Answer:
[tex]K = 2.037*10^{-3} m/s[/tex]
[tex]V_s = 0.0122 \ m/s[/tex]
Explanation:
Given that;
diameter (d) = 10cm/2 = 0.1m/2 = 0.05 m
length (l) = 10 cm = 0.1 m
porosity = 50%
height (h) = 30 cm = 0.3 m
time (t) = 5 s
volume (v) = 60 cm³ = 60 × 10⁻⁶ m³
Q (flow rate) = [tex]\frac{v}{t}[/tex]
Q = [tex]\frac{60*10^{-6} m^3}{5}[/tex]
Q = [tex]12*10^{-6} m^3 /sec[/tex]
From constant head method, we use the relation K = [tex](\frac{Q*L}{A*h})[/tex] to determine the hydraulic conductivity ; we have:
[tex]K = \frac{12*10^{-6}*0.1}{\frac{\pi}{4}0.05^2*0.3}[/tex]
[tex]K = 0.002037\\\\K = 2.037*10^{-3} m/s[/tex]
Seepage velocity [tex]V_s = \frac{velocity }{porosity}[/tex]
where; velocity = [tex]K*i[/tex]
=[tex](2.037*10^{-3}*)(\frac{0.3}{0.1})[/tex]
= [tex]6.111*10^{-3} m/s[/tex]
[tex]V_s = \frac{6.111*10^{-3}}{0.5}[/tex]
[tex]V_s = 0.0122 \ m/s[/tex]
