Cippoletti notch; tank inflow and area

Cippoletti notch

A Cippoletti notch (weir) is a trapezoidal notch whose sides have a slope of 1 horizontal to 4 vertical. The slope is chosen so that the extra discharge through the triangular sides offsets the discharge lost due to end contractions on a rectangular weir. Its discharge is $Q = \tfrac23 C_d L\sqrt{2g}\,H^{3/2}$, with no end-contraction correction.

Tank with orifice and inflow

Given: orifice $d = 40$ mm, $a = \dfrac{\pi}{4}(0.04)^2 = 1.2566\times 10^{-3}$ m², $C_d = 0.62$. Let the inflow be $Q_{in}$ and the tank area $A$. The level is rising, so inflow exceeds outflow:

$$Q_{in} - Q_{out} = A\frac{dh}{dt},\qquad Q_{out} = C_d\,a\sqrt{2gh}$$

Outflows:

$$\begin{aligned} Q_{out1} &= 0.62(1.2566\times 10^{-3})\sqrt{2\times 9.81\times 0.68} = 2.8458\times 10^{-3}\ \text{m}^3/\text{s}\\ Q_{out2} &= 0.62(1.2566\times 10^{-3})\sqrt{2\times 9.81\times 1.24} = 3.8429\times 10^{-3}\ \text{m}^3/\text{s} \end{aligned}$$

Equations:

$$\begin{aligned} Q_{in} - 2.8458\times 10^{-3} &= A(0.0014)\\ Q_{in} - 3.8429\times 10^{-3} &= A(0.00062) \end{aligned}$$

Subtracting:

$$(3.8429 - 2.8458)\times 10^{-3} = A(0.0014 - 0.00062)\;\Rightarrow\; A = \frac{0.9971\times 10^{-3}}{0.00078} = 1.278\ \text{m}^2$$ $$Q_{in} = 2.8458\times 10^{-3} + 1.278(0.0014) = 4.636\times 10^{-3}\ \text{m}^3/\text{s}$$

Check with the second equation: $3.8429\times 10^{-3} + 1.278(0.00062) = 4.636\times 10^{-3}$. Correct.

Answer: rate of inflow $= 4.64\times 10^{-3}$ m³/s (4.64 L/s); cross-sectional area of the tank $A = 1.28$ m².