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For medium silt whose average grain size is 0.16 mm, Lacey's silt factor is likely to be
- a)0.30
- b)0.45
- c)0.70
- d)1.32
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
For medium silt whose average grain size is 0.16 mm, Lacey's silt ...
Solution:
Lacey’s silt factor, f = 1.76dm−−−√
= 1.76x0.16−−−−√ = 0.704.
Most Upvoted Answer
For medium silt whose average grain size is 0.16 mm, Lacey's silt ...
Lacey's Silt Factor for Medium Silt
Lacey's silt factor is a coefficient used in hydraulic engineering to estimate the settling velocity of silt particles in water. It is defined as the ratio of the settling velocity of silt to that of sand of the same size.
Formula for Lacey's Silt Factor
Lacey's silt factor (Fs) can be calculated using the following formula:
Fs = (Ds/D50)^0.5
Where,
Ds = Specific gravity of silt
D50 = Average grain size of silt
Solution
Given, average grain size of medium silt = 0.16 mm
Assuming the specific gravity of silt to be 2.65, we can calculate the value of Ds as follows:
Ds = Specific gravity of silt / Specific gravity of water
Ds = 2.65 / 1
Ds = 2.65
Now, we can substitute the values of Ds and D50 in the formula for Lacey's silt factor:
Fs = (Ds/D50)^0.5
Fs = (2.65/0.16)^0.5
Fs = 0.70
Therefore, the correct answer is option C, 0.70.
Lacey's silt factor is a coefficient used in hydraulic engineering to estimate the settling velocity of silt particles in water. It is defined as the ratio of the settling velocity of silt to that of sand of the same size.
Formula for Lacey's Silt Factor
Lacey's silt factor (Fs) can be calculated using the following formula:
Fs = (Ds/D50)^0.5
Where,
Ds = Specific gravity of silt
D50 = Average grain size of silt
Solution
Given, average grain size of medium silt = 0.16 mm
Assuming the specific gravity of silt to be 2.65, we can calculate the value of Ds as follows:
Ds = Specific gravity of silt / Specific gravity of water
Ds = 2.65 / 1
Ds = 2.65
Now, we can substitute the values of Ds and D50 in the formula for Lacey's silt factor:
Fs = (Ds/D50)^0.5
Fs = (2.65/0.16)^0.5
Fs = 0.70
Therefore, the correct answer is option C, 0.70.
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For medium silt whose average grain size is 0.16 mm, Lacey's silt ...
0.70
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For medium silt whose average grain size is 0.16 mm, Lacey's silt factor is likely to bea)0.30b)0.45c)0.70d)1.32Correct answer is option 'C'. Can you explain this answer?
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