Percentage error in notch discharge

Given: $Q = 20$ L/s $= 0.02$ m³/s; error in head $dH = 1$ mm $= 0.001$ m. The coefficients are not given, so take $C_d = 0.62$ for the rectangular notch and $C_d = 0.60$ for the right-angled triangular notch.

Percentage error formula: since $Q \propto H^n$,

$$\frac{dQ}{Q} = n\frac{dH}{H}$$

Rectangular notch ($L = 0.30$ m, $n = 3/2$)

$$H = \left[\frac{Q}{\frac{2}{3}C_dL\sqrt{2g}}\right]^{2/3} = \left[\frac{0.02}{\frac{2}{3}(0.62)(0.3)(4.4294)}\right]^{2/3} = 0.1099\ \text{m}$$ $$\frac{dQ}{Q} = 1.5\times\frac{0.001}{0.1099} = 0.01365 = 1.37\ \%$$

Right-angled triangular notch ($\theta = 90^\circ$, $\tan 45^\circ = 1$, $n = 5/2$)

$$H = \left[\frac{Q}{\frac{8}{15}C_d\sqrt{2g}}\right]^{2/5} = \left[\frac{0.02}{\frac{8}{15}(0.6)(4.4294)}\right]^{0.4} = 0.1819\ \text{m}$$ $$\frac{dQ}{Q} = 2.5\times\frac{0.001}{0.1819} = 0.01374 = 1.37\ \%$$

Answer: the error in discharge is about 1.37 % for the rectangular notch and about 1.37 % for the triangular notch. Both give nearly the same error for 1 mm error in head; the result depends only slightly on the assumed $C_d$.