07-Str-B11 · May 2016

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

Paper format. National Exams, May 2016 — 07-Str-B11, Hydraulic Engineering. Three hours, closed book; one 8.5 × 11 inch aid sheet (both sides may be used) and any non-communicating calculator are permitted. The paper prints six questions of equal value, and a candidate completes any five of them; where a question has parts, the parts also carry equal value. All six are worked below, so the set is useful whichever five a reader chooses.

Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, branched and looped network analysis, control valves and extended-period simulation; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American distribution practice, maximum-day-plus-fire-flow loading and service-pressure criteria; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning’s equation, specific energy and channel transitions; Henderson, F.M., Open Channel Flow (Macmillan) — critical flow, controls and surface-profile classification; White, F.M., Fluid Mechanics (7th ed., McGraw-Hill) — the pipe force balance, the friction-factor definition and laminar film flow.

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. Following the cover-page Note 6, local losses and velocity head are neglected, 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 is taken as $\rho = 1000$ kg/m³ with $\nu = 1.31\times10^{-6}$ m²/s (the cover page prints the exponent without its minus sign, which is an obvious typesetting slip — $1.31\times10^{6}$ m²/s is not a physical viscosity).

Source note — the Question 4 schematic is not reproduced in the examination paper. Question 4 refers to “the figure below”, but no figure is printed between Question 4 and Question 5 on page 3 of the paper, and page 4 opens with the continuation of Question 5. The network is therefore assumed to be a single loop: a source reservoir feeding node N1 through pipe P1 and node N2 through pipe P2, with pipe P3 closing the loop between N1 and N2. That topology is adopted here and is stated explicitly in the Given.

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, circular pipe carrying a steady incompressible flow, with a measured pressure drop over a known length.

Given data — Question 2
Quantity	Symbol	Value
Pressure difference over the test length	$\Delta p$	21 kPa = 21 000 Pa
Test length	$L$	2 m
Pipe diameter	$D$	200 mm = 0.200 m
Fluid density (cover-page Note 6)	$\rho$	1000 kg/m³

Find. A closed-form expression linking the wall shear stress to the average velocity that holds in either flow regime, and the numerical value of the wall shear stress for the stated pressure drop.

The derivation is a statics argument before it is a hydraulics one. Take as the control volume the whole cylinder of fluid of diameter $D$ and length $L$. In steady, fully developed flow through a straight horizontal pipe the fluid neither accelerates nor decelerates, so the net momentum flux through the two end faces is zero and the sum of the surface forces on that cylinder must vanish. Exactly two surface forces act: the pressure difference pushing on the two circular end faces, and the shear the wall exerts on the curved surface, which resists the motion. Nothing in that statement refers to the internal structure of the flow, which is why the resulting expression is as valid for a turbulent pipe at a Reynolds number of a million as it is for creeping laminar flow.

Balance pressure force against wall shear on a cylindrical control volume. The driving force is the pressure difference acting over the pipe cross-section, and the resisting force is the mean wall shear acting over the wetted perimeter times the length: $$\Delta p \cdot \frac{\pi D^2}{4} = \tau_0 \cdot \pi D L$$ Cancelling $\pi D$ and rearranging gives the first closed-form result, $$\boxed{\tau_0 = \frac{\Delta p\, D}{4L} = \frac{\gamma\, h_f\, D}{4L} = \gamma R_h S_f}$$ where $R_h = D/4$ is the hydraulic radius of a full circular pipe and $S_f = h_f/L$ the friction slope. The last form is the general one: wall shear equals unit weight times hydraulic radius times friction slope, and it holds for any conduit shape.
Bring in the average velocity through the friction-factor definition. The force balance is exact but contains no velocity. The link is the Darcy-Weisbach equation, which defines the friction factor $f$ by writing the same head loss in terms of the velocity head: $$h_f = f\,\frac{L}{D}\,\frac{\bar V^2}{2g}$$ Substituting $h_f$ into $\tau_0 = \gamma h_f D/(4L)$, the length and diameter cancel completely and so does $g$: $$\tau_0 = \frac{\rho g D}{4L}\cdot f\,\frac{L}{D}\,\frac{\bar V^2}{2g} = \boxed{\tau_0 = \frac{f}{8}\,\rho\,\bar V^{\,2}}$$ This is the required closed-form relation between wall shear stress and average velocity. It is regime-independent because all of the regime dependence has been pushed into $f$, which is a function of Reynolds number and relative roughness: $f = 64/Re$ in laminar flow, and the Colebrook or Swamee-Jain value in turbulent flow.
Evaluate the wall shear stress for the stated data. The numerical part needs only the first boxed relation, since the pressure drop is measured directly: $$\tau_0 = \frac{21\,000 \times 0.200}{4 \times 2} = \frac{4200}{8}$$ $$\boxed{\tau_0 = 525\ \text{Pa}}$$ Equivalently the friction slope is $S_f = h_f/L = (21\,000/9810)/2 = 1.070$ m per metre, and $\tau_0 = \gamma R_h S_f = 9810 \times 0.05 \times 1.070 = 525$ Pa, which is the same number reached by the general route.
Express the result as a friction velocity, and sanity-check the data. The shear velocity that scales all turbulent wall layers follows directly: $$u_* = \sqrt{\frac{\tau_0}{\rho}} = \sqrt{\frac{525}{1000}} = 0.725\ \text{m/s}$$ It is worth noticing what the stated pressure drop implies. Inverting $\tau_0 = (f/8)\rho \bar V^2$ with a typical turbulent $f \approx 0.02$ gives $\bar V \approx 14.5$ m/s, an order of magnitude above normal pipeline practice. A friction slope of 1.07 m of head per metre of pipe is likewise about a hundred times a design gradient. The arithmetic asked for is unaffected — the force balance needs no velocity — but it is worth saying in an answer that the figures describe a laboratory or an emergency condition rather than a transmission main.

Check: the pipe is taken as horizontal, so that the piezometric-head difference equals the pressure difference. If the 2 m length were inclined, $\Delta p$ in the force balance must be replaced by the difference in piezometric pressure $\Delta(p + \gamma z)$; the boxed relations are otherwise unchanged.

Final results — Question 2
Quantity	Result
Force balance on the fluid cylinder	$\tau_0 = \Delta p\,D/(4L) = \gamma R_h S_f$
Closed form in terms of average velocity	$\boldsymbol{\tau_0 = (f/8)\,\rho\,\bar V^{\,2}}$
Friction slope implied by the data	1.070 m/m
Wall shear stress	525 Pa
Corresponding friction velocity	0.725 m/s

Topic. The pipe-flow force balance, the hydraulic-radius form of wall shear, and the definition of the Darcy friction factor.

Key relations. $\tau_0 = \Delta p\,D/(4L)$; the general conduit form $\tau_0 = \gamma R_h S_f$ with $R_h = A/P$; Darcy-Weisbach $h_f = f(L/D)\bar V^2/(2g)$; and their combination $\tau_0 = (f/8)\rho\bar V^2$, with $u_* = \sqrt{\tau_0/\rho}$.

Why this works. Fully developed flow means the velocity profile is identical at every section, so the momentum flux entering the control volume equals the momentum flux leaving it and Newton’s second law degenerates into a statement of static equilibrium among the surface forces. Only pressure and wall shear survive, and their ratio is fixed by geometry alone, namely the ratio of cross-sectional area to wetted perimeter. The friction factor is then not a physical law but a bookkeeping device: it is defined so that the head loss can be written as a multiple of the velocity head, which is why $\tau_0 = (f/8)\rho\bar V^2$ is an identity rather than an approximation, and why the same equation serves both regimes.

Common pitfall. The factor of four is the usual casualty: the area-to-perimeter ratio for a full circular pipe is $D/4$, not $D/2$, so $\tau_0 = \Delta p D/(4L)$, and a student who writes $\Delta p D/(2L)$ has confused the pipe force balance with the thin-walled-cylinder hoop-stress formula. The second trap is quoting $\tau_0 = (f/4)\rho \bar V^2/2$, which is right for the Fanning friction factor but wrong by a factor of four for the Darcy factor used in Darcy-Weisbach.

Source. White, Fluid Mechanics 7e, §6.2 to §6.4; Mays, Water Resources Engineering 3e, Ch. 4; Chin, Water-Resources Engineering 3e, §2.2.

← Question 1 Next: Question 3 →

Quantity	Symbol	Value
Pressure difference over the test length	\(\Delta p\)	21 kPa = 21 000 Pa
Test length	\(L\)	2 m
Pipe diameter	\(D\)	200 mm = 0.200 m
Fluid density (cover-page Note 6)	\(\rho\)	1000 kg/m³

Quantity	Result
Force balance on the fluid cylinder	\(\tau_0 = \Delta p\,D/(4L) = \gamma R_h S_f\)
Closed form in terms of average velocity	\(\boldsymbol{\tau_0 = (f/8)\,\rho\,\bar V^{\,2}}\)
Friction slope implied by the data	1.070 m/m
Wall shear stress	525 Pa
Corresponding friction velocity	0.725 m/s