Question

A cube of sandstone, 3 cm on each side, is made up of three layers that are each 1 cm thick and have permeability values of 1, 5 and 10 darcys.
For the linear flow of an incompressible fluid (μ = 1 cp) with a potential difference of 1 atm, determine the flow rates and apparent permeability for both the series and parallel flow directions.

Answer 1

Answer:

The answers to the questions are;

Series flow

Flow rate =  $\frac{90}{13}$ cm³/s

Apparent permeability = 30/13 Darcy.

Parallel flow

Flow rates for each layer

q₁ =3 cm²

q₂ = 5 cm³/s

q₃ =  10 cm³/s

Total flow rate = q = 16  cm³/s  

Apparent permeability = 16/3 Darcy.

Explanation:

Here we have the flow rate given by

q = $-\frac{KA}{\mu} \frac{dp}{dL}$

Where:

q = Flow rate

K = Permeability

dP = Potential difference = 1 atm

μ = Viscosity

Series flow

 We note that the average permeability is given by

k' =$\frac{L}{Sum \frac{L_i}{K_i}}$ =$\frac{3}{\frac{1}{1}+\frac{1}{5} +\frac{1}{10} }$  = 30/13 Darcy

P₁ - P₂ =$\frac{q\mu}{A} \frac{L}{k'}$

Therefore, q = $\frac{K'A}{\mu} \frac{dp}{L}$,

here  A = 3 cm × 3 cm = 9 cm²

Hence q = $\frac{\frac{30}{13} *9}{1} \frac{1}{3}$ = $\frac{90}{13}$ cm³/s.

Parallel flow  

The apparent permeability k' = ∑(k$_i$h$_i$)/∑h$_i$ = (1 + 5 + 10)/(1 + 1 + 1) = 16/3 Darcy

q = $Sum\frac{K_iA_i}{\mu} \frac{P_1-P_2}{L}$

A$_i$ = W × h$_i$  = 3 cm²

∴ we have q = $(\frac{1*3}{1} +\frac{5*3}{1} +\frac{10*3}{1} )*\frac{P_1-P_2}{L}$ = $48*\frac{1}{3}$ = 16 cm³/s.

Aliter

Therefore for each of the three layers, we have

q₁ = $-\frac{K_1A_1}{\mu_1} \frac{dp}{dL}$  A₁ =3 cm²

q₁ = $-\frac{1*3}{1} \frac{1}{3}$ = $\frac{3}{3}$ cm³/s     = 1 cm³/s

For q₂, we have

q₂ = $-\frac{5*3}{1} \frac{1}{3}$ =  5 cm³/s   and q₃ =  $-\frac{10*3}{1} \frac{1}{3}$ = 10 cm³/s

Therefore we have the total flow rate ∑q = q₁ + q₂ + q₃

=  1 cm³/s +5 cm³/s +10 cm³/s =  16  cm³/s.