Fluid Properties — Viscosity, Density, Surface Tension | EngineeringHulk

Fluid Properties — Viscosity, Density & Surface Tension

Fundamental physical properties of fluids essential for every Fluid Mechanics problem in GATE CE

Last Updated: April 2026

Key Takeaways
  • A fluid is a substance that continuously deforms (flows) under any shear stress, however small.
  • Dynamic viscosity (μ) relates shear stress to velocity gradient; kinematic viscosity ν = μ/ρ.
  • For liquids, viscosity decreases with temperature; for gases, viscosity increases with temperature.
  • Specific gravity (SG) = ρfluid / ρwater; water at 4 °C has ρ = 1000 kg/m³.
  • Surface tension (σ) causes capillary rise in tubes; capillary height h = 4σ cosθ / (ρgd).
  • Bulk modulus K = –V (dp/dV) measures compressibility; water is nearly incompressible (K ≈ 2.1 GPa).
  • Newton’s law of viscosity: τ = μ (du/dy) — GATE CE favourite for 1–2 mark questions every year.

1. Definition & Classification of Fluids

A fluid is defined as a substance that cannot sustain a shear stress in static equilibrium. Unlike a solid (which deforms a fixed amount under shear and then stops), a fluid deforms continuously as long as any shear stress — even infinitesimally small — acts on it.

1.1 Liquids vs Gases

PropertyLiquidGas
Molecular spacingSmall (cohesive)Large (widely spaced)
Definite volume?YesNo
Definite shape?No (takes shape of container)No (fills container)
CompressibilityVery low (nearly incompressible)High
Free surface formed?YesNo

1.2 Types of Fluids (based on viscosity behaviour)

Fluid TypeRelationship τ vs du/dyExample
NewtonianLinear through origin (μ = constant)Water, air, thin oils
Pseudo-plastic (shear thinning)Concave curve; μ decreases with strain rateBlood, polymer solutions, paints
Dilatant (shear thickening)Convex curve; μ increases with strain rateCornstarch suspension, wet sand
Bingham PlasticRequires yield stress τ₀ before flowToothpaste, drilling mud, concrete
Ideal Fluidτ = 0 always (μ = 0); theoreticalInviscid flow assumption in theory

For GATE CE, Newtonian fluids dominate all numerical problems. The Bingham plastic model appears as conceptual/MCQ questions.

2. Density, Specific Weight & Specific Gravity

2.1 Mass Density (ρ)

ρ = m / V

Units: kg/m³

Water at 4 °C: ρw = 1000 kg/m³
Seawater: ≈ 1025 kg/m³
Air at STP: ≈ 1.2 kg/m³
Mercury: ≈ 13,600 kg/m³

2.2 Specific Weight (γ)

γ = ρg = W / V

Units: N/m³

Water: γw = 1000 × 9.81 = 9810 N/m³ ≈ 9.81 kN/m³

2.3 Specific Gravity (SG) / Relative Density

SG = ρfluid / ρwater = γfluid / γwater

Dimensionless. SG of water = 1.0; Mercury = 13.6; Petrol ≈ 0.72

2.4 Effect of Temperature on Density

For liquids: density decreases with rising temperature (thermal expansion).
For ideal gases: ρ = p / (RT) — density decreases if pressure is constant and temperature rises.

Temp (°C)ρ of Water (kg/m³)μ of Water (×10⁻³ Pa·s)
0999.81.792
41000.01.567
20998.21.002
40992.20.653
100958.40.282

3. Viscosity — Newton’s Law & Types

Viscosity is the measure of a fluid’s internal resistance to flow (resistance to deformation by shear stress).

3.1 Newton’s Law of Viscosity

τ = μ (du/dy)

where:

  • τ = shear stress (Pa or N/m²)
  • μ = dynamic (absolute) viscosity (Pa·s or N·s/m²)
  • du/dy = velocity gradient (rate of shear strain) in s⁻¹

SI unit of μ: Pa·s (also written as N·s/m²)
CGS unit: Poise (P); 1 P = 0.1 Pa·s

3.2 Kinematic Viscosity (ν)

ν = μ / ρ

SI unit: m²/s
CGS unit: Stoke (St); 1 St = 10⁻⁴ m²/s
Water at 20 °C: μ = 1.002 × 10⁻³ Pa·s; ν = 1.004 × 10⁻⁶ m²/s

3.3 Variation of Viscosity with Temperature

FluidEffect of ↑ Temperature on μReason
Liquidsμ decreases (fluid becomes less viscous)Reduced intermolecular cohesion at higher temp
Gasesμ increases (fluid becomes more viscous)Increased molecular momentum exchange (kinetic theory)

Empirical formulae:

Liquids (Andrade’s equation): μ = A · eB/T

Gases (Sutherland’s law): μ/μ₀ = (T/T₀)3/2 × (T₀+S)/(T+S)

where T is in Kelvin and S is Sutherland’s constant (≈ 110 K for air)

3.4 No-Slip Condition

At a solid boundary, fluid velocity equals the wall velocity (zero for a stationary wall). This is the no-slip condition — a direct consequence of viscosity and is the basis for the velocity gradient (du/dy) between layers.

4. Surface Tension & Capillarity

4.1 Surface Tension (σ)

Surface tension is the force per unit length acting along the liquid surface due to unbalanced molecular cohesion at the interface.

σ = F / L

Units: N/m
Water at 20 °C: σ ≈ 0.0728 N/m
Mercury at 20 °C: σ ≈ 0.4800 N/m

4.2 Pressure Inside a Droplet / Bubble

Liquid droplet: Δp = 4σ / d (one interface)

Soap bubble: Δp = 8σ / d (two interfaces — inner & outer)

Liquid jet (cylinder): Δp = 2σ / d

where d = diameter

4.3 Capillarity

Capillary action occurs in narrow tubes due to the interplay of cohesion (within the liquid) and adhesion (between liquid and tube wall).

h = 4σ cosθ / (ρ g d)

where:

  • h = capillary rise (+) or depression (–) in metres
  • σ = surface tension (N/m)
  • θ = contact angle with tube wall
  • ρ = fluid density (kg/m³)
  • g = 9.81 m/s²
  • d = tube diameter (m)

Water in glass: θ ≈ 0° → cosθ = 1 → capillary RISE
Mercury in glass: θ ≈ 135° → cosθ = –0.707 → capillary DEPRESSION

Liquid–Tube CombinationContact Angle θMeniscusCapillary Effect
Water – glassConcaveRise
Mercury – glass135°ConvexDepression
Water – paraffin~107°ConvexDepression

5. Compressibility & Bulk Modulus

5.1 Bulk Modulus of Elasticity (K)

K = – V (dp / dV) = ρ (dp / dρ)

Units: Pa or N/m²

Water: K ≈ 2.1 × 10⁹ Pa = 2.1 GPa
Air (isothermal): K = p (atmospheric ≈ 10⁵ Pa)
Air (adiabatic): K = γp where γ = 1.4

5.2 Compressibility (β)

β = 1/K = –(1/V)(dV/dp)

High K → Low compressibility (stiff fluid like water)
Low K → High compressibility (like gases)

5.3 Speed of Sound in a Fluid

c = √(K/ρ)

Water: c ≈ √(2.1×10⁹ / 1000) ≈ 1450 m/s
Air (20 °C): c ≈ 343 m/s

For most civil engineering hydraulics (velocities much less than 1450 m/s), water is treated as incompressible (ρ = constant). Compressibility is important only in water hammer analysis.

6. Vapour Pressure & Cavitation

6.1 Vapour Pressure (pv)

Vapour pressure is the pressure at which a liquid begins to vaporise at a given temperature. It increases with temperature.

Temperature (°C)pv of Water (kPa absolute)
00.611
202.337
407.375
100101.325 (boiling)

6.2 Cavitation

Cavitation occurs when the local pressure in a flowing liquid drops to or below the vapour pressure. Vapour bubbles form and then implode violently when they move to a higher-pressure region, causing noise, vibration, erosion, and damage to pump impellers and pipe walls.

Prevention: Ensure the minimum pressure in the system stays well above pv. In pump design, this is quantified by the Net Positive Suction Head (NPSH).

7. Worked Examples (GATE CE Level)

Example 1 — Newton’s Law of Viscosity (GATE CE 2019 type)

Problem: A flat plate moves over a fixed bottom plate with a gap of 2 mm filled with oil (μ = 0.8 Pa·s). If the upper plate moves at 0.5 m/s, find the shear stress on the upper plate.

Given:
Gap y = 2 mm = 0.002 m
Velocity of upper plate u = 0.5 m/s (lower plate fixed → u at y = 0 is 0)
μ = 0.8 Pa·s

Velocity gradient (linear profile for Newtonian fluid):

du/dy = Δu / Δy = (0.5 – 0) / 0.002 = 250 s⁻¹

Shear stress:

τ = μ × (du/dy) = 0.8 × 250 = 200 Pa

Answer: τ = 200 Pa

Example 2 — Capillary Rise (GATE CE level)

Problem: Water (σ = 0.073 N/m, ρ = 1000 kg/m³, θ = 0°) rises in a glass tube of diameter 0.5 mm. Find the capillary rise h.

Given:
σ = 0.073 N/m, θ = 0° → cosθ = 1
ρ = 1000 kg/m³, g = 9.81 m/s²
d = 0.5 mm = 0.0005 m

Formula:

h = 4σ cosθ / (ρ g d)

h = (4 × 0.073 × 1) / (1000 × 9.81 × 0.0005)

h = 0.292 / 4.905

h = 0.0595 m ≈ 59.5 mm

Answer: h ≈ 59.5 mm

Example 3 — Kinematic Viscosity & Specific Gravity

Problem: A liquid has a dynamic viscosity of 5 × 10⁻³ Pa·s and a specific gravity of 0.85. Find its kinematic viscosity in m²/s and in Stokes.

Given:
μ = 5 × 10⁻³ Pa·s
SG = 0.85 → ρ = 0.85 × 1000 = 850 kg/m³

Kinematic viscosity:

ν = μ / ρ = (5 × 10⁻³) / 850 = 5.882 × 10⁻⁶ m²/s

Convert to Stokes:

1 St = 10⁻⁴ m²/s
ν = 5.882 × 10⁻⁶ / 10⁻⁴ = 0.05882 St = 5.882 cSt

Answer: ν = 5.88 × 10⁻⁶ m²/s = 5.88 cSt

Example 4 — Pressure Inside a Soap Bubble

Problem: Find the pressure difference inside a soap bubble of diameter 4 cm. Surface tension of soap solution = 0.025 N/m.

Given:
d = 4 cm = 0.04 m; σ = 0.025 N/m

Soap bubble has TWO surfaces:

Δp = 8σ / d = (8 × 0.025) / 0.04 = 0.2 / 0.04 = 5 Pa

Answer: Δp = 5 Pa

Example 5 — Bulk Modulus & Volumetric Compression

Problem: A 2 m³ volume of water is subjected to a pressure increase of 5 MPa. Find the decrease in volume. (Kwater = 2.1 GPa)

Given:
V = 2 m³, Δp = 5 × 10⁶ Pa, K = 2.1 × 10⁹ Pa

Volumetric strain:

ΔV / V = –Δp / K = –(5 × 10⁶) / (2.1 × 10⁹) = –2.381 × 10⁻³

Change in volume:

ΔV = –2.381 × 10⁻³ × 2 = –4.76 × 10⁻³ m³ = –4.76 litres

Answer: Volume decreases by ≈ 4.76 × 10⁻³ m³ (4.76 litres)

8. Common Mistakes

Mistake 1 — Confusing Dynamic and Kinematic Viscosity

Error: Using μ and ν interchangeably in the Reynolds number formula or Darcy-Weisbach calculations.
Root Cause: Both have “viscosity” in the name. The Reynolds number Re = VD/ν uses kinematic viscosity, while shear stress τ = μ(du/dy) uses dynamic viscosity.
Fix: Always check units. μ has units Pa·s; ν has units m²/s. Re = VD/ν (no density term needed because ν = μ/ρ already incorporates it).

Mistake 2 — Wrong Surface Tension Formula for Bubbles vs Droplets

Error: Applying Δp = 4σ/d for a soap bubble (should be 8σ/d).
Root Cause: Forgetting that a soap bubble has TWO liquid–air interfaces (inner wall + outer wall), while a liquid droplet has only ONE.
Fix: Droplet = 4σ/d; Soap bubble = 8σ/d; Liquid jet = 2σ/d. Memorise this trio.

Mistake 3 — Capillary Rise Formula: Using Radius Instead of Diameter

Error: Writing h = 4σ cosθ / (ρgr) or getting confused about whether to use r or d.
Root Cause: Some textbooks derive the formula in terms of radius r (giving h = 2σ cosθ / ρgr); others state it in terms of diameter d (h = 4σ cosθ / ρgd). Both are correct but must not be mixed.
Fix: Derive once from force balance: upward force = σ(πd)cosθ; weight of liquid = ρg(πd²/4)h. Equating gives h = 4σ cosθ/(ρgd).

Mistake 4 — Assuming Viscosity of Liquids Increases with Temperature

Error: Believing all fluids become more viscous when heated (correct only for gases).
Root Cause: Everyday intuition — hot things feel different. But liquid viscosity is dominated by intermolecular cohesion which weakens at higher temperatures, making liquids flow more easily.
Fix: Remember the rule: Liquids → μ ↓ with ↑T; Gases → μ ↑ with ↑T.

Mistake 5 — Treating Density of Water as 1000 kg/m³ at All Temperatures

Error: Using ρ = 1000 kg/m³ for water problems regardless of temperature.
Root Cause: Memorising a single standard value without noting its temperature dependence (maximum density occurs at 4 °C).
Fix: For GATE, 1000 kg/m³ is accepted unless a specific temperature is given. If T = 20 °C or 100 °C, use the table value. Always note if the problem specifies temperature.

9. Frequently Asked Questions

Q1. What is the physical significance of kinematic viscosity?

Kinematic viscosity ν = μ/ρ represents the ratio of viscous forces to inertial forces per unit volume — or equivalently, the momentum diffusivity of the fluid. It determines how quickly momentum is diffused through the fluid (how fast a velocity perturbation spreads). Fluids with high ν (like thick oil) diffuse momentum rapidly, dampening flow disturbances; fluids with low ν (like water) resist momentum diffusion, making turbulence easier to sustain. In the Reynolds number Re = VD/ν, a higher ν for the same velocity and geometry means lower Re (tendency toward laminar flow).

Q2. Why does mercury show capillary depression in glass while water shows rise?

The direction of capillary action depends on the contact angle θ between the liquid and the tube material. For water–glass, adhesion (water molecules attracted to glass, which is silica-based and polar) dominates over cohesion, so θ < 90° (≈ 0°), cos θ is positive, and the liquid rises. For mercury–glass, cohesion among mercury atoms (strong metallic bonds) dominates over adhesion to glass, so θ > 90° (≈ 135°), cos θ is negative, and the formula gives a negative h — i.e., capillary depression. This is why mercury thermometers have a convex meniscus while water-filled manometers have a concave one.

Q3. What is cavitation and why is it important in civil engineering hydraulics?

Cavitation is the formation of vapour cavities (bubbles) in a flowing liquid when the local absolute pressure drops to or below the vapour pressure at that temperature. These bubbles collapse violently when they travel into higher-pressure zones, generating intense local shock waves (pressure up to several hundred MPa). In civil engineering, cavitation is critical in:

  • Pump design: Cavitation on impeller blades causes pitting, vibration, noise, and efficiency loss. NPSH (Net Positive Suction Head) criteria are used to prevent it.
  • Hydraulic turbines: Francis and Kaplan turbines are susceptible near the draft tube exit.
  • High-velocity spillways: Chute blocks and flip buckets can experience cavitation at flow velocities above ~12 m/s, requiring aerators or special surface finishes.

Q4. How is Newton’s law of viscosity different for a Bingham plastic?

For a Newtonian fluid: τ = μ (du/dy) — flow begins the moment any shear stress is applied, and the relationship is linear with zero intercept. For a Bingham plastic: τ = τ₀ + μpl(du/dy), but only when τ > τ₀. Below the yield stress τ₀, the material behaves like a rigid solid and does not flow. This is why toothpaste or fresh concrete sits still on a nearly flat surface but flows when sufficient force is applied. In geotechnical engineering, this model also applies to soft clays, mud slurries, and grout flow in grouting operations.