Question

A Newtonian fluid of viscosity \( \mu \) is undergoing laminar flow in a tube of radius \( R \). For a constant pressure drop per unit length along the tube, which of the following statement(s) is/are CORRECT?
Note: \( r \) represents the radial position; assume no slip condition at the boundaries.

• A. The radial profile of the flow velocity is proportional to \( \left( 1 - \frac{r^2}{R^2} \right) \)
• B. The radial profile of the flow velocity is proportional to \( \left( 1 - \frac{r^3}{R^3} \right) \)
• C. Shear stress is zero at \( r = R \)
• D. Shear stress is zero at \( r = 0 \)

Hint:

In laminar flow of a Newtonian fluid, the velocity profile is parabolic, and the shear stress is zero both at the wall (\( r = R \)) and at the center of the tube (\( r = 0 \)).

Answer 1

The Correct Option is A, D

For laminar flow of a Newtonian fluid in a tube, the velocity profile follows the equation: \[ v(r) = \frac{\Delta P}{4 \mu L} \left( R^2 - r^2 \right) \] Where:
\( v(r) \) is the velocity at a radial position \( r \),
\( \Delta P \) is the pressure drop per unit length,
\( \mu \) is the dynamic viscosity,
\( L \) is the length of the tube,
\( R \) is the radius of the tube,
\( r \) is the radial position.
Step 1: Radial velocity profile. 
The flow velocity profile follows the form \( v(r) \propto \left( 1 - \frac{r^2}{R^2} \right) \), which corresponds to option (A). This is the typical parabolic velocity profile in laminar flow for a Newtonian fluid. 
Step 2: Shear stress. 
The shear stress in a laminar flow is given by: \[ \tau(r) = \mu \frac{dv}{dr} \] At \( r = R \), where the fluid touches the boundary (wall), the shear stress is zero because the velocity gradient becomes zero. This makes option (C) correct, but the **shear stress is zero at the center of the tube** (\( r = 0 \)) as well, as there is no velocity change at the center of the tube, making option (D) correct. 
Step 3: Conclusion. 
The correct answers are (A) and (D):
Option (A) is correct because the velocity profile is proportional to \( \left( 1 - \frac{r^2}{R^2} \right) \).
Option (D) is correct because the shear stress is zero at the center of the tube, \( r = 0 \).