22-Mec-A6 Fluid Machinery · December 2019

Question 3 of 6: Turbulent Flow through an Annular Pipe from a Reservoir

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams — December 2019, 16-Mec-A6 Advanced Fluid Mechanics. Open book, 3 hours. Six questions of equal value (20 marks); any five constitute a complete paper. All six are solved here as a study resource.

Reference texts. J.D. Anderson, Modern Compressible Flow, 3rd ed. (Q1); Kundu, Cohen & Dowling, Fluid Mechanics, 6th ed. and F.M. White, Fluid Mechanics, 8th ed. (Q2, Q5); F.M. White, Fluid Mechanics, 8th ed. (Q3); F.M. White, Viscous Fluid Flow, 3rd ed. (Q4, Q6); Schlichting & Gersten, Boundary-Layer Theory, 8th ed. (Q6); Fox & McDonald, Introduction to Fluid Mechanics, 10th ed. (general).

Question 3: Turbulent Flow through an Annular Pipe from a Reservoir (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 concentric annular duct discharges water from a reservoir to atmosphere; the head $h$ supplies the exit velocity head plus the pipe friction.

Given data
Inner / outer radii	$a=0.04\ \text{m}$, $b=0.06\ \text{m}$
Length	$L=40\ \text{m}$
Wall roughness	$e=0.046\ \text{mm}$ (commercial steel)
Water properties	$\rho=1000\ \text{kg/m}^3$, $\nu=1.02\times10^{-6}\ \text{m}^2/\text{s}$
Flow rate	$Q=0.01\ \text{m}^3/\text{s}$
Effective-diameter factor	$D_{\text{eff}}/D_h=0.670$

Find. (a) the reservoir head $h$ for the target $Q$, neglecting entrance losses; (b) the head penalty of a sharp-edged ($K=0.5$) versus a well-designed entrance.

Figure 3. Reservoir head $h$ drives water through a $40\ \text{m}$ concentric annulus ($a=4$ cm, $b=6$ cm) to atmosphere.

Approach. The energy equation from the reservoir surface (velocity $\approx0$, gauge $0$) to the pipe exit (velocity $V$, gauge $0$) gives $h=V^2/2g+h_f$. For a non-circular duct White’s method uses the hydraulic diameter $D_h$ for the friction formula and roughness, but an effective diameter $D_{\text{eff}}$ for the Reynolds number that enters the Moody/Colebrook correlation.

Velocity and diameters. Flow area $A_c=\pi(b^2-a^2)=\pi(0.06^2-0.04^2)=6.283\times10^{-3}\ \text{m}^2$, so $V=Q/A_c=1.592\ \text{m/s}$. The hydraulic diameter of a concentric annulus is $D_h=2(b-a)=0.04\ \text{m}$, and the effective diameter $D_{\text{eff}}=0.670\,D_h=0.0268\ \text{m}$.
Reynolds number and friction factor. Using $D_{\text{eff}}$ for the Reynolds number, $Re=\dfrac{V D_{\text{eff}}}{\nu}=\dfrac{(1.592)(0.0268)}{1.02\times10^{-6}}=4.18\times10^{4}$ (turbulent). With relative roughness $e/D_h=0.046/40=1.15\times10^{-3}$, the Colebrook equation $$\frac{1}{\sqrt f}=-2\log_{10}\!\Big(\frac{e/D_h}{3.7}+\frac{2.51}{Re\sqrt f}\Big)\ \Rightarrow\ \boxed{f=0.0250 .}$$
(a) Required head. The velocity head is $V^2/2g=1.592^2/(2\cdot9.81)=0.1291\ \text{m}$; the friction head is $h_f=f\dfrac{L}{D_h}\dfrac{V^2}{2g}=0.0250\cdot\dfrac{40}{0.04}\cdot0.1291=3.23\ \text{m}$. Hence $$h=\frac{V^2}{2g}\Big(1+f\frac{L}{D_h}\Big)=0.1291\,(1+25.0)=\boxed{3.36\ \text{m}.}$$
(b) Entrance effect. A sharp-edged entrance adds $h_{\text{ent}}=K\dfrac{V^2}{2g}=0.5\times0.1291=0.0646\ \text{m}$; a well-designed (bell-mouth) entrance has $K\approx0.04$, essentially zero. The difference is $$\boxed{\Delta h\approx0.065\ \text{m}\ \ (\approx1.9\%\ \text{of }h).}$$ The entrance choice is therefore a minor correction — friction over the long duct dominates the head requirement.

Check: the effective-diameter method uses $D_{\text{eff}}$ only in $Re$ (Colebrook), and $D_h$ in $e/D_h$ and in $h_f=f(L/D_h)(V^2/2g)$. Applying $D_h$ throughout instead gives $Re=6.24\times10^4$, $f=0.0245$ and $h\approx3.20\ \text{m}$ — about $5\%$ lower; the effective-diameter value $3.36\ \text{m}$ is the recommended estimate.

Question 3 — results
Quantity	Value
Mean velocity $V$	$1.59\ \text{m/s}$
$D_h$ / $D_{\text{eff}}$	$0.040\ \text{m}$ / $0.0268\ \text{m}$
$Re$ (on $D_{\text{eff}}$) / $f$	$4.18\times10^4$ / $0.0250$
(a) Required head $h$	$3.36\ \text{m}$
(b) Sharp-edge penalty	$0.065\ \text{m}$ ($\approx1.9\%$)

Topic. Fully-developed turbulent flow in a non-circular (annular) duct; hydraulic and effective diameters, the Colebrook/Moody friction factor and reservoir–pipe energy balance.

Key relations. $D_h=4A_c/P=2(b-a)$ for a concentric annulus; effective diameter $D_{\text{eff}}=(D_{\text{eff}}/D_h)\,D_h$ set so laminar $f\,Re=64$ still holds; Colebrook $f(Re,e/D_h)$; energy $h=\dfrac{V^2}{2g}\big(1+K+f L/D_h\big)$.

Why this works. A hydraulic diameter alone mis-predicts friction for annuli because the velocity profile differs from a pipe’s. White’s effective diameter absorbs that geometric factor into a single corrected Reynolds number, so the ordinary Moody chart can be used with confidence; the roughness ratio and the loss formula still reference the physical $D_h$.

Common pitfall. Using $D_{\text{eff}}$ everywhere (including $L/D_h$ and $e/D_h$). It belongs only in the Reynolds number that enters the friction correlation. Forgetting the exit velocity head is another slip — here it is $0.13\ \text{m}$, small but not zero.

Source. F.M. White, Fluid Mechanics, 8th ed., §6.6 & Example 6.14 (flow in non-circular ducts; effective diameter for the annulus).

← Question 2 Next: Question 4 →