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

Authors: B. J. Gireesha; G. S. Roopa; C. S. Bagewadi
Journal: Mathematical Problems in Engineering
 (2010)

---

## 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.


---

## Body

## 1. Introduction

The concept of an unsteady flow and heat transfer of a dusty fluidhas 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,(2.1)∇⋅u⃗=0,∂u⃗∂t+(u⃗⋅∇)u⃗=-1ρ∇p+ν∇2u⃗+fτv(v⃗-u⃗)-σB02ρu⃗,ρ{∂E∂t+(u⃗⋅∇E)}=Q+(v⃗-u⃗)⋅F+k∇⋅(∇T)+Φf.For dust phase,(2.2)∇⋅v⃗=0,∂v⃗∂t+(v⃗⋅∇)v⃗=1τv(u⃗-v⃗),N{∂Ep∂t+(v⃗⋅∇Ep)}=-Q-Φd.
We have following nomenclature. E=cpT, Ep=cmTp, Q=Ncp(Tp-T)/τT is the thermal interaction between fluid and dust particle phases, F=N(v⃗-u⃗)/τv is the velocity interaction force between the fluid and dust particle phase, τv=m/6πaμ=m/K is the velocity relaxation time of the dust particles, τT=mcp/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, ΦfandΦd are the viscous dissipation of fluid and dust particles. u⃗, ρ, p, ν, T, cp, and k are, respectively, the velocity vector, density, pressure, kinematic viscosity, temperature, specific heat, and thermal conductivity of the fluid, v⃗, N, Tp, cm, 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 alongz-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 (3.1)u⃗=u(x,y,t)k̂,v⃗=v(x,y,t)k̂.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,(4.1)∂u∂t=-1ρ∂p∂z+ν(∂2u∂x2+∂2u∂y2)+fτv(v-u)-σB02ρu,ρcp∂T∂t=NcpτT(Tp-T)+Nτv(v-u)2+k(∂2T∂x2+∂2T∂y2)+μ[43(∂u∂x)2+(∂u∂y)2].For dust phase,(4.2)∂v∂t=1τv(u-v),cm∂Tp∂t=cpτT(T-Tp)-μpN[43(∂v∂x)2+(∂v∂y)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 as(4.3)u=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.
Since we have assumed that the pulsatile pressure gradient has influence on the flow, we have that(4.4)-1ρ∂p∂z=A[1+ϵeiωt],
where ϵ is a small quantity and A and ω are constants.To make the above system dimensionless, introduce the following nondimensional variables:(4.5)u¯=uωA,v¯=vωA,t¯=tω,θ=T-T0T1-T0,ξ=xh,η=yh,ψ=zh,p¯=pAρh,θp=Tp-T0T1-T0,Pr=μcpk,Re=ωh2ν,Rep=ωh2νp,Ec=A2ω2cp(T1-T0),
where h is the distance between plates, Ec the Eckert number, Pr the Prandtl number, Re and Rep the Reynolds numbers of fluid and dust phases, x and y the space coordinates along and perpendicular to the plates, θ and θp the dimensionless fluid and dust phase temperatures, μ the viscosity of fluid, and νp and ρp 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(4.6)∂u∂t=-∂p∂ψ+1Re(∂2u∂ξ2+∂2u∂η2)+fα(v-u)-βu,∂θ∂t=1RePr(∂2θ∂ξ2+∂2θ∂η2)+α1β1(θp-θ)+αβ1Ec(v-u)2+EcRe[43(∂u∂ξ)2+(∂u∂η)2],∂v∂t=α(u-v),∂θp∂t=γα1(θ-θp)-γEcRepβ2[43(∂v∂ξ)2+(∂v∂η)2],
where α=1/ωτv, β=σB02/ρω, α1=1/ωτT, β1=N/ρ, β2=N/ρp, and  γ=cp/cm.The dimensionless boundary conditions are(4.7)u=0,v=0,θ=0atη=0,u=0,v=0,θ=1atη=1,u=0,v=0,θ=Taatξ=-r,u=0,v=0,θ=Tbatξ=r,
where Ta=(T2-T0)/(T1-T0), Tb=(T3-T0)/(T1-T0), and r=d/h.The nondimensional form of pressure gradient is given by(4.8)-∂p∂ψ=[1+ϵeit].
Now assume the solutions of the velocities and temperature of both fluid and dust phases as(4.9)u(ξ,η,t)=u0(ξ,η)+ϵu1(ξ,η)eit,v(ξ,η,t)=v0(ξ,η)+ϵv1(ξ,η)eit,θ(ξ,η,t)=θ0(ξ,η)+ϵθ1(ξ,η)eit+ϵ2θ2(ξ,η)e2it,θp(ξ,η,t)=θp0(ξ,η)+ϵθp1(ξ,η)eit+ϵ2θp2(ξ,η)e2it.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):(4.10)1Re[∂2u0∂ξ2+∂2u0∂η2]+fα(v0-u0)-βu0=-1,(4.11)α(u0-v0)=0,(4.12)1RePr[∂2θ0∂ξ2+∂2θ0∂η2]+α1β1(θp0-θ0)+αβ1Ec(v0-u0)2+EcRe[43(∂u0∂ξ)2+(∂u0∂η)2]=0,γα1(θ0-θp0)-γEcRepβ2[43(∂v0∂ξ)2+(∂v0∂η)2]=0.Unsteady part (coefficient of ϵ):(4.13)1Re[∂2u1∂ξ2+∂2u1∂η2]+fα(v1-u1)-(β+i)u1=-1,α(u1-v1)-iv1=0,(4.14)1RePr[∂2θ1∂ξ2+∂2θ1∂η2]+α1β1(θp1-θ1)+2αβ1Ec(v0-u0)(v1-u1)+2EcRe[43∂u0∂ξ∂u1∂ξ+∂u0∂η∂u1∂η]-iθ1=0,(4.15)γα1[θ1-θp1]-2γEcRepβ2[43∂v0∂ξ∂v1∂ξ+∂v0∂η∂v1∂η]-iθp1=0.Unsteady part (coefficient of ϵ2):(4.16)1RePr[∂2θ2∂ξ2+∂2θ2∂η2]+α1β1(θp2-θ2)+αβ1Ec(v1-u1)2+EcRe[43(∂u1∂ξ)2+(∂u1∂η)2]-2iθ2=0,(4.17)γα1[θ2-θp2]-γEcRepβ2[43(∂v1∂ξ)2+(∂v1∂η)2]-2iθp2=0.
Now, the corresponding dimensionless boundary conditions are as follows:(4.18)u0=0,v0=0,θ0=Taatξ=-r,u0=0,v0=0,θ0=Tbatξ=r,u0=0,v0=0,θ0=0atη=0,u0=0,v0=0,θ0=1atη=1;(4.19)u1=0,v1=0,θ1=0atξ=-r,u1=0,v1=0,θ1=0atξ=r,u1=0,v1=0,θ1=0atη=0,u1=0,v1=0,θ1=0atη=1;(4.20)θ2=0atξ=-r,θ2=0atξ=r,θ2=0atη=0,θ2=0atη=1.
By substituting (4.11) in (4.10), one can get(4.21)∂2u0∂ξ2+∂2u0∂η2-βReu0=-Re.
To solve (4.21), we assume that the solution is in the form(4.22)u0(ξ,η)=X(ξ,η)+Y(ξ).
Substituting u0(ξ,η) in (4.21), then we obtain that(4.23)∂2X∂ξ2+∂2Y∂η2+∂2X∂η2-βRe(X+Y)=-Re.
so that(4.24)∂2Y∂η2-βReY+Re=0,(4.25)∂2X∂ξ2+∂2X∂η2-βReX=0.
The corresponding boundary conditions will become(4.26)u0(-r,η)=X(-r,η)+Y(-r)=0,u0(r,η)=X(r,η)+Y(r)=0,u0(ξ,0)=X(ξ,0)+Y(ξ)=0,u0(ξ,1)=X(ξ,0)+Y(ξ)=0.
By solving (4.24) and using the method of separation of variables to (4.25), we obtain the solution in the form(4.27)u0(ξ,η)=1β[1-cosh⁡(βReξ)cosh⁡(βRer)]+2β∑n=1∞sin(nπrξ)[sinh⁡A1(η-1)-sinh⁡(A1η)sinh⁡A1]×{1nπ-nπr2A12cosh⁡(βRer)-βRenπA12(-1)n},v0(ξ,η)=u0(ξ,η),
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 that(4.28)u1(ξ,η)=ReQ2[1-cosh⁡(Qξ)cosh⁡(Qr)]+2ReQ2∑n=1∞sin(nπrξ)[sinh⁡B(η-1)-sinh⁡(Bη)sinh⁡B]×{1nπ-nπr2B2cosh⁡(Qr)-Q2nπB2(-1)n},v1=(αα+i)u1,
where Q2=Re[fαi/(α+i)+(β+i)] and B=(Q2r2+n2π2)/r2.We define the finite Fourier sine transform ofθ and θp as(4.29)Fs(θ)=∫-rrθ(ξ,η)sin(nπrξ),Fs(θp)=∫-rrθp(ξ,η)sin(nπrξ).
Eliminating θp0 from (4.12), we get that(4.30)∂2θ0∂ξ2+∂2θ0∂η2=H1[43(∂u0∂ξ)2+(∂u0∂η)2],(4.31)θp0(ξ,η)=θ0-EcRepα1β2[43(∂v0∂ξ)2+(∂v0∂η)2],
where H1=EcRePr(β1/Repβ2-1/Re).Applying the finite fourier sine transform to (4.30) with respect to the variable ξ and to boundary conditions, one obtains that(4.32)d2Fs(θ0)dη2-n2π2r2Fs(θ0)=∑n=1∞nπr[(Tb-Ta)(-1)n]+4H138π2βReβ2×∑n=1∞n2βRer2+4n2π2sinh⁡(βRer)cosh⁡(βRer)×(1nπ-nπr2A12cosh⁡(βRer)-βRe(-1)nnπA12)×[sinh⁡A1(η-1)-sinh⁡(A1η)sinh⁡A1],(4.33)Fs(θ0(ξ,0))=0,Fs(θ0(ξ,1))=0.
The temperature of fluid is obtained by solving (4.32) with the help of boundary conditions (4.33) as(4.34)Fs(θ0)=[sinh⁡(βRer)cosh⁡(βRer)∑n=1∞rnπ[(Tb-Ta)(-1)n]+32π2βReH13β2∑n=1∞n2βRer2+4n2π2×r2A12r2-n2π2sinh⁡(βRer)cosh⁡(βRer)(1nπ-nπr2A12cosh⁡(βRer)-βRe(-1)nnπA12)]×{cosh⁡(nπrη)+1-cosh⁡((nπ/r)η)sinh⁡(nπ/r)sinh⁡(nπrη)}-∑n=1∞rnπ[(Tb-Ta)(-1)n]+32π2βReH13β2∑n=1∞n2βRer2+4n2π2×r2A12r2-n2π2sinh⁡(βRer)cosh⁡(βRer)(1nπ-nπr2A12cosh⁡(βRer)-βRe(-1)nnπA12)×[sinh⁡A1(η-1)-sinh⁡(A1η)sinh⁡A1].
Now taking the inverse finite Fourier sine transform to (4.34), one can obtain that(4.35)θ0(ξ,η)=∑r=1∞2r{[βRe(-1)nnπA12(1nπ-nπr2A12cosh⁡(βRer)-βRe(-1)nnπA12)∑n=1∞rnπ[(Tb-Ta)(-1)n]+32π2βReH13β2×∑n=1∞n2βRer2+4n2π2r2A12r2-n2π2sinh⁡(βRer)cosh⁡(βRer)×(1nπ-nπr2A12cosh⁡(βRer)-βRe(-1)nnπA12)]×{cosh⁡(nπrη)+1-cosh⁡((nπ/r)η)sinh⁡(nπ/r)sinh⁡(nπrη)}-∑n=1∞rnπ[(Tb-Ta)(-1)n]+32π2βReH13β2∑n=1∞n2βRer2+4n2π2×r2A12r2-n2π2sinh⁡(βRer)cosh⁡(βRer)[sinh⁡A1(η-1)-sinh⁡(A1η)sinh⁡A1]×(1nπ-nπr2A12cosh⁡(βRer)-βRe(-1)nnπA12)}sin(nπrξ).
The temperature of dust θp0 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 (4.36)θ1(ξ,η)=∑r=1∞2r{16Reπ2H33β[1Q∑n=1∞n2A12-q121Q2r2+4n2π2sinh⁡(Qr)cosh⁡(Qr)×(1nπ-nπr2A12cosh⁡(βRer)-βRe(-1)nnπA12)+βReQ2∑n=1∞n2(B2-q12)1βRer2+4n2π2×(1nπ-nπ(Q2r2+n2π2)cosh⁡(Qr)-Q2(-1)nnπB2)sinh⁡βRercosh⁡βRer]×{cosh⁡(q1η)+(1-cos⁡hq1)sinh⁡q1sinh⁡(q1η)}+16Reπ2H33β[1Q∑n=1∞n2A12-q121Q2r2+4n2π2×(1nπ-nπr2A12cosh⁡(βRer)-βRe(-1)nnπA12)sinh⁡(Qr)cosh⁡(Qr)×[sinh⁡A1(η-1)-sinh⁡(A1η)sinh⁡A1]+(βRe)Q2∑n=1∞n2(B2-q12)1βRer2+4n2π2×(1nπ-nπ(Q2r2+n2π2)cosh⁡(Qr)-Q2(-1)nnπB2)sinh⁡βRercosh⁡βRer×(sinh⁡B(η-1)-sinh⁡(Bη)sinh⁡B)]}sin(nπrξ).
Using θ1, we get the expression for θp1 as (4.37)θp1(ξ,η)=(γα1γα1+i)θ1-2γEcRepβ2(γα1+i)[43∂v0∂ξ∂v1∂ξ+∂v0∂η∂v1∂η],
where (4.38)q1=H2+n2π2r2,H2=RePr[α1β1iγα1+i+i],H3=2EcRePr[α1β1γRepβ2(γα1+i)α(α+i)-1Re].Similarly, the solutions of (4.16) and (4.17) using boundary conditions (4.20) are obtained as (4.39)θ2(ξ,η)=∑r=1∞2r{8Re2Q3[H5Q∑n=1∞(r-1Q)+4π2H63∑n=1∞n2Q2r2+4n2π2]1B2-q22×sinh⁡(Qr)cosh⁡(Qr)(1nπ-nπQ2r2+n2π21cosh⁡(Qr)-Q2(-1)nnπB2)+[cosh⁡(q2η)(1-cosh⁡q2)sinh⁡q2sinh⁡(q2η)+sinh⁡B(η-1)-sinh⁡(Bη)sinh⁡B]}×sin(nπrξ).
From (4.17), one can get that (4.40)θp2(ξ,η)=(γα1γα1+2i)θ2-γEcRepβ2(γα1+2i)[43(∂v1∂ξ)2+(∂v1∂η)2],
where (4.41)q2=H4+n2π2r2,H4=RePr[2iα1β1γα1+2i+2i],H5=-Ec RePrαβ1(α+i)2,H6=Ec RePr[α1β1γRepβ2(γα1+2i)(αα+i)2-1Re].


## 5. Results and Discussion

Figures2, 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 τv→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.Variation of fluid velocityu0(ξ,η) with ξ and η (for Re=1 and Re=10).
(a)(b)Variation of dust velocityv0(ξ,η) with ξ and η (for Re=1 and Re=10).
(a)(b)Variation of fluid velocityu1(ξ,η) with ξ and η (for Re=1 and Re=10).
(a)(b)Variation of dust velocityv1(ξ,η) with ξ and η  (for Re=1 and Re=10).
(a)(b)Steady part of the Nusselt number (Nu0) versus ξandη (for ξ=r and η=0 and ξ=-r and η=1).
(a)(b)Unsteady part of Nusselt number (Nu1) versus ξandη (for ξ=r and η=0) and ξ=-r and η=1).
(a)(b)Unsteady part of Nusselt number (Nu2) versus ξandη (for ξ=r and η=0 and ξ=-r and η=1).
(a)(b)The graphs are drawn for the following values:ω=0.5, N=0.4, β=2, σ=1, Ec=0.02, Pr=0.72, T0=0.5, T1=1, T2=1.5, T3=2, τv=τT=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 (5.1)Nu=-∂θ∂ξ|atξ=rorξ=-r=-[dθ0dξ+ϵeitdθ1dξ+ϵ2e2itdθ2dξ]atξ=rorξ=-r=-[Nu0+ϵeitNu1+ϵ2e2itNu2]atξ=rorξ=-r.Next to discuss is the heat transfer at the horizontal walls, so we consider the Nusselt number(Nu) of the fluid as (5.2)Nu=-∂θ∂η|atη=0orη=1=-[dθ0dη+ϵeitdθ1dη+ϵ2e2itdθ2dη]atη=0orη=1=-[Nu0+ϵeitNu1+ϵ2e2itNu2]atη=0orη=1,
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.Figure7 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.



---

*Source: 898720-2011-02-22.xml*