Question 9 of 13: Falling-Sphere Viscometer — Oil Viscosity
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-7 Mechanics of Fluids — National Examination, 2017-May. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (Graphical & 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.
Reference texts: White, F.M., Fluid Mechanics (8th ed.) — fluid statics and capillarity (Ch. 2), integral analysis and buoyancy (Ch. 3), viscous flow in ducts, pipe friction and the Moody chart (Ch. 6), flow past immersed bodies and drag (Ch. 7), open-channel flow and the hydraulic jump (Ch. 10).
Check — assumptions used across this paper:
Q2's figure shows the vertical leaf of the gate rising above the free surface, so water acts over the full depth x = 2.0 m; both "0.3 m" dimensions locate the centre of gravity (0.3 m right of the vertical leaf, 0.3 m above the arm). Reservoir water also fills the space beneath the 1.2 m arm (the vertical lines below O are dimension extension lines, not a wall), so the arm carries uplift at pressure ρgx, and the arm tip bears up against a lip on the spillway crest — the gate can only open by rotating clockwise, vertical leaf toward the spillway.
Q4's flow coefficient K ≈ 0.69 is read from the attached VDI chart on the Do/D1 = 0.50 curve (third from the top, the one that levels off at 0.62) at the approach Reynolds number R ≈ 2550. The printed curve gives 0.695 at R = 2000 and 0.684 at R = 3000; a reading of 0.68–0.70 moves the answer by only ±0.3 kPa.
Q7 and Q8 friction factors are obtained from the Colebrook–White/Haaland equation the Moody chart itself plots (smooth-wall case, as both problems specify or imply smooth surfaces).
Q9's sphere drag coefficient vs. Reynolds number is obtained from the Schiller–Naumann correlation $C_D=\tfrac{24}{Re}\left(1+0.15\,Re^{0.687}\right)$, which reproduces the published sphere-drag curve (the same curve reprinted in the attachment) to within a few percent for $Re<1000$ — used here because the resulting Reynolds number (≈19–20) is well above the range where the simple Stokes'-law formula ($C_D=24/Re$) alone is valid.
Find. The oil's absolute (dynamic) viscosity $\mu$.
Approach. At terminal velocity, net weight (weight minus buoyancy) equals drag; this fixes the required drag coefficient $C_D$, but since $C_D$ itself depends on the (unknown) Reynolds number through $\mu$, iterate: guess $Re$, read/compute $C_D(Re)$ from the sphere drag curve, check the force balance, and refine $Re$ until it converges — then back out $\mu$.
Check: $C_D(Re)$ obtained from the Schiller–Naumann correlation (see paper-level check note), used because the converged Reynolds number turns out to be ≈19–20 — well outside the strict $Re<1$ validity range of simple Stokes' Law, so reading (or computing) the actual drag curve is necessary, not optional.
Required drag coefficient from the force balance. Weight $-$ buoyancy $=$ drag: $(\rho_s-\rho_f)g\left(\tfrac{\pi d^3}{6}\right)=C_D\left(\tfrac12\rho_fV_t^2\right)\left(\tfrac{\pi d^2}{4}\right)$, which simplifies to
$$C_D=\frac{4(\rho_s-\rho_f)gd}{3\rho_fV_t^2}=\frac{4\times6975\times9.81\times0.006}{3\times825\times0.5^2}=\boxed{2.654}$$
Check the naive Stokes'-Law estimate first. Stokes' Law ($C_D=24/Re$) gives $\mu_{Stokes}=(\rho_s-\rho_f)gd^2/(18V_t)=0.274\ \text{Pa}\cdot\text{s}$, for which $Re=\rho_fV_td/\mu_{Stokes}=9.04$ — already above the $Re\approx1$ limit where Stokes' Law is reliable, so the chart-based (Schiller–Naumann) curve must be used instead.
Iterate on the sphere drag curve. Solving $C_D(Re)=\tfrac{24}{Re}\left(1+0.15\,Re^{0.687}\right)=2.654$ numerically converges at
$$Re\approx19.5$$
Recover the viscosity. $Re=\rho_fV_td/\mu\;\Rightarrow$
$$\mu=\frac{\rho_fV_td}{Re}=\frac{825\times0.5\times0.006}{19.5}=\boxed{0.127\ \text{Pa}\cdot\text{s}}$$
