Which expression gives the duty at the field head and its value?

D = 8.64 B / delta = 288 ha/cumec

D = 8.64 delta / B = 25.9 ha/cumec

How do the 15% losses change the duty at the head of the channel?

Duty decreases to 288 x 0.85 = 244.8 ha/cumec

Duty increases to 288/0.85 = 338.8

2078 Poush · 6 marks

Duty and head discharge, 4000 ha channel

Short answer

Net irrigation depth is 45 cm. Duty at field head is 288 ha/cumec, at channel head (after 15% losses) is 244.8 ha/cumec. Head discharge required is 11.44 m³/s.

The command area of a channel is 4000 hectares. The intensity of irrigation of a crop is 70%. The crop requires 60 cm of water in 15 days, when the effective rainfall is recorded as 15 cm during that period. Find: (i) The duty at the head of field. (ii) The duty at the head of channel (iii) The head discharge at the head of channel. Assume total losses as 15%.

Given: CCA = 4000 ha, intensity = 70%, crop water need = 60 cm in $B=15$ days, effective rainfall = 15 cm, total losses = 15%.

Area irrigated: $A = 0.70\times 4000 = 2800$ ha.

Net depth from irrigation (delta at the field)

\Delta = 60 - 15 = 45\ \text{cm} = 0.45\ \text{m}

(i) Duty at the head of the field

D_{field} = \frac{8.64\,B}{\Delta} = \frac{8.64\times 15}{0.45} = 288.0\ \text{ha/cumec}

(ii) Duty at the head of the channel. Losses of 15% reduce the area served per cumec:

D_{head} = 288.0\times(1-0.15) = 244.8\ \text{ha/cumec}

(iii) Head discharge of the channel

Q = \frac{A}{D_{head}} = \frac{2800}{244.8} = 11.44\ \text{cumec}

Answer: (i) 288 ha/cumec; (ii) 244.8 ha/cumec; (iii) 11.44 cumec.

✓ Answer checked by Claude Opus 5.5 (high effort)

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