Fluid Mechanics · Chapter 2
Fluid Mechanics · Chapter 2

Properties of Fluids

Before a flow can be analyzed, the fluid carrying it must be characterized. This chapter defines the properties that recur throughout the subject — density and specific gravity, vapor pressure and the cavitation it threatens, compressibility, the all-important viscosity, and the surface tension responsible for capillary rise.

Chapter Objectives

  • Work confidently with the basic properties of fluids and the continuum approximation.
  • Understand viscosity and the frictional effects it produces in a flow.
  • Calculate capillary rise and drop caused by surface tension.
  • Select state-consistent property data and verify the applicability of each model.
Section 2–1

Introduction: Intensive & Extensive Properties

Any characteristic of a system is a property. Properties split into two kinds depending on whether they scale with the amount of matter present.

Intensive vs. Extensive Properties

Intensive properties (temperature, pressure, density) are independent of system size. Extensive properties (total mass, total volume, total momentum) scale with the extent of the system. Divide a system in two and the intensive values are unchanged while the extensive values halve.

Specific Properties & the Continuum

An extensive property expressed per unit mass is a specific property (e.g. specific volume \(v = V/m\)). Treating a fluid as a continuum — a smooth distribution of matter with no gaps — lets us speak of properties "at a point," valid whenever the system is large compared with molecular spacing.

Section 2–2

Density & Specific Gravity

Density & Specific Volume

Density is mass per unit volume; its reciprocal, specific volume, is volume per unit mass.

Density and Specific Volume
\[ \rho = \frac{m}{V}, \qquad v = \frac{V}{m} = \frac{1}{\rho} \]

Specific Gravity (Relative Density)

The ratio of a substance's density to that of water at 4°C (1000 kg/m³). Being a ratio, it is dimensionless.

Specific Gravity
\[ \mathrm{SG} = \frac{\rho}{\rho_{\mathrm{H_2O}}} \]

Ideal-Gas Equation of State

Any relation among pressure, temperature, and density is an equation of state. The simplest is the ideal-gas law, accurate for many gases at low density (high temperature, low pressure). Here \(R = R_u/M\) is the specific gas constant.

Ideal-Gas Law
\[ P = \rho R T \qquad\text{or}\qquad P v = R T \]
Section 2–3

Vapor Pressure & Cavitation

Saturation Temperature & Pressure

For a pure substance there is a one-to-one link between the temperature and pressure at which phase change occurs. At a given pressure the phase-change temperature is the saturation temperature \(T_{sat}\); at a given temperature the phase-change pressure is the saturation pressure \(P_{sat}\).

Vapor Pressure

The pressure exerted by a substance's vapor in phase equilibrium with its liquid at a given temperature. For pure liquid–vapor equilibrium it equals the saturation pressure, \(P_v = P_{sat}\), and it rises with temperature.

Cavitation

If the local liquid pressure in a flow drops below the vapor pressure, the liquid flashes into vapor bubbles. Swept into higher-pressure regions, these bubbles collapse violently, producing destructive pressure spikes, noise, vibration, and erosion — a key design concern for pumps and turbines.

Section 2–4

Energy & Specific Heats

Total Energy and Its Forms

The total energy \(E\) (or \(e\) per unit mass) is the sum of all macroscopic and microscopic forms. The microscopic forms together make up the internal energy \(U\). Kinetic energy is the energy of motion; potential energy is the energy of elevation.

Kinetic and Potential Energy (per unit mass)
\[ \mathrm{ke} = \frac{V^2}{2}, \qquad \mathrm{pe} = gz \]

Flow Energy & Enthalpy

A flowing fluid also carries flow energy \(P/\rho\), the work needed to push it across a boundary. Internal energy plus flow energy defines enthalpy, \(h = u + P/\rho\) — the natural energy grouping for flow systems. A fluid at rest has no flow energy.

Specific Heats

For ideal gases, internal energy and enthalpy changes follow \(\Delta u = c_v\,\Delta T\) and \(\Delta h = c_p\,\Delta T\). For incompressible substances the two specific heats are essentially equal, \(c_p \approx c_v = c\).

Section 2–5

Coefficient of Compressibility

Coefficient of Compressibility (Bulk Modulus)

Measures a fluid's resistance to volume change under pressure — how much pressure rise is needed to compress it. A large value means the fluid is nearly incompressible. It has the units of pressure.

Coefficient of Compressibility
\[ \kappa = -v\left(\frac{\partial P}{\partial v}\right)_T = \rho\left(\frac{\partial P}{\partial \rho}\right)_T \]

Coefficient of Volume Expansion

Measures the fractional change in volume (or density) with temperature at constant pressure. For an ideal gas it equals the reciprocal of absolute temperature, \(\beta_{\text{ideal gas}} = 1/T\).

Coefficient of Volume Expansion
\[ \beta = \frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_P = -\frac{1}{\rho}\left(\frac{\partial \rho}{\partial T}\right)_P \]
Combined Change in Density (approximate)
\[ \frac{\Delta \rho}{\rho} \approx \frac{\Delta P}{\kappa} - \beta\,\Delta T \]
Section 2–6

Viscosity

Viscosity is the property that quantifies a fluid's internal resistance to flow — its "stickiness." It is the reason a glass ball sinks slowly through oil and quickly through water.

Dynamic (Absolute) Viscosity

In simple shear between two plates, the velocity profile is linear and the rate of deformation equals the velocity gradient \(du/dy\). For most common fluids the shear stress is directly proportional to this gradient, with the proportionality constant being the dynamic viscosity \(\mu\).

Newton's Law of Viscosity

In one-dimensional shear flow of a Newtonian fluid, the shear stress is proportional to the velocity gradient (rate of deformation):

\[ \tau = \mu\,\frac{du}{dy} \]

Fluids obeying this linear relation are Newtonian (water, air, oils). Fluids whose apparent viscosity changes with shear rate are non-Newtonian — shear-thinning (pseudoplastic), shear-thickening (dilatant), or Bingham plastics that need a yield stress before flowing.

moving plate, velocity V → fixed plate (u = 0) u(y) = (y/ℓ)V linear profile
Figure 2–1. In steady laminar flow between two plates, the velocity varies linearly from zero at the fixed wall to V at the moving wall; the slope is the velocity gradient that sets the shear stress. (Original schematic.)

Kinematic Viscosity

The ratio of dynamic viscosity to density, which appears constantly in fluid mechanics and heat transfer.

Kinematic Viscosity
\[ \nu = \frac{\mu}{\rho} \]
Temperature dependence: viscosity of liquids decreases with temperature (energized molecules overcome cohesion), while viscosity of gases increases with temperature (more molecular collisions). Pressure has little effect on dynamic viscosity.

Concentric-Cylinder Viscometer

Measuring torque on a rotating inner cylinder separated from an outer one by a thin fluid film yields viscosity directly, since the gap behaves like two parallel plates.

Viscometer Torque (thin gap)
\[ T = \frac{4\pi^2 R^3 \dot{n}\,L}{\ell}\,\mu \]
Section 2–7

Surface Tension & the Capillary Effect

Surface Tension

At a liquid–gas (or liquid–liquid) interface, unbalanced molecular attraction pulls surface molecules inward, making the interface behave like a stretched elastic membrane. The tension per unit length along the surface is the surface tension \(\sigma_s\) (units N/m).

Excess Pressure Inside a Droplet and a Bubble
\[ \Delta P_{\text{droplet}} = \frac{2\sigma_s}{R}, \qquad \Delta P_{\text{bubble}} = \frac{4\sigma_s}{R} \]

A soap bubble carries the factor 4 because it has two surfaces (inner and outer). In both cases the excess pressure is inversely proportional to radius — smaller bubbles are at higher pressure.

Capillary Effect & Contact Angle

The rise or fall of a liquid in a narrow tube, driven by the balance of cohesive (liquid–liquid) and adhesive (liquid–solid) forces. The contact angle \(\phi\) measures wetting: a liquid wets the surface when \(\phi < 90^\circ\) (water in glass, \(\phi\approx0\)) and does not when \(\phi > 90^\circ\) (mercury in glass, \(\phi\approx130^\circ\)).

Capillary Rise in a Circular Tube
\[ h = \frac{2\sigma_s}{\rho g R}\cos\phi \]

A negative \(h\) (when \(\phi>90^\circ\)) is a capillary drop. Rise is inversely proportional to tube radius, so the effect is significant only in fine tubes (diameter under about 1 cm) — important when sizing manometer tubes.

WATER (wets, φ < 90°) h > 0 MERCURY (φ > 90°) h < 0
Figure 2–2. Water rises in a glass capillary (it wets the surface) while mercury is depressed (it does not). The sign of the contact angle sets the sign of the capillary rise. (Original schematic.)
Instructor Note

Two properties dominate this chapter in practice: viscosity (it controls pumping power and drag) and vapor pressure (it controls cavitation). If you remember only Newton's law of viscosity and the cavitation condition \(P \geq P_{sat}\), you'll handle most real design questions.

Physical meaning

Fluid properties connect molecular behavior to continuum balances. Density links volume and mass, viscosity links deformation rate and shear, bulk modulus links pressure and volume, and vapor pressure marks a phase-change boundary. Every value belongs to a stated temperature, pressure, composition, and reference basis.

Problem-solving strategy

  1. Identify which property closes the governing mass, momentum, or energy relation.
  2. Define the material or control volume, coordinate direction, pressure reference, and thermodynamic state.
  3. Select supplied or traceable property data at that state; interpolate only within a justified range.
  4. Choose an equation of state, Newtonian law, compressibility approximation, or interfacial balance consistent with the regime.
  5. Check units, signs, conservation, phase limits, sensitivity, and whether iteration is needed because properties depend on the result.
EXAMPLE 1Nitrogen mass in a test accumulator
A rigid \(0.350\ \text{m}^3\) accumulator contains nitrogen at \(180\ \text{kPa}\) absolute and \(315\ \text{K}\). Using the supplied \(R=296.8\ \text{J/(kg}\!\cdot\!\text{K)}\), estimate density and mass.

Recognition strategy

Absolute pressure, absolute temperature, and a specific gas constant signal an ideal-gas property calculation followed by the density definition.

Given information

  • Closed rigid system: \(V=0.350\ \text{m}^3\), \(P_{\rm abs}=180000\ \text{Pa}\), \(T=315\ \text{K}\), supplied nitrogen \(R=296.8\ \text{J/(kg}\!\cdot\!\text{K)}\).
  • No coordinate direction affects scalar density; pressure is explicitly absolute and temperature is the property basis.

Model and interpretation

Assume a uniform equilibrium state and ideal-gas behavior. \(P=\rho RT\) is approximate for real nitrogen; \(m=\rho V\) is exact for uniform density.

Step-by-step solution
1. Select the state relation. The supplied gas constant is mass based, so use \(P=\rho RT\) with SI units.
2. Solve symbolically. Isolate density before substitution: \(\rho=P_{\rm abs}/(RT)\).
3. Evaluate density. \(\rho=180000/[(296.8)(315)]=1.925\ \text{kg/m}^3\).
4. Audit mass conservation. The sealed inventory is \(m=\rho V=(1.925)(0.350)=0.674\ \text{kg}\).

Verification

Units: Since \(\text{Pa}=\text{J/m}^3\), dividing by \(\text{J/kg}\) leaves \(\text{kg/m}^3\); multiplying by volume leaves kilograms.

Signs and direction: Absolute pressure and kelvins are positive state variables, so positive density and mass follow; spatial direction is not applicable.

Plausibility: The gas is moderately pressurized and warm, so density somewhat above typical near-atmospheric gas density is credible; a real-gas compressibility factor would refine the estimate.

Final result: \(\rho\approx1.925\ \text{kg/m}^3\) and \(m\approx0.674\ \text{kg}\).

Engineering interpretation

The ideal-gas estimate supports preliminary inventory work. Certification would require composition, calibrated state measurements, a suitable real-gas model, and uncertainty propagation.

EXAMPLE 2Shear load on a translating coating plate
A \(0.600\ \text{m}\) by \(0.400\ \text{m}\) plate moves at \(1.80\ \text{m/s}\) over a fixed wall across a \(3.00\ \text{mm}\) liquid film. Given \(\mu=0.420\ \text{Pa}\!\cdot\!\text{s}\) at \(25^\circ\text{C}\), find drag.

Recognition strategy

A thin Newtonian film between parallel surfaces with stated speed and gap indicates Couette shear and Newton's viscosity law.

Given information

  • Supplied property \(\mu=0.420\ \text{Pa}\!\cdot\!\text{s}\) at \(25^\circ\text{C}\); \(U=1.80\ \text{m/s}\), \(h=0.00300\ \text{m}\), \(A=0.240\ \text{m}^2\).
  • Let \(+x\) follow plate motion and \(+y\) point from the wall to the plate; pressure reference is irrelevant.

Model and interpretation

Assume steady, laminar, fully developed, incompressible, isothermal Newtonian flow with no slip and negligible edge and pressure-gradient effects. The linear profile is approximate for the finite plate.

Step-by-step solution
1. Establish boundary velocities. No slip gives \(u(0)=0\) at the wall and \(u(h)=U\) at the plate.
2. Obtain the gradient. The assumed linear profile gives \(du/dy=U/h=600\ \text{s}^{-1}\).
3. Calculate shear. \(\tau=\mu U/h=(0.420)(600)=252\ \text{Pa}\).
4. Resolve plate drag. \(F=\tau A=(252)(0.240)=60.5\ \text{N}\), opposing \(+x\).

Verification

Units: \(\text{Pa}\!\cdot\!\text{s}\) times \(\text{s}^{-1}\) gives pascals, and pascals times square metres gives newtons.

Signs and direction: The fluid traction on the moving plate opposes its velocity, so its \(x\)-component is \(-60.5\ \text{N}\).

Plausibility: A viscous liquid, millimetre gap, and metre-per-second motion create a large gradient; doubling the gap would halve both shear and drag in this model.

Final result: Drag magnitude \(F\approx60.5\ \text{N}\), acting opposite plate motion.

Engineering interpretation

The result is sensitive to gap and viscosity. A production estimate needs temperature control, measured rheology, edge corrections, and confirmation that the flow remains laminar and fully developed.

EXAMPLE 3Local cavitation head margin
At a pump-eye survey point, a liquid at \(50^\circ\text{C}\) has measured local pressure \(18.0\ \text{kPa}\) absolute. Given \(\rho=980\ \text{kg/m}^3\) and vapor pressure \(12.3\ \text{kPa}\) absolute at \(50^\circ\text{C}\), compute pressure and head margins.

Recognition strategy

Cavitation screening compares local absolute pressure with vapor pressure at the same temperature; gauge pressure cannot be used directly.

Given information

  • \(P_{\rm local,abs}=18.0\ \text{kPa}\), supplied \(P_{v,\rm abs}=12.3\ \text{kPa}\), supplied \(\rho=980\ \text{kg/m}^3\), all at \(50^\circ\text{C}\); \(g=9.81\ \text{m/s}^2\).
  • Take available head upward-positive as a positive margin; no pump curve or vendor-required NPSH is supplied.

Model and interpretation

The local thermodynamic screen \(P_{\rm local}>P_v\) is exact at equilibrium; converting pressure margin with constant density is approximate and is not a complete system NPSH analysis.

Step-by-step solution
1. Align references. Both pressures are absolute and correspond to the same \(50^\circ\text{C}\) property state.
2. Compute pressure margin. \(\Delta P=P_{\rm local}-P_v=18.0-12.3=5.7\ \text{kPa}\).
3. Convert to liquid head. \(H=\Delta P/(\rho g)=5700/[(980)(9.81)]=0.593\ \text{m}\).
4. Make the decision. The measured point is above vapor pressure, but adequacy cannot be certified without required NPSH and uncertainty.

Verification

Units: Pascals divided by \(\text{kg/m}^3\) and \(\text{m/s}^2\) reduce to metres of liquid head.

Signs and direction: The positive margin means local absolute pressure exceeds vapor pressure; a negative value would indicate local flashing is thermodynamically possible.

Plausibility: About \(9.61\ \text{kPa}\) corresponds to one metre of this liquid, so \(5.7\ \text{kPa}\) corresponding to \(0.59\ \text{m}\) is consistent.

Final result: Pressure margin \(5.7\ \text{kPa}\); equivalent head margin \(0.593\ \text{m}\).

Engineering interpretation

This positive local margin is not proof against cavitation. Compare NPSH available with the vendor's supplied NPSH required plus project uncertainty and operating allowance.

Chapter recap

  • Match each property to its physical role and valid state range.
  • Use absolute pressure and absolute temperature in equations of state.
  • Compare local absolute pressure with same-temperature vapor pressure for cavitation screening.

Before moving on: you should be able to obtain density, distinguish viscosity measures, apply Newtonian shear, assess compressibility and capillarity assumptions, and document property uncertainty.