Oil Viscosity from a Sphere's Terminal Velocity: 04-BS-7, December 2014, Q9

Question 9 — Oil Viscosity from a Sphere's Terminal Velocity (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
Specific gravity of steel, SG_s	7.8
Sphere diameter, d	6 mm
Specific gravity of oil, SG_f	0.825
Terminal velocity, V_t	0.5 m/s

Find. The absolute (dynamic) viscosity of the oil, μ.

Fig. Q9 — steel sphere falling at constant (terminal) velocity: weight W balances buoyancy F_B plus drag F_D.

Approach. At terminal velocity, net weight (weight minus buoyancy) balances drag — this fixes the required drag coefficient C_D directly, with no viscosity needed yet. The attached drag-coefficient chart (equivalently, its algebraic sphere-drag correlation) is then searched for the Reynolds number that reproduces this C_D, from which the viscosity follows.

Required drag coefficient from the force balance. ρ_s = 7800 kg/m³, ρ_f = 825 kg/m³: $$(\rho_s-\rho_f)g\left(\frac{\pi d^3}{6}\right) = C_D\left(\tfrac12\rho_f V_t^2\right)\!\left(\frac{\pi d^2}{4}\right) \;\Rightarrow\; C_D = \frac{4(\rho_s-\rho_f)gd}{3\rho_f V_t^2} = \boxed{2.654}$$
Find Re from the sphere drag-coefficient chart. A first Stokes-law guess (C_D=24/Re) gives Re≈9.0 — too high for pure Stokes flow to be accurate, so the chart's actual curve (fit here by the standard sphere-drag correlation C_D=(24/Re)(1+0.15Re^0.687), valid well beyond the Stokes range) must be searched by iteration for the Re giving C_D = 2.654: $$\boxed{Re \approx 19.5}$$
Back out the viscosity. $$\mu = \frac{\rho_f V_t d}{Re} = \frac{825\times0.5\times0.006}{19.5} = \boxed{0.127\text{ Pa}\cdot\text{s} \;(127\text{ cP})}$$

Quantity	Result
Required drag coefficient, C_D	2.65
Reynolds number, Re	19.5
Oil viscosity, μ	0.127 Pa·s

Topic: Terminal (settling) velocity of a sphere — drag-coefficient chart method.

Key relations: force balance (ρ_s−ρ_f)g(πd³/6) = C_D·½ρ_fV_t²(πd²/4); sphere drag curve C_D≈(24/Re)(1+0.15Re^0.687), reducing to Stokes' law C_D=24/Re as Re→0.

Why this works: the force balance fixes C_D from known/measured quantities alone; viscosity enters only through Re, so the chart (or its algebraic fit) must be read/iterated for the Re consistent with that C_D whenever the flow is not deep in the Stokes regime.

Common pitfall: applying Stokes' law directly without checking Re≪1 — here Re≈19.5, well outside strict Stokes validity, so the naive Stokes estimate (0.274 Pa·s) overstates the true viscosity by more than double.

Source: White, Fluid Mechanics, 8th ed., §7.3–7.4 (drag on immersed bodies, sphere drag-coefficient chart, terminal velocity).

Question 9 of 13: Oil Viscosity from a Sphere's Terminal Velocity

Question 9 — Oil Viscosity from a Sphere's Terminal Velocity (5 marks)