
Concept explainers
Find the hydraulic conductivity of the sand.

Answer to Problem 6.12P
The hydraulic conductivity of the sand is
Explanation of Solution
Given information:
The shape factor (SF) is 6.5.
The void ratio of clay e is 0.5.
Calculation:
Find the opening size of the sieve:
Refer Table 2.6, “US sieve sizes with number designation” in the textbook.
For sieve no 30, the opening size of the sieve is 0.06 cm.
For sieve no 40, the opening size of the sieve is 0.0425 cm.
For sieve no 60, the opening size of the sieve is 0.02 cm.
For sieve no 100, the opening size of the sieve is 0.015 cm.
For sieve no 200, the opening size of the sieve is 0.075 cm.
For sieve no 40:
Find the fraction between two consecutive sieves:
Substitute 100 % for percent passing of sieve number 30 and 80 % for percent passing of sieve number 40.
For sieve no 60:
Find the fraction between two consecutive sieves:
Substitute 80 % for percent passing of sieve number 40 and 68 % for percent passing of sieve number 60.
Similarly calculate the fraction between two consecutive sieves.
Summarize the calculated values of opening size of the sieve and fraction between two consecutive sieves as in Table 1.
Sieve Number | Opening (cm) | Percent passing | Fraction between two consecutive sieves |
30 | 0.06 | 100 | |
40 | 0.0425 | 80 | 20 |
60 | 0.02 | 68 | 12 |
100 | 0.015 | 28 | 40 |
200 | 0.0075 | 0 | 28 |
Table 1
Refer Table 1;
For fraction between sieve numbers 30 and 40:
Find the term
Substitute 20% for
For fraction between sieve numbers 40 and 60:
Find the term
Substitute 12% for
For fraction between sieve numbers 60 and 100:
Find the term
Substitute 40% for
For fraction between sieve numbers 100 and 200:
Find the term
Substitute 28% for
Find the term
Substitute 408 for
Find the hydraulic conductivity of the sand at void ratio of 0.5 (k) using the formula:
Substitute 0.0166 for
Thus, the hydraulic conductivity of the sand is
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Chapter 6 Solutions
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