Critical depth and specific energy in open-channel flow At the critical depth of flow in a prismatic open channel, the specific energy (per unit weight) is at which extremum value?

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:

  • Prismatic channel, steady flow, negligible energy losses over the control section.
  • Specific energy E = y + V^2/(2g) for unit discharge q.
  • Critical depth yc occurs when the Froude number Fr = 1.


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

More Questions from Hydraulics

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion