# Irrigation Efficiencies

Efficiency is the ratio of the water output to the water input, and is usually expressed as percentage.

Irrigation Efficiencies

Efficiency is the ratio of the water output to the water input, and is usually expressed as percentage. Input minus output is nothing but losses, and hence, if losses are more, output is es and, therefore, efficiency is less. Hence, efficiency is inversely proportional to the losses. Water is lost in irrigation during various processes and, therefore, there are different kinds of irrigation efficiencies, as given below:

(i)  Efficiency of waterconveyance (ηc): It is a ratio of the water delivered into the fields from the outlet point of the channel, to the water pumped into the channel at the starting point. It may be represented by ηc. It takes the conveyance or transit losses into account.

(ii)  Efficiency of waterapplication (ηa): It is the ratio of the quantity of water stored into the root zone of the crops to the quantity of water actually delivered into the field. it may be represented by ηa. It may also be termed as farm efficiency, as it takes into account the water is lost in the farm.

(iii)   Efficiency of waterstorage (ηs): It is the ratio of the water stored in the root zone during irrigation to the water needed in the root zone prior to irrigation (i.e., field capacity �'existing moisture content). It may be represented by ηs.

Efficiency of water use (ηu): It is the ratio of the water beneficially used, including leaching water, to the quantity of water delivered. It may be represented by ηu.

Example: Once cumec of water is pumped into a farm distribution system. 0.8 cumec is delivered to a turn out, 0.9 kilometres from the well. Compute the conveyance efficiency.

Solution: By definition

ηc = Output/ Input  x 100 = 0.8/1.0 . 100 = 80%

Example: 10 cumecs of water is delivered to a 32 hectare field, for 4 hours. Soil probing after the indicated that 0.3 metre of water has been stored in the root zone. Compute the water application efficiency.

Solution:

Volume of water supplied by 10 cumecs of water applied for 4 hours (10 =(4 60x 60)m3 = 1,44,000 m3

= 14.4  x104 m3 =   14.4m  x 104m2 =   14. 4ha.m.

Depth of water applied =        volume/area =      1,44, 000/32,0, 000  =  144         /320

= .45

 Input = 14.4 ha.m

Output = 32 hectares land is storing water upto 0.3 m depth,

Output = 32 x0.3 ha.m = 9.6 ha.m

Water application efficiency (ηa) = Output/ Input   x 100

96/14.4 = 67%

(v) Uniformity coefficient or Water distribution efficiency:

The effectiveness of irrigation may also be measured by its water distribution efficiency (ηd), which is defined below:

ηd = |1d(D)

Where ηd = Water distribution efficiency

D = Mean depth of water stored during irrigation.

d = Average of the absolute values of deviations from the mean.

The water distribution efficiency represents the extent to which the water has penetrated to a uniform depth, throughout the field. When the water has penetrated uniformly throughout the field, the deviation from the mean depth is zero and water distribution efficiency is 1.0.

Examples: The depths of penetrations along the length of a boarder strip at points 30 metres aprt from probed. Their observed values are 2.0, 1.9, 1.8, 1.6 and 1.5 metres. Compute the water distribution efficiency.

Solution:

The observed depths at five stations are 2.0, 1.9, 1.8, 1.6 and 1.5 metres, respectively.

Mean depth = D = 2.0+1.9+1.8+1.6+1.5 / 5 = 8.8/5 = 1.76metres

Values of deviations from the means are (2.0 �'1.76), (1.9 �'1.76), (1.8 �'1.76). (1.6 �'1.76), (1.5

�'1.76) i.e.,

0.24, 0.14, 0.04,  0.16 and �'0.26.

The absolute values of these deviations from the mean are 0.24, 0.14, 0.04, 0.16 and 0.26.

The average of these absolute values of deviations from the

Mean = d=  (0.24 + 0.14 + 0.04 + 0.16 + .26. )/5

= 0.84/5 = 0.168 metre

The water distribution efficiency  Hence, the water distribution efficiency = 0.905. Ans.

Example: A stream of 130 litres per second was diverted from a canal and 100 litres per second were delivered to the field. An area of 1.6 hectares was irrigated in 8 hours. The effective depth of root zone was 1.7 m. The runoff loss in the field was 420 cu. M. The depth of water penetration varied linearly from 1.7 m at the head end of the field to1.1 m at the tail end. Available moisture holding capacity of the soil is 20 cm per metre depth of soil. It is required to determine the water conveyance efficiency, water application efficiency, water storage efficiency, and water distribution efficiency. Irrigation was started at a moisture extraction level of 50% of the available moisture.

Solution:

(i)                Water conveyance efficiency (C)

=( Water delivered to the fields/ Water supplied into the canal at the head) x 100

= 100/130 x 100 =77%

(ii)Water application efficiency (ηa)

Water stored in the root zone during irrigation / Water delivered to the field  x 100

Water supplied to field during 8 hours @ 100 litres per second

= 100x8  x60 x 60 litres = 2880 cu. m.

Runoff loss in the field = 420 cu. M.

the water stored in the root zone

= 2880 �'420 = 2460 cu. m.

Water application efficiency (ηa)

=   2460 /100 = 85.4% Ans. 2880

orage efficiency

= (Water stored in the root zone during irrigation /

Water needed in the root zone prior to irrigation)  x 100

Moisture holding capacity of soil

= 20 cm per m depth x1.7 m depth of root zone = 34 cm

Moisture already available in the root zone at the time of start of irrigation

50/100  x  34 =17cm.

Additional water required in the root zone

= 34 �'17 = 17 cm.

= 2720 cu. m.

But actual water stored in root zone = 2460 cu. m.

Water storage efficiency (ηs)

=2460 /2720  x 100 90% (say)

(iv) Water distribution efficiency

Where D = mean depth of water stored in the root zone

D = ( 1.7+1.1 )/2  = 1.4m

d is computed as below:

Deviation from the mean at upper end (absolute value) =  |1.7 -1.4| = 0.3

Deviation from the mean at lower end = | 1.1 -1.4 | =0.3

d = Average of the absolute values of deviations from mean =

0.4 +0.3/2 = 0.35

Using equations, we have,

ηd = 75 or 75%    Ans.

Consumptive Use or Evapotranspiration (Cu)

Consumptive use for a particular crop may be defined as the total amount of water used by the plant in transpiration (building of plant tissues, etc.) and evaporation from adjacent soils or from plant leaves, in any specified time. The values of consumptive use (Cu) may be different for different crops, and may be different for the same crop at different times and places.

In fact, the consumptive use for a given crop at a given place may vary throughout the day, throughout the month and throughout the crop period. Values of daily consumptive use or monthly consumptive use, are generally determined for a given crop and at a given place. Values of monthly consumptive use over the entire crop period are then used to determine the irrigation requirement of the crop.

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

Related Topics