How is the volume for each crop calculated?

Q = A/D, then V = Q x B x 86400

V = A/D x B only in cumec-days without converting

How is the 25% canal loss allowed for?

Divide the field volume by (1 - 0.25)

Multiply the field volume by 0.75

What is the final step and result?

Divide canal-head volume by (1 - 0.15), giving about 632.96 Mm^3

Multiply 538.0 by 0.85, giving 457 Mm^3

Add 15% to the field volume 403.5, giving 464 Mm^3

Take 538.0 Mm^3 as the capacity

2078 Poush · 6 marks

Reservoir capacity with intensity table

Short answer

Total water volume required at field is about 403.5 million m³. After adding 25% canal losses and 15% reservoir losses, the required reservoir capacity is about 633 million m³.

The base period, the intensity of irrigation and duty of water for various crops under the canal system are given. Determine the reservoir capacity if the cultivable command area is 40,000 ha, canal losses are 25% and reservoir losses are 15%.

Crop	Base Period (Days)	Duty of water at the field (ha/m³/sec)	Irrigation Intensity (%)
Wheat	120	1,800	20
Sugarcane	360	1,700	20
Cotton	180	1,400	10
Rice	120	800	15
Vegetable	120	700	15

Given: CCA = 40,000 ha, canal loss = 25%, reservoir loss = 15%. Area of each crop = intensity x CCA (wheat 8000 ha, sugarcane 8000 ha, cotton 4000 ha, rice 6000 ha, vegetable 6000 ha).

Method. For each crop, discharge needed at the field $Q = A/D$ ; volume for the base period $V = Q\times B\times 86400$ m $^3$ . The losses are taken as a fraction of the water entering that part of the system, so field volume is divided by $(1-\text{loss})$ .

Crop	Area (ha)	Duty (ha/cumec)	Q = A/D (cumec)	B (days)	Volume = Q x B x 86400 (Mm $^3$ )
Wheat	8,000	1,800	4.444	120	46.080
Sugarcane	8,000	1,700	4.706	360	146.372
Cotton	4,000	1,400	2.857	180	44.434
Rice	6,000	800	7.500	120	77.760
Vegetable	6,000	700	8.571	120	88.869

Total volume at the field: $V_f = 403.515\times10^6$ m $^3$

Allow canal loss (25%):

V_{canal\ head} = \frac{403.515}{1-0.25} = 538.019\times10^6\ \text{m}^3

Allow reservoir loss (15%):

V_{reservoir} = \frac{538.019}{1-0.15} = 632.964\times10^6\ \text{m}^3

Answer: reservoir capacity $\approx$ 632.96 million m $^3$ (0.6330 Gm $^3$ ).

✓ Answer checked by Claude Opus 5.5 (high effort)

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