Question

The equation below represents a steady-state laminar flow of a fluid through a horizontal tube:
\[ \mu\,\frac{1}{r}\frac{d}{dr}\left(r\frac{dv_z}{dr}\right) - \frac{dP}{dz} = 0 \]

Figure: a horizontal cylindrical tube of radius R with fluid flowing along the z-axis; the radial coordinate r is measured outward from the tube's central axis and the axial coordinate z runs along the length of the tube.
Here \(P\) is fluid pressure, \(\mu\) is dynamic viscosity, \((r, z)\) are the coordinates of a cylindrical polar system, and \(v_z\) is the axial velocity in the z-direction.
Which one of the following statements related to the above case is NOT correct?

• A. The fluid is Newtonian
• B. Shear stress is maximum at the center \((r = 0)\)
• C. Radial velocity is zero
• D. There is no variation of \(v_z\) in z-direction

Hint:

Integrate the given equation once to get the shear stress as a function of r, then check where it is zero and where it is largest.

Answer 1

The Correct Option is B

Step 1: Understand the physical situation.
The equation \(\mu\,\dfrac{1}{r}\dfrac{d}{dr}\left(r\dfrac{dv_z}{dr}\right) - \dfrac{dP}{dz} = 0\) is the reduced Navier-Stokes equation for steady, fully developed, laminar flow of an incompressible fluid through a horizontal circular tube of radius \(R\). Only the axial velocity \(v_z(r)\) exists, and it varies with the radial position \(r\) alone.

Step 2: Solve for the velocity profile.
Rearranging,
\[ \frac{1}{r}\frac{d}{dr}\left(r\frac{dv_z}{dr}\right) = \frac{1}{\mu}\frac{dP}{dz} = \text{constant} \]
because for fully developed flow \(dP/dz\) does not change along the tube. Integrating once,
\[ r\frac{dv_z}{dr} = \frac{1}{2\mu}\frac{dP}{dz}\,r^2 + C_1 \]
The velocity gradient must stay finite at the centerline \(r = 0\), so \(C_1 = 0\). Dividing by \(r\),
\[ \frac{dv_z}{dr} = \frac{r}{2\mu}\frac{dP}{dz} \]
Integrating again and applying the no-slip condition \(v_z = 0\) at the wall \(r = R\) gives the parabolic profile
\[ v_z(r) = \frac{1}{4\mu}\frac{dP}{dz}\left(r^2 - R^2\right) \]

Step 3: Find the shear stress.
For a Newtonian fluid, shear stress is proportional to the velocity gradient, \(\tau_{rz} = \mu\,dv_z/dr\). Using Step 2,
\[ \tau_{rz} = \mu\cdot\frac{r}{2\mu}\frac{dP}{dz} = \frac{r}{2}\frac{dP}{dz} \]
This is directly proportional to \(r\). At the centerline, \(r = 0\), so \(\tau_{rz} = 0\). At the wall, \(r = R\), so \(\tau_{rz}\) reaches its largest size. Shear stress is therefore zero at the center and maximum at the wall, the exact opposite of what statement (B) claims.

Step 4: Check the remaining statements.
(A) The fluid is Newtonian: the governing equation itself uses a constant viscosity \(\mu\) relating shear stress linearly to the velocity gradient, which is the defining property of a Newtonian fluid. Correct statement.
(C) Radial velocity is zero: the flow is assumed purely axial and fully developed, so \(v_r = 0\) everywhere; only \(v_z\) exists. Correct statement.
(D) There is no variation of \(v_z\) in the z-direction: fully developed flow means the velocity profile no longer changes as the fluid moves along the tube, so \(\partial v_z/\partial z = 0\), matching the equation, which contains only \(r\)-derivatives of \(v_z\). Correct statement.

Step 5: Final conclusion.
Statements (A), (C) and (D) are all true for this flow, but statement (B) reverses the actual shear stress distribution, since shear stress is zero at the center and maximum at the wall, not the other way round.
\[ \boxed{\text{(B) Shear stress is maximum at the center } (r=0) \text{ is NOT correct}} \]