If Q = discharge into the channel, a = cross sectional area at the entrance, and Dn is depth at the channel entrance, what is the equation for depth of the reservoir (Dr) when the flow in the channel is subcritical ?
- ADr = Dn − Q / 2ga²
- BDr = Dn − Q² / 2ga²
- CDr = Dn + Q / 2ga²
- DDr = Dn + Q² / 2ga²Correct
Solution
Concept
Energy equation at the entrance of a channel
Formula
Dr = Dn + V²/2g, V = Q/a
Step-by-step solution
1. Energy is conserved from the reservoir to the entrance: Dr = Dn + V²/2g. 2. V = Q / a, so Dr = Dn + Q² / (2 g a²).
Explanation
The reservoir level is the sum of the depth at the entrance and the velocity head there.
Why the other options are wrong
Minus signs would put the reservoir level below the entrance depth. A term with Q (not Q²) is dimensionally wrong.
Shortcut
Reservoir level = depth + velocity head.
Common mistake
Forgetting to square Q.
Correct answer: D) Dr = Dn + Q² / 2ga²