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 $(\vec{V})$ are given for a flow field as: \[ T = x^2 + y t + 35 \] \[ \vec{V} = (4xy)\mathbf{i} + (xt - 2y^2)\mathbf{j} \] The total rate of change of temperature in the flow field (in integer) for time $t=2$ at a point $(2,3)$ is ............

Hint:

Always compute the substantial derivative using local derivative + convective terms: \(\tfrac{DT}{Dt} = \tfrac{\partial T}{\partial t} + \vec{V}\cdot\nabla T.\)

Answer 1

Correct Answer: 71

Step 1: Recall material derivative.
The total (substantial) derivative of temperature is: \[ \frac{DT}{Dt} = \frac{\partial T}{\partial t} + u \frac{\partial T}{\partial x} + v \frac{\partial T}{\partial y}, \] where $u,v$ are the velocity components in $x$ and $y$ directions.

Step 2: Compute partial derivatives of $T$.
\[ T = x^2 + y t + 35 \] \[ \frac{\partial T}{\partial t} = y, \frac{\partial T}{\partial x} = 2x, \frac{\partial T}{\partial y} = t. \]

Step 3: Velocity components.
From $\vec{V}$: \[ u = 4xy, v = xt - 2y^2. \]

Step 4: Evaluate at $(x,y,t)=(2,3,2)$.
- Partial derivatives: \[ \frac{\partial T}{\partial t} = y = 3, \frac{\partial T}{\partial x} = 2(2)=4, \frac{\partial T}{\partial y} = t=2. \] - Velocity: \[ u = 4(2)(3)=24, v = (2)(2) - 2(3^2) = 4 - 18 = -14. \]



Step 5: Substitute into material derivative.
\[ \frac{DT}{Dt} = \frac{\partial T}{\partial t} + u \frac{\partial T}{\partial x} + v \frac{\partial T}{\partial y} = 3 + (24)(4) + (-14)(2). \] \[ = 3 + 96 - 28 = 71. \] Final Answer:
\[ \boxed{71} \]