Fluid Properties, Pressure and Manometers
Density, viscosity and Newton's law of viscosity, surface tension and capillarity, hydrostatic pressure, gauge and absolute pressure, and manometer problems with solved examples.
Fluid mechanics begins with the properties of fluids and with fluids at rest, where pressure depends only on depth.
Fluid properties
- Density ρ (kg/m³); specific weight γ = ρg; specific gravity = ρ/ρ_water (ρ_water = 1000 kg/m³).
- Viscosity: Newton's law of viscosity, τ = μ du/dy. μ is the dynamic viscosity (Pa·s); ν = μ/ρ is the kinematic viscosity (m²/s).
- Liquids get less viscous when heated; gases get more viscous.
- Non-Newtonian fluids: shear-thinning (paint), shear-thickening, Bingham plastic (toothpaste).
- Surface tension σ (N/m): pressure inside a droplet Δp = 4σ/d; inside a soap bubble (two surfaces) Δp = 8σ/d.
- Capillary rise in a tube of diameter d: h = 4σ cos θ/(ρ g d). Water rises (θ ≈ 0°); mercury falls (θ > 90°).
- Bulk modulus K = −dp/(dV/V); compressibility = 1/K.
Hydrostatic pressure
In a fluid at rest, pressure increases with depth:
p = p₀ + ρ g h
- Pressure at a point is the same in all directions (Pascal's law).
- Absolute pressure = gauge pressure + atmospheric pressure (about 101.3 kPa).
- Points at the same level in the same continuous fluid at rest have the same pressure.
Manometers
Start at one end, add ρgh going down and subtract going up, and set the sum equal to the pressure at the other end.
- Piezometer: p = ρgh (only for small positive gauge pressures of liquids).
- U-tube: p_A + ρ₁ g h₁ − ρ_m g h_m = p_atm.
- Differential manometer between two pipes A and B with the same fluid ρ and manometer fluid ρ_m, reading x: p_A − p_B = (ρ_m − ρ) g x (when A and B are at the same level).
Worked example
Water flows in two pipes at the same level connected by a mercury differential manometer (SG 13.6) with a reading of 100 mm. Find p_A − p_B.
- p_A − p_B = (13600 − 1000) × 9.81 × 0.1 = 12.36 kPa.
- In metres of water: 12.6 × 0.1 = 1.26 m.
Practice
What is the gauge pressure (kPa) at 5 m depth in oil of specific gravity 0.8? (g = 9.81 m/s², give 2 decimal places.)
Show answer and explanation
Answer: 39.24p = ρgh = 800 × 9.81 × 5 = 39,240 Pa = 39.24 kPa.
The velocity gradient in a fluid of μ = 0.1 Pa·s is 200 s⁻¹. What is the shear stress in Pa?
Show answer and explanation
Answer: 20τ = μ du/dy = 0.1 × 200 = 20 Pa.
Water (σ = 0.072 N/m, θ = 0°) is in a glass tube of diameter 1 mm. What is the capillary rise in mm? (g = 9.81 m/s², give 1 decimal place.)
Show answer and explanation
Answer: 29.4h = 4σ/(ρgd) = 0.288/9.81 = 0.0294 m = 29.4 mm.
When a liquid is heated, its viscosity generally:
Show answer and explanation
Answer: AHeat weakens the cohesive forces between liquid molecules, so liquids flow more easily. Gases behave the opposite way.