Critical velocity in terms of critical depth in open-channel flow If h_c is the critical depth, what is the expression for the critical velocity V_c in a wide channel (hydraulic depth ≈ depth)?
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AV_c = sqrt(g * h_c)
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BV_c = g * h_c
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CV_c = h_c / sqrt(g)
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DNone of these
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EV_c = sqrt(g / h_c)
Answer
Correct Answer: V_c = sqrt(g * h_c)
Explanation
Introduction / Context:Critical flow concepts are used to size control sections and predict transitions such as hydraulic jumps. The relation between critical depth and critical velocity provides a quick check in design and analysis for wide, prismatic channels.
Given Data / Assumptions:
- Wide channel so hydraulic depth D ≈ flow depth y.
- Critical state satisfies Fr = 1.
- Acceleration due to gravity = g.
Concept / Approach:
Froude number Fr = V / sqrt(g * D). At critical flow, Fr = 1, hence V_c = sqrt(g * D_c). For wide channels, D_c ≈ h_c, yielding the familiar square-root relation.
Step-by-Step Solution:
Write Fr = V / sqrt(g * D) = 1 at critical.Replace D by h_c in wide-channel approximation.Obtain V_c = sqrt(g * h_c).Verification / Alternative check:
The result is consistent with specific energy theory where minimum specific energy occurs at Fr = 1 and V_c relates to depth through the same relation.
Why Other Options Are Wrong:
g * h_c or h_c / sqrt(g) have incorrect dimensions; sqrt(g / h_c) is dimensionally inconsistent for velocity.
Common Pitfalls:
Using hydraulic radius or area/width inconsistently; the wide-channel simplification avoids those complications.
Final Answer:
V_c = sqrt(g * h_c)