Download Example 8: A smooth sphere with density ρs = 1500 kg/m3 (sand) is

Example 8:
A smooth sphere with density ρs = 1500 kg/m3 (sand) is falling in water. Recall that for a
sphere volume is V = π 6 D3 and surface area is As = π D2 where D is diameter.
(a) Find terminal velocity for D = 0.10 mm.
(b) Find terminal velocity for D = 5.0 cm.
€ flow fields for parts (a) and (b)
€ for both a smooth and rough surface and
(c) Sketch
indicate how your answers to (a) and (b) change.
Known: ρs = 1500 kg/m3, (a) D = 0.10 mm, (b) D = 5.0 cm
Assumptions: Steady flow, water is a Newtonian fluid with constant properties at STP,
quiescent water far from boundaries
Find: Terminal velocity, U, for (a) and (b), (c) sketch flow fields for smooth and rough
Solution:
Properties: for water at STP
ρ = 999 kg/m3
µ = 1.12 x 10-3 N•s/m2
Answers: (a) 2.4 mm/s, (b) 0.81 m/s
Streamlines for cylinder in uniform cross flow for a range of Reynolds numbers:
CD =
1
2
D
ρ U 2 Ap
€
Drag coefficient, CD, versus Reynolds number, Re, where D is total drag, U is uniform
velocity, D is diameter, Ap is projected frontal area, ρ is density, and µ is viscosity.