Question 5.2: A soil specimen 10 cm (3.9 in.) in diameter and 5 cm (2.0 in...
A soil specimen 10 cm (3.9 in.) in diameter and 5 cm (2.0 in.) in length was tested in a falling-head permeameter. The head dropped from 45 cm (17.7 in.) to 30 cm (11.8 in.) in 4 min and 32 s. The area of the standpipe was 0.5 cm² (0.078 in.²)
a What was the capillary rise of water in the standpipe?
b Considering the effect of capillary rise in the standpipe, compute the coefficient of permeability of the soil in units of cm/s.
c The soil sample had a specific gravity of solids equal to 2.67 and the sample weighed 10 g (0.22 lb) dry. Compute the average seepage velocity.
A=\Pi r^{2} =\left(3.1416\right) \left(5.0\right) ^{2} =78.5 cm^{2} ; a=0.5 cm^{2} ;
r_{pipe} =\left(a/\Pi \right) ^{0.5}= \left(0.5/3.1416 \right) ^{0.5}=0.40 cm
a Capillary rise from Equation (5.2)
h_{c} =\frac{2T_{s}\cos \alpha }{gr\left(\rho _{1}- \rho _{g} \right) } (5.2)
h_{c} =\frac{2\left(0.00075 N/c m\right) cos(0)}{(981 cm/s^{2} )\left(1g/cm^{3} \right)\left(0.40 cm\right) } \times \frac{\left(1 kg m/s^{2} \right) }{1 N} \frac{100 cm}{1 m} \frac{1000 g}{1 kg} =0.38 cm \approx 0.40b Coefficient of permeability computed from Equation (5.7)
Q=\nu A=k i A (5.7)
h_{0} =45-0.4=44.6 h_{1} =30-0.4=29.6
4 min 32 s.=272 s.
c Seepage velocity computed from Equation (5.9)
\nu _{s} =\nu A/A_{V} =\nu V/V_{V} =\nu /n (5.9)
v = \text{ discharge velocity }= ki
k = 2.1 \times 10^{-5} cm/s \text{ from step }(b)
i= \text{ hydraulic gradient } = h/L (44.6–29.6)/2 = 7.5
V = \text{ total volume }= Ah = (78.5)(5) = 392.5 cm^{3}
n = \text{ porosity } = V_{v}/V = (V -V_{s})/V = (V- \rho _{s}M_{s})/V = [392.5- (2.67) \times(100)]/392.5 = 0.32
The afore mentioned relationships for hydraulic conductivity in soil correspond to pressure gradients. Other gradients or potentials such as thermal, electric and electromagnetic also lead to the movement of fluid or mass and will be discussed in Chapter 6.