Avg. Permeability
3.59
Permeability
Avg. Permeability Falling Head
$=\frac{\mathrm{K1(TestI)+K2(TestII)+K3(TestIII)}}{3}$
$=\frac{\mathrm{3.97+3.19+3.60}}{3}$
$={\mathrm{3.59\; \%}}^{}$
Result of Test-I
3.97
Permeability (K)
Permeability Falling Head
$=2.303\times \frac{a\mathrm{(L)}}{A\left(tf-ti\right)}\times {\mathrm{log}}_{10}\left(\frac{h1}{h2}\right)$
$=2.303\times \frac{\mathrm{706.86}\mathrm{(10)}}{\mathrm{314.16}\left(\mathrm{30}\right)}\times {\mathrm{log}}_{10}\left(\frac{\mathrm{10}}{\mathrm{20}}\right)$
$={\mathrm{3.97}}^{}$
Result of Test-II
3.19
Permeability (K)
Permeability Falling Head
$=2.303\times \frac{a\mathrm{(L)}}{A\left(tf-ti\right)}\times {\mathrm{log}}_{10}\left(\frac{h1}{h2}\right)$
$=2.303\times \frac{\mathrm{706.86}\mathrm{(10)}}{\mathrm{314.16}\left(\mathrm{50}\right)}\times {\mathrm{log}}_{10}\left(\frac{\mathrm{30}}{\mathrm{40}}\right)$
$={\mathrm{3.19}}^{}$
Result of Test-III
3.60
Permeability (K)
Permeability Falling Head
$=2.303\times \frac{a\mathrm{(L)}}{A\left(tf-ti\right)}\times {\mathrm{log}}_{10}\left(\frac{h1}{h2}\right)$
$=2.303\times \frac{\mathrm{706.86}\mathrm{(10)}}{\mathrm{314.16}\left(\mathrm{40}\right)}\times {\mathrm{log}}_{10}\left(\frac{\mathrm{20}}{\mathrm{30}}\right)$
$={\mathrm{3.60}}^{}$