Analytical solution for fully developed channel and pipe flow of Phan ...

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274 P. J. Oliveira and F. T. PinhoThe pressure gradient −p ,x is usually unknown but can be related to the crosssectionalaverage velocity through the definition of the flow rate following Bird,Armstrong & Hassager (1987)ū ≡ 1 ∫ H2 j y j u(y)dyH j+1 0and the integration yields the solution of the direct problem, the flux for a givenpressure gradientū =()−p ,x H 21+ ɛλ2 p 2 ,xH 2 (j +3). (12)η(j + 1)(j +3) 2 (2j−1) η 2 (j +5)For the solution of the inverse problem, the determination of the pressure gradientfor a given flux, it is advantageous to work with the normalized velocity profile (10b)which needs to be integrated to yield the following non-dimensional cubic equationfor ū N /ū:1=ūN ū( ) 2 ) (ūN1+būwith8(3 + j)κ2b ≡ ɛDe 2 .(5 + j)This equation shows that ū N < ū, i.e. for an identical longitudinal pressure gradient ahigher flow rate results for the PTT fluid than for the Newtonian fluid on account ofthe shear-thinning behaviour. From the Cardan–Tartaglia formula for the solution ofalgebraic cubic equations it can be readily shown that the real solution of (13) isū Nū = (432)1/6 (δ 2/3 − 2 2/3 )(14)6b 1/2 δ 1/3with the following definitions used to simplify the notation:α =3 3 b +4; β =3 3/2 b 1/2 ; δ = α 1/2 + β.Equation (14) gives the explicit relation for the pressure gradient as a function of thecross-sectional average velocity ū, once ū N is substituted by its definition (equation(11)). The main results of the analysis for the linearized PTT fluid are therefore thevelocity profile (10), the flux equation (12) and the unknown driving pressure gradientat given flow rate obtained from (14). The maximum velocity at the centreline (y =0)is also useful and is given byu 0ū = κ1+4κ2 ɛDe 2 (ū N /ū) 21+b(ū N /ū) 2showing that it is smaller than in the Newtonian case. Expressions for the normalizedstress components are readily obtained after scaling with the wall shear stress forthe Newtonian (or UCM) fluid. From equation (8), the non-dimensional normalstress is) 2 ( ) 2τ xx(ūN y2κη ū/H =4κDe (15)ū H(13)
Fully developed flow of Phan-Thien–Tanner fluids 275and the normalized shear stress component is calculated from (7):)( )τ xy(ūN y2κη ū/H = − . (16)ū HThe shear strain rate, from (9), becomes˙γ(y)2κ ū/H = −ūN ūand the viscosity profile isµ(˙γ) ≡ τ xy˙γ⇒ µ(˙γ)η( yH)(1+8κ 2 ɛDe 2 (ūNū=) 2 ( yH) 2 )(17)() 2 ( ) 2 ) −1 (ūN y1+8κ 2 ɛDe 2 . (18)ū HWall values for these quantities are useful to define non-dimensional quantities andare obtained after setting y = H:) 2 ) −1µ w(ūN(1+8κη = 2 ɛDe 2 (τ xy ) w,ū 2κη ū/H = ūN andū) 2(τ xx ) w(ūN2κη ū/H =4κDe ,ū(19)where (τ xy ) w is defined positive. These values are smaller than the corresponding valuesfor the UCM fluid, a point to be taken into account when comparing non-dimensionalvalues.2.2. Exponential PTT modelThe normal and shear stress profiles are independent of the function f(trτ), thereforethey are still given by equations (15) and (16), respectively. However, the ratio ū N /ūin those expressions is different since it depends on the new velocity distribution. Thisdistribution is obtained in a similar way by inserting the new f function (equation(3)) into equation (9) followed by integration, to yield the dimensional and thecorresponding non-dimensional forms of the velocity profile, respectively:andu(y) = exp(ɛλ2 H 2 p 2 ,x/(2 (2j−1) η 2 ))−p ,x ɛλ 2 /(2 j−2 η)u(y)ū= κūN ū(1 − exp(− ɛλ2 p 2 ,xη 2 2 2j−1 (H2 − y 2 )))(20a)exp (b(ū N /ū) 2 )b(ū N /ū) 2 (1 − exp(−b(ū N /ū) 2 (1 − (y/H) 2 ))), (20b)where now b ≡ 8κ 2 ɛDe 2 . This is also valid for the axisymmetric case, after effectingthe appropriate changes. Appropriately, in the limit of small b(ū N /ū) 2 , equation (20b)tends to the parabolic profile, as it should.The equation relating the flow rate and the pressure gradient, required for thesolution of the direct problem, is no longer common to both geometries. For thechannel, we obtain( 2ɛλ 2 p 2 ,xH 2ū =−η (exp4ɛλ 2 p ,xη 2 )−η2λp ,x Hπ 1/2 erf (i(λp ,x H/η) √ )2ɛ)i √ 2ɛ(21a)and for the pipeū =−η ( ɛλ 2 p 2 ,xH 2 )(exp1+ exp(−ɛλ2 p 2 ,xH 2 /(2η 2 )))−1. (21b)2ɛλ 2 p ,x 2η 2 ɛλ 2 p 2 ,xH 2 /(2η 2 )
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