The document provides an overview of the key concepts and principles covered in a Drainage Engineering syllabus. It discusses Darcy's law and the fundamental equations governing groundwater flow. It also addresses topics like waterlogging, salinity, drainage system design, land reclamation, canal lining, and cross drainage structures. Major drainage projects in Pakistan are also introduced. Recommended textbooks on drainage and irrigation engineering are listed.

1. DRAINAGE ENGINEERING
Syllabus
Basic Principles and Fundamental Equations
Darcy's law. Volume elasticity of aquifer. Differential-equation governing
groundwater flow. Hydraulic boundaries. Flow from and to stream. Flow net and
numerical analysis of water levels.
Water logging and Salinity
Definition of water logging. Salinity. Environmental impacts of water logging and
salinity. Mechanism of destruction and remedial measures.
Drainage
Purpose of drainage. Drainage needs. Water table. Water movements in subsoil,
permeability and methods of determination of permeability.
Design of Drainage Systems
Surface drainage. Design of open drains. Maintenance, Alignment of drainage
system. Methods of construction. Subsurface drainage: tile drains, mole drains,
determining the depth and spacing of drains. Drainage coefficient, size of the tile
drain, outlets for drains, envelope material, maintenance of tile drains and
interceptor drains.

2. Land Reclamation
Soil fertility. Factors affecting soil fertility. Nutrient elements in the soil. Land
reclamation of agricultural lands, coastal areas and strip-mines, methods of land
reclamation. Harmful effects of land reclamation
Canal Lining
Lining and its types. Financial justification and economics of canal lining. Design
of lined irrigation channels, permissible velocities in lined channels
Cross Drainage Structures
Introduction. Classification. Design of cross drainage structures
Major Drainage Projects of Pakistan
Introduction to various drainage projects of Pakistan

3. Books recommended :
Land Drainage
Cambert K. Smendena and David W. Rycroft
Cornell University Press, New York
Drainage Engineering
James N. Luthin
Rober E Krieger Publishers Company New York
Drainage of Agricultural Land in Pakistan
Dr. Nazir Ahmed
Shahzad Nazeer, Gulberg-1I1 Lahore
Irrigation and Drainage Engineering
I. H. Siddiqui
Oxford University Press

5. the velocity or flow rate moving within the aquifer
the average time of travel from the head of the aquifer
to a point located downstream

6. Darcy’s law provides an
accurate description of the flow
of ground water in almost all
hydrogeologic environments.

8. Flow rate determined by Head loss dh = h1 - h2

9. Henri Darcy established empirically that the flux
of water through a permeable formation is
proportional to the distance between top and
bottom of the soil column.
The constant of proportionality is called the
hydraulic conductivity (K).
V = Q/A, V α – ∆h, and V α 1/∆L

10. V = – K (∆h/∆L)
and since
Q = VA (A = total area)
Q = – KA (dh/dL)

11. K represents a measure of the ability for flow through
porous media:
Gravels - 0.1 to 1 cm/sec
Sands - 10-2
to 10-3
cm/sec
Silts - 10-4
to 10-5
cm/sec
Clays - 10-7
to 10-9
cm/sec

12. Darcy’s Law holds for:
1. Saturated flow and unsaturated flow
2. Steady-state and transient flow
3. Flow in aquifers and aquitards
4. Flow in homogeneous and heterogeneous systems
5. Flow in isotropic or anisotropic media
6. Flow in rocks and granular media

13. V is the specific discharge (Darcy velocity).
(–) indicates that V occurs in the direction of the decreasing
head.
Specific discharge has units of velocity.
The specific discharge is a macroscopic concept, and is easily
measured. It should be noted that Darcy’s velocity is different
from the microscopic velocities associated with the actual paths
of individual particles of water as they wind their way through
the grains of sand.
The microscopic velocities are real, but are probably impossible
to measure.

14. Darcy velocity is a fictitious velocity since it
assumes that flow occurs across the entire cross-
section of the soil sample. Flow actually takes
place only through interconnected pore channels.
A = total area
Av voids

15. From the Continuity Eqn:
Q = A VD = AV Vs
Where:
Q = flow rate
A = total cross-sectional area of material
AV = area of voids
Vs = seepage velocity
VD = Darcy velocity

16. Therefore: VS = VD (A/AV)
Multiplying both sides by the length of the medium (L)
VS = VD ( AL / AVL ) = VD ( VT / VV )
Where:
VT = total volume VV =
void volume
By Definition, Vv / VT = n, the soil porosity
Thus VS = VD / n

17. Example 1:
 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
of aquifer is 5 km.
Compute: (a) the rate of flow through the aquifer; and (b) the average
time of travel from the head of the aquifer to a point 4 km
downstream.

18. Cross-Sectional area, A = (30) (5 x 1000) = 15 x 104
m2
Hydraulic gradient, i = (55-50)/1000 = 5 x 10-3
Rate of Flow for K = 50 m/day
Q = K.i.A = (50 m/day) (15 x 104
m2
) (5 x 10-3
)
= 37,500 m3
/day
Darcy Velocity: VD = Q/A
= (37,500 m3
/day) / (15 x 104
m2
)
= 0.25 m/day

19. Seepage Velocity:
Vs = VD/n = (0.25) / (0.2)
= 1.25 m/day (about 4.1 ft/day)
Time to travel 4 km downstream:
since, Velocity = Distance/Time
or Time = Distance/Velocity
Time = (4 x1000 m) / (1.25 m/day)
= 3200 days or 8.77 years
This example shows that water moves very slowly
underground.

20. 1. For Reynold’s Number, Re > 10 or where the flow is
turbulent, as in the immediate vicinity of pumped wells.
2. Where water flows through extremely fine-grained
materials (colloidal clay)

21. Confining Layer Aquifer
30 ft
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 110 ft 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.

22. 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
The solution

23. Constant Head Falling Head

24. Apply Darcy’s Law to find K:
V/t = Q = KA(h/L)
or:
K = (VL) / (Ath)
Where:
V = volume flowing in time t
A = cross-sectional area of the sample
L = length of sample
h = constant head
t = time of flow