Surface Tension and Capillarity

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

11 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 Surface Tension and Capillarity 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 surface tension and capillarity describes physically and when it applies.
State every one of the 11 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. Surface Tension and Capillarity 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: surface tension and capillarity.
Capstone Studio instructional photograph

Fluid Mechanics — Surface Tension and Capillarity: 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 11 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

σ	Quantity produced by "σ = F/L" — read its definition and unit from the handbook line directly above the equation.
F	Quantity produced by "F = surface force at the interface" — read its definition and unit from the handbook line directly above the equation.
L	Quantity produced by "L = length of interface" — read its definition and unit from the handbook line directly above the equation.
h	Quantity produced by "h = (4σ cos β)/(γd)" — read its definition and unit from the handbook line directly above the equation.
β	Quantity produced by "β = angle made by the liquid with the wetted tube wall" — read its definition and unit from the handbook line directly above the equation.
γ	Quantity produced by "γ = specific weight of the liquid" — read its definition and unit from the handbook line directly above the equation.
d	Quantity produced by "d = diameter of the capillary tube" — 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.

where
The capillary rise h is approximated by
where

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

Density, specific weight, specific gravity and viscosity — Surface Tension and Capillarity

A liquid has a specific gravity of 0.88 and a dynamic viscosity of 0.00050 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 0.5 m³.

Given

SG = 0.88
μ = 0.00050 Pa·s
Volume = 0.5 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 0.88 (1000) = 880.0 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 880.0 (9.81) = 8, 633 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/880.0 = 1.136 e - 3 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00050/880.0 = 5.682 e - 7 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 880.0 (0.5) = 440.0 k g$

Answer: ρ = 880.0 kg/m³, γ = 8,633 N/m³, ν = 5.68e-7 m²/s, m = 440.0 kg

Why the other options are there

ρ = 0.88 kg/m³ (SG reported as density)
ν = 0.440 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Surface Tension and Capillarity

Example 3

Density, specific weight, specific gravity and viscosity — Surface Tension and Capillarity (2)

A liquid has a specific gravity of 1.41 and a dynamic viscosity of 0.00160 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 2.0 m³.

Given

SG = 1.41
μ = 0.00160 Pa·s
Volume = 2.0 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 1.41 (1000) = 1, 410 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 1, 410 (9.81) = 13, 832 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/1, 410 = 7.092 e - 4 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00160/1, 410 = 1.135 e - 6 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 1, 410 (2.0) = 2, 820 k g$

Answer: ρ = 1,410 kg/m³, γ = 13,832 N/m³, ν = 1.13e-6 m²/s, m = 2,820 kg

Why the other options are there

ρ = 1.41 kg/m³ (SG reported as density)
ν = 2.256 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Surface Tension and Capillarity

Example 4

Density, specific weight, specific gravity and viscosity — Surface Tension and Capillarity (3)

A liquid has a specific gravity of 1.34 and a dynamic viscosity of 0.00180 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 4.5 m³.

Given

SG = 1.34
μ = 0.00180 Pa·s
Volume = 4.5 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 1.34 (1000) = 1, 340 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 1, 340 (9.81) = 13, 145 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/1, 340 = 7.463 e - 4 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00180/1, 340 = 1.343 e - 6 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 1, 340 (4.5) = 6, 030 k g$

Answer: ρ = 1,340 kg/m³, γ = 13,145 N/m³, ν = 1.34e-6 m²/s, m = 6,030 kg

Why the other options are there

ρ = 1.34 kg/m³ (SG reported as density)
ν = 2.412 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Surface Tension and Capillarity

Example 5

Density, specific weight, specific gravity and viscosity — Surface Tension and Capillarity (4)

A liquid has a specific gravity of 1.51 and a dynamic viscosity of 0.00140 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 3.5 m³.

Given

SG = 1.51
μ = 0.00140 Pa·s
Volume = 3.5 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 1.51 (1000) = 1, 510 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 1, 510 (9.81) = 14, 813 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/1, 510 = 6.623 e - 4 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00140/1, 510 = 9.272 e - 7 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 1, 510 (3.5) = 5, 285 k g$

Answer: ρ = 1,510 kg/m³, γ = 14,813 N/m³, ν = 9.27e-7 m²/s, m = 5,285 kg

Why the other options are there

ρ = 1.51 kg/m³ (SG reported as density)
ν = 2.114 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Surface Tension and Capillarity

Example 6

Density, specific weight, specific gravity and viscosity — Surface Tension and Capillarity (5)

A liquid has a specific gravity of 1.57 and a dynamic viscosity of 0.00050 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 0.5 m³.

Given

SG = 1.57
μ = 0.00050 Pa·s
Volume = 0.5 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 1.57 (1000) = 1, 570 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 1, 570 (9.81) = 15, 402 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/1, 570 = 6.369 e - 4 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00050/1, 570 = 3.185 e - 7 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 1, 570 (0.5) = 785.0 k g$

Answer: ρ = 1,570 kg/m³, γ = 15,402 N/m³, ν = 3.18e-7 m²/s, m = 785.0 kg

Why the other options are there

ρ = 1.57 kg/m³ (SG reported as density)
ν = 0.785 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Surface Tension and Capillarity

Example 7

Density, specific weight, specific gravity and viscosity — Surface Tension and Capillarity (6)

A liquid has a specific gravity of 1.41 and a dynamic viscosity of 0.00080 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 4.0 m³.

Given

SG = 1.41
μ = 0.00080 Pa·s
Volume = 4.0 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 1.41 (1000) = 1, 410 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 1, 410 (9.81) = 13, 832 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/1, 410 = 7.092 e - 4 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00080/1, 410 = 5.674 e - 7 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 1, 410 (4.0) = 5, 640 k g$

Answer: ρ = 1,410 kg/m³, γ = 13,832 N/m³, ν = 5.67e-7 m²/s, m = 5,640 kg

Why the other options are there

ρ = 1.41 kg/m³ (SG reported as density)
ν = 1.128 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Surface Tension and Capillarity

Example 8

Density, specific weight, specific gravity and viscosity — Surface Tension and Capillarity (7)

A liquid has a specific gravity of 1.03 and a dynamic viscosity of 0.00170 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 1.0 m³.

Given

SG = 1.03
μ = 0.00170 Pa·s
Volume = 1.0 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 1.03 (1000) = 1, 030 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 1, 030 (9.81) = 10, 104 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/1, 030 = 9.709 e - 4 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00170/1, 030 = 1.650 e - 6 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 1, 030 (1.0) = 1, 030 k g$

Answer: ρ = 1,030 kg/m³, γ = 10,104 N/m³, ν = 1.65e-6 m²/s, m = 1,030 kg

Why the other options are there

ρ = 1.03 kg/m³ (SG reported as density)
ν = 1.751 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Surface Tension and Capillarity

Example 9

Density, specific weight, specific gravity and viscosity — Surface Tension and Capillarity (8)

A liquid has a specific gravity of 1.08 and a dynamic viscosity of 0.00150 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 5.0 m³.

Given

SG = 1.08
μ = 0.00150 Pa·s
Volume = 5.0 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 1.08 (1000) = 1, 080 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 1, 080 (9.81) = 10, 595 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/1, 080 = 9.259 e - 4 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00150/1, 080 = 1.389 e - 6 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 1, 080 (5.0) = 5, 400 k g$

Answer: ρ = 1,080 kg/m³, γ = 10,595 N/m³, ν = 1.39e-6 m²/s, m = 5,400 kg

Why the other options are there

ρ = 1.08 kg/m³ (SG reported as density)
ν = 1.620 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Surface Tension and Capillarity

Example 10

Density, specific weight, specific gravity and viscosity — Surface Tension and Capillarity (9)

A liquid has a specific gravity of 0.91 and a dynamic viscosity of 0.00200 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 1.0 m³.

Given

SG = 0.91
μ = 0.00200 Pa·s
Volume = 1.0 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 0.91 (1000) = 910.0 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 910.0 (9.81) = 8, 927 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/910.0 = 1.099 e - 3 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00200/910.0 = 2.198 e - 6 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 910.0 (1.0) = 910.0 k g$

Answer: ρ = 910.0 kg/m³, γ = 8,927 N/m³, ν = 2.20e-6 m²/s, m = 910.0 kg

Why the other options are there

ρ = 0.91 kg/m³ (SG reported as density)
ν = 1.820 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Surface Tension and Capillarity

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

Surface Tension and Capillarity contains 11 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.