Open-channel rating — estimating discharge from slope change at same stage A river reach has discharge Q1 = 173 m^3/s when the water-surface slope S1 = 1/6000 and the stage at the gauge is 10.0 m. During a flood, the stage remains 10.0 m but the water-surface slope increases to S2 = 1/2000. Assuming the cross-section and roughness are unchanged (uniform flow at the same stage), estimate the new discharge.

Introduction / Context:
For steady uniform flow, Manning’s or Chezy’s equation shows discharge depends on channel geometry, roughness, and energy slope (approximately equal to the water-surface slope in prismatic reaches). When stage is unchanged, the flow area and hydraulic radius remain essentially the same. Thus, discharge scales with the square root of the slope.

Given Data / Assumptions:

• Initial discharge Q1 = 173 m^3/s at slope S1 = 1/6000.
• New slope S2 = 1/2000; stage constant at 10.0 m → same A and R.
• Roughness unchanged; uniform-flow assumption applies for scaling.

Concept / Approach:
Using Manning: Q = (1/n) * A * R^(2/3) * S^(1/2). With A, R, n constant at the same stage, the ratio of discharges reduces to Q2/Q1 = (S2/S1)^(1/2).

Step-by-Step Solution:
Compute slope ratio: S2/S1 = (1/2000) / (1/6000) = 6000/2000 = 3.Take square root: (S2/S1)^(1/2) = √3 ≈ 1.732.New discharge: Q2 = Q1 * √3 ≈ 173 * 1.732 ≈ 299.6 m^3/s ≈ 300 m^3/s.

Verification / Alternative check:
Chezy's equation Q ∝ A * √(R * S) leads to the same square-root dependence on S when A and R are fixed, corroborating the estimate.

Why Other Options Are Wrong:
371 and 519 m^3/s correspond to multiplying by larger factors; 100 and 250 m^3/s underestimate the effect of tripling the slope.

Common Pitfalls:

• Assuming linear proportionality with S instead of square-root dependence.
• Ignoring that unchanged stage implies unchanged area and hydraulic radius.

Final Answer:
300 m^3/s