Given the velocity field: V=(6+2xy+t^2 )i-(xy^2+10t)j+25k What is the acceleration of a particle at (3, 0, 2) at time t = 1?

Problem Statement

Given the velocity field: $$ V = (6 + 2xy + t^2) \hat{i} – (xy^2 + 10t) \hat{j} + 25 \hat{k} $$ Compute the acceleration at the point \( (3, 0, 2) \) at \( t = 1 \).

Solution

1. Local Acceleration:

   $$ \frac{\partial u}{\partial t} = 2t = 2(1) = 2 $$
   $$ \frac{\partial v}{\partial t} = -10 $$
   $$ \frac{\partial w}{\partial t} = 0 $$

2. Convective Acceleration:

   For \( a_x \):

   $$ \frac{\partial u}{\partial x} = 2y = 2(0) = 0 $$
   $$ \frac{\partial u}{\partial y} = 2x = 2(3) = 6 $$
   $$ u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} + w \frac{\partial u}{\partial z} $$
   $$ = 7(0) + (-10)(6) + 25(0) = -60 $$

   For \( a_y \):

   $$ \frac{\partial v}{\partial x} = -y^2 = 0 $$
   $$ \frac{\partial v}{\partial y} = -2xy = 0 $$
   $$ u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} + w \frac{\partial v}{\partial z} $$
   $$ = 7(0) + (-10)(0) + 25(0) = 0 $$

3. Total Acceleration:

   $$ a_x = 2 + (-60) = -58 $$
   $$ a_y = -10 + 0 = -10 $$
   $$ a_z = 0 $$

4. Final Acceleration Vector:

   $$ \mathbf{a} = -58 \hat{i} – 10 \hat{j} + 0 \hat{k} $$

   Magnitude of Acceleration:

   $$ |\mathbf{a}| = \sqrt{(-58)^2 + (-10)^2} \approx 58.86 \text{ units} $$

Physical Significance & Engineering Context

Fundamental Concepts

• Acceleration Field: Describes how fluid particles change velocity over time and space.
• Local Acceleration: Occurs due to time-dependent velocity components.
• Convective Acceleration: Arises from spatial variations in velocity and the motion of fluid particles.

Practical Interpretation

• Acceleration in fluid flow affects pressure variations and turbulence.
• Understanding acceleration fields helps in CFD simulations for aerodynamics and hydraulic designs.
• Applied in studying river flows, wind dynamics, and high-speed jets.