Question

In a groundwater model, the hydraulic conductivity of the aquifer is 10 m/day and the cross-sectional area perpendicular to the flow is 20 m$^2$. The hydraulic gradient is 0.006. Calculate the flow rate (Q) using the Darcy's Law.

• A. 1 m$^3$/day
• B. 1.2 m$^3$/day
• C. 2 m$^3$/day
• D. 0.12 m$^3$/day

Hint:

• Darcy's Law for groundwater flow: Q = K $\times$ A $\times$ i
• Q = Flow rate
• K = Hydraulic conductivity
• A = Cross-sectional area of flow
• i = Hydraulic gradient (change in head per unit distance)
• Substitute the given values: K = 10 m/day, A = 20 m$^2$, i = 0.006.
• Q = 10 $\times$ 20 $\times$ 0.006 = 200 $\times$ 0.006 = 1.2 m$^3$/day.

Answer 1

The Correct Option is B

Darcy's Law describes the flow of fluid through a porous medium. The formula for discharge (flow rate, Q) is: Q = K $\times$ A $\times$ i Where:

• Q = Discharge or flow rate (e.g., m$^3$/day)
• K = Hydraulic conductivity of the aquifer (e.g., m/day)
• A = Cross-sectional area perpendicular to the flow (e.g., m$^2$)
• i = Hydraulic gradient (dimensionless, e.g., m/m)

We are given: Hydraulic conductivity (K) = 10 m/day Cross-sectional area (A) = 20 m$^2$ Hydraulic gradient (i) = 0.006 Substitute these values into Darcy's Law: Q = (10 m/day) $\times$ (20 m$^2$) $\times$ (0.006) Q = (200 m$^3$/day) $\times$ (0.006) Q = 200 $\times$ (6 / 1000) m$^3$/day Q = 1200 / 1000 m$^3$/day Q = 1.2 m$^3$/day Therefore, the flow rate (Q) is 1.2 m$^3$/day. \[ \boxed{\text{1.2 m}^3\text{/day}} \]