Sharp-Edged Weir Design Analysis
Problem Statement
A sharp-edged weir is to be constructed across a stream in which the normal flow is 200 litres/sec. If the maximum flow likely to occur in the stream is 5 times the normal flow, determine the length of weir necessary to limit the rise in water level to 40cm above that for normal flow. The coefficient of discharge (Cd) is 0.61.
Given Data
| Normal Discharge (Q1) | 200 litres/sec = 0.2 m3/s |
| Maximum Discharge (Q2) | 5 × 0.2 = 1.0 m3/s |
| Coefficient of Discharge (Cd) | 0.61 |
| Water Level Rise Limit | 40 cm = 0.4 m |
| Acceleration due to Gravity (g) | 9.81 m/s2 |
1. Establishing the Discharge Equations
Let y = height of water above weir during normal flow
For a sharp-edged weir, the general discharge equation is:
Equation for normal flow (Q1):
Equation for maximum flow (Q2):
2. Solving for Normal Water Height (y)
Equating equations (a) and (b):
Taking the 2/3 power of both sides:
3. Determining the Required Weir Length
Substituting the found value of y into equation (a):
Verification
Let’s verify our solution by checking both flow conditions:
For normal flow:
Q1 = (2/3) × 0.61 × 1.17 × (0.208)3/2 × √(2 × 9.81) = 1.801 × 1.17 × 0.0949 = 0.2 m3/s ✓
For maximum flow:
Q2 = (2/3) × 0.61 × 1.17 × (0.608)3/2 × √(2 × 9.81) = 1.801 × 1.17 × 0.474 = 1.0 m3/s ✓
Rise in water level:
0.608 – 0.208 = 0.4 m ✓
Conclusion
For the given stream conditions:
1. The normal water height above the weir is 0.208 m (20.8 cm).
2. A sharp-edged weir with a length of 1.17 m is required to limit the rise in water level to 40 cm when the flow increases from the normal 0.2 m3/s to the maximum 1.0 m3/s.
3. During maximum flow conditions, the water level will be 0.608 m above the weir crest.
