Orifice plate flow in hydropower canal

Given: orifice $b = 0.30$ m wide, $d = 0.10$ m high, bottom edge 5 cm above the bed, so the top edge is 0.15 m above the bed. Upstream depth $H_1 = 0.300$ m, downstream depth $H_2 = 0.225$ m, $C_d = 0.63$.

Both depths are above the top edge of the orifice (0.15 m), so the orifice is fully submerged (drowned): the discharge is governed by the difference of the water levels, not by the depth below the surface.

 upstream 0.300 m  |  |  downstream 0.225 m
 ~~~~~~~~~~~~~~~~~~|  |~~~~~~~~
                   |__|  orifice  (0.15 m top)
                   |  |  (0.05 m bottom)
 ------------------------ bed

Head difference:

$$H = H_1 - H_2 = 0.300 - 0.225 = 0.075\ \text{m}$$

Area of orifice $a = 0.30\times 0.10 = 0.03\ \text{m}^2$.

$$\begin{aligned} Q &= C_d\,a\sqrt{2gH}\\ &= 0.63\times 0.03\times\sqrt{2\times 9.81\times 0.075}\\ &= 0.63\times 0.03\times 1.2130\\ &= 0.02293\ \text{m}^3/\text{s} \end{aligned}$$

The velocity of approach is small (it is $0.153$ m/s if the canal is 0.5 m wide), so it is neglected. The canal width does not enter the calculation.

Answer: $Q \approx 0.0229$ m³/s (about 22.9 L/s).