Question

A tiny temperature probe is fully immersed in a flowing fluid and is moving with zero relative velocity with respect to the fluid. The velocity field in the fluid is \[ \mathbf{V} = (2x) \hat{i} + (y + 3t) \hat{j}, \] and the temperature field in the fluid is \[ T = 2x^2 + xy + 4t, \] where \(x\) and \(y\) are the spatial coordinates, and \(t\) is the time. The time rate of change of temperature recorded by the probe at \( (x = 1, y = 1, t = 1) \) is ________________.

• A. 4
• B. 0
• C. 18
• D. 14

Hint:

The material derivative of temperature takes into account both the time rate of change and the convective transport of temperature in the flow. Remember to compute both components for an accurate result.

Answer 1

The Correct Option is C

We are given the velocity field \( \mathbf{V} = (2x) \hat{i} + (y + 3t) \hat{j} \) and the temperature field \( T = 2x^2 + xy + 4t \). The rate of change of temperature at the probe, moving with zero relative velocity with respect to the fluid, is given by the material derivative of temperature, which is: \[ \frac{D T}{D t} = \frac{\partial T}{\partial t} + \mathbf{V} \cdot \nabla T. \] Step 1: Compute the time derivative of temperature First, compute the partial derivative of \( T \) with respect to \( t \): \[ \frac{\partial T}{\partial t} = \frac{\partial}{\partial t} (2x^2 + xy + 4t) = 4. \] Step 2: Compute the spatial derivative of temperature Next, compute the gradient \( \nabla T \): \[ \nabla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \right) = \left( 4x + y, x \right). \] Step 3: Compute the convective term Now, compute the convective term \( \mathbf{V} \cdot \nabla T \): \[ \mathbf{V} = (2x) \hat{i} + (y + 3t) \hat{j}, \quad \nabla T = (4x + y, x). \] So, \[ \mathbf{V} \cdot \nabla T = (2x)(4x + y) + (y + 3t)(x). \] At \( x = 1, y = 1, t = 1 \), substitute these values into the equation: \[ \mathbf{V} \cdot \nabla T = (2)(4 + 1) + (1 + 3)(1) = 10 + 4 = 14. \] Step 4: Combine the terms The total rate of change of temperature is: \[ \frac{D T}{D t} = \frac{\partial T}{\partial t} + \mathbf{V} \cdot \nabla T = 4 + 14 = 18. \] Thus, the time rate of change of temperature is 18.