GROUNDWATER

INTRODUCTION
•Groundwater is water stored under the surface of the ground in the tiny pore spaces between rock, sand, soil, and gravel. It occurs in two “zones”

•An upper, unsaturated zone where most of the pore spaces are filled with air, and a deeper, saturated zone in which all the pore spaces are filled with water. To picture this, think of filling a cup with gravel. Then add enough water to half fill the cup. The top of the water layer represents the water table.

•Below it, where the gravel is covered with water, is the saturated zone. Above it, where there is just gravel, would be the unsaturated zone. The bottom of the cup forms a barrier to the water continuing to move downward. This barrier can occur in nature as bedrock, clay, or permafrost and is called an aquitard.

DEFINITION OF GROUNDWATER HYDROLOGY

•figure shows the waters found in limestones, crystalline rocks, and two types of fine-grained rocks

•These waters derive their compositions from a variety of processes, including dissolution, hydrolysis, and precipitation reactions; adsorption and ion exchange; oxidation and reduction; gas exchange between groundwater and the atmosphere; and biological processes . The biological processes of greatest importance are microbial metabolism, organic production, and respiration (oxidation). By far the most important overall process for the major constituents of groundwater is that of mineral–water reactions, which were briefly described above in River and ocean waters. Thus, the composition of groundwaters strongly reflects the types of rock minerals that the waters have encountered in their movement through the subsurface..

ROCK PROPERTIES AFFECTING GROUNDWATER

• POROSITY

• AQUIFER

TYPES OF AQUIFER

• Confined or Artesian Aquifer
  
• Unconfined or Water Table Aquifer
  

CONFINED AND UNCONFINED AQUIFERA

WATER MOVEMENT IN AQUIFERS

AQUIFERS

PARAMETERS OF GROUNDWATER STORAGE

ONE DIMENSIONAL GROUNDWATER FLOW IN STEADY CONDITION

Groundwater flows underground in response to elevation differences (downwards) and pressure differences (from areas of high pressure to areas of low pressure).

DARCY’S LAW RELATES

•       The rate of the groundwater movement (Q)
•       To the hydraulic conductivity (K)
•       The cross-sectional area (A)
•       And to the hydraulic gradient or slope of the water table (h/l)
•       The hydraulic conductivity depends on the permeability of the rock and on the properties of the water. Water generally flows in the direction of the hydraulic gradient and slope of the water table

	Q = Kah/l

 

DETERMINATION OF PERMEABILITY

• PERMEABILITY

The ability of substance to allow another substance to pass through it, especially the ability of a porous rock, sediment, or soil to transmit fluid through pores and cracks.

• PERMEAMETERS

Device use to determine permeability.

EXAMPLE OF DARCY’S LAW

• A confined aquifer has a source of recharge.
• K for the aquifer is 50 m/day, and n is 0.2.
• The piezometric head in two wells 1000 m apart is 55 m and 50 m respectively, from a common datum.
• The average thickness of the aquifer is 30 m, and the average width is 5 km.

CALCULATE:

a) The rate of flow through the aquifer

b) The time of travel from the head of the aquifer to a point 4 km downstream      

*assume no dispersion or diffusion

SOLUTION

Cross-Sectional area    = 30(5)(1000) = 15 x 104 m2 Hydraulic gradient      = (55-50)/1000 = 5 x 10-3 Rate of Flow for K     = 50 m/day                                         Q         = (50 m/day) (75 x 101 m2)                                                 = 37,500 m3/day Darcy Velocity:                                               V = Q/A = (37,500m3/day) / (15 x 104 m2) = 0.25m/day Seepage Velocity:                               Vs        = V/n   = (0.25) / (0.2) = 1.25 m/day (about 4.1 ft/day) Time to travel 4 km downstream:                   T = 4(1000m) / (1.25m/day) = 3200 days or 8.77 years This example shows that water moves very slowly underground.

 

DARCY’S LAW:
EXAMPLE 2

•         A channel runs almost parallel to a river, and they are 2000 ft apart.

•         The water level in the river is at an elevation of 120 ft and 110ft in the channel.

•         A pervious formation averaging 30 ft thick and with K of 0.25 ft/hr joins them.

•         Determine the rate of seepage or flow from the river to the channel.

SOLUTION:

•         Consider a 1-ft length of river (and channel).                        Q = KA [(h1 – h2) / L] •         Where: A = (30 x 1) = 30 ft2                                                                                                                                                                                                                     K = (0.25 ft/hr) (24 hr/day) = 6 ft/day •         Therefore, Q = [6 (30) (120 – 110)] / 2000                                                                                                                                             = 0.9 ft3/day/ft length =  0.9 ft2/day

 

STEADY STATE FLOW EQUATION

• Steady-state flow refers to the condition where the fluid properties at any single point in the system do not change over time. These fluid properties include temperature, pressure and velocity. One of the most significant properties that is constant in a steady-state flow system is the system mass flow rate. This means that there is no accumulation of mass within any component in the system.
• CONFINED FLOW

CONFINED FLOW

EQUATION : q = Kb[(h0-h)/x]

• UNCONFINED FLOW

UNCONFINED FLOW

EQUATION: q = K[(h0-h1)/x][(h0-h1)/2]

EXAMPLE:

A channel runs parallel to a river. The water level in the river at an elevation of 200 ft and in the channel at an elevation of  180 ft. the river and the channel are 2500 ft apart and a pervious formation of average 20 ft thickness and hydraulic conductivity of  0.10 ft/hr joint  them together. Determine the rate of seepage glow from the river to the channel.

Solution:

This is a confined flow, use equation q = Kb[(h0-h)/x]

h0 = 200 ft

h = 180 ft

x = 2500 ft

K = 0.10 ft/hr

b = 20 ft

q = Kb[(h0-h)/x]

q = (0.10)(20)[(200-180)/2500]

q = 0.016 ft^3/hr or 0.384 ft^3/day

CONFINED FLOW TO WELL

CONFINED FLOW TO WELL

EQUATION: (Q/2πbK)In(R/r) = H-h 

WHERE:

• K= TRANSMISSIVITY OR PERMEABLITY OF AQUIFER
• b = THICKNESS OF CONFINED AQUIFER
• Q = DISCHARGE FROM THE WELL
• H = PIEZOMETRIC HEAD AT RADIAL DISTANCE R
• h = PIEZOMETRIC HEAD AT ANY DISTANCE r

EXAMPLE:

An aquifer of 15m average thickness is overlain by an impermeable layer of 50m thickness. A test well of 1m diameter and two observation wells at a distance of 30m and 80m  from the test well are drilled through the aquifer. After pumping at a rate of 0.5 m^3/s for a long time the following  draw-downs are stabilized in these wells: first observation well, 10m; second observation well, 8m. Determine the hydraulic conductivity.

Solution:

Use equation (Q/2πbK)In(R/r) = H-h 

Q = 0.5 m^3/s

b = 15m

R = 80m

r = 30m

H = 10m

h = 8m

 (Q/2πbK)In(R/r) = H-h 

 K = [Q/2πb(H-h) ]In(R/r)

 K = [0.5/2π(15)(10-8)]In(80/30)

 K = 2.65 *10^-3 m/s

UNCONFINED FLOW TO A WELL

An unconfined aquifer (Figure 9.1) is close to the land surface, being under the direct influence of the climatic factors (precipitations mainly, but temperature also). The groundwater fluctuations follow with a certain lag, depending on the depth and the nature of the unsaturated zone, the variation of the fallen precipitations. The unconfined aquifers extend from thewater table to the base of the aquifer, represented by an impermeable boundary. Most of the unconfined aquifers are formed by highly permeable layers (gravel, coarse or medium sand) and less permeable formations (silt or clay) that do not cut the hydraulic continuity of the permeable layers on a regional level.

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Figure 1: Unconfined aquifer

Figure 2: Steady Flow to a Well in an Unconfined Aquifer

From Figure 2, consider that the well is pumping continuously with a constant rate ,Q in an unconfined aquifer with homogenous hydraulic conductivity. Originally, when the well was not pumping the groundwater table was at a level h_o or H_o above the base of the aqufier., which can be assumed more or less horizontal in the vicinity of the well. When the well is pumping a cone of depression is formed that enables groundawter flow towards the well. This also causes a decrease of the water table such that the position H of the water table measured from the base of the aquifer becomes variable. The mass balance equation now becomes:

H² - h² =  In

Example for calculation

1.      A 200 mm diameter well which fully penetrating a confined aquifer of thickness 25 m is pumped at a constant rate of 2000 m3/day. The steady state drawdown in the well is 8 m and the drawdown in a piezometer 100 m from the well is 1.4 m.

H² - h² =  In

where h₁ & h₂ are measured from the base of the aquifer.

So,                               (25 - 1.4)² - (25 - 8)² = (2000/πK) In(100/0.1)

So,                                K =(2000/πK) In(100/0.1)=16.4m/day

and                                T = K.h = 16.4 × 25 = 410 m²/day

at radius of influence (R), h = 25.

So,                                 25² - (25-8)² =  (2000/16.4π)In(R/0.1)

                                                        http://easycalculation.com/physics/fluid-mechanics/unconfined-aquifer-well.php                      

R = 0.1e[(16.4π* ( 25² - (25-8)²)/2000] = 574 m

Unconfined Aquifer Well Flow Rate Calculator

PUMPING TEST

Typical Orifice Meter Used For Aquifer Pumping Tests

System Geometry For Unconfined Aquifer Pumping Test 

System Geometry For A Confined Aquifer