EM 1110-2-1100 (Part III)
30 Apr 02
Example Problem III-1-3 (Concluded)
c) The fall velocity is read from Figure III-1-6 as 150 cm/sec. Re is calculated from its
definition as:
Re = (150 * 0.02)/0.15 = 20
CD is calculated from 1-7 as:
CD = 4/3 * 980*0.02/(150)2 * ((2.65/0.0012)-1) = 2.5
This agrees reasonably well with Figure III-1-5.
d) Expecting to find Re < 0.5, Equation 1-9 can be used to find Wf.
Wf = 980/18 (0.0001)2/0.011 (2.594/1-1) = 7.9 x 10-5 cm/sec
From its definition:
Re = 7.9 x10-5 * 0.0001 / 0.011 = 7.2 x 10-7
Thus, use of Equation 1-9 is well justified. From Equation 1-8:
CD = 24/7.2 x 10-7 = 3.3 x 107
e) Estimating that Re will be in the range 400 < Re < 200,000, Equation 1-10 can be used to find
an estimate of Wf.
Wf = 1.6 (980 * 0.45 ((7.8/0.0012)-1))1/2 = 2,700 cm/sec
Re = .45 * 2700 / 0.15 = 8100
Thus, use of Equation 1-10 is justified. From Figure III-1-5, it is seen that a CD value of 0.4 is better
than the value of 0.5 used in Equation III-1-10. Iterating using this value in Equation 1-7 yields:
Wf = (4/3 * 980 * 0.45/0.4 * (7.8/0.0012-1))1/2 = 3,100 cm/sec
Re = .45 * 3,100 / 0.15 = 9,300
Alternatively, Equation III-1-12 could have been used to obtain:
π/8 CD Re2 = 3.14*(0.45)3*(7.8/0.0012-1)*980/(6 * (0.15)2)=1.3x107
Using Figure III-1-5, this yields similar values of 12,000 and 0.4 for Re and CD, respectively.
f) It is important to realize that this particle is lighter than water and its direction of motion will be
upward. This is implied in the result by the fall velocity having a negative value. Estimating that Re will
be in the range Re > 200,000, Equation 1-11 can be used to find an estimate of Wf.
Wf = 2.6 (980 * 16.5 ((0.48/1)-1))1/2 = -240 cm/sec
Re = 240 * 15.2 / 0.011 = 360,000
Thus, use of Equation 1-11 is justified. Alternatively, Equation 1-12 could have been used to obtain:
π/8 CD Re2 = 3.14*(16.5)3*(0.48/1-1)*980/(6 * (0.011)2)=(-)9.9x109
Using Figure III-1-5, this yields similar values of 4x105 and 0.2 for Re and CD, respectively.
Coastal Sediment Properties
III-1-29