Mathematical Relation Between Duty, Delta And Base Period

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 Period is the period from the first to the last watering of 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

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”.

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 1m³sec in one day = 1 × (24 × 60 × 60) = 86400 m²

Volume of water at 1m³sec in “B” days = 1 × (24 × 60 × 60) = 86400B m³ = 86400 m²m — (i)

As, 1 Hectare = 10000 m²

1 m² = 1104 H

Then, equation (i) becomes,

Volume of water at 1 m³ sec in “B” days = 86400B m³ = 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