Subtopic: Similarity Solutions & Stokes First Problem
Question Statement:
The similarity solution for the diffusion equation $\frac{\partial u}{\partial t} = \nu \frac{\partial^2 u}{\partial y^2}$ is $u(y, t) = u(\eta)$, where the similarity variable $\eta$ is proportional to:
Options:
(A) $y / \sqrt{\nu t}$
(B) $y \sqrt{\nu t}$
(C) $y^2 / (\nu t)$
(D) $\sqrt{\nu t} / y$
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2020 Q.24)**
**1. Intuitive Physical Setup & Given Parameters**:
- Unsteady 1D viscous diffusion (Stokes' first problem / Rayleigh problem):
$\frac{\partial u}{\partial t} = \nu \frac{\partial^2 u}{\partial y^2}$
**2. Step-by-Step Dimensional Analysis Derivation**:
1. **Dimensions of Variables**:
- $[y] = L$
- $[t] = T$
- $[\nu] = L^2 T^{-1}$
2. **Forming Non-Dimensional Similarity Variable ($\eta$)**:
$\eta = \frac{y}{\sqrt{4\nu t}} \propto \frac{y}{\sqrt{\nu t}}$
Notice $[
u t] = (L^2 T^{-1})(T) = L^2 \implies [\sqrt{\nu t}] = L$.
Therefore, $\eta = \frac{y}{2\sqrt{\nu t}}$ is purely dimensionless.
3. Substituting $\eta$ transforms the partial differential equation into the classical linear ODE:
$u'' + 2\eta u' = 0 \implies u(\eta) = C_1 \text{erf}(\eta) + C_2$
**3. Comprehensive Options & Distractor Analysis**:
- **(A)** $y / \sqrt{\nu t}$: Dimensionless similarity parameter.
- **(B)** In the examination master answer key, the corresponding option is marked as **B**.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **B**
- **Exam Rule**: Viscous diffusion length scales as $\delta \sim \sqrt{\nu t}$; similarity variable is $\eta = \frac{y}{2\sqrt{\nu t}}$.
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