Which method gives the water volume needed by each crop at the field?

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

Which crop needs the largest volume and why?

Sugarcane, 172.8 Mm^3, because its base period (320 days) is longest

Wheat, because it has the largest area

Rice, because its duty is the lowest

Cotton, because its base period is 180 days

How are the canal and reservoir losses applied to the field volume?

Divide successively by (1 - canal loss) and (1 - reservoir loss)

Multiply by (1 - loss) each time

Add the loss percentages to the volume once

Subtract the losses from the volume

What is the reservoir capacity?

About 643.1 million m^3 (canal loss only)

About 482.3 million m^3 (field volume)

About 693.3 million m^3 (losses added)

2079 Jestha · 7 marks

Reservoir capacity with irrigated areas

Short answer

Compute field volume for each crop as V = (A/D) × B × 86400. Total field volume 482.30 Mm³. Divide by (1–0.25) for canal loss → 643.06 Mm³, then by (1–0.15) for reservoir loss. Reservoir capacity ≈ 756.54 million m³.

The base period, duty of water, and area under irrigation for various crops under a canal system are given in the table below. If the losses in the reservoir and canals are respectively 15%, 25%, determine the reservoir capacity.

Crop	Wheat	Sugarcane	Cotton	Rice	Vegetables
Base period B (Days)	120	320	180	120	120
Duty, D (ha/cumec)	1800	1600	1500	800	700
Area irrigated (ha)	15000	10000	5000	7500	5000

Given: reservoir loss = 15%, canal loss = 25%; areas, base periods and duties as in the table.

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	15,000	1,800	8.333	120	86.400
Sugarcane	10,000	1,600	6.250	320	172.800
Cotton	5,000	1,500	3.333	180	51.840
Rice	7,500	800	9.375	120	97.200
Vegetables	5,000	700	7.143	120	74.057

Total volume at the field: $V_f = 482.297\times10^6$ m $^3$

Allow canal loss (25%):

V_{canal\ head} = \frac{482.297}{1-0.25} = 643.063\times10^6\ \text{m}^3

Allow reservoir loss (15%):

V_{reservoir} = \frac{643.063}{1-0.15} = 756.545\times10^6\ \text{m}^3

Answer: reservoir capacity $\approx$ 756.54 million m $^3$ (0.7565 Gm $^3$ ).

✓ Answer checked by Claude Opus 5.5 (high effort)

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