Respuesta :
Answer:
The answers to the questions are;
Series flow
Flow rate = [tex]\frac{90}{13}[/tex] 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 = [tex]-\frac{KA}{\mu} \frac{dp}{dL}[/tex]
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' =[tex]\frac{L}{Sum \frac{L_i}{K_i}}[/tex] =[tex]\frac{3}{\frac{1}{1}+\frac{1}{5} +\frac{1}{10} }[/tex] = 30/13 Darcy
P₁ - P₂ =[tex]\frac{q\mu}{A} \frac{L}{k'}[/tex]
Therefore, q = [tex]\frac{K'A}{\mu} \frac{dp}{L}[/tex],
here A = 3 cm × 3 cm = 9 cm²
Hence q = [tex]\frac{\frac{30}{13} *9}{1} \frac{1}{3}[/tex] = [tex]\frac{90}{13}[/tex] cm³/s.
Parallel flow
The apparent permeability k' = ∑(k[tex]_i[/tex]h[tex]_i[/tex])/∑h[tex]_i[/tex] = (1 + 5 + 10)/(1 + 1 + 1) = 16/3 Darcy
q = [tex]Sum\frac{K_iA_i}{\mu} \frac{P_1-P_2}{L}[/tex]
A[tex]_i[/tex] = W × h[tex]_i[/tex] = 3 cm²
∴ we have q = [tex](\frac{1*3}{1} +\frac{5*3}{1} +\frac{10*3}{1} )*\frac{P_1-P_2}{L}[/tex] = [tex]48*\frac{1}{3}[/tex] = 16 cm³/s.
Aliter
Therefore for each of the three layers, we have
q₁ = [tex]-\frac{K_1A_1}{\mu_1} \frac{dp}{dL}[/tex] A₁ =3 cm²
q₁ = [tex]-\frac{1*3}{1} \frac{1}{3}[/tex] = [tex]\frac{3}{3}[/tex] cm³/s = 1 cm³/s
For q₂, we have
q₂ = [tex]-\frac{5*3}{1} \frac{1}{3}[/tex] = 5 cm³/s and q₃ = [tex]-\frac{10*3}{1} \frac{1}{3}[/tex] = 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.
