---
title: "Piezometer levels in layered fluids"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_3/fluid_mechanics:piezometer-levels-layered-fluids"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/piezometer-levels-layered-fluids
site: BCT Engineering
---

# Piezometer levels in layered fluids

Fluid Mechanics · Chapter 2: Pressure and Head

- Asked in: 2075 Asoj, 8 marks
- Video: [piezometer || piezometer tube || piezometer in hindi || piezometer fluid mechanics](https://www.youtube.com/watch?v=l-kzHJKyKwQ) — AB CLASSES
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/piezometer-levels-layered-fluids
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2

## Question

In Fig. below, sensor A reads 1.5 kPa (gage). All fluids are at 20°C. Determine the elevations Z in meters of the liquid levels in the open piezometer tubes B and C. [Figure: closed tank with air (2 m deep) on top, gasoline (1.5 m) and glycerin (1 m) below; sensor A on the air; piezometer tube B connected at the gasoline layer and tube C at the glycerin layer; z = 0 at the tank bottom]

## Short answer

Using hydrostatic balance from the 1.5 kPa sensor, gasoline (γ = 6671 N/m³) rises to Z_B ≈ 2.72 m and glycerin (γ = 12 361 N/m³) to Z_C ≈ 1.93 m.

## Answer

**Fluid properties at 20 °C (White's table):** gasoline $\rho=680\ \text{kg/m}^3$, $\gamma_g=680\times9.81=6671\ \text{N/m}^3$; glycerin $\rho=1260\ \text{kg/m}^3$, $\gamma_{gl}=12\,361\ \text{N/m}^3$. Air weight is neglected.

**Reading of the figure:** $z=0$ at the tank bottom; glycerin 1 m deep ($z=0$ to 1), gasoline 1.5 m deep ($z=1$ to 2.5), air above; $p_A=1.5$ kPa gauge acts on the gasoline surface at $z=2.5$ m. Piezometer B joins the gasoline layer, C the glycerin layer.

```text
   A (1.5 kPa)  air
  ------------------ z = 2.5 m
   gasoline  <- tube B
  ------------------ z = 1.0 m
   glycerin  <- tube C
  ------------------ z = 0
```

#### Piezometer B (gasoline)
The pressure at the top of the gasoline is 1500 Pa. The gasoline rises in tube B to a height where the pressure is zero:

$$
Z_B=2.5+\frac{1500}{6671}=2.5+0.225=2.725\ \text{m}
$$

#### Piezometer C (glycerin)
Pressure at the gasoline-glycerin interface ($z=1$ m):

$$
p=1500+6671(1.5)=11\,507\ \text{Pa}
$$

The glycerin in tube C rises above $z=1$ m until the pressure is zero:

$$
Z_C=1+\frac{11\,507}{12\,361}=1+0.931=1.931\ \text{m}
$$

**Answer: $Z_B\approx2.72$ m and $Z_C\approx1.93$ m.**

## Check yourself

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

1. How high does liquid rise in an open piezometer?
   A. To where the pressure equals the pressure at the tapping
   B. To the top of the tank
   C. To the tank bottom
   D. To where the pressure is zero gauge, so height = p/gamma above the tapping
   Answer: D. The liquid column in the open tube must have zero gauge pressure at its top, so rise = p/gamma. (Key point: Piezometer principle)

2. What is Z_B for gasoline with the surface at 2.5 m and p = 1.5 kPa?
   A. 2.725 m
   B. 2.5 m
   C. 1.5 m
   D. 2.275 m
   Answer: A. Z_B = 2.5 + 1500/6671 = 2.725 m. (Key point: Tube B)

3. What is the pressure at the gasoline-glycerin interface?
   A. 1500 Pa
   B. 11 507 Pa
   C. 10 007 Pa
   D. 12 361 Pa
   Answer: B. 1500 + 6671 x 1.5 = 11 507 Pa. (Key point: Interface pressure)

4. How is Z_C found?
   A. Z_C = 1 + 11507/6671
   B. Z_C = 11507/12361
   C. Z_C = 1 + 11507/12361 = 1.93 m
   D. Z_C = 2.5 + 11507/12361
   Answer: C. Glycerin rises above z = 1 m by p/gamma_gl = 0.931 m. (Key point: Tube C)

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