Advanced Weir Design & Flood Routing Tool

Advanced Weir Design with Tailwater and Flood Routing

Rainfall Intensity Table (mm/hr)

Duration / RP 2 yr 10 yr 25 yr
30 min
1 hour
2 hour

Weir Design & Flood Routing Calculation (Sample Step-by-Step)

Design Sample Target: 60 min Duration | 10-Year Return Period

Input Parameters:
  • Catchment Area ($A$): $2.0\text{ km}^2 = 200\text{ ha}$
  • Runoff Coefficient ($C$): $0.6$
  • Weir Crest Height ($P$): $1.2\text{ m}$
  • Discharge Coefficient ($C_d$): $0.6$
  • Assumed Head over Weir ($H$): $0.5\text{ m}$
  • Tailwater Depth above Crest ($H_t$): $0.3\text{ m}$
  • Rainfall Intensity ($I$): $75\text{ mm/hr}$
  • Gravitational Acceleration ($g$): $9.81\text{ m/s}^2$

Step 1: Calculate Peak Inflow Discharge ($Q_{in}$)

Using the Rational Method formula:

$$Q_{in} = \frac{C \cdot I \cdot A_{ha}}{360}$$ $$Q_{in} = \frac{0.6 \times 75 \times 200}{360} = \frac{9000}{360} = 25.00\text{ m}^3/\text{s}$$

Step 2: Calculate Free Discharge Unit Capacity ($Q_{weir}$)

Using the standard rectangular weir equation per unit length ($m^3/s/m$):

$$Q_{weir} = \frac{2}{3} \cdot C_d \cdot \sqrt{2g} \cdot H^{1.5}$$ $$Q_{weir} = \frac{2}{3} \times 0.6 \times \sqrt{2 \times 9.81} \times (0.5)^{1.5}$$ $$Q_{weir} = 0.4 \times 4.4294 \times 0.35355 = 0.626\text{ m}^3/\text{s/m}$$

Step 3: Tailwater Submergence Adjustment ($Q_{sub}$)

  1. Tailwater Ratio: $$\text{Ratio} = \frac{H_t}{H} = \frac{0.3}{0.5} = 0.6$$
  2. Submergence Factor ($\phi$):
    Since $\text{Ratio} (0.6) < 0.7$, submergence reduction is not applied ($\phi = 1.0$).
  3. Submerged Discharge per unit length: $$Q_{sub} = Q_{weir} \times \phi = 0.626 \times 1.0 = 0.63\text{ m}^3/\text{s/m}$$

Step 4: Calculate Required Crest Length ($L_{req}$)

$$L_{req} = \frac{Q_{in}}{Q_{sub}} = \frac{25.00}{0.626} = 39.93\text{ m}$$

Step 5: Storage & Risk Assessment

  • Risk Evaluation: Since $Q_{sub} (0.63\text{ m}^3/\text{s/m}) < Q_{in} (25.00\text{ m}^3/\text{s})$, a 1-meter crest unit is insufficient, flagging ⚠️ Risk of Ponding.
  • Flood Storage Estimate: Assuming a triangular hydrograph shape: $$\text{Storage} = \frac{(Q_{in} - Q_{sub}) \times (d \times 60)}{2}$$ $$\text{Storage} = \frac{(25.00 - 0.626) \times (60 \times 60)}{2} = \frac{24.374 \times 3600}{2} = 43,873.20\text{ m}^3$$

Calculation Summary

Parameter Formula / Source Result
Event Scenario Duration / Return Period 60 min / 10 yr
Peak Inflow ($Q_{in}$) $(C \cdot I \cdot A) / 360$ 25.00 m³/s
Free Weir Unit Q ($Q_{weir}$) $\frac{2}{3} C_d \sqrt{2g} H^{1.5}$ 0.63 m³/s/m
Submerged Unit Q ($Q_{sub}$) $Q_{weir} \times \phi$ 0.63 m³/s/m
Required Crest Length ($L_{req}$) $Q_{in} / Q_{sub}$ 39.93 m
Flood Storage Estimate $(Q_{in} - Q_{sub}) \cdot t / 2$ 43873.20 m³
Risk Assessment $Q_{sub} < Q_{in}$ ⚠️ Risk of Ponding
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