Question 3 of 6: Head to Drive Flow Through an Annulus

Question 3: Head to Drive Flow Through an 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.

Find. The reservoir level $h$ (part a, entrance neglected) and the incremental effect of a sharp vs well-designed entrance (part b).

Approach. Get the mean velocity from continuity, form the hydraulic and effective diameters, obtain the friction factor from the Colebrook equation at the effective Reynolds number, then apply the steady-flow energy equation from the reservoir surface to the annulus exit.

1. Mean velocity. The annular flow area and mean velocity are $$A=\pi(b^2-a^2)=\pi(0.06^2-0.04^2)=6.283\times10^{-3}\ \text{m}^2,\quad V=\frac{Q}{A}=\frac{0.01}{6.283\times10^{-3}}=1.592\ \text{m/s}.$$
2. Hydraulic and effective diameters. For a concentric annulus $D_h=4A/P_{\text{wet}}=2(b-a)=0.040$ m, and the given ratio sets $D_{\text{eff}}=0.670\,D_h=0.0268$ m.
3. Reynolds number and relative roughness. Using the effective diameter to enter the Moody/Colebrook chart (this accounts for the annular shape), $$Re_{\text{eff}}=\frac{V\,D_{\text{eff}}}{\nu}=\frac{1.592\times0.0268}{1.02\times10^{-6}}=4.18\times10^4,\qquad \frac{e}{D_h}=\frac{0.046\times10^{-3}}{0.040}=1.15\times10^{-3}.$$
4. Friction factor (Colebrook). Solving $$\frac{1}{\sqrt{f}}=-2\log_{10}\!\left(\frac{e/D_h}{3.7}+\frac{2.51}{Re_{\text{eff}}\sqrt{f}}\right)\ \Rightarrow\ f=0.0250.$$
5. Energy equation, reservoir surface to exit. With both ends at atmospheric pressure, the surface at rest and the exit carrying velocity $V$, neglecting entrance loss, $$h=\underbrace{\frac{V^2}{2g}}_{\text{exit KE}}+\underbrace{f\frac{L}{D_h}\frac{V^2}{2g}}_{\text{friction}}.$$ With $V^2/2g=0.129$ m and $f\,L/D_h=0.0250\times40/0.040=25.0$, $$\boxed{h=0.129+25.0\times0.129=0.129+3.23=3.36\ \text{m}.}$$
6. (b) Entrance effect. A sharp-edged entrance adds $h_{\text{ent}}=K\,V^2/2g=0.5\times0.129=0.065$ m to the required level; a well-designed (rounded) entrance has $K\approx0.05$, adding only $\approx0.006$ m. Hence $$\boxed{\Delta h_{\text{sharp}}\approx0.065\ \text{m}\ (\sim2\%\text{ of }h),\quad \Delta h_{\text{rounded}}\approx0.006\ \text{m}.}$$ The entrance geometry matters little here because the 40 m of wall friction dominates the head budget.