CELE Hydraulics & Fluid Mechanics — Properties of FluidsDetailed Explanation
This is the "office hours" version of Properties of Fluids for the CELE 2026. No shortcuts, no hand-waving — just a full unpacking of why Professional Regulation Commission (PRC) — Board of Civil Engineering cares about each concept and how the Hydraulics & Fluid Mechanics section items tend to play out on exam day. Read this once, then hit the practice questions with real understanding.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Hydraulics & Fluid Mechanics section sits under a "Core" weighting, and Properties of Fluids is the 1st chapter in the 10-chapter CELE Hydraulics & Fluid Mechanics rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Hydraulics & Fluid Mechanics.
Properties of Fluids - Detailed Explanation
The physical properties of fluids are the foundation of every topic in Hydraulics and Fluid Mechanics. Whether you are computing hydrostatic pressure, designing a water supply network, or sizing a pump, you must first understand density, specific weight, viscosity, surface tension, compressibility, and vapor pressure. In the PRC Civil Engineer Licensure Examination, questions on fluid properties appear both as standalone items and as embedded sub-problems in flow and pressure topics. This chapter establishes the precise definitions, governing equations, and board-exam strategies you need to score on these items. All values are in SI units; standard water properties used throughout are ρ = 1000 kg/m³, γ = 9.81 kN/m³, and g = 9.81 m/s².
Concepts
Density, Specific Weight, and Specific Gravity
Three interrelated scalar properties describe how much matter or weight a fluid contains per unit volume. **Density (ρ)** is the mass per unit volume: ρ = m / V (SI unit: kg/m³) For liquid water at 4 °C: ρ = 1000 kg/m³. For seawater: ρ ≈ 1025 kg/m³. **Specific weight (γ)**, also called unit weight, is the gravitational force per unit volume: γ = ρ · g (SI unit: N/m³ or kN/m³) For water: γ = 1000 × 9.81 = 9810 N/m³ = 9.81 kN/m³. **Specific volume (v)** is the reciprocal of density: v = 1 / ρ (SI unit: m³/kg) **Specific gravity (s)**, also called relative density, is the dimensionless ratio of the fluid density to the density of pure water at 4 °C: s = ρ_fluid / ρ_water = γ_fluid / γ_water Because it is dimensionless, specific gravity is the same in any unit system — a very common board-exam advantage. Key relationships: γ = s · γ_water = s × 9.81 kN/m³ ρ = s · ρ_water = s × 1000 kg/m³ In Philippine practice, oil is frequently specified by its specific gravity (e.g., s = 0.85 for a light crude), and you back-calculate ρ and γ from it.
Examples
Starting from specific gravity is the fastest route. Multiply s by the water reference value for either ρ or γ directly.
Scenario
An oil has specific gravity s = 0.88. Determine its density ρ and specific weight γ.
Solution
ρ = s × ρ_water = 0.88 × 1000 = 880 kg/m³ γ = ρ × g = 880 × 9.81 = 8632.8 N/m³ ≈ 8.63 kN/m³ Alternatively: γ = s × γ_water = 0.88 × 9.81 kN/m³ = 8.63 kN/m³
Density is computed from fundamentals (mass/volume), then all other properties follow from the standard relationships.
Scenario
A 0.50-m³ tank is filled with a liquid whose mass is 420 kg. Find ρ, s, and γ.
Solution
ρ = m / V = 420 / 0.50 = 840 kg/m³ s = ρ / ρ_water = 840 / 1000 = 0.84 γ = ρ × g = 840 × 9.81 = 8240.4 N/m³ = 8.24 kN/m³
Applications
- Hydrostatic pressure calculations: p = γh requires γ of the fluid.
- Buoyancy analysis: net upward force = γ_fluid × V_displaced.
- Mixture problems: computing average density or specific gravity of blended liquids.
- Pipeline design: checking whether a fluid (e.g., brine or crude oil) is heavier or lighter than water to set pump heads.
- Sedimentation and slurry transport in Philippine dredging projects.
Misconceptions
- Confusing specific weight γ (N/m³) with density ρ (kg/m³) — they differ by factor g = 9.81.
- Using g = 10 m/s² — always use 9.81 m/s² unless the problem explicitly states 10.
- Thinking specific gravity equals specific weight — s is dimensionless; γ is not.
- Applying water density 1000 kg/m³ to seawater — seawater is denser (≈ 1025 kg/m³).
Related Concepts
- Hydrostatic pressure (p = γh)
- Buoyancy and Archimedes' principle
- Fluid statics on submerged surfaces
- Manometry (uses specific gravities of multiple fluids)
Common Exam Questions
Example
A liquid occupies 2 L and weighs 18 N. Find its specific gravity. Solution: γ = W/V = 18/0.002 = 9000 N/m³; s = 9000/9810 = 0.917.
Approach
Given mass, volume, or specific gravity, apply ρ = m/V, γ = ρg, and s = ρ/ρ_water directly.
Question Type
Direct property calculation
Example
Mercury has s = 13.6. Its specific weight = 13.6 × 9.81 = 133.4 kN/m³.
Approach
s × ρ_water or s × γ_water to get ρ or γ of the unknown fluid.
Question Type
Back-calculation from specific gravity
Key Points To Remember
- ρ (kg/m³) × g (m/s²) = γ (N/m³) — always multiply by g = 9.81 m/s², not 9.8 or 10.
- Specific gravity is dimensionless and equals the ratio to water density or specific weight.
- For water: ρ = 1000 kg/m³, γ = 9.81 kN/m³, s = 1.0 exactly.
- For seawater: s ≈ 1.025; for mercury: s ≈ 13.6; for typical oil: s ≈ 0.80–0.90.
- Specific volume v = 1/ρ — rarely asked directly but appears in thermodynamics-linked problems.
- γ and ρ both vary with temperature; water is densest at 4 °C.
Viscosity
Viscosity is the property that quantifies a fluid's resistance to shear deformation — its internal 'stickiness'. When adjacent fluid layers move at different velocities, shear stress develops between them. **Newton's Law of Viscosity:** τ = μ (dv/dy) where: τ = shear stress (Pa = N/m²) μ = dynamic (absolute) viscosity (Pa·s) dv/dy = velocity gradient perpendicular to flow (s⁻¹), also called the rate of shear deformation. **Kinematic viscosity (ν):** ν = μ / ρ (SI unit: m²/s) Kinematic viscosity appears in Reynolds number, pipe-flow, and boundary-layer equations. The older CGS unit stokes (St) = 1 cm²/s = 10⁻⁴ m²/s; 1 centistokes = 10⁻⁶ m²/s. **Newtonian vs. Non-Newtonian fluids:** A Newtonian fluid (water, air, light oil) obeys Newton's law exactly — at a given temperature, μ is constant regardless of the shear rate. A non-Newtonian fluid (paint, blood, drilling mud, ketchup) has an apparent viscosity that depends on the shear rate. The PRC board exam deals almost exclusively with Newtonian fluids. **Temperature effect:** For liquids, μ decreases as temperature rises (molecules have more energy, bonds loosen). For gases, μ increases with temperature. This is a classic exam trick question. **Practical note for the board exam:** When a problem gives a film of liquid between two plates, assume a **linear velocity profile** (dv/dy = V/h, where V = plate velocity and h = film thickness). The shear force on the plate = τ × A.
Examples
The film thickness must be in meters (0.0015 m, not 1.5). The velocity gradient is simply the plate velocity divided by the film thickness under the linear-profile assumption.
Scenario
A flat plate slides over an oil film 1.5 mm thick at a velocity of 3 m/s. The oil has dynamic viscosity μ = 0.002 Pa·s. Find the shear stress in the film.
Solution
Velocity gradient: dv/dy = V/h = 3 / 0.0015 = 2000 s⁻¹ Shear stress: τ = μ(dv/dy) = 0.002 × 2000 = 4.0 Pa
Board exams frequently ask for force (not just stress). Multiply shear stress by the plate area.
Scenario
A 200 mm × 300 mm plate is pulled at 1.2 m/s over a 0.4 mm oil film (μ = 0.05 Pa·s). Calculate the shear force on the plate.
Solution
A = 0.200 × 0.300 = 0.06 m² dv/dy = 1.2 / 0.0004 = 3000 s⁻¹ τ = 0.05 × 3000 = 150 Pa F = τ × A = 150 × 0.06 = 9.0 N
Kinematic viscosity is the ratio of dynamic viscosity to density. It appears in dimensionless groups like Reynolds number (Re = VD/ν).
Scenario
An oil has μ = 0.0008 Pa·s and ρ = 860 kg/m³. Compute kinematic viscosity ν in m²/s.
Solution
ν = μ / ρ = 0.0008 / 860 = 9.30 × 10⁻⁷ m²/s
Applications
- Pipe-flow friction (Hagen-Poiseuille equation for laminar flow: ΔP = 128μLQ/(πD⁴)).
- Reynolds number Re = ρVD/μ = VD/ν — determines laminar vs. turbulent regime.
- Journal bearing and lubrication analysis in mechanical systems.
- Viscous drag on submerged bodies (ships, piles, bridge piers).
- Oil selection for hydraulic systems in Philippine infrastructure machinery.
Misconceptions
- Thinking viscosity is a property of the flow, not the fluid — it depends on temperature and fluid type only.
- Forgetting to convert mm to m for film thickness — the most common arithmetic error in viscosity problems.
- Confusing μ (Pa·s) with ν (m²/s) — dynamic and kinematic are not interchangeable.
- Assuming all fluids are Newtonian — valid for water and most engineering oils, but not for slurries or polymers.
- Believing that a more viscous liquid is always denser — viscosity and density are independent properties.
Related Concepts
- Reynolds number and flow regime classification
- Hagen-Poiseuille laminar flow in pipes
- Boundary layer theory
- Turbulence and the Darcy-Weisbach friction factor
Common Exam Questions
Example
Plate 0.5 m² moves at 2 m/s over a 2 mm film (μ = 0.003 Pa·s). τ = 0.003(2/0.002) = 3 Pa; F = 3 × 0.5 = 1.5 N.
Approach
Identify V (velocity), h (film thickness in meters), μ; apply τ = μ(V/h). Then F = τA if force is asked.
Question Type
Shear stress in a fluid film
Example
μ = 1.002 × 10⁻³ Pa·s for water at 20 °C; ρ = 998 kg/m³; ν = 1.004 × 10⁻⁶ m²/s.
Approach
ν = μ/ρ. Watch units — μ in Pa·s, ρ in kg/m³ gives ν in m²/s.
Question Type
Dynamic vs. kinematic viscosity conversion
Key Points To Remember
- τ = μ(dv/dy) — Newton's Law of Viscosity. Know which symbol means what.
- Dynamic viscosity μ has units Pa·s (N·s/m²). Kinematic viscosity ν has units m²/s.
- ν = μ/ρ — divide dynamic by density to get kinematic.
- Newtonian fluids: μ is independent of shear rate (constant at a given temperature).
- For liquids: μ decreases with increasing temperature. For gases: μ increases with temperature.
- Linear velocity profile assumption: dv/dy = V/h when a plate slides over a thin oil film.
- Shear force on a plate: F = τ × A = μ(V/h) × A.
Surface Tension and Capillarity
**Surface tension (σ)** arises because molecules at a liquid surface have a net inward cohesive force — they lack neighbors on one side. This creates a 'skin' effect measurable as force per unit length or energy per unit area: σ (SI unit: N/m) For water at 20 °C: σ ≈ 0.0728 N/m. Surface tension decreases with increasing temperature. **Contact angle (θ):** When a liquid meets a solid surface, the contact angle between the liquid surface and the solid governs whether the liquid wets the surface: • θ < 90°: wetting liquid (water on clean glass) → capillary rise. • θ > 90°: non-wetting liquid (mercury on glass) → capillary depression. • θ = 0° is assumed for water on clean glass in most board problems. **Capillary rise (or depression):** h = 4σ cosθ / (γ · d) where d = internal tube diameter (m), γ = specific weight of the liquid (N/m³), and h is positive for rise and negative for depression. Alternate form using radius r = d/2: h = 2σ cosθ / (γ · r) **Key insight:** Smaller tube diameter → larger capillary rise. This is why water rises higher in silt or clay soils (very fine pores) than in coarse sand. **Pressure inside a droplet or bubble:** Droplet: Δp = 4σ / d Bubble (two surfaces): Δp = 8σ / d In the PRC board exam, capillary rise problems almost always assume θ = 0° (water, clean glass), so cosθ = 1 and the formula simplifies to h = 4σ/(γd).
Examples
Always convert diameter to meters. With θ = 0°, cosθ = 1, so the formula simplifies. This is the standard board-exam setup.
Scenario
Water (σ = 0.0728 N/m, γ = 9810 N/m³, θ = 0°) rises in a clean glass capillary tube of internal diameter 2 mm. Find the height of capillary rise.
Solution
h = 4σ cosθ / (γ d) h = 4 × 0.0728 × cos(0°) / (9810 × 0.002) h = 0.2912 / 19.62 h = 0.01484 m ≈ 14.8 mm
Halving the tube diameter doubles the capillary rise — confirm: at 2 mm the rise was ~14.8 mm; at 1 mm it is ~29.7 mm (exactly double), consistent with h ∝ 1/d.
Scenario
Find the capillary rise of water (σ = 0.0728 N/m, θ = 0°) in a 1 mm diameter tube.
Solution
h = 4 × 0.0728 × 1 / (9810 × 0.001) h = 0.2912 / 9.81 h = 0.02969 m ≈ 29.7 mm
A droplet has one air-water interface. The pressure inside is higher than ambient by 4σ/d. A soap bubble would have twice this because it has two surfaces.
Scenario
A spherical water droplet has diameter 0.5 mm. Compute the pressure difference between the inside and outside of the droplet (σ = 0.0728 N/m).
Solution
Δp = 4σ / d = 4 × 0.0728 / 0.0005 = 582.4 Pa
Applications
- Capillary rise in fine-grained soils (silts, clays) — relevant to soil mechanics and groundwater.
- Interpretation of piezometer readings — capillary effects cause errors in small-diameter tubes.
- Droplet formation in spray nozzles for irrigation systems.
- Adhesion of concrete to formwork (wetting behavior).
- Design of wicks and drainage geotextiles in Philippine infrastructure projects.
Misconceptions
- Using radius r instead of diameter d in h = 4σcosθ/(γd) — double-check the formula each time.
- Forgetting to convert mm to m for tube diameter — the single most common error in this topic.
- Thinking σ has units N/m² (pressure) rather than N/m (line force per unit length).
- Assuming capillary rise occurs for all liquids — mercury depresses (θ > 90°) rather than rising.
Related Concepts
- Piezometric measurements (capillary effects in manometers)
- Soil capillarity and seepage in geotechnical engineering
- Wettability in concrete and waterproofing
- Droplet dynamics in hydraulic spray systems
Common Exam Questions
Example
Water in a 3 mm tube: h = 4(0.0728)(1)/(9810 × 0.003) = 0.2912/29.43 = 9.9 mm.
Approach
Identify d (in meters), σ, θ (usually 0° for water), and γ. Substitute directly into h = 4σcosθ/(γd).
Question Type
Capillary rise calculation
Example
To achieve h = 20 mm: d = 4(0.0728)(1)/(9810 × 0.020) = 0.0015 m = 1.5 mm.
Approach
Rearrange: d = 4σcosθ/(γh). Solve for d in meters, then convert to mm.
Question Type
Determining tube diameter for target rise
Key Points To Remember
- σ has units N/m (not N/m²) — it is a line force, not a pressure.
- Capillary rise formula: h = 4σ cosθ / (γd) — uses diameter d, not radius.
- For water on glass: θ ≈ 0°, cosθ = 1 (maximum rise).
- For mercury in glass: θ ≈ 140°, cosθ is negative → capillary depression.
- Rise increases as tube diameter decreases (inversely proportional).
- Surface tension decreases as temperature increases.
- Pressure excess in a droplet: Δp = 4σ/d; in a bubble (two interfaces): Δp = 8σ/d.
Compressibility and Bulk Modulus of Elasticity
**Compressibility** measures how much a fluid volume changes under applied pressure. For most liquids, including water, this change is very small — liquids are treated as incompressible in virtually all hydraulics problems at the board-exam level. **Bulk modulus of elasticity (E_B):** E_B = -dp / (dV/V) = dp / (dρ/ρ) SI unit: Pa (or GPa for liquids) The negative sign in the first form appears because pressure increases when volume decreases (dV is negative when dp is positive). The second form in terms of density change is more convenient: E_B = dp / (dρ/ρ) For water at ambient conditions: E_B ≈ 2.2 GPa = 2.2 × 10⁹ Pa. This is extremely large — it means you need enormous pressures to change water's volume noticeably. **Compressibility (β):** The reciprocal of bulk modulus: β = 1 / E_B (Pa⁻¹) **Speed of sound in a fluid:** c = √(E_B / ρ) (m/s) For water: c ≈ √(2.2 × 10⁹ / 1000) ≈ 1483 m/s — useful in water-hammer analysis. **Practical application — Water hammer:** When a valve is suddenly closed in a pipeline, the kinetic energy of flowing water creates a pressure surge. The bulk modulus governs the speed and magnitude of this pressure wave. Ignoring water's small but finite compressibility would give an infinite wave speed, which is unphysical. **Board-exam calculation:** Given: pressure increase Δp, find fractional volume change ΔV/V: ΔV/V = -Δp / E_B or equivalently, fractional density change: Δρ/ρ = Δp / E_B
Examples
The negative sign indicates volume compression. The tiny fraction (less than 0.1%) justifies treating water as incompressible in standard hydraulics.
Scenario
A pressure increase of 2 MPa is applied to water (E_B = 2.2 GPa). Find the fractional change in volume.
Solution
ΔV/V = −Δp / E_B = −2 × 10⁶ / (2.2 × 10⁹) = −9.09 × 10⁻⁴ The volume decreases by approximately 0.091% — confirming water's near-incompressibility.
Sound travels roughly 4.3 times faster in water than in air (≈ 343 m/s), which is why underwater communication using acoustics is feasible over long distances.
Scenario
Water has E_B = 2.2 × 10⁹ Pa and ρ = 1000 kg/m³. Compute the speed of sound in water.
Solution
c = √(E_B / ρ) = √(2.2 × 10⁹ / 1000) = √(2.2 × 10⁶) = 1483 m/s
Applications
- Water-hammer analysis in long pipelines (e.g., MWSS transmission mains in Metro Manila).
- Hydraulic pressure testing of pipelines — uses E_B to relate pressure drop to micro-leakage.
- Acoustic depth sounding (sonar) — based on speed of sound in water.
- Deep-sea pressure effects on density and buoyancy calculations.
- Hydraulic shock absorbers and accumulators in heavy construction equipment.
Misconceptions
- Thinking incompressible means E_B = ∞ — it means E_B is very large (finite but huge for liquids).
- Mixing up ΔV/V (negative for compression) and Δρ/ρ (positive for compression) — the signs are opposite.
- Confusing bulk modulus with Young's modulus (for solids) — they are analogous but applied to different media.
- Neglecting units — Δp must be in Pa if E_B is in Pa.
Related Concepts
- Water-hammer pressure surges (Joukowski equation)
- Speed of sound in fluids
- Pressure waves and transient flow
- Ideal gas law (compressibility of gases)
Common Exam Questions
Example
Δp = 5 MPa, E_B = 2.2 GPa: Δρ/ρ = 5×10⁶ / 2.2×10⁹ = 2.27 × 10⁻³ (0.227% density increase).
Approach
Use ΔV/V = −Δp/E_B or Δρ/ρ = Δp/E_B. Express Δp and E_B in the same pressure units.
Question Type
Fractional volume or density change
Example
E_B = 2.2 GPa, ρ = 1000 kg/m³: c = 1483 m/s.
Approach
c = √(E_B/ρ). Convert E_B to Pa and ρ to kg/m³.
Question Type
Speed of sound in a liquid
Key Points To Remember
- E_B for water ≈ 2.2 GPa — very large, confirming near-incompressibility.
- E_B = dp/(dρ/ρ) — pressure change divided by fractional density change.
- Fractional volume change: ΔV/V = −Δp/E_B (negative = compression).
- Fractional density change: Δρ/ρ = Δp/E_B (positive = density increases under compression).
- Speed of sound: c = √(E_B/ρ) — relevant in water-hammer and acoustic problems.
- Liquids: large E_B → nearly incompressible. Gases: small E_B → highly compressible.
- Compressibility β = 1/E_B is the inverse of bulk modulus.
Vapor Pressure and Cavitation
**Vapor pressure (p_v)** is the pressure at which a liquid is in thermodynamic equilibrium with its vapor at a given temperature — in other words, the pressure at which the liquid begins to vaporize (boil) at that temperature. Key points: • p_v increases with increasing temperature. • At 20 °C, p_v for water ≈ 2.34 kPa (absolute). • At 100 °C, p_v = 101.325 kPa (standard atmospheric pressure) — that is why water boils at 100 °C at sea level. **Cavitation:** When the local absolute pressure in a flowing fluid drops to or below p_v, the liquid flashes into vapor bubbles. These bubbles then collapse violently when they move to a higher-pressure region, generating intense localized pressure spikes — cavitation. **Effects of cavitation:** • Pitting and erosion of metal surfaces (pump impellers, turbine runners, valve seats). • Vibration and noise in hydraulic machinery. • Significant reduction in pump and turbine efficiency. • In severe cases, complete pump failure. **Where cavitation occurs:** • Pump suction (inlet) side — especially in high-elevation installations. • Constrictions and throttling valves (venturi throats, orifices). • Sharp bends and rapid velocity changes in pipelines. **NPSH (Net Positive Suction Head):** The available NPSH at a pump inlet must exceed the required NPSH (from the pump manufacturer) to prevent cavitation. This is a classic board-exam topic linking vapor pressure to pump selection: NPSH_available = (p_atm - p_v)/γ + z_s - h_f where z_s = suction head and h_f = friction head losses on suction side.
Examples
Each meter of suction lift and each meter of friction head loss reduces available NPSH. This is why pump stations for Philippine water utilities place pumps as close to the water source as practicable.
Scenario
A pump draws water at 20 °C (p_v = 2.34 kPa abs). Atmospheric pressure is 101.325 kPa. If the suction pipe has 4 m of friction head loss and the pump is located 3 m above the suction reservoir, compute the NPSH available.
Solution
NPSH_available = (p_atm − p_v) / γ + z_s − h_f Convert pressures to head: (101,325 − 2,340) / 9810 = 98,985 / 9810 = 10.09 m Note: z_s is negative because the pump is above the reservoir (suction lift = −3 m using the sign convention where rise reduces NPSH): NPSH_avail = 10.09 − 3.0 − 4.0 = 3.09 m If the pump's required NPSH is 2.5 m, cavitation will NOT occur (3.09 > 2.5). If required NPSH is 3.5 m, cavitation WILL occur.
Applications
- Pump station design — limiting suction lift to prevent cavitation.
- Turbine runner design for hydroelectric plants (e.g., Angat, Magat dams).
- Pipeline system design — avoiding low-pressure zones at pipe summits.
- Hydraulic model testing — using vapor pressure to define scale limits.
- Selection of pump materials (cavitation-resistant alloys and coatings).
Misconceptions
- Using gauge pressure instead of absolute pressure when comparing to vapor pressure — always use absolute.
- Thinking cavitation only occurs in pumps — it also occurs in turbines, valves, and any high-velocity constriction.
- Assuming p_v is negligible — at high temperatures or low pressures, p_v becomes significant.
- Confusing vapor pressure with atmospheric pressure — p_v is a fluid property dependent on temperature.
Related Concepts
- NPSH in centrifugal pump design
- Bernoulli's equation and pressure at constrictions
- Pipeline hydraulic grade line analysis
- Water supply system design under NAMPAP/LWUA standards
Common Exam Questions
Example
Pressure at pipe throat = −95 kPa gauge = 101.3 − 95 = 6.3 kPa abs. Since 6.3 kPa > 2.34 kPa (p_v at 20 °C), no cavitation.
Approach
Convert gauge pressure to absolute. If absolute pressure ≤ p_v at the given temperature, cavitation occurs.
Question Type
Cavitation identification
Example
Suction lift 5 m, h_f = 2 m, T = 20 °C: NPSH = 10.09 − 5 − 2 = 3.09 m.
Approach
Apply NPSH = (p_atm − p_v)/γ − z_s(lift) − h_f. Compare with required NPSH from pump data.
Question Type
NPSH computation
Key Points To Remember
- Vapor pressure p_v increases with temperature — always specify the temperature.
- Cavitation occurs when local absolute pressure ≤ p_v.
- Standard value: p_v (water, 20 °C) ≈ 2.34 kPa absolute.
- Cavitation causes erosion, noise, vibration, and efficiency loss in pumps and turbines.
- NPSH must be checked in all pump installation problems involving suction lift.
- Lowering pump elevation (reducing suction lift) is the primary remedy for cavitation.
- Absolute pressure = gauge pressure + atmospheric pressure (101.325 kPa at sea level).
Practice Problems
The weight in Newtons divided by volume gives specific weight directly. Divide by g = 9.81 for density. Specific gravity is the ratio to water density. This liquid is a medium-density oil (s ≈ 0.87 is typical of diesel fuel).
Problem
Problem 1. An unknown liquid fills a 0.80 m³ container. A weight measurement shows the liquid weighs 6.86 kN. Determine: (a) specific weight γ, (b) density ρ, (c) specific gravity s.
Solution
(a) γ = W / V = 6860 N / 0.80 m³ = 8575 N/m³ = 8.575 kN/m³ (b) ρ = γ / g = 8575 / 9.81 = 874.1 kg/m³ (c) s = ρ / ρ_water = 874.1 / 1000 = 0.874
Step (a) finds shear stress from force and area. Step (b) back-solves Newton's law for μ — note h must be in meters. Step (c) converts to kinematic viscosity. This is a complete viscosity problem covering all three sub-quantities the board exam may ask.
Problem
Problem 2. Two parallel plates are separated by an oil film of thickness 3 mm. The lower plate is fixed; the upper plate (area = 0.25 m²) is moved at a constant velocity of 2.5 m/s by a horizontal force of 6.25 N. Find: (a) shear stress τ, (b) dynamic viscosity μ, (c) kinematic viscosity ν if ρ = 870 kg/m³.
Solution
(a) τ = F / A = 6.25 / 0.25 = 25.0 Pa (b) From Newton's law: τ = μ (V/h) μ = τ · h / V = 25.0 × 0.003 / 2.5 = 0.030 Pa·s (c) ν = μ / ρ = 0.030 / 870 = 3.45 × 10⁻⁵ m²/s
Convert diameter 0.8 mm → 0.0008 m before substituting. The rise is significant — 37 mm in a sub-millimetre tube. This illustrates why glass capillary tubes must have diameters above ~6 mm to keep capillary effects below 5 mm in precision manometry.
Problem
Problem 3. Determine the height of capillary rise of water at 20 °C in a clean glass tube of diameter 0.8 mm. Use σ = 0.0728 N/m, θ = 0°, γ = 9810 N/m³.
Solution
h = 4σ cosθ / (γ d) h = 4 × 0.0728 × cos(0°) / (9810 × 0.0008) h = 0.2912 / 7.848 h = 0.03711 m = 37.1 mm
Seawater is 2.5% denser than fresh water — the difference matters in harbor and offshore structure design. Specific volume is rarely asked alone but may appear in thermodynamic contexts.
Problem
Problem 4. Seawater has specific gravity s = 1.025. Compute: (a) density ρ, (b) specific weight γ, (c) specific volume v.
Solution
(a) ρ = s × ρ_water = 1.025 × 1000 = 1025 kg/m³ (b) γ = ρ × g = 1025 × 9.81 = 10,055 N/m³ = 10.06 kN/m³ (c) v = 1 / ρ = 1 / 1025 = 9.756 × 10⁻⁴ m³/kg
The magnitudes of ΔV/V and Δρ/ρ are equal (both 0.20%) but opposite in sign — compression decreases volume and increases density. The tiny fraction confirms that assuming water is incompressible introduces only 0.20% error even at 4.4 MPa, far beyond typical pipe pressures.
Problem
Problem 5. Water at 20 °C (E_B = 2.2 GPa) is subjected to a pressure increase of 4.4 MPa. Find: (a) fractional change in volume ΔV/V, (b) fractional change in density Δρ/ρ.
Solution
(a) ΔV/V = −Δp / E_B = −4.4 × 10⁶ / (2.2 × 10⁹) = −2.0 × 10⁻³ The volume decreases by 0.20%. (b) Δρ/ρ = Δp / E_B = 4.4 × 10⁶ / (2.2 × 10⁹) = 2.0 × 10⁻³ The density increases by 0.20%.
At 25 °C, vapor pressure is slightly higher than at 20 °C (3.17 vs 2.34 kPa), reducing available NPSH slightly. Always use the actual operating temperature. A margin of less than 1 m is a design caution flag in practice.
Problem
Problem 6 (Board-style integrative). A pump lifts water at 25 °C (p_v = 3.17 kPa abs) from a reservoir 4.5 m below the pump centerline. The suction pipe has a total head loss of 1.8 m. Atmospheric pressure = 101.325 kPa. The pump manufacturer specifies a required NPSH of 3.0 m. Will cavitation occur?
Solution
NPSH_available = (p_atm − p_v) / γ − z_s − h_f where z_s = 4.5 m (suction lift, pump above reservoir) = (101,325 − 3,170) / 9810 − 4.5 − 1.8 = 98,155 / 9810 − 4.5 − 1.8 = 10.005 − 4.5 − 1.8 = 3.705 m Since NPSH_available (3.705 m) > NPSH_required (3.0 m), cavitation will NOT occur. The margin is 0.705 m — acceptable but relatively tight.
Exam Preparation Tips
- Memorize the four water reference values: ρ = 1000 kg/m³, γ = 9.81 kN/m³, s = 1.0, and p_v(20 °C) = 2.34 kPa abs. Everything else derives from these.
- Always use g = 9.81 m/s² — never round to 10 unless the problem explicitly states it. A 2% error from rounding can shift you between answer choices.
- Unit discipline is critical in fluid properties: μ is Pa·s (N·s/m²); ν is m²/s; σ is N/m; E_B is Pa or GPa; p_v is Pa or kPa absolute.
- For viscosity problems: convert all thicknesses from mm to m before computing dv/dy = V/h. This single conversion eliminates the most common board-exam arithmetic error.
- For capillary problems: h = 4σcosθ/(γd) uses diameter d, not radius. Write out the formula from scratch each time to avoid the r-vs-d confusion.
- For compressibility: Δρ/ρ = +Δp/E_B (density rises); ΔV/V = −Δp/E_B (volume falls). Signs are opposite.
- For cavitation/NPSH: ALWAYS convert gauge pressure to absolute pressure before comparing to vapor pressure. The formula NPSH = (p_atm − p_v)/γ − z_suction_lift − h_f_suction is board-exam standard.
- Specific gravity problems are the fastest to solve — multiply s by the water reference value and you are done. Identify these problems early in the exam to bank quick points.
- Temperature effects: liquid viscosity decreases with rising temperature; gas viscosity increases; surface tension decreases; vapor pressure increases. These pattern-recognition questions require no calculation.
- In multiple-choice items, eliminate choices that mix up units (e.g., answer choices with γ in kg/m³ or μ in m²/s are immediately wrong). Use dimensional analysis as a filter.
- Practice the three-step cycle: (1) write the governing formula, (2) substitute with consistent SI units, (3) check the answer's magnitude against reference values (e.g., h_capillary for a 1 mm tube should be ~30 mm — if you get 30 m, you forgot to convert mm to m).
- Review the PRC board exam syllabus topic 'Properties of Fluids' — it consistently appears in Hydraulics Set A or Set B, typically 3–5 items per 100-item examination.
In summary
The properties of fluids — density, specific weight, specific gravity, viscosity, surface tension, compressibility, and vapor pressure — are not isolated facts to be memorized. They are the physical language through which every hydraulics equation is written. Mastery of this chapter means you can instantly recognize which property a problem is testing, recall the correct formula with proper SI units, and execute the solution without unit-conversion errors. For the PRC Civil Engineer Licensure Examination, keep these anchors fixed in your memory: water has ρ = 1000 kg/m³ and γ = 9.81 kN/m³; Newton's viscosity law is τ = μ(dv/dy) with μ in Pa·s; capillary rise uses diameter d and gives h = 4σcosθ/(γd); water's bulk modulus is 2.2 GPa confirming near-incompressibility; and cavitation begins when absolute pressure reaches vapor pressure. These five anchors, combined with disciplined unit conversion and the solution flowcharts presented here, will handle the vast majority of board-exam fluid-properties items. Proceed to the next chapters — Fluid Statics, Fluid Kinematics, and Pipe Flow — with confidence, knowing that the property relationships established here underpin every pressure, velocity, and energy equation you will encounter throughout the examination.
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