Question 4 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 Examinations — May 2014 — 07-Str-B11 Hydraulic Engineering. Three-hour, CLOSED-BOOK examination; one 8.5 × 11 in aid sheet (both sides) and any non-communicating calculator are permitted. The paper prints six questions of 20 marks each and instructs the candidate to complete any five; where a question has more than one part, the parts carry equal marks. Candidates are urged to submit a clear statement of any interpretive assumptions with the answer paper. All six questions are worked below, because this set is intended as a study resource.
Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, branched and looped network analysis, transmission mains and valve characteristics; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American distribution practice, fire-flow criteria, minimum service pressures and gutter hydraulics; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning’s equation and the momentum treatment of the hydraulic jump; Henderson, F.M., Open Channel Flow (Macmillan) — specific force, conjugate depths and jump classification; White, F.M., Fluid Mechanics (7th ed., McGraw-Hill) — the control-volume force balance that links wall shear stress to the Darcy friction factor.
Notation and conventions used throughout. The paper supplies the SI Hazen-Williams form \(Q = 0.278\,C\,D^{2.63}S^{0.54}\) with \(Q\) in m3/s, \(D\) in metres and \(S = h_f/L\); it is used exactly as printed. Writing \(K = 0.278\,C\,D^{2.63}\) inverts it to the head-loss form \(h_f = L\,(Q/K)^{1.852}\), the exponent \(n = 1.852\) being the value the paper itself quotes in its loop-correction note. Local losses and velocity head are neglected, as instructed in Note 6, so the hydraulic grade line (HGL) and the energy grade line coincide and “pressure head at a node” means \(p/\gamma = \mathrm{HGL} - z\). Water properties are \(\rho = 1000\) kg/m3 and \(\nu = 1.31\times10^{-6}\) m2/s.
Question 4: Wall Shear Stress from a Pipe Force Balance (20 marks)
Given. A straight, horizontal, full-flowing circular pipe carrying a steady incompressible flow, with a measured pressure drop over a known length.
Given data — Question 4
Quantity
Symbol
Value
Pressure difference over the reach
\(\Delta p\)
15 kPa = 15,000 Pa
Length of the reach
\(L\)
2 m
Pipe diameter
\(D\)
150 mm = 0.150 m
Water density (paper, Note 6)
\(\rho\)
1000 kg/m3
Find. A closed-form relation between wall shear stress and mean velocity valid in both flow regimes, and the numerical wall shear stress for the stated data.
Figure 4.1 — Free-body diagram of the fluid cylinder occupying the full pipe bore over a length \(L\). Pressure acts on the two end faces of area \(\pi D^2/4\); the wall shear \(\tau_0\) acts over the curved surface \(\pi D L\) and opposes the motion. Steady flow means the two must balance exactly.
Approach. Take the fluid filling the pipe over a length \(L\) as a control volume; because the flow is steady and fully developed the momentum flux in equals the momentum flux out, so the pressure force must be balanced entirely by wall shear — and substituting Darcy-Weisbach for the pressure drop introduces the mean velocity.
Write the axial force balance on the fluid cylinder. For steady, fully developed flow in a horizontal pipe the momentum flux entering and leaving the control volume are identical, so the net force vanishes:
$$\left(p_1 - p_2\right)\frac{\pi D^{2}}{4} = \tau_0\,\pi D L$$
The left side is the pressure force on the two end faces; the right side is the retarding shear on the curved wall.
Solve for the wall shear stress in terms of the measured pressure drop. Cancelling \(\pi D\) from both sides,
$$\boxed{\tau_0 = \frac{\Delta p\,D}{4L}}$$
Written with the hydraulic radius \(R_h = D/4\) and the friction slope \(S_f = \Delta p/(\rho g L)\), this is the general result \(\tau_0 = \rho g R_h S_f\), which holds for any duct shape and any flow regime because it uses only equilibrium, not a turbulence model.
Introduce the mean velocity through Darcy-Weisbach. The Darcy-Weisbach equation, quoted in Note 5 of the paper, expresses the same pressure drop in terms of the mean velocity and the friction factor:
$$\Delta h = f\frac{L}{D}\frac{V^{2}}{2g} \quad\Rightarrow\quad \Delta p = \rho g \Delta h = f\frac{L}{D}\frac{\rho V^{2}}{2}$$
This is the only step that brings the velocity into play, and it is valid in both regimes because \(f\) is simply defined so as to make it true.
Combine the two to obtain the required closed-form relation. Substituting the Darcy-Weisbach pressure drop into the force balance,
$$\tau_0 = \frac{D}{4L}\left(f\frac{L}{D}\frac{\rho V^{2}}{2}\right) \quad\Rightarrow\quad \boxed{\tau_0 = \frac{f\,\rho\,V^{2}}{8}}$$
This is the closed-form equation the question asks for. It is regime-independent in form: for laminar flow \(f = 64/Re\), which reduces it to \(\tau_0 = 8\mu V/D\), while for turbulent flow \(f\) comes from the Colebrook equation or the Moody chart. It is also the definition behind the shear velocity, \(u_* = \sqrt{\tau_0/\rho} = V\sqrt{f/8}\).
Evaluate the numerical wall shear stress. The data give the pressure drop directly, so the first boxed relation is used without needing \(f\) or \(V\):
$$\tau_0 = \frac{\Delta p\,D}{4L} = \frac{15{,}000 \times 0.150}{4 \times 2} = \frac{2250}{8} = \boxed{281.25\ \mathrm{Pa}}$$
For scale, the corresponding shear velocity is \(u_* = \sqrt{281.25/1000} = 0.53\) m/s, and a pressure gradient of 7.5 kPa/m over 150 mm pipe is a severe one — this is a high-velocity or high-viscosity duct, not a municipal main.
