---
title: "U-tube gauge with enlarged ends"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_3/fluid_mechanics:u-tube-enlarged-ends-pressure-gauge"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/u-tube-enlarged-ends-pressure-gauge
site: BCT Engineering
---

# U-tube gauge with enlarged ends

Fluid Mechanics · Chapter 2: Pressure and Head

- Asked in: 2076 Chaitra, 6 marks
- Video: [Pressure Measurement Manometers](https://www.youtube.com/watch?v=oXXGIFSmZRE) — TutorialsPoint
- 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-pressure-gauge
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2

## Question

A pressure gauge consists of U tube with equal enlarged ends and is filled with water on one side and oil of specific gravity 0.97 on the other, the surface of separation being in the tube below the enlarged ends. Calculate the diameter of each enlarged end if the tube diameter is 5 mm and the surface of separation moves 25 mm for a difference in pressure head of 1 mm of water.

## Short answer

From volume displacement, (d/D)² = 0.005076, so D = 5/0.07125 = 70.2 mm. Each enlarged end diameter is about 70 mm.

## Answer

**Data:** tube $d=5$ mm, enlarged ends $D$ (equal), oil $S=0.97$, interface moves $x=25$ mm for a pressure difference of 1 mm of water.

**Set-up.** Water is in one limb and oil in the other, the interface lying in the tube. When a pressure difference $\Delta p$ is applied, the interface moves a distance $x$ in the tube. The displaced volume $\dfrac{\pi}{4}d^{2}x$ makes both free surfaces in the enlarged ends move by

$$
y=x\left(\frac{d}{D}\right)^{2}
$$

(the water surface falls $y$, the oil surface rises $y$). Before the pressure change the pressures balance at the initial interface level, so only the changes count. Balancing at the new interface level, with $s=0.97$:

$$
\frac{\Delta p}{\gamma_w}=y(1+s)+x(1-s)
$$

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

Substitute $\dfrac{\Delta p}{\gamma_w}=1$ mm, $x=25$ mm:

$$
1=25\left[1.97\left(\frac{d}{D}\right)^{2}+0.03\right]\ \Rightarrow\ \left(\frac{d}{D}\right)^{2}=\frac{0.04-0.03}{1.97}=0.005076
$$

$$
\frac{d}{D}=0.07125\ \Rightarrow\ D=\frac{5}{0.07125}=70.2\ \text{mm}
$$

**Answer: diameter of each enlarged end $\approx70$ mm.**

## Check yourself

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

1. If the interface moves x in the tube (diameter d), how far do the free surfaces move in the enlarged ends (diameter D)?
   A. y = x (d/D)^2
   B. y = x (D/d)^2
   C. y = x d/D
   D. y = x
   Answer: A. Equal volumes: (pi/4) d^2 x = (pi/4) D^2 y. (Key point: Surface movement in ends)

2. Which equation relates pressure head to the movements (s = oil S.G.)?
   A. Delta p/gamma_w = y(1 - s) + x(1 + s)
   B. Delta p/gamma_w = y(1 + s) + x(1 - s)
   C. Delta p/gamma_w = (x + y)(1 + s)
   D. Delta p/gamma_w = x y s
   Answer: B. The free surfaces move y apart through both liquids, and the interface shifts x with density difference (1 - s). (Key point: Balance equation)

3. With 1 = 25[1.97 (d/D)^2 + 0.03], what is (d/D)^2?
   A. 0.0051
   B. 0.0407
   C. 0.0051 x 25
   D. 0.51
   Answer: A. (0.04 - 0.03)/1.97 = 0.005076. (Key point: Diameter ratio)

4. What is D for d = 5 mm and d/D = 0.07125?
   A. 7.1 mm
   B. 35 mm
   C. 140 mm
   D. 70 mm
   Answer: D. D = 5/0.07125 = 70.2 mm. (Key point: Enlarged end diameter)

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