1. Chapter 2: Properties of Fluids

2. Introduction Any characteristic of a system is called a property.
Familiar: pressure P, temperature T, volume V, and mass m.
Less familiar: viscosity, thermal conductivity, modulus of elasticity, thermal expansion coefficient, vapor pressure, surface tension.
Intensive properties are independent of the mass of the system. Examples: temperature, pressure, and density.
Extensive properties are those whose value depends on the size of the system. Examples: Total mass, total volume, and total momentum.
Extensive properties per unit mass are called specific properties. Examples include specific volume v = V/m and specific total energy e=E/m.

3. Continuum Atoms are widely spaced in the gas phase.
However, we can disregard the atomic nature of a substance.
View it as a continuous, homogeneous matter with no holes, that is, a continuum.
This allows us to treat properties as smoothly varying quantities.
Continuum is valid as long as size of the system is large in comparison to distance between molecules.

4. Density and Specific Gravity
Density is defined as the mass per unit volume r = m/V. Density has units of kg/m3
Specific volume is defined as v = 1/r = V/m.
For a gas, density depends on temperature and pressure.
Specific gravity, or relative density is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4°C), i.e., SG=r/rH20. SG is a dimensionless quantity.
The specific weight is defined as the weight per unit volume, i.e., gs = rg where g is the gravitational acceleration. gs has units of N/m3.

5. Density of Ideal Gases
Equation of State: equation for the relationship between pressure, temperature, and density.
The simplest and best-known equation of state is the ideal-gas equation P v = R T or P = r R T
Ideal-gas equation holds for most gases.
However, dense gases such as water vapor and refrigerant vapor should not be treated as ideal gases. Tables should be consulted for their properties, e.g., Tables A-3E through A-6E in textbook.

6. Vapor Pressure and Cavitation
Vapor Pressure Pv is defined as the pressure exerted by its vapor in phase equilibrium with its liquid at a given temperature
If P drops below Pv, liquid is locally vaporized, creating cavities of vapor.
Vapor cavities collapse when local P rises above Pv.
Collapse of cavities is a violent process which can damage machinery.
Cavitation is noisy, and can cause structural vibrations.

7. Energy and Specific Heats
Total energy E is comprised of numerous forms: thermal, mechanical, kinetic, potential, electrical, magnetic, chemical, and nuclear.
Units of energy are joule (J) or British thermal unit (BTU).
Microscopic energy
Internal energy u is for a non-flowing fluid and is due to molecular activity.
Enthalpy h=u+Pv is for a flowing fluid and includes flow energy (Pv).
Macroscopic energy
Kinetic energy ke=V2/2
Potential energy pe=gz
In the absence of electrical, magnetic, chemical, and nuclear energy, the total energy is eflowing=h+V2/2+gz.

8. Coefficient of Compressibility
How does fluid volume change with P and T?
Fluids expand as T ↑ or P ↓
Fluids contract as T ↓ or P ↑
Need fluid properties that relate volume changes to changes in P and T.
Coefficient of compressibility (also called as Bulk Modulus of Elasticity)
Coefficient of volume expansion
Combined effects of P and T can be written as

9. Values for Bulk Modulus for selected liquids at atmospheric pressure and 20 deg C
BULK MODULUS (MPa)
Ethyl Alcohol
896
Benzene
1 062
Machine Oil
1 303
Water
2 179
Glycerine
4 509
Mercury
24 750

10. Example:
Compute the change in pressure that must be applied to water to change its volume by 1.0 per cent.
Solution:
The 1.0 % volume change means ∆V/V =
From Table : Bulk Modulus of Water is κ = MPa
Therefore, from the definition of the Bulk Modulus,
i.e., κ = - ∆P/(∆V/V), we get
∆P = - κ * ∆V/V = 2 179*0.01 = MPa. ANS.
Compute the pressure change required to cause a decrease in the volume of mercury by 1.0 per cent. Express the result in MPa.
[Answer: MPa]

11. Isothermal Compressibility is the inverse of the coefficient of compressibility. It represents the fractional change in volume or density corresponding to a unit change in pressure and is denoted by α. Thus,
The combined effects of pressure and temperature changes on the volume change of a fluid can be determined by taking the specific volume to be function of T and P. Differentiating V = V(T,P) and using the definitions of the compression and expansion coefficients α and β give

12. Example: Variation of Density with Temperature and Pressure
Consider water initially at 20 deg C and 1 atm. Determine the final density of water (a) if it is heated to 50 deg C at a constant pressure of 1 atm, and (b) if it is compressed to 100-atm pressure at a constant temperature of 20 deg C. Take the isothermal compressibility of water to be α = 4.80x10-5 atm-1.
Soln.
The density of water at 20 deg C and 1 atm is ρ1=998.0 kg/m3.
The average temperature is (20+50)/2 = 35 0C
The value of β (for water) at this average temperature is = 0.337x10-3 K-1 (From Table)
The value of α = 4.80x10-5 atm-1 (Given).
The change in density due to change of temperature from 20 0C to 50 0C at constant pressure is
∆ρ = -βρ∆T = -(0.337x10-3)(998)(50-20) = kg/m3
Noting that ∆ρ = ρ2 – ρ1, the density of water at 50 0C and 1 atm is:
ρ2 = ρ1 + ∆ρ = ( -10.0) = kg/m3 ANSWER

13. (b) The change in density due to change of pressure from 1 atm to 100 atm at constant temperature is given by
∆ρ = αρ∆P = (4.80x10-5)(998.0)(100 – 1)
= 4.7 kg/m3
Then the density of water at 100 atm and 20 0C becomes
ρ2 = ρ1 + ∆ρ = = kg/m3 ANSWER
Discussion: The density of water decreases while being heated and increases while being compressed, as expected.

14. Viscosity
Viscosity is a property that represents the internal resistance of a fluid to motion.
The force a flowing fluid exerts on a body in the flow direction is called the drag force, and the magnitude of this force depends, in part, on viscosity.

15. Viscosity
To obtain a relation for viscosity, consider a fluid layer between two very large parallel plates separated by a distance ℓ
Definition of shear stress is t = F/A.
Using the no-slip condition, u(0) = 0 and u(ℓ) = V, the velocity profile and gradient are u(y)= Vy/ℓ and du/dy=V/ℓ
Shear stress for Newtonian fluid: t = mdu/dy
m is the dynamic viscosity and has units of kg/m·s, Pa·s, or poise.

16. Values of Dynamic viscosity
Liquid
Temperature deg C
Dynamic Visco-sity N-s/m2
Water 20
1.0x10-3
Gasoline
3.1x10-4
SAE 30 Oil
3.5x10-1 80
1.9x10-2

17. Viscometry How is viscosity measured? A rotating viscometer.
Two concentric cylinders with a fluid in the small gap ℓ.
Inner cylinder is rotating, outer one is fixed.
Use definition of shear force:
If ℓ/R << 1, then cylinders can be modeled as flat plates.
Torque T = FR, and tangential velocity V=wR
Wetted surface area A=2pRL.
Measure T and w to compute m

18. MEASUREMENT OF VISCOSITY
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19. Example:
The viscosity of a fluid is to be measured by a viscometer constructed of two 40-cm-long concentric cylinders. The outer diameter of the inner cylinder is 12 cm, and the gap between the two cylinders is 0.15 cm. The inner cylinder is rotated at 300 rev/min, and the torque is measured to be 1.8 N-m. Determine the viscosity of the fluid.
Soln:
The velocity profile is linear only when the curvature effects are negligible, and the profile can be approximated as being linear in this case since l/R << 1. Therefore,
μ = Tl/(4 π2R3NL), gives
μ = N-s/m2 ANS.

20. Problem
The viscosity of a fluid is to be measured by a viscometer constructed of two 75-cm-long concentric cylinders. The outer diameter of the inner cylinder is 15 cm, and the gap between the two cylinders is 0.12 cm. The inner cylinder is rotated at 200 rev/min, and the torque is measured to be 0.8 N.m. Determine the viscosity of the fluid.

21. Problem: In regions far from the entrance, fluid flow through a circular pipe is on-dimensional, and the velocity profile for laminar flow is given by u(r) = umax(1-r2/R2), where R is the radius of the pipe, r is the radial distance from the center of the pipe, and umax is the maximum flow velocity, which occurs at the center.
Obtain (a) a relation for the drag force applied by the fluid on a section of the pipe of length L and (b) the value of the drag force for water flow at 20 0C with R = 0.08 m, L = 15 m, umax = 3 m/s and μ = kg/m.s

22. Prob: A flat plate 0.1 m2 area is pulled at 30 cm/s relative to another plate located at a distance of 0.01 cm from it, the fluid separating them being water with dynamic viscosity of Pa-s. Find the force and power required to maintain the velocity.
Prob: Determine the torque and power required to turn a 10 cm long, 5 cm diameter shaft at 500 rev/min in a 5.1 cm diameter concentric bearing flooded with a lubricating oil of viscosity 100 centipoise.
Note that: I poise = 0.1 Pa-s.
Prob: A skater weighing 800 N skates at 54 km/hr on ice at 0 deg C. The average skating area supporting him is 10 cm2 and the effective dynamic coefficient of friction between the skates and the ice is If there is actually a thin film of water between the skates and the ice, determine the average thickness.
Note that the friction force is given by friction coefficient*Weight of the skater. Dynamic viscosity is taken to be Pa-s.

23. Surface Tension
Liquid droplets behave like small spherical balloons filled with liquid, and the surface of the liquid acts like a stretched elastic membrane under tension.
The pulling force that causes this is
due to the attractive forces between molecules
called surface tension ss.
Attractive force on surface molecule is not symmetric.
Repulsive forces from interior molecules causes the liquid to minimize its surface area and attain a spherical shape.

24. Capillary Effect
Capillary effect is the rise or fall of a liquid in a small-diameter tube.
The curved free surface in the tube is call the meniscus.
Water meniscus curves up because water is a wetting fluid.
Mercury meniscus curves down because mercury is a nonwetting fluid.
Force balance can describe magnitude of capillary rise.