How are the crop areas found for the 40,000 ha command?

Intensity x CCA: wheat 8000, rice 6000, maize 8000 ha

No losses are given. How does the answer treat them?

Losses are neglected and duties are taken as measured at the canal head/reservoir

A 25% loss is assumed by default

Losses are added to the base period

How is the total storage obtained and what is it?

Sum of the three crop volumes, about 145.54 million m^3

Largest crop volume only, 72.58 million m^3

Average of the three volumes, 48.5 million m^3

Sum divided by 0.85, 171 million m^3

2076 Baisakh · 8 marks

Reservoir storage for wheat, rice, maize

Short answer

Area = intensity × 40000 ha. Q = A/D. Compute volume = Q × B × 86400 for wheat (34.56 Mm³), rice (72.58 Mm³), maize (38.40 Mm³). Sum is 145.54 Mm³. No losses given. Storage required ≈ 145.54 million m³.

A reservoir has to supply irrigation water to 40,000 hectares land. Calculate the storage required in the reservoir to meet the irrigation demand of various crops as detailed below:

Crop	Base period (days)	Duty (ha/cumecs)	Intensity of irrigation
Wheat	120	2400	20%
Rice	140	1000	15%
Maize	100	1800	20%

Given: CCA = 40,000 ha. Area of each crop = intensity x CCA. No canal or reservoir losses are given, so the duties are taken as measured at the reservoir (canal head); losses are neglected.

Method: $Q = A/D$ and $V = Q\times B\times 86400$ .

Crop	Area (ha)	Duty (ha/cumec)	Q = A/D (cumec)	B (days)	Volume (Mm $^3$ )
Wheat	8,000	2,400	3.333	120	34.560
Rice	6,000	1,000	6.000	140	72.576
Maize	8,000	1,800	4.444	100	38.400

V_{total} = 145.536\times10^6\ \text{m}^3

Answer: storage required in the reservoir $\approx$ 145.54 million m $^3$ . (If canal and reservoir losses were to be included, the volume would be divided by $(1-\text{loss})$ for each.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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