EM 1110-2-1100 (Part III)
30 Apr 02
EXAMPLE PROBLEM III-1-3
FIND:
Estimate Wf, Re, and CD for the following particles falling through 20o C fluids.
a) 0.2-mm (=0.02-cm) quartz grain in fresh water
b) 0.2-mm quartz grain in salt water
c) 0.2-mm quartz grain in air
d) 0.001-mm-(=0.0001-cm-) diam kaolinite clay particle (unflocculated) in fresh water
e) 4.5-mm (=0.45-cm) steel BB pellet in air
f) 6-1/2-(=16.5-cm-) diam pine sphere in fresh water
GIVEN:
g = 980 cm/sec2
νwater = 0.011 cm2/sec
νair = 0.15 cm2/sec
ρfresh water = 1.00 gm/cm3
ρsalt water = 1.03 gm/cm3
ρair = 0.0012 gm/cm3
ρquartz = 2648 kg/m3 from Table III-1-4 (=2.648 gm/cm3)
ρkaolinite = 2594 kg/m3 from Table III-1-4 (=2.594 gm/cm3)
ρsteel = 7800 kg/m3 (=7.8 gm/cm3)
ρpine = 480 kg/m3 (=0.48 gm/cm3)
SOLUTION:
For convenience, all calculations will be done in the cgs (centimeters, grams, seconds) system.
a) The fall velocity is read from Figure III-1-6 as 2.3 cm/sec. Re is calculated from its definition
as:
Re = (2.3 * 0.02)/0.011 = 4.2
CD can be found from Equation III-1-7 when rearranged as:
CD = (4/3) * 980*0.02/(2.3)2 * ((2.65/1) - 1) = 8.1
This agrees reasonably well with Figure III-1-5.
b) From section III-1-4-b, the fall velocity should be decreased from its value in part (a) by the factor
1.25/1.28 = 0.977, thus:
Wf = 2.3 * 0.977 = 2.2 cm/sec
Re = (2.2 * 0.02)/0.011 = 4.0
CD = 4/3 * 980*0.02/(2.2)2 * (2.65/1-1) = 8.9
(Continued)
III-1-28
Coastal Sediment Properties
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