Question

In a laboratory constant head permeability test, a cylindrical sample 100 mm in diameter and 150 mm high is subjected to an upward flow of 540 cm3/min. the head loss over the length of the sample is measured to be 360 mm. calculate the coefficient of permeability in m/s.

Answer 1

use the formula,
$k=\frac{QL}{\Delta{h}.At}$


where, k is the coefficient of permeability.
Q is the rate of upward flow of liquid { cm³/min}
∆h is head loss
A is the cross section area
t is time

here,
Q = 540 cm³/min = 540 × 10^-6/60 m³/s
∆h = 360 mm = 360 × 10^-3 m
L = 150mm = 150 × 10^-3 m
A = πr² = π(100/2)² = 2500π × 10^-6 m³
t for 1 sec

k = 540 × 10^-6 × 150 × 10^-3/(360 × 10^-3 × 2500π × 10^-6 × 1 × 60 )
= 4.8 × 10^-4 m/s

hence, coefficient of permeability is 4.8 × 10^-4 m/s

Answer 2

K = coefficient of permeability. Q = rate of upward flow of liquid { cm³/min} ∆h = head loss A = cross section area t = time The solution therefore is simplified below Q = 540 cm³/min = 540 × 10^-6/60 m³/s ∆h = 360 mm = 360 × 10^-3 m L = 150mm = 150 × 10^-3 m A = πr² = π(100/2)² = 2500π × 10^-6 m³ t for 1 sec k = 540 × 10^-6 × 150 × 10^-3/(360 × 10^-3 × 2500π × 10^-6 × 1 × 60 ) = 4.8 × 10^-4 m/s Therefore the permeability coefficient is equal to 4.8 × 10^-4 m/s