7.1 Fluid Properties, Units, and Pressure Measurement

Key Takeaways

  • Fluid mass density (ρ\rho), specific weight (γ=ρg\gamma = \rho g), and specific gravity (SG=γ/γwSG = \gamma / \gamma_w) form the foundational weight-volume metrics, with standard water at 4°C exhibiting γw=9.81 kN/m3\gamma_w = 9.81\text{ kN/m}^3 (62.4 lb/ft362.4\text{ lb/ft}^3) and ρw=1000 kg/m3\rho_w = 1000\text{ kg/m}^3.

  • Newton's law of viscosity (τ=μdvdy\tau = \mu \frac{dv}{dy}) defines dynamic shear resistance (Pa⋅s\text{Pa}\cdot\text{s}), whereas kinematic viscosity (ν=μ/ρ\nu = \mu / \rho, m2/s\text{m}^2/\text{s}) quantifies momentum diffusivity; liquid viscosity decreases with temperature, while gas viscosity increases.

  • Capillary rise or depression in clean cylindrical tubes follows the Jurin relation h=4σcos⁡θγdh = \frac{4\sigma \cos \theta}{\gamma d}, producing liquid elevation for wetting fluids (θ<90∘\theta < 90^\circ) and depression for non-wetting fluids like mercury (θ≈130∘–140∘\theta \approx 130^\circ\text{–}140^\circ).

  • Hydrostatic pressure increases linearly with downward vertical depth according to dP=γdhdP = \gamma dh, governed by the absolute-gauge relationship Pabs=Pgauge+PatmP_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}}, where standard sea-level atmospheric pressure equals 101.325 kPa101.325\text{ kPa} (10.33 m of water, 760 mm Hg).

  • Multi-fluid manometer deflection problems are solved systematically via the continuous meniscus method: traversing downward adds pressure (+γihi+\gamma_i h_i), traversing upward subtracts pressure (−γihi-\gamma_i h_i), and points at identical horizontal elevations within the same continuous static fluid share equal pressure.

Last updated: October 2026

7.1 Fluid Properties, Units, and Pressure Measurement

In the Philippine Civil Engineering Licensure Examination (CELE), fluid mechanics constitutes a core pillar of the Hydraulics and Principles of Geotechnical Engineering (HGE) cluster (30% total examination weight). Mastery of fluid statics begins with an exact understanding of physical fluid properties, thermodynamic compressibility, and the mechanics of static pressure distribution across multi-fluid manometric systems.


1. Primary Physical Properties of Fluids

A fluid is defined as a substance that deforms continuously when subjected to shear (tangential) stress of any magnitude, encompassing both liquids (which possess a definite volume and form a free surface) and gases (which expand indefinitely to occupy their container). Under the continuum hypothesis, fluid properties are treated as point functions across continuous space.

PropertySymbolDefining EquationSI UnitStandard Value (Pure Water at 4°C)
Mass Densityρ\rhoρ=mV\rho = \frac{m}{V}kg/m3\text{kg/m}^31,000 kg/m31,000\text{ kg/m}^3 (1.0 g/cm31.0\text{ g/cm}^3, 1.94 slugs/ft31.94\text{ slugs/ft}^3)
Specific Weight (Unit Weight)γ\gammaγ=ρg\gamma = \rho gN/m3\text{N/m}^3 or kN/m3\text{kN/m}^39,810 N/m3=9.81 kN/m39,810\text{ N/m}^3 = 9.81\text{ kN/m}^3 (62.4 lb/ft362.4\text{ lb/ft}^3)
Specific Volumevsv_svs=Vm=1ρv_s = \frac{V}{m} = \frac{1}{\rho}m3/kg\text{m}^3/\text{kg}0.001 m3/kg0.001\text{ m}^3/\text{kg}
Specific GravitySGSG or ssSG=ρρwater=γγwaterSG = \frac{\rho}{\rho_{\text{water}}} = \frac{\gamma}{\gamma_{\text{water}}}Dimensionless1.0001.000 (dimensionless)

Note

In PRC board examination problems, standard gravitational acceleration is taken as g=9.81 m/s2g = 9.81\text{ m/s}^2 unless specified otherwise. For mercury (Hg\text{Hg}), SG=13.6SG = 13.6, yielding γHg=13.6×9.81=133.42 kN/m3\gamma_{\text{Hg}} = 13.6 \times 9.81 = 133.42\text{ kN/m}^3.


2. Viscosity and Newton's Law of Viscosity

Viscosity represents the internal molecular resistance of a fluid to shear deformation or flow. Consider two parallel plates separated by a small clearance yy filled with fluid, where the lower plate is stationary and the upper plate moves with constant velocity vv.

Dynamic Viscosity (μ\mu)

Under laminar conditions, the shearing stress τ\tau between fluid layers is directly proportional to the transverse velocity gradient dvdy\frac{dv}{dy} according to Newton's Law of Viscosity: τ=μdvdy\tau = \mu \frac{dv}{dy} where:

  • τ=FA\tau = \frac{F}{A} is the shearing stress (Pa=N/m2\text{Pa} = \text{N/m}^2).
  • μ\mu is the dynamic (absolute) viscosity (Pa⋅s=N⋅s/m2=kg/(m⋅s)\text{Pa}\cdot\text{s} = \text{N}\cdot\text{s/m}^2 = \text{kg}/(\text{m}\cdot\text{s})).
  • In the CGS system, dynamic viscosity is expressed in Poise (P\text{P}) or centipoise (cP\text{cP}), where 1 Poise=0.1 Pa⋅s1\text{ Poise} = 0.1\text{ Pa}\cdot\text{s} and 1 cP=10−3 Pa⋅s=1 mPa⋅s1\text{ cP} = 10^{-3}\text{ Pa}\cdot\text{s} = 1\text{ mPa}\cdot\text{s}. Water at 20∘C20^\circ\text{C} has μ≈1.002 cP\mu \approx 1.002\text{ cP}.

Kinematic Viscosity (ν\nu)

Kinematic viscosity is the ratio of dynamic viscosity to fluid mass density, representing momentum diffusivity: ν=μρ\nu = \frac{\mu}{\rho}

  • SI unit: m2/s\text{m}^2/\text{s}.
  • CGS unit: Stokes (St\text{St}), where 1 St=1 cm2/s=10−4 m2/s1\text{ St} = 1\text{ cm}^2/\text{s} = 10^{-4}\text{ m}^2/\text{s}, and 1 cSt=10−6 m2/s1\text{ cSt} = 10^{-6}\text{ m}^2/\text{s}. Water at 20∘C20^\circ\text{C} has ν≈1.0×10−6 m2/s\nu \approx 1.0 \times 10^{-6}\text{ m}^2/\text{s}.

Temperature Dependence of Viscosity

  • Liquids: Viscosity is dominated by cohesive intermolecular forces. As temperature increases, intermolecular bonds weaken, causing dynamic viscosity to decrease significantly.
  • Gases: Viscosity is dominated by molecular collision and momentum exchange across streamlines. As temperature increases, molecular kinetic energy increases, causing gas viscosity to increase.

3. Surface Tension, Capillarity, and Bulk Modulus

Surface Tension (σ\sigma)

Surface tension is the tensile force per unit length acting at the interface between two immiscible fluids or between a liquid and a gas, arising from unbalanced cohesive molecular attractions at the surface (SI unit: N/m\text{N/m}). For clean water in air at 20∘C20^\circ\text{C}, σ≈0.0728 N/m\sigma \approx 0.0728\text{ N/m}.

Internal pressure intensity ΔP=Pinternal−Pexternal\Delta P = P_{\text{internal}} - P_{\text{external}} depends on geometric curvature:

  1. Spherical Liquid Droplet (1 interface): ΔP=4σd=2σR\Delta P = \frac{4\sigma}{d} = \frac{2\sigma}{R}
  2. Soap Bubble / Hollow Sphere (2 interfaces): ΔP=8σd=4σR\Delta P = \frac{8\sigma}{d} = \frac{4\sigma}{R}
  3. Cylindrical Liquid Jet: ΔP=2σd=σR\Delta P = \frac{2\sigma}{d} = \frac{\sigma}{R}

Capillary Rise and Depression

When a narrow glass tube of internal diameter dd is inserted into a liquid, the meniscus elevates or depresses due to the balance between adhesive forces (liquid-to-solid) and cohesive forces (liquid-to-liquid). By equating the vertical surface tension component around the tube perimeter to the weight of the liquid column: (σcos⁡θ)(πd)=γ(πd24)h  ⟹  h=4σcos⁡θγd(\sigma \cos \theta)(\pi d) = \gamma \left(\frac{\pi d^2}{4}\right) h \implies h = \frac{4\sigma \cos \theta}{\gamma d} where:

  • hh is capillary height (m\text{m}).
  • θ\theta is the contact angle (wetting angle). For clean glass and pure water, θ≈0∘\theta \approx 0^\circ (cos⁡θ=1.0\cos \theta = 1.0), resulting in capillary rise (h>0h > 0).
  • For clean glass and mercury, θ≈130∘–140∘\theta \approx 130^\circ\text{–}140^\circ (cos⁡θ<0\cos \theta < 0), resulting in capillary depression (h<0h < 0).

Bulk Modulus of Elasticity (EvE_v) & Compressibility (β\beta)

The bulk modulus of elasticity quantifies a fluid's resistance to volumetric compression under uniform hydrostatic pressure: Ev=−ΔPΔV/V=ρdPdρE_v = -\frac{\Delta P}{\Delta V / V} = \rho \frac{dP}{d\rho}

  • SI unit: Pa\text{Pa} or GPa\text{GPa}. For water at standard atmospheric conditions, Ev≈2.20 GPa=2.20×109 PaE_v \approx 2.20\text{ GPa} = 2.20 \times 10^9\text{ Pa}.
  • Compressibility is the reciprocal of the bulk modulus: β=1Ev\beta = \frac{1}{E_v} (Pa−1\text{Pa}^{-1}). Because EvE_v is extremely high, liquids are treated as practically incompressible in standard civil engineering statics.

Vapor Pressure and Cavitation Risk

The saturation vapor pressure (PvP_v) is the partial pressure exerted by vapor molecules in thermodynamic equilibrium with liquid at a given temperature. If the local absolute pressure within a hydraulic system (such as at the crest of a siphon, a pipe bend, or near pump suction impellers) drops to or below the fluid's vapor pressure (Pv=2.34 kPaP_v = 2.34\text{ kPa} absolute for water at 20∘C20^\circ\text{C}), the liquid rapidly vaporizes at ambient temperature, producing vapor bubbles. When these bubbles travel into higher-pressure zones, they collapse violently, generating microjets and localized shock pressures exceeding hundreds of megapascals—a damaging phenomenon known as cavitation.


4. Hydrostatic Pressure Fundamentals

Pascal's Principle and Pressure Direction

Pascal's Principle dictates that external pressure applied to an enclosed static fluid is transmitted undiminished in all directions. Fluid pressure at any point is isotropic (Px=Py=PzP_x = P_y = P_z) and always acts perpendicular (normal) to any contacting boundary surface.

Absolute vs. Gauge vs. Vacuum Pressure

Static pressure is referenced against two distinct baselines:

  1. Absolute Pressure (PabsP_{\text{abs}}): Measured relative to an absolute perfect vacuum (zero pressure datum).
  2. Gauge Pressure (PgaugeP_{\text{gauge}}): Measured relative to local atmospheric pressure (PatmP_{\text{atm}}).
  3. Vacuum (Negative Gauge) Pressure (PvacP_{\text{vac}}): Pressure below local atmospheric pressure.

Pabs=Pgauge+PatmP_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}} Pvac=Patm−Pabs=−PgaugeP_{\text{vac}} = P_{\text{atm}} - P_{\text{abs}} = -P_{\text{gauge}}

Unit / SystemStandard Atmospheric Value at Sea Level
Kilopascals (SI)101.325 kPa=101,325 N/m2101.325\text{ kPa} = 101,325\text{ N/m}^2
Bar1.01325 bar1.01325\text{ bar} (1 bar=100 kPa1\text{ bar} = 100\text{ kPa})
Water Column Height10.33 m of freshwater10.33\text{ m of freshwater} ("10.33 m H2O""10.33\text{ m H}_2\text{O}")
Mercury Column Height760 mm Hg=76 cm Hg=29.92 in Hg760\text{ mm Hg} = 76\text{ cm Hg} = 29.92\text{ in Hg}
Pounds per Square Inch14.696 psi≈14.7 lb/in214.696\text{ psi} \approx 14.7\text{ lb/in}^2

Pressure Variation with Depth

Equating vertical static forces on an elemental fluid prism of cross-sectional area dAdA and height dhdh: dPdh=γ=ρg  ⟹  dP=γdh\frac{dP}{dh} = \gamma = \rho g \implies dP = \gamma dh Integrating for an incompressible homogeneous fluid with constant specific weight γ\gamma: P2−P1=γ(h2−h1)=γhP_2 - P_1 = \gamma (h_2 - h_1) = \gamma h where hh is the vertical depth below the reference level. The quantity Pγ=h\frac{P}{\gamma} = h is designated as the pressure head (expressed in meters of fluid).


5. Manometry and Pressure Measurement Devices

Manometers utilize columns of static liquid to measure pressure differences.

Classification of Manometers

  1. Piezometer Tube: A simple vertical glass tube open to the atmosphere at the top, tapped directly into a pipe or container. It measures positive liquid gauge pressure directly: P=γhP = \gamma h. It cannot measure gas pressures (gas escapes) or vacuum pressures (air is sucked into the vessel).
  2. Open U-Tube Manometer: A U-shaped tube containing a heavy, immiscible gage fluid (typically mercury, SG=13.6SG = 13.6). It can measure high liquid or gas pressures as well as vacuum pressures.
  3. Differential Manometer: Connects two points in a piping system or two different fluid tanks to measure the differential pressure ΔP=PA−PB\Delta P = P_A - P_B.
  4. Inverted U-Tube Manometer: An upside-down U-tube containing a lighter fluid (such as air or light oil) at the top, used for measuring small pressure differences between two liquid lines with high precision.

The Step-by-Step Continuous Meniscus Method

To write the governing hydrostatic equation for any multi-fluid manometer:

  1. Start at one terminal point (e.g., container AA or open atmosphere), writing down its pressure symbol (PAP_A or Patm=0 kPa gaugeP_{\text{atm}} = 0\text{ kPa gauge}).
  2. Traverse through the tube step-by-step from interface to interface:
    • Moving downward through a vertical column of fluid of specific weight γi\gamma_i and height hih_i adds pressure: +γihi+ \gamma_i h_i.
    • Moving upward through a vertical column subtracts pressure: −γihi- \gamma_i h_i.
  3. Jump horizontally across identical continuous fluids: Two points at the exact same horizontal elevation in the same continuous, homogeneous, static liquid have identical pressures. You can transition directly between legs at this datum.
  4. Equate the running sum to the pressure at the final terminal point (e.g., =PB= P_B or =Patm= P_{\text{atm}}).

6. Worked Example: Multi-Fluid Differential Manometer

Problem Statement: A differential U-tube manometer connects Pipe A carrying freshwater (γw=9.81 kN/m3\gamma_w = 9.81\text{ kN/m}^3) and Pipe B carrying hydraulic oil (SGo=0.82SG_o = 0.82, γo=0.82×9.81=8.044 kN/m3\gamma_o = 0.82 \times 9.81 = 8.044\text{ kN/m}^3). The gage liquid in the bottom loop is mercury (SGm=13.6SG_m = 13.6, γm=13.6×9.81=133.42 kN/m3\gamma_m = 13.6 \times 9.81 = 133.42\text{ kN/m}^3).

  • Centerline elevation of Pipe A: Elev 12.00 m\text{Elev } 12.00\text{ m}
  • Centerline elevation of Pipe B: Elev 13.50 m\text{Elev } 13.50\text{ m}
  • Water descends from Pipe A to the water-mercury interface in the left leg at Elev 10.80 m\text{Elev } 10.80\text{ m}.
  • The mercury-oil interface in the right leg rests at Elev 11.45 m\text{Elev } 11.45\text{ m}.
  • Oil fills the right leg from the mercury interface up to Pipe B.

Determine the pressure difference PA−PBP_A - P_B in kilopascals.

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Step-by-Step Solution:

  1. Calculate vertical column heights:

    • Freshwater column in left leg: hw=12.00 m−10.80 m=1.20 mh_w = 12.00\text{ m} - 10.80\text{ m} = 1.20\text{ m} (downward traverse from Pipe A).
    • Mercury differential column: hm=11.45 m−10.80 m=0.65 mh_m = 11.45\text{ m} - 10.80\text{ m} = 0.65\text{ m} (upward traverse in right leg above the left meniscus level).
    • Hydraulic oil column in right leg: ho=13.50 m−11.45 m=2.05 mh_o = 13.50\text{ m} - 11.45\text{ m} = 2.05\text{ m} (upward traverse to Pipe B).
  2. Set up the continuous meniscus manometer equation: PA+γwhw−γmhm−γoho=PBP_A + \gamma_w h_w - \gamma_m h_m - \gamma_o h_o = P_B

  3. Isolate differential pressure (PA−PB)(P_A - P_B): PA−PB=γmhm+γoho−γwhwP_A - P_B = \gamma_m h_m + \gamma_o h_o - \gamma_w h_w

  4. Substitute numerical values: PA−PB=(133.42 kN/m3)(0.65 m)+(8.044 kN/m3)(2.05 m)−(9.81 kN/m3)(1.20 m)P_A - P_B = (133.42\text{ kN/m}^3)(0.65\text{ m}) + (8.044\text{ kN/m}^3)(2.05\text{ m}) - (9.81\text{ kN/m}^3)(1.20\text{ m}) PA−PB=86.723 kPa+16.490 kPa−11.772 kPa=91.44 kPaP_A - P_B = 86.723\text{ kPa} + 16.490\text{ kPa} - 11.772\text{ kPa} = 91.44\text{ kPa}


7. CELE Exam Traps & Common Computational Errors

Warning

Trap 1: Horizontal Jump Across Discontinuous Fluids: You may ONLY equate pressures at equal elevations when remaining within the same continuous homogeneous liquid. Jumping horizontally between the left and right legs at an elevation where the left contains water and the right contains oil violates Pascal's law.

Warning

Trap 2: Absolute vs. Gauge Inconsistency: When computing vacuum pressures, bulk modulus compression, or cavitation thresholds, always work in absolute pressure. If atmospheric pressure is 101.325 kPa101.325\text{ kPa}, a gauge pressure reading of −35.0 kPa-35.0\text{ kPa} translates to an absolute pressure of 101.325−35.0=66.325 kPa abs101.325 - 35.0 = 66.325\text{ kPa abs}.

Warning

Trap 3: Viscosity Unit Confusion: Do not confuse dynamic viscosity μ\mu (Pa⋅s\text{Pa}\cdot\text{s}) with kinematic viscosity ν\nu (m2/s\text{m}^2/\text{s}). In shear stress calculations (τ=μdvdy\tau = \mu \frac{dv}{dy}), using ν\nu instead of μ\mu causes an error proportional to the fluid density ρ\rho (1000×1000\times error for water!).

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Classification of Fluid Properties and Pressure Measurement Hierarchy
Test Your Knowledge

A sample of water initially occupying a volume of 1.000 m³ is subjected to a pressure increase under laboratory conditions. If the bulk modulus of elasticity of the water is 2.20 GPa, what pressure increase ΔP is required to compress the water volume by exactly 0.75%?

A

8.25 MPa

B

33.00 MPa

C

22.00 MPa

D

16.50 MPa

Test Your Knowledge

An open vertical U-tube manometer containing mercury (SG = 13.6, specific weight γ = 133.42 kN/m³) is connected to a vacuum chamber. The mercury level in the leg open to the atmosphere (where atmospheric pressure is 101.30 kPa) is found to be 320 mm lower than the mercury level in the leg connected directly to the chamber. What is the absolute pressure inside the vacuum chamber?

A

144.0 kPa

B

82.4 kPa

C

58.6 kPa

D

68.5 kPa

Test Your Knowledge

A clean glass capillary tube with an internal diameter of 2.0 mm is dipped vertically into a bath of liquid mercury at 20°C (surface tension σ = 0.514 N/m, contact angle θ = 140°, specific weight γ = 133.4 kN/m³). Which of the following describes the capillary meniscus behavior inside the tube?

A

A capillary depression of 5.90 mm

B

A capillary rise of 2.95 mm

C

A capillary depression of 11.81 mm

D

A capillary rise of 5.90 mm

Sections you finish are checked off in the contents.