Mechanical Properties of Fluids · Lesson 2 of 8
Measuring Pressure and Hydraulic Machines
“Measure absolute and gauge pressure, balance liquid columns, and explain force, movement and work in hydraulic systems.”
• Distinguish absolute pressure, gauge pressure and pressure difference. • Explain a mercury barometer and solve manometer and layered-liquid balances. • Apply Pascal’s transmission principle to hydraulic machines. • Relate hydraulic force multiplication to displacement and work conservation.
Atmospheric pressure and choosing a reference
Air has weight. At a given level, a unit-area vertical column of air above that level must be supported by the pressure below it. Sea-level atmospheric pressure is approximately 1.013 × 10⁵ Pa, called one standard atmosphere. Local atmospheric pressure changes with weather and altitude. Air is denser near the ground than high above it, so its pressure cannot be found accurately over the whole atmosphere by multiplying a single sea-level density by the full atmospheric height.
A tyre gauge generally reads the pressure above the surrounding atmosphere. Blood-pressure instruments also report gauge pressure. An open container’s surface has zero gauge pressure but a nonzero absolute pressure. For a net force across a window, use the pressure on one side minus that on the other. Multiplying outside absolute pressure by area alone ignores the force exerted from inside.
| Unit | Approximate pressure |
|---|---|
| 1 atm | 1.013 × 10⁵ Pa |
| 1 bar | 10⁵ Pa |
| 1 millibar | 100 Pa |
| 1 torr, approximately 1 mm Hg | 133 Pa |
| 1 Pa | 1 N/m² |
Problem
Assume air density stays at its sea-level value 1.29 kg/m³. What thickness of this hypothetical atmosphere produces pressure 1.01 × 10⁵ Pa? Take g = 9.8 m/s².
- 1.Set the pressure at the hypothetical top to zero. Hydrostatic balance gives P_atm = ρgh, so h = P_atm/(ρg).
- 2.h = 101000/(1.29 × 9.8) = 7989 m, approximately 8.0 km. This is an equivalent uniform-density height, not the actual atmosphere’s upper boundary.
- 3.In reality air density falls with altitude and a thin atmosphere extends beyond 100 km. A large fraction of its mass lies near Earth. Pressure can therefore fall to about half its sea-level value around 6 km without implying that half the atmosphere’s geometric height has been traversed.
A barometer balances atmosphere against a liquid column
Torricelli filled a closed-end glass tube with mercury and inverted its open end into a mercury trough. The column falls a little, leaving a nearly empty space above it. Compare a point in the tube at the trough’s free-surface level with a neighbouring point in the trough: equal-height points in the connected mercury have equal pressure. The column is supported by atmospheric pressure on the trough, not by a mysterious upward pull of vacuum.
Mercury’s high density gives a convenient column of about 0.76 m at sea level. A much less dense liquid needs a much taller column. If the tube is wide enough for capillary effects to be negligible, changing its bore does not change the hydrostatic height. A real barometer needs attention to liquid density, temperature, vapour pressure and local g when high accuracy matters.
Problem
Find the column height for wine of density 984 kg/m³ under normal atmospheric pressure 1.013 × 10⁵ Pa. Take g = 9.8 m/s² and neglect vapour pressure.
- 1.Use h = P_atm/(ρg), the same balance used for mercury.
- 2.h = 101300/(984 × 9.8) = 10.50 m, or approximately 10.5 m. The tall column follows from wine’s low density, not a larger atmospheric pressure.
- 3.Mercury is about fourteen times denser than this wine, so a mercury column is correspondingly much shorter. The chosen liquid’s vapour pressure limits the accuracy of this simplified comparison.
A manometer compares pressures by following liquid levels
An open-tube manometer is a U-tube containing liquid, with one arm connected to a gas and the other open to atmosphere. If the gas pushes its liquid surface lower by a vertical difference Δh relative to the open arm, compare pressures along a horizontal line in the same connected liquid. Moving downward increases pressure by ρg times the depth; moving upward decreases it by the same amount.
Use a low-density liquid for a small pressure difference: it creates a larger, easier-to-read height difference. Mercury permits large pressure differences without an extremely tall apparatus. With several immiscible liquids, follow a single continuous path from one known pressure to the other. Add ρg times each downward change and subtract it for each upward change, using the density of the layer actually traversed. Equal height alone does not guarantee equal pressures in disconnected liquids or in different liquids separated by an interface.
Problem
Both arms of an open U-tube contain mercury below a liquid. One arm has a 10.0 cm water layer, the other a 12.5 cm spirit layer. The mercury interfaces are level. Find spirit relative density; then find the mercury-level separation if a further 15.0 cm of each liquid is added. Use mercury relative density 13.6 and equal uniform arm areas.
- 1.Initially the two mercury interfaces have the same pressure. The common atmospheric pressure cancels, leaving ρ_water × 0.100 = ρ_spirit × 0.125. Thus ρ_spirit/ρ_water = 0.800.
- 2.After addition the liquid-layer thicknesses are 0.250 m water and 0.275 m spirit. Each layer’s thickness is fixed by its volume and the uniform tube area, even as the mercury interfaces shift.
- 3.The water-side interface pressure exceeds the spirit-side interface pressure by g(1000 × 0.250 − 800 × 0.275) = 30g Pa. The water side pushes mercury down and the spirit side rises.
- 4.If δ is the separation of the mercury interfaces, the mercury pressure difference is 13600gδ. Hence δ = 30/13600 = 0.002206 m = 0.221 cm. Each interface moves half this distance for equal arm areas; 0.221 cm is their separation, not each movement.
Submerged structures: outside pressure versus net loading
Problem
A submarine is 1000 m below sea level in water of density 1.03 × 10³ kg/m³. Its cabin remains at 1.01 × 10⁵ Pa. Find outside absolute pressure, outside gauge pressure and net inward force on a 20 cm by 20 cm window. Use g = 10 m/s².
- 1.Gauge pressure is ρgh = 1.03 × 10³ × 10 × 1000 = 1.03 × 10⁷ Pa.
- 2.Outside absolute pressure = 1.03 × 10⁷ + 1.01 × 10⁵ = 1.0401 × 10⁷ Pa. The numbers correspond roughly to 103 and 104 atmospheres when 10⁵ Pa is used for a quick atmosphere estimate.
- 3.The cabin pressure equals the outside surface pressure. The net pressure across the window is therefore 1.03 × 10⁷ Pa, not the full outside absolute pressure.
- 4.Window area = 0.20 × 0.20 = 0.040 m². Net inward force = 1.03 × 10⁷ × 0.040 = 4.12 × 10⁵ N. Constant sea-water density and uniform pressure over the small window are approximations.
A 3 km ocean depth similarly gives a hydrostatic pressure of order 3 × 10⁷ Pa. Compared only with an allowed stress of 10⁹ Pa, this is much smaller. Such a textbook comparison can answer a simplified suitability question. Real engineering also needs geometry, stability, corrosion, currents and appropriate stress analysis; pressure and the maximum stress in a structure are not automatically the same everywhere.
Pascal’s transmission principle and hydraulic force multiplication
Push on the piston of a confined resting liquid. After the fluid settles, the added pressure is transmitted throughout the liquid and to its container walls without diminution in the ideal static model. This does not remove the original depth-dependent pressure difference. It means that an imposed pressure increment is added equally. In a cylinder connected to several vertical indicating tubes, pushing the piston raises the equal-level columns together.
A hydraulic lift connects a small piston of area A₁ to a larger piston of area A₂. For pistons at equal elevation, negligible piston weights and friction, and the same outside atmosphere, applied forces produce equal gauge pressures. Atmospheric forces cancel when one writes the net piston balance. If the piston elevations differ appreciably, include the hydrostatic correction rather than blindly equating their total pressures.
Problem
Water-filled syringes have piston diameters 1.0 cm and 3.0 cm. A 10 N force pushes the smaller piston in by 6.0 cm. Find the larger piston’s force and outward displacement.
- 1.For circular pistons A₂/A₁ = (d₂/d₁)² = (3.0/1.0)² = 9.
- 2.The transmitted pressure gives F₂ = 9 × 10 = 90 N.
- 3.Volume conservation gives x₂ = x₁/9 = 6.0/9 cm = 0.67 cm. The volume moved into one syringe leaves the other.
- 4.Input work is 10 × 0.060 = 0.60 J, equal to 90 × 0.00667 ≈ 0.60 J. The gain in force is accompanied by a loss in distance.
Problem
A car of mass 1350 kg rests on a piston of radius 15 cm. A connected input piston has radius 5.0 cm. Find the required input force and hydraulic gauge pressure, using g = 9.8 m/s².
- 1.The output piston must support F₂ = 1350 × 9.8 = 13230 N. The area ratio is (15/5)² = 9.
- 2.Required input force F₁ = 13230/9 = 1470 N, approximately 1.5 × 10³ N.
- 3.Gauge pressure ΔP = 1470/[π(0.050)²] = 1.87 × 10⁵ Pa, approximately 1.9 × 10⁵ Pa. It is also F₂/[π(0.15)²].
- 4.This is the required pressure above the common outside atmosphere. If atmospheric pressure is 1.01 × 10⁵ Pa, the corresponding fluid absolute pressure is approximately 2.88 × 10⁵ Pa. Keep the reference explicit.
Hydraulic brakes use the same idea. The pedal drives a master piston; brake fluid transmits pressure to wheel pistons, which press friction surfaces together. Equal pressure is available through the connected fluid, while wheel force depends on each piston’s area. Equal braking forces require appropriate matching of piston areas and mechanisms; equal transmitted pressure by itself does not guarantee every possible brake design has identical wheel forces.
Quick Quiz
Quiz
A gauge reads 2.0 × 10⁵ Pa above an atmosphere of 1.0 × 10⁵ Pa. What is absolute pressure?
Absolute pressure is atmospheric plus gauge pressure.
A manometer’s gas-side liquid surface is lower than its atmospheric-side surface. Which is true?
Higher gas pressure pushes its liquid surface down; the vertical liquid head balances the pressure excess.
A lift’s output piston diameter is four times its input diameter. Its ideal force gain is what?
Area is proportional to diameter squared: the area and force ratio is 4² = 16.
Why does a wine barometer need a taller column than a mercury barometer?
For a given pressure, h = P/(ρg); smaller density requires greater height.
An ideal hydraulic output force is ten times the input force. How far does the output move for an input displacement of 20 cm?
Incompressible-volume conservation makes the displacement ratio inverse to the force ratio: 20/10 = 2 cm.
Practice Problems
- An open water manometer shows a 12.0 cm level difference with the gas-side surface higher. Find gas gauge and absolute pressures for ρ = 1000 kg/m³, g = 9.8 m/s² and P_atm = 1.01 × 10⁵ Pa. Explain the sign.
- A barometer contains a liquid of density 800 kg/m³. With negligible pressure above the column, what vertical height balances 1.00 × 10⁵ Pa? Use g = 10 m/s². Explain whether tilting a sufficiently long tube changes the vertical height.
- A hydraulic lift supports a 3000 kg car on an area of 425 cm². Find the required gauge pressure using g = 9.8 m/s². If the input piston area is 5.0 cm², find the input force.
- A submarine cabin is at 1.01 × 10⁵ Pa. It is submerged 200 m in water of density 1030 kg/m³. For a flat 0.030 m² window, find outside absolute pressure and net inward force, using g = 9.8 m/s². State the approximation concerning pressure variation over the window.
- Open arms of a U-tube contain level mercury interfaces below 20.0 cm of water and a spirit column. Spirit density is 800 kg/m³. Find the spirit depth. Then explain how to determine which mercury arm rises after adding 10.0 cm of each liquid, and calculate their separation using mercury density 13600 kg/m³.
Key Takeaways
• Use vacuum or local atmosphere consistently as the pressure zero. • Follow vertical steps through each liquid in manometer problems; keep the correct density and sign. • Pascal’s law transmits pressure increments; hydraulic force gain is an area ratio. • Ideal force multiplication costs displacement and preserves work; include both sides’ pressures when calculating net force.