Pressure measurement: manometers, Bourdon tube, bellows, diaphragms

Why it matters

Almost every process plant variable is either a pressure or is inferred from one: flow from a differential pressure, level from a hydrostatic head, and many temperatures from filled-system pressure. Boilers, reactors, compressors and pipelines are protected by pressure gauges and switches, so knowing how each primary element works, its range and its errors is basic plant knowledge. Manometers are still the reference standard for low pressures, and the elastic elements (Bourdon tube, bellows, diaphragm) are the sensing heart of most gauges and transmitters.

Key ideas

Pressure references.

• Absolute pressure is measured from perfect vacuum; gauge pressure from local atmospheric pressure; differential pressure is the difference between two process points.
• p_abs = p_gauge + p_atm. A negative gauge pressure is called vacuum (suction).
• 1 bar = 10⁵ Pa, 1 atm = 101.325 kPa, 1 psi ≈ 6.895 kPa, 1 mm Hg ≈ 133.3 Pa (at standard density and g).

Manometers balance the unknown pressure against a liquid column (hydrostatics, p = ρgh).

• U-tube: the reading is the vertical difference h between the two menisci. If the pipe fluid (density ρf) fills the legs above the manometer liquid (ρm), the effective density is (ρm − ρf), not ρm.
• Well (cistern) type: one leg is a large well of area A, the other a narrow tube of area a. Only one level is read; the small fall in the well is corrected by the factor (1 + a/A) or, when a/A is small, ignored.
• Inclined manometer: the reading leg is tilted at angle θ to the horizontal. A vertical rise h becomes a scale length L = h/sinθ, so resolution improves by 1/sinθ. Used for draft and duct pressures of a few hundred pascals.
• Manometer liquids: mercury for large ranges, water or coloured oil for small ones. Errors come from temperature (density changes), capillarity (use tubes wider than about 6 mm or read both menisci), and local g.

Elastic elements convert pressure into displacement (or force), which then drives a pointer, a potentiometer, an LVDT, a strain gauge or a capacitance cell.

• Bourdon tube: a tube of flattened (oval) cross-section bent into a C, spiral or helix, sealed at the free tip. Internal pressure tends to make the cross-section circular, which tends to straighten the tube; the tip moves roughly in proportion to gauge pressure. A linkage, sector and pinion amplify the tip motion. Ranges from about 0.5 bar to several thousand bar (C-type); spiral and helical forms give more tip travel. Not suited to very low pressures.
• Bellows: a thin-walled corrugated cylinder. Pressure acting on its effective area produces a force; the bellows (often with a range spring) deflects in proportion. Large travel at low pressure, so used for low pressures and differential pressures (a few kPa to a few bar). Sensitive to overpressure and to fatigue.
• Diaphragm: a thin flat or corrugated plate clamped at its rim. Deflection is small, so it is normally read electrically (strain gauge, capacitance, piezoresistive). Corrugations increase linear travel. Diaphragms also serve as seals that isolate the sensing element from corrosive or viscous process fluid.
• Linearity: elastic elements are linear only for small deflections. A flat diaphragm is reasonably linear while its centre deflection stays below about 30 % of its thickness; beyond that membrane stretching stiffens it.
• Elastic errors: hysteresis, drift (creep), temperature effect on modulus, and permanent set after overpressure. Gauges are normally chosen so the working pressure sits in the middle third of the scale.

Formulas

p_abs = p_gauge + p_atm

Δp = ρ·g·h (manometer liquid with gas above both legs)

Δp = (ρm − ρf)·g·h (U-tube with process liquid of density ρf above the manometer liquid)

Δp = ρ·g·h·(1 + a/A) (well-type, h = rise in the narrow tube)

Δp = ρ·g·L·(sinθ + a/A) (inclined well-type; L = length moved along the inclined tube; drop the a/A term for an ideal large well)

x = p·Ae / k (bellows or spring-opposed element)

y₀ = 3·p·R⁴·(1 − ν²) / (16·E·t³) (centre deflection of a flat circular diaphragm, clamped edge)

Symbols: p, Δp = pressure or pressure difference (Pa); ρ, ρm, ρf = densities (kg/m³); g = 9.81 m/s²; h = vertical column difference (m); L = inclined length (m); θ = tube angle from the horizontal; a, A = tube and well areas (m²); x = deflection (m); Ae = effective area of bellows (m²); k = spring rate (N/m); y₀ = centre deflection (m); R = diaphragm radius (m); t = thickness (m); E = Young's modulus (Pa); ν = Poisson's ratio. The diaphragm formula assumes small deflection (y₀ < about 0.3 t), uniform pressure and a fully clamped edge.

Worked examples

Example 1 (standard): U-tube with water above mercury. Given: a mercury U-tube is connected across two points of a water pipe; reading h = 150 mm; ρm = 13 600 kg/m³, ρf = 1000 kg/m³, g = 9.81 m/s².

1. Water fills both legs above the mercury, so Δp = (ρm − ρf)·g·h.
2. ρm − ρf = 13 600 − 1000 = 12 600 kg/m³.
3. Δp = 12 600 × 9.81 × 0.150 = 18 540.9 Pa.
4. Answer: Δp ≈ 18.5 kPa. Using ρm alone would give 20.0 kPa, an error of about 8 %.

Example 2 (GATE level): diaphragm deflection. Given: a flat steel diaphragm, clamped edge, radius R = 10 mm, thickness t = 0.5 mm, E = 200 GPa, ν = 0.3, pressure p = 100 kPa. Find the centre deflection and check the small-deflection assumption.

1. y₀ = 3·p·R⁴·(1 − ν²) / (16·E·t³).
2. Numerator: 3 × 1.0 × 10⁵ × (0.010)⁴ × 0.91 = 3 × 10⁵ × 1.0 × 10⁻⁸ × 0.91 = 2.73 × 10⁻³ N·m².
3. Denominator: 16 × 200 × 10⁹ × (0.5 × 10⁻³)³ = 3.2 × 10¹² × 1.25 × 10⁻¹⁰ = 400 N·m.
4. y₀ = 2.73 × 10⁻³ / 400 = 6.83 × 10⁻⁶ m.
5. Check: y₀/t = 6.83 × 10⁻⁶ / 0.5 × 10⁻³ ≈ 0.014, well below 0.3, so the linear formula is valid.
6. Answer: y₀ ≈ 6.8 µm. Such a small motion is why diaphragms are read by strain gauges or capacitance rather than a pointer.

Example 3 (inclined manometer). Given: inclined well-type manometer, θ = 20°, oil ρ = 850 kg/m³, a/A = 0.01, the liquid moves L = 120 mm along the tube.

1. Δp = ρ·g·L·(sinθ + a/A); sin 20° = 0.342.
2. Δp = 850 × 9.81 × 0.120 × (0.342 + 0.010) = 1000.6 × 0.352 = 352 Pa.
3. Answer: Δp ≈ 352 Pa (342 Pa if the well fall is ignored, about 3 % low).

Example 4 (bellows). A bellows of effective area 5 cm² works against a spring of 2500 N/m. For p = 20 kPa: force = 20 000 × 5 × 10⁻⁴ = 10 N; x = 10/2500 = 4 mm.

Common mistakes

• Using ρm instead of (ρm − ρf) when the manometer legs are filled with a liquid process fluid.
• Reading the length along an inclined tube as if it were the vertical height.
• Mixing gauge and absolute pressure, especially in vacuum problems and gas-law calculations.
• Forgetting the R⁴ and t³ dependence of diaphragm deflection: doubling the thickness reduces deflection eightfold.
• Thinking a Bourdon tube curls tighter under pressure; it tends to straighten.
• Choosing a gauge whose working point is near full scale, which shortens its life and increases hysteresis error.
• Ignoring temperature: manometer liquid density and elastic modulus both change with temperature.

For GATE IN

Expect short numericals on manometers (U-tube with two fluids, inclined and well-type corrections), unit conversions between bar, Pa, mm Hg and gauge/absolute, and diaphragm or bellows deflection with a given formula. Conceptual questions ask which element suits which range, what makes a Bourdon tube move, and how elastic elements are coupled to secondary transducers (LVDT, strain gauge, capacitance). Practise writing the hydrostatic balance carefully from one meniscus to the other.

Quick check

1. A mercury manometer has water above the mercury in both legs and reads 100 mm. What is Δp?
2. Why does an inclined manometer improve resolution?
3. What happens to diaphragm centre deflection if its radius is doubled?
4. Which elastic element is preferred for a 0–5 kPa differential pressure: Bourdon tube or bellows?
5. A gauge reads 250 kPa where p_atm = 101 kPa. What is the absolute pressure?

Answers: 1. 12 600 × 9.81 × 0.1 ≈ 12.4 kPa. 2. The scale length is h/sinθ, larger than h for θ < 90°. 3. It increases 16 times (R⁴). 4. Bellows. 5. 351 kPa.