The critical depth (yc) for a triangular channel with a side slope of 1H:1V having a specific energy of 2.5 m will be
- A3 m
- B2.5 m
- C2 mCorrect
- D1 m
Solution
Concept
Critical depth in a triangular channel
Formula
For triangular section: Emin = 5 yc / 4
Step-by-step solution
1. Critical condition: A³/T = Q²/g with A = z y², T = 2 z y. 2. This gives V²/2g = yc/4. 3. E = yc + yc/4 = 5 yc/4. 4. yc = 4E / 5 = 4 × 2.5 / 5 = 2 m.
Explanation
For a triangular channel the hydraulic depth is y/2, so at critical flow V²/2g = yc/4. Hence the minimum specific energy is 1.25 yc, independent of the side slope.
Shortcut
Rectangular 3/2 yc; triangular 5/4 yc; parabolic 4/3 yc.
Common mistake
Applying the rectangular channel relation yc = 2E/3.
Correct answer: C) 2 m