Hydrostatic Pressure & Manometers

A fluid at rest cannot resist shear, so the only forces on a small piece of it are its weight and the pressure on its faces. Take a thin horizontal slab of area A and thickness dy, with y measured upward. The fluid below pushes up with P(y)A, the fluid above pushes down with P(y + dy)A, and the slab weighs ρgA dy. Equilibrium gives P(y) − P(y + dy) = ρg dy, which is the differential equation dP/dy = −ρg. Pressure falls as you rise and grows as you descend, at a rate set only by the local density. For a liquid of constant density, integrate from the free surface, where the pressure is P₀, down to depth h: P = P₀ + ρgh. Fresh water adds 9.81 kPa per metre, seawater at 1025 kg m⁻³ adds 10.1 kPa per metre, and mercury adds 133 kPa per metre.

Now take a thin horizontal cylinder of fluid. Its weight has no horizontal component, and the pressure on its curved side acts perpendicular to the axis, so the only forces along the axis are the pressures on its two end caps. Equilibrium makes them equal: two points at the same height, joined horizontally through the same fluid at rest, share one pressure. Chain horizontal and vertical steps and any path works, however it winds, through a U-bend, under a floor or round a corner, so long as it stays inside one continuous fluid. That is why connected open vessels of one liquid fill to the same level, and why P₂ − P₁ = ρg(y₁ − y₂) needs no information about the route. Both conditions matter. A closed valve breaks the path outright, and an interface with a second fluid means level points on either side need not agree; the path can still be walked, but each stretch uses its own density.

The P₀ in P = P₀ + ρgh is whatever presses on the free surface, usually the atmosphere at about 101.3 kPa. Most instruments read the excess over their surroundings, the gauge pressure, so P_abs = Pₐₜₘ + P_gauge; a tyre gauge showing 230 kPa means 331 kPa absolute. Which one a problem wants is a modelling decision with a rule behind it. When air presses on both faces of a wall, window or hatch, the atmospheric terms cancel and the net force comes from gauge pressure alone. When a question compares pressures as ratios, or involves something that responds to the total pressure, such as the volume of a gas bubble, use absolute. The two diverge sharply: in fresh water the absolute pressure doubles at a depth of Pₐₜₘ/ρg = 10.3 m, while the gauge pressure has only risen from zero to one atmosphere. Gauge pressure can be negative, as inside a drinking straw being sucked; the absolute pressure in an ordinary liquid cannot drop far below its vapour pressure before bubbles form.

dP/dh = ρg is a local statement, so when density changes with depth the pressure becomes an integral, P(h) = P₀ + g∫₀ʰ ρ(h′) dh′. For immiscible layers the integral is a sum. Float 0.60 m of oil of density 900 kg m⁻³ on 1.40 m of water: the oil contributes 900 × 9.81 × 0.60 = 5.30 kPa, the water 1000 × 9.81 × 1.40 = 13.73 kPa, and the floor sits at 19.03 kPa gauge. A pressure-depth graph shows the structure: a straight segment in each layer with slope ρg, and a kink at the interface where the density jumps. The pressure itself does not jump, because a flat interface has no mass to carry a net force. The floor value also follows from the depth-averaged density, (900 × 0.60 + 1000 × 1.40)/2.00 = 970 kg m⁻³, over the full 2.00 m. That shortcut is exact at the floor and wrong in between: 1.00 m down it gives 9.52 kPa, where the layers give 9.22 kPa.

A manometer turns the depth rule around: measure a height, infer a pressure. The reliable method is a walk. Start where the pressure is known, usually an open surface at Pₐₜₘ, and follow a continuous path to the unknown point, adding ρgΔh for each stretch you descend and subtracting it for each stretch you climb, always with the density of the fluid that stretch passes through. Cross between arms only through one liquid at one level. For an open U-tube with gas above the liquid on one side, the gas weighs almost nothing and the walk gives P_gas − Pₐₜₘ = ρₘ gΔh. The liquid sets the sensitivity: 1.00 kPa holds up 102 mm of water but only 7.5 mm of mercury. When a liquid of density ρ fills the lines above the manometer liquid, it sits in both arms and the reading becomes (ρₘ − ρ)gΔh: 50.0 mm of mercury under water means 6.18 kPa, not the 6.67 kPa you get by forgetting the water.

Every line above assumed a fluid at rest in a uniform g with a known density. Accelerate the container and each parcel needs a net force, so the free surface tilts and g must be replaced by the effective field g − a. Let the fluid flow and pressure differences also accelerate it, which is Bernoulli's territory. Constant density is excellent for liquids: even 4 km down in the ocean, compression raises seawater's density by only about 1.7%, and the pressure error stays under 1%. It fails for gases over large heights. Air at 1.20 kg m⁻³ loses 1.18 kPa over 100 m, a 1.2% change a building can ignore and a mountain cannot. Narrow tubes add surface tension: water climbs about 15 mm up a clean 2 mm bore, the size of a small manometer reading, so manometers use wide tubes, or identical bores in both arms so the rises cancel.