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Pressure Measurement

Chapter 3: Pressure Measurement

Course: Prep4Uni Fluid Mechanics 1

Chapter 1: Pressure
Chapter 2: Variation of Pressure with Depth
Chapter 3: Pressure Measurement
Chapter 4: Buoyant Forces & Archimedes’ Principle
Chapter 5: Fluid Dynamics
Chapter 6: Bernoulli’s Equation
Chapter 7: Applications of Fluid Dynamics

🚁Overview

Overview:
Explore common pressure‐measurement devices—manometers, Bourdon gauges, and piezoelectric sensors—and how to read them.

📖Contents

  1. Definition
  2. SI Unit
  3. Gauge Vs Absolute Pressure
  4. Characteristics of Pressure in Fluids
  5. Example Calculation
  6. Key Formula Recap

🎯Learning Outcomes

Learning Outcomes: Chapter 3 – Pressure Measurement

Learning Outcomes

By the end of Chapter 3: Pressure Measurement, students will be able to:

  1. Describe the working principles of manometers: Explain how U-tube and differential manometers use hydrostatic balance to measure pressure differences.
  2. Read and interpret manometer readings: Calculate pressure differences from measured height differences (Δh) using Δp = ρ·g·Δh.
  3. Explain Bourdon gauge operation: Describe how a curved, flattened tube responds to internal pressure and translates motion into a pointer reading.
  4. Understand electronic (piezoelectric) sensors: Explain how mechanical stress on a piezoelectric crystal generates an electrical signal proportional to pressure.
  5. Compare mechanical vs. electronic methods: Discuss advantages, limitations, accuracy, and response time of Bourdon gauges versus piezoelectric and other electronic sensors.
  6. Summarize key points (Section 3.5): Recall major considerations in pressure measurement, including fluid selection, device calibration, and sources of error.
  7. Select appropriate devices: Choose between manometers, Bourdon gauges, and electronic sensors for given pressure ranges and application requirements.
  8. Perform unit conversions: Convert between common pressure units (Pa, kPa, bar, psi, mmHg) in the context of device readings.
 

Table of Contents

3.1 Manometers

Manometers measure pressure by the height difference of a liquid column in a tube:

  • U-tube manometer:
    A U-shaped tube partially filled with a reference fluid.Pressure difference:
    Δp = ρ g Δh,
    where Δh is the vertical height difference between the two arms.

Here are the details on Working Principle of a U-Tube Manometer

Device description

A U-shaped tube is partially filled with a reference fluid of known density ρ.
One limb is connected to the pressure source P1, and the other limb to a second pressure source P2 (often atmospheric pressure).

Hydrostatic equilibrium

In steady state, the pressure at any horizontal level within the same continuous fluid is the same.
Let the fluid columns stand at heights h1 (under P1) and h2 (under P2) measured from a common datum.

Pressure balance

Writing a hydrostatic balance from the free surface under P1 down to the datum and then up to the free surface under P2 gives:

P1 + ρ·g·(h1 – 0) = P2 + ρ·g·(h2 – 0)

Rearranging, the pressure difference is:

P1 – P2 = ρ·g·(h2 – h1) = ρ·g·Δh

where Δh = h2 – h1 is the measurable height difference.

Gauge vs. differential use

    • If P2 = atmospheric pressure Patm, the device reads gauge pressure:
      Pgauge = P1 – Patm = ρ·g·Δh
    • If both ends see unknown pressures, the same relation gives the differential pressure P1 – P2.
      Reading the manometer
  1. Observe the two fluid surfaces and measure the vertical height difference Δh.
  2. Calculate the pressure difference: ΔP = ρ·g·Δh.
  3. Convert the result into desired pressure units (e.g. Pa, psi).
U-Tube Manometer
U-Tube Manometer

Key points

    • Density dependence: A denser reference fluid (e.g., mercury) yields a smaller Δh for the same pressure difference.
    • Fluid immiscibility: In differential manometers using two fluids, account for both densities in the hydrostatic balance.
    • Accuracy: Temperature changes and capillary effects can introduce errors. Use wider tubes and temperature control for high-precision measurements.

Summary on Principles of U-tube Manometer

A U-tube manometer converts a pressure difference into a measurable height difference via hydrostatic equilibrium.
The fundamental relation is:

ΔP = ρ·g·Δh

This simple yet reliable device is widely used for measuring gauge and differential pressures in laboratories and industrial applications.

  • Inclined-tube manometer:
    One limb is tilted to increase sensitivity.For a tilt angle θ,
    Δp = ρ g (Δh cos θ),
    using the inclined‐column reading.

Example of Pressure Measurement using a U-Tube Manometer

Application of U-Tube Manometer
Application of U-Tube Manometer

This is an U-tube (or open tube) manometer. One end (shown on the right) of the U-shaped tube is open to the atmosphere. The other end (shown on the left) is connected to the pressure to be measured. The Pressure at point B is P0 + ρgh.

The pressures at points A and B must be the same. The pressure at A is the unknown pressure of the gas. Equating the unknown pressure P to the pressure at point B: P = P0 + ρgh. The difference in pressure P – P0  is equal to ρgh. Pressure P is called the absolute pressure, and the difference P – P0 is called the gauge pressure.

3.2 Bourdon Gauges

Bourdon gauges convert internal pressure into mechanical pointer movement:

  • A curved, hollow metal tube straightens as internal pressure rises.
  • The tube’s end is linked to a geared pointer that sweeps over a calibrated dial.
  • Suitable for industrial ranges from kPa to several MPa.
Bourdon Tube Pressure Gauge
Bourdon Tube Pressure Gauge

When the pressure increases, the circular tube opens and pulls the base of the y-shape level up, causes it to rotate anti-clockwise, turns the dial clockwise, indicating an increase in pressure.

Method Variation

Instead of using the spring effect of a Bourdon tube, we can also use spring compression to measure pressure.

Spring Manometer

The spring manometer shown above measures liquid pressure by balancing the fluid‐induced force against the restoring force of a helical spring.

When a pressure P is applied over a cross-sectional area A, it produces a force
Ffluid = P·A. The spring exerts an opposing force
Fspring = k·x,
where k is the spring constant and x is the compression length. At equilibrium:

P·A = k·x

Rearranging gives the theoretical expression for pressure:

P = (k·x) / A

In practice, the manometer is calibrated so that each measured compression x corresponds directly to a pressure reading on the graduated scale, accounting for any non-ideal effects.

3.3 Electronic Sensors

Electronic sensors use electrical transduction for precise, digital pressure readings:

  • Piezoelectric sensors:
    Certain crystals (e.g. PZT which stands for lead zirconate titanate) generate a voltage when stressed by pressure.
  • Strain-gauge sensors:
    A diaphragm’s flex changes the resistance of bonded strain gauges, producing a voltage change.
Strain Gauge
Strain Gauge

3.4 Mechanical vs. Electronic

  • Mechanical devices (manometers, Bourdon):
    • No power required
    • Very robust
    • Readout is analog (liquid column or dial)
  • Electronic sensors (piezo, strain gauge):
    • Require excitation voltage
    • High precision & repeatability
    • Digital output & easy integration

3.5 Key Points Recap

DeviceOperating PrincipleTypical Use
U-tube manometerFluid‐column height ΔhLab & low‐range measurements
Inclined manometerInclined‐column Δh cos θHigh‐sensitivity readings
Bourdon gaugeTube‐deflection → pointerIndustrial & portable gauges
Piezoelectric sensorCrystal voltage under stressDynamic & transient pressures
Strain-gauge sensorResistance change in diaphragmDigital systems & data logging

Proceed to Chapter 4: Buoyant Forces & Archimedes’ Principle

Return to: Prep4Uni Fluid Mechanics 1

📝EXERCISES

Pressure Measurement Exercises

Conceptual Questions

  1. What is pressure and what is its SI unit?

    Pressure is the normal force F applied per unit area A, i.e. p = F/A. Its SI unit is the Pascal (Pa), where 1 Pa = 1 N/m2.

  2. Distinguish between absolute pressure and gauge pressure.

    Absolute pressure (pabs) is measured relative to a perfect vacuum. Gauge pressure (pgauge) is measured relative to local atmospheric pressure: pgauge = pabs – patm.

  3. State the hydrostatic pressure equation.

    p = p0 + ρ · g · h, where ρ is fluid density, g is gravitational acceleration, and h is depth below the reference surface.

  4. How does a U-tube manometer measure a pressure difference?

    By reading the height difference Δh between the two fluid columns and using p = ρ · g · Δh.

  5. Why is mercury often used in barometers instead of water?

    Mercury’s high density (~13 600 kg/m3) yields a manageable column height (~760 mm) for atmospheric pressure, whereas water would require ~10 m.

  6. Define differential pressure and give one example of its measurement.

    Differential pressure is the difference between two pressures, e.g., a U-tube manometer measuring p1 – p2.

  7. What is the principle of a Bourdon gauge?

    A curved, flattened tube straightens when internal pressure rises; its tip movement is linked to a pointer indicating pressure.

  8. Convert 2 bar into Pascals and psi.

    2 bar = 2 × 105 Pa; 2 bar ≈ 29.0 psi.

  9. Explain why capillary effects can affect manometer readings.

    Surface tension in narrow tubes causes the fluid meniscus to rise or fall, altering the apparent Δh.

  10. When does a pressure transducer measure a negative gauge pressure?

    When the measured pressure is below atmospheric pressure, e.g., in vacuum systems.

  11. Express 1 atm in kPa and mmHg.

    1 atm = 101.325 kPa ≈ 760 mmHg.

  12. How does temperature affect pressure measurement in a closed system?

    For a fixed-volume ideal gas, p ∝ T (Kelvin), so temperature changes cause pressure variations.

  13. List two advantages of digital pressure sensors over mechanical gauges.

    Higher precision; easy integration with data-logging and control systems.

  14. What is a dead-weight tester and how is it used?

    A device applying known masses on a piston of known area to generate a reference pressure: p = weight/area.

  15. Why must manometric fluids be immiscible with the measured fluid?

    To maintain a sharp interface and accurate height reading without mixing.

  16. Define vacuum in terms of pressure.

    Any pressure below atmospheric; often expressed as a negative gauge value.

  17. What role does fine calibration play in precision manometers?

    Provides fine gradations to read small Δh changes accurately.

  18. How does a piezoelectric pressure sensor generate a signal?

    Mechanical stress on a piezoelectric crystal produces an electric charge proportional to pressure.

  19. In what scenario would you prefer a differential pressure sensor over two independent gauges?

    When measuring small pressure drops (e.g., across a filter) to improve sensitivity and reject common-mode variations.

  20. Why is it important to zero a gauge before measurement?

    To remove any offset (spring preload or drift) and ensure accuracy.

Calculation Problems

  1. A U-tube manometer with water (ρ = 1000 kg/m3) shows Δh = 0.15 m. Find the pressure difference Δp.

    Δp = ρ · g · Δh = 1000 × 9.81 × 0.15 = 1471.5 Pa.

  2. Convert a gauge reading of 250 kPa to absolute pressure at sea level (patm=101.325 kPa).

    pabs = 250 + 101.325 = 351.325 kPa.

  3. A mercury barometer reads 0.760 m of Hg (ρHg=13 600 kg/m3). Compute patm.

    p = ρ · g · h = 13600 × 9.81 × 0.760 ≈ 101300 Pa.

  4. A pressure gauge reads 50 psi. Convert this to kPa (1 psi = 6.895 kPa).

    50 × 6.895 = 344.75 kPa.

  5. In a differential manometer, one limb contains oil (ρ=850 kg/m3) and the other water. The water side is 0.10 m higher than the oil side. Find Δp.

    Δp = 850 × 9.81 × 0.10 = 833.85 Pa.

  6. A tank has water depth of 6 m. Find the pressure at the bottom.

    p = 1000 × 9.81 × 6 = 58860 Pa.

  7. A dead-weight tester uses a 5 kg mass on a 2 cm2 piston. Compute the applied pressure.

    Area = 2 cm2 = 2 × 10–4 m2; force = 5 × 9.81 = 49.05 N; p = 49.05 / 2×10–4 = 245250 Pa.

  8. A sensor outputs 0 V at 0 kPa and 5 V at 500 kPa. What pressure corresponds to 3 V?

    p = (3/5) × 500 = 300 kPa.

  9. A U-tube manometer with ρ=1200 kg/m3 shows hleft=0.12 m and hright=0.18 m. Compute Δp.

    Δh = 0.18 – 0.12 = 0.06 m; Δp = 1200 × 9.81 × 0.06 = 706.32 Pa.

  10. A gauge reads –30 kPa (gauge). What is the absolute pressure?

    pabs = 101.325 – 30 = 71.325 kPa.

  11. Convert 760 mmHg into meters of water (ρ=1000 kg/m3).

    First, p = 0.760 mHg × 13600 × 9.81 = 101300 Pa; hwater = 101300/(1000 × 9.81) ≈ 10.33 m.

  12. A U-tube uses mercury and water. The mercury column difference is 0.010 m. Find Δp.

    Δp = 13600 × 9.81 × 0.010 = 1334.16 Pa.

  13. A 1 bar-span gauge outputs 4–20 mA. What current corresponds to 0.5 bar gauge?

    Slope = 16 mA/bar; I = 4 + 16 × 0.5 = 12 mA.

  14. Atmospheric pressure drops to 95 kPa. What height of water column is equivalent?

    h = 95000 / (1000 × 9.81) ≈ 9.69 m.

  15. Estimate capillary rise in a 2 mm tube (γ=0.072 N/m, θ=0°).

    h = 4γ/(ρ g D) = 4×0.072/(1000×9.81×0.002) ≈ 0.0147 m (14.7 mm).

  16. A piston-cylinder sees 200 kPa gauge. If piston area is 0.05 m2, find the force.

    F = p·A = 200000 × 0.05 = 10000 N.

  17. Water at 20 °C (ρ=998 kg/m3) fills a manometer with Δh=0.25 m. Compute Δp.

    Δp = 998 × 9.81 × 0.25 ≈ 2448 Pa.

  18. A Bourdon gauge shifts zero by –2 kPa after test. Explain.

    The pointer reads 2 kPa low when vented, indicating a zero-offset error of –2 kPa.

  19. A chamber evacuated to –90 kPa gauge. What fraction of atm remains?

    (101.325 – 90)/101.325 ≈ 0.111 or 11.1 %.

  20. A differential piezo sensor: 30 kPa → 0.6 V. What is sensitivity?

    Sensitivity = 600 mV/30 kPa = 20 mV/kPa.