𝜂  (for Re=1 and Re=10).

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(b)

Figure 6 

Steady part of the Nusselt number (Nu0) versus 𝜉and𝜂 (for 𝜉=𝑟 and 𝜂=0 and 𝜉=−𝑟 and 𝜂=1).

(a)

(b)

Figure 7 

Unsteady part of Nusselt number (Nu1) versus 𝜉and𝜂 (for 𝜉=𝑟 and 𝜂=0) and 𝜉=−𝑟 and 𝜂=1).

(a)

(b)

Figure 8 

Unsteady part of Nusselt number (Nu2) versus 𝜉and𝜂 (for 𝜉=𝑟 and 𝜂=0 and 𝜉=−𝑟 and 𝜂=1).

The graphs are drawn for the following values: 𝜔=0.5, 𝑁=0.4, 𝛽=2, 𝜎=1, Ec=0.02, 𝑃_𝑟=0.72, 𝑇₀=0.5, 𝑇₁=1, 𝑇₂=1.5, 𝑇₃=2, 𝜏_𝑣=𝜏_𝑇=0.15, and 𝛾=1.4.

Now we discuss the heat transfer at the vertical walls, so we consider the Nusselt number (Nu) of the fluid as Nu=−𝜕𝜃𝜕𝜉_{at𝜉=𝑟or𝜉=−𝑟}=−𝑑𝜃₀𝑑𝜉+𝜖𝑒^𝑖𝑡𝑑𝜃₁𝑑𝜉+𝜖²𝑒^2𝑖𝑡𝑑𝜃₂𝑑𝜉_{at𝜉=𝑟or𝜉=−𝑟}=−Nu₀+𝜖𝑒^𝑖𝑡Nu₁+𝜖²𝑒^2𝑖𝑡Nu_{2at𝜉=𝑟or𝜉=−𝑟}. (5.1)

Next to discuss is the heat transfer at the horizontal walls, so we consider the Nusselt number (Nu) of the fluid as Nu=−𝜕𝜃𝜕𝜂_{at𝜂=0or𝜂=1}=−𝑑𝜃₀𝑑𝜂+𝜖𝑒^𝑖𝑡𝑑𝜃₁𝑑𝜂+𝜖²𝑒^2𝑖𝑡𝑑𝜃₂𝑑𝜂_{at𝜂=0𝑜r𝜂=1}=−Nu₀+𝜖𝑒^𝑖𝑡Nu₁+𝜖²𝑒^2𝑖𝑡Nu_{2at𝜂=0or𝜂=1}, (5.2) where Nu₀, Nu₁, and Nu₂ denote the Nusselt number for steady part, unsteady part for coefficient of 𝜖, and unsteady part for coefficient of 𝜖², respectively.

The graphs of steady part of Nusselt number (Nu₀) against 𝜉 and 𝜂 (at 𝜂=0 and 𝜉=𝑟 or at 𝜂=1 and 𝜉=−𝑟) has been drawn in Figure 6. It shows that for different values of Prandtl number, Nusselt number increases with increase in 𝜉 and 𝜂.

Figure 7 shows the unsteady part of amplitude of Nusselt number (Nu₁) against 𝜉 and 𝜂 (at 𝜂=0 and 𝜉=𝑟 or at 𝜂=1 and 𝜉=−𝑟). It reveals that for different values of Prandtl number, amplitude of Nusselt number increases with increase of 𝜉 and 𝜂. The unsteady part of the amplitude of Nusselt number (Nu₂) against 𝜉 and 𝜂 has been drawn in Figure 8. Here, one can see that the amplitude of Nusselt number increases with increase of 𝜉 and 𝜂 for different values of Prandtl number.

6. Conclusions

A detailed analytical study has been carried out for the unsteady flow and heat transfer of a dusty fluid through a rectangular channel. Here, one can see that the flow of fluid particles is parallel to that of dust. Further, one can see that the fluid particles will reach the steady state earlier than the dust particles. From the graphs the impressive effect of Reynolds number on the velocity fields of both fluid and dust phases is evident. It is clear that the effect of Reynolds number on velocity fields is favorable, that is, the velocity profiles for both fluid and dust particles increase as Reynolds number increases.

Further, one can observe the changes in the steady and unsteady parts of amplitude of Nusselt number. It is clear that for different values of Prandtl number steady part of Nusselt number (Nu₀) increases with increase of 𝜉 and 𝜂. In the same manner unsteady parts of amplitude of Nusselt number (Nu₁) and (Nu₂) increases with increase of 𝜉 and 𝜂 for different values of Prandtl number.

Acknowledgments

The authors wish to express their thanks to DST (Department of Science and Technology), india for granting a Major Research Project (no: SR/S4/MS: 470/07, 25-08-2008). They are also grateful to the referees for useful comments.