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Water Supply and Treatment Water Supply and Treatment

Water Supply and Treatment. Average Precipitation

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Page 1: Water Supply and Treatment. Average Precipitation

Water Supply and TreatmentWater Supply and Treatment

Page 2: Water Supply and Treatment. Average Precipitation
Page 3: Water Supply and Treatment. Average Precipitation

Average PrecipitationAverage Precipitation

Page 4: Water Supply and Treatment. Average Precipitation

Water DeficitsWater Deficits

Page 5: Water Supply and Treatment. Average Precipitation

Hydrologic CycleHydrologic Cycle

Page 6: Water Supply and Treatment. Average Precipitation

Groundwater ProblemsGroundwater Problems

Page 7: Water Supply and Treatment. Average Precipitation

Groundwater SuppliesGroundwater Supplies

Page 8: Water Supply and Treatment. Average Precipitation

The amount of water stored in an aquifer is equal to the volume of void The amount of water stored in an aquifer is equal to the volume of void spaces between the soil grains, the porosityspaces between the soil grains, the porosity

Porosity = volume of voids/total Porosity = volume of voids/total volumevolume

The amount of water that can be extracted from the aquifer is called the The amount of water that can be extracted from the aquifer is called the specific yield:specific yield:

Specific Yield = Specific Yield = Volume of water that will drain freely from a soilVolume of water that will drain freely from a soil

Total volume of water in a Total volume of water in a soilsoil

Page 9: Water Supply and Treatment. Average Precipitation

Q = A vQ = A v

v = superficial v = superficial velocityvelocity

Q = A v = a v’Q = A v = a v’

a = area available for flow (pores)a = area available for flow (pores)v’ = actual velocity of flow v’ = actual velocity of flow

v’ = Av/a = A v L / a L = v / porosityv’ = Av/a = A v L / a L = v / porosity

When water flows through a soil it loses energy just like when it When water flows through a soil it loses energy just like when it flows through a pipe (friction, etc.)flows through a pipe (friction, etc.)

dh/dLdh/dLh = energy, measured as elevation of the water table in an unconfined h = energy, measured as elevation of the water table in an unconfined aquifer or as the pressure in a confined aquiferaquifer or as the pressure in a confined aquiferL = horizontal distance in direction of flowL = horizontal distance in direction of flow

In an unconfined aquifer, the drop in elevation of the water table with distance is dh/dL.In an unconfined aquifer, the drop in elevation of the water table with distance is dh/dL.The elevation of the water surface is the potential energy of the water. Water flows The elevation of the water surface is the potential energy of the water. Water flows from higher elevation to lower elevation, losing energy.from higher elevation to lower elevation, losing energy.

Page 10: Water Supply and Treatment. Average Precipitation

Flow through a porous medium like a soil is related to the energy loss Flow through a porous medium like a soil is related to the energy loss using the Darcy equation:using the Darcy equation:

Q = K A dh/dLQ = K A dh/dL

K = coefficient of permeability, m/secK = coefficient of permeability, m/sec

A = cross-sectional area, mA = cross-sectional area, m22

Page 11: Water Supply and Treatment. Average Precipitation

A soil sample is installed in a permeameter as in the figure below. The sample A soil sample is installed in a permeameter as in the figure below. The sample length is 0.1 m with a cross-sectional area of 0.05 mlength is 0.1 m with a cross-sectional area of 0.05 m22. The pressure on the . The pressure on the upstream side is 2.5 m and on the downstream side is 0.5 m. A flow rate of 2.0 upstream side is 2.5 m and on the downstream side is 0.5 m. A flow rate of 2.0 mm33/day is observed. What is the coefficient of permeability, K?/day is observed. What is the coefficient of permeability, K?

dh/dL = (2.5 – 0.5)/ 0.1 = 20dh/dL = (2.5 – 0.5)/ 0.1 = 20

K = Q / (A dh/dL) = 2.0 / (0.05 x 20) = 2 m/dayK = Q / (A dh/dL) = 2.0 / (0.05 x 20) = 2 m/day

Page 12: Water Supply and Treatment. Average Precipitation

DrawdownDrawdown

This is an unconfined aquifer being pumped. As water flows to the well the area This is an unconfined aquifer being pumped. As water flows to the well the area through which it can travel becomes smaller and smaller, therefore the velocity gets through which it can travel becomes smaller and smaller, therefore the velocity gets larger and larger. This means more loss of energy (the gradient dh/dL increases) larger and larger. This means more loss of energy (the gradient dh/dL increases) and the slope of the water surface increases, resulting in a “cone of depression”and the slope of the water surface increases, resulting in a “cone of depression”

Page 13: Water Supply and Treatment. Average Precipitation

Given the cylinder shown at the right,Given the cylinder shown at the right,Darcy’s law becomes:Darcy’s law becomes:

Q = K (2Q = K (2rw) dh/dLrw) dh/dL

Separate the variables and integrate:Separate the variables and integrate:

Q = [Q = [K(hK(h1122 – h – h22

22)]/ ln(r)]/ ln(r11/r/r22))

Page 14: Water Supply and Treatment. Average Precipitation

ExampleExample

A well is 0.2 m in diameter and pumps from an unconfined aquifer 30 m deep at A well is 0.2 m in diameter and pumps from an unconfined aquifer 30 m deep at an equilibrium rate of 1000 man equilibrium rate of 1000 m33/day. Two observation wells are located at /day. Two observation wells are located at distances of 50 and 100 m and have drawdowns of 0.3 and 0.2 m, respectively. distances of 50 and 100 m and have drawdowns of 0.3 and 0.2 m, respectively. What is the coefficient of permeability and estimated drawdown at the well?What is the coefficient of permeability and estimated drawdown at the well?

K = [Q ln(rK = [Q ln(r11/r/r22)]/[)]/[(h(h1122 – h – h22

22)] = [1000 ln(100/50)]/[3.14(29.8)] = [1000 ln(100/50)]/[3.14(29.822 – 29.7 – 29.722]]

= 37.1 m/day= 37.1 m/day

Page 15: Water Supply and Treatment. Average Precipitation

The radius of the well is 0.2/2 = 0.1 m. Now:The radius of the well is 0.2/2 = 0.1 m. Now:

Q = [(Q = [(K(hK(h1122 – h – h22

22)] / [ln(r)] / [ln(r11/r/r22)] = [3.14 x 37.1 x (29.7)] = [3.14 x 37.1 x (29.722 – h – h2222)]/[ln(50/0.1)])]/[ln(50/0.1)]

= 1000 m= 1000 m33/day/day

hh22 = 28.8 m = 28.8 m

Since the aquifer is 30 m deep, the drawdown is:Since the aquifer is 30 m deep, the drawdown is:

30 – 28.8 = 1.2 m30 – 28.8 = 1.2 m

Page 16: Water Supply and Treatment. Average Precipitation

Multiple WellsMultiple Wells