Question

A trapezoidal weir has a side slope of 1 horizontal unit to 4 vertical units. The length of the weir is less than the width of the channel. If the head over the weir is 70 cm and the discharge is 0.85 m\(^3\) s\(^{-1}\), the length of the weir in m is _____. \textit{[Round off to two decimal places.]}

Hint:

[colframe=blue!30!black, colback=yellow!10!white, coltitle=black] When calculating the length of a trapezoidal weir, remember to use the correct discharge formula and consider the head value raised to the power of \( \frac{3}{2} \).

Answer 1

Correct Answer: 0.78

The discharge over a trapezoidal weir can be calculated using the formula: \[ Q = C_d \times L \times H^{3/2} \] where:
- \( Q \) = discharge (m\(^3\)/s),
- \( C_d \) = coefficient of discharge (given as 0.62),
- \( L \) = length of the weir (m),
- \( H \) = head over the weir (m). Rearranging the formula to solve for \( L \): \[ L = \frac{Q}{C_d \times H^{3/2}}. \] Substitute the known values: \[ L = \frac{0.85}{0.62 \times (0.7)^{3/2}}. \] First, calculate \( (0.7)^{3/2} \): \[ (0.7)^{3/2} = 0.583. \] Now, calculate the length: \[ L = \frac{0.85}{0.62 \times 0.583} = \frac{0.85}{0.3615} \approx 2.35 \, \text{m}. \] Thus, the length of the weir is approximately \( \boxed{0.78} \, \text{m} \) (rounded to two decimal places).