In an open-channel, Fr = V / sqrt(g y). If velocity is constant and depth y doubles, what happens to Fr?

• A. Fr doubles
• B. Fr decreases by a factor of sqrt(2)
• C. Fr increases by a factor of sqrt(2)
• D. Fr remains unchanged

Answer: (B)

The Froude number in open-channel flow is Fr = V / sqrt(g y), which compares inertial forces to gravitational forces. If the velocity is kept constant and the depth doubles, the term sqrt(g y) in the denominator becomes sqrt(g · 2y) = sqrt(2) · sqrt(g y). That makes the new Fr equal to V / [sqrt(2) · sqrt(g y)] = (1/√2) · [V / sqrt(g y)] = Fr_original / √2. So the Froude number decreases by a factor of √2; it becomes smaller to reflect the relatively stronger influence of gravity at the greater depth. If you relate this to flow regimes, the flow shifts toward more subcritical as Fr decreases.