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Irrigation demand from ETo and efficiency

Short answer

Net irrigation requirement = 150–30 = 120 mm/month. Gross requirement = 120 / 0.30 = 400 mm/month. Volume = 0.40 m × 6000×10⁴ m² = 2.4×10⁷ m³/month; discharge = 2.4×10⁷ / (30×86400) = 9.26 cumec.

An irrigation project has 6000 ha of CCA and ETo is 150 mm/day, effective rainfall is 30 mm/month and the overall efficiency of the project is 30%. Calculate the irrigation demand in cumec.

Assumptions. The value of 150 mm for EToET_o cannot be a daily rate, so it is taken as mm per month, the same time unit as the effective rainfall (30 mm/month). Crop coefficient Kc=1K_c=1 (crop water requirement = EToET_o), and a month = 30 days.

Given: CCA = 6000 ha, ETo=150ET_o = 150 mm/month, Pe=30P_e = 30 mm/month, overall efficiency =30%= 30\%.

Step 1: Net irrigation requirement

NIR=ETo−Pe=150−30=120 mm/monthNIR = ET_o - P_e = 150 - 30 = 120\ \text{mm/month}

Step 2: Gross irrigation requirement at the head

GIR=NIRη=1200.30=400 mm/monthGIR = \frac{NIR}{\eta} = \frac{120}{0.30} = 400\ \text{mm/month}

Step 3: Volume of water per month

V=0.40 m×6000×104 m2=2.4×107 m3V = 0.40\ \text{m}\times 6000\times 10^4\ \text{m}^2 = 2.4\times 10^7\ \text{m}^3

Step 4: Discharge

Q=V30×86400=2.4×1072 592 000=9.26 m3/sQ = \frac{V}{30\times 86400} = \frac{2.4\times 10^7}{2\,592\,000} = 9.26\ \text{m}^3/\text{s}

Answer: irrigation demand ≈9.26\approx 9.26 cumec.

✓ Answer checked by Claude Opus 5.5 (high effort)

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