Question

2. suppose the owners of the reservoir described in the previous problem decide to drain the reservoirand put down a thin pre-fabricated bentonite mat on top of the compacted clay liner to improve theperformance of the liner. the bentonite mat is 1.0 cm thick when fully hydrated and the hydraulicconductivity of the bentonite is 1.0e-8 cm/s. a. what is the equivalent hydraulic conductivity of the new liner system (clay bentonite mat) [cm/s]?b. how much water would be lost per month with the new liner system [m^3]?

Answer 1

A. To calculate the equivalent hydraulic conductivity of the new liner system (clay + bentonite mat), we use the formula:

Keq = (K1d1 + K2d2) / (d1 + d2)

Where:

Keq = equivalent hydraulic conductivity of the new liner system

K1 = hydraulic conductivity of the clay liner (provided in the problem statement)

d1 = thickness of the clay liner (provided in the problem statement)

K2 = hydraulic conductivity of the bentonite mat (provided in the problem statement)

d2 = thickness of the bentonite mat (provided in the problem statement)

Plugging in the given values:

Keq = (1.0e-9 cm/s * 0.6 m + 1.0e-8 cm/s * 0.001 m) / (0.6 m + 0.001 m)

Keq = (6.0e-10 + 1.0e-8) cm/s

B. To calculate the amount of water lost per month with the new liner system, we use the formula:

V = Keq * A * (H2 - H1) / (12 * 3600 * 24 * 30)

Where:

V = volume of water lost per month

Keq = equivalent hydraulic conductivity of the new liner system (calculated in part A)

A = area of the reservoir (provided in the problem statement)

H2 = height of the water in the reservoir (provided in the problem statement)

H1 = height of the water table (provided in the problem statement)

Step-by-step explanation:

Plugging in the given values and converting Keq to m/s:

V = (6.0e-10 + 1.0e-8 cm/s) * 1.0e+8 m2 * (15 m - 1 m) / (12 * 3600 * 24 * 30)

V = (6.0e-10 + 1.0e-8) * 1.0e+8 * 14 m3/month

The amount of water lost per month with the new liner system is 14 m^3