Question

At critical state of flow, if the velocity of flow is 981 cm/sec, then the hydraulic depth and Froude number will be, respectively:

• A. 1.0 m : 1.2
• B. 981 cm : 0.98
• C. 9.81 m : 1.0
• D. 1.0 m : 1.0

Hint:

Critical flow: \(Fr = 1\), \(V = \sqrt{g \times y}\).
For V = 9.81 m/s, y = 9.81 m.
Hydraulic depth = A/T (area/top width).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Critical flow in an open channel occurs when the Froude number is equal to unity. The hydraulic depth at this condition is obtained from the Froude number equation.

Step 2: Key Formula:
For critical flow, \[ Fr=\frac{V}{\sqrt{gy}}=1 \] where \[ V=\text{velocity},\qquad g=\text{acceleration due to gravity},\qquad y=\text{hydraulic depth}. \]

Step 3: Calculation:
Given, \[ V=981~\text{cm/s}=9.81~\text{m/s} \] \[ g=9.81~\text{m/s}^2 \] Since \(Fr=1\), \[ V=\sqrt{gy} \] Squaring both sides, \[ V^2=gy \] \[ y=\frac{V^2}{g} =\frac{(9.81)^2}{9.81} =9.81~\text{m} \] Thus, the hydraulic depth is \(9.81~\text{m}\), and the Froude number is \(1.0\).

Step 4: Final Answer:
Hydraulic depth=9.81~m,Fr=1.0} Hence, the correct option is (C).
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