Question

Consider, \( \mathbf{i} \) and \( \mathbf{j} \) are unit vectors along x and y directions of a Cartesian (x, y) coordinate system, respectively and \( t \) is time. Temperature (T) and fluid velocity (\( \mathbf{V} \)) are given for a flow field as: \[ T = x^2 + yt + 35 {and} \mathbf{V} = (4xy)\mathbf{i} + (xt - 2y^2)\mathbf{j} \] The total rate of change of temperature in the flow field (in integer) for time \( t = 2 \) at a point (2, 3) is .........

Hint:

The material derivative accounts for both local and convective changes in a flow field. Always use:
\[ \frac{dT}{dt} = \frac{\partial T}{\partial t} + \vec{V} \cdot \nabla T \]

Answer 1

The total derivative of temperature \( T \) with respect to time in a flow field is given by the material derivative:
\[ \frac{dT}{dt} = \frac{\partial T}{\partial t} + \vec{V} \cdot \nabla T \]
Step 1: Compute partial derivatives of \( T \):
\[ T = x^2 + y t + 35 \Rightarrow \frac{\partial T}{\partial t} = y, \frac{\partial T}{\partial x} = 2x, \frac{\partial T}{\partial y} = t \]
Step 2: Find \( \nabla T = \left( \frac{\partial T}{\partial x} \right) \vec{i} + \left( \frac{\partial T}{\partial y} \right) \vec{j} = (2x)\vec{i} + (t)\vec{j} \)
Step 3: At point \( (x, y, t) = (2, 3, 2) \):
\[ \frac{\partial T}{\partial t} = 3, \nabla T = 4\vec{i} + 2\vec{j} \]
Step 4: Evaluate \( \vec{V} \) at (2, 3, 2):
\[ \vec{V} = (4 \cdot 2 \cdot 3)\vec{i} + (2 \cdot 2 - 2 \cdot 3^2)\vec{j} = 24\vec{i} + (4 - 18)\vec{j} = 24\vec{i} - 14\vec{j} \]
Step 5: Compute dot product \( \vec{V} \cdot \nabla T \):
\[ (24\vec{i} - 14\vec{j}) \cdot (4\vec{i} + 2\vec{j}) = 24 \cdot 4 + (-14) \cdot 2 = 96 - 28 = 68 \]
Step 6: Compute total rate of change:
\[ \frac{dT}{dt} = \frac{\partial T}{\partial t} + \vec{V} \cdot \nabla T = 3 + 68 = 71 \]