Explain Mathematical **Relation Between Duty, Delta And Base Period** with definition and duty and delta in irrigation formula, duty and delta in irrigation, the base period **definition in irrigation**, the relation between duty, delta and base period so read the full article and help other by sharing education information

## Base Period Definition in Irrigation

Base Periodis the period from thefirst to the last wateringof the crop just before its maturity.

Base Period is denoted by “**B**“.

The Base Period is expressed in a **number of days**.

## Delta Definition in Irrigation

Deltais thetotal depth of waterrequired by a crop during the entire base period.

Delta is also called consumptive use.

Delta lies in the base period.

Delta is expressed in terms of depth and denoted by “**Δ”**.

## Duty Definition in Irrigation

The **duty of water** is defined as a number of hectares that can be irrigated by a constant supply of water at the rate of one cumec throughout the base period.

**Duty** is expressed in hectares/cumec and is denoted by “D”.

For example, if 3 cumecs of water are required for the crop sown in, an area of 5100 hectares, the duty of the irrigation will be 51003 = **1700 hectares/cumecs** and the discharge of 3 cumecs is required throughout the base period.

The Mathematical Relation Between Duty, Delta And Base Period in both systems is described below

## Mathematical Relation Between Duty, Delta and Base Period In M.K.S System

Let assume the following data as per definition,

**Duty = D** (hectares/cumecs)

**Delta = A** meters

**Base period = B** days

One cumec of water flowing continuously for “B” days gives a depth of water “A” over an area of “D” hectares.

**Volume of water** at 1m^{3}sec in one day = 1 × (24 × 60 × 60) = 86400 m^{2}

**Volume of water** at 1m^{3}sec in “B” days = 1 × (24 × 60 × 60) = 86400B m^{3} = 86400 m^{2}m — (i)

As, 1 Hectare = 10000 m^{2}

1 m^{2} = 1104 H

Then, equation (i) becomes,

Volume of water at 1 m^{3 }sec in “B” days = 86400B m^{3} = 86400B × 1104 H-m Volume of water @ 1 m3sec in “B” days = 8.64 x B H-m ==> (ii)

Depth of water required by crop, A = Volume Area A = 8.64 × B H-mD H A = 8.64 × B D m

In F.P.S System:

Let,

Duty = D (Acres/cusecs)

Delta = A feet Base period = B days By definition,

One cusec of water flowing continuously for “B” days gives a depth of water “A” over an area of “D” acres.

Volume of water @ 1 ft3sec in one day = 1×24*60*60 = 86400 3

Volume of water @ 1 ft3sec in “B” days = 1×24*60*60 = 86400B ft 3 = 86400 ft2ft —(i)

As, 1 Acre = 43560 ft2 1 ft2 = 143560 Acre Then, equation i becomes,

Volume of water @ 1 ft3sec in “B” days = 86400B ft3 = 86400B*143560 Acre-ft Volume of water @ 1 ft3sec in “B” days = 1.983*B Acre-ft —(ii)

Depth of water required by crop,A = Volume Area A = 1.983 B Acre-ftD Acre A = 1.983xB D ft

## FAQ 1: What is Delta Definition in Irrigation Engineering?

Delta is the total depth of water required by a crop during the entire base period.Delta is also called consumptive use.Delta lies in the base period.Delta is expressed in terms of depth and denoted by “Δ”.

## FAQ 2: What is Duty Definition in Irrigation Engineering?

The duty of water is defined as a number of hectares that can be irrigated by a constant supply of water at the rate of one cumec throughout the base period.Duty is expressed in hectares/cumec and is denoted by “D”.For example, if 3 cumecs of water are required for the crop sown in, an area of 5100 hectares, the duty of the irrigation will be 51003 = 1700 hectares/cumecs and the discharge of 3 cumecs is required throughout the base period.