The document contains examples of groundwater problems involving calculations of specific yield, groundwater storage, discharge, and permeability. Example 1 examines how the annual drinking water needs of a local population can be met from groundwater resources in an area. It is determined that there is enough groundwater available in the area to meet local needs. Example 2 involves calculations to determine the specific yield of an aquifer and the soil moisture deficit before irrigation based on water table fluctuations from irrigation and pumping. Example 3 calculates the specific yield and change in groundwater storage given information about the water table drop, porosity, and specific retention in an area.

1. Example 1:
The following data is obtained from the difficult rockyareas of Southern U.P.
(India):
Area (rocky) – 1 km2
Normal rainfall – 700 mm
Normal fluctuation of water table before and after rains – 3.2 m
Specific yield of the rock – 2%
Population – 154/km2
Examine how far the drinking water needs of the local population can be met.
Solution:
Ground water storage available annually-
Q = Area × Depth of fluctuation of g.w.t. × Specific yield 2
= 106 × 3.2 × 2/100 = 64,000 m3
Which can be replenished by normal rainfall whose volume, assuming an
infiltration rate of 10% = (1 × 106) × 700/1000 × 10/100 = 70,000 m3 and also as
observed by the normal fluctuation of water table. Assuming a per capita
consumption of 180 lpd annual drinking water supply required = 154 × 180 × 365
= 10,120,000 litres or 10,120 m3.

2. The annual drinking water supply required is 10,120 m3 against an availability of
64,000 m3. Thus, there is enough replenishable ground water resource available in
the area to meet the drinking water needs of the local population. The only problem
is location and economic construction of potential wells.
Example 2:
In a phreatic aquifer extending over 1 km2 the water table was initially at 25 m
below ground level. Sometime after irrigation with a depth of 20 cm of water, the
water table rose to a depth of 24 m below ground level. Later 3 × 105 m3 of water
was pumped out and the water table dropped to 26.2 m below ground levels.
Determine- (i) specific yield of the aquifer, (ii) deficit in soil moisture (below field
capacity) before irrigation.
Solution:
Volume of water pumped out = Area of aquifer × drop in g.w.t. × specific yield
3 × 105 = 106 × 2.2 × Sy
Sy = 0.136, or 13.6%
Volume of irrigation water recharging the aquifer = Area of aquifer × rise in g.w.t
× Sy Considering an area of 1 m2 of aquifer,
1 × y = 1 × 1 × 0.136
Recharge volume (depth) y = 0.136 m, or 136 mm === assignment question
Soil moisture deficit (below field capacity) before irrigation = 200 – 136 = 64 mm

3. Example 3:
In an area of 100 ha, the water table dropped by4.5 m. If the porosity is 30% and
the specific retention is 10% determine- (i) the specific yield of the aquifer, (ii)
change in ground water storage.
Solution:
Porosity = Sy + Sr
30% = Sy + 10% Sy = 30 – 10 = 20% or 0.2
Change in ground water storage = Area of aquifer × drop in g.w.t. × Sy
= 100 × 4.5 × 0.2
= 90 ha-m, or 90 × 104 m3
3.5.1 Example Problem 1
In an unconfined aquifer extending over 4 km2, the water table was initially at 26
m below the ground surface. Sometime after an irrigation of 20 cm (full irrigation),
the water table rises to a depth of 25.5 m below the ground surface. Afterward 1.5 ´
106 m3 of groundwater was withdrawn from this aquifer, which lowered the water
table to 27.5 m below the ground surface. Determine: (i) specific yield of the
aquifer, and (ii) soil moisture deficit (SMD) before irrigation.
Solution:

4. (i) Volume of groundwater withdrawn from the unconfined aquifer = Area of
the aquifer * Drop in the water table * Specific yield
Substituting the values, we have,
(ii) Volume of water recharged due to irrigation (VR) = Area of the aquifer
influenced by irrigation * Rise in the water table * Sy
Let us consider the aquifer area influenced by irrigation to be 140 m2, then the
volume of water recharged (VR) will be:
VR = 140´X(26.0-25.5)´0.19 = 13.3 m3
Volume of irrigation water (VI) = 140 × 0.20 = 28.0 m3
Now, Soil moisture deficit (SMD) before irrigation = VI - VR = 28.0-13.3 = 14.7
m3.
Or, SMD = = 0.105 m = 10.5 cm, Ans.
3.5.2 Example Problem 2

5. In an area of 200 ha, the water table declines by 3.5 m. If the porosity of the
aquifer material is 30% and the specific retention is 15%, determine: (i) specific
yield of the aquifer, and (ii) change in groundwater storage.
Solution:
(i) We know, Porosity = Specific yield (Sy) + Specific retention (Sr)
(ii) Change in groundwater storage = Area of the aquifer * Drop in the water table
*Specific yield
= (200X104) X3.5X0.15
= 105X104 m3, Ans.
3.5.3 Example Problem 3
The average thickness of a confined aquifer extending over an area of 500 km2 is
25 m. The piezometric level of this aquifer fluctuates annually from 10 m to 22 m
above the top of the aquifer. Assuming a storage coefficient of the aquifer as
0.0006. Estimate annual groundwater storage in the aquifer.
Solution:

6. Annual groundwater storage (GWS) in the confined aquifer is given as:
GWS = Area of the aquifer ´ Rise in the piezometric level ´ Storage coefficient
= (500X106)´(22-10)X0.0006
= 3.6X106 m3, Ans.
20cm long filed sample of silt, fine sand with a diameter of 10cmis tested using
falling head permeater. The falling head tube has the diameter of 3.0cm and the
initial head is
Problem 1
A confined aquifer is 3.0 m thick. The piezometric level drops 0.15 m between two
observation wells which are located 238 m apart. The hydraulic conductivity of the
aquifer is 6.5 m/day and the effective porosity is 0.15. Determine the following: (a)
Discharge of groundwater through a strip of the aquifer having 10 m width, and (b)
Average linear velocity of groundwater.
Solution: From the question, we have:
Thickness of aquifer, b = 3.0 m
Difference in piezometric levels, Δh = 0.15 m
Distance between the observation wells, ΔL = 238 m

7. Hydraulic conductivity of the aquifer, K = 6.5 m/day
Effective porosity, ne = 0.15
Width of the aquifer strip, W = 10 m
(a) Groundwater discharge per unit width of the confined aquifer (q) is given as:
Groundwater discharge through the 10 m aquifer strip = W × q = 10 × 0.012
= 0.12 m3/day, Ans.
(b) Average linear velocity of groundwater =
= 0.027 m/day, Ans.

8. 6.3.2 Problem 2
An unconfined aquifer has a hydraulic conductivity of 1.2×10-2 cm/s. There are
two fully penetrating observation wells installed in this aquifer, which are
separated by a distance of 98.5 m from each other. In the upstream observation
well, the water level is 7.5 m above the aquifer bottom, and in the downstream
observation well, it is 6.0 m above the aquifer bottom.
(i) What is the groundwater discharge per 40 m-wide strip of the aquifer? Express
your answer in cubic meters per day.
(ii) What is the water-table elevation at a point midway between the two
observation wells?
Solution: From the question, we have:
Hydraulic conductivity of the aquifer, (K) = 1.2 × 10-2 cm/s = 1.2 × 10-4 m/s
Distance between the two observation wells, L = 98.5 m.
Considering bottom of the aquifer as a datum, hydraulic head at the upstream
observation well (h1) = 7.5 m, and hydraulic head at the downstream observation
well (h2) = 6.0 m.
Width of the aquifer strip, W = 40 m

9. (i) Groundwater discharge per unit width of the unconfined aquifer (q) is given as:
= 1.23 × 10-5 m2/s
Groundwater discharge per 40 m-wide aquifer strip = W × q
= 40 × (1.23 × 105)
= 49.2 × 10-5 m3/s
= 42.51 m3/day, Ans.
(ii) Distance of the point midway between the two observation wells (x) = 98.5/2 =
49.25 m. Water-table elevation at the point midway between two observation wells
(hmid) can be calculated from the following equation:
, where x = 49.25 m

10. Ans.
Example 23.1:
If the elevation of h1 is 35m and the elevation of h2 is 0m, what is the hydraulic
gradient if the distance from h1 to h2 is 5.6 km? (Answer in m/km).
Solution:
Given, h2-h1= 35m and L=5.6 km
We know: i= (h2-h1)/L
i= 35/5.6 = 6.25m/km. Ans.
Example 23.2: Find the velocity of the water flow between two wells located at a
distance of 1000 m and the hydraulic conductivity is 114m/day. Drop in elevation
between two well is given as 60 m.

11. Solution:
Given: K=114m/day, h2-h1=60m, L=1000m
We know, Hydraulic gradient, i = h2-h1/L = 60/1000
= 0.06
We know,
V=KI or
V=K(h2-h1/L)
V=114m/day * 0.06
V=6.84 m/day. Ans.
Example 23.3
An aquifer is 2045 m wide and 28 m thick. Hydraulic gradient across it is 0.05 and
its hydraulic conductivity is145m/day. Calculate the velocity of the groundwater as
well as the amount of water that passes through the end of the aquifer in a day if
the porosity of the aquifer is 32%.
Solution:
Given: K=145m/day, i= 0.05, W=2045m, D=28m, Porosity =32%
First we must solve for V. We know,

12. V= Ki =145m/day X 0.05
=7.25m/day
Now that we know V we can determine the discharge (Q) of water through the end
of the aquifer
Q= Area. Velocity = A. V= (2045m x 28m) x 7.25m/day
Q=415,135 m3/day.
This means that each day, if the aquifer had a porosity of 100%, like a river, would
have discharge of 415,135 m3/day.
However, the aquifer has porosity of 32 % and hence discharge through aquifer
would be
415,135 m3/day X0.32 = 132843.2 m3/day. Ans
Example 23.4:
A constant head permeability test was performed on a medium dense sand sample
of diameter 60 mm and height 150 mm. The water was allowed to flow under a

13. head of 600 mm. The permeability of sand was 4 x 10-1 mm/s. Determine (a) the
discharge (mm3/s), (b) the discharge velocity.
Solution:
(a) We know,
Discharge
(b) Discharge velocity 1.60 mm/s
Example 23.5:
During a falling head permeability test, the head fell from 600 mm to 300 mm in
540 s. the specimen was 50 mm in diameter and had a length of 100 mm. The cross
– sectional areaof the stand pipe was 60 mm2. Compute the coefficient of
permeability of the soil.
Solution:
Given, a = 60 mm2, L= 100 mm, t = 540
s, h1= 600 mm, h2= 300 mm.
We know,