Manual 1 - Introduction to Irrigation

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rrigation Water Management: TrainingManual No. 1 - Introduction to Irrigation

Table of Cont

Provisional edition

FAO - FOOD AND AGRICULTURE ORGANIZATION OFHE UNITED NATIONS

a manual prepared jointlyby

C. Brouwer

International Institute for Land Reclamation and Improvementand

A. GoffeauM. Heibloem

FAO Land and WaterDevelopment Division

he designations employed and the presentation of material in this publication do not imply thexpression of any opinion whatsoever on the part of the Food and Agriculture Organization of the

United Nations concerning the legal status of any country, territory, city or area or of its authorities,

oncerning the delimitation of its frontiers or boundaries.

ll rights reserved. No part of this publication may be reproduced, stored in a retrieval system, oransmitted in any form or by any means, electronic, mechanical, photocopying or otherwise, withouior permission of the copyright owner. Applications for such permission, with a statement of the

urpose and extent of the reproduction, should be addressed to the Director, Publications Division, Fd Agriculture Organization of the United Nations, Via delle Terme di Caracalla, 00100 Rome, Italy

FAO 1985

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his electronic document has been scanned using optical character recognition (OCR) software and

reful manual recorrection. Even if the quality of digitalisation is high, the FAO declines all

sponsibility for any discrepancies that may exist between the present document and its original pri

rsion.

Table of Contents

REFACE

HAPTER 1 - BASIC TERMS AND CALCULATIONS

1.1 Introduction to surface area

1.1.1 Triangles

1.1.2 Squares and Rectangles

1.1.3 Rhombuses and Parallelograms1.1.4 Trapeziums

1.1.5 Circles

1.1.6 Metric Conversions

1.2 Surface areas of canal cross-sections and farms

1.2.1 Determination of the surface areas of canal cross-sections

1.2.2 Determination of the surface area of a farm

1.3 Introduction to volume

1.3.1 Units of volume

1.3.2 Volume of water on a field

1.4 Introduction to flow-rate

1.4.1 Definition

1.4.2 Calculation and Units

1.5 Introduction to percentage and per mil1.5.1 Percentage

1.5.2 Per mil

1.6 Introduction to graphs

1.6.1 Example 1

1.6.2 Example 2

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1.7 Test your knowledge

1.7.1 Questions

1.7.2 Answers

HAPTER 2 - SOIL AND WATER

2.1 The soil

2.1.1 Soil composition

2.1.2 Soil profile

2.1.3 Soil texture

2.1.4 Soil structure

2.2 Entry of water into the soil

2.2.1 The infiltration process

2.2.2 Infiltration rate

2.2.3 Factors influencing the infiltration rate

2.3 Soil moisture conditions

2.3.1 Soil moisture content

2.3.2 Saturation

2.3.3 Field capacity

2.3.4 Permanent wilting point

2.4 Available water content

2.5 Groundwater table

2.5.1 Depth of the groundwater table

2.5.2 Perched groundwater table

2.5.3 Capillary rise

2.6 Soil erosion by water

2.6.1 Sheet erosion

2.6.2 Gully erosion

HAPTER 3 - ELEMENTS OF TOPOGRAPHY

3.1 Slopes

3.1.1 Definition

3.1.2 Method of expressing slopes

3.1.3 Cross slopes

3.2 Elevation of a point

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3.2.1 Definition

3.2.2 Bench mark and mean sea level

3.3 Contour lines

3.4 Maps

3.4.1 Description of a map

3.4.2 Interpretation of contour lines on a map

3.4.3 Mistakes in the contour lines3.4.4 Scale of a map

HAPTER 4 - RAINFALL AND EVAPOTRANSPIRATION

4.1 Rainfall

4.1.1 Amount of rainfall

4.1.2 Rainfall intensity

4.1.3 Rainfall Distribution

4.1.4 Effective Rainfall

4.2 Evapotranspiration

4.2.1 Evaporation

4.2.2 Transpiration

4.2.3 Evapotranspiration

4.2.4 Factors influencing crop evapotranspiration

HAPTER 5 - IRRIGATION SYSTEM

5.1 Main intake structure and pumping station

5.1.1 Main intake structure

5.1.2 Pumping station

5.2 Conveyance and distribution system

5.2.1 Open canals

5.2.2 Canal structures

5.3 Field application systems

5.3.1 Surface irrigation

5.3.2 Sprinkler irrigation

5.3.3 Drip irrigation

5.4 Drainage system

HAPTER 6 - DRAINAGE

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6.1 Need for drainage

6.2 Different types of drainage

6.2.1 Surface drainage

6.2.2 Subsurface drainage

HAPTER 7 - SALTY SOILS

7.1 Salinization7.2 Salinity

7.2.1 Water salinity

7.2.2 Soil salinity

7.3 Crops and saline soils

7.4 Sodicity

7.5 Improvement of saline and sodic soils

7.5.1 Improvement of saline soils7.5.2 Improvement of sodic soils

7.6 Prevention of salinization

7.6.1 Irrigation water quality

7.6.2 Irrigation management and drainage

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PREFACE

his is one in a series of training manuals on subjects related to irrigation that will be issued in 1985 986.

he papers are intended for use by field assistants in agricultural extension services and irrigationchnicians at the village and district levels who want to increase their ability to deal with farm-leveligation issues.

he papers contain material that is intended to provide support for irrigation training courses and tocilitate their conduct. Thus, taken together, they do not present a complete course in themselves, bustructors may find it helpful to use those papers or sections that are relevant to the specific irrigationditions under discussion. The material may also be useful to individual students who want to revi

rticular subject without a teacher.

ollowing an introductory discussion of various aspects of irrigation in the first paper, subsequentbjects discussed will be:

- topographic surveying- crop water needs- irrigation scheduling- irrigation methods- irrigation system design

- land grading and levelling- canals and structures- drainage- salinity- irrigation management

t this stage, all the papers will be marked as draft because experience with the preparation of irrigataining material for use at the village level is limited. After a trial period of a few years, when there en time to evaluate the information and the use of methods outlined in the draft papers, a definitiversion can then be issued.

or further information and any comments you may wish to make please write to:

The International Support Programme for Irrigation Water ManagementLand and Water Development DivisionFAOVia delle Terme di Caracalla00100 RomeItaly

BOUT THIS PAPER

PREFACE

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troduction to Irrigation is the first in a series of training manuals on irrigation. As the title suggestsanual contains an introductory discussion of irrigation topics that will be dealt with in greater detaie subsequent elements of the series: it brings together explanatory notes on concepts, terms, method calculations that are basic to the discussion of the subject matter. In doing so this manual may sean easy reference in the study of irrigation.

PREFACE

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HAPTER 1 - BASIC TERMS AND CALCULATIONS

1.1 Introduction to surface area

1.2 Surface areas of canal cross-sections and farms

1.3 Introduction to volume1.4 Introduction to flow-rate

1.5 Introduction to percentage and per mil

1.6 Introduction to graphs

1.7 Test your knowledge

1 Introduction to surface area

1.1.1 Triangles

1.1.2 Squares and Rectangles

1.1.3 Rhombuses and Parallelograms

1.1.4 Trapeziums

1.1.5 Circles

1.1.6 Metric Conversions

important to be able to measure and calculate surface areas. It might be necessary to calculate, for example, the surface area of thess-section of a canal or the surface area of a farm.

s Section will discuss the calculation of some of the most common surface areas: the triangle, the square, the rectangle, the rhombus, theallelogram, the trapezium and the circle (see Fig. 1a).

Fig. 1a. The most common surface areas

height (h) of a triangle, a rhombus, a parallelogram or a trapezium, is the distance from a top corner to the opposite side called base (b)ght is always perpendicular to the base; in other words, the height makes a "right angle" with the base. An example of a right angle is thener of this page.

he case of a square or a rectangle, the expression length (1) is commonly used instead of base and width (w) instead of height. In the casle the expression diametre (d) is used (see Fig. 1b).

Fig. 1b. The height (h), base (b), width (w), length (1) and diametre (d) of the most common surface areas

CHAPTER 1 - BASIC TERMS AND CALCULATIONS

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.1 Triangles

surface area or surface (A) of a triangle is calculated by the formula:

triangle) = 0.5 x base x height = 0.5 x b x h ..... (1)

angles can have many shapes (see Fig. 2) but the same formula is used for all of them.

Fig. 2. Some examples of triangles

AMPLE

culate the surface area of the triangles no. 1, no. 1a and no. 2

Given Answer

angles no. 1 and no. 1a: base = 3 cmheight = 2 cm

Formula: A = 0.5 x base x height

= 0.5 x 3 cm x 2 cm = 3 cm2

angle no. 2: base =3 cmheight = 2 cm

A = 0.5 x 3 cm x 2 cm = 3 cm2

an be seen that triangles no. 1, no. 1a and no. 2 have the same surface; the shapes of the triangles are different, but the base and the heighll three cases the same, so the surface is the same.

surface of these triangles is expressed in square centimetres (written as cm 2). Surface areas can also be expressed in square decimetres




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are metres (m2), etc...

ESTION

culate the surface areas of the triangles nos. 3, 4, 5 and 6.

Given Answer

angle no. 3: base =3 cmheight = 2 cm

Formula: A = 0.5 x base x height

= 0.5 x 3 cm x 2 cm = 3 cm2

angle no. 4: base = 4 cmheight = 1 cm

A = 0.5 x 4 cm x 1 cm = 2 cm2

angle no. 5: base = 2 cm

height = 3 cm

A = 0.5 x 2 cm x 3 cm = 3 cm2

angle no. 6: base = 4 cmheight = 3 cm

A = 0.5 x 4 cm x 3 cm = 6 cm2

.2 Squares and Rectangles

surface area or surface (A) of a square or a rectangle is calculated by the formula:

square or rectangle) = length x width = l x w ..... (2)

square the lengths of all four sides are equal and all four angles are right angles.

rectangle, the lengths of the opposite sides are equal and all four angles are right angles.

Fig. 3. A square and a rectangle

e that in a square the length and width are equal and that in a rectangle the length and width are not equal (see Fig. 3).

ESTION

culate the surface areas of the rectangle and of the square (see Fig. 3).

Given Answer

uare: length = 2 cmwidth = 2 cm

Formula: A = length x width

= 2 cm x 2 cm = 4 cm2

ctangle: length = 5 cmwidth = 3 cm

Formula: A = length x width

= 5 cm x 3 cm = 15 cm2

ated to irrigation, you will often come across the expression hectare (ha), which is a surface area unit. By definition, 1 hectare equals 10

For example, a field with a length of 100 m and a width of 100 m2(see Fig. 4) has a surface area of 100 m x 100 m = 10 000 m 2= 1 ha

Fig. 4. One hectare equals 10 000 m2




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.3 Rhombuses and Parallelograms

surface area or surface (A) of a rhombus or a parallelogram is calculated by the formula:

rhombus or parallelogram) = base x height = b x h ..... (3)

rhombus the lengths of all four sides are equal; none of the angles are right angles; opposite sides run parallel.

parallelogram the lengths of the opposite sides are equal; none of the angles are right angles; opposite sides run parallel (see Fig. 5).

Fig. 5. A rhombus and a parallelogram




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ESTION

culate the surface areas of the rhombus and the parallelogram (see Fig. 5).

Given Answer

ombus: base = 3 cmheight = 2 cm

Formula: A = base x height

= 3 cm x 2 cm = 6 cm2

allelogram: base = 3.5 cmheight = 3 cm

Formula: A = base x height

= 3.5 cm x 3 cm = 10.5 cm2

.4 Trapeziums

surface area or surface (A) of a trapezium is calculated by the formula:

trapezium) = 0.5 (base + top) x height =0.5 (b + a) x h ..... (4)

top (a) is the side opposite and parallel to the base (b). In a trapezium only the base and the top run parallel.

me examples are shown in Fig. 6:

Fig. 6. Some examples of trapeziums

AMPLE

culate the surface area of trapezium no. 1.

Given Answer

pezium no. 1: base = 4 cmtop = 2 cmheight = 2 cm

Formula: A =0.5 x (base x top) x height= 0.5 x (4 cm + 2 cm) x 2 cm

= 0.5 x 6 cm x 2 cm = 6 cm2

ESTION

culate the surface areas trapeziums nos. 2, 3 and 4.

Given Answer


Formula: A = 0.5 x (base + top) x height= 0.5 x (5 cm + 1 cm) x 2 cm

= 0.5 x 6 cm x 2 cm = 6 cm2




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A = 0.5 x (3 cm + 1 cm) x 2 cm

= 0.5 x 4 cm x 2 cm = 4 cm2


A = 0.5 x (2 cm + 4 cm) x 2 cm

= 0.5 x 6 cm x 2 cm = 6 cm2

e that the surface areas of the trapeziums 1 and 4 are equal. Number 4 is the same as number 1 but upside down.

other method to calculate the surface area of a trapezium is to divide the trapezium into a rectangle and two triangles, to measure their sito determine separately the surface areas of the rectangle and the two triangles (see Fig. 7).

Fig. 7. Splitting a trapezium into one rectangle and two triangles. Note that A = A1+ A2+ A3= 1 + 6 + 2 =9 cm2

.5 Circles

surface area or surface (A) of a circle is calculated by the formula:

circle) = 1/4 (x d x d) = 1/4 (x d2) = 1/4 (3.14 x d2) ..... (5)

ereby d is the diameter of the circle and (a Greek letter, pronounced Pi) a constant (= 3.14). A diameter (d) is a straight line which dicircle in two equal parts.

Fig. 8. A circle




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AMPLE

Given Answercle: d = 4.5 cm Formula: A = 1/4 (x d)

= 1/4 (3.14 x d x d)= 1/4 (3.14 x 4.5 cm x 4.5 cm)

= 15.9 cm2

ESTION

culate the surface area of a circle with a diameter of 3 m.

Given Answer

cle: d = 3 m Formula: A = 1/4 (x d) = 1/4 (3.14 x d x d)

= 1/4 (3.14 x 3 m x 3 m) = 7.07 m2

.6 Metric Conversions

nits of length

basic unit of length in the metric system is the metre (m). One metre can be divided into 10 decimetres (dm), 100 centimetres (cm) or 1limetres (mm); 100 m equals to 1 hectometre (hm); while 1000 m is 1 kilometre (km).

1 m = 10 dm = 100 cm = 1000 mm0.1 m = 1 dm = 10 cm = 100 mm0.01 m = 0.1 dm = 1 cm = 10 mm0.001 m = 0.01 dm = 0.1 cm = 1 mm

1 km = 10 hm = 1000 m0.1 km = 1 hm = 100 m

0.01 km = 0.1 hm = 10 m0.001 km = 0.01 hm = 1 m

Units of surface

basic unit of area in the metric system is the square metre (m), which is obtained by multiplying a length of 1 metre by a width of 1 mee Fig. 9).

Fig. 9. A square metre




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1 m2= 100 dm2= 10 000 cm2= 1 000 000 mm2

0.01 m2= 1 dm2= 100 cm2= 10 000 mm2

0.0001 m2= 0.01 dm2= 1 cm2= 100 mm2

0.000001 m2= 0.0001 dm2= 0.01 cm2= 1 mm2

1 km2= 100 ha2= 1 000 000 m2

0.01 km2= 1 ha2= 10 000 m2

0.000001 km2= 0.0001 ha2= 1 m2

TE:

a =100 m x 100 m = 10 000 m2

2 Surface areas of canal cross-sections and farms

1.2.1 Determination of the surface areas of canal cross-sections

1.2.2 Determination of the surface area of a farm

s Section explains how to apply the surface area formulas to two common practical problems that will often be met in the field.

2.1 Determination of the surface areas of canal cross-sections

most common shape of a canal cross-section is a trapezium or, more truly, an "up-side-down" trapezium (see Fig. 10).

. 10. Canal cross-section

area (A B C D), hatched on the above drawing, is called the canal cross-section and has a trapezium shape (compare with trapezium nous, the formula to calculate its surface is similar to the formula used to calculate the surface area of a trapezium (formula 4):

rface area of the canal cross-section = 0.5 (base + top line) x canal depth = 0.5 (b + a) x h ..... (6)

ereby:

base (b) = bottom width of the canal

top line (a) = top width of the canal

canal depth (h) = height of the canal (from the bottom of the canal to the top of the embankment)

pose that the canal contains water, as shown in Fig. 11.

. 11. Wetted cross-section of a canal




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area (A B C D), hatched on the above drawing, is called the wetted canal cross-section or wetted cross-section. It also has a trapezium sthe formula to calculate its surface area is:

rface area of the wetted canal cross-section = 0.5 (base + top line) x water depth = 0.5 (b + a 1) x h1..... (7)

ereby:

base (b) = bottom width of the canal

top line (a1) = top width of the water level

water depth (h1) = the height or depth of the water in the canal (from the bottom of the canal to the water level).

AMPLE

culate the surface area of the cross-section and the wetted cross-section, of the canal shown in Fig. 12 below.

. 12. Dimensions of the cross-section

Given Answer

nal cross-section:

e (b) =1.25 mline (a) =3.75 mal depth (h) = 1.25 m

Formula: A = 0.5 x (b + a) x h= 0.5 x (1.25 m + 3.75 m) x 1.25 m

= 3.125 m2

nal wetted cross-section:

e (b) = 1.25 mline (a1) = 3.25 m

ter depth (h1) =1.00 m

Formula: A = 0.5 x (b + a1) x h

= 0.5 x (1.25 m + 3.25 m) x 1.00 m

= 2.25 m2

2.2 Determination of the surface area of a farm

may be necessary to determine the surface area of a farmer's field. For example, when calculating how much irrigation water should be girtain field, the size of the field must be known.

en the shape of the field is regular and has, for example, a rectangular shape, it should not be too difficult to calculate the surface area ongth of the field (that is the base of its regular shape) and the width of the field have been measured (see Fig. 13).

Fig. 13. Field of regular shape

AMPLE

Given Answer

ngth of the field =50 mdth of the field = 30 m

Formula: A = length x width (formula 2)

= 50 m x 30 m = 1500 m2

ESTION

at is the area of the same field, expressed in hectares?




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SWER

tion 1.1.2 explained that a hectare is equal to 10 000 m. Thus, the formula to calculate a surface area in hectares is:

..... (8)

his case: area of the field in

re often, however, the field shape is not regular, as shown in Fig. 14a.

Fig. 14a. Field of irregular shape

his case, the field should be divided in several regular areas (square, rectangle, triangle, etc.), as has been done in Fig. 14b.

Fig. 14b. Division of irregular field into regular areas

face area of the square: As= length x width = 30 m x 30 m = 900 m2

face area of the rectangle: Ar= length x width = 50 m x 15 m = 750 m2

face area of the triangle: At= 0.5 x base x height = 0.5 x 20 m x 30 m = 300 m2

al surface area of the field: A = As+ Ar+ At= 900 m2+ 750 m2+ 300 m2= 1950 m2




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3 Introduction to volume

1.3.1 Units of volume

1.3.2 Volume of water on a field

olume (V) is the content of a body or object. Take for example a block (Fig 15). A block has a certain length (l), width (w) and height (hh these three data, the volume of the block can be calculated using the formula:

block) = length x width x height = l x w x h ..... (9)

Fig. 15. A block

AMPLE

culate the volume of the above block.

Given Answergth = 4 cmdth = 3 cmght = 2 cm

Formula: V = length x width x height= 4 cm x 3 cm x 2 cm

= 24 cm3

volume of this block is expressed in cubic centimetres (written as cm). Volumes can also be expressed in cubic decimetres (dm 3), cubi

res (m3), etc.

ESTION

culate the volume in m3of a block with a length of 4 m, a width of 50 cm and a height of 200 mm.

Given Answer

data must be converted in metres (m)

gth = 4 mdth = 50 cm = 0.50 mght = 200 mm = 0.20 m

Formula: V = length x width x height= 4 m x 0.50 m x 0.20 m

= 0.40 m3

ESTION

culate the volume of the same block, this time in cubic centimetres (cm 3)

Given Answer

data must be converted in centimetres (cm)

gth = 4 m = 400 cmdth = 50 cmght = 200 mm = 20 cm

Formula: V = length x width x height= 400 cm x 50 cm x 20 cm

= 400 000 cm3




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course, the result is the same: 0.4 m3= 400 000 cm3

3.1 Units of volume

basic unit of volume in the metric system is the cubic metre (m 3) which is obtained by multiplying a length of 1 metre, by a width of 1a height of 1 metre (see Fig. 16).

Fig. 16. One cubic metre

1 m3= 1.000 dm3= 1 000 000 cm3= 1 000 000 000 mm3

0.001 m3= 1 dm3= 1 000 cm3= 1 000 000 mm3

0.000001 m3= 0.001 dm3= 1 cm3= 1 000 mm3

0.000000001 m3= 0.000001 dm3= 0.001 cm3= 1 mm3

TE

m3= 1 litre

m3= 1000 litres

3.2 Volume of water on a field

pose a one-litre bottle is filled with water. The volume of the water is thus 1 litre or 1 dm3. When the bottle of water is emptied on a taber will spread out over the table and form a thin water layer. The amount of water on the table is the same as the amount of water that wbottle; being 1 litre.

volume of water remains the same; only the shape of the "water body" changes (see Fig. 17).

Fig. 17. One litre of water spread over a table




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imilar process happens if you spread irrigation water from a storage reservoir over a farmer's field.

ESTION

pose there is a reservoir, filled with water, with a length of 5 m, a width of 10 m and a depth of 2 m. All the water from the reservoir is sr a field of 1 hectare. Calculate the water depth (which is the thickness of the water layer) on the field, see Fig. 18.

. 18. A volume of 100 m3of water spread over an area of one hectare

formula to use is:

..... (10)

the first step, the volume of water must be calculated. It is the volume of the filled reservoir, calculated with formula (9):

Volume (V) = length x width x height = 5 m x 10 m x 2 m = 100 m 3

the second step, the thickness of the water layer is calculated using formula (10):

Given Answer

face of the field = 10 000 m2

lume of water = 100 m3 Formula:

d = 0.01 md = 10 mm

ESTION

water layer 1 mm thick is spread over a field of 1 ha. Calculate the volume of the water (in m 3), with the help of Fig. 19.

Fig. 19. One millimetre water depth on a field of one hectare




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formula to use is:

lume of water (V) = Surface of the field (A) x Water depth (d) ..... (11)

Given Answer

face of the field = 10 000 m2

ter depth = 1 mm =1/1 000 = 0.001 m

Formula: Volume (m) = surface of the field (m) x water depth (m)

V = 10 000 m2x 0.001 m

V = 10 m3or 10 000 litres

4 Introduction to flow-rate

1.4.1 Definition

1.4.2 Calculation and Units

4.1 Definition

flow-rate of a river, or of a canal, is the volume of water discharged through this river, or this canal, during a given period of time. Rela

gation, the volume of water is usually expressed in litres (l) or cubic metres (m3) and the time in seconds (s) or hours (h). The flow-rate ed discharge-rate.

4.2 Calculation and Unitswater running out of a tap fills a one litre bottle in one second. Thus the flow rate (Q) is one litre per second (1 l/s) (see Fig. 20).

Fig. 20. A flow-rate of one litre per second




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ESTION

water supplied by a pump fills a drum of 200 litres in 20 seconds. What is the flow rate of this pump?

formula used is:

..... (12a)

Given Answer

lume of water: 200 lme: 20 s Formula:

unit "litre per second" is commonly used for small flows, e.g. a tap or a small ditch. For larger flows, e.g. a river or a main canal, the unbic metre per second" (m3/s) is more conveniently used.

ESTION

ver discharges 100 m3of water to the sea every 2 seconds. What is the flow-rate of this river expressed in m 3/s?

formula used is:

..... (12b)

Given Answer

lume of water: 100 m3

me: 2 s Formula:

discharge rate of a pump is often expressed in m3per hour (m3/h) or in litres per minute (l/min).

..... (12c)

..... (12d)

TE: Formula 12a, 12b, 12c and 12d are the same; only the units change




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5 Introduction to percentage and per mil

1.5.1 Percentage

1.5.2 Per mil

elation to agriculture, the words percentage and per mil will be met regularly. For instance "60 percent of the total area is irrigated durinseason". In this Section the meaning of the words "percentage" and "per mil" will be discussed.

5.1 Percentage

word "percentage" means literally "per hundred"; in other words one percent is the one hundredth part of the total. You can either writecent, or %, or 1/100, or 0.01.

me examples are:

5 percent = 5% =5/100 = 0.0520 percent = 20% = 20/100= 0.2025 percent = 25% = 25/100 = 0.2550 percent = 50% = 50/100 =0.50100 percent = 100% = 100/100 = 1150 percent = 150% = 150/100 = 1.5

ESTIONw many oranges are 1% of a total of 300 oranges? (see Fig. 21)

Fig. 21. Three oranges are 1% of 300 oranges

SWER

of 300 oranges = 1/100 x 300 = 3 oranges

QUESTIONS ANSWERS

of 100 cows 6/100 x 100 = 6 cows

% of 28 hectares 15/100 x 28 = 4.2 ha

% of 90 irrigation projects 80/100 x 90 = 72 projects

0% of a monthly salary of $100 150/100 x 100 = 1.5 x 100 = $150% of 194.5 litres 0.5/100 x 194.5 = 0.005 x 194.5 = 0.9725 litres

5.2 Per mil

word "per mil" means literally "per thousand"; in other words one per mil is one thousandth part of the total.

u can either write: per mil, or , or 1/1000, or 0.001.

me examples are:

5 per mil = 5 =5/1000= 0.00550 per mil = 50 = 50/1000 = 0.05095 per mil = 95 = 95/1000 = 0.095




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ESTION

w many oranges are 4 of 1000 oranges? (see Fig. 22)

Fig. 22. Four oranges are 4 of 1000 oranges

SWER

of 1000 oranges = 4/1000 x 1000 = 4 oranges

TE

= 1%

ause 10 = 10/1000 = 1/10 = 1%

QUESTIONS ANSWERS

of 3 000 oranges 3/1000 x 3 000 = 9 oranges

of 10 000 ha 35/1000 x 10 000 = 350 ha

of 750 km2 0.5/1000 x 750 =0.375 km2

6 Introduction to graphs

1.6.1 Example 1

1.6.2 Example 2

raph is a drawing in which the relationship between two (or more) items of information (e.g. time and plant growth) is shown in a symby.

this end, two lines are drawn at a right angle. The horizontal one is called the x axis and the vertical one is called the y axis.

ere the x axis and the y axis intersect is the "0" (zero) point (see Fig. 23).

plotting of the information on the graph is discussed in the following examples.

Fig. 23. A graph




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6.1 Example 1

pose it is necessary to make a graph of the growth rate of a maize plant. Each week the height of the plant is measured. One week afternting the seed, the plant measures 2 cm in height, two weeks after planting it measures 5 cm and 3 weeks after planting the height is 10 cstrated in Fig. 24a.

. 24a. Measuring the growth rate of a maize plant

se results can be plotted on a graph. The time (in weeks) will be indicated on the x axis; 2 cm on the axis represents 1 week. The plant hcentimetres) will be indicated on the y axis; 1 cm on the axis represents 1 cm of plant height.

er 1 week the height is 2 cm; this is indicated on the graph with A; after 2 weeks the height is 5 cm, see B, and after 3 weeks the height isee C, as shown in Fig. 24b.

planting (Time = 0) the height was zero, see D.

w connect the crosses (see Fig. 24c) with a straight line. The line indicates the growth rate of the plant; this is the height increase over tim

Fig. 24b. Growth rate of a maize plant




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an be seen from the graph that the plant is growing faster and faster (during the first week 2 cm and during the third week 5 cm); the lineo C is steeper than the line from D to A.

m the graph can be read what the height of the plant was after, say 2 1/2 weeks; see the dotted line (Fig. 24c). Locate on the horizontal aweeks and follow the dotted line upwards until the dotted line crosses the graph. From this crossing follow the dotted line to the left untical axis is reached. Now take the reading: 7.5 cm, which means that the plant had a height of 7.5 cm after 2 1/2 weeks. This height has n measured in reality, but with the graph the height can be determined anyway.

ESTION

at was the height of the plant after 1 1/2 weeks?

SWER

height of the plant after 1 1/2 weeks was 3.5 cm (see Fig. 24c).




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Fig. 24c. Graph of the growth rate of a maize plant

6.2 Example 2

other example to illustrate how a graph should be made is the variation of the temperature over one full day (24 hours). Suppose the outsperature (always in the shade) is measured, with a thermometer, every two hours, starting at midnight and ending the following midnigh

pose the following results are found:

me (hr) Temperature (C)

0 16

2 13

4 6

6 8




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8 13

10 19

12 24

14 28

16 2

18 27

20 22

22 19

24 16

the x axis indicate the time in hours, whereby 1 cm on the graph is 2 hours. On the y axis indicate the temperature in degrees Celsius (Cereby 1 cm on the graph is 5C.

w indicate (with crosses) the values from the table (above) on the graph paper and connect the crosses with straight dotted lines (see Fig

. 25a. Graph showing temperature over 24 hours; mistake 16 hour reading

his stage, if you look attentively at the graph, you will note that there is a very abrupt change in its shape around the sixteenth hour. Theside temperature seems to have fallen from 28C to 2C in two hours time! That does not make sense, and the reading of the thermometesixteenth hour must have been wrong. This cross cannot be taken in consideration for the graph and should be rejected. The only dottedcan accept is the straight one in between the reading at the fourteenth hour and the reading at the eighteenth hour (see Fig. 25b).

. 25b. Graph showing temperature over 24 hours; estimated correction of mistake

eality the temperature will change more gradually than indicated by the dotted line; that is why a smooth curve is made (continuous lineooth curve represents the most realistic approximation of the temperature over 24 hours (see Fig. 25c).

. 25c. Graph showing temperature over 24 hours; smooth curve

m the graph it can be seen that the minimum or lowest temperature was reached around 4 o'clock in the morning and was about 6C. Thhest temperature was reached at 4 o'clock in the afternoon and was approximately 29C.

ESTION

at was the temperature at 7, 15 and 23 hours? (Always use the smooth curve to take the readings).

SWER (see Fig. 25c)

mperature at 7 hours: 10C

mperature at 15 hours: 29Cmperature at 23 hours: 17C

7 Test your knowledge

1.7.1 Questions

1.7.2 Answers

7.1 Questions

Calculate the surface areas of the following triangles:

a. height = 6 cm, base = 12 cm = = = A = .....cm 2

b. height = 22 cm, base = 48 cm = = = A = .....cm 2

c. height = 16 cm, base = 24 cm = = = A = .....cm 2

d. height = 0.8 m, base = 0.3 m = = = A = .....m 2

Calculate the surface areas of the following trapeziums:

a. height = 12 cm, base = 52 cm, top = 16 cm = = = A = .....cm 2

b. height = 20 cm, base = 108 dm, top = 16 cm = = = A = .....cm 2

c. height = 0.3 m, base = 1.8 m, top = 1.5 m = = = A = .....m 2

Calculate the cross-section of the canal when given:

height = 1 m




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top width = 2.6 mbottom width = 1.2 m

Calculate the wetted cross-section when in addition to 3) is given that the water height is 0.8 m and the top width of the water surface is 2

A rectangular field has a length of 120 m and a width of 85 m. What is the area of the field in hectares?

a. 25% of 1820 metres = .....metresb. 13% of 971 cm = .....cmc. 83% of 8000 apples = .....apples

d. 7 of 18 060 metres = .....metrese. 13% of 26 hectares = .....hectaresf. 1.5 of 28 000 metres = .....metres

Calculate the volume of the following blocks, when given:

a. length = 75 cm, width = 3 m, height = 6 cm = = = V = .....m 3

b. length = 0.5 cm, width = 1 dm, height = 20 cm = = = V = .....m 3

c. length = 15 cm, width = 2 dm, height = 0.5 m = = = V = .....litres

Calculate the volume of water (in m3) on a field, when given: the length = 150 m, the width = 56 m and the water layer = 70 mm.

Calculate the minimum depth of a reservoir, which has: a length of 15 m and a width of 10 m and which has to provide 50 mm water for 75 m long and 95 m wide.

Make a graph of the monthly rainfall over a period of 1 year, when given:

onth Rain (mm/month)

an. 42

eb. 65

Mar. 140

pr. 120

May 76

une 24

uly 6

ug. 0

ept. 0

ct. 10

ov. 17

ec. 27

7.2 Answers

a. A = 0.5 x b x h = 0.5 x 6 cm x 12 cm = 36 cm 2

b. A = 0.5 x 22 cm x 48 cm = 528 cm2

c. A = 0.5 x 16 cm x 24 cm = 192 cm2

d. A = 0.5 x 0.8 m x 0.3 m = 0.12 m2

a. A = 0.5 x (b + a) x h = 0.5 x (52 cm + 16 cm) x 12 cm = 408 cm 2

b. A = 0.5 x (108 cm + 16 cm) x 20 cm = 1240 cm2

c. A = 0.5 x (1.8 m + 1.5 m) x 0.3 m = 0.495 m2

A = 0.5 x (b + a) x h = 0.5 x (1.2 m + 2.6 m) x 1 m = 1.9 m 2

A = 0.5 x (b + a1) x h1- 0.5 x (1.2 m + 2.32 m) x 0.8 m = 1.408 m2

Area of the field in square metres = l (m) x w (m) = 120 m x 85 m = 10 200 m 2




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a. 1820 m x 25/100 = 455 mb. 971 cm x 13/100 = 126.23 cmc. 8000 apples x 83/100 = 6640 applesd. 18 060 m x 7/1000 = 126.42 me. 26 ha x 13/100= 3.38 haf. 28 000 m x 1.5/1000 = 42 m

a. V = l x w x h = 0.75 m x 3 m x 0.06 m = 0.135 m 3

b. V = 0.005 m x 0.10 m x 0.20 m = 0.0001 m3

c. V = 1.5 dm x 2 dm x 5 dm = 15 dm3= 15 litres

V = l x w x h = 150 m x 56 m x 0.070 m = 588 m 3

Volume of water required on a field: V = length of field (m) x width of field (m) x thickness of water layer (m) = 175 m x 95 m x

0.050 m = 831.25 m3

Volume of the reservoir: V = 831.25 m3= length of reservoir (m) x width of reservoir (m) x depth of reservoir (m)

aph of monthly rainfall over a period of one year


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CHAPTER 2 - SOIL AND WATER

2.1 The soil2.2 Entry of water into the soil

2.3 Soil moisture conditions

2.4 Available water content

2.5 Groundwater table

2.6 Soil erosion by water

.1 The soil

2.1.1 Soil composition

2.1.2 Soil profile

2.1.3 Soil texture

2.1.4 Soil structure

1.1 Soil composition

hen dry soil is crushed in the hand, it can be seen that it is composed of all kinds of particles of different s

ost of these particles originate from the degradation of rocks; they are called mineral particles. Some origiom residues of plants or animals (rotting leaves, pieces of bone, etc.), these are called organic particles (organic matter). The soil particles seem to touch each other, but in reality have spaces in between. These spae called pores. When the soil is "dry", the pores are mainly filled with air. After irrigation or rainfall, the pe mainly filled with water. Living material is found in the soil. It can be live roots as well as beetles, wormvae etc. They help to aerate the soil and thus create favourable growing conditions for the plant roots (Fig).

Fig. 26. The composition of the soil


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1.2 Soil profile

a pit is dug in the soil, at least 1 m deep, various layers, different in colour and composition can be seen. Tyers are called horizons. This succession of horizons is called the profile of the soil (Fig. 27).

Fig. 27. The soil profile




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very general and simplified soil profile can be described as follows:

a. The plough layer (20 to 30 cm thick): is rich in organic matter and contains many live roots. Thislayer is subject to land preparation (e.g. ploughing, harrowing etc.) and often has a dark colour(brown to black).

b. The deep plough layer: contains much less organic matter and live roots. This layer is hardlyaffected by normal land preparation activities. The colour is lighter, often grey, and sometimesmottled with yellowish or reddish spots.

c. The subsoil layer: hardly any organic matter or live roots are to be found. This layer is not veryimportant for plant growth as only a few roots will reach it.

d. The parent rock layer: consists of rock, from the degradation of which the soil was formed. Thisrock is sometimes called parent material.

he depth of the different layers varies widely: some layers may be missing altogether.




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1.3 Soil texture

he mineral particles of the soil differ widely in size and can be classified as follows:

ame of the particles Size limits in mm Distinguisable with naked eye

ravel larger than 1 obviously

and 1 to 0.5 easily

lt 0.5 to 0.002 barely

ay less than 0.002 impossible

he amount of sand, silt and clay present in the soil determines the soil texture.

In coarse textured soils: sand is predominant (sandy soils).In medium textured soils: silt is predominant (loamy soils).In fine textured soils: clay is predominant (clayey soils).

the field, soil texture can be determined by rubbing the soil between the fingers (see Fig. 28).

rmers often talk of light soil and heavy soil. A coarse-textured soil is light because it is easy to work, whilne-textured soil is heavy because it is hard to work.

xpression used by the farmer Expression used in literature

ght sandy coarse

edium loamy medium

eavy clayey fine

he texture of a soil is permanent, the farmer is unable to modify or change it.

g. 28a. Coarse textured soil is gritty. Individual particules are loose and fall apart in the hand, even

hen moist.

g. 28b. Medium textured soil feels very soft (like flour) when dry. It can be easily be pressed when w

d then feels silky.

g. 28c. Fine textured soil sticks to the fingers when wet and can form a ball when pressed.

1.4 Soil structure

il structure refers to the grouping of soil particles (sand, silt, clay, organic matter and fertilizers) into porompounds. These are called aggregates. Soil structure also refers to the arrangement of these aggregatesparated by pores and cracks (Fig. 29).

he basic types of aggregate arrangements are shown in Fig. 30, granular, blocky, prismatic, and massiveucture.

Fig. 29. The soil structure




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hen present in the topsoil, a massive structure blocks the entrance of water; seed germination is difficult dor aeration. On the other hand, if the topsoil is granular, the water enters easily and the seed germination itter.

a prismatic structure, movement of the water in the soil is predominantly vertical and therefore the supplyater to the plant roots is usually poor.

nlike texture, soil structure is not permanent. By means of cultivation practices (ploughing, ridging, etc.), trmer tries to obtain a granular topsoil structure for his fields.

Fig. 30. Some examples of soil structures

GRANULAR BLOCKY




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PRISMATIC MASSIVE




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.2 Entry of water into the soil

2.2.1 The infiltration process

2.2.2 Infiltration rate

2.2.3 Factors influencing the infiltration rate

2.1 The infiltration process

hen rain or irrigation water is supplied to a field, it seeps into the soil. This process is called infiltration.

filtration can be visualized by pouring water into a glass filled with dry powdered soil, slightly tamped. Thater seeps into the soil; the colour of the soil becomes darker as it is wetted (see Fig. 31).

Fig. 31. Infiltration of water into the soil


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2.2 Infiltration rate

epeat the previous test, this time with two glasses. One is filled with dry sand and the other is filled with dray (see Fig. 32a and b).

he infiltration of water into the sand is faster than into the clay. The sand is said to have a higher infiltratioe.

Fig. 32a. The same amount of water is supplied to each glass

g. 32b. After one hour the water has infiltrated in the sand, while some water is still ponding on the




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he infiltration rate of a soil is the velocity at which water can seep into it. It is commonly measured by thepth (in mm) of the water layer that the soil can absorb in an hour.

n infiltration rate of 15 mm/hour means that a water layer of 15 mm on the surface of the soil, will take onur to infiltrate (see fig. 33).

Fig. 33. Soil with an infiltration rate of 15 mm/hour

range of values for infiltration rates is given below:

ow infiltration rate less than 15 mm/hour

edium infiltration rate 15 to 50 mm/hourgh infiltration rate more than 50 mm/hour

2.3 Factors influencing the infiltration rate

he infiltration rate of a soil depends on factors that are constant, such as the soil texture. It also depends onctors that vary, such as the soil moisture content.

i. Soil Texture

Coarse textured soils have mainly large particles in between which there are large pores.

On the other hand, fine textured soils have mainly small particles in between which there are smallpores (see Fig. 34).

Fig. 34. Infiltration rate and soil texture




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In coarse soils, the rain or irrigation water enters and moves more easily into larger pores; it takesless time for the water to infiltrate into the soil. In other words, infiltration rate is higher for coarsetextured soils than for fine textured soils.

ii. The soil moisture content

The water infiltrates faster (higher infiltration rate) when the soil is dry, than when it is wet (see Fig.35). As a consequence, when irrigation water is applied to a field, the water at first infiltrates easily,but as the soil becomes wet, the infiltration rate decreases.

Fig. 35. Infiltration rate and soil moisture content




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iii. The soil structure

Generally speaking, water infiltrates quickly (high infiltration rate) into granular soils but veryslowly (low infiltration rate) into massive and compact soils.

Because the farmer can influence the soil structure (by means of cultural practices), he can alsochange the infiltration rate of his soil.




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.3 Soil moisture conditions

2.3.1 Soil moisture content

2.3.2 Saturation

2.3.3 Field capacity

2.3.4 Permanent wilting point

3.1 Soil moisture content

he soil moisture content indicates the amount of water present in the soil.

is commonly expressed as the amount of water (in mm of water depth) present in a depth of one metre of r example: when an amount of water (in mm of water depth) of 150 mm is present in a depth of one metreil, the soil moisture content is 150 mm/m (see Fig. 36).

Fig. 36. A soil moisture content of 150 mm/m

he soil moisture content can also be expressed in percent of volume. In the example above, 1 m3of soil (e.

th a depth of 1 m, and a surface area of 1 m2) contains 0.150 m3of water (e.g. with a depth of 150 mm = 0

and a surface area of 1 m2). This results in a soil moisture content in volume percent of:

hus, a moisture content of 100 mm/m corresponds to a moisture content of 10 volume percent.


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ote: The amount of water stored in the soil is not constant with time, but may vary.

3.2 Saturation

uring a rain shower or irrigation application, the soil pores will fill with water. If all soil pores are filled water the soil is said to be saturated. There is no air left in the soil (see Fig. 37a). It is easy to determine in thld if a soil is saturated. If a handful of saturated soil is squeezed, some (muddy) water will run between th

ngers.

ants need air and water in the soil. At saturation, no air is present and the plant will suffer. Many crops canthstand saturated soil conditions for a period of more than 2-5 days. Rice is one of the exceptions to this r

he period of saturation of the topsoil usually does not last long. After the rain or the irrigation has stopped,the water present in the larger pores will move downward. This process is called drainage or percolation.

he water drained from the pores is replaced by air. In coarse textured (sandy) soils, drainage is completedthin a period of a few hours. In fine textured (clayey) soils, drainage may take some (2-3) days.

3.3 Field capacity

fter the drainage has stopped, the large soil pores are filled with both air and water while the smaller poresll full of water. At this stage, the soil is said to be at field capacity. At field capacity, the water and air conthe soil are considered to be ideal for crop growth (see Fig. 37b).

3.4 Permanent wilting point

ttle by little, the water stored in the soil is taken up by the plant roots or evaporated from the topsoil into thmosphere. If no additional water is supplied to the soil, it gradually dries out.

he dryer the soil becomes, the more tightly the remaining water is retained and the more difficult it is for thant roots to extract it. At a certain stage, the uptake of water is not sufficient to meet the plant's needs. The

ant looses freshness and wilts; the leaves change colour from green to yellow. Finally the plant dies.

he soil water content at the stage where the plant dies, is called permanent wilting point. The soil still contme water, but it is too difficult for the roots to suck it from the soil (see Fig. 37c).

Fig. 37. Some soil moisture characteristics




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.4 Available water content

he soil can be compared to a water reservoir for the plants. When the soil is saturated, the reservoir is full.owever, some water drains rapidly below the rootzone before the plant can use it (see Fig. 38a).

Fig. 38a. Saturation




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hen this water has drained away, the soil is at field capacity. The plant roots draw water from what remaine reservoir (see Fig. 38b).

Fig. 38b. Field capacity




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hen the soil reaches permanent wilting point, the remaining water is no longer available to the plant (see Fc).

Fig. 38c. Permanent wilting point




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he amount of water actually available to the plant is the amount of water stored in the soil at field capacitynus the water that will remain in the soil at permanent wilting point. This is illustrated in Fig. 39.

Fig. 39. The available soil moisture or water content

vailable water content = water content at field capacity - water content at permanent wilting point ..... (13

he available water content depends greatly on the soil texture and structure. A range of values for differentpes of soil is given in the following table.

oil Available water content in mm water depth per m soil depth (mm/m)

and 25 to 100

am 100 to 175

ay 175 to 250

he field capacity, permanent wilting point (PWP) and available water content are called the soil moisturearacteristics. They are constant for a given soil, but vary widely from one type of soil to another.

.5 Groundwater table

2.5.1 Depth of the groundwater table

2.5.2 Perched groundwater table

2.5.3 Capillary rise

rt of the water applied to the soil surface drains below the rootzone and feeds deeper soil layers which arermanently saturated; the top of the saturated layer is called groundwater table or sometimes just water tabee Fig. 40).

Fig. 40. The groundwater table


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5.1 Depth of the groundwater table

he depth of the groundwater table varies greatly from place to place, mainly due to changes in topography e area (see Fig. 41).

Fig. 41. Variations in depth of the groundwater table




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one particular place or field, the depth of the groundwater table may vary in time.

llowing heavy rainfall or irrigation, the groundwater table rises. It may even reach and saturate the rootzoolonged, this situation can be disastrous for crops which cannot resist "wet feet" for a long period. Where oundwater table appears at the surface, it is called an open groundwater table. This is the case in swampyeas.

he groundwater table can also be very deep and distant from the rootzone, for example following a prolongy period. To keep the rootzone moist, irrigation is then necessary.

5.2 Perched groundwater table

perched groundwater layer can be found on top of an impermeable layer rather close to the surface (20 to m). It covers usually a limited area. The top of the perched water layer is called the perched groundwater ta

he impermeable layer separates the perched groundwater layer from the more deeply located groundwater ee Fig. 42).

Fig. 42. A perched groundwater table




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il with an impermeable layer not far below the rootzone should be irrigated with precaution, because in thse of over irrigation (too much irrigation), the perched water table may rise rapidly.

5.3 Capillary rise

far, it has been explained that water can move downward, as well as horizontally (or laterally). In additioater can move upward.

a piece of tissue is dipped in water (Fig. 43), the water is sucked upward by the tissue.

g. 43. Upward movement of water or capillary rise

he same process happens with a groundwater table and the soil above it. The groundwater can be suckedward by the soil through very small pores that are called capillars. This process is called capillary rise.

fine textured soil (clay), the upward movement of water is slow but covers a long distance. On the other hcoarse textured soil (sand), the upward movement of the water is quick but covers only a short distance.

oil texture Capillary rise (in cm)

oarse (sand) 20 to 50 cm

edium 50 to 80 cm

ne (clay) more than 80 cm up to several metres




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.6 Soil erosion by water

2.6.1 Sheet erosion

2.6.2 Gully erosion

osion is the transport of soil from one place to another. Climatic factors such as wind and rain can cause

osion, but also under irrigation it may occur.

ver a short period, the process of erosion is almost invisible. However, it can be continuous and the wholertile top layer of a field may disappear within a few years.

il erosion by water depends on:

- the slope: steep, sloping fields are more exposed to erosion;- the soil structure: light soils are more sensitive to erosion;- the volume or rate of flow of surface runoff water: larger or rapid flows induce more erosion.

osion is usually heaviest during the early part of irrigation, especially when irrigating on slopes. The dry

rface soil, sometimes loosened by cultivation, is easily removed by flowing water. After the first irrigationil is moist and settles down, so erosion is reduced. Newly irrigated areas are more sensitive to erosion,pecially in their early stages.

here are two main types of erosion caused by water: sheet erosion and gully erosion. They are often combi

6.1 Sheet erosion

eet erosion is the even removal of a very thin layer or "sheet" of topsoil from sloping land. It occurs over eas of land and causes most of the soil losses (see Fig. 44).

g. 44. Sheet erosion

he signs of sheet erosion are:

- only a thin layer of topsoil; or the subsoil is partly exposed; sometimes even parent rock isexposed;

- quite large amounts of coarse sand, gravel and pebbles in the arable layer, the finer material hasbeen removed;

- exposure of the roots;

- deposit of eroded material at the foot of the slope.

6.2 Gully erosion

ully erosion is defined as the removal of soil by a concentrated water flow, large enough to form channels llies.

hese gullies carry water during heavy rain or irrigation and gradually become wider and deeper (see Fig. 4

g. 45. Gully erosion


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he signs of gully erosion on an irrigated field are:

- irregular changes in the shape and length of the furrows;- accumulation of eroded material at the bottom of the furrows;- exposure of plant roots.


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CHAPTER 3 - ELEMENTS OFTOPOGRAPHY

3.1 Slopes

3.2 Elevation of a point

3.3 Contour lines

3.4 Maps

.1 Slopes

3.1.1 Definition

3.1.2 Method of expressing slopes

3.1.3 Cross slopes

.1.1 Definition

slope is the rise or fall of the land surface. It is important for the farmer or irrigator to identify the sln the land.

slope is easy to recognize in a hilly area. Start climbing from the foot of a hill toward the top, this islled a rising slope (see Fig. 46, black arrow). Go downhill, this is a falling slope (see Fig. 46, whiterow).

Fig. 46. A rising and a falling slope

CHAPTER 3 - ELEMENTS OF TOPOGRAPHY

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at areas are never strictly horizontal; there are gentle slopes in a seemingly flat area, but they are oftrdly noticeable to the naked eye. An accurate survey of the land is necessary to identify these so callat slopes".

.1.2 Method of expressing slopeshe slope of a field is expressed as a ratio. It is the vertical distance, or difference in height, between tints in a field, divided by the horizontal distance between these two points. The formula is:

..... (14a)

n example is given in Fig. 47.

g. 47. The dimensions of a slope

he slope can also be expressed in percent; the formula used is then:

..... (14b)

sing the same measurements shown in Fig. 47:

nally, the slope can be expressed in per mil; the formula used is then:




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..... (14c)

ith the figures from the same example:

OTE:

lope in = slope in % x 10

UESTION

hat is the slope in percent and in per mil of a field with a horizontal length of 200 m and a heightfference of 1.5 m between the top and the bottom?

NSWER

eld slope in = field slope in % x 10 = 0.75 x 10 = 7.5

UESTION

hat is the difference in height between the top and the bottom of a field when the horizontal length oe field is 300 m and the slope is 2.

NSWER

us: height difference (m) = 0.002 x 300 m = 0.6 m.

he following table shows a range of slopes commonly referred to in irrigated fields.

Slope % Horizontal 0 - 0.2 0 - 2

Very flat 0.2 - 0.5 2 - 5

lat 0.5 - 1 5 - 10

Moderate 1 - 2.5 10 - 25

teep more than 2.5 more than 25

g. 48a. A steep slope




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g. 48b. A flat slope

.1.3 Cross slopes

ace a book on a table and lift one side of it 4 centimetres from the table (Fig. 49a). Now, tilt the boodeways (6 cm) so that only one corner of it touches the table (Fig. 49b).

Fig. 49a. Main slope

Fig. 49b. Main slope and cross slope




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he thick arrow indicates the direction of what can be called the main slope; the thin arrow indicates trection of the cross slope, the latter crosses the direction of the main slope.

n illustration of the main slope and the cross slope of an irrigated field is shown in Fig. 50.

Fig. 50. The main slope and cross slope of an irrigated field




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.2 Elevation of a point

3.2.1 Definition3.2.2 Bench mark and mean sea level

.2.1 Definition

figure 51, point A is at the top of a concrete bridge. Any other point in the surrounding area is highewer than A, and the vertical distance between the two can be determined. For example, B is higher t and the vertical distance between A and B is 2 m. Point C, is lower than A and the vertical distancetween A and C is 1 m. If point A is chosen as a reference point or datum, the elevation of any other

int in the field can be defined as the vertical distance between this point and A.

Fig. 51. Reference point or datum "A"


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hus, the height or elevation of B, in relation to the datum A, is 2 m and the elevation of C, also relate

e datum A, is 1 m.s a reminder that a point is above or below the datum, its elevation is prefixed by the sign + (plus) ifove the datum, or - (minus) if it is below the datum.

herefore, in relation to the datum A, the elevation of B is +2 m and the elevation of C is -1 m.

.2.2 Bench mark and mean sea level

bench mark is a permanent mark established in a field to use as a reference point. A bench mark canconcrete base in which an iron bar is fixed, indicating the exact place of the reference point.

bench mark can also be a permanent object on the farm, such as the top of a concrete structure.

most countries the topographical departments have established a national network of bench marks wficially registered elevations. All bench mark heights are given in relationship to the one national daane which in general is the mean sea level (MSL) (see Fig. 52).

Fig. 52. A bench mark (B.M.) and mean sea level (M.S.L.)




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XAMPLE

Figure 52, the elevation of point A in relation to the bench mark (BM) is 5 metres. The BM elevatiolative to the mean sea level (MSL) is 10 m. Thus, the elevation of point A relative to the MSL is 5 m

0 m = 15 m and is called the reduced level (RL) of A.

UESTION

hat is the reduced level of point B in Figure 52.

NSWER

he elevation of B relative to BM = 3 m

he elevation of BM relative to MSL = 10 m

hus, the reduced level of B = 3 m + 10 m = 13 m

UESTION

hat is the difference in elevation between A and B? What does it represent?

NSWER

he difference in elevation between A and B is the reduced level of A minus the reduced level of B = - 13 m = 2 m, which represents the vertical distance between A and B.




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.3 Contour lines

contour line is the imaginary horizontal line that connects all points in a field which have the sameevation. A contour line is imaginary but can be visualized by taking the example of a lake.

he water level of a lake may move up and down, but the water surface always remains horizontal. Thvel of the water on the shore line of the lake makes a contour line because it reaches points which arthe same elevation (Fig. 53a).

g. 53a. The shore line of the lake forms a contour line

uppose the water level of the lake rises 50 cm above its original level. The contour line, formed by thore line, changes and takes a new shape, now joining all the points 50 cm higher than the original lavel (Fig. 53b).

g. 53b. When the water level rises, a new contour line is formed

ontour lines are useful means to illustrate the topography of a field on a flat map; the height of eachntour line is indicated on the map so that the hills or depressions can be identified.

.4 Maps

3.4.1 Description of a map

3.4.2 Interpretation of contour lines on a map

3.4.3 Mistakes in the contour lines

3.4.4 Scale of a map

.4.1 Description of a map

g. 54 represents a three-dimensional view of a field with its hills, valleys and depressions; the contones have also been indicated.

g. 54. A three-dimensional view

uch a representation gives a very good idea of what the field looks like in reality. Unfortunately, it

quires much skill to draw and is almost useless for the designing of roads, irrigation and drainagefrastructures. A much more accurate and convenient representation of the field, on which all dataferring to topography can be plotted, is a map (Fig. 55). The map is what you see when looking at three-dimensional view (Fig. 54) from the top.

.4.2 Interpretation of contour lines on a map

he arrangement of the contour lines on a map gives a direct indication of the changes in the field'spography (Fig. 55).


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g. 55. A two-dimensional view or map

hilly areas, the contour lines are close together while they are wider apart on flat slopes. The closer ntour lines, the steeper the slope. The wider the contour lines, the flatter the slopes.

n a hill, the contour lines form circles; whereby the values of their elevation increase from the edge te centre.

a depression, the contour lines also form circles; the values of their elevation, however, decrease fro

e edge to the centre.

.4.3 Mistakes in the contour lines

ontour lines of different heights can never cross each other. Crossing countour lines would mean thatersection point has two different elevations, which is impossible (see Fig. 56).

g. 56. WRONG; crossing contour lines

contour line is continuous; there can never be an isolated piece of contour line somewhere on the m

shown in Figure 57.

g. 57. WRONG; an isolated piece of contour line

.4.4 Scale of a map

o be complete and really useful, a map must have a defined scale. The scale is the ratio of the distanctween two points on a map and their real distance on the field. A scale of 1 in 5000 (1:5000) means cm measured on the map corresponds to 5000 cm (or converted into metres, 50 m) on the field.

UESTIONhat is the real distance between points A and B on the field when these two points are 3.5 cm apart oap whose scale is 1 to 2 500? (see Fig. 58)

g. 58. Measuring the distance between A and B

NSWER

he scale is 1:2 500, which means that 1 cm on the map represents 2 500 cm in reality. Thus, 3.5 cmtween A and B on the map corresponds to 3.5 x 2 500 cm = 8 750 cm or 87.5 m on the field.


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CHAPTER 4 - RAINFALL ANDEVAPOTRANSPIRATION

4.1 Rainfall

4.2 Evapotranspiration

l crops need water to grow and produce yields. The most important source of water for crop growth is rainhen rainfall is insufficient, irrigation water may be supplied to guarantee a good harvest.

ne of the main problems of the irrigator is to know the amount of water that has to be applied to the field t

eet the water needs of the crops; in other words the irrigation requirement needs to be determined. Too muater means a waste of water which is so precious in arid countries. It can also lead to a rise of the groundwble and an undesirable saturation of the rootzone. Too little water during the growing season causes the plawilt. Long periods during which the water supply is insufficient, result in loss of yield or even crop failuredition, the irrigation requirement needs to be determined for proper design of the irrigation system and fortablishment of the irrigation schedules.

.1 Rainfall

4.1.1 Amount of rainfall4.1.2 Rainfall intensity

4.1.3 Rainfall Distribution

4.1.4 Effective Rainfall

he primary source of water for agricultural production, for large parts of the world, is rainfall or precipitatiainfall is characterized by its amount, intensity and distribution in time.

1.1 Amount of rainfall

magine an open square container, 1 m wide, 1 m long and 0.5 m high (see Fig. 59a).

Fig. 59a. An open container to collect rainwater

CHAPTER 4 - RAINFALL AND EVAPOTRANSPIRATION

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his container is placed horizontally on an open area in a field (see Fig. 59b).

Fig. 59b. Container placed in the field

uring a rain shower, the container collects the water.

ppose that when the rain stops, the depth of water contained in the pan is 10 mm (see Fig. 59c).

Fig. 59c. 10 mm rainwater collected in the container




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he volume of water collected in the pan is:

V (m3) = l (m) x w (m) x d (m) = 1 m x 1 m x 0.010 m = 0.01 m 3or 10 litres

can be assumed that the surrounding field has also received an uniform water depth of 10 mm (see Fig. 59

Fig. 59d. 10 mm rainfall on the field




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terms of volume, with a rainfall of 10 mm, every square metre of the field receives 0.01 m, or 10 litres, ofater. With a rainfall of 1 mm, every square metre receives 1 litre of rain water.

rainfall of 1 mm supplies 0.001 m3, or 1 litre of water to each square metre of the field. Thus 1 ha receive0 litres.

UESTION

hat is the total amount of water received by a field of 5 ha under a rainfall of 15 mm?

NSWER

ch hectare (10 000 m2) receives 10 000 m2x 0.015 m = 150 m3of water. Thus the total amount of waterceived by the 5 hectares is: 5 x 150 m = 750 m

ainfall is often expressed in millimetres per day (mm/day) which represents the total depth of rainwater (mring 24 hours. It is the sum of all the rain showers which occurred during these 24 hours.

1.2 Rainfall intensityhe rainfall intensity is the depth of water (in mm) received during a shower divided by the duration of theower (in hours). It is expressed in millimetres of water depth per hour (mm/hour).

..... (15)

r example, a rain shower lasts 3.5 hours and supplies 35 mm of water. The intensity of this shower is




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(see Fig. 60).

g. 60. Rainfall intensity

OW INTENSITY

IGH INTENSITY

ppose the same amount of water (35 mm) is supplied in one hour only, thus by a shower of higher intensi

(see Fig. 60).

though the same amount of water (35 mm) has been supplied by both showers, the high intensity shower iss profitable to the crops. The high intensity rainfall usually has big drops that fall with more force on the rface. In fine textured soil especially, the soil aggregates break down rapidly into fine particles that seal thil surface (see Fig. 61). The infiltration is then reduced, and surface runoff increases (see Section 4.1.4).

Fig. 61. Sealing of the soil surface by raindrops

he low intensity rainfall has finer drops. The soil surface is not sealed, the rainwater infiltrates more easilyrface runoff is limited (see Section 4.1.4).

1.3 Rainfall Distribution

ppose that during one month, a certain area receives a total amount of rain water of 100 mm (100 mm/mo

r crop growth, the distribution of the various showers during this month is important.

ppose that the rainwater falls during two showers of 50 mm each, one at the beginning of the month and ther one at the end of the month (see Fig. 62a). In between these two showers, the crop undergoes a long drriod and may even wilt. Irrigation during this period is then required.

g. 62a. 100 mm rainfall, poorly distributed over one month

n the other hand, if the rainwater is supplied regularly by little showers, evenly distributed over the monthg. 62b), adequate soil moisture is continuously maintained and irrigation might not be required.




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g. 62b. 100 mm rainfall, evenly distributed over one month

ot only the rainfall distribution within a month is important. It is also important to look into the rainfallstribution over the years.

ppose that in a certain area the average rainfall in May is 150 mm and that this amount is just sufficient totisfy the water need of the crops during this month. You may however find that, in this area, the rainfall inceptionally dry year is only 75 mm, while in a wet year the rainfall is 225 mm. In a dry year it would thuscessary to irrigate the crops in May, while in an average year or a wet year, irrigation is not needed.

1.4 Effective Rainfall

ntroduction

hen rain water ((1) in Fig. 63) falls on the soil surface, some of it infiltrates into the soil (2), some stagnate surface (3), while some flows over the surface as runoff (4).

hen the rainfall stops, some of the water stagnating on the surface (3) evaporates to the atmosphere (5), whe rest slowly infiltrates into the soil (6).

om all the water that infiltrates into the soil ((2) and (6)), some percolates below the rootzone (7), while thst remains stored in the rootzone (8).

Fig. 63. Effective rainfall (8) = (1) - (4) - (5) - (7)

other words, the effective rainfall (8) is the total rainfall (1) minus runoff (4) minus evaporation (5) and mep percolation (7); only the water retained in the root zone (8) can be used by the plants, and represents wlled the effective part of the rainwater. The term effective rainfall is used to define this fraction of the tota

mount of rainwater useful for meeting the water need of the crops.




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Factors influencing effective rainfall

any factors influence the amount of the effective rainfall. There are factors which the farmer cannot influeg. the climate and the soil texture) and those which the farmer can influence (e.g. the soil structure).

a. Climate

The climate determines the amount, intensity and distribution of rainfall which have direct influenceon the effective rainfall (see 4.1.3 and 4.1.4).

b. Soil textureIn coarse textured soil, water infiltrates quickly but a large part of it percolates below the rootzone.In fine textured soil, the water infiltrates slowly, but much more water is kept in the rootzone thanin coarse textured soil.

c. Soil structure

The condition of the soil structure greatly influences the infiltration rate and therefore the effectiverainfall. A favourable soil structure can be obtained by cultural practices (e.g. ploughing, mulching,ridging, etc.).

d. Depth of the rootzone

Soil water stored in deep layers can be used by the plants only when roots penetrate to that depth.The depth of root penetration is primarily dependent on the type of crop, but also on the type of soil.The thicker the rootzone, the more water available to the plant. (see Fig. 64).

Fig. 64. Effective rainfall and depth of the rootzone




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e. Topography

On steep sloping areas, because of high runoff, the water has less time to infiltrate than in rather flatareas (see Fig. 65). The effective rainfall is thus lower in sloping areas.

Fig. 65. Effective rainfall and topography




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f. Initial soil moisture content

Chapter 2.2.3 ii, explained that for a given soil, the infiltration rate is higher when the soil is drythan when it is moist. This means that for a rain shower occurring shortly after a previous shower orirrigation, the infiltration rate is lower and the surface runoff higher (see Fig. 66).

Fig. 66. Effective rainfall and initial soil moisture content




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g. Irrigation methods

There are different methods of irrigation which will be explained in Chapter V, and each methodhas a specific influence on the effective rainfall.

In basin irrigation there is no surface runoff. All the rainwater is trapped in the basin and has time to

infiltrate (Fig. 67a).

In inclined border and furrow irrigation, the runof

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