Step by step please 1.00.58 0.0-0.5-1.0-3-2-112X2. A very narrow pipe of length L and cross-sectional area A is filled with water.At time t = 0, a certain amount of salt, of total mass M, is placed at one end thepipe, reaching a length d. The pipe is then sealed at both ends. We wish to knowthe concentration of salt, C, at any time. Since the pipe is narrow, the variationsof concentration of salt over the cross section of the pipe can be neglected, and theconcentration is given by C = C (x, t). The concentration satisfies the 1-D diffusionequation:D-ƏtThe initial concentration profile isMAdC(x, 0) = {0 < x < d0, d a) Start with a solution of the form C(x, t)equation and show that the problem reduces to two ordinary differential equations:X (x)T(t), replace in the diffusionX"(x) = AX(x); T'(t) = ADT(t)where A is an arbitrary constant.b) We seek time-decaying solutions, show that implies A = -k², where k is an arbi-trary constant.c) Solve the two ODES and show that after applying the two boundary conditions:C(x, t) = an cos(-Σn=0d) Find C(x, t) so that it satisfies the initial concentration profile.e) Show that your expression for C(x, t) satisfies the equilibrium profile C(x, t –)expected from conservation of mass.3

Hi! As per norms, we will be solving only the first three subparts. If you need an answer to others then kindly re-post the question by specifying it. Given: A very narrow pipe of length L and cross-sectional area A is filled with water. At time t = 0, a certain amount of salt, of total mass M, is placed at one end the pipe, reaching a length d. The pipe is then sealed at both ends. We wish to know the concentration of salt, C, at any time. Since the pipe is narrow, the variations of concentration of salt over the cross section of the pipe can be neglected, and the concentration is given by C = C(x, t). The concentration satisfies the 1-D diffusion equation: ∂C∂t=D∂2C∂x2. -----(1) The initial concentration profile is C(x, 0)=MAd, 0≤x<d0, d≤x≤L. -------(2) And since the pipe is closed at both ends: ∂C(0, t)∂x=∂C(L, t)∂x=0. --------(3)For (a), We will start with a solution of the form C(x, t)=X(x)T(t), replace in the diffusion equation and show that the problem reduces to the mentioned two ordinary differential equations. So, let us put C(x, t)=X(x)T(t) in equation (1). Now, ∂C∂t=D∂2C∂x2⇒XT'=DX''T⇒T'DT=X''X=λ (Say)⇒T'(t)=λDT(t), X''(x)=λX(x) Note: Here, λ is an arbitrary constant.For (b), The boundary condition of after the substitution are X(0)=X(L)=0. Now, let us first solve the following differential equation with the boundary conditions. X''(x)-λX(x)=0, X(0)=X(L)=0 The characteristics equation of the above differential equation is m2-λ=0. So, m=±λ. Case 1 λ=k2>0: Then m=±k and the general solution of the differential equation is X(x)=c1ekx+c2e-kx. For X(0)=0, X(0)=c1ek(0)+c2e-k(0)⇒0=c1+c2 And for X(L)=0, X(L)=c1ekL+c2e-kL⇒0=c1ekL+c2e-kL After solving the above two equations we get that c1=0=c2. Thus, we get a trivial solution.Case 2 λ=0: Then m=0 and the general solution of the differential equation is X(x)=c1+c2x. For X(0)=0, X(0)=c1⇒0=c1 And for X(L)=0, X(L)=c2e-kL⇒0=c2e-kL⇒c2=0 So, we get that c1=0=c2. Thus, we get a trivial solution. Case 3 λ=-k2<0: Then m=±ki and the general solution of the differential equation is X(x)=c1 cos(kx)+c2 sin(kx). For X(0)=0, X(0)=c1 cos(0)+c2 sin(0)⇒0=c1 And for X(L)=0, X(L)=c2 sin(kL)⇒0=c2 sin(kL)⇒sin(kL)=0⇒kL=nπ, n∈ℤ⇒k=nπL Thus, we get non-trivial solutions for λ=-k2, where k is an arbitrary constant.For (c), Now, we will solve the following differential equation. T'(t)-λDT(t)=0. The general solution of the above is T(t)=c3eλDt. Using λ=-k2=-nπL2 we get T(t)=c3e-nπL2Dt. Thus, the solution is Cn(x, t)=an sinnπLxe-nπL2Dt where an is the constant. And by the super position principle we get, C(x, t)=∑n=0∞an sinnπLxe-nπL2Dt. Note: Here, we get C(x, t)=∑n=0∞an sinnπLxe-nπL2Dt not C(x, t)=∑n=0∞an cosnπLxe-nπL2Dt.

Answered: 2. A very narrow pipe of length L and…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Transcribed Image Text:1.0 0.5 8 0.0 -0.5 -1.0 -3 -2 -1 1 2 X 2. A very narrow pipe of length L and cross-sectional area A is filled with water. At time t = 0, a certain amount of salt, of total mass M, is placed at one end the pipe, reaching a length d. The pipe is then sealed at both ends. We wish to know the concentration of salt, C, at any time. Since the pipe is narrow, the variations of concentration of salt over the cross section of the pipe can be neglected, and the concentration is given by C = C (x, t). The concentration satisfies the 1-D diffusion equation: D- Ət The initial concentration profile is M Ad C(x, 0) = { 0 < x < d 0, d<x < L and since the pipe is closed at both ends: ƏC(0, t) ƏC(L,t) = 0 2

Transcribed Image Text:a) Start with a solution of the form C(x, t) equation and show that the problem reduces to two ordinary differential equations: X (x)T(t), replace in the diffusion X"(x) = AX(x); T'(t) = ADT(t) where A is an arbitrary constant. b) We seek time-decaying solutions, show that implies A = -k², where k is an arbi- trary constant. c) Solve the two ODES and show that after applying the two boundary conditions: C(x, t) = an cos(- Σ n=0 d) Find C(x, t) so that it satisfies the initial concentration profile. e) Show that your expression for C(x, t) satisfies the equilibrium profile C(x, t –) expected from conservation of mass. 3

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