Head required to drive 10 L/s through a 30 m annulus: 22-Mec-B6 Advanced Fluid Mechanics, December 2016, Q3

Question 3: Head required to drive 10 L/s through a 30 m annulus (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.

Quantity	Symbol	Value
Inner radius of the annulus	a	3 cm = 0.03 m
Outer radius of the annulus	b	5 cm = 0.05 m
Length	L	30 m
Wall roughness (commercial steel)	e	0.046 mm
Water density	ρ	1000 kg/m³
Kinematic viscosity	ν	1.02 × 10⁻⁶ m²/s
Volume flow rate	Q	0.01 m³/s
Effective-diameter ratio	D_eff/D_h	0.670
Sharp-edged entrance loss coefficient	K	0.5

Find. The reservoir surface elevation above the annulus centreline needed to drive 10 L/s when entrance effects are ignored, and the extra head a sharp-edged entrance would demand relative to a well-designed one.

Figure 3 — The reservoir, the 30 m annular run, and an end view of the annular section. The driving head h is measured from the free surface to the pipe centreline; the flow leaves to atmosphere, so the exit kinetic energy is part of the head budget.

Approach. Reduce the annulus to an equivalent circular pipe using the hydraulic diameter for the head-loss formula and the effective diameter for the Reynolds number, get the friction factor from Colebrook, and then write the steady energy equation from the reservoir surface to the discharge.

Compute the flow area and the hydraulic diameter. For a concentric annulus $$A=\pi\left(b^{2}-a^{2}\right)=\pi\left(0.05^{2}-0.03^{2}\right)=5.027\times10^{-3}\ \text{m}^{2},$$ $$D_h=\frac{4A}{P}=\frac{4\pi(b^{2}-a^{2})}{2\pi(b+a)}=2(b-a)=2(0.05-0.03)=0.040\ \text{m}.$$ The wetted perimeter includes both the outer tube and the inner core, which is what collapses the hydraulic diameter to twice the gap.
Get the bulk velocity. $$V=\frac{Q}{A}=\frac{0.01}{5.027\times10^{-3}}=1.989\ \text{m/s},$$ so the velocity head is $V^{2}/2g=(1.989)^{2}/(2\times9.81)=0.2017$ m.
Check the flow regime. $$\mathrm{Re}_{D_h}=\frac{VD_h}{\nu}=\frac{1.989(0.040)}{1.02\times10^{-6}}=7.80\times10^{4},$$ far above the transitional value of about 2300, so the flow is fully turbulent and a Colebrook friction factor is appropriate.
Apply the effective-diameter correction. A hydraulic diameter alone is only accurate to about ±15 per cent for non-circular ducts. The standard refinement replaces the Reynolds number by one built on the effective diameter, which for this annulus is $$D_{\text{eff}}=0.670\,D_h=0.670(0.040)=0.0268\ \text{m},\qquad \mathrm{Re}_{\text{eff}}=\frac{VD_{\text{eff}}}{\nu}=5.23\times10^{4}.$$ The relative roughness stays referred to the hydraulic diameter, $e/D_h=0.046\times10^{-3}/0.040=1.15\times10^{-3}$.
Solve Colebrook for the friction factor. $$\frac{1}{\sqrt f}=-2\log_{10}\!\left(\frac{e/D_h}{3.7}+\frac{2.51}{\mathrm{Re}_{\text{eff}}\sqrt f}\right) =-2\log_{10}\!\left(3.11\times10^{-4}+\frac{2.51}{5.23\times10^{4}\sqrt f}\right),$$ which iterates to $$\boxed{f=0.0243}$$ and therefore $fL/D_h=0.0243(30/0.040)=18.22$ velocity heads of friction.
Write the energy equation from the reservoir surface to the discharge. Taking the datum at the pipe centreline, with the reservoir surface and the jet both at atmospheric pressure and the reservoir velocity negligible, $$h=\frac{V^{2}}{2g}+f\frac{L}{D_h}\frac{V^{2}}{2g} =\left(1+f\frac{L}{D_h}\right)\frac{V^{2}}{2g}=(1+18.22)(0.2017),$$ $$\boxed{h=3.88\ \text{m}}$$ which answers part (a). Of that total, 3.67 m is wall friction and only 0.20 m is the kinetic energy carried away by the jet.
Part (b) — add the sharp-edged entrance loss. A sharp-edged inlet costs K = 0.5 velocity heads, whereas a well-designed bellmouth costs essentially nothing, so $$h_{\text{sharp}}=\left(1+K+f\frac{L}{D_h}\right)\frac{V^{2}}{2g}=(1+0.5+18.22)(0.2017) =3.98\ \text{m}.$$
Quantify the penalty. The difference is $$\Delta h=K\frac{V^{2}}{2g}=0.5(0.2017),$$ $$\boxed{\Delta h=0.101\ \text{m}\ \ (2.6\ \text{per cent of }h)}$$ so rounding the entrance saves about 10 cm of reservoir level. That is a real but minor saving here, and the reason is structural: with 750 diameters of pipe the friction term is worth 18.2 velocity heads, so any fixed-K minor loss is diluted almost to insignificance. On a short run of, say, 2 m the same entrance would be worth roughly 23 per cent of the head and the bellmouth would be the single most cost-effective change available.

Check: the effective-diameter route is the more accurate of the two standard treatments and is the one the question steers toward by supplying D_eff/D_h. Entering Colebrook with Re_Dh = 7.80 × 10⁴ instead gives f = 0.0232 and h = 3.71 m, about 4 per cent lower. The design conclusion — friction dominates and the entrance detail is worth about a tenth of a metre — is unchanged either way.

Quantity	Symbol	Result
Flow area	A	5.027 × 10⁻³ m²
Hydraulic diameter	D_h	0.040 m
Bulk velocity	V	1.989 m/s
Reynolds number on D_h / on D_eff	Re	7.80 × 10⁴ / 5.23 × 10⁴
Friction factor	f	0.0243
(a) Required reservoir level	h	3.88 m
(b) Level with a sharp-edged entrance	h_sharp	3.98 m
(b) Penalty of the sharp entrance	Δh	0.101 m, i.e. 2.6 per cent

Topic. Turbulent flow in non-circular ducts: hydraulic diameter, effective diameter and minor losses.

Key relations. $D_h=4A/P$; for a concentric annulus $D_h=2(b-a)$; Colebrook $1/\sqrt f=-2\log_{10}\!\left(\tfrac{e/D_h}{3.7}+\tfrac{2.51}{\mathrm{Re}\sqrt f}\right)$; Darcy–Weisbach $h_f=f\tfrac{L}{D_h}\tfrac{V^{2}}{2g}$; minor loss $h_m=K\tfrac{V^{2}}{2g}$; energy equation $h=\tfrac{V^{2}}{2g}+h_f+\sum h_m$.

Why this works. The hydraulic diameter is an area-to-perimeter argument: it makes a non-circular duct pass the same wall-shear-to-momentum ratio as a round pipe of that diameter, and for turbulent flow, where the shear is concentrated in a thin wall layer and the core is nearly uniform, that argument is good to about 15 per cent. It is much poorer in laminar flow, where the whole cross-section participates, and the effective diameter is the fix: D_eff is defined so that f = 64/Re would be exact in the laminar limit for that particular shape, and using it in the Reynolds number carries most of the shape information into the turbulent range as well. For this annulus a/b = 0.6 and the tabulated laminar product f Re_Dh is about 95.6, so D_eff/D_h = 64/95.6 = 0.670, which is exactly the number the question supplies.

Common pitfall. The classic error is to build the hydraulic diameter from the outer pipe alone, D_h = 2b = 0.10 m, forgetting that the inner core is also a wetted surface. That mistake more than halves fL/D_h and would predict a head of about 1.6 m instead of 3.9 m. A second slip is to omit the exit velocity head: the jet leaves with kinetic energy that the reservoir must supply, so the “1” in (1 + fL/D_h) is not optional.

Source. White, Fluid Mechanics, 8th ed., §6.6 (flow in non-circular ducts, incl. the annulus table) and §6.7 (minor losses); Fox, McDonald & Mitchell, Introduction to Fluid Mechanics, 10th ed., Ch. 8.

Question 3 of 6: Head required to drive 10 L/s through a 30 m annulus

Question 3: Head required to drive 10 L/s through a 30 m annulus (20 marks)