Sieve analysis calculation, particle size distribution curve and hydraulic conductivity --Exam question

 

                                  Question

1a. The result of the sieve analysis test on soil is shown below

1b. Plot the grading curve and report the result quantitatively

1c. Given the average grain size coefficient of 100 and the weight of the sample as 350g, determine the hydraulic conductivity

                                                                          

                    Solution

To solve this problem, first, you need to convert the sieve size unit from micrometers (710 µm, 500 µm, and 300 µm) to millimeters. Do that by simply multiplying each of the values by 0.001.

This is because 1 µm = 0.001mm. 

Watch a video on how to do sieve test calculation here

1b.

    Figure 1: sieve test calculation table

 

Next, we are going to plot a semi-log graph of sieve size(mm) against cumulative passing % ( figure 2).

In the exam hall, you are required to plot the graph using a semi-log graph, but here I will demonstrate how to do it using Microsoft excel.  

 Watch a video on how to plot a particle size distribution curve here using Excel

 Figure 2: particle size distribution graph (semi-log graph)

 

1c.  To determine the hydraulic conductivity, we are going to use the following formula

  Hydraulic conductivity (K) = C × (D10)²

                      

we need to extrapolate the D10 value from the graph. D10 is also called effective grain size.

From the curve, the value of D10 = 0.46476 mm, and, (grain size coefficient) is given in the question as  100

Hence,

                       =

                        =     100 × (0.465)²

              =   0.2209 cm/s     or 2.209×10^-1 cm/s

Therefore, the hydraulic conductivity is 0.2209 cm/s