1. Basic Irrigation Hydraulics
N.L Mufute

2. Introduction
Hydraulics is the study of water at rest and in motion.
Applied hydraulics is concerned primarily with the computation of flow rates, pressures and forces in water storage and conveyance systems.
For practical purposes water is considered to be an incompressible liquid.
Pressure
Water exerts forces against the walls of its container, whether it is stored in a reservoir or flowing in a pipeline or canal. We can say that it exerts a pressure.
Pressure is defined as a force per unit area:
P= F/A
Where P = Pressure
F = Force
A= Area
In SI metric units pressure is expressed in terms of newtons per square metre (N/m2); 1 N/m2 = 1pascal (pa).
Common pressure units: 1Kpa = 1000pa, 1atmosphere approx = 100kpa (=101kpa), 1bar = 100kpa, 1bar= 10m, 1000Kpa= 0.145psi .

3. Hydrostatic Pressure It is the pressure that water exerts at rest.
The following very important principles always apply for hydrostatic pressure.
The pressure depends only on the height of water above the point in question ( not on the surface area). P=gh (where P = hydrostatic pressure in kpa, h = water depth from the surface, m, g = acceleration due to gravity= 9.8 m/s2)
The pressure increases in direct proportion to the depth.
The pressure in a continuous volume of water is the same at all points that are at the same depth.
The pressure at any point in the water acts in all directions at the same magnitude.

4. Example 1.
An elevated water storage tank and connecting pipeline are shown in figure 1 below. Compute the hydrostatic pressure at points A, B,C,D and E .
Figure 1
Elevation m
Ground Surface
Closed valve
D. Elevation 55.00m
A. Elevation 95.00m
Water storage tank
C. Elevation 70.00m
B , Elevation m
Water main E

5. Solution
Point A: The height of the water above point A in the tank is equal to the difference in elevation between the water surface and the tank bottom, or = 5m. Therefore
PA = 9.8 x 5 = 49kpa.
Point B: The total height of water above point B is =30.00m. The pressure at that point is
PB= 9.8x30.00 = 290kpa.
Point C : The pressure at point C is equal to the pressure at point B because these points are the same elevation. Thus,
PC = PB= 290kpa
Point D: The total height of water above point D is = 45.00m. The pressure at that point is
PD = 9.8x45.00 = 440kpa
Point E: There is not enough information to determine the pressure at this point because it is isolated from the system above it by the closed valve. Remember pressures are transmitted only in a continuous fluid; the water at point E is not continuous with the water on the other side of the valve.

6. Pressure Head
It is often convenient to express pressure in terms of the height of a column of water, in meters, instead of in kpa.
This is what is called the pressure head.
Measurement of Pressure
Several methods are available; some of the main ones are:
Piezometer tube
Manometer
Bourdon tube gauge /pressure gauge
Pressure transducers

7. Measurement of Pressure Cont.
Piezometer tube
The simplest way to determine pressure.
For example , if a narrow transparent tube is attached to a pipeline under pressure as shown in figure 2 below, the water in the pipe will rise in the tube until the pressure head caused by the column of water is equal to the pressure in the pipe.
By measuring the height of the column in meters and using the equation P =gh one can find the pressure in kpa.

8. Measurement of Pressure Cont.
Figure 2: A Piezometer Tube
15.3m
Piezometer Tube
Water under pressure
150kpa
Although they are simple, piezometers are not very practical for field use e.g. to measure a pressure of 150kpa the piezometer will have to be 15m high.
In ground water, the term piezometric surface is used to indicate pressure in water that is confined under the ground.

9. Measurement of Pressure Cont.
Manometer
It is a more practical device for measuring pressure using the height of a column of liquid.
The liquid in the manometer tube is different from that in the system being measured.
A well-type manometer using mercury as the manometer fluid, is illustrated in figure3 below.
Mercury is a heavy metal that is liquid at room temperature; it is 13.6 times as heavy as water.
In a well-type mercury manometer, the equivalent head of water in the system is 13.6 times the measured height of the column of mercury.

10. Measurement of Pressure Cont.
Figure 3 . A well-type manometer
Mercury
40 cm
Pressure
53kpa
Manometer
Fluid Well
As an example, if the column of mercury shown in figure 3 is 40cm high, the pressure in the system is P=13.6 x 9.8x 0.4m =53kpa

11. Measurement of Pressure Cont.
Bourdon Pressure gauge
It is one of the most commonly used pressure measuring devices.
It works on the principle that a flattened hollow metal tube, curved in the form of a spiral or circular arc, tends to uncurl as pressure is applied inside the tube.
As the tube uncurls, a pointer linked to it indicates the pressure on a calibrated scale.

12. Measurement of Pressure Cont.
Pressure transducers
These are devices that sense changes in pressure and convert them to pneumatic or electrical signals are installed in water treatment plants and in pumping stations.
They transmit signals to a central control panel, where the operator can see the pressure readings conveniently displayed.

13. WATER FLOW
Most applications of hydraulics in irrigation involve water in motion – in pipes under pressure or in open channels under the force of gravity.
The volume of water flowing past any given point in the pipe or channel per unit time is called the flow rate or discharge.
In the SI system, the unit of flow rate is cubic metres per second (m3/s).
Other Units:
litres per second (L/s)
mega litres per day (ML/d)
Note: 1m3 = 1000l and 1ML = 106L .

14. Example 2
Convert a flow of 50m3/s to its equivalent value expressed in terms of L/s and ML/d.
Solution : 50m3/s x 1000L /m3 = L/s
50 000L/s x 3600s/hr x 24hr/day x 10-6ML/L = 4320ML/d

15. Flow rate Should not to be confused with velocity of flow.
Flow rate represents volume per unit time, whereas velocity represents distance per unit time.
Q= AV
Where Q = flow rate or discharge (m3/s)
A = cross sectional flow area (m2)
V = velocity of flow (m/s)

16. Example 3
Determine the required diameter of a pipe that will carry a discharge of 50ML/d of water at a velocity of 3m/s.
Solution: The pipe diameter D can be determined from the required flow area A. From Q=AV; re arranging; A = Q/V.
Given: Q = 50 x 106 L/d x 1 d/24h x 1h/3600s x 1m3/1000L = 0.58m3/s.
V = 3m/s
Therefore substituting; = 0.19m2
Re arranging the terms in the formula A =πD2/4 therefore D = 0.49m
Or D = 0.49m x 1000mm/m = 490mm.

17. Continuity flow
Water is considered to be incompressible fluid. i.e. , its volume does not change significantly with changing pressure.
Thus, for a steady discharge in a pipe, the flow rate Q must be constant at any section in the pipe, no matter how the flow area or velocity may change.
Figure 4.
Section 1
Section 2
Qout
Q2, A2 ,V2
Q1, A1 ,V1
Qin
Qin = Qout
V2> V1

18. Continuity equation
Referring to figure 4, it can be said that the flow rate Q1 at section 1 must equal the flow rate Q2 at section 2, since water is neither added to nor removed from the pipe between those two sections.
But the path of flow is constricted at section 2 of the pipe. At this section therefore, the flow velocity is increasing as the water moves from section 1 to 2.
Since Q is constant, when A gets smaller, V must get larger. Conversely, if the area of flow increases, the velocity of flow must decrease.
This principle is sometimes referred to as continuity of flow. A common formula used to express this is;
Q = A1xV1 =A2xV2
The equation above is called the continuity equation.

19. Example 4.
In the pipeline shown in the figure 4 above, the area at section 1 is 0.50m^2 and the area at section 2 is 0.25m^2 .
For Qin = 1000L/s , determine the velocities at sections 1 and 2.
Solution: First convert 1000L/s to 1.0m^3 /s ;
Q = A1xV1 =A2xV2
1m^3/s = 0.50x V1= 0.25x V2 
V1= 2m/s
V2= 4m/s
Note that because the area decreased by a factor of ½, the velocity increased by a factor of 2.
The velocity is inversely proportional to the area. Also, the velocity is inversely proportional to the square of the diameter.

20. Conservation of Energy
It is a basic principle in physics that energy can neither be neither created nor destroyed, but it can be converted from one form to another.
In a given closed system, the total energy is constant. This is the law of conservation of energy.
Applied to problems involving the flow of water, it proves to be a useful principle.
In hydraulic systems, there exist three forms of mechanical energy: potential energy due to elevation, potential energy due to pressure, and kinetic energy due to motion.
In a hydraulic system therefore, there are elevation head, pressure head, and velocity head.
The total energy head in a hydraulic system is equal to the sum of these individual energy heads:

21. Conservation of Energy
E = z p/w v2/2g
Total
Head
Elevation Head
Pressure head
Velocity head
Where:
E = Total energy head (kpa or m)
z= height of the water above a reference plane, (m)
p= pressure in kpa or m
w=unit weight of water, 9.8kN/m3
v = flow velocity, m/s
g = acceleration due to gravity, 9.8m/s2

22. Bernoulli’s Equation Figure 5. z1 + p1/w + v12/2g = z2 + p2/w + v22/2g
Total energy line
Pressure drop 1 Q 2
Constriction
Reference plane Z2
P2/w
V22/2g E2
Consider the constricted section of pipe shown in figure 5 above.
From the law of conservation of energy, the total energy head at section 1, E1, must equal the total energy head at section 2, E2. Setting E1 = E2 and using equation above;
z1 + p1/w + v12/2g = z2 + p2/w + v22/2g
This equation is called Bernoulli’s equation and is one of the most useful equations in hydraulics. As written here, it applies to ideal fluids because viscosity and energy loss due to friction are neglected.

23. FLOW IN PIPES UNDER PRESSURE
When water flows in a pipeline, there is friction acting between the flowing water and the pipe wall, and between the layers of water moving at different velocities in the pipe.
The flow velocity is actually zero at the pipe wall and maximum along the centre line of the pipe.
When the term velocity of flow is used in this module, it means the average velocity over the cross section of flow.
The frictional resistance to flow causes a loss of energy in the system.
This loss of energy is manifested as a continuous pressure drop along the path of flow.
It is often necessary to be able to compute the expected pressure drop in a given system.

24. Hazen- Williams Equation
To be able to design new water distribution pipelines or to analyse existing pipe networks, it is necessary to be able compute head losses, pressures and flows throughout the system.
There are several formulae in hydraulics to do this, but one of the most commonly used is the Hazen-Williams equation:

25. The equation can be used in the form of a Nomograph to facilitate quicker calculations.

27. Example 5:
What is the minimum pipe diameter that is required to carry a flow of 30L/s without causing the pressure to drop more than 10kpa per kilometre of pipeline?
Solution: Convert the pressure drop to an equivalent pressure head using P=gh;
hL = P/9.8 = 0.10 x 10 = 1.0m/km.
Therefore head loss is 1m per 1000m
Entering the nomograph with Q = 30L/s and head loss of 1m per 1000, read 310mm on the D or diameter axis.
The energy loss due to viscosity and friction along the straight length of the pipeline accounts for most of the pressure drop.
This loss is called the major loss. As the water flows through valves, bends, and other pipe fittings, there are additional losses due to turbulence.
These losses are called minor losses.

28. Flow Measurement in closed pipelines
The rate at which water is pumped into a distribution system must be known for proper control and operation of the system.
Several types of flow meters can be used:
Venturi meter
Magnetic flow meter
Pitot static tube ( measuring the flow discharged by an open hydrant/ outlet)

29. GRAVITY FLOW IN PIPES
When water flows in a pipe or channel with a free surface exposed to the atmosphere, it is called open channel or gravity flow.
In most routine problems in the design involving gravity flow, a condition called steady uniform flow is assumed.
Steady flow means that the discharge is constant with time.
Uniform flow means that the slope of the water surface and the cross sectional flow area are also constant.
A length of a stream, channel or pipeline that has a relatively constant slope and cross section is called a reach.
Under steady uniform flow conditions, the slope of the water surface is the same as the slope of the channel bottom.

30. Figure 7 Steady Uniform Open Channel Flow
Channel Bottom
Water surface = HGL Q hL
S =hL/L

31. Figure 8: Flow with a free surface exposed to the atmosphere : Open Channel Flow
Pipe
Crown
Ground Surface
Air
b.) Stream
Buried pipe
Partial flow
Invert
The length of wetted surface on the pipe or stream cross section is called the wetted perimeter.
The size of the channel, as well as the slope and the wetted perimeter, are important factors related to its discharge capacity.

32. Manning’s formula
A common formula for solving open channel flow problems.
Q =1/n AR2/3S1/2
It is an empirical or experimentally derived equation, where
Q = Channel discharge capacity , m3/s
n = Manning channel roughness coefficient
A = Cross-sectional flow area, m2
R= hydraulic radius of the channel, m
S= Slope of the channel bottom.
The hydraulic radius of a channel is defined as the ratio of the flow area to the wetted perimeter P.
R=A/P.
The roughness coefficient n depends on material and edge of pipe or lined channel and on topographic features for a natural stream bed.
It can range from a value of 0.01 for a smooth clay pipe to 0.1 for a small natural stream.
A value of n commonly assumed for concrete pipes or lined channels is

33. Example 6
A rectangular drainage channel is 1m wide and is lined with concrete as illustrated in figure 9 below.
The bottom of the channel drops in elevation at a rate of 0.15m per 30m.
What is the discharge in the channel when the depth of water is 0.45m? Assume n =

34. Figure 9
Wetted perimeter
Free water surface
0.45m 1m

35. Solution: The cross sectional flow area A = 1m x 0.45m = 0.45m2
P = 0.45m+1m+0.45m =1.9m.
R= A/P = 0.45m2/1.9m = 0.237m. The slope S = 0.15/30 =0.005
Applying the Manning’s formula we get
Q= 1/0.013(0.45x /3 x /2)
= 0.937m3/s

36. Circular Pipes Flowing Full
In a circular pipe carrying water such that the pipe is just full to the crown but still under atmospheric pressure and gravity flow, the manning formula takes the following form;
Q= πD2/4n x (D/4)2/3 x S1/2
For a given value of n, only the pipe diameter and slope are needed to solve for discharge in a circular pipe flowing full.
To facilitate the application of the Manning formula, particularly routine problems with circular pipes, charts or nomographs have also been developed.
To use the nomograph, two of the four variables (D,S,Q and V) must be known.

37. Example 7: A 450mm diameter storm sewer is built on grade of 2%.
What are the discharge capacity and velocity of flow when the pipe is full?
Solution: A nomograph can be used; but using the Manning’s gives
Q = (1/0.013) X π0.452/4 x (0.45/4)2/3 x0.021/2
And V=Q/A = 0.4/( π x 0.452/4) =2.5m/s

38. Open Channel Flow Measurement
The following methods are used;
Floating object method
Weirs (rectangular, triangular/V-notch and trapezoidal/Cipolletti)
Flumes ( Parshall , cut throat etc)
Propeller powered flow meters,

39. NON UNIFORM OPEN CHANNEL FLOW
Open channel flow is not always steady over time or uniform in cross sectional area.
Such gradually varied flow occurs in natural streams and rivers.
The physical principles and mathematical methods needed to fully analyse non-uniform flow problems are beyond the scope of this module.
This presentation provides a brief descriptive overview of non uniform and gradually varied flow, to introduce students to hydraulic conditions that are often encountered in practice.
Under uniform and steady flow the depth of flow is called normal depth.

40. Specific Energy
Channels do not always flow at normal depth, particularly in the vicinity of changes of slope or channel cross sectional changes.
An important concept related to the type of flow in a channel is called the specific energy, which is defined as the sum of the depth of flow and the velocity head in a channel;
E = y + v2/2g
Where E = specific energy , m
y = depth of flow , m,
v = average velocity, m/s
g= acceleration due to gravity, m/s2
v2/2g = “velocity head”

41. Specific Energy cont.
If friction losses and change in elevation are neglected for a very short section of a channel, the specific energy must be constant for two adjacent channel cross sections, or E1 = E2, and thus
Y1 +v12/2g = Y2 +v22/2g
From the principle of continuity of flow, the velocity of flow is inversely proportional to the area of flow, with constant discharge.
Also area of flow is a function of channel depth.
Therefore for a given discharge the specific energy is solely a function of depth at each point in a channel, and there may be more than one depth having the same specific energy.
This is depicted in figure 10 below, which is a plot of channel depth y versus specific energy E for a constant flow rate.

42. Figure 10: A graph of the flow in an open channel versus the specific energy of flow.
450
Channel
Depth, y
Specific Energy, E
Supercritical flow
Critical Depth
Subcritical flow
The critical depth occurs when the specific energy is minimum.

43. Critical Flow
It can be seen from the figure 10 above that there is one flow depth at which the specific energy reaches a minimum value.
This is called the critical depth, and the velocity at which critical depth occurs is called the critical velocity.
If the actual depth of flow is higher than the critical depth, the type of flow is characterised as sub-critical; flow velocity in subcritical flow is slower than the critical velocity.
If the actual depth of flow is lower than the critical depth , the flow is characterised as supercritical ; flow velocity in super critical flow is faster than the critical velocity.
Typically water flowing rapidly down a steeply sloped, shallow channel is undergoing supercritical flow, whereas water flowing slowly in a relatively deep channel on a gentle or mild slope is in subcritical flow conditions.

44. The hydraulic jump
It occurs when flow passes from supercritical to subcritical conditions.
There are significant energy losses associated with hydraulic jumps due to the amount of turbulence (rapid mixing) that occurs.
The forces caused by the jump can cause significant erosion, so engineers may try to prevent hydraulic jumps from occurring in sewer systems for e.g.
As an alternative they can calculate the expected location of the jump in order to provide adequate channel, pipe, or structure protection .
For instance when a jump occurs just downstream of a dam spillway (figure 11 below)

45. Figure 11.
Flow
Water
surface
Spillway
of dam
Supercritical
Concrete apron
Subcritical
Hydraulic
jump
A suitable apron must be designed and built so that excessive channel bottom erosion does not occur.

46. References
FAO, Irrigation Manual: Planning, Development, Monitoring & Evaluation of Irrigated Agriculture with Farmer Participation. FAO, Harare.
Nathanson J.A., Basic Environmental Technology; Water Supply, Waste Management and Pollution Control, 4th edition. Prentice Hall of India, New Delhi.
Shammas, N.K. and L.K. Wang , Water Supply and Wastewater Removal, 3rd ed., John Wiley and Sons,