Polymer Processing by Calendering
Author's Data: Housam BINOUS
Department of Chemical Engineering
National Institute of Applied Sciences and Technology
Tunis, TUNISIA
Email: binoushousam@yahoo.com

Acknowledgement :
Analytical expression and numerical values were obtained from the excellent book Fluid Mechanics for Chemical Engineers, James O. Wilkes, Prentice Hall, 1999.

Off[General::"spell1"]

Numerical solution:
We evaluate the point of departure (x2=0.0225494) and the pressure distribution in a Calendered Sheet.

x1 = -0.05; α = 100;

\(Sol[x2_] := 
      NDSolve[{D[\(p\^*\)[x], {x, 1}] == \ 
            8\ α^1.5\ \((x^2 - 
                    x2^2)\)/\((1 + α\ x^2)\)^3, 
\(p\^*\)[x1] == \ 0}, \(p\^*\)[x], {x, x1, x2}]\)

\(pend[
        x2_?NumericQ] := \((\(\(p\^*\)[x] /. Sol[x2]\) /. 
          x -> x2)\)\)

sol = FindRoot[pend[ξ] ==  0, {ξ, 0.1}]

\(plt1 = 
      Plot[Evaluate[\(p\^*\)[x] /. Sol[ξ /. sol]], {x, x1, 
          Evaluate[ξ /. sol]}, PlotStyle -> RGBColor[1, 0, 0], 
        PlotRange -> {{\(-0.06\), 0.04}, {0, 0.125}}, 
        Frame -> True]\)

Analytical solution:
We get the same result as the numerical method for the point of departure and the pressure distribution in a Calendered Sheet.

FindRoot[((x1 (Sqrt[α]) (α x1^2 (1 - 3  α x2^2) - 1 - 5 α x2^2))/((1 + α x1^2)))^2 + ((x2 (Sqrt[ α]) (1 + 3  α x2^2))/(1 + α x2^2)) + (1 - 
                3  α x2^2) (ArcTan[x1 (Sqrt[ α])] - 
                ArcTan[x2 (Sqrt[ α])]) ==  0, {x2, 0.1}]

(Subscript[p, anal])[
        x_] := ((x (Sqrt[α]) (α x^2 (1 - 3  α x2^2) - 1 - 5 α x2^2))/((1 + α x^2)))^2 + ((x2 (Sqrt[ α]) (1 + 3  α x2^2))/(1 + α x2^2)) + (1 - 
              3  α x2^2) (ArcTan[x (Sqrt[ α])] - 
              ArcTan[x2 (Sqrt[ α])])

plt2 = 
      Plot[(Subscript[p, anal])[x] /. x2 -> (ξ /. sol), {x, x1, 
          Evaluate[ξ /. sol]}, PlotStyle -> RGBColor[0, 1, 0], 
        PlotRange -> {{-0.06, 0.04}, {0, 0.125}}, 
        Frame -> True]

Analytical solution and numerical solution agree.

Show[plt1, plt2]