Problem1.2.4 - Problem 1.2.4 (a) We consider an arbitrary...

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ü Problem 1.2.4 (a) We consider an arbitrary slice of thickness D x located between x = a and x = b . If u H x, t L is the density or concentaration per unit volume in that slice and j ê is the mass flux per unit area, then the material balance is (1) d ÅÅÅÅÅ dt Ÿ a b u H x, t L A H x L x = - j ê ÿ n ê a A H a L - j ê ÿ n ê b A H b L where n ê a is the outward directed normal to the face of the slice, pointing in the negative x -direction at x = 0 ; n ê b is the outward directed normal at the face located at x = b , pointing in the positive x -direction; A H x L is the surface area of the slice. Now since (2) - j ê ÿ n ê a A H a L - j ê ÿ n ê b A H b L = j H a L A H a L - j H b L A H b L = - Ÿ a b ÅÅÅÅÅ x H jA L x We can rewrite the above balance as (3) Ÿ a b ÅÅÅÅÅ t @ u H x, t L A H x LD x + Ÿ a b ÅÅÅÅÅ x H jA L x = 0 Since our slice is arbitrary, the integrand under the common integral must vanish to give: (4) ÅÅÅÅÅ t @ u H x, t L A H x LD =- ÅÅÅÅÅ x H j A L But, (5) j H x, t L = - ! u ÅÅÅÅÅÅÅ x Thus our material balance becomes (6) ÅÅÅÅÅ t @ u H x, t L A H x LD = ÅÅÅÅÅ x H
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Problem1.2.4 - Problem 1.2.4 (a) We consider an arbitrary...

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