---
title: "Oil-water interface in enlarged-end U-tube"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_3/fluid_mechanics:u-tube-enlarged-ends-oil-water-interface"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/u-tube-enlarged-ends-oil-water-interface
site: BCT Engineering
---

# Oil-water interface in enlarged-end U-tube

Fluid Mechanics · Chapter 2: Pressure and Head

- Asked in: 2076 Asoj, 10 marks
- Video: [How to solve manometer problems](https://www.youtube.com/watch?v=oT29E4FNx1g) — Engineer4Free
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/u-tube-enlarged-ends-oil-water-interface
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2

## Question

A manometer consists of a U-tube, 7 mm internal diameter, with vertical limbs each with an enlarged upper end 44 mm diameter. The left hand limb and the bottom of the tube is filled with water and the top of the right-hand limb is filled with oil of specific gravity 0.83. The free surfaces of the liquids are in the enlarged ends and the interface between the oil and water is in the tube below the enlarged end. What would be the difference in pressures applied to the free surfaces which would cause the oil/water interface to move 1 cm.

## Short answer

Interface movement x relates to pressure difference by p₁−p₂ = γw x[(1+s)(d/D)²+(1−s)]. With x=1 cm, this gives 21.2 Pa (2.16 mm water).

## Answer

**Data:** tube $d=7$ mm, enlarged ends $D=44$ mm, oil $S=0.83$ (right limb), water in the left limb and base, interface moves $x=1$ cm $=0.01$ m.

```text
   p1 (water side)       p2 (oil side)
   [D]                   [D]
    |  water      oil     |
    |                     |
    |__________ interface moves x
```

Raise the pressure $p_1$ on the water side. The interface moves up the right limb by $x$. The volume displaced in the narrow tube moves the free surfaces in the enlarged ends by

$$
y=x\left(\frac{d}{D}\right)^{2}=0.01\left(\frac{7}{44}\right)^{2}=2.53\times10^{-4}\ \text{m}
$$

(the water surface falls $y$, the oil surface rises $y$). Initially the pressures balance at the interface level, so the change in the pressure difference is balanced by the change in column heights:

$$
p_1-p_2=\gamma_w y+\gamma_o y+(\gamma_w-\gamma_o)x
$$

$$
p_1-p_2=\gamma_w\,x\left[(1+s)\left(\frac{d}{D}\right)^{2}+(1-s)\right]
$$

With $s=0.83$:

$$
p_1-p_2=9810(0.01)\left[1.83(0.02531)+0.17\right]=98.1\times0.2163
$$

$$
p_1-p_2=21.2\ \text{Pa}
$$

**Answer: a pressure difference of about 21.2 Pa (2.16 mm of water) moves the interface by 1 cm.** The small value shows how sensitive the enlarged-end manometer is.

## Check yourself

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

1. For d = 7 mm, D = 44 mm and interface movement x = 1 cm, how far do free surfaces move?
   A. y = 0.01 x (44/7)^2
   B. y = 0.01 x (7/44)^2 = 2.53e-4 m
   C. y = 0.01 x 7/44
   D. y = 0.01 m
   Answer: B. y = x (d/D)^2 = 2.53 x 10^-4 m. (Key point: Movement in enlarged ends)

2. Which equation gives p1 - p2?
   A. gamma_w y + gamma_o y + (gamma_w - gamma_o) x
   B. (gamma_w + gamma_o) x
   C. gamma_w y - gamma_o y
   D. (gamma_w - gamma_o) y only
   Answer: A. The free surfaces' relative shift through both liquids plus the interface shift with density difference. (Key point: Pressure balance)

3. Which simplified form results?
   A. gamma_w x [(1 - s)(d/D)^2 + (1 + s)]
   B. gamma_w x [(1 + s)(d/D)^2 + (1 - s)]
   C. gamma_w x (1 - s)
   D. gamma_w x (1 + s)
   Answer: B. Substituting y gives gamma_w x[(1 + s)(d/D)^2 + (1 - s)]. (Key point: Simplified form)

4. What pressure difference moves the interface 1 cm?
   A. About 212 Pa
   B. About 2.1 Pa
   C. About 21.2 Pa
   D. About 2120 Pa
   Answer: C. 98.1 x [1.83 x 0.02531 + 0.17] = 21.2 Pa, about 2.16 mm of water. (Key point: Result)

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