---
title: "Minimum throat diameter of venturimeter"
subject: "Fluid Mechanics"
chapter: 6
key: "semester_3/fluid_mechanics:venturimeter-minimum-throat-diameter"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/venturimeter-minimum-throat-diameter
site: BCT Engineering
---

# Minimum throat diameter of venturimeter

Fluid Mechanics · Chapter 6: Flow measurement

- Asked in: 2073 Shrawan, 6 marks
- Video: [How To Solve Venturimeter Problems Fluid dynamics](https://www.youtube.com/watch?v=iPrCIYTkPWk) — ENG-School
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/venturimeter-minimum-throat-diameter
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-6

## Question

A venturimeter is to be fitted in a horizontal pipe of 0.15 m diameter to measure a flow of water which may be anything up to 240 m³/hour. The pressure head at the inlet for this flow is 18 m above atmospheric and the pressure head at the throat must not be lower than 7 m below atmospheric. Between the inlet and the throat there is an estimated frictional loss of 10% of the difference in pressure head between these points. Calculate the minimum allowable diameter for the throat.

## Short answer

With inlet pressure head 18 m, min throat -7 m, and 10% friction loss (2.5 m), Bernoulli gives V2^2/(2g)=23.23 m, V2=21.35 m/s. For Q=240 m^3/h, min throat area=0.00312 m^2, diameter≈63 mm.

## Answer

**Given:** $d_1 = 0.15$ m, $Q = 240$ m³/h $= 0.06667$ m³/s, $p_1/\gamma = 18$ m (gauge), minimum throat pressure head $p_2/\gamma = -7$ m (gauge). Friction loss $= 10\%$ of the pressure head difference.

Smaller throat gives higher velocity and lower pressure, so the **minimum** throat diameter corresponds to $p_2/\gamma = -7$ m.

$$
\begin{aligned}
\Delta\left(\frac{p}{\gamma}\right) &= 18 - (-7) = 25\ \text{m}\\
h_L &= 0.1\times 25 = 2.5\ \text{m}
\end{aligned}
$$

**Inlet velocity:**

$$
A_1 = \frac{\pi}{4}(0.15)^2 = 0.017671\ \text{m}^2,\qquad V_1 = \frac{0.06667}{0.017671} = 3.7726\ \text{m/s},\qquad \frac{V_1^2}{2g} = 0.7254\ \text{m}
$$

**Bernoulli (horizontal pipe):**

$$
\frac{p_1}{\gamma} + \frac{V_1^2}{2g} = \frac{p_2}{\gamma} + \frac{V_2^2}{2g} + h_L
$$

$$
\begin{aligned}
\frac{V_2^2}{2g} &= 18 + 0.7254 - (-7) - 2.5 = 23.2254\ \text{m}\\
V_2 &= \sqrt{2\times 9.81\times 23.2254} = 21.347\ \text{m/s}
\end{aligned}
$$

**Throat area and diameter:**

$$
A_2 = \frac{Q}{V_2} = \frac{0.06667}{21.347} = 3.123\times 10^{-3}\ \text{m}^2
$$

$$
d_2 = \sqrt{\frac{4A_2}{\pi}} = 0.0631\ \text{m}
$$

**Answer: minimum throat diameter $\approx 63$ mm** (0.0631 m). The throat velocity is 21.35 m/s and the loss is 2.5 m.

## Check yourself

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

1. Why does the minimum throat diameter correspond to p2/gamma = -7 m?
   A. A larger throat gives lower pressure
   B. The inlet pressure fixes the throat size
   C. A smaller throat gives higher velocity and lower pressure, so -7 m is the limit
   D. Atmospheric pressure is needed at the throat
   Answer: C. Decreasing throat diameter raises V2 and lowers p2, so the smallest allowable throat gives the lowest allowed pressure. (Key point: Which pressure governs)

2. What is the head loss between inlet and throat?
   A. 2.5 m
   B. 25 m
   C. 1.8 m
   D. 0.25 m
   Answer: A. Pressure head difference is 18 - (-7) = 25 m and the loss is 10% of it. (Key point: Friction loss)

3. What does Bernoulli give for V2^2/2g?
   A. 18 + 0.7254 + 7 - 2.5 = 23.2254 m
   B. 18 - 7 + 2.5
   C. 18 + 7 + 2.5
   D. 18 + 0.7254 - 7 - 2.5
   Answer: A. Loss is subtracted and the throat pressure head -7 m gives +7 on the other side. (Key point: Throat velocity head)

4. What is the minimum throat diameter?
   A. About 120 mm
   B. About 31 mm
   C. About 150 mm
   D. About 63 mm
   Answer: D. A2 = 0.06667/21.347 = 3.123e-3 m^2, so d2 = 0.0631 m. (Key point: Throat diameter)

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