Summary notes: Properties of Fluids

Everything in Chapter 1 of A Textbook of Fluid Mechanics, in the order the chapter teaches it. Read this first, then take the questions. Boxes marked Correction show where the textbook is wrong or unclear and what is right.

Formula sheet

1.1 Introduction

Fluid mechanics is the branch of science that deals with the behaviour of fluids (liquids or gases) at rest as well as in motion. It therefore covers the static, kinematic and dynamic aspects of fluids.

• Fluid statics: the study of fluids at rest.
• Fluid kinematics: the study of fluids in motion without considering the forces that cause the motion (only velocities, accelerations and flow patterns).
• Fluid kinetics: the study of fluids in motion when the forces causing the motion, such as pressure forces, are also considered.
• Fluid dynamics: the study of fluids in motion as a whole. It is made up of kinematics and kinetics.

1.2 Basic properties of fluids

1.2.1 Density or mass density

Density is the ratio of the mass of a fluid to its volume, so it is the mass per unit volume. It is written ρ (rho): ρ = mass of fluid ÷ volume of fluid. The SI unit is kg/m³. The density of a liquid can be treated as constant, while the density of a gas changes with pressure and temperature. Water has a density of 1 gm/cm³, which is 1000 kg/m³.

1.2.2 Specific weight or weight density

Specific weight is weight per unit volume, written w: w = (mass × g) ÷ volume = ρg (equation 1.1). For water, w = 9.81 × 1000 N/m³ in SI units.

1.2.3 Specific volume

Specific volume is the volume occupied by a unit mass, or volume per unit mass: volume ÷ mass = 1/ρ. It is the reciprocal of mass density, it is measured in m³/kg, and it is commonly applied to gases.

1.2.4 Specific gravity

Specific gravity (also called relative density) is the ratio of the weight density (or density) of a fluid to that of a standard fluid. The standard is water for liquids and air for gases. It has no dimensions and is written S.

• Weight density of a liquid = S × 1000 × 9.81 N/m³.
• Density of a liquid = S × 1000 kg/m³ (equation 1.1A).
• Example: mercury has S = 13.6, so its density is 13.6 × 1000 = 13600 kg/m³.

Problem 1.1 (one litre weighing 7 N): the volume is 1/1000 m³, so w = 7000 N/m³, ρ = 7000 ÷ 9.81 = 713.5 kg/m³ and S = 0.7135. Problem 1.2 (petrol, S = 0.7, one litre): ρ = 700 kg/m³, w = 6867 N/m³ and the weight is 6867 × 0.001 = 6.867 N.

1.3 Viscosity

Viscosity is the property of a fluid that resists the movement of one layer of fluid over an adjacent layer. Take two layers a distance dy apart, moving at velocities u and u + du (Fig. 1.1, the velocity profile near a solid boundary). The viscosity and this relative velocity cause a shear stress between the layers. The top layer drags the lower layer forward, and the lower layer holds the top layer back. The shear stress τ (tau) is proportional to the rate of change of velocity with respect to y:

μ (mu) is the constant of proportionality and is called the coefficient of dynamic viscosity, or simply viscosity. du/dy is the rate of shear strain, the rate of shear deformation, or the velocity gradient. Viscosity can also be defined as the shear stress needed to produce unit rate of shear strain.

1.3.1 Units of viscosity (textbook page 4)

Put dimensions into equation 1.3:

Converting MKS to CGS: 1 kgf s/m² = 9.81 N s/m². Since 1 N = 1 kg × 1 m/s² = (1000 gm)(100 cm)/s² = 1000 × 100 dyne, we get 1 kgf s/m² = 9.81 × 100000 dyne s/(100 × 100 cm²) = 98.1 poise. So a viscosity in poise must be divided by 98.1 to get its MKS value. Likewise 1 N s/m² = 98.1/9.81 = 10 poise, so one poise = 1/10 N s/m².

Alternate method. One poise = dyne s/cm² = (1 gm × 1 cm/s²)(s/cm²) = 1 gm/(s cm) = (1/1000 kg) ÷ (s × 1/100 m) = 1/10 kg/(s m). So 1 kg/(s m) = 10 poise.

• In SI units the second is written 's', not 'sec'.
• If a viscosity is given in poise, divide by 10 to get the SI number.
• 1 centipoise (cP) = 1/100 poise.
• The viscosity of water at 20°C is 0.01 poise, or 1.0 centipoise.

1.3.2 Kinematic viscosity

Kinematic viscosity is the ratio of dynamic viscosity to density. It is written ν (nu): ν = μ/ρ (equation 1.4). Its dimensions are (Force × Time/Length²) ÷ (Mass/Length³) = Length²/Time, using force = mass × length/time². In MKS and SI the unit is m²/s. In CGS it is cm²/s, which is called the stoke. One stoke = cm²/s = (1/100)² m²/s = 10⁻⁴ m²/s. A centistoke is 1/100 stoke.

1.3.3 Newton's law of viscosity

The shear stress on a fluid layer is directly proportional to the rate of shear strain. The constant of proportionality is the coefficient of viscosity: τ = μ du/dy. Fluids that obey this relation are Newtonian, and fluids that do not obey it are non-Newtonian.

1.3.4 Variation of viscosity with temperature

Viscous forces in a fluid come from cohesive forces and from molecular momentum transfer.

• Liquids: the molecules are closely packed, so cohesive forces dominate. A rise in temperature weakens cohesion, so viscosity decreases.
• Gases: cohesive forces are small and molecular momentum transfer dominates. A rise in temperature increases the momentum transfer, so viscosity increases.

Here μ is the viscosity at t°C, μ₀ is the viscosity at 0°C, and α and β are constants of the fluid. For water the book gives μ₀ = 1.79 × 10⁻³, α = 0.03368 and β = 0.000221. For air it gives μ₀ = 0.000017, α = 0.000000056 and β = 0.1189 × 10⁻⁹.

1.3.5 Types of fluids

Fluids are classified into five types (Fig. 1.2 plots shear stress against velocity gradient du/dy):

The "ideal solid" in Fig. 1.2 lies along the shear stress axis: it takes stress with no deformation rate.

Worked problems 1.3 to 1.19 (viscosity)

• 1.3: u = ⅔y − y², μ = 8.63 poise = 0.863 N s/m². du/dy = ⅔ − 2y, which is 0.667 at y = 0 and 0.367 at y = 0.15. So τ = 0.5756 and 0.3167 N/m².
• 1.4: plate 0.025 mm from a fixed plate, 0.6 m/s, 2 N/m². μ = τ·dy/du = 8.33 × 10⁻⁵ N s/m² = 8.33 × 10⁻⁴ poise.
• 1.5: area 1.5 m², du = 0.4 m/s, dy = 0.15 mm, μ = 0.1 N s/m². τ = 266.66 N/m², F = τ × area = 400 N, and power = F × u = 160 W (Nm/s = watt).
• 1.6: shaft D = 0.1 m at 150 r.p.m. gives u = πDN/60 = 0.785 m/s. With clearance 1.5 mm, τ = 52.33 N/m².
• 1.7: square plate on a 30° incline. The shear force equals the component of weight along the plane, 300 cos 60° = 150 N, so τ = 150/0.64. This gives μ = 1.17 N s/m² = 11.7 poise.
• 1.8: plates 1.25 cm apart, μ = 14 poise, u = 2.5 m/s. τ = 1.4 × 2.5/0.0125 = 280 N/m².
• 1.9: square plates of side 0.6 m, film 12.5 mm, F = 98.1 N, u = 2.5 m/s. This gives μ = 1.3625 N s/m² = 13.625 poise. With S = 0.95, ρ = 950 and ν = 0.001434 m²/s = 14.34 stokes.
• 1.10: ρ = 981, τ = 0.2452, du/dy = 0.2 gives μ = 1.226, so ν = 0.125 × 10⁻² m²/s = 12.5 stokes.
• 1.11: μ = 0.05 poise, ν = 0.035 stokes gives ρ = 1428.5 kg/m³, so S = 1.43.
• 1.12: ν = 6 stokes, S = 1.9, so ρ = 1900 and μ = 1.14 N s/m² = 11.40 poise.
• 1.13: u = ¾y − y² gives du/dy = 0.45 at y = 0.15. The book then uses 8.5 poise and gets 0.3825 N/m².
• 1.14: shaft D = 0.4 m, N = 190 r.p.m., L = 90 mm, film 1.5 mm, μ = 0.6 N s/m². u = 3.98 m/s, τ = 1592 N/m², F = τπDL = 180.05 N, torque T = F × D/2 = 36.01 N m, and power lost = 2πNT/60 = 716.48 W.
• 1.15: a parabolic profile u = ay² + by + c with u = 0 at y = 0 (so c = 0), u = 120 cm/s at y = 20 cm, and du/dy = 0 at y = 20 cm. This gives a = −0.3, b = 12 and u = −0.3y² + 12y. Then du/dy = −0.6y + 12, which is 12, 6 and 0 per second at y = 0, 10 and 20 cm. With μ = 0.85, τ = 10.2, 5.1 and 0 N/m².
• 1.16: F = Aμu/y, so F ∝ u (A, μ and y are constant). F₁/u₁ = F₂/u₂ gives 40/50 = 200/u₂, so u₂ = 250 cm/s.
• 1.17: coaxial cylinders, D = 0.15 m, outer 0.151 m, L = 0.25 m, T = 12 N m, N = 100 r.p.m. u = 0.7854 m/s, area = 0.1178 m², gap dy = 0.0005 m. Torque = F × D/2 gives μ = 0.864 N s/m² = 8.64 poise.
• 1.18: thin plate between two large surfaces 2.4 cm apart (glycerine, μ = 0.81 N s/m², u = 0.6 m/s, area 0.5 m²). Force F = F₁ + F₂, one shear force on each side. In the middle, each gap is 1.2 cm and F₁ = F₂ = 20.25 N, so F = 40.5 N. With the plate 0.8 cm from one surface, F₁ = 15.18 N (gap 1.6 cm) and F₂ = 30.36 N (gap 0.8 cm), so F = 45.54 N.
• 1.19: a 1.2 m × 1.2 m × 0.2 cm plate lifted at 0.15 m/s in the middle of a 2.2 cm vertical gap. Each side gap is (2.2 − 0.2)/2 = 1 cm. Shear force per side = 2.0 × (0.15/0.01) × 1.44 = 43.2 N, so the total shear is 86.4 N. The plate displaces 0.00288 m³ of fluid, so the upthrust is 9.81 × 900 × 0.00288 = 25.43 N. The net weight is 40 − 25.43 = 14.57 N, so the force needed is 86.4 + 14.57 = 100.97 N.

1.4 Thermodynamic properties

Gases are compressible, so with changes of pressure and temperature their density varies a great deal. The relation between absolute pressure, specific volume and absolute temperature is the equation of state: pV = RT or p/ρ = RT (equation 1.5). Here p is the absolute pressure in N/m², V is the specific volume (1/ρ), R is the gas constant, T is the absolute temperature and ρ is the density.

1.4.1 Dimension of R

R = p/(ρT). It depends on the particular gas. In MKS units R has the unit kgf-m/(kg K). In SI units R = (N/m²) ÷ [(kg/m³) K] = N m/(kg K) = J/(kg K). For air, R = 29.3 kgf-m/(kg K) in MKS, which is 29.3 × 9.81 = 287 J/(kg K) in SI.

1.4.2 Isothermal process

A change of density that occurs at constant temperature is isothermal: p/ρ = constant (1.6).

1.4.3 Adiabatic process

A change of density with no heat exchange to or from the gas is adiabatic. If no heat is generated within the gas by friction either, then p/ρᵏ = constant (1.7). Here k is the ratio of the specific heat at constant pressure to the specific heat at constant volume, and k = 1.4 for air. For a mass of gas, pV = constant (isothermal) and pVᵏ = constant (adiabatic), and combining with pV = mRT gives TVᵏ⁻¹ = constant.

1.4.4 Universal gas constant

For a mass m (kg) of gas of volume V: pV = mRT (1.8). The equation becomes universal when written on a mole basis. Let n be the number of moles and M the molecular weight, so that n × M = m. Then pV = nMRT (1.9). The product M × R is the universal gas constant: 848 kgf-m/(kg-mole K) in MKS and 8314 J/(kg-mole K) in SI. One kilogram mole is the mass in kg equal to the molecular weight of the gas. To get R for a particular gas, divide 8314 by its molecular weight (nitrogen: 8314/28 = 296.9 J/kg K).

Worked problems

• 1.20: a gas weighs 16 N/m³ at 25°C and 0.25 N/mm² absolute. ρ = w/g = 1.63 kg/m³. R = p/(ρT) = 25×10⁴/(1.63 × T).
• 1.21: 0.6 m³ of air at 50°C (323 K) and 0.3 N/mm², compressed to 0.3 m³. Isothermal: p₂ = p₁V₁/V₂ = 0.6 N/mm². Adiabatic: p₂ = 30×10⁴ × 2¹·⁴ = 0.791 N/mm², and T₂ = T₁(V₁/V₂)ᵏ⁻¹ = 323 × 2⁰·⁴ = 426.2 K, so t₂ = 153.2°C.
• 1.22: 5 kg of nitrogen at 10°C (283 K), 0.4 m³. R = 8314/28 = 296.9, so p = mRT/V = 1050283.7 N/m² = 1.05 N/mm².

1.5 Compressibility and bulk modulus

The bulk modulus of elasticity K is the ratio of compressive stress to volumetric strain. Picture a cylinder with a piston (Fig. 1.9) holding a volume V of gas. Raise the pressure from p to p + dp and the volume drops from V to V − dV. The volumetric strain is −dV/V (the minus sign shows the volume falls as pressure rises). So:

K for a gas. Isothermal: pV = constant, so p dV + V dp = 0, which gives K = p (1.12). Adiabatic: pVᵏ = constant, so p k Vᵏ⁻¹ dV + Vᵏ dp = 0, which gives pk dV = −V dp and K = kp (1.13). So the compressibility of a gas is 1/p (isothermal) or 1/(kp) (adiabatic).

• 1.23: pressure raised from 70 to 130 N/cm² (dp = 60), volume decreases 0.15%. K = 60 ÷ (0.15/100) = 4 × 10⁴ N/cm².
• 1.24: volume 0.0125 → 0.0124 m³ (dV = 0.0001), pressure 80 → 150 N/cm² (dp = 70). K = 70 ÷ (0.0001/0.0125) = 8.75 × 10³ N/cm².

1.6 Surface tension and capillarity

Surface tension is the tensile force acting on the surface of a liquid in contact with a gas, or on the surface between two immiscible liquids, so that the contact surface behaves like a membrane under tension. The force per unit length of the free surface equals the surface energy per unit area. It is written σ (sigma). Units: N/m in SI, kgf/m in MKS, dyne/cm in CGS.

Why it happens (Fig. 1.10). Take three molecules A, B and C of a liquid. Molecule A, deep inside, is attracted equally in all directions, so the resultant force on it is zero. Molecule B, near the free surface, has unbalanced upward and downward forces, so a net force acts downward. Molecule C, at the free surface, feels a resultant downward force. So the free surface acts like a thin elastic film under tension.

1.6.1 Liquid droplet

Let σ be surface tension, p the pressure inside the droplet in excess of outside, and d the diameter. Cut the droplet into two halves (Fig. 1.11). On one half the surface tension force around the cut circumference is σ × πd. The pressure force on the projected area is p × (π/4)d². These are equal and opposite at equilibrium:

As the droplet diameter decreases, the pressure inside it increases.

1.6.2 Hollow bubble

A soap bubble has two surfaces in contact with air, so p × (π/4)d² = 2 × (σπd), which gives p = 8σ/d (1.15).

1.6.3 Liquid jet

For a jet of diameter d and length L (Fig. 1.12), take a semi-jet. Pressure force = p × L × d and surface tension force = σ × 2L. Equating gives p = σ × 2L/(L × d) = 2σ/d (1.16).

1.6.4 Capillarity

Capillarity is the rise or fall of a liquid surface in a small tube relative to the adjacent general level of liquid when the tube is held vertically in the liquid. The rise is capillary rise and the fall is capillary depression. It is expressed in cm or mm of liquid and depends on the specific weight of the liquid, the diameter of the tube and the surface tension.

Capillary rise (Fig. 1.13). Take a glass tube of small diameter d open at both ends, inserted in a liquid such as water. Let h be the rise, σ the surface tension and θ the angle of contact. At equilibrium the weight of the liquid column is balanced by the surface tension force:

For water and clean glass θ ≈ 0, so cosθ = 1 and h = 4σ/(ρ g d) (1.20).

Capillary fall (Fig. 1.14). For mercury in a glass tube, the level inside is lower than outside. Two forces act: the surface tension force σ × πd × cosθ acting downward, and the hydrostatic force acting upward, which is p × (π/4)d² = ρgh × (π/4)d². Equating gives h = 4σ cosθ/(ρ g d) (1.21), and with θ obtuse cosθ is negative, so h is negative (a depression). The book gives θ = 128° for mercury and glass.

• 1.25: p = 0.02 N/cm² = 200 N/m², σ = 0.0725. d = 4σ/p = 0.00145 m = 1.45 mm.
• 1.26: soap bubble of 40 mm, p = 2.5 N/m². σ = pd/8 = 0.0125 N/m.
• 1.27: droplet of 0.04 mm, σ = 0.0725. Excess pressure 4σ/d = 7250 N/m² = 0.725 N/cm², so the pressure inside is 10.32 + 0.725 = 11.045 N/cm².
• 1.28: tube of 2.5 mm. Water rises 1.18 cm. Mercury (σ = 0.52, θ = 130°, density 13600) falls 0.4 cm.
• 1.29: tube of 4 mm. Water (σ = 0.073575, ρ = 998) rises 7.51 mm. Mercury (σ = 0.51, θ = 130°) falls 2.46 mm.
• 1.30: rise not to exceed 0.2 mm, so d = 4σ/(ρgh) = 0.148 m = 14.8 cm.
• 1.31: rise restricted to 2 mm (σ = 0.073575), so d = 0.015 m = 1.5 cm.
• 1.32: shaft D = 0.5 m, 200 r.p.m., L = 0.1 m, film 1 mm, μ = 0.5 N s/m². u = 5.235 m/s, τ = 2617.5 N/m², F = 410.95 N, T = 102.74 N m, power = 2π × 200 × 102.74/60 = 2150 W = 2.15 kW.

1.7 Vapour pressure and cavitation

Vaporisation is the change from the liquid state to the gaseous state. It depends on the prevailing pressure and temperature, and happens because molecules continually escape through the free liquid surface. Take water at 20°C in a closed vessel at atmospheric pressure: it vaporises at 100°C. As vaporisation goes on, vapour molecules collect in the space above the liquid and exert a pressure on the liquid surface. This is the vapour pressure of the liquid, the pressure at which the liquid is converted into vapour.

If the pressure above the liquid is reduced, the boiling temperature also falls. Once the pressure falls to or below the vapour pressure, boiling starts even at 20°C. So a liquid can boil at ordinary temperature if the pressure above it is reduced to its vapour pressure at that temperature.

Cavitation. In a flowing liquid, if the pressure at any point falls to or below the vapour pressure, the liquid vaporises there. The vapour bubbles are carried by the flow into a region of higher pressure, where they collapse suddenly and create very high impact pressure. The metal surfaces the liquid is flowing over are hit by these pressures, which cause a pitting action, so material is eroded and cavities form. Hence the name cavitation: the formation of vapour bubbles in a flowing liquid where the pressure falls below the vapour pressure, followed by their sudden collapse in a region of higher pressure.

Solving the problems: four set-ups

Unit traps: poise ÷ 10 for SI, but ÷ 98.1 for MKS. Stokes × 10⁻⁴ for m²/s. 1 N/cm² = 10⁴ N/m². 1 N/mm² = 10⁶ N/m². Specific weight is in N/m³, not N/m².

All 32 worked problems at a glance

The chapter's highlights

1. Weight density (specific weight) is weight per unit volume: w = ρg.
2. Specific volume is the reciprocal of mass density.
3. Shear stress is proportional to the velocity gradient du/dy: τ = μ du/dy.
4. Kinematic viscosity ν = μ/ρ.
5. Poise and stokes are the units of viscosity and kinematic viscosity respectively.
6. To convert viscosity from poise to MKS units, divide by 98.1. To convert to SI units, divide by 10. The SI unit is N s/m² or Pa s, where N/m² = Pa = pascal.
7. For a perfect gas, p/ρ = RT. For air R = 29.3 kgf-m/(kg K) = 287 J/(kg K).
8. Isothermal: p/ρ = constant. Adiabatic: p/ρᵏ = constant.
9. Bulk modulus K = −dp ÷ (dV/V).
10. Compressibility = 1/K.
11. Surface tension is in N/m or dyne/cm. Excess pressure: p = 4σ/d (droplet), 8σ/d (soap bubble), 2σ/d (liquid jet).
12. Capillary rise or fall h = 4σ cosθ/(wd), where w = ρg. θ is taken as zero for water and 128° for mercury.

The exercises

(A) Theoretical problems (17): definitions of density, weight density, specific volume and specific gravity; differences between liquids and gases, real and ideal fluids, specific weight and specific volume; dynamic versus kinematic viscosity and their units and dimensions; Newton's law of viscosity with applications and the effect of temperature; Newtonian and non-Newtonian fluids; isothermal, adiabatic and universal gas constant; compressibility (proving 1/p and 1/(kp)); surface tension and the proof of p = 4σ/d; capillarity and capillary rise; vacuum pressure and compressible versus incompressible fluids; converting 1 kg/s-m to poise; why gas viscosity rises with temperature and liquid viscosity falls; surface tension examples; vapour pressure.

(B) Numerical problems (34): the same families as the worked problems (specific weight, density and gravity; shear stress from velocity profiles; plates, shafts and bearings; kinematic viscosity and specific gravity; gas laws; bulk modulus; droplets, bubbles and capillary rise; inclined planes).

Every correction in one place

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