Let's solve the problem step by step **using the separation of variables technique**. --- ## **Problem Restatement** - Slab thickness: \( a \) - Initial temperature: \( T_\theta \) - Face at \( x = 0 \): contacts coolant at temperature \( T_I \) with **convective boundary** (Newton's law, coefficient \( h \)) - Face at \( x = a \): **kept at constant** temperature \( T_\theta \) - **Find:** Unsteady temperature profile \( T(x, t) \) - **Method:** Separation of variables --- ## **Step 1: Write the Governing Equation and Conditions** ### **Heat Equation (1D, No Source):** \[ \frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2} \] where \( \alpha \) = thermal diffusivity. ### **Initial Condition:** \[ T(x, 0) = T_\theta \] ### **Boundary Conditions:** - At \( x = 0 \): Convective boundary (Newton's Law of Cooling) \[ -k \frac{\partial T}{\partial x}\Big|_{x=0} = h [T(0, t) - T_I] \] - At \( x = a \): Constant temperature \[ T(a, t) = T_\theta \] --- ## **Step 2: Non-dimensionalization (Optional but Simplifies Algebra)** Let's define dimensionless temperature: \[ \theta(x, t) = T(x, t) - T_\theta \] So initial condition: \(\theta(x, 0) = 0\). Let \( \theta_I = T_I - T_\theta \), so that at \( x=0 \), the boundary is to the coolant at temperature difference \( \theta_I \). Now, the equation is: \[ \frac{\partial \theta}{\partial t} = \alpha \frac{\partial^2 \theta}{\partial x^2} \] IC: \(\theta(x, 0) = 0\) BCs: - At \( x=0 \): \[ -k \frac{\partial \theta}{\partial x}\bigg|_{x=0} = h (\theta(0, t) - \theta_I) \] - At \( x=a \): \(\theta(a, t) = 0\) --- ## **Step 3: Steady-State Solution** First, find the steady-state solution (\(\theta_s(x)\)), where \(\frac{\partial \theta}{\partial t} = 0\): \[ 0 = \alpha \frac{d^2 \theta_s}{dx^2} \implies \theta_s(x) = A x + B \] Apply BC at \( x = a \): \( \theta_s(a) = 0 \implies 0 = A a + B \implies B = -A a \) So: \[ \theta_s(x) = A (x - a) \] Apply BC at \( x = 0 \) (convective): \[ -k \frac{d\theta_s}{dx}\bigg|_{x=0} = h [\theta_s(0) - \theta_I] \] \[ -k A = h [A (0 - a) - \theta_I] \] \[ -k A = h [-A a - \theta_I] \] \[ -k A = -h A a - h \theta_I \] \[ -k A + h A a = -h \theta_I \] \[ A (-k + h a) = -h \theta_I \] \[ A = \frac{-h \theta_I}{-k + h a} = \frac{h \theta_I}{k - h a} \] So: \[ \theta_s(x) = \frac{h \theta_I}{k - h a} (x - a) \] --- ## **Step 4: Define the Transient Part** Let \[ \theta(x, t) = \theta_s(x) + \phi(x, t) \] So, \[ \frac{\partial \phi}{\partial t} = \alpha \frac{\partial^2 \phi}{\partial x^2} \] **IC:** At \( t=0 \), from above: \[ \theta(x, 0) = 0 \implies \phi(x, 0) = -\theta_s(x) \] **BCs:** At \( x=a \): \[ \theta(a, t) = 0 \implies \phi(a, t) = -\theta_s(a) = 0 \] At \( x=0 \): \[ -k \frac{\partial \theta}{\partial x}\bigg|_{x=0} = h[\theta(0, t) - \theta_I] \] But \[ \frac{\partial \theta}{\partial x} = \frac{d \theta_s}{dx} + \frac{\partial \phi}{\partial x} \] and \[ \theta(0, t) = \theta_s(0) + \phi(0, t) \] Substitute: \[ -k \left(\frac{d \theta_s}{dx}\bigg|_{x=0} + \frac{\partial \phi}{\partial x}\bigg|_{x=0}\right) = h [\theta_s(0) + \phi(0, t) - \theta_I] \] But \[ -k \frac{d \theta_s}{dx}\bigg|_{x=0} = h [\theta_s(0) - \theta_I] \] So \[ -k \frac{\partial \phi}{\partial x}\bigg|_{x=0} = h \phi(0, t) \] --- ## **Step 5: Separation of Variables** Assume \( \phi(x, t) = X(x) T(t) \): \[ X(x) \frac{dT}{dt} = \alpha T(t) \frac{d^2 X}{dx^2} \implies \frac{1}{\alpha T} \frac{dT}{dt} = \frac{1}{X} \frac{d^2 X}{dx^2} = -\lambda \] So, \[ T(t) = e^{-\alpha \lambda t} \] \[ \frac{d^2 X}{dx^2} + \lambda X = 0 \] General solution: \[ X(x) = A \sin(\sqrt{\lambda} x) + B \cos(\sqrt{\lambda} x) \] **Apply BCs:** At \( x = a \): \( X(a) = 0 \Rightarrow \) \[ A \sin(\sqrt{\lambda} a) + B \cos(\sqrt{\lambda} a) = 0 \] At \( x = 0 \): \[ -k X'(0) = h X(0) \] \[ X'(x) = A \sqrt{\lambda} \cos(\sqrt{\lambda} x) - B \sqrt{\lambda} \sin(\sqrt{\lambda} x) \] At \( x = 0 \): \[ X'(0) = A \sqrt{\lambda} \] \[ X(0) = B \] So, \[ -k A \sqrt{\lambda} = h B \] Now, solve for the allowed values of \( \lambda \): From \( X(a) = 0 \): \[ A \sin(\sqrt{\lambda} a) + B \cos(\sqrt{\lambda} a) = 0 \implies A = -B \cot(\sqrt{\lambda} a) \] But from before: \[ -k A \sqrt{\lambda} = h B \implies A = -\frac{h}{k \sqrt{\lambda}} B \] Set equal: \[ -\frac{h}{k \sqrt{\lambda}} B = -B \cot(\sqrt{\lambda} a) \implies \frac{h}{k \sqrt{\lambda}} = \cot(\sqrt{\lambda} a) \] Or, \[ \tan(\sqrt{\lambda} a) = \frac{k \sqrt{\lambda}}{h} \] Let \( \beta = \sqrt{\lambda} \), so: **Eigenvalue equation:** \[ \tan(\beta a) = \frac{k \beta}{h} \] The allowed \( \beta_n \) are roots of this equation. --- ## **Step 6: General Solution** Thus, \[ X_n(x) = \sin(\beta_n x) - \frac{h}{k \beta_n} \cos(\beta_n x) \] (You can check this by plugging the value \( B=1 \), \( A = -\frac{h}{k \beta_n} \)) So, \[ \phi(x, t) = \sum_{n=1}^\infty C_n [\sin(\beta_n x) - \frac{h}{k \beta_n} \cos(\beta_n x)] e^{-\alpha \beta_n^2 t} \] --- ## **Step 7: Initial Condition and Coefficient Calculation** Recall: \[ \phi(x, 0) = -\theta_s(x) \] So, \[ -\theta_s(x) = \sum_{n=1}^\infty C_n [\sin(\beta_n x) - \frac{h}{k \beta_n} \cos(\beta_n x)] \] The coefficients \( C_n \) are found by multiplying both sides by the eigenfunction and integrating over \( x \) from \( 0 \) to \( a \), using the orthogonality property (for Sturm-Liouville problems with Robin BCs). That is, \[ C_n = \frac{ \int_0^a -\theta_s(x) [\sin(\beta_n x) - \frac{h}{k \beta_n} \cos(\beta_n x)] dx }{\int_0^a [\sin(\beta_n x) - \frac{h}{k \beta_n} \cos(\beta_n x)]^2 dx } \] --- ## **Step 8: Final Answer** ### **The temperature profile is:** \[ \boxed{ T(x, t) = T_\theta + \frac{h (T_I - T_\theta)}{k - h a} (x-a) + \sum_{n=1}^{\infty} C_n \left[ \sin(\beta_n x) - \frac{h}{k \beta_n} \cos(\beta_n x) \right] e^{-\alpha \beta_n^2 t} } \] where: - \( \beta_n \) are the positive roots of: \[ \tan(\beta_n a) = \frac{k \beta_n}{h} \] - The coefficients \( C_n \) are: \[ C_n = \frac{ \int_0^a -\frac{h (T_I - T_\theta)}{k - h a} (x-a) \left[ \sin(\beta_n x) - \frac{h}{k \beta_n} \cos(\beta_n x) \right] dx }{\int_0^a \left[ \sin(\beta_n x) - \frac{h}{k \beta_n} \cos(\beta_n x) \right]^2 dx } \] --- ## **Summary of Steps** 1. **Find steady-state solution** for boundary conditions. 2. **Subtract steady state** to get homogeneous PDE and homogeneous BCs. 3. **Use separation of variables** for the transient part. 4. **Apply boundary conditions** to get eigenvalue equation (\( \tan(\beta a) = \frac{k \beta}{h} \)). 5. **Construct general solution** as a sum over eigenfunctions. 6. **Use initial condition** to find coefficients \( C_n \) via orthogonality. --- ### This is the **complete separation of variables solution** to the unsteady temperature profile for a slab with one face convectively cooled and the other kept at constant temperature.