Difficulty: Easy
Correct Answer: minimum
Explanation:
Introduction / Context:Specific energy diagrams are central to open-channel hydraulics for analyzing transitions, controls, and critical flow conditions. Recognizing the extremum at critical depth enables design of flumes and weirs for flow measurement and control.
Given Data / Assumptions:
Concept / Approach:
For a given discharge, the specific energy has a minimum at the critical depth. Mathematically, dE/dy = 0 yields the critical condition q^2 = g * A^3 / T for general sections, and for a rectangular section yc = (q^2/g)^(1/3). At this point, E is minimized, not maximized.
Step-by-Step Solution:
Express V = q / A; for a rectangular channel A = b y, T = b.E(y) = y + (q^2)/(2g b^2 y^2).Set dE/dy = 0 → 1 − (q^2)/(g b^2 y^3) = 0 → y = yc = (q^2/g)^(1/3).At yc, E is at a minimum for fixed q → defines critical flow (Fr = 1).Verification / Alternative check:
On the specific-energy diagram, subcritical and supercritical branches meet at the lowest point where Fr = 1; any deviation in y increases E for the same discharge.
Why Other Options Are Wrong:
(b) Maximum occurs as y → ∞ for fixed q (E grows with y). (c) No averaging principle applies. (d) A definite extremum exists. (e) Bed slope does not set the extremum of E at a control section for a given q.
Common Pitfalls:
Confusing energy grade line (head losses) with specific energy; mixing critical depth with normal depth (which depends on slope and roughness).
Final Answer:
minimum
Discussion & Comments