# Mathematical Relation Between Duty, Delta and Base Period

## Definitions

### Base Period:

It is the period from the first to the last watering of the crop just before its maturity. It is denoted by “B” and expressed in number of days.

### Delta:

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

### Duty:

The duty of water is defined as number of hectares that can be irrigated by constant supply of water at the rate of one cumec throughout the base period. It is expressed in hectares/cumec and is denoted by “D”. For example if 3 cumecs of water is required for the crop sown in, an area 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 explained as follows:

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

Let,

Duty = D (hectares/cumecs)

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

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

Volume of water @ 1m^{3}sec in one day = 1x24*60*60 = 86400 m^{3}

Volume of water @ 1m^{3}sec in "B" days = 1x24*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 @ 1 m^{3}sec in "B" days = 86400B m^{3} = 86400B*1104 H-m Volume of water @ 1 m^{3}sec in "B" days = 8.64 x B H-m **— (ii)**

Depth of water required by crop, A = Volume Area A = 8.64xB 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 ft^{3}sec in one day = 1x24*60*60 = 86400 ^{3}

Volume of water @ 1 ft^{3}sec in "B" days = 1x24*60*60 = 86400B ft ^{3} = 86400 ft^{2}ft** —(i)**

As, 1 Acre = 43560 ft^{2} 1 ft^{2} = 143560 Acre Then, equation i becomes,

Volume of water @ 1 ft^{3}sec in "B" days = 86400B ft^{3} = 86400B*143560 Acre-ft Volume of water @ 1 ft^{3}sec 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