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?
Show 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.
Step 1: Picture the flow. A fluid moves steadily along a horizontal circular tube, driven only by a pressure drop, with no motion in the radial direction and a velocity pattern that no longer changes further along the tube. This is the classic Hagen-Poiseuille flow, and the given equation is Newton's second law written for a thin cylindrical shell of fluid inside the tube.
Step 2: Do a force balance on a fluid shell. For a thin cylindrical shell of radius $r$ and thickness $dr$, the net pressure force pushing it along the tube must balance the net viscous shear force on its curved surfaces. Writing that balance and letting the shell shrink to zero thickness gives exactly $\mu\dfrac{1}{r}\dfrac{d}{dr}\left(r\dfrac{dv_z}{dr}\right) = \dfrac{dP}{dz}$, the equation stated in the question.
Step 3: Get the shear stress directly. Multiplying through by $r$ and integrating once, \[ r\mu\frac{dv_z}{dr} = \frac{r^2}{2}\frac{dP}{dz}+C_1 \] At $r=0$ the term $\mu\,dv_z/dr$ is just the shear stress $\tau_{rz}$, and it must stay finite there, so $C_1=0$. This leaves \[ \tau_{rz}=\mu\frac{dv_z}{dr}=\frac{r}{2}\frac{dP}{dz} \] a straight line in $r$, starting at zero when $r=0$ and reaching its biggest size at the tube wall $r=R$.
Step 4: Read off the velocity profile too. Dividing by $\mu$ and integrating once more, with $v_z=0$ enforced at the wall $r=R$ (no-slip), gives the familiar parabola \[ v_z(r)=\frac{1}{4\mu}\frac{dP}{dz}\left(r^2-R^2\right) \] which peaks at the centerline and falls to zero at the wall, the opposite trend to the shear stress.
Step 5: Judge each statement. The fluid is treated as Newtonian throughout, since $\tau=\mu\,dv_z/dr$ is used, so (A) holds. The flow is assumed purely axial with $v_r=0$, so (C) holds. Being fully developed, $v_z$ does not change with $z$, so (D) holds. Only (B) fails, since the shear stress found in Step 3 is zero at the center and largest at the wall, not the reverse.
Step 6: Conclude. \[ \boxed{\text{(B) is NOT correct: shear stress is zero at the center and maximum at the wall}} \]
