Question 2 of 6: Wall shear stress from a pipe force balance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — 98-Civ-A5 Hydraulic Engineering — May 2016.
Closed book; one aid sheet and a non-communicating calculator permitted. Six questions of equal value (20 marks each);
candidates answer any five. All six are solved here as a study resource. Exam-supplied relations: Hazen–Williams
$Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with $S = h_f/L$ (SI, $Q$ in m$^3$/s, $D,L$ in m), Manning
$Q=\tfrac{1}{n}A R^{2/3}S^{1/2}$, Darcy–Weisbach $\Delta h = 0.0826\,f\,L\,Q^2/D^5$, valve law
$Q=\tau E_s\sqrt{H_{u/s}-H_{d/s}}$, and $\mathrm{TDH}=H_s+H_f$. Water: $\rho=1000$ kg/m$^3$,
$\nu=1.31\times10^{-6}$ m$^2$/s.
Reference texts. Mays, Water Resources Engineering, 2nd ed. (pipe
networks, branched systems, valves, transmission mains); Chow, Open-Channel Hydraulics (Manning uniform flow,
critical depth, specific energy); Crowe, Elger & Roberson, Engineering Fluid Mechanics (pipe force balance,
laminar film flow).
Question 2: Wall shear stress from a pipe force balance (20 marks)
Given. Steady, fully-developed flow in a circular pipe; pressure drop $\Delta p=21$ kPa over
length $L=2$ m; diameter $D=200\text{ mm}=0.200$ m.
Find. A closed-form wall-shear relation valid for laminar or turbulent flow, and the numerical wall
shear stress $\tau_w$.
Approach. Balance the net pressure force pushing a cylindrical slug of fluid forward against
the wall-shear force resisting it; the result is geometry-only, hence regime-independent. Then substitute the
Darcy–Weisbach pressure drop to expose the dependence on average velocity.
Force balance on a fluid cylinder. For a slug of diameter $D$ and length $L$ in steady flow the net
pressure force equals the wall drag:
$$\Delta p\cdot\frac{\pi D^2}{4}=\tau_w\cdot(\pi D L).$$
Solve for the wall shear. Cancelling $\pi D$,
$$\boxed{\ \tau_w=\dfrac{\Delta p\,D}{4L}\ }$$
which holds for laminar or turbulent flow because no constitutive (viscosity) assumption entered — only equilibrium.
Relate to average velocity. Darcy–Weisbach writes the same pressure drop as
$\Delta p=f\dfrac{L}{D}\dfrac{\rho V^2}{2}$. Substituting into the boxed result gives
$$\tau_w=\frac{D}{4L}\cdot f\frac{L}{D}\frac{\rho V^2}{2}=\frac{f}{8}\,\rho V^2,$$
the closed-form link between wall shear and the average velocity $V$ (the friction factor $f$ carries the
laminar/turbulent behaviour).
Numerical value. With $\Delta p=21\,000$ Pa, $D=0.200$ m, $L=2$ m:
$$\tau_w=\frac{21\,000\times0.200}{4\times2}=\boxed{525\ \text{Pa}}.$$
