A steady discharge of 1.0 m^3/s flows uniformly in a rectangular channel of width 1.0 m and flow depth 0.25 m. Based on this uniform flow regime, the channel bed slope is best classified as which type?

Introduction / Context:
In open-channel hydraulics, flow regime (subcritical, critical, supercritical) relates to the Froude number. For a given discharge and geometry, uniform flow at a particular depth occurs only when the bed slope equals the normal slope for that regime. Recognizing whether a slope is mild, steep, critical, or adverse follows from checking the Froude number at the uniform depth.

Given Data / Assumptions:

• Discharge Q = 1.0 m^3/s.
• Rectangular channel width b = 1.0 m.
• Uniform depth y = 0.25 m.
• Gravity g ≈ 9.81 m/s^2; friction and roughness chosen so this depth is the normal depth.

Concept / Approach:

Compute mean velocity v = Q / (b * y) = 1 / (1 * 0.25) = 4 m/s. The critical speed scale is sqrt(g * y) = sqrt(9.81 * 0.25) ≈ 1.566 m/s. The Froude number Fr = v / sqrt(g y) = 4 / 1.566 ≈ 2.56 > 1, indicating supercritical flow. For uniform supercritical flow to exist, the bed slope must be steeper than the critical slope; such channels are termed “steep” slopes. Mild slopes correspond to subcritical normal depths, while adverse slopes rise in the flow direction and cannot sustain the given uniform regime at this depth.

Step-by-Step Solution:

Verification / Alternative check:

If the slope were mild, the normal depth would be subcritical (Fr < 1). Because the given uniform depth produces Fr > 1, only a steep slope is consistent with uniform flow at that depth.

Why Other Options Are Wrong:

• Mild: Would yield subcritical normal depth (Fr < 1).
• Critical: Corresponds to Fr = 1, not 2.56.
• Adverse/horizontal: Not compatible with the stated uniform, supercritical regime.

Common Pitfalls:

• Mixing up regime names with bed slope labels (mild vs. steep).
• Forgetting that regime is evaluated at normal (uniform) conditions.

Final Answer:

steep