Wall shear stress from a pipe force balance: 16-Civ-A5 Hydraulic Engineering, December 2019, Q2

Question 2: 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. A straight, full pipe running steadily. Pressure drop $\Delta P = 15\ \text{kPa}$ over length $L = 2\ \text{m}$; diameter $D = 0.150\ \text{m}$.

Find. A general relation $\tau_w$–to–$\overline{V}$ (valid for any regime), then the numerical wall shear stress.

Figure 2-1. Axial force balance on the pipe control volume: pressure force on the ends is resisted by wall shear on the circumference.

Approach. Balance axial forces on a cylindrical control volume of the full pipe cross-section; the pressure force on the end areas equals the shear force on the wetted wall. This is a momentum statement, so it holds regardless of whether the flow is laminar or turbulent.

Axial force balance. For steady flow the net force on the slug is zero. Pressure acts on the end area $\tfrac{\pi}{4}D^2$; wall shear $\tau_w$ acts on the surface $\pi D L$: $$\big(P_1-P_2\big)\frac{\pi D^2}{4} = \tau_w\,(\pi D L).$$
Solve for wall shear. Cancelling $\pi D$ and writing $\Delta P = P_1-P_2$, $$\boxed{\tau_w = \frac{\Delta P\,D}{4L}}$$ This is exact from equilibrium alone — no assumption about the velocity profile.
Link to average velocity. Introduce the Darcy friction factor $f$ through $\Delta P = f\,\dfrac{L}{D}\,\dfrac{\rho \overline{V}^2}{2}$. Substituting, $$\tau_w = \frac{D}{4L}\cdot f\frac{L}{D}\frac{\rho\overline{V}^2}{2} = \frac{f}{8}\,\rho\,\overline{V}^2 .$$ The same $\tau_w\!-\!\overline V$ form holds for laminar flow (where $f=64/Re$, giving $\tau_w=8\mu\overline V/D$) and for turbulent flow (where $f$ comes from Colebrook/Moody).
Numerical value. With $\Delta P = 15\,000\ \text{Pa}$, $D=0.150\ \text{m}$, $L=2\ \text{m}$: $$\tau_w = \frac{15\,000\,(0.150)}{4\,(2)} = \boxed{281\ \text{Pa}\ (281.25\ \text{Pa}).}$$

Question 2 — results
Quantity	Value
General relation (equilibrium)	$\tau_w = \Delta P\,D/(4L)$
Relation to average velocity	$\tau_w = (f/8)\,\rho\,\overline{V}^2$
Wall shear stress	$\tau_w = 281\ \text{Pa}$

Topic. Pipe-flow force balance; wall shear, pressure gradient and the friction factor.

Key relations. $\tau_w=\Delta P\,D/(4L)$ (equilibrium); $\Delta P = f\tfrac{L}{D}\tfrac{\rho\overline V^2}{2}$ (Darcy–Weisbach); hence $\tau_w=\tfrac{f}{8}\rho\overline V^2$.

Why this works. A momentum balance on the whole cross-section only counts external forces on the control volume — end pressures and wall friction — and says nothing about the internal velocity distribution. That is precisely why the $\tau_w=\Delta P D/4L$ result is regime-independent: it is kinematics-free. The velocity profile only re-enters when we express the value of $f$ (or of $\Delta P$) in terms of $\overline V$.

Common pitfall. Using the hydraulic radius $R=D/4$ carelessly: $\tau_w=\rho g R S_f = \gamma (D/4)(\Delta h/L)$ is the same result written with head loss — mixing the two forms double-counts the factor of four.

Source. Munson, Young & Okiishi, Fundamentals of Fluid Mechanics, §8.1–8.4.

Question 2 of 6: Wall shear stress from a pipe force balance

Question 2: Wall shear stress from a pipe force balance (20 marks)