Math Problem Statement

  1. Find out the capacity of a reservoir from the following data. The culturable command area is 80,000

hectares.

Crop

Base in days

Duty in hectares/cumec

Irrigation Intensity (%)

Rice

120

1800

25

Wheat

120

2000

30

Sugarcane

320

2500

20

Assume the canal and reservoir losses as 5% and 10% respectively

Solution

To find the capacity of the reservoir, we use the concept of Delta, which is the total depth of water required by a crop for its entire growing period. The steps are outlined below:

1. Key Formula:

Δ=8.64BD\Delta = \frac{8.64 \cdot B}{D} where:

  • Δ\Delta = Delta in meters (depth of water required for the crop)
  • BB = Base period in days
  • DD = Duty in hectares/cumec

2. Net Water Requirement:

The Net Water Requirement (NWRNWR) for each crop is: NWR=ΔCulturable Command Area (CCA)Irrigation Intensity100NWR = \Delta \cdot \text{Culturable Command Area (CCA)} \cdot \frac{\text{Irrigation Intensity}}{100}

3. Total Water Requirement:

The Total Water Requirement accounts for canal and reservoir losses: TWR=NWR(1+Canal Loss Fraction)(1+Reservoir Loss Fraction)TWR = NWR \cdot (1 + \text{Canal Loss Fraction}) \cdot (1 + \text{Reservoir Loss Fraction})

4. Calculate for Each Crop:

a. Rice:

ΔRice=8.641201800=0.576m\Delta_{\text{Rice}} = \frac{8.64 \cdot 120}{1800} = 0.576 \, \text{m} NWRRice=0.5768000025100=11520hectare-metersNWR_{\text{Rice}} = 0.576 \cdot 80000 \cdot \frac{25}{100} = 11520 \, \text{hectare-meters}

b. Wheat:

ΔWheat=8.641202000=0.5184m\Delta_{\text{Wheat}} = \frac{8.64 \cdot 120}{2000} = 0.5184 \, \text{m} NWRWheat=0.51848000030100=12441.6hectare-metersNWR_{\text{Wheat}} = 0.5184 \cdot 80000 \cdot \frac{30}{100} = 12441.6 \, \text{hectare-meters}

c. Sugarcane:

ΔSugarcane=8.643202500=1.10592m\Delta_{\text{Sugarcane}} = \frac{8.64 \cdot 320}{2500} = 1.10592 \, \text{m} NWRSugarcane=1.105928000020100=17694.72hectare-metersNWR_{\text{Sugarcane}} = 1.10592 \cdot 80000 \cdot \frac{20}{100} = 17694.72 \, \text{hectare-meters}

5. Add Total Net Water Requirement:

Total NWR=11520+12441.6+17694.72=41656.32hectare-meters\text{Total NWR} = 11520 + 12441.6 + 17694.72 = 41656.32 \, \text{hectare-meters}

6. Apply Losses to Find Total Water Requirement:

Canal Loss Fraction=5%=0.05,Reservoir Loss Fraction=10%=0.10\text{Canal Loss Fraction} = 5\% = 0.05, \quad \text{Reservoir Loss Fraction} = 10\% = 0.10 TWR=41656.32(1+0.05)(1+0.10)=41656.321.051.10=48157.42hectare-metersTWR = 41656.32 \cdot (1 + 0.05) \cdot (1 + 0.10) = 41656.32 \cdot 1.05 \cdot 1.10 = 48157.42 \, \text{hectare-meters}

7. Convert to Cubic Meters:

1 hectare-meter = 104m310^4 \, \text{m}^3 TWR in cubic meters=48157.42104=481574200m3\text{TWR in cubic meters} = 48157.42 \cdot 10^4 = 481574200 \, \text{m}^3

Final Answer:

The capacity of the reservoir is approximately 481.57 million cubic meters (MCM).


Do you want detailed calculations or clarifications?

Related Questions:

  1. How is the duty of water affected by soil type or crop variety?
  2. What are the assumptions behind irrigation intensity calculations?
  3. How can the reservoir losses be minimized in real-world applications?
  4. What is the impact of climate change on irrigation and water management systems?
  5. Why is it essential to include losses in reservoir capacity calculations?

Tip:

Always consider local climatic and soil conditions while estimating crop water requirements to avoid over- or under-designing reservoirs.

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Math Problem Analysis

Mathematical Concepts

Irrigation water management
Reservoir capacity estimation
Percentage and loss calculations
Unit conversions (hectare-meters to cubic meters)

Formulas

Delta (∆) = (8.64 × Base period in days) / Duty
Net Water Requirement (NWR) = ∆ × CCA × (Irrigation Intensity / 100)
Total Water Requirement (TWR) = NWR × (1 + Canal Loss Fraction) × (1 + Reservoir Loss Fraction)
1 hectare-meter = 10,000 cubic meters

Theorems

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Suitable Grade Level

Undergraduate (Civil Engineering, Agriculture, or Water Resource Studies)