---
title: "Cippoletti notch; tank inflow and area"
subject: "Fluid Mechanics"
chapter: 6
key: "semester_3/fluid_mechanics:cippoletti-notch-tank-inflow"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/cippoletti-notch-tank-inflow
site: BCT Engineering
---

# Cippoletti notch; tank inflow and area

Fluid Mechanics · Chapter 6: Flow measurement

- Asked in: 2073 Shrawan, 2+4 marks
- Video: [Notches and Weir](https://www.youtube.com/watch?v=a5YsNtFItFQ) — Gear Institute Mechanical Engineering Videos and Physics by Gear Institute
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/cippoletti-notch-tank-inflow
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-6

## Question

What is Cippoletti notch? A tank of area A is provided with an orifice 40 mm in diameter at its bottom. Water flows into tank at a uniform rate from the top and is discharged through the orifice. It is found that when the head of the water over the orifice is 0.68 m, the water surface rose at 0.0014 m/sec, but, when the head of the water over the orifice is 1.24 m, the water surface rose at 0.00062 m/sec. Find the rate of inflow and the cross-sectional area of the tank. Take $C_d$ = 0.62.

## Short answer

Cippoletti notch: trapezoidal with 1:4 slopes, Q=(2/3)Cd L sqrt(2g) H^(3/2). Tank inflow Q_in and area A found by solving Q_in - Cd a sqrt(2gh)=A dh/dt at h=0.68 m (0.0014 m/s) and h=1.24 m (0.00062 m/s): Q_in=4.64e-3 m^3/s, A=1.28 m^2.

## Answer

#### 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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is a Cippoletti notch?
   A. A triangular notch of 90 degrees
   B. A trapezoidal notch with side slope 1 horizontal to 4 vertical
   C. A broad-crested weir
   D. A rectangular notch with end contractions
   Answer: B. Its slope offsets the discharge lost to end contractions, so no contraction correction is needed. (Key point: Cippoletti notch)

2. Which equation applies when the water level rises in the tank?
   A. Qout = A dh/dt
   B. Qin - Qout = A dh/dt
   C. Qin + Qout = A dh/dt
   D. Qin = A dh/dt
   Answer: B. Net inflow stores water, raising the level at rate dh/dt. (Key point: Mass balance)

3. How is A found from the two observations?
   A. Assume Qin equals Qout
   B. Subtract the two equations so Qin cancels, A = (Qout2 - Qout1)/(difference in rise rates)
   C. Add the equations
   D. Use only the first reading
   Answer: B. Subtracting eliminates Qin: A = 0.9971e-3/0.00078 = 1.278 m^2. (Key point: Solving for A)

## Related in Fluid Mechanics

- [Jet from tank on plate over second tank](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Atwo-tanks-jet-plate-height) — Fluid Mechanics, chapter 7
- [Cippoletti weir slope and discharge formula](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Acipolletti-weir-trapezoidal-notch-discharge) — Fluid Mechanics, chapter 6
- [Time to lower level in two-orifice tank](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Atwo-orifice-tank-lowering-time) — Fluid Mechanics, chapter 6
- [Flow through partially and fully submerged orifice](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Asubmerged-orifice-flow-expression) — Fluid Mechanics, chapter 6
- [Percentage error in notch discharge](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Anotch-head-error-percentage) — Fluid Mechanics, chapter 6
- [Pressure diagram in rotating closed tank](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Aclosed-rotating-tank-pressure-diagram) — Fluid Mechanics, chapter 2

## Related in other subjects

- [Storage for uniform inflow over 24 hours](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ares-a74) — Water Supply Engineering, chapter 7
- [Sampler clearances and area ratio (68 mm)](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Asampler-clearance-68-70) — Foundation Engineering, chapter 2
- [Water surface over hump with contraction](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahump-contraction-surface-elevations-1-6m) — Hydraulics, chapter 7
- [Reservoir routing through spillway, 6-hr inflow](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Areservoir-routing-spillway-6hr) — Engineering Hydrology, chapter 7