Find the discharge of water flowing over a rectangular notch of length 1 m when the constant head over the notch is 100 mm. Take C_D = 0.60. Given \(\rm \sqrt{2g}\) = 4.43.

• Weir or notch is a physical structure of masonry constructed across the channel width to calculate the discharge of the channel section.

• The discharge through a rectangular notch weir is,

\(Q = \frac{2}{3}C_{d}L\sqrt{2g}H^{\frac{3}{2}}\)

Where,

Q = discharge of fluid,

L =  length of weir

C_d = Coefficient of discharge and

H = height of water above the notch

Given

C_d = 0.6

L = 1 m

H = 100 mm = 0.1 m

Calculation

\(Q = \frac{2}{3}.(0.6).1\sqrt{2g}.0.1^{\frac{3}{2}} \)
Q = 0.0568 m³/s

Q = 56.8 litres/sec   (1 m³ = 1000 litres)

 Additional Information
Triangular Notch:

• A V-notch weir is also called the triangular notch or weir. The discharge over a triangular weir or notch is given by the:

\(Q = \frac{8}{15}C_{d}tan\frac{\theta }{2}\sqrt{2g}H^{\frac{5}{2}} \)

Where,
Q = discharge of fluid,

C_d = Coefficient of discharge,

θ = Notch angle and

H = height of water above the notch
Trapezoidal weir (or) Notch:

• The discharge over a Trapezoidal weir or notch is given by the:

\(Q = \frac{2}{3}C_{d1}L\sqrt{2g}H^{\frac{3}{2}} + \frac{8}{15}C_{d2}tan\frac{\theta }{}\sqrt{2g}H^{\frac{5}{2}}\)

Where,

θ = weir angle of inclination with the vertical.
C_d1 = Coefficient of discharge for rectangular portion.
C_d2 = Coefficient of discharge for the triangular portion