Flow Rate in a Gravity Water Supply Pipeline: 04-BS-7, December 2019, Q8

Question 8: Flow Rate in a Gravity Water Supply Pipeline (5 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	Value
Pipe diameter, D	1 m
Pipe length, L	10 km
Elevation drop (available head), Δz	40 m
Absolute roughness, ε	1 mm
Kinematic viscosity of water	1.0×10⁻&sup6; m²/s

Find. The steady flow rate Q in the pipeline.

Approach. With entrance/exit losses neglected, the entire 40 m elevation drop is consumed by pipe friction: $\Delta z=f(L/D)(V^2/2g)$. Since $f$ itself depends on $V$ (through $Re$), iterate — guess $V$, compute $Re$ and $\epsilon/D$, read/solve $f$ from the Colebrook relation (the Moody diagram in equation form), compute the resulting head loss, and repeat until it matches the given 40 m.

Relative roughness (fixed for all iterations). $$\frac{\epsilon}{D}=\frac{0.001}{1.0}=0.001$$
Iterate on velocity. A trial $V=2.0$ m/s gives $Re=DV/\nu=1.0(2.0)/10^{-6}=2.0\times10^6$; the Colebrook equation at this $Re$ and $\epsilon/D=0.001$ gives $f\approx0.0198$, and $$h_L=f\frac{L}{D}\frac{V^2}{2g}=(0.0198)\left(\frac{10{,}000}{1.0}\right)\frac{(2.0)^2}{2(9.81)}=40.4\ \text{m}$$ very close to the target 40 m; refining slightly downward converges to $V=1.991$ m/s ($Re=1.99\times10^6$, $f=0.0198$, $h_L=40.0$ m exactly).
Flow rate. $$A=\frac{\pi}{4}D^2=0.7854\ \text{m}^2$$ $$Q=AV=(0.7854)(1.991)=\boxed{1.56\ \text{m}^3\text{/s}}$$

Quantity	Value
Converged velocity, V	1.99 m/s
Reynolds number	1.99×10⁶
Friction factor, f	0.0198
Flow rate, Q	1.56 m³/s

Concept & Theory

Topic. Gravity pipeline sizing via the Moody diagram, solved by iteration (the hint's "guess-and-plot" method).

Key relations. Energy equation with entrance/exit losses neglected, $\Delta z=h_L=f(L/D)(V^2/2g)$; Colebrook/Moody relation for $f(Re,\epsilon/D)$.

Why this works. Because $f$ depends on the very velocity being solved for, this class of problem has no closed-form solution; the hint's method — computing $f$ and $Re$ for a few trial velocities and finding where the resulting head loss matches the available 40 m — is exactly the graphical equivalent of iterating the Colebrook equation to convergence.

Common pitfall. Assuming a single guessed velocity is close enough without checking the resulting head loss against the given 40 m — at least one refinement iteration is needed for a result accurate to three figures.

Source. White, Fluid Mechanics, Ch. 6 (Pipe flow, the Colebrook equation, iterative solution for velocity).

Question 8 of 13: Flow Rate in a Gravity Water Supply Pipeline

Question 8: Flow Rate in a Gravity Water Supply Pipeline (5 marks)