---
title: "Venturimeter with reservoir suction tube"
subject: "Fluid Mechanics"
chapter: 6
key: "semester_3/fluid_mechanics:venturimeter-reservoir-lift-height"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/venturimeter-reservoir-lift-height
site: BCT Engineering
---

# Venturimeter with reservoir suction tube

Fluid Mechanics · Chapter 6: Flow measurement

- Asked in: 2079 Baisakh, 5+5 marks
- Video: [Venturi Meter Problems, Bernolli's Principle, Equation of Continuity - Fluid Dynamics](https://www.youtube.com/watch?v=lyk_HloXz-E) — The Organic Chemistry Tutor
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/venturimeter-reservoir-lift-height
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-6

## Question

The figure shows below a venturimeter where the reservoir open to atmosphere is connected to the throat by a tube. i) What is the fluid velocity in the smaller diameter section of pipe? ii) What is maximum height of fluid that can be lifted from reservoir (h)? Assume the fluid in lifting pipe is not moving. [Figure: horizontal venturimeter carrying water with piezometer tubes (differences of 1 cm and 5 cm marked) and throat diameter $D_2$ = 9 cm [?]; a vertical tube from the throat dips into a reservoir at the bottom]

## Short answer

Apply continuity and Bernoullli between inlet and throat: V2 = √[2gΔh/(1-(D2/D1)⁴)] = √(2×9.81×0.05 / 0.8704) ≈ 1.06 m/s. Throat gauge pressure head is –0.04 m, so the maximum lift height h = 0.04 m (4 cm).

## Answer

The figure values are only partly readable, so the following reading is assumed: the pipe diameters are $D_1 = 15$ cm and $D_2 = 9$ cm (throat); the piezometer at the inlet stands 1 cm above the pipe axis ($p_1/\gamma = 0.01$ m), and the piezometer difference between the inlet and the throat is 5 cm of water. The venturimeter is horizontal and the throat is connected by a tube to a reservoir open to the atmosphere.

#### (i) Velocity in the smaller diameter section (throat)

Bernoulli between 1 and 2 (horizontal, no loss) and continuity $A_1V_1 = A_2V_2$:

$$
\frac{p_1 - p_2}{\gamma} = \frac{V_2^2 - V_1^2}{2g} = \frac{V_2^2}{2g}\left[1 - \left(\frac{D_2}{D_1}\right)^4\right]
$$

$$
\begin{aligned}
\left(\frac{9}{15}\right)^4 &= 0.1296\\
V_2 &= \sqrt{\frac{2g\,(0.05)}{1 - 0.1296}} = \sqrt{\frac{0.981}{0.8704}} = 1.062\ \text{m/s}
\end{aligned}
$$

Flow $Q = A_2V_2 = \dfrac{\pi}{4}(0.09)^2(1.062) = 6.76\times 10^{-3}$ m³/s.

#### (ii) Maximum height h lifted from the reservoir

Pressure head at the throat:

$$
\frac{p_2}{\gamma} = \frac{p_1}{\gamma} - 0.05 = 0.01 - 0.05 = -0.04\ \text{m (gauge)}
$$

The fluid in the lifting tube is at rest, so the throat pressure just balances a column of fluid of height $h$ above the reservoir surface (open to atmosphere): $p_2 = -\gamma h$.

$$
h_{max} = \left|\frac{p_2}{\gamma}\right| = 0.04\ \text{m} = 4\ \text{cm}
$$

Fluid rises up the tube until it reaches the height at which the throat suction is balanced.

```text
   inlet 1          throat 2
 ----------\      /-----------
            \____/
                |   <- tube, h = 4 cm lift
 ~~~~~~~~~~~~~~~|~~~ reservoir (open to air)
```

**Answer: (i) $V_2 \approx 1.06$ m/s (so $Q \approx 6.8$ L/s); (ii) $h_{max} = 0.04$ m.** (Method: $V_2 = \sqrt{2g\Delta h/(1-(D_2/D_1)^4)}$ and $h = -p_2/\gamma$.)

## Check yourself

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

1. V2 =
   A. sqrt(2g dh)
   B. sqrt(2g dh (1-(D2/D1)^4))
   C. sqrt(2g dh/(1-(D2/D1)^4))
   D. dh/(D1-D2)
   Answer: C. Bernoulli plus continuity. (Key point: Throat velocity formula)

2. V2 for dh = 0.05 m, D2/D1=0.6?
   A. 0.5 m/s
   B. 2.1 m/s
   C. 3.1 m/s
   D. 1.06 m/s
   Answer: D. sqrt(0.981/0.8704)=1.062. (Key point: Value)

3. Throat pressure head is:
   A. -0.04 m gauge
   B. +0.04 m
   C. -0.05 m
   D. 0
   Answer: A. 0.01 - 0.05. (Key point: Throat pressure)

4. Why is hmax = |p2/gamma|?
   A. Velocity head
   B. Suction balances a column of fluid at rest
   C. Friction
   D. Surface tension
   Answer: B. p2 = -gamma h. (Key point: Lift height)

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