Calculate pressure through a series of static fluid columns. Follow each manometer leg and find the final gauge and absolute pressure.
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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.
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.
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.
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