Wall Shear Stress from a Pipe Force Balance: 16-Civ-A5 Hydraulic Engineering, May 2014, Q4

Question 4: Wall Shear Stress from a Pipe Force Balance (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Steady flow in a circular pipe; pressure drop $\Delta p=15\ \text{kPa}=15000\ \text{Pa}$ over length $L=2\ \text{m}$; diameter $D=150\ \text{mm}=0.150\ \text{m}$.

Find. A closed-form relation between wall shear stress and average velocity, then the numerical wall shear stress.

Approach. Apply Newton’s second law to a cylindrical control volume of fluid filling the pipe: the net pressure force on its ends is resisted by the wall shear on its curved surface; then substitute Darcy–Weisbach to introduce the average velocity.

Force balance on a fluid cylinder. For a plug of length $L$ and diameter $D$ in steady flow (no acceleration), pressure force = wall shear force: $\Delta p\left(\dfrac{\pi D^2}{4}\right)=\tau_w(\pi D L)$.
Solve for wall shear. $\boxed{\tau_w=\dfrac{\Delta p\,D}{4L}}$ — the closed-form result, valid for laminar or turbulent flow since no constitutive law was assumed.
Introduce average velocity. Darcy–Weisbach writes the same pressure drop as $\Delta p=\rho g\,h_f=f\dfrac{L}{D}\dfrac{\rho V^2}{2}$. Substituting into the boxed relation gives $\tau_w=\dfrac{f\rho V^2}{8}$, the shear–velocity relation.
Numerical wall shear. $\tau_w=\dfrac{\Delta p\,D}{4L}=\dfrac{15000\times0.150}{4\times2}=\boxed{281.25\ \text{Pa}}$.

Question 4 results
Quantity	Value
Closed-form relation	$\tau_w=\Delta p\,D/(4L)=f\rho V^2/8$
Wall shear stress	281.25 Pa

Topic: momentum (force) balance on pipe flow; wall shear and the friction factor.

Key relations: $\tau_w=\Delta p\,D/(4L)$ (equivalently $\tau_w=\gamma R\,S$ with $R=D/4$); Darcy–Weisbach $\Delta p=f\tfrac{L}{D}\tfrac{\rho V^2}{2}$; hence $\tau_w=f\rho V^2/8$.

Why this works: in fully developed steady flow there is no net momentum change along the pipe, so the driving pressure force is balanced entirely by wall friction. Because the balance is purely mechanical it holds regardless of whether the flow is laminar or turbulent—the flow regime only enters through the value of $f$ (or the velocity profile). The hydraulic-radius form $\tau_w=\gamma R S$ generalises the same idea to open channels.

Common pitfall: using radius instead of diameter (the area/perimeter ratio gives $D/4$, not $D/2$), or forgetting that the result is independent of the fluid model.

Source: Crowe, Elger & Roberson, Engineering Fluid Mechanics, Ch. 10 (flow in conduits; wall shear and friction factor).

Question 4 of 6: Wall Shear Stress from a Pipe Force Balance

Question 4: Wall Shear Stress from a Pipe Force Balance (20 marks)