Lined canal capacity and irrigable area

The printed data look like misprints, so these are assumed: depth $D = 2.5$ m (not 25 m) and $n = 0.016$ (not 0.16; typical for concrete lining). $B = 5$ m, $S = 1/1000$, side slope 1.5H:1V, so $\cot\theta = 1.5$ and $\theta = \tan^{-1}(1/1.5) = 0.588$ rad.

Carrying capacity (Manning's formula)

Giving $\theta$ in radians shows that the standard lined trapezoidal section with rounded bottom corners (radius = $D$) is intended, for which

$$A = BD + D^2(\theta + \cot\theta),\qquad P = B + 2D(\theta + \cot\theta)$$

Here $\theta + \cot\theta = 0.588 + 1.5 = 2.088$.

$$\begin{aligned} A &= 5\times 2.5 + 2.5^2\times 2.088 = 12.50 + 13.05 = 25.55\ \text{m}^2 \\ P &= 5 + 2\times 2.5\times 2.088 = 15.44\ \text{m} \\ R &= A/P = 1.655\ \text{m} \\ V &= \frac{1}{n}R^{2/3}S^{1/2} = \frac{1}{0.016}\times 1.655^{2/3}\times 0.001^{1/2} = 2.765\ \text{m/s} \\ Q &= AV = 25.55\times 2.765 = 70.65\ \text{m}^3/\text{s} \end{aligned}$$

Maximum carrying capacity $Q \approx 70.6$ m³/s (full supply depth 2.5 m).

Irrigable area

Water delivered during one kor period of 10 days:

$$\text{Volume} = Q\,t = 70.65\times 10\times 86400 = 6.104\times10^{7}\ \text{m}^3$$

Field irrigation requirement (depth) $= 150$ mm $= 0.15$ m. Neglecting losses beyond the stated field requirement:

$$\text{Area} = \frac{\text{Volume}}{\text{Depth}} = \frac{6.104\times10^{7}}{0.15} = 4.069\times10^{8}\ \text{m}^2 \approx 40,690\ \text{ha}$$

Answer: $Q \approx 70.6$ m³/s; irrigable area $\approx 40,700$ ha. (Velocity 2.77 m/s is within the safe limit for concrete lining.) If the section is treated as a plain trapezoid with sharp corners, $A = 21.875$ m², $P = 14.01$ m, $V = 2.66$ m/s, $Q \approx 58.2$ m³/s and the area $\approx 33,510$ ha.