---
title: "Venturimeter with CO2 and meriam oil"
subject: "Fluid Mechanics"
chapter: 6
key: "semester_3/fluid_mechanics:venturimeter-co2-manometer"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/venturimeter-co2-manometer
site: BCT Engineering
---

# Venturimeter with CO2 and meriam oil

Fluid Mechanics · Chapter 6: Flow measurement

- Asked in: 2074 Asoj, 4+5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/venturimeter-co2-manometer
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-6

## Question

In figure below the flowing fluid is CO₂ (density = 3 kg/m³). Neglect losses. If $p_1$ = 170 kPa and the manometer fluid is meriam red oil (S.G = 0.827). Estimate: (a) $p_2$ and (b) the gas rate in m³/h. [Figure: horizontal venturimeter with $d_1$ = 10 cm, $d_2$ = 6 cm and a U-tube manometer reading 8 cm]

## Short answer

p1-p2 = (827-3)×9.81×0.08 = 647 Pa, so p2≈169.35 kPa. Throat velocity V2 = sqrt(2×647/(3×0.8704))≈22.26 m/s, Q=A2 V2≈0.0629 m^3/s (226.5 m^3/h).

## Answer

**Given:** CO₂ density $\rho = 3$ kg/m³, $p_1 = 170$ kPa, manometer fluid S.G. = 0.827 ($\rho_m = 827$ kg/m³), manometer reading $h = 8$ cm $= 0.08$ m, $d_1 = 10$ cm, $d_2 = 6$ cm. The gas is treated as incompressible (small pressure change), horizontal meter, no losses.

#### (a) Pressure at the throat

The U-tube manometer measures the pressure difference between sections 1 and 2:

$$
p_1 - p_2 = (\rho_m - \rho)\,g\,h = (827 - 3)(9.81)(0.08) = 646.7\ \text{Pa}
$$

$$
p_2 = 170\,000 - 646.7 = 169\,353\ \text{Pa} \approx 169.35\ \text{kPa}
$$

#### (b) Gas rate

$$
\beta = \frac{d_2}{d_1} = 0.6,\qquad \beta^4 = 0.1296
$$

Bernoulli with continuity: $p_1 - p_2 = \dfrac{\rho}{2}V_2^2\left(1 - \beta^4\right)$.

$$
V_2 = \sqrt{\frac{2(646.7)}{3\times 0.8704}} = \sqrt{495.3} = 22.26\ \text{m/s}
$$

$$
\begin{aligned}
A_2 &= \frac{\pi}{4}(0.06)^2 = 2.827\times 10^{-3}\ \text{m}^2\\
Q &= A_2V_2 = 0.06293\ \text{m}^3/\text{s} = 226.5\ \text{m}^3/\text{h}
\end{aligned}
$$

**Answer: (a) $p_2 \approx 169.35$ kPa; (b) $Q \approx 0.0629$ m³/s $\approx 226.5$ m³/h.**

## Check yourself

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

1. What expression gives p1 - p2 for a gas flowing with an oil U-tube manometer?
   A. (rho_m - rho) g h
   B. rho_m g h only
   C. (rho_m + rho) g h
   D. rho g h only
   Answer: A. The manometer fluid minus the gas density is used: (827 - 3)(9.81)(0.08) = 646.7 Pa. (Key point: Manometer pressure difference)

2. What is the throat pressure p2?
   A. About 646.7 Pa
   B. About 100 kPa
   C. About 170.65 kPa
   D. About 169.35 kPa
   Answer: D. p2 = 170000 - 646.7 = 169353 Pa. (Key point: Value of p2)

3. Which relation links the pressure drop to the throat velocity?
   A. p1 - p2 = (rho/2) V2^2 beta^4
   B. p1 - p2 = (rho/2) V2^2 (1 - beta^4)
   C. p1 - p2 = rho V2^2
   D. p1 - p2 = (rho/2) V2^2 (1 + beta^4)
   Answer: B. With V1 = V2 beta^2, Bernoulli gives (rho/2)V2^2(1 - beta^4). (Key point: Bernoulli with continuity)

4. What is the gas rate?
   A. About 2265 m3/h
   B. About 22.65 m3/h
   C. About 62.9 m3/h
   D. About 226.5 m3/h
   Answer: D. V2 = 22.26 m/s, A2 = 2.827e-3 m^2, Q = 0.0629 m3/s = 226.5 m3/h. (Key point: Gas rate)

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