-the thiele modulus
- how to use eq (1) to get eq (2)
- how to get the end of reaction (2.2)
- how to get equation H1
The Thiele Modulus How To Use Eq 1 To Get Eq 2 How To Get The End Of Reaction 2 2 How To Get Equation H1 1 (45.19 KiB) Viewed 26 times
The Thiele Modulus How To Use Eq 1 To Get Eq 2 How To Get The End Of Reaction 2 2 How To Get Equation H1 2 (39.17 KiB) Viewed 26 times
Dividing by (A.Az), we get: [A. (-D. d), -A.-D. d)] (-De dCA e dz dz Α. Δz Taking the limit as Az → 0, gives: d (D. dC) - P. K. C₁ = 0 De CA dz dz For De = f(z) = constant → d²CA Pp. k.CA = (1) dz² De The boundary conditions are given by @z = 0 →>> CA = CAS (surface concentration) (1.1) dCA @z=L→ (1.2) = 0 (end of reaction) dz 2/1/2022 Advar Rendering Equations (1, 1.1, and 1.2) dimensionless by defining: 0.5 CA Pp. k. 1² X = ; y = ; Thiele modulus = 0 = De CAS d²x dL² L 0²X = 0 z+Az - Pp.k. CA = 0 (2) 6
The same profiles could have been produced had we used Equ. (4) and 10 boundary conditions (2.1 and 2.2) X = C cosh(Ø y) + D sinh(Ø y) (4) @y=0X = 1 (surface concentration) (2.1) dX @ y = 1 = 0 (end of reaction) (2.2) dy The derivative of Equ. (4) can be obtained by using → d[cosh(y)] =¢ sinh(¢ y) = dy d[sinh(¢ y)] dy = cosh ( y) = ø Applying these we get: cosh([Ø (1y)]) (H1) cosh (Ø) X = е фу 2 ey + ey 2 y -
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