Fluid Mechanics & Hydraulics

Hydrostatic Pressure & Manometers

Pressure change through static fluid columns by the manometer traverse method: start at a known pressure and add γ·Δz moving down, subtract γ·Δz moving up, leg by leg (each leg its own fluid — water, mercury, oil by SG, unit weight, or density). Reports the end pressure in gage AND absolute terms plus each leg's contribution. Use for: U-tube and differential manometers, 'pressure at depth h', piezometers, tank pressure traverses. Not for: accelerating fluids or compressible gas columns.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Units

Load a sample problem:

start pressureKnown pressure at the point where the traverse starts (often zero gage at an open surface) [dimension: pressure]reference (optional)Whether startPressure is a gage or an absolute pressureatmospheric pressure (optional)Local atmospheric pressure; omit for the standard atmosphere, 101.325 kPa [dimension: pressure]

legsTraverse legs in order from the start point to the end point, one per continuous fluid column

legs 1

fluidFluid definition — give at most ONE of specificGravity, unitWeight, or density; omit all three for water (γ = 9.81 kN/m³)

specific gravity (optional)Specific gravity SG relative to water (γw = 9.81 kN/m³): 13.6 mercury, ~0.85 oil, 1 water. Give at most one of specificGravity, unitWeight, density.unit weight (optional)Fluid unit weight γ [dimension: unitweight]density (optional)Fluid mass density ρ (converted internally via γ = ρ·g) [dimension: density]

delta zVertical elevation change of this leg (positive magnitude; the direction field carries the sense) [dimension: length]direction'down' = the traverse moves deeper, pressure increases; 'up' = it rises, pressure decreases

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

SI: gage pressure 3 m below a free water surface

Given

start pressure: 0 kPa
reference: gage
legs: 1.fluid delta z 3 mdirection down

Assumptions

Fluids are static and incompressible, so pressure varies only with elevation: dp/dz = −γ within each fluid column.
Each leg is a continuous column of one fluid; pressure is continuous across fluid interfaces.

Solution steps

Manometer traverse convention
Starting from the point of known pressure, walk the gage path leg by leg: moving DOWN through a fluid adds γ·Δz to the pressure, moving UP subtracts γ·Δz. This traverse is exactly the classical manometer method — writing the pressure at each interface until the end point is reached.
p_end = p_start + Σ(±γᵢ·Δzᵢ) (+ for down, − for up)
Traverse leg — moving down
Moving down through water, the pressure increases by γ·Δz.
Δp = +γ·Δz
Δp = +(9.81 kN/m^3 × 3 m) = 29.4 kPa; running pressure = 29.4 kPa
= 29.4 kPa
End pressure (gage)
Sum the starting gage pressure and every leg contribution.
p_end = p_start + ΣΔpᵢ
p_end = 0 kPa + (29.4 kPa) = 29.4 kPa
= 29.4 kPa
End pressure (absolute)
Absolute pressure adds the local atmospheric pressure to the gage value.
p(abs) = p(gage) + p(atm)
p_abs = 29.4 kPa + 101 kPa = 131 kPa
= 131 kPa

Results

Pressure at the end point (gage)

29.43kPa

Pressure at the end point (absolute)

130.8kPa

Pressure change across leg 1

29.43kPa

Where this answer was checked

source: Hydrostatic pressure p = γ·h (NCEES FE Reference Handbook) — single downward leg of water, start at 0 gage
verified by: hand-recomputed
derivation: p = 9810 · 3 = 29 430 Pa = 29.43 kPa gage; absolute = 29 430 + 101 325 = 130 755 Pa = 130.755 kPa.

Example 2

US: 10 ft of water at 62.4 pcf gives 624 psf

Given

start pressure: 0 psf
reference: gage
legs: 1.fluid unit weight 62.4 pcfdelta z 10 ftdirection down

Assumptions

Fluids are static and incompressible, so pressure varies only with elevation: dp/dz = −γ within each fluid column.
Each leg is a continuous column of one fluid; pressure is continuous across fluid interfaces.

Solution steps

Manometer traverse convention
Starting from the point of known pressure, walk the gage path leg by leg: moving DOWN through a fluid adds γ·Δz to the pressure, moving UP subtracts γ·Δz. This traverse is exactly the classical manometer method — writing the pressure at each interface until the end point is reached.
p_end = p_start + Σ(±γᵢ·Δzᵢ) (+ for down, − for up)
Traverse leg — moving down
Moving down through the fluid (from its unit weight), the pressure increases by γ·Δz.
Δp = +γ·Δz
Δp = +(62.4 pcf × 10 ft) = 624 psf; running pressure = 624 psf
= 624 psf
End pressure (gage)
Sum the starting gage pressure and every leg contribution.
p_end = p_start + ΣΔpᵢ
p_end = 0 psf + (624 psf) = 624 psf
= 624 psf
End pressure (absolute)
Absolute pressure adds the local atmospheric pressure to the gage value.
p(abs) = p(gage) + p(atm)
p_abs = 624 psf + 2120 psf = 2740 psf
= 2740 psf

Results

Pressure at the end point (gage)

624psf

Pressure at the end point (absolute)

2740psf

Pressure change across leg 1

624psf

Where this answer was checked

source: Hydrostatic pressure p = γ·h in US customary units — γ = 62.4 lbf/ft³ over h = 10 ft
verified by: hand-recomputed
derivation: p = 62.4 pcf · 10 ft = 624 psf exactly by unit algebra (SI check: γ = 62.4·157.087 = 9802.26 N/m³, h = 3.048 m, p = 29 877 Pa = 624.0 psf).

Example 3

U-tube manometer: open end to pipe through mercury then water

Given

start pressure: 0 kPa
reference: gage
legs: 1.fluid specific gravity 13.6delta z 0.25 mdirection down
2.fluid specific gravity 1delta z 0.6 mdirection up

Assumptions

Fluids are static and incompressible, so pressure varies only with elevation: dp/dz = −γ within each fluid column.
Each leg is a continuous column of one fluid; pressure is continuous across fluid interfaces.

Solution steps

Manometer traverse convention
Starting from the point of known pressure, walk the gage path leg by leg: moving DOWN through a fluid adds γ·Δz to the pressure, moving UP subtracts γ·Δz. This traverse is exactly the classical manometer method — writing the pressure at each interface until the end point is reached.
p_end = p_start + Σ(±γᵢ·Δzᵢ) (+ for down, − for up)
Traverse leg — moving down
Moving down through the fluid (from its specific gravity), the pressure increases by γ·Δz.
Δp = +γ·Δz
Δp = +(133 kN/m^3 × 0.25 m) = 33.4 kPa; running pressure = 33.4 kPa
= 33.4 kPa
Traverse leg — moving up
Moving up through the fluid (from its specific gravity), the pressure decreases by γ·Δz.
Δp = −γ·Δz
Δp = −(9.81 kN/m^3 × 0.6 m) = -5.89 kPa; running pressure = 27.5 kPa
= -5.89 kPa
End pressure (gage)
Sum the starting gage pressure and every leg contribution.
p_end = p_start + ΣΔpᵢ
p_end = 0 kPa + (33.4 kPa) + (-5.89 kPa) = 27.5 kPa
= 27.5 kPa
End pressure (absolute)
Absolute pressure adds the local atmospheric pressure to the gage value.
p(abs) = p(gage) + p(atm)
p_abs = 27.5 kPa + 101 kPa = 129 kPa
= 129 kPa

Results

Pressure at the end point (gage)

27.47kPa

Pressure at the end point (absolute)

128.8kPa

Pressure change across leg 1

33.35kPa

Pressure change across leg 2

-5.886kPa

Where this answer was checked

source: Classic U-tube manometer traverse (any fluid mechanics text's manometer method) — open end (0 gage) → down 0.25 m mercury (SG 13.6) → up 0.6 m water → pipe centreline
verified by: hand-recomputed
derivation: p_pipe = 0 + 13.6·9810·0.25 − 9810·0.6 = 9810·(3.4 − 0.6) = 9810·2.8 = 27 468 Pa = 27.468 kPa gage.