---
title: "U-tube manometer in accelerating vehicle"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_3/fluid_mechanics:vehicle-accelerating-u-tube-pressure"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/vehicle-accelerating-u-tube-pressure
site: BCT Engineering
---

# U-tube manometer in accelerating vehicle

Fluid Mechanics · Chapter 2: Pressure and Head

- Asked in: 2076 Chaitra, 8 marks
- Video: [Calculation for Acceleration Components](https://www.youtube.com/watch?v=k6xp6WzVJwQ) — Chemical Engineering Education
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/vehicle-accelerating-u-tube-pressure
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2

## Question

A test vehicle contains a U-tube manometer for measuring differences of air pressure. The manometer is so mounted that, when the vehicle is on level ground, the plane of the U is vertical and in the fore-and-aft direction. The arms of the U are 60 mm apart, and contain alcohol of relative density 0.79. When the vehicle is accelerated forwards down an incline at 20° to the horizontal at 2 m/s² the difference in alcohol levels (measured parallel to the arms of the U) is 73 mm, that nearer the front of the vehicle being the higher. What is the difference of air pressure to which this reading corresponds?

## Short answer

Using effective body force components in the accelerating frame, the rear-to-front pressure difference is ρ[(g sinθ – a)Δx – g cosθ Δy] = 467 Pa, with the rear limb at higher pressure.

## Answer

**Data:** arms $b=60$ mm apart (fore-and-aft), alcohol density $\rho=0.79\times1000=790\ \text{kg/m}^3$, slope $\theta=20^\circ$ down, acceleration $a=2\ \text{m/s}^2$ forward (down the slope), level difference along the arms $=73$ mm, front limb higher.

Use axes fixed to the vehicle: $x$ along the slope (forward), $y$ perpendicular to the slope (along the arms). In this frame the fluid feels an effective body force per unit mass $\vec g-\vec a$:

- $x$ component: $g\sin\theta-a=9.81\sin20^\circ-2=1.355\ \text{m/s}^2$ (forward)
- $y$ component: $-g\cos\theta=-9.81\cos20^\circ=-9.218\ \text{m/s}^2$

Pressure varies as $dp=\rho(b_x\,dx+b_y\,dy)$ within the liquid. Go from the surface in the rear limb to the surface in the front limb: $\Delta x=+0.060$ m forward and $\Delta y=+0.073$ m (the front level is higher):

$$
p_{front}-p_{rear}=790\left[1.355(0.060)-9.218(0.073)\right]=790(0.0813-0.6729)=-467.4\ \text{Pa}
$$

The air pressure acting on each liquid surface is therefore lower in the front limb:

$$
p_{rear}-p_{front}=467\ \text{Pa}
$$

**Answer: the air pressure difference is about 467 Pa (0.467 kPa), the rear limb being at the higher pressure.**

## Check yourself

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

1. In axes fixed to the accelerating vehicle, what effective body force per unit mass acts on the fluid?
   A. g + a
   B. g - a (vector)
   C. g only
   D. a only
   Answer: B. The fluid feels the vector g - a, which has components along and perpendicular to the slope. (Key point: Effective body force)

2. What is the forward (x) component of effective acceleration?
   A. g cos(20) - 2
   B. g sin(20) + 2
   C. g sin(20) - 2 = 1.355 m/s^2
   D. g sin(20) = 3.36 m/s^2
   Answer: C. The slope component of g is g sin(theta) = 3.355 and the forward acceleration a = 2 subtracts. (Key point: Component along slope)

3. Which equation gives p_front - p_rear?
   A. rho[b_x Delta x + b_y Delta y] with Delta x = 0.060, Delta y = 0.073
   B. rho g x 0.073
   C. rho a x 0.060
   D. rho g cos(20) x 0.060
   Answer: A. dp = rho(b_x dx + b_y dy) integrated from rear to front surface. (Key point: Pressure change between limbs)

4. The result p_front - p_rear = -467 Pa means what?
   A. Front limb air pressure is higher
   B. Both pressures equal
   C. Rear limb air pressure is higher by 467 Pa
   D. The alcohol has zero density
   Answer: C. The liquid pressure falls going to the front, so the rear limb air is 467 Pa higher. (Key point: Which limb is higher in pressure)

## Related in Fluid Mechanics

- [Double U-tube manometer, oil and brine pipes](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Adouble-u-tube-oil-brine-pressure) — Fluid Mechanics, chapter 2
- [Absolute and gauge pressure, mercury manometer](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Aabsolute-gauge-pressure-vapour-mercury) — Fluid Mechanics, chapter 2
- [Inverted U-tube and inclined manometer use](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Ainverted-u-tube-single-column-condition) — Fluid Mechanics, chapter 2
- [Inverted manometer differential level](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Ainverted-manometer-differential-level) — Fluid Mechanics, chapter 2
- [Oil-water manometer with gas cylinder](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Aoil-water-manometer-gas-cylinder) — Fluid Mechanics, chapter 2
- [Rotating U-tube pressure and water lost](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Arotating-u-tube-suction-leak) — Fluid Mechanics, chapter 2

## Related in other subjects

- [Differential manometer pressure at point B](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Amanometer-air-water-oil) — Hydro Power, chapter 1
- [U-tube manometer with oil, mercury and water](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Au-tube-oil-mercury-water) — Hydro Power, chapter 1
- [Pressure difference using compound manometer](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Amanometer-water-oil-mercury) — Hydro Power, chapter 1
- [Differential U-tube manometer between two pipes](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Adifferential-manometer-pipes) — Fluid Mechanics, chapter 2