Discharge over rectangular and V-notches

Rectangular notch

Let $L$ be the crest length and $H$ the head over the crest. Consider a horizontal strip of thickness $dh$ at depth $h$ below the free surface.

   ~~~~~~~~~~~~~~~~~~~~~~   free surface
    |        ^
    |        | h   strip dh
    |  ______|_________
    | |   L           |  H
    |_|_______________|_ crest

Velocity through the strip: $\sqrt{2gh}$. Area of strip: $L\,dh$.

$$dQ = C_d\,L\,dh\,\sqrt{2gh}$$ $$Q = C_d\,L\sqrt{2g}\int_0^H h^{1/2}dh = \frac23\,C_d\,L\sqrt{2g}\;H^{3/2}$$

If the velocity of approach $V_a$ is significant, $H$ is replaced by $(H + h_a)$ with $h_a = V_a^2/2g$ and the equation becomes $Q = \tfrac23 C_d L\sqrt{2g}\,[(H + h_a)^{3/2} - h_a^{3/2}]$. For end contractions, Francis's formula uses an effective length $L - 0.1nH$.

Triangular (V) notch

Let the notch angle be $\theta$. The width of the strip at depth $h$ is $2(H - h)\tan(\theta/2)$.

$$dQ = C_d\,2(H - h)\tan\frac{\theta}{2}\,\sqrt{2gh}\,dh$$ $$Q = 2C_d\tan\frac{\theta}{2}\sqrt{2g}\int_0^H (H - h)h^{1/2}dh = \frac{8}{15}\,C_d\tan\frac{\theta}{2}\sqrt{2g}\;H^{5/2}$$

Advantage of V-notch

• For small discharges, the head $H$ is large enough to be measured accurately, since $Q \propto H^{5/2}$.
• A single notch covers a wide range of flows.
• No end contractions, and no ventilation problem.

Numerical

Rectangular weir:

$$Q = \frac23 \times 0.62 \times 4.429 \times 1.5 \times (0.25)^{1.5} = \frac23 \times 0.62 \times 4.429 \times 1.5 \times 0.125 = 0.343\ \text{m}^3/\text{s}$$

90° V-notch ($\tan 45^\circ = 1$):

$$Q = \frac{8}{15} \times 0.6 \times 1 \times 4.429 \times (0.15)^{2.5} = 0.01235\ \text{m}^3/\text{s} = 12.35\ \text{L/s}$$

Answer: rectangular weir $Q = 0.343\ \text{m}^3/\text{s}$; V-notch $Q = 12.35\ \text{L/s}$