Field channel discharge for gram and wheat

Formula: $Q = \dfrac{A\,d}{T\times 86400}$ ($A$ in m$^2$, $d$ in m, $T$ = kor period in days).

Areas: gram $= 0.30\times 2000 = 600$ ha; wheat $= 0.50\times 2000 = 1000$ ha.

Gram ($T = 18$ days, $d = 0.12$ m):

$$Q_g = \frac{6000000\times 0.12}{18\times 86400} = 0.463\ \text{m}^3/\text{s}$$

Wheat ($T = 15$ days, $d = 0.15$ m):

$$Q_w = \frac{10000000\times 0.15}{15\times 86400} = 1.157\ \text{m}^3/\text{s}$$

Gram and wheat are both rabi crops, grown at the same time on different parts of the command area (30% + 50% = 80% of the CCA). Their kor waterings can therefore coincide, so the field channel must be able to carry both demands together:

$$Q = Q_g + Q_w = 0.463 + 1.157 = 1.620\ \text{m}^3/\text{s}$$

Answer: design discharge of the field channel = 1.62 cumec (gram 0.463 cumec + wheat 1.157 cumec).