Let's break this down step by step. ## **Given:** - Slab thickness = \( a \) - Initial temperature = \( T_\theta \) - One face at \( x=0 \) is **convectively cooled** by fluid at \( T_\infty \), heat transfer coefficient \( h \) - Other face at \( x=a \) is kept at constant temperature \( T_\theta \) - \( T(x, t) \) is the temperature profile to be found. - **Use separation of variables.** --- ## **1. Governing Equation** For 1D heat conduction (no internal heat generation): \[ \frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2} \] where \(\alpha\) = thermal diffusivity. --- ## **2. Initial and Boundary Conditions** - **Initial condition:** \[ T(x, 0) = T_\theta, \quad 0 \leq x \leq a \] - **Boundary conditions:** 1. At \( x = 0 \): Convective boundary \[ -k \left. \frac{\partial T}{\partial x} \right|_{x=0} = h [T(0, t) - T_\infty] \] 2. At \( x = a \): Dirichlet (fixed temperature) \[ T(a, t) = T_\theta \] --- ## **3. Non-dimensionalization (for simplification)** Let us define the dimensionless temperature: \[ \theta(x, t) = \frac{T(x, t) - T_\infty}{T_\theta - T_\infty} \] So, the conditions become: - \(\theta(x, 0) = 1\) - \(\theta(a, t) = 1\) - At \(x=0\): \( -k \frac{\partial \theta}{\partial x} = h (T_\theta - T_\infty) [\theta(0, t)] \implies -k \frac{\partial \theta}{\partial x} = h (T_\theta - T_\infty) \theta(0, t)\) - Or, dividing both sides by \(k (T_\theta - T_\infty)\): \[ -\left. \frac{\partial \theta}{\partial x} \right|_{x=0} = \frac{h}{k} \theta(0, t) \] Let \( Bi = \frac{h a}{k} \) (Biot number), and rescale \( x \) by \( a \): \( \xi = \frac{x}{a} \), \( 0 \leq \xi \leq 1 \). So, the PDE becomes: \[ \frac{\partial \theta}{\partial t} = \alpha \frac{\partial^2 \theta}{\partial x^2} \] with - \( \theta(x, 0) = 1 \) - \( \theta(a, t) = 1 \) - \( -\frac{\partial \theta}{\partial x}|_{x=0} = \frac{h}{k} \theta(0, t) \) --- ## **4. Separation of Variables** Let \( \theta(x, t) = X(x) \cdot T(t) \). Plug into the PDE: \[ X(x) \frac{dT}{dt} = \alpha T(t) \frac{d^2 X}{dx^2} \] \[ \Rightarrow \frac{1}{\alpha T(t)} \frac{dT}{dt} = \frac{1}{X(x)} \frac{d^2 X}{dx^2} = -\lambda^2 \] So: - \( \frac{dT}{dt} + \alpha \lambda^2 T = 0 \implies T(t) = A e^{-\alpha \lambda^2 t} \) - \( \frac{d^2 X}{dx^2} + \lambda^2 X = 0 \implies X(x) = B \cos(\lambda x) + C \sin(\lambda x) \) --- ## **5. Apply Boundary Conditions to \(X(x)\)** **At \( x = a \):** \[ X(a) = 0 \implies B \cos(\lambda a) + C \sin(\lambda a) = 0 \implies B \cos(\lambda a) = -C \sin(\lambda a) \] **At \( x = 0 \):** \[ -k \frac{dX}{dx}\bigg|_{x=0} = h X(0) \] \[ \frac{dX}{dx} = -B \lambda \sin(\lambda x) + C \lambda \cos(\lambda x) \] At \(x=0\): \[ \frac{dX}{dx}|_{x=0} = C \lambda \] So: \[ -k C \lambda = h B \] Or: \[ C \lambda = -\frac{h}{k} B \implies C = -\frac{h}{k \lambda} B \] --- Now plug this for \(C\) into the expression from the Dirichlet BC at \(x=a\): \[ B \cos(\lambda a) + C \sin(\lambda a) = 0 \] \[ B \cos(\lambda a) - \frac{h}{k \lambda} B \sin(\lambda a) = 0 \] \[ \cos(\lambda a) - \frac{h}{k \lambda} \sin(\lambda a) = 0 \] \[ \frac{h}{k \lambda} \sin(\lambda a) = \cos(\lambda a) \] \[ \frac{\sin(\lambda a)}{\cos(\lambda a)} = \frac{k \lambda}{h} \] \[ \tan(\lambda a) = \frac{k \lambda}{h} = \frac{\lambda a}{Bi} \] So, the **eigenvalue equation** is: \[ \tan(\lambda_n a) = \frac{\lambda_n a}{Bi} \] where \( Bi = \frac{h a}{k} \). --- ## **6. General Solution** Given the eigenvalues \( \lambda_n \) (solutions of the transcendental equation above), the general solution is: \[ \theta(x, t) = \sum_{n=1}^{\infty} A_n X_n(x) e^{-\alpha \lambda_n^2 t} \] with \[ X_n(x) = \cos(\lambda_n x) - \frac{h}{k \lambda_n} \sin(\lambda_n x) \] or, factoring \(B\) (which can be absorbed into \(A_n\)): \[ X_n(x) = \cos(\lambda_n x) - \frac{Bi}{\lambda_n a} \sin(\lambda_n x) \] --- ## **7. Determine Coefficients \(A_n\)** From the initial condition: \[ \theta(x, 0) = 1 = \sum_{n=1}^{\infty} A_n X_n(x) \] The coefficients \(A_n\) can be found by using the orthogonality of the eigenfunctions (see standard methods for Sturm-Liouville problems). For such mixed boundary conditions, the eigenfunctions are orthogonal with weight 1. So: \[ A_n = \frac{ \int_{0}^{a} 1 \cdot X_n(x) dx }{ \int_{0}^{a} X_n^2(x) dx } \] --- ## **8. Final Answer** \[ \boxed{ T(x, t) = T_\infty + (T_\theta - T_\infty) \sum_{n=1}^{\infty} A_n \left[ \cos(\lambda_n x) - \frac{Bi}{\lambda_n a} \sin(\lambda_n x) \right] e^{-\alpha \lambda_n^2 t} } \] Where: - \( \lambda_n \) are the positive roots of \( \tan(\lambda_n a) = \frac{\lambda_n a}{Bi} \) - \( Bi = \frac{h a}{k} \) - \( A_n = \frac{ \int_0^a X_n(x) dx }{ \int_0^a [X_n(x)]^2 dx } \) - \( X_n(x) = \cos(\lambda_n x) - \frac{Bi}{\lambda_n a} \sin(\lambda_n x) \) --- ### **Summary Table** | Step | Description | Equation(s) | |------|-------------|-------------| | 1 | Heat eqn | \( \frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2} \) | | 2 | IC | \( T(x,0) = T_\theta \) | | 3 | BC at \( x=0 \) | \( -k \frac{\partial T}{\partial x} = h(T - T_\infty) \) | | 4 | BC at \( x=a \) | \( T(a, t) = T_\theta \) | | 5 | Eigenvalue eqn | \( \tan(\lambda_n a) = \frac{\lambda_n a}{Bi} \) | | 6 | General soln | See boxed formula above | | 7 | Coefficient | \( A_n = \frac{ \int_0^a X_n(x) dx }{ \int_0^a [X_n(x)]^2 dx } \) | --- ### **References** - Carslaw and Jaeger, "Conduction of Heat in Solids" (Section 3.3) - Incropera & DeWitt, "Fundamentals of Heat and Mass Transfer" (Transient Conduction) --- **If you need help evaluating \(A_n\) or want a more explicit form, let me know!**