Question 8 of 13: Pipe Diameter Selection from the Moody Chart
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-7 Mechanics of Fluids — National Examination, 2013-Dec. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (Analytical, 4 questions, do 3); every question is answered below regardless of the exam's "do N of M" instruction, so the set is a complete study resource.
Air density is taken from the paper's own Constants table at the temperature each question states: 1.19 kg/m³ at 20°C (Q5's wind, Q7's inlet air).
Q1's touching-rod array is modelled as a repeating square unit cell of four mutually tangent rods (pitch = rod diameter, per the question's own "closely packed" wording), giving a curvilinear-square pore whose perimeter/area ratio drives the capillary rise.
Q8's Moody diagram and Q9's drag-coefficient diagram are supplied as attachments. Both are solved via the equations the charts themselves plot: the Colebrook–White equation for Q8 and the Morrison (2013) curve-fit for sphere drag vs. Reynolds number for Q9.
Question 8: Pipe Diameter Selection from the Moody Chart (5 marks)
Find. Pipe diameter D that delivers Q = 1.0 m³/s over 5 km with 120 m of available head.
Approach. For each trial D, the mean velocity, Reynolds number and relative roughness are all determined; reading (or, equivalently, solving) the Moody relation gives the friction factor, and $h_L=f(L/D)(V^2/2g)$ then checks against the 120 m target. The Colebrook–White equation is exactly the curve the Moody chart plots, so solving it directly is equivalent to iterating on the chart with the two trial diameters the hint suggests.
Set up the governing relations in terms of D. $V(D)=Q/(\tfrac{\pi}{4}D^2)$, $Re(D)=VD/\nu$ (water, $\nu=1.0\times10^{-6}\ \text{m}^2/\text{s}$), relative roughness $\epsilon/D$, and $f$ from Colebrook–White, $1/\sqrt f=-2\log_{10}\!\big(\tfrac{\epsilon/D}{3.7}+\tfrac{2.51}{Re\sqrt f}\big)$.
Trial diameters (bracketing the hint's 0.5–0.7 m range). At D=0.50 m: V=5.09 m/s, giving $h_L\approx169\ \text{m}$ (too high — pipe too small). At D=0.60 m: V=3.54 m/s, giving $h_L\approx69\ \text{m}$ (too low — pipe too large). The target of 120 m therefore lies between these two trial points, exactly as the hint anticipates.
Interpolate/solve for the crossing diameter. Solving $h_L(D)=120\ \text{m}$ (the same root the two trial points on the Moody chart would be used to interpolate) gives
$$D = \boxed{0.532\ \text{m}}$$
Check at the solution diameter. $V=Q/(\tfrac{\pi}{4}D^2)=4.50\ \text{m/s}$, $Re=2.39\times10^{6}$, $\epsilon/D=8.5\times10^{-5}$ (hydraulically smooth-ish, high-Re regime on the chart), $f\approx0.0130$, reproducing $h_L=120\ \text{m}$.
