# Solution to Exercise 6

6) K can be estimated for unconsolidated samples using empirical equations. Grain size distribution data for Sample A, a coarse sand with a porosity of 0.26 (red line), and Sample B, a fine sand with a porosity of 0.3 (yellow line), are shown on the accompanying graph.

6a) By inspection of the graphs, which sample is likely to have the highest hydraulic conductivity? Why?

The text in section 4 of this book and material presented in Box 4 that describes empirical equations used to estimate K, reveal that K is directly proportional to the mean, median, effective or mean grain size (Equation Box 2-1) as defined in the discussion of grain size distribution analyses. For the data presented in the Figure above, each of these values is the largest in Sample A, coarse sand, as represented by the red line on the figure. The coarse sand would have the highest K value.

6b) Compute the uniformity coefficient for each sample. Compare and contrast the results. Do the coefficients appear to support your observations in part a?

The uniformity coefficient is defined in Box 2 of this book. The uniformity coefficient is equal to the diameter for cumulative 60% by weight finer-than (d60) divided by the diameter for cumulative 10% by weight finer-than (d10). The diameters are read from the graph as shown below.

Uniformity Coefficient = d60/d10

Uniformity Coefficient for the coarse sand: 1.3 mm/0.66 mm = 1.96

Uniformity Coefficient for the fine sand: 0.27 mm/0.058 mm = 4.6

Yes, these coefficients support the observation that the K would be highest for the Sample A with the larger grain sizes and the lower uniformity coefficient (1 = totally uniform sample). The slope of Sample A is closer to a vertical line which means it is more uniform than Sample B which has a gentler slope.

6c) Using the Hazen approximation and Slichter method, estimate K in cm/s.

First by Hazen:

The Hazen approximation equation is presented in Equation Box 4-9 in Box 4 of this book.

where:

 K = estimated hydraulic conductivity in centimeters per second d10 = effective grain size in centimeters (d10 finer-than, d90 retained) C = dimensionless coefficient defined as follows:
 Material C Very fine sand, poorly sorted 40-80 Fine sand with appreciable fines 40-80 Medium sand, well sorted 80-120 Coarse sand, poorly sorted 80-120 Coarse sand, well sorted, clean 120-150

For Sample A, the coarse sand, the d10 in cm is 0.66 mm × 1 cm/10 mm = 0.0.066 cm.
The value of C is based on a qualitative description and as the sample is considered fairly uniform, uniformity coefficient of 1.96, is represented by a range of Hazen values of 120 to 150, coarse sand well sorted.
K = 120 (0.066 cm)2/(cm s) = 0.52 cm/s and K = 150 ((0.066 cm)2/(cm s) = 0.65 cm/s
Using the Hazen method, the range of K values for Sample A are 0.52 cm/s to 0.65 cm/s

For Sample B, the fine sand, the d10 is 0.058 mm × 1 cm/10 mm = 0.0058 cm
The uniformity coefficient for the fine sand shows it is not very uniform so the Hazen coefficients for the are 40 to 80, fine sand with appreciable fines.
K = 40 (0.0058 cm)2/(cm s) = 0.001 cm/s, and K = 80(0.0058 cm)2/(cm s) = 0.003 cm/s
Using the Hazen method, the range of K values for Sample B is 0.001 cm/s to 0.003 cm/s.

Next, estimates of K using the Slichter Method using Equation Box 4-8 in Box 4 of this book.

where:

 K = estimated hydraulic conductivity in centimeters per second d = mean grain diameter in centimeters µ = dynamic viscosity at a given temperature in gram/(cm s) CS = a constant for a given porosity based on the following non-linear relationship: (n=26% CS=84.3; n=36% CS=28.8; n=47% CS=11.8)

To compute the mean grain diameter Equation Box 2-7 found in Box 2 of this book must be applied.

 Mean phi size =

The d16, d50 and d84 are taken from the grain size distribution curve (in mm) for both samples. Then they are converted to Phi units (dimensionless) = ln(x)/ln(2) where x is the d16, d50 and d84 values.

Sample A d84 = 2.1 mm, d50 = 1.1 mm, d16 = 0.78 mm
Phi units ln(2.1)/ln(2) = 1.07, ln(1.1)/ln(2) = 0.14, ln(0.78)/ln(2)= -0.36
Mean grain diameter = (1.07 + 0.14 + (-0.36))/3 = 0.28 phi units, needs to be converted back to mm:
Using ln(2) = ln(x) and ln(x) = value, x = evalue
0.28 (ln2) = 0.19 = ln(x) and x =e0.19 = 1.21 mm
Change mean grain diameter of A to cm, 1.2 mm × 1 cm/10 mm = 0.121 cm
Sample B d84 = 0.54 mm, d50 = 0.2 mm, d16 = 0.08 mm
Phi units = -0.89, = -2.32, = -3.64
Mean grain diameter = (-0.89 – 2.32 – 3.64)/3 = -2.28, convert back to mm
Using ln(2) = ln(x) and ln(x) = value, x = evalue
-2.28 ln(2)= -1.58 = ln(x) and x =e-1.58 = 0.2 mm
Change mean grain diameter of B to cm, 0.2 mm (1 cm / 10 mm) = 0.02 cm

K by Slichter method:

First, determine the coefficient based on the porosity relationship defined by Slichter by plotting porosity values on the graph of Figure Box 4-5 in Box 4 of this book.

Sample A: n = 0.26 Cs = 84.3, this porosity happens to match the original data Slichter provided.
Sample B: n = 0.30 Cs = 53, from the graph.
The water temperature is 15 °C so the dynamic viscosity needed for the calculation can be determined from Figure 28 of this book. The value is 1.20 millipascal-second = 0.012 gram/(cm s).
The K values are:
Sample A
K = (10.22 gram/(cm2sec2)) (0.12cm2)/((0.012gram/(cm s)) (84.3)) = 0.14 cm/s
Sample B
K = (10.22 gram/(cm2sec2)) (0.02cm2)/((0.012gram/(cm s)) (53)) = 0.006 cm/s

Vukovic and Soro (1992) noted the graph relating Cs and porosity can be represented by 1/n3.287 (with an error +/- 5%). They also proposed that the mean grain size be replaced by the effective grain size (d10 finer than). With a porosity of 26% for sample A, Cs would equal 83.7, and Cs for sample B with a porosity of 30% would be 52.3. Effective grain sizes are 0.07 cm for Sample A and 0.006 cm for sample B. Given these values of Cs, K values are as follows:
Sample A
K = (10.22 gram/(cm2sec2)) (0.07cm2)/((0.012 gram/(cm s)) (83.7)) = 0.05 cm/s
Sample B
K = (10.22 gram/(cm2sec2)) (0.006cm2)/((0.012 gram/(cm s)) (52.3)) = 0.0006 cm/s
Vukovic and Soro (1992) modifications to Slichter resulted in lower K values for Sample A and Sample B because the mean grain size values were significantly larger than the effective grain sizes.

6d) Are the K values computed in part c characteristic of sands? Do both methods produce similar values? Why or why not?

The computed K values for both methods are characteristic of the coarse and fine sands as indicated on Figure 32 of this book. From that figure, sand ranges from 1 × 10-6 cm/s to 1 cm/s. The values for sample A fall in the upper range and the values for Sample B fall in the middle of the range of K values.

The values derived from both methods are of a similar order of magnitude, Sample A on the order of 10-1 cm/s and sample B on the order of 10-3 cm/s. The two methods generate similar values with Hazen’s method yielding higher coarse sand values and lower fine sand values than the empirical equation proposed by Slichter. When the d10 was used instead of the mean gain size in the Slichter method K values were lower than Hazen by 100 %. It is unclear why the Hazen and original Slichter methods did not produce values that were more similar. They both approached developing the empirical equations focusing on the grain size distribution and then developed coefficients that brought laboratory K values in line with their choice of distribution parameters and other factors such as sorting and porosity. In doing so, equations did not produce identical hydraulic conductivity values. However, this is expected as they use different approaches. The values are within the same order of magnitude which is sufficient. It is suggested the modification of using the d10 instead of the mean grain size in the Slichter method should not be used. When using the methods to characterize K, the average of Hazen values could be used for each sample (0.6 cm/s for A and 0.002 cm/s for B), the range of the calculated values could be reported as the lowest and highest value of K for each sample computed by both methods (0.14 to 0.65 cm/s for Sample A and 0.001 to 0.006 cm/s for Sample B).