Chapter 1–5 Fluid Mechanics Vocabulary
CHAPTER 1 The Nature of Fluids and the Study of Fluid Mechanics
Big picture: Fluid mechanics studies fluids at rest (fluid statics) and in motion (fluid dynamics); complex piping and fluid power systems are everywhere in daily life and industry.
Key takeaway: Fluids include liquids and gases; their behavior determined by properties (density, specific weight, specific gravity, surface tension, viscosity) and by energy changes as fluids move through systems.
Daily-life connections to observe fluids: home water systems, car fuel and cooling circuits, machine actuation with pressurized fluids, water transport in buildings, open-channel flow (gutters), etc.
1.1 Learning objectives (highlights)
Differentiate gas vs liquid behavior; define pressure; understand SI and U.S. Customary units; ensure unit consistency in equations; relate force and mass; define density, specific weight, and specific gravity; define surface tension; set up equations with consistent units; understand energy aspects and energy losses/additions in flowing fluids; recognize the role of pumps, valves, and measurements in fluid systems.
1.2 Basic introductory concepts
Pressure is force per unit area: p = \frac{F}{A}.
Fluids can be liquids or gases and may compressible (gases readily) or nearly incompressible (liquids).
The top surface of a liquid in a container tends to be at atmospheric pressure; piping systems pressurize fluids to move them.
Buoyancy and stability of floating or submerged bodies come from hydrostatic pressure distributions.
Energy in a flowing fluid includes kinetic energy (velocity), potential energy ( elevation), and pressure energy; energy can be added by pumps or removed by work-producing devices; energy is lost to friction and obstructions.
Measurements of pressure, temperature, and flow rate are essential to system performance.
1.3 The International System of Units (SI)
Base quantities and units:
Length: L = \text{m}
Time: t = \text{s}
Mass: m = \text{kg}
Derived units:
Force: F = ma = \text{N} = \text{kg} \cdot \text{m}/\text{s}^2
1 N is the force required to accelerate 1 kg at 1 m/s^2.
Equivalents: mass can be expressed as \text{kg} or as \text{N}\cdot \text{s}^2/\text{m}; force can be expressed as \text{N} or \text{kg}\cdot \text{m}/\text{s}^2.
SI prefixes (Table 1.1): \text{T} = 10^{12}, \text{G} = 10^9, \text{M} = 10^6, \text{k} = 10^3, \text{m} = 10^{-3}, \mu = 10^{-6}, \text{n} = 10^{-9}, \text{p} = 10^{-12}
Note on units: problems usually require keeping numbers between 0.1 and 10,000 times a power of 10; plan unit choices accordingly.
1.3.1 SI unit prefixes (summary)
Prefixes and factors (example): \text{k} = 10^3, \ \text{m} = 10^{-3}, \ \text{µ} = 10^{-6}, \ \text{n} = 10^{-9}, \ \text{p} = 10^{-12}
Practical guidance: adjust results so the numeric value lies in a convenient range with appropriate prefix.
1.4 The U.S. Customary System (USCS)
Basic quantities differ: length in ft, mass often in slug, force in lb, time in s; 1 slug is a mass unit such that F = m a = (\text{slug})(\text{ft}/\text{s}^2).
Relationship between weight and mass: weight w = mg with g = 32.2 (\text{ft}/\text{s}^2) on Earth.
A common bridge between mass and force uses the constant g_c = 32.2\ \text{lbm}\,\text{ft}/(\text{lbf}\,\text{s}^2), so that
In USCS, to relate mass to force: F = m a / g_c or, for weight atEarth gravity, w = m g = \text{lbf} when using consistent units.
Mass units: slug or lb_s^2/ft (lb·s^2/ft) depending on context; 1 slug = 32.174 lb·s^2/ft.
Mass expressed as lbm (pounds-mass) vs lbf (pounds-force); care needed to keep consistent with gc.
Practical convention in this text: use slugs for mass in USCS; lbm and lbf can be related through gc and g.
1.5 Weight and Mass
Mass vs weight definitions:
Mass: quantity of fluid, inertia w.r.t. motion; symbol m.
Weight: gravitational force on the fluid, symbol w.
In SI: w = m g with g = 9.81\ \text{m}/\text{s}^2.
In USCS: for Earth, w = m g with g = 32.2\ \text{ft}/\text{s}^2; mass can be expressed via slug or via m = w / g_c when needed.
Example (SI): a 5.60 kg rock weighs w = 5.60\times 9.81 = 54.9\ \text{N}.
Example (USCS): a container of oil weighs 84.6 lb; mass is m = \frac{w}{g} = \frac{84.6}{32.2} = 2.63\ \text{slug} (or equivalently use lbf–lbm formulations with gc).
Pounds-mass vs slugs: mass expressed as slug in problems; kgm or slug-based mass; conversions using g_c where needed.
Mass expressed as lbm vs weight in lbf: a helpful relation is weight in lbf equals mass in lbm times g in ft/s^2 divided by gc when appropriate; on Earth, weight (lbf) ≈ mass (lbm) when g equals standard 32.174 ft/s^2 if gc is used consistently.
1.6 Temperature
Temperature scales: Celsius (°C), Fahrenheit (°F); freezing/boiling points for water: 0°C and 100°C; 32°F and 212°F.
Celsius to Kelvin (absolute): TK = TC + 273.15.
Fahrenheit to Rankine and Kelvin: TR = TF + 459.67, TK = (TF + 459.67)/1.8.
Absolute zero: 0 K, -273.15°C, -459.67°F.
In this book: problems in SI use Kelvin; problems in USCS use Rankine or Fahrenheit as appropriate.
1.6.1 Absolute Temperature
The Kelvin scale has the same interval as Celsius; point of absolute zero is 0 K.
For converting from Celsius to Kelvin: TK = TC + 273.15.
Rankine: TR = TF + 459.67; relationship to Kelvin: TK = TR/1.8.
1.7 Consistent Units in an Equation (Unit-Cancellation Method)
Six-step procedure to ensure units cancel properly and results are dimensionally correct:
1) Solve algebraically for the desired term.
2) Decide the proper units for the result.
3) Substitute known values with units.
4) Cancel units across numerator/denominator.
5) Use conversion factors to remove unwanted units.
6) Perform the calculation.Example problem: travel time s = y t with y as speed; convert from hours to seconds using factor 3600 s/h and cancel units accordingly to obtain answer in seconds.
Table 1.2 lists common SI and USCS units for quantities (length, time, mass, force, pressure, energy, power, volume, area, Q, W, density) with appropriate conversion factors.
1.8 The definition of pressure
Pressure defined as p= rac{F}{A} ( Pascals in SI, psi or psf in USCS ).
Pascal’s laws (two key ideas):
Pressure acts uniformly in all directions on a small fluid volume.
Pressure on a surface within a fluid acts normal to the surface.
Example: Given force on a piston, pressure under piston is p = F/A.
Example (SI): if a 500 N force acts on a piston area of 2500 mm^2, pressure is p = \frac{500\text{ N}}{2500\text{ mm}^2} = 0.20\text{ N/mm}^2 = 0.20\text{ MPa}.
USCS example: 200 lb on 4.91 in^2 area gives p = \frac{200}{4.91} = 40.7\ \text{psi}.
Bar and pascal units: 1 bar = 10^5 Pa; 1 bar ≈ 100 kPa; atmospheric pressure near sea level ~ 1 bar.
1.9 Compressibility
Liquids: compressible but very slightly; bulk modulus $E$ relates pressure change to relative volume change: E = -\Delta p \frac{\Delta V}{V} ^{-1} or equivalently \Delta p = -E\frac{\Delta V}{V}.
Typical bulk moduli (examples): water ~ 2.2 GPa (316,000 psi-ish); other liquids have high bulk moduli; gases are highly compressible under pressure.
Practical note: liquids are treated as incompressible in this book unless stated otherwise.
Example (1.4): change in pressure to compress water volume by 1%: \Delta p = -E\left(\frac{\Delta V}{V}\right) = -316000\, \text{psi} \cdot (-0.01) = 3160\ \text{psi}.
1.10 Density, Specific Weight, and Specific Gravity
Density: \rho = \frac{m}{V} (kg/m^3 in SI; slug/ft^3 in USCS).
Specific weight: \gamma = \frac{w}{V} (N/m^3 in SI; lb/ft^3 in USCS).
Specific gravity: ratio to water at 4°C (density or weight):
\text{sg} = \frac{\rho}{\rhow(4°C)} = \frac{\gamma}{\gammaw(4°C)}.
Water properties at 4°C: \gammaw(4°C) = 9.81\ \text{kN/m}^3; \ \rhow(4°C) = 1000\ \text{kg/m}^3; in USCS, $\gammaw(4°C) = 62.4\ \text{lb/ft}^3$; mass density $\rhow(4°C) = 1.94\ \text{slugs/ft}^3$.
Example: oil with mass 825 kg in volume 0.917 m^3 has density \rho = 825/0.917 = 900\ \text{kg/m}^3.
Relation between density and specific weight: \gamma = \rho g, with $g$ taken as 9.81 m/s^2 (SI) or 32.2 ft/s^2 (USCS).
Baumé and API scales for SG depend on reference water temperature; in this book, SG is based on water at 4°C unless noted otherwise.
1.11 Surface tension
Surface tension reflects energy required to create new surface; arises from molecular cohesive forces.
For a liquid in capillary tubes, the liquid surface forms a curved meniscus depending on the liquid’s surface tension and interaction with the tube walls (adhesion and cohesion).
Units: N/m or lb/ft; related to work per unit area. The general expression: surface tension = work/area = N·m / m^2 = N/m.
Table 1.4 gives surface tension of water vs temperature; Table 1.5 lists other liquids (water, methanol, ethanol, benzene, mercury, etc.).
Examples: water droplets bead on waxed surfaces; capillary rise (or drop) in small tubes is driven by the balance of cohesive and adhesive forces.
Practical implication: capillary action drives wicking and fluid movement in small spaces and porous media.
Specific gravity references and temperature effects
Specific gravity can be reported in Baumé or API scales; API gravity relates to density by reference water at 60°F for petroleum industry data.
Practice problems and programs (overview)
The textbook uses programmed-example problems to teach unit cancellation and dimensional analysis (e.g., Problem 1.1 through 1.7). These illustrate how to set up unit-consistent calculations and convert between unit systems.
Temperature conversion, pressure conversions, and the use of standard atmosphere data are provided for practice.
Quick takeaways for quick reference
Pressure: p = F/A; units: Pa (SI), psi (USCS).
Density: \rho = m/V; Specific weight: \gamma = w/V = \rho g.
Specific gravity: \text{sg} = \dfrac{\rho}{\rhow(4°C)} = \dfrac{\gamma}{\gammaw(4°C)}.
Dynamic viscosity: defined via shear stress and velocity gradient (Chapter 2).
Kinematic viscosity: \nu = \dfrac{\mu}{\rho}; units m^2/s (SI).
Compressibility: bulk modulus links pressure change to volume change; liquids are nearly incompressible in this text.
Temperature scales and absolute temperatures: Kelvin and Rankine scales for engineering calculations.
CHAPTER 2 Viscosity of Fluids
Big picture: Viscosity characterizes how easily a fluid flows; distinguishes Newtonian vs non-Newtonian behavior; introduces dynamic viscosity and kinematic viscosity; discusses how viscosity varies with temperature, additives, and polymeric fluids; surveys measurement methods.
2.1 Learning objectives (highlights)
Define dynamic viscosity, kinematic viscosity; identify units; distinguish Newtonian vs non-Newtonian fluids; discuss viscosity variation with temperature; understand viscosity indices and standard grades for lubricants (SAE, ISO, ISO VG).
2.2 Dynamic viscosity ((\eta))
Definition via shear stress and shear rate: \tau = \eta \left( \frac{\Delta y}{\Delta y} \right) where the velocity gradient is\n \frac{\Delta y}{\Delta y} in a simple Couette-type flow.
In the standard form, dynamic viscosity can be written as \eta = \frac{\tau}{\dot{\gamma}} where \dot{\gamma} = \frac{\Delta y}{\Delta y} is the shear rate.
SI units: \eta in Pa·s (N·s/m^2); can also be written as kg/(m·s).
USCS units: lb_s·s/ft^2 or slug/(ft·s) depending on convention; viscosity units can be Pa·s or N·s/m^2, etc.
Obsolete CGS units: poise (P), centipoise (cP).
2.3 Kinematic viscosity
Definition: \nu = \frac{\eta}{\rho}
Units: SI: \text{m}^2/\text{s};
USCS: \text{ft}^2/\text{s}.Relationship to dynamic viscosity: heavier fluids with higher density have lower (\nu) for the same (\eta).
2.4 Newtonian vs Non-Newtonian Fluids
Newtonian fluids: viscosity is constant for a given temperature; shear stress is proportional to shear rate: \tau = \eta \dot{\gamma} (constant slope in a plot of (\tau) vs. (\dot{\gamma})).
Non-Newtonian fluids: viscosity depends on shear rate (and possibly time or history).
Time-independent non-Newtonian classes:
Pseudoplastic (shear-thinning): viscosity decreases with shear rate.
Dilatant (shear-thickening): viscosity increases with shear rate.
Bingham: yield stress must be exceeded before flow begins; once flowing, behaves like a Newtonian fluid with a finite slope.
Time-dependent fluids: viscosity changes with time even at constant shear; examples include thixotropic (viscosity decreases with time under constant shear) and rheopectic fluids (viscosity increases with time).
Polymers and nanofluids may exhibit non-Newtonian and time-dependent behavior; magnetic-field- or electrically-controlled fluids (MRF, ERF) can have tunable viscosity.
2.5 Variation of viscosity with temperature; Viscosity index (VI)
Generally, for liquids, viscosity decreases as temperature increases; for gases, viscosity increases with temperature.
Viscosity index (VI) measures how strongly viscosity changes with temperature; higher VI means smaller changes in viscosity with temperature.
VI is determined from viscosity at 40°C and 100°C and compared to reference oils (ASTM D2270).
Practical importance: high VI fluids (lubricants) maintain better lubricating film across operating temperatures.
2.6 Viscosity measurement methods (overview)
Rotating-drum viscometer: measures dynamic viscosity by drag torque on a drum, using a known velocity gradient in the fluid.
Capillary-tube viscometer: uses pressure drop (via flow through a capillary) to infer viscosity; the equation is \eta = \frac{(p1 - p2)D^2}{32 y L} where D is capillary inside diameter, y is velocity, L is capillary length, p1-p2 is pressure drop.
Capillary viscometers include standard calibrated glass viscometers (Ubbelohde, Cannon-Fenske, etc.).
Falling-ball viscometer: ball falls through a liquid; at terminal velocity the balance of forces gives viscosity via \eta = (\rhos - \rhof)\frac{D^2}{18 v} g (for Newtonian fluids and small analogies; the exact form given in text is \eta = (\rhos - \rhof) \frac{D^2}{18 y} in terms of specific weights).
Saybolt Universal viscometer (SUS): measures time to flow 60 mL through an orifice; SUS is a relative measure and is correlated to kinematic/dynamic viscosity via tables/curves; newer standards favor direct kinematic viscosity in SI units.
Other standards: ISO VG grades, SAE viscosity grades for engine oils; ISO grade values refer to viscosity at 40°C (mm^2/s).
2.7 SAE viscosity grades
SAE grades classify engine oils by viscosity at low (cold-start) and high temperatures; suffix “W” denotes winter (low-temperature) viscosity.
Examples: 0W, 5W, 10W, 15W, 20W, 30, 40, 50, etc. The “W” grades must meet low-temperature viscosity standards; high-temperature viscosity is specified for the hot regime.
2.8 ISO viscosity grades
ISO VG grades provide a standard classification of lubricants by their kinematic viscosity at 40°C; ISO VG 32, 46, 68, 100, 150, 220, 320, etc. Values have a nominal target with allowable tolerance ±10%.
2.9 Hydraulic fluids and ISO viscosity grades
Fluids used in hydraulic systems must be stable across temperatures; ISO VG grades are common for specifying hydraulic oils; oils may be formulated to achieve a high VI for broad operating temperature ranges.
2.9–2.14 Supplemental and practice problems (overview)
The chapter includes numerous practice problems converting between SUS, cSt (mm^2/s), Pa·s, and ft^2/s; problems also connect ISO VG and SAE grades to viscosity values.
Example problems illustrate converting viscosity values across measurement systems, estimating viscosity from tables, and applying different viscometer formulas.
CHAPTER-3. Pressure Measurement
Big picture: Pressure concepts; absolute vs gage pressure; atmospheric pressure; pressure variation with elevation; barometers and manometers; pressure transducers and gages; pressure expressed as height of a column; Pasal’s paradox.
3.1 Learning objectives
Distinguish absolute vs gage pressure; describe atmospheric pressure properties; relate elevation to pressure; describe how manometers, barometers, gages, and transducers work; understand piezometric head.
3.2 Absolute and gage pressure
Relationship: p{abs} = p{gage} + p_{atm}.
Examples: convert between gage and absolute using local atmospheric pressure (assumed 101 kPa abs, unless stated otherwise, or use measured atm).
Example 3.1: 155 kPa gage with local atmosphere 98 kPa abs → p_{abs}=155+98=253\text{ kPa}.
Example 3.2: 225 kPa abs with atm 101 kPa → p_{gage}=225-101=124\text{ kPa}.
Example 3.3: 10.9 psia with atm 15.0 psia → p_{gage}=10.9-15.0=-4.1\text{ psig} (vacuum).
Example 3.4: -6.2 psig with atm 14.7 psia → p_{abs}= -6.2+14.7=8.5\text{ psia}.
3.3 Pressure–Elevation relationship (hydrostatics for liquids)
Change in pressure with change in elevation: \Delta p = g h where $g$ is the specific weight of the liquid and $h$ is the elevation change.
For a homogeneous, incompressible liquid, this reduces to \Delta p = \gamma \Delta z with $\gamma = w/V$ (specific weight).
Key conclusions: pressure at a given horizontal level is the same; pressure increases with depth; elevation changes linearly affect pressure.
For gases, the relation is more complex because gas density/weight varies with pressure; problem contexts often assume uniform pressure in gases unless large elevations are involved.
3.4 Standard atmosphere and piezometric head
Appendix E provides standard atmosphere properties; barometric pressure varies with altitude; at sea level ~ 101 kPa abs (~14.7 psi).
Pascal’s paradox: pressure at a given depth depends only on depth and the fluid, not on container shape; used to design water towers and standpipes.
Piezometric head concept (recipe to convert pressure pa into an equivalent fluid depth ha):
ha = \frac{pa}{g} where $p_a$ is the pressure above the free surface.
This is used to convert nonzero pressure at the surface to an equivalent water depth for hydrostatic calculations.
3.4.1 Liquids vs 3.4.2 Gases
Liquids: incompressible (for the range considered here); 3.4.3 provides standard atmospheric properties and tables.
Gases: density and specific weight vary with pressure; hydrostatic relations require thermodynamics for exact treatment.
3.5 Pascal’s Paradox
The bottom pressure in a fluid depends only on fluid depth and fluid properties, not on the container shape or amount of fluid (as long as the fluid is the same).
3.6 Manometers and related devices
The simplest manometer is a U-tube with a fluid of known density; pressure difference between two points is translated into height differences via \Delta p = \gamma_{manometer} \Delta h.
Steps for solving manometer problems:
Start at a point with known pressure; traverse the manometer using \Delta p = \gamma \Delta h for each column.
Follow the sign convention (pressure increases downward, decreases upward).
Solve for the desired pressure difference.
Example 3.8–3.13 illustrate calculations with fluids of various specific weight (oil, water) and with pressure at different points; problems include open-to-atmosphere ends and differential pressure readings.
3.7 Barometers
Barometer types: mercury column devices (liquid mercury column supported by atmospheric pressure) and aneroid barometers (mechanical dials).
The barometer relates atmospheric pressure to the height of a mercury column: p{atm} = \gamma{Hg} h with temperature corrections for accuracy.
Standard values for mercury density and gravitational acceleration used to compute heights:
In SI: $\gamma{Hg}\approx 13.3\ \text{kN/m}^3$; in USCS: $\gamma{Hg}\approx 848.7\ \text{lb/ft}^3$.
3.8 Pressure expressed as height of a column
Pressures in ducts or low-pressure readings are often given as inches of water (in H2O) or inches of mercury (inHg); converting to psi or Pa uses p = \gamma h with $h$ in length units compatible with the chosen $\gamma$.
Example: 1.0 in H2O ≈ 0.0361 psi (via hydrostatic conversion).
Conversion factors to Pa and psi are provided in Appendix K.
3.9 Pressure gages and transducers
Bourdon-tube gauges: mechanical device that deflects with pressure; scales read gauge pressure, typically relative to atmosphere.
Magnehelic gauge: magnetic indicator with a sealed mechanism; measures small pressure differences.
Pressure transducers: convert pressure into electrical signal (strain-gage, piezoelectric sensors, etc.); used for remote monitoring and automatic control.
3.10 Problem set and references (overview)
The chapter includes a wide range of problems converting between absolute and gauge pressures, calculating heights corresponding to given pressures, and applying standard atmosphere data.
CHAPTER- 4. Forces Due to Static Fluids
Big picture: When fluids exert pressure on surfaces, forces can be computed for planar and curved surfaces; cable design, dam walls, gates, and submerged surfaces require evaluating resultant forces and centers of pressure.
4.1 Learning objectives (highlights)
Compute force on a plane area due to a static gas; compute force on a horizontal plane in a liquid; determine resultant forces on submerged planes; define center of pressure; determine line of action; analyze forces on curved surfaces; account for pressure head (piezometric effects).
4.2 Gases under pressure
For a gas in a cylinder, pressure acts nearly uniformly on the piston face; the resulting force is F = p A with the pressure given in the same units as the gas pressure.
The outer shell must withstand the end pressure as well; the internal pressure acts on the piston face and the piston rod.
4.3 Horizontal flat surfaces under liquids (examples)
For a cylindrical drum with liquid inside and the top open to atmosphere, bottom pressure is uniform across the bottom and the force is F = pB A with pB = p{atm} + \gammaw hw + \gammal h_l depending on how the vertical arrangement contributes.
Example: bottom pressure of a drum with oil and water layers, and area calculation yields total force on the bottom.
4.4 Rectangular walls and triangular pressure distribution
For a vertical rectangular wall with liquid height $h$, pressure varies linearly with depth: p(z) = \gamma z; the resultant force on the wall isFR = p{avg} A = \gamma\frac{h}{2} A, where $A$ is wall area and $p_{avg}$ is the average pressure (the centroid of the triangular distribution).
Center of pressure on a vertical wall is located at one-third the height from the bottom (for uniform, linear variation).
For inclined walls or walls with different geometry, the same approach applies but the geometry (area, centroid) must be computed relative to the wall orientation.
4.5–4.6 Submerged plane areas—general procedure
Identifies the steps to compute the resultant force on any submerged plane, including inclined planes:
1) Identify the point where the fluid surface intersects the plane.
2) Locate the area centroid and the corresponding vertical depth hc to the centroid.
3) Compute the area A and the resultant force FR = γ hc A.
4) Determine the centroidal moment of inertia Ic about the centroid axis of the area.
5) Determine the center of pressure distance Lp via Lp = Lc + Ic/(Lc A).
6) Sketch the resultant force acting at the center of pressure, perpendicular to the plane.The vertical depth to the center of pressure can be found via hp = Lp sin u (or hp = hc + Ic sin^2 u/(hc A)).
4.7 Piezometric head (extension of hydrostatics to nonzero surface pressure)
If the surface pressure pa is not atmospheric, define a piezometric head ha = pa/g; then use an equivalent depth he to compute forces, FR = g hce A, where hce = hc + ha.
4.8 Forces on curved surfaces (two components)
Horizontal component FH: computed from the projection of the curved surface onto a vertical plane; for a curved surface with a rectangular projection, FH = γ w (hc + s/2) where w is width and hc is the centroid depth of the projected area.
Vertical component FV: equals the weight of the fluid above the curved surface (if the fluid is statically above the surface).
Resultant FR is obtained from the orthogonal components FH and FV; its line of action passes through the center of curvature and its angle relative to the horizontal is f = arctan(FV/FH).
Summary formulas (for a typical curved surface with liquid above):
FR = \sqrt{FH^2 + FV^2}, with the orientation given by f = \tan^{-1}(FV/FH).
4.9 Effect of a pressure above the surface (piezometric head in curved surfaces)
When surface pressure pa is applied above the fluid, convert to equivalent depth ha and use he = hc + ha for depth calculations of both horizontal and vertical components.
4.10–4.11 Forces on curved surfaces with fluid below, or with fluid both above and below
The same concepts apply with careful free-body isolation of the fluid above or below the surface; the horizontal component is calculated via projection; the vertical component is the weight of the fluid that would be displaced by the curved surface.
4.12–4.14 Application problems (overview)
A large set of problems illustrates doors, gates, tanks, and submerged/partially submerged surfaces; emphasize Finding FR, hp, Lp, and direction of resultant.
CHAPTER-5. Buoyancy and Stability
Big picture: Buoyant force arises from displaced fluid; stability of floating and submerged bodies depends on the relative positions of gravity, buoyancy, and metacenter; metacentric height MG is a measure of stability for floating bodies.
5.1 Learning objectives
Write the buoyant force equation; analyze floating vs submerged bodies; apply static equilibrium in vertical direction; define and compute metacenter; analyze stability for submerged and floating bodies; compute center of buoyancy and center of gravity; compute stability curves.
5.2 Buoyancy (Archimedes’ principle)
Buoyant force: Fb = \gammaf Vd where $\gammaf$ is the specific weight of the fluid and $V_d$ is the displaced volume.
For floating bodies, the displaced volume adjusts so that weight equals buoyant force: w = F_b.
Buoyant force acts upward through the centroid of the displaced volume.
5.3–5.5 Free-body analysis of buoyant problems
Free-body diagrams include weight w, buoyant force Fb, and possibly external forces (e.g., moorings or tethers).
Example 5.1: bronze cube, completed submerged in water and in mercury; compute required external force Fe to hold equilibrium. In water: Fe = w − Fb. In mercury: Fe = w − Fb (with larger Fb due to higher density of mercury). The result demonstrates that a more dense surrounding liquid increases buoyant force; the external force may be upward or downward depending on relative densities.
Example 5.2: find displaced volume given a submerged apparent weight; use Fb = \gammaw Vd and w = m g to solve for Vd, then compute density from w/Vd.
Example 5.3 and 5.4 illustrate submerged-in-glycerin problems and neutrally buoyant configurations.
5.6 Degree of stability and metacenter concept
Submerged bodies: stable if the center of gravity (cg) is below the center of buoyancy (cb).
Floating bodies: define the metacenter (mc) as the intersection of the vertical axis in equilibrium with a vertical line through the new center of buoyancy when the body tilts slightly.
A buoyant floating body is stable if the metacenter is above the cg: y{mc} > y{cg}.
The distance MB between cb and mc is MB = I / Vd, where I is the least moment of inertia of the waterline (horizontal cross-section) about the vertical axis, and Vd is displaced volume.
Metacentric height MG = ymc − ycg; stability increases with MG but too large MG can produce uncomfortable rolling.
Practical guidelines: MG around 1.5 ft (0.46 m) is a minimum for small ships; much larger MG can lead to uncomfortable motion.
5.7 Example problems (illustrative results)
Example 5.5: stability of a flatboat in freshwater; compute submerged depth, center of buoyancy, and MB; determine stability via ymc vs ycg.
Example 5.6: stability of a solid cylinder in oil; compute submerged depth X, center of buoyancy cb, center of gravity cg, and metacenter mc; determine stability (MC below CG → unstable).
Example 5.7: compute metacentric height MG for the flatboat using ymc and ycg values; MG = ymc − ycg.
5.6 Static stability curve
A stability curve (GH vs angle) plots the righting arm (GH) as a function of tilt angle; stability requires GH > 0 over the range of operation; as tilt increases, righting arm decreases and may vanish or reverse, indicating instability.
5.8–5.9 Buoyancy problems and design considerations
A broad set of practice problems covers neutrally buoyant objects, stability in various liquids, and design of buoyant components (foam, syntactic foams, etc.).
Practical takeaways
Buoyant force depends only on displaced fluid volume and fluid weight; stability depends on relative positions of cg, cb, and mc; the metacentric height MG is a crucial stability metric for floating bodies.
In many engineering applications (submarines, light craft, offshore platforms), designers use buoyant materials (buoyancy foams, syntactic foams) to achieve neutral buoyancy or desired stability characteristics.
Overall cross-chapter connections
Chapter 1 lays the foundation for units, pressure, density and surface tension; Chapter 2 introduces viscosity which affects energy losses and pressure drop in pipes; Chapter 3 and 4 build on pressure measurement and hydrostatics for design of components; Chapter 5 uses buoyancy and stability concepts to analyze floating/submerged bodies, tying together density, pressure, and geometry.
Quick reference formulas (compiled)
Pressure: p=\frac{F}{A} (SI: Pa; USCS: psi or psf)
Weight-mass relation: w=mg; in USCS: use g\approx 32.2\ \text{ft}/\text{s}^2; mass-unit conversions via g_c\,.
Density: \rho = \frac{m}{V}; Specific weight: \gamma = \frac{w}{V} = \rho g.
Specific gravity: \text{sg} = \frac{\rho}{\rhow(4^{\circ}C)} = \frac{\gamma}{\gammaw(4^{\circ}C)}.
Pressure-elevation: for liquids, \Delta p = \gamma h; for a height increase: increase in depth increases pressure.
Buoyant force: Fb = \gammaf V_d.
Metacentric height: MG = y{mc} - y{cg}; MB = \frac{I}{Vd}; "y{mc}=y_{cb}+MB".
Piezometric head: ha = \frac{pa}{g}; equivalent depth for pressure at surface.
Viscosity (dynamic): \tau = \eta \dot{\gamma}; kinematic: \nu = \eta/\rho.
Elementary viscosity units: SI: Pa·s; cSt, mPa·s; SUS (Saybolt) relates to kinematic viscosity via tables.
For exam preparation
Be able to classify fluids as Newtonian vs non-Newtonian and describe their basic behaviors.
Be able to convert between pressure types (pabs, pgage, patm) and perform simple height-to-pressure conversions (inH2O, inHg).
Be able to compute forces on flat and curved submerged surfaces; locate center of pressure; perform simple stability checks for floating/submerged bodies using the MG and MB concepts.
Be comfortable with SI and USCS unit systems, and perform unit cancellation using the six-step procedure.
Understand the role of piezometric head in problems where surface pressure is not atmospheric.
Notes on how to study effectively
Draw free-body diagrams for buoyancy and submerged-surface problems; separately compute centers of pressure, projected areas, and centroids.
Practice unit-cancellation steps on mixed-unit problems; always verify that units cancel to the desired remaining units.
Use the standard reference tables (Appendix K) for conversion factors (pressure units, viscosity units, density, etc.).
Familiarize yourself with the typical problem structure: identify system, compute necessary depths/areas, compute forces, check directions, verify with alternative method when possible.
Final reminder
The book interlinks static and dynamic fluid concepts, showing how pressure, density, and energy interact in real systems (pipes, pumps, dams, hydraulic actuators). Use the core ideas of hydrostatics, viscous flows, and buoyancy to analyze both everyday and complex engineering problems.