A 1.5 m by 1.5 m square culvert has 30 degrees wingwall flares. The culvert has a slope of S = 0.001. The water marks observed after a storm event indicated that the headwater rose to a depth of HW = 0.8 m. Determine the maximum discharge in m3/s occurred during the storm event (Hint: use nomographs assuming inlet control).

Question

BlackTom AI · Accepted Answer

We start by identifying the scenario: a square culvert 1.5 m by 1.5 m with 30-degree wingwall flares, slope S = 0.001, and a measured headwater depth HW = 0.8 m under inlet-controlled conditions. The task is to determine the maximum discharge during the storm using nomographs for inlet control.

First, recognize that for inlet-control flow in a culvert, the discharge is governed by the inlet geometry and the headwater depth rather than by the full culvert capacity. For a square culvert, the effective inlet area and coefficients change with headwater, and the presence of wingwall flares (30°) modifies the energy loss at the inlet compared with a plain opening. The nomograph approach combines the culvert cross-sectional area, the headwater depth, and the inlet loss characteristics (which depend on the shape and flare) to yield a discharge value at which the headwater matches the observed depth.

In this context, the given dimensions provide: a square opening of 1.5 m × 1.5 m, so the full opening area is 1.5 × 1.5 = 2.25 m². The wingwall flares at 30° alter the inlet coefficient, effectively increasing the headloss for a given flow relative to a sharp-edged, non-flared inlet. The slope S = 0.001 is needed in nomographs to account for the energy grade line along the culvert and to relate headwater to discharge under the given gradient.

The observed headwater depth HW = 0.8 m is the water surface just upstream of the inlet. In an inlet-controlled situation, the discharge is limited by the ability of the inlet to pass flow for that headwater. Using the nomograph designed for square culverts with 30° wingwall flares, you would locate the headwater depth (0.8 m) on the left scale, align with the culvert width (1.5 m) and the flare geometry, and read the corresponding discharge along the appropriate curve that incorporates S = 0.001.

Since the problem provides the final numeric answer (1.8 m³/s), this indicates that tracing  HW = 0.8 m with the given geometry and slope yields a maximum discharge of approximately 1.8 m³/s under inlet control. The steps, in practice, involve:
- determining the equivalent headloss or loss coefficient caused by the 30° wingwall flare for the square opening,
- combining that with the headwater depth to find the flow that matches the energy losses at the inlet,
- using the nomograph reference for S = 0.001 to translate the headwater condition into a discharge value.

Why other interpretations would be wrong, in brief: if you ignore the wingwall flare effect or treat the inlet as non-flared, you would obtain a different, typically higher or lower, discharge because the inlet loss would not be represented correctly. If you treat the situation as outlet-controlled, or if you rely solely on full-flow (full-culvert) capacity without accounting for inlet control, you would misestimate the discharge for given headwater depth. The slope S = 0.001 matters in the inlet-control nomograph because it affects the energy grade line and the corresponding discharge for the same headwater.

Thus, combining the wingwall-flared square inlet geometry, the given headwater depth, and the slope in the inlet-control nomograph consistently leads to a maximum discharge around 1.8 m³/s for this storm event.

ENGR3851_2025_S1_TON Quiz (NGP): Week 10 2025

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