Manometers

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

28 formulas

10 exam-style examples

~60 min

All Fluid Mechanics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Manometers within Fluid Mechanics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what manometers describes physically and when it applies.
State every one of the 28 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.

Lecture

Why this section exists. Manometers is the part of Fluid Mechanics that lets you connect a pipeline, jet or submerged surface to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as continuity plus energy, with one head-loss or force term. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: manometers.
Capstone Studio instructional photograph

Fluid Mechanics — Manometers: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a pipeline, jet or submerged surface. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 28 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

P0	Quantity produced by "P0 = P2 + γ2h2 – γ1h1 = P2 + g (ρ2 h2– ρ1 h1)" — read its definition and unit from the handbook line directly above the equation.
If h1	Quantity produced by "If h1 = h2 = h" — read its definition and unit from the handbook line directly above the equation.
P	Quantity produced by "P = pressure" — read its definition and unit from the handbook line directly above the equation.
γ	Quantity produced by "γ = specific weight of fluid" — read its definition and unit from the handbook line directly above the equation.
h	Quantity produced by "h = height" — read its definition and unit from the handbook line directly above the equation.
g	Quantity produced by "g = acceleration of gravity" — read its definition and unit from the handbook line directly above the equation.
ρ	Quantity produced by "ρ = fluid density" — read its definition and unit from the handbook line directly above the equation.
Patm	Quantity produced by "Patm = PA = Pv + γh = PB + γh = PB + ρgh" — read its definition and unit from the handbook line directly above the equation.
Pv	Quantity produced by "Pv = vapor pressure of the barometer fluid" — read its definition and unit from the handbook line directly above the equation.
θ	Quantity produced by "θ" — read its definition and unit from the handbook line directly above the equation.
PC	Quantity produced by "PC = pressure at the centroid of area" — read its definition and unit from the handbook line directly above the equation.
PCP	Quantity produced by "PCP = pressure at center of pressure" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Bober, W., and R.A. Kenyon, Fluid Mechanics, Wiley, 1980. Diagrams reprinted by permission of William Bober and Richard A. Kenyon.
For a simple manometer,
Note that the difference between the two densities is used.
Another device that works on the same principle as the manometer is the simple barometer.
Patm
Bober, W., and R.A. Kenyon, Fluid Mechanics, Wiley, 1980. Diagrams reprinted by permission of William Bober and Richard A. Kenyon.
Forces on Submerged Surfaces and the Center of Pressure
Patm
LIQUID h Patm z PLANAR VIEW FROM ABOVE
SIDE VIEW
(y-z PLANE) P dF (x-y PLANE)
y yCP
y CENTROID (C)
CENTER OF PRESSURE (CP)
SUBMERGED PLANE SURFACE
Elger, Donald F., et al, Engineering Fluid Mechanics, 10th ed., 2012. Reproduced with permission of John Wiley & Sons, Inc.
The pressure on a point at a vertical distance h below the surface is:
where
If atmospheric pressure acts above the liquid surface and on the non-wetted side of the submerged surface:

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Hydrostatic force on a vertical gate

A vertical rectangular gate 2.0 m wide and 3.0 m tall has its top at the water surface. Find the resultant force and its depth of application.

Given

b = 2.0 m
h = 3.0 m
Top at the free surface
γ = 9.81 kN/m³

Find

F and its line of action

Start with the thinking

Resultant equals pressure at the centroid times the area.
For a surface-piercing rectangle the resultant acts at 2h/3.

Step-by-step solution

Centroid depth
$h ˉ = 3.0/2 = 1.50 m$
Area
$A = 2.0 (3.0) = 6.00 m^{2}$
Resultant — F = γh̄A = 9.81(1.50)(6.00)
Evaluate
$F = 88.3 k N$
Line of action
$y_{p} = 2 h /3 = 2.00 mb e l o w t h es u r f a ce$

Answer: F = 88.3 kN acting 2.00 m below the surface

Why the other options are there

177 kN (bottom pressure used over the whole area)
y_p = 1.50 m (centroid taken as the pressure centre)

Reference: FE Reference Handbook — Fluid Mechanics — Hydrostatic forces

Example 2

Hydrostatic pressure at depth — Manometers

Find the gauge pressure 4.5 ft below the surface of a fluid with specific gravity 12.50.

Given

h = 4.5 ft
SG = 12.50
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Manometers

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 12.50 (62.4) = 780.0 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 780.0 (4.5) = 3, 510 p s f$
Convert
$p = 3, 510/144 = 24.38 p s i$

Answer: p ≈ 3,510 psf (24.38 psi)

Why the other options are there

280.8 psf (specific gravity ignored)
505,440 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Manometers

Example 3

Hydrostatic pressure at depth — Manometers (2)

Find the gauge pressure 19.0 ft below the surface of a fluid with specific gravity 4.00.

Given

h = 19.0 ft
SG = 4.00
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Manometers (2)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 4.00 (62.4) = 249.6 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 249.6 (19.0) = 4, 742 p s f$
Convert
$p = 4, 742/144 = 32.93 p s i$

Answer: p ≈ 4,742 psf (32.93 psi)

Why the other options are there

1,186 psf (specific gravity ignored)
682,906 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Manometers

Example 4

Hydrostatic pressure at depth — Manometers (3)

Find the gauge pressure 10.0 ft below the surface of a fluid with specific gravity 10.80.

Given

h = 10.0 ft
SG = 10.80
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Manometers (3)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 10.80 (62.4) = 673.9 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 673.9 (10.0) = 6, 739 p s f$
Convert
$p = 6, 739/144 = 46.80 p s i$

Answer: p ≈ 6,739 psf (46.80 psi)

Why the other options are there

624.0 psf (specific gravity ignored)
970,445 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Manometers

Example 5

Hydrostatic pressure at depth — Manometers (4)

Find the gauge pressure 9.0 ft below the surface of a fluid with specific gravity 11.50.

Given

h = 9.0 ft
SG = 11.50
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Manometers (4)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 11.50 (62.4) = 717.6 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 717.6 (9.0) = 6, 458 p s f$
Convert
$p = 6, 458/144 = 44.85 p s i$

Answer: p ≈ 6,458 psf (44.85 psi)

Why the other options are there

561.6 psf (specific gravity ignored)
930,010 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Manometers

Example 6

Hydrostatic pressure at depth — Manometers (5)

Find the gauge pressure 13.5 ft below the surface of a fluid with specific gravity 3.90.

Given

h = 13.5 ft
SG = 3.90
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Manometers (5)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 3.90 (62.4) = 243.4 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 243.4 (13.5) = 3, 285 p s f$
Convert
$p = 3, 285/144 = 22.81 p s i$

Answer: p ≈ 3,285 psf (22.81 psi)

Why the other options are there

842.4 psf (specific gravity ignored)
473,092 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Manometers

Example 7

Hydrostatic pressure at depth — Manometers (6)

Find the gauge pressure 6.0 ft below the surface of a fluid with specific gravity 9.60.

Given

h = 6.0 ft
SG = 9.60
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Manometers (6)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 9.60 (62.4) = 599.0 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 599.0 (6.0) = 3, 594 p s f$
Convert
$p = 3, 594/144 = 24.96 p s i$

Answer: p ≈ 3,594 psf (24.96 psi)

Why the other options are there

374.4 psf (specific gravity ignored)
517,571 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Manometers

Example 8

Hydrostatic pressure at depth — Manometers (7)

Find the gauge pressure 28.5 ft below the surface of a fluid with specific gravity 11.20.

Given

h = 28.5 ft
SG = 11.20
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Manometers (7)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 11.20 (62.4) = 698.9 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 698.9 (28.5) = 19, 918 p s f$
Convert
$p = 19, 918/144 = 138.3 p s i$

Answer: p ≈ 19,918 psf (138.3 psi)

Why the other options are there

1,778 psf (specific gravity ignored)
2,868,204 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Manometers

Example 9

Hydrostatic pressure at depth — Manometers (8)

Find the gauge pressure 26.0 ft below the surface of a fluid with specific gravity 11.60.

Given

h = 26.0 ft
SG = 11.60
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Manometers (8)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 11.60 (62.4) = 723.8 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 723.8 (26.0) = 18, 820 p s f$
Convert
$p = 18, 820/144 = 130.7 p s i$

Answer: p ≈ 18,820 psf (130.7 psi)

Why the other options are there

1,622 psf (specific gravity ignored)
2,710,057 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Manometers

Example 10

Hydrostatic pressure at depth — Manometers (9)

Find the gauge pressure 3.5 ft below the surface of a fluid with specific gravity 12.20.

Given

h = 3.5 ft
SG = 12.20
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Manometers (9)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 12.20 (62.4) = 761.3 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 761.3 (3.5) = 2, 664 p s f$
Convert
$p = 2, 664/144 = 18.50 p s i$

Answer: p ≈ 2,664 psf (18.50 psi)

Why the other options are there

218.4 psf (specific gravity ignored)
383,685 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Manometers

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a pipeline, jet or submerged surface, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Manometers contains 28 relations; you must be able to find this page in under 15 seconds.
Exam style: continuity plus energy, with one head-loss or force term.
Unit rule: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Manometers

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

P0​=P2​+γ2​h2​−γ1​h1​=P2​+g(ρ2​h2​−ρ1​h1​)

Ifh1​=h2​=h

P0​=P2​+(γ2​−γ1​)h=P2​+(ρ2​−ρ1​)gh

P=pressure

γ=specificweightoffluid

h=height

g=accelerationofgravity

ρ=fluiddensity

Patm = PA = Pv + γh = PB + γh = PB + ρghNeeds technical review

Pv=vaporpressureofthebarometerfluid

h = y sin θ h θyNeeds technical review

θ

P = Patm + ρgh, for h ≥ 0Needs technical review

P=pressure

Patm=atmosphericpressure

PC=pressureatthecentroidofarea

PCP=pressureatcenterofpressure

yC=slantdistancefromliquidsurfacetothecentroidofarea

yC=hC/sinθ

hC=verticaldistancefromliquidsurfacetocentroidofarea

yCP=slantdistancefromliquidsurfacetocenterofpressure

hCP=verticaldistancefromliquidsurfacetocenterofpressure

θ=anglebetweenliquidsurfaceandedgeofsubmergedsurface

IxC=momentofinertiaaboutthecentroidalx−axis

yCP=yC+IxC/yCA

yCP = yC + ρg sin θ IxC/PCANeeds technical review

Wetted side: FR = (Patm+ ρgyC sin θ)ANeeds technical review

Patm acting both sides: FR = (ρgyC sin θ)ANeeds technical review

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section