For open-channel specific energy, which expression gives E in terms of depth y and velocity V?

• A. E = Q^2/(2gA^2)
• B. E = γh
• C. E = V^2/(2g) + y
• D. E = y + V^2/(2g)

Answer: (D)

Specific energy in open-channel flow is the energy per unit weight carried by the water, and it combines the potential energy from the depth with the kinetic energy from the flow. That gives the energy head as E = y + V^2/(2g). Here y is the depth, and V is the flow velocity, with g the acceleration due to gravity. If the velocity distribution is uniform (or we use the simple form), the velocity head is V^2/(2g) and the total specific energy is the sum of depth and velocity head.

So, E = y + V^2/(2g) expresses open-channel specific energy in terms of depth and velocity. The other forms either give only the velocity head (missing the depth term) or represent pressure head rather than total energy head.