Design of Concrete Lined Channel

Discharge and Slope

- Discharge (Q) = 250 cumecs
- Slope (S) = 1 in 6000

Trapezoidal Section

- Side slopes (z) = 1.5 : 1
- Limiting depth (y) = 3 m

Design Steps

1. Determine the cross-sectional dimensions of the channel

- Cross-sectional area (A) = Q / V
- Where, V = velocity of flow
- Assume V = 1.8 m/s (typical range for lined channels)
- A = 250 / 1.8 = 138.89 sq.m

- Channel width (T) = A / y
- T = 138.89 / 3 = 46.3 m

- Bottom width (B) = T - zy
- B = 46.3 - (1.5 x 3) = 41.8 m

2. Calculate the hydraulic radius (R)

- R = A / (B + y/2)
- R = 138.89 / (41.8 + 1.5) = 3.22 m

3. Determine the design flow depth (D)

- Use Manning's equation to determine the required depth for a given slope and discharge
- Q = (1.49 / N) R^(2/3) S^(1/2) D^(5/3)
- Where, N = Manning's roughness coefficient
- Rearranging the equation to solve for D:
- D = [(Q / (1.49 / N) R^(2/3) S^(1/2))]^(3/5)
- D = [(250 / (1.49 / 0.015) 3.22^(2/3) (1/6000)^(1/2))]^(3/5)
- D = 2.47 m

4. Check for velocity limitation

- Calculate the velocity of flow using the design flow depth
- V = Q / A
- V = 250 / 138.89 = 1.8 m/s (same as assumed velocity)
- Check if the velocity is within the allowable range of 0.6-3.0 m/s
- Since the calculated velocity is within the range, the design is acceptable.

5. Check for stability and erosion protection

- Verify that the lining material is suitable for the flow conditions and the slope of the channel
- Check if the velocity and shear stresses are within the allowable limits for the lining material
- Provide erosion protection measures if necessary (e.g. riprap, concrete armor units, etc.)

Conclusion

- The concrete lined trapezoidal channel with a discharge of 250 cumecs and a slope of 1 in 6000 is designed with a bottom width of 41.8 m, a depth of 2.47 m, and a limiting depth of 3 m.
- The design flow depth is within the limiting depth, and the velocity is within the allowable range.
- The lining material should be verified for stability and erosion protection.

No