Question

In a Cartesian coordinate system, a steady, incompressible velocity field of a Newtonian fluid is given by \[ V = u_0 \left( 1 - a y^2 \right) \hat{i} \] Here, \( V \) is the velocity vector in m/s, \( \hat{i} \) is the unit vector in the x-direction, \( u_0 \) is a positive, real constant in m/s, and \( a \) is a positive, real constant in m\(^{-2}\). The viscosity of the fluid is \( \mu \) in Pa-s. The absolute value of the pressure gradient (in Pa/m) is

• A. \( a \mu u_0 \)
• B. \( 2 a \mu u_0 \)
• C. \( 3 a \mu u_0 \)
• D. \( 4 a \mu u_0 \)

Hint:

The pressure gradient in a Newtonian fluid can be calculated by taking the second derivative of the velocity profile and multiplying by the fluid's viscosity.

Answer 1

The Correct Option is B

Given the velocity field \( V = u_0 (1 - a y^2) \hat{i} \), we need to find the pressure gradient.

Step 1: Calculate the velocity gradient.
The velocity gradient is the rate of change of the velocity with respect to the spatial coordinate \( y \): \[ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y} \left( u_0 (1 - a y^2) \right) = -2 a u_0 y \]

Step 2: Use the relation for the pressure gradient.
For a Newtonian fluid, the pressure gradient is related to the velocity gradient by the following relation: \[ \frac{\partial P}{\partial x} = - \mu \frac{\partial^2 V}{\partial y^2} \]

Step 3: Calculate the second derivative of the velocity.
Taking the second derivative of \( V \) with respect to \( y \): \[ \frac{\partial^2 V}{\partial y^2} = \frac{\partial}{\partial y} (-2 a u_0 y) = -2 a u_0 \]

Step 4: Find the pressure gradient.
Now, we can substitute the value of \( \frac{\partial^2 V}{\partial y^2} \) into the pressure gradient equation: \[ \frac{\partial P}{\partial x} = - \mu (-2 a u_0) = 2 a \mu u_0 \] Thus, the absolute value of the pressure gradient is \( 2 a \mu u_0 \), and the correct answer is (B).

Final Answer: (B) \( 2 a \mu u_0 \)