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Application of differential transformation method (DTM) for heat and mass transfer in a porous channel

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Application of differential transformation method (DTM) for heat and mass transfer in a porous channel Q2 Q1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 3[.]

Trang 1

H O S T E D B Y

ORIGINAL ARTICLE

Application of differential transformation method

(DTM) for heat and mass transfer in a porous

channel

Q2

S Sepasgozara,n, M Farajib, P Valipourc

Q1

a

Department of Civil Engineering, Babol University of Technology, Babol, Iran

b

Department of Mechanical Engineering, Babol University of Technology, Babol, Iran

c

Department of Textile and Apparel, Qaemshahr Branch, Islamic Azad University, Qaemshahr, Iran

Received 10 April 2015; accepted 4 March 2016

KEYWORDS

Differential

transfor-mation method

(DTM);

Axisymmetric channel;

Rotating disk;

Porous media;

Non-Newtonianfluid

Abstract In the present paper a differential transformation method (DTM) is used to obtain the solution of momentum and heat transfer equations of non-Newtonian fluid flow in an axisymmetric channel with porous wall The comparison between the results from the differential transformation method and numerical method are in well agreement which proofs the capability of this method for solving such problems After this validity, results are investigated for the velocity and temperature for various values of Reynolds number, Prandtl number and power law index

& 2017 National Laboratory for Aeronautics and Astronautics Production and hosting by Elsevier B.V.

This is an open access article under the CC BY-NC-ND license

( http://creativecommons.org/licenses/by-nc-nd/4.0/ ).

1 Introduction

The problem of non-Newtonianfluid flow has been under a

lot of attention in recent years Various applications in different

fields of engineering specially the interest in heat transfer

problems of non-Newtonian fluid flow, such as hot rolling, lubrication, cooling problems and drag reduction was the main reason for this considerable attention Debruge and Han [1] Q3

studied a problem concerning heat transfer in channel flow, which can be considered as an application of the previous works reported by Yuan [2], White [3] and Treill [4] Q4

Increasing the resistance of the blades against the hot stream around the blades for cooling was interested However the cooling process gives rise to excess energy consumption which leads to largely decrease of turbine efficiency

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http://ppr.buaa.edu.cn/

www.sciencedirect.com

Propulsion and Power Research

http://dx.doi.org/10.1016/j.jppr.2017.01.001

2212-540X & 2017 National Laboratory for Aeronautics and Astronautics Production and hosting by Elsevier B.V This is an open access article under the

CC BY-NC-ND license ( http://creativecommons.org/licenses/by-nc-nd/4.0/ ).

n Corresponding author.

E-mail address: s.sepasgozar@yahoo.com (S Sepasgozar).

Peer review under responsibility of National Laboratory for Aeronautics

and Astronautics, China.

Propulsion and Power Research ]]]];](]):]]]–]]]

Trang 2

Most of phenomena in our world are essentially nonlinear and

are described by nonlinear equations Nonlinear differential

equations usually arise from mathematical modeling of many

physical systems Some of them are solved using numerical

methods and some are solved using analytic methods such as

perturbation Perturbation techniques are based on the existence

of small or large parameters, this is called perturbation quantity

Unfortunately, many nonlinear problems in science and

engi-neering do not contain such kind of perturbation quantities

Therefore, same as the HAM[5,6], HPM[7–10], ADM[11,12]

and OHAM[13,14]can overcome the foregoing restrictions and

limitations of perturbation methods One of the semi-exact

methods which does not need small parameters is the differential

transformation method This method constructs an analytical

solution in the form of a polynomial It is different from the

traditional higher-order Taylor series method The Taylor series

method is computationally expensive for large orders The

differential transform method is an alternative procedure for

obtaining an analytic Taylor series solution of differential

equations The main advantage of this method is that it can be

applied directly to nonlinear differential equations without

requiring linearization, discretization and therefore, it is not

affected by errors associated to discretization The concept of

DTM wasfirst introduced by Zhou[15], who solved linear and

nonlinear problems in electrical circuits Chen and Ho [16]

developed this method for partial differential equations and Ayaz

[17]applied it to the system of differential equations Jang et al

[18]applied the two-dimensional differential transform method

to the solution of partial differential equations Sheikholeslami

et al [19]used this method to solve the problem of nanofluid

flow between parallel plates considering Thermophoresis and

Brownian effects Sheikholeslami and Ganji[20]applied DTM

to solve the problem of nanofluid flow and heat transfer between

parallel plates considering Brownian motion They concluded

that Nusselt number increases with augment of nanoparticle

volume fraction, Hartmann number while it decreases with

increase of the squeeze number Natural convection of a

non-Newtonian copper–water nanofluid between two infinite parallel

verticalflat plates is investigated by Domairry et al.[21] They conclude that as the nanoparticle volume fraction increases, the momentum boundary layer thickness increases, whereas the thermal boundary layer thickness decreases New analytical and numerical method has been developed in recent year in different field of science[22–56]

In this study, the purpose is to solve nonlinear equations via DTM It can be seen that this method is strongly capable for solving a large class of coupled and nonlinear differ-ential equations without tangible restriction of sensitivity to the degree of the nonlinear term

2 Mathematical formulation

2.1 Flow analysis

This study is concerned with simultaneous development of flow and heat transfer for non-Newtonian viscoelastic fluid flow on the turbine disc for cooling purposes The problem to

be considered is depicted schematically inFigure 1 The r-axis

is parallel to the surface of disk and the z-axis is normal to it

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Nomenclature

A, B symmetric kinematic matrices

C specific heat

Cn blade-wall temperature coefficients

f velocity function

F transformation ofƒ

xk general coordinates

Kr rotation parameter

k fluid thermal conductivity

n power law index in temperature distribution

p fluid pressure

qnð Þη temperature function

Pr Prandtl number

Re injection Reynolds number

T temperature

ur; uz velocity components in r, z directions, respectively

V injection velocity r; θ; z cylindrical coordinate symbols

δv m

δx n velocity gradients

δa m

δx n acceleration gradients

Greek symbols

ρ fluid density

τij stress tensor component

ϕk viscosity coefficients

φ dissipation function

η dimension less coordinates in z direction

ψ stream function

Trang 3

The porous disc of the channel is at z¼þL The wall that

coincides with the r-axis is heated externally and from the

other perforated wall non-Newtonian fluid is injected

uni-formly in order to cool the heated wall

As can observed inFigure 1the cooling problem of the disk

can be considered as a stagnation pointflow with injection For

a steady, ax symmetric, non-newtonianfluid flow the following

equations can be written in cylindrical coordinates

The continuity equation:

∂ðrurÞ

∂r þ

∂ðruzÞ

And the momentum equations:

ur

∂ðurÞ

∂r þ uz

∂ðurÞ

∂z

¼  1ρ∂P∂rþ1ρ ∂τrr

∂r þ

1

rðτrrτθθÞ þ∂τrz

∂z

ð2Þ

ur∂ðuzÞ

∂r þ uz∂ðuzÞ

∂z

¼  1

ρ

∂P

∂zþ

1 ρ

∂τzr

∂r þ

1

rτrzþ∂τzz

∂z

ð3Þ

Hereτrr; τrz; τzr; τzz are the components of stress matrix

The analytical model under consideration leads to the

following boundary conditions:

@z ¼ 0 ur¼ uz¼ 0 ð4Þ

@z ¼ L ur¼ 0; uz¼ V ð5Þ

Here ur; uz are the velocity components in the r and z

directions and V is the injection velocity;ρ; P are the density,

and pressure For particular class of viscoelastic and

viscoi-nelasticfluids Rivlin[36]showed that if the stress components

τij at a point xk ðk ¼ 1; 2; 3Þ and time t are assumed to be

polynomials in the velocity gradientδv m

δx n ðm; n ¼ 1; 2; 3Þ and the acceleration gradients δa m

δx n ðm; n ¼ 1; 2; 3Þ, and if in addition the medium is assumed to

Q5 be isotropic the stress

matrix can be expressed in the form

‖τij‖ ¼ ϕ0Iþ ϕ1A þ ϕ2Bþ ϕ3A2þ ::: ð6Þ

Here I is the unit matrices, A and B are symmetric

kinematic matrixes defined by:

A¼ δvi

δxj

þδvj

δxi



;









B¼ δai

δxj

þδaj

δxi

þ 2δvm

δxi

δvm

δxj











And ϕk ðk ¼ 0; 1; 2; 3Þ are polynomials in the invariants

of A, B, A2 This study is restricted to second orderfluids for

which ϕk ðk ¼ 0; 1; 2; 3Þ are constant and ϕk ðk ¼ 4; 5; …Þ

are zero So that the stress components are as follows:

τrr¼ ϕ1Arrþ ϕ2Arr2þ ϕ3Brr ð8Þ

τzz¼ ϕ1Azzþ ϕ2Azz2þ ϕ3Bzz ð9Þ

τθθ¼ ϕ1Aθθþ ϕ2Aθθ2þ ϕ3Bθθ ð10Þ

τrz¼ ϕ1Arzþ ϕ2Arz2þ ϕ3Brz ð11Þ For the solution of the problem depicted inFigure 1 in the case of axially symmetricflow it is convenient to define

a stream function so that the continuity equation is satisfied:

ψ ¼ Vr2fðηÞ ð12Þ Where η ¼ z

L and the velocity components can be derived as:

ur¼V r

L f

0

uz¼ 2Vf ðηÞ ð14Þ Using Eqs.(12)–(14)the equations of motion reduce to:

f022f f00¼  L2

ρV2r

∂P

∂rþ

ϕ1

ρVLf

000

þ ϕ2

ρL2ð f0022f0f000Þ þ ϕ3

ρL2ð f0022f fivÞ

ð15Þ 4f f0¼  L2

ρV2

∂P

∂z 2

ϕ1

ρVLf

00

þ 2 ϕ2

ρL2 14f0f00þr2

Lf

00

f000

þ 4 ϕ3

ρL2 11f0f00þ f f000þr2

Lf

00

f000

ð16Þ The pressure term can be eliminated by differentiating Eq.(15)

with respect to z and Eq.(16)with respect to r and subtracting the resulting equations This gives the following equations:

 2f f000¼f iv

ReK1 ð4f00f000þ 2f0fivÞ

 K2 ð4f00f000þ 2f0fivþ 2f fvÞ ð17Þ Where K1¼ Φ 2

ρL 2 is the injection Reynolds number For K2¼ 0,the equation turned to:

fivþ 2 Re f f000 K1 Reð4f00f000þ 2f0fivÞ ¼ 0 ð18Þ The boundary conditions are:

fð0Þ ¼ 0; f0ð0Þ ¼ 0;

fð1Þ ¼ 1; f0ð1Þ ¼ 0: ð19Þ

2.2 Heat transfer analysis

The energy equation for the present problem with viscous dissipation in non-dimensional form is given by:

ρC ur

∂T

∂r þ uz

∂T

∂z

¼ k∇2

Tþ φ ð20Þ

φ ¼ τrr∂ur

∂r þ τθθ

ur

r þ τzz∂uz

∂z þ τrz ∂ur

∂z þ

∂uz

∂r

ð21Þ

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Trang 4

Here ur; uz are the velocity components in the r and z

directions and V is the injection velocity;ρ; P; T; C; k are the

density, pressure, temperature, specific heat, and heat conduction

coefficient of fluid, respectively φ is the dissipation function

Letting the blade wall (z¼0) temperature distribution be

Tw¼ T0þ P1

r L

 n

And assuming thefluid temperature to have the form of

Ref.[1]

T¼ T0þX1

Cn

r L

n

qnðηÞ ð22Þ Where T0 is the temperature of the incoming coolant

(z¼L) and neglecting dissipation effect the following

equations and boundary conditions are obtained:

qn00  Pr Re ð f0

qn2f qn

0

Þ ¼ 0;

ðn ¼ 0; 2; 3; 4; :::Þ ð23Þ

qnð0Þ ¼ 1; qnð1Þ ¼ 0 ð24Þ

3 Fundamentals of differential transform

method [20]

Basic definitions and operations of differential

transfor-mation are introduced as follows Differential

transforma-tion of the functransforma-tion fð Þ is defined as follows:η

F kð Þ ¼ 1

k!

dkfð Þη

dηk

 

η ¼ η 0

ð25Þ

In Eq (6), fð Þ is the original function and F kη ð Þ is the

transformed function which is called the T-function (it is also

called the spectrum of the fð Þ at η ¼ ηη 0, in the k domain) The

differential inverse transformation of F kð Þ is defined as:

fð Þ ¼η X1

FðkÞðηη0Þk ð26Þ

by combining Eq.(25)and Eq.(26), fð Þ can be obtained:η

fð Þ ¼η X1

dkfð Þη

dηk

 

η ¼ η 0

ð ηη0Þk

k! ð27Þ

Eq.(27)implies that the concept of the differential transforma-tion is derived from Taylor's series expansion, but the method does not evaluate the derivatives symbolically However, relative derivatives are calculated by an iterative procedure that is described by the transformed equations of the original functions

From the definitions of Eqs.(25)and(26), it is easily proven that the transformed functions comply with the basic mathematical operations shown in below In real applications, the function fð Þη

in Eq.(27)is expressed by afinite series and can be written as:

fð Þ ¼η X

N

FðkÞðηη0Þk ð28Þ

Eq (9) implies that fð Þ ¼η P1k¼ Nþ1FðkÞðηη0Þk

is negligibly small, where N is series size

Theorems to be used in the transformation procedure, which can be evaluated from Eqs.(25)and(26), are given below (Table 1)

4 Solution with DTM

Now differential transformation method into governing equations has been applied Taking the differential trans-forms of Eqs (18) and (23) with respect to χ and considering H¼1 gives:

ðk þ 1Þðk þ 2Þðk þ 3Þðk þ 4ÞF½k þ 4

þ 2 Re X

k

F½k mðm þ 1Þ

ðm þ 2Þðm þ 3ÞF½m þ 3

!

 4 Re K1

Xk

ðk m þ 1Þðk m þ 2Þ

F½k m þ 2ðm þ 1Þ

ðm þ 2Þðm þ 3ÞF½m þ 3

0 B

1 C

 2 Re K1

Xk

ðk m þ 1ÞF½k m þ 1

ðm þ 1Þðm þ 2Þðm þ 3Þ

ðm þ 4ÞF½m þ 4

0 B

1 C

A ¼ 0

ð29Þ

F½0 ¼ 0; F½1 ¼ 0; F½2 ¼ α; F½3 ¼ β ð30Þ

ðk þ 1Þðk þ 2ÞQn½k þ 2

 nPrReX

k

ðm þ 1ÞF½m þ 1Qn½k m

þ 2PrReX

k

ðm þ 1Þ Qn½m þ 1F½k m

ð31Þ

Qn½0 ¼ 1; Qn½1 ¼ λ ð32Þ where FðkÞ; QnðkÞ are the differential transforms of

fðηÞ; qnðηÞ and α; β; λ are constants which can be obtained through boundary condition, Eqs (19) and (24) This problem can be solved as followed:

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method.

Original function Transformed function

f ðηÞ ¼ αgðηÞ7βhðηÞ F ½k ¼ αG½k7βH½k

f ðηÞ ¼ d n g ðηÞ

dη n F ½k ¼ ðkþnÞ!

k! G½k þ n

f ðηÞ ¼ gðηÞhðηÞ F ½k ¼Pk

m  0F½mH½k m

f ðτÞ ¼ sin ðϖη þ αÞ F ½k ¼ ϖ k

k! sinð πk

2 þ αÞ

f ðτÞ ¼ cos ðϖη þ αÞ F½k ¼ϖ k

k ! cosð πk

2 þ αÞ

f ðηÞ ¼ e λη F½k ¼ λ k

k !

FðηÞ ¼ 1 þ η ð Þ m

F ½k ¼ m ðm  1Þ:::ðm  kþ1Þ

k!

f ðηÞ ¼ η m

F ½k ¼ δðk mÞ ¼ 10 ; k ¼ m; k am

(

Trang 5

F½0 ¼ 0

F½1 ¼ 0

F½2 ¼ α

F½3 ¼ β

F½4 ¼ 2 Re K1αβ

F½5 ¼ 6

F½6 ¼  1

þ 64

F½7 ¼  1

105α2Re2K1β

þ 132

þ 265

F½8 ¼ 11

35Re3K1 α3β

þ 1752

þ 768

ð33Þ

Qn½0 ¼ 1

Qn½1 ¼ λ

Qn½2 ¼ 0

Qn½3 ¼ 1

3PrRenα

Qn½4 ¼ 1

6PrRenαλ þ1

4PrRenβ1

Qn½5 ¼ 3

20PrRenβλ þ2

 1

Qn½6 ¼ 1

45Pr2Re2n2α2

þ 4

15PrRe2nK1αβλ þ1

þ 4

 2

15Pr2Re2α2n

Qn½7 ¼ 1

126Pr2Re2n2α2λ þ1

28Pr2Re2n2βα

 5

126Pr2Re2nα2λ þ1

7PrRe2nλK1β2

þ 4

þ 44

 2

 2

21Pr2Re2βnα þ 2

63Pr2Re2α2λ

Qn½8 ¼ 1

70Pr2Re2n2αβλ þ13

210Pr2Re3n2K1α2β

 53

840Pr2Re2nαβλ þ 3

224Pr2Re2n2β2

 1

280PrRe2nλαβ þ33

þ 48

 11

þ 204

þ 1

 16

7Pr2Re3K1α2βn

 1

ð34Þ

The above process is continuous By substituting Eqs

(33)and (34)into the main equation based on DTM, it can

be obtained that the closed form of the solutions is:

FðηÞ ¼ 1 þ α η2þ βη3þ 2 Re Kð 1αβÞ η4

þ 6

η5

þ 

1

þ64

!

η6þ :::

ð35Þ

Qnð Þ ¼ 1 þ λη þη 1

3Pr Re nα

η3

þ 1

6Pr Re nαλ þ1

4Pr Re nβ1

6Pr Reαλ

η4

þ

3

20Pr Re nβλ þ2

5Pr Re2nK1αβ

 1

10Pr Reβλ

!

η5

þ

1

15Pr Re2nK1αβλ

þ1

5Pr Re2n K1β2

þ4

5Pr Re3nK1 α2β 2

15Pr Re2K1αβλ

 1

!

η6

by substituting the boundary condition from Eqs.(19)and

(24) into Eqs (35) and (36) in point η ¼ 1, it can be obtained the values ofα; β; λ

Fð1Þ ¼ 1 þ α þ β þ 2 Re Kð 1αβÞ

þ 6

þ  1

F0ð1Þ ¼ 1 þ 2α þ 3β þ 4 2 Re Kð 1αβÞ η4

þ 5 6

þ 6 

1

þ64

!

Qnð Þ ¼ 1 þ λ þ1 1

3Pr Re nα

þ 1

6Pr Re nαλ þ1

4Pr Re nβ1

6Pr Reαλ

þ 3

20Pr Re nβλ þ2

5Pr Re2nK1αβ 1

10Pr Reβλ

þ 1

15Pr Re2nK1αβλ



þ 1

5Pr Re

5Pr Re

 2

15Pr Re

15Pr

2Re2α2n



By solving Eqs.(37)–(39)gives the values ofα; β; λ By substituting obtainedα; β; λ into Eqs (35)and (36), it can

be obtained the expression of FðηÞ; QnðηÞ

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Trang 6

5 Results and discussion

The objective of the present study was to apply the differential

transformation method to obtain an explicit analytic solution of

heat transfer equation of a non-Newtonian fluid flow in an

axisymmetric channel with a porous wall for turbine cooling

applications (Figure 1) The results that obtained by the

differential transformation method were well matched with the

results carried out by the numerical solution obtained by a

four-1

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η N ¼4 Error N ¼6 Error

0.1 0.031767 0.00214 0.031767 0.002141

0.2 0.116853 0.007411 0.116874 0.007432

0.3 0.239782 0.013885 0.239975 0.014078

0.4 0.384867 0.01927 0.385834 0.020237

0.5 0.536216 0.020762 0.539633 0.024179

0.6 0.677728 0.014878 0.68739 0.024541

0.7 0.793091 0.002704 0.816482 0.020687

0.8 0.865789 0.037278 0.916281 0.013214

0.9 0.879096 0.095201 0.978892 0.004596

η N ¼8 Error N ¼10 Error

0.1 0.029687 0.000059 0.02955 0.000076

0.2 0.109737 0.000295 0.109271 0.000171

0.3 0.226588 0.000692 0.225724 0.000173

0.4 0.366737 0.00114 0.365524 0.000073

0.5 0.516936 0.001481 0.515532 0.000077

0.6 0.664418 0.001569 0.66306 0.000211

0.7 0.797125 0.00133 0.796061 0.000267

0.8 0.903891 0.000823 0.903288 0.00022

0.9 0.974563 0.000266 0.974391 0.000094

 

Trang 7

order Runge–Kutta method as shown inTable 2 In these tables,

the error is introduced as followed:

Error¼ f ðηÞ NMf ðηÞDTM ð40Þ

As shown in Figures 2 and 3 for constant value of K1,

velocity values increase as a result of Reynolds number

increase At low Reynolds numbers the velocity profile exhibit

center line symmetry indicating a Poiseuille flow for

non-Newtonianfluids At higher Reynolds numbers the maximum

velocity point is shifted to the solid wall where shear stress

becomes larger as the Reynolds number grows Since f00ð0Þ is

measure of friction force, it is advisable to use viscoinelastic

fluids as a coolant fluid for industrial gas turbine engines

In Figure 4 temperature profiles for different values of

power law index (n) are shown It shows that for constant

value of η temperature increase if power law index

decrease Increasing Reynolds number leads to increasing

the curve of temperature profile and decreasing of qnðηÞ

values as shown in Figure 5

6 Conclusion

In this paper, non-Newtonianfluid flow in an axisymmetric

channel with a porous wall for turbine cooling applications

problem (Figure 1) has been solved via a sort of analytical

method, differential transformation method (DTM), and this

problem has been also solved by a numerical method (the

Runge–Kutta method of order 4) The observed good

agree-ment between the present method and numerical results shows

that differential transformation method is a powerful approach

for solving nonlinear differential equation such as this problem

Some facts were observed through the results Friction force

increased as a result of Reynolds numbers increase Increasing

power law index leads to decreasing temperature value

between two plates Nusselt number increases as a result of

Reynolds number increase Prandtl number increase leads to

Nusselt number increase as well Also Nusselt number increase

as an effect made by power law increase

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