Canal design discharge at 0, 1, 2, 3 km

Assumptions

• Wheat and rice are grown in different seasons, so the canal is designed for the crop that needs the larger discharge (not their sum).
• Intensity is on CCA. The GCA at a km is the area still commanded below that point. Total loss below 3 km $= 0.4$ m³/s is a fixed loss added at 3 km; the channel loss in each km is 4% of the discharge at the beginning of that km.

Discharge per hectare of crop

$$\begin{aligned} \text{Wheat}: q_w &= \frac{0.14\times 10^4}{28\times 86400} = 0.5787\times 10^{-3}\ \text{m}^3/\text{s per ha} \\ \text{Rice}: q_r &= \frac{0.20\times 10^4}{21\times 86400} = 1.1023\times 10^{-3}\ \text{m}^3/\text{s per ha} \end{aligned}$$

Demand per hectare of GCA: wheat $= 0.7\times 0.40\times q_w = 0.1620\times 10^{-3}$; rice $= 0.7\times 0.25\times q_r = 0.1929\times 10^{-3}$ m³/s. Rice governs at every section.

Demand of outlets (rice) and discharge

Working from the tail end

Area served between two consecutive km marks $= 6000$ ha GCA, with outlet demand $6000\times 0.1929\times 10^{-3} = 1.157$ m³/s.

$$\begin{aligned} Q_{3} &= 4.244 + 0.4 = 4.644\ \text{m}^3/\text{s} \\ Q_{begin} &= Q_{end} + \text{outlet demand} + 0.04\,Q_{begin} \Rightarrow Q_{begin} = \frac{Q_{end} + 1.157}{0.96} \\ Q_2 &= \frac{4.644 + 1.157}{0.96} = 6.043\ \text{m}^3/\text{s} \\ Q_1 &= \frac{6.043 + 1.157}{0.96} = 7.500\ \text{m}^3/\text{s} \\ Q_0 &= \frac{7.500 + 1.157}{0.96} = 9.019\ \text{m}^3/\text{s} \end{aligned}$$

Answer: Design discharge at 0, 1, 2, 3 km = 9.02, 7.50, 6.04, 4.64 m³/s (rice season governs).