Notch discharge coefficient error

Given: $Q = 12\ \text{L/s} = 0.012\ \text{m}^3/\text{s}$.

Coefficient of discharge, triangular notch

Included angle $\theta = 45^\circ$, so $\tan(\theta/2) = \tan 22.5^\circ = 0.4142$; $H = 0.21$ m.

$$Q = \frac{8}{15}C_d\tan\frac{\theta}{2}\sqrt{2g}\,H^{5/2}$$ $$\begin{aligned} H^{5/2} &= 0.21^{2.5} = 0.020208\\ 0.012 &= \frac{8}{15}(C_d)(0.4142)(4.4294)(0.020208) = 0.019774\,C_d\\ C_d &= 0.607 \end{aligned}$$

Coefficient of discharge, rectangular notch

$L = 0.30$ m, $H = 0.078$ m.

$$Q = \frac{2}{3}C_dL\sqrt{2g}\,H^{3/2}$$ $$\begin{aligned} H^{3/2} &= 0.078^{1.5} = 0.021784\\ 0.012 &= \frac{2}{3}(C_d)(0.3)(4.4294)(0.021784) = 0.019299\,C_d\\ C_d &= 0.622 \end{aligned}$$

Error in head for 2 % error in discharge

Differentiating $Q \propto H^n$: $\dfrac{dQ}{Q} = n\dfrac{dH}{H}$.

• Triangular notch ($n = 5/2$): $\dfrac{dH}{H} = \dfrac{0.02}{2.5} = 0.008$, so $dH = 0.008\times 210 = 1.68$ mm.
• Rectangular notch ($n = 3/2$): $\dfrac{dH}{H} = \dfrac{0.02}{1.5} = 0.01333$, so $dH = 0.01333\times 78 = 1.04$ mm.

Answer: $C_d = 0.607$ (triangular) and $0.622$ (rectangular); a 2 % error in discharge corresponds to an error in head of about 1.68 mm (triangular) and 1.04 mm (rectangular), i.e. 0.8 % and 1.33 % of the head.