A channel of bed slope 0.0009 carries a discharge 300m3/s when the dep...
- **Given Data:**
- Bed slope of the channel, S₁ = 0.0009
- Discharge of the channel, Q₁ = 300 m³/s
- Depth of flow, y₁ = 1 m
- **Calculation for the new discharge:**
- Using the Chezy formula, Q = A * V = A * C * √(R * S), where:
- Q = Discharge
- A = Cross-sectional area of flow
- V = Velocity of flow
- C = Chezy coefficient
- R = Hydraulic radius
- S = Bed slope
- Since the channels are exactly similar:
- Cross-sectional area, A ∝ y * b
- Hydraulic radius, R = A / P ∝ y * b / 2y = b / 2
- Velocity, V ∝ √(R * S)
- Therefore, the new discharge Q₂ can be expressed as:
- Q₂ = A₂ * C * √(R * S₂)
- Q₂ = k * b * y * C * √(b / 2 * S₂), where k is a constant
- Given Q₁ = Q₂ when y = 1 m and S₁ = 0.0009, we can solve for k:
- 300 = k * 1 * b * C * √(b / 2 * 0.0009)
- k = 300 / (b * C * √(b / 2 * 0.0009))
- Substituting k back into the equation for Q₂:
- Q₂ = (300 / (b * C * √(b / 2 * 0.0009))) * b * 1 * C * √(b / 2 * 0.0001)
- Q₂ = 300 * √(0.0001 / 0.0009)
- Q₂ = 100 m³/s
- **Conclusion:**
- The discharge carried by an exactly similar channel at the same depth of flow, but with an increased slope of 0.0001, will be 100 m³/s.