Question 4 of 6: Wall shear from a pipe force balance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2014 — 98-Civ-A5 Hydraulic Engineering, 3 hours, closed book (one aid sheet). Six questions of equal value (20 marks each); candidates complete any five. All six are solved here as a study resource. ‘cms’ = m³/s; water ρ = 1000 kg/m³, ν = 1.31×10−6 m²/s; local losses and velocity head neglected unless stated.
Reference texts. Mays, Water Resources Engineering, 3rd ed. (pipe systems, valves, parallel/loop networks); Chow, Open-Channel Hydraulics (Manning uniform flow, critical depth, specific energy); Crowe, Elger & Roberson, Engineering Fluid Mechanics (pipe force balance, St-Venant equations). Exam-supplied relations: Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with $S = h_f/L$ (SI), Manning $Q = \tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$, Darcy–Weisbach, and $\text{TDH} = H_s + H_f$. In inverted form the friction loss carried by a known discharge is $h_f = L\left(\dfrac{Q}{0.278\,C\,D^{2.63}}\right)^{1/0.54}$.
Question 4: Wall shear from a pipe force balance (20 marks)
Given. Steady flow in a circular pipe; a pressure drop $\Delta p = 15\ \text{kPa}$ over a length $L = 2\ \text{m}$ in a pipe of diameter $D = 150\ \text{mm}$.
Find. A closed-form relation between wall shear stress $\tau_w$ and average velocity $V$ (valid laminar or turbulent), and the numerical $\tau_w$ for the stated data.
Cylindrical control volume of fluid filling the pipe: net pressure force is resisted by wall shear over the surface πDL.
Approach. Take the fluid cylinder that fills the pipe over length $L$ as a control volume; under steady state its momentum flux in equals out, so the net pressure force balances the wall shear force. Then link the pressure drop to velocity through Darcy–Weisbach.
Force balance on the cylinder. Pressure acts on the end area $A = \pi D^2/4$; wall shear $\tau_w$ acts over the lateral area $\pi D L$. Steady, fully developed flow has no net momentum change, so $$(p_1 - p_2)\dfrac{\pi D^2}{4} = \tau_w\,(\pi D L).$$
Solve for wall shear. Cancelling $\pi D$: $$\boxed{\tau_w = \dfrac{(p_1 - p_2)\,D}{4L} = \dfrac{\Delta p\,D}{4L}.}$$ This is exact for steady flow regardless of regime (it is pure force balance, no constitutive law invoked).
Relate to average velocity. Darcy–Weisbach gives $\Delta p = \rho g\,h_f = f\dfrac{L}{D}\dfrac{\rho V^2}{2}$. Substituting into the box above, the length and diameter cancel to a closed form in $V$: $$\tau_w = \dfrac{D}{4L}\!\left(f\dfrac{L}{D}\dfrac{\rho V^2}{2}\right) = \dfrac{f}{8}\,\rho V^2.$$ Because $f$ covers both laminar ($f = 64/Re$) and turbulent (Colebrook) regimes, $\tau_w = \tfrac{f}{8}\rho V^2$ applies to either.
