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

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 )