The discharge passing over an ogee spillway is given by (where, L is effective length of spillway crest and H is the total head over the spillway crest including velocity head.) A. $${\text{CL}}{{\text{H}}^{\frac{3}{2}}}$$ B. $${\text{CH}}{{\text{L}}^{\frac{3}{2}}}$$ C. $${\text{CL}}{{\text{H}}^{\frac{5}{2}}}$$ D. $${\text{CL}}{{\text{H}}^{\frac{1}{2}}}$$

The correct answer is $\boxed{\text{CL}H^{\frac{3}{2}}}$.

The discharge passing over an ogee spillway is given by the following equation:

$$Q = CLH^{\frac{3}{2}}$$

where:

• $Q$ is the discharge (volume of water per unit time)
• $C$ is a discharge coefficient that depends on the shape of the spillway
• $L$ is the effective length of the spillway crest
• $H$ is the total head over the spillway crest including velocity head

The discharge coefficient $C$ is a dimensionless number that depends on the shape of the spillway. For an ogee spillway, the discharge coefficient is typically in the range of 0.6 to 0.8.

The effective length of the spillway crest $L$ is the length of the spillway crest that is actually submerged by water. The effective length of the spillway crest is typically less than the total length of the spillway crest.

The total head over the spillway crest $H$ is the height of the water above the spillway crest. The total head includes the static head (the height of the water above the spillway crest) and the velocity head (the kinetic energy of the water).

The discharge passing over an ogee spillway is proportional to the square root of the total head. This is because the higher the head, the faster the water will flow over the spillway. The discharge passing over an ogee spillway is also proportional to the effective length of the spillway crest. This is because the longer the spillway crest, the more water can flow over the spillway.