1. Chapter 2 Properties of Fluids

2. WHAT IS A FLUID?
Substance exits in three primary phases: solid, liquid, and gas. A substance in the liquid or gas phase is referred to as a fluid.
Distinction between a solid and a fluid is made on the basis of the substance’s ability to resist an applied shear (or tangential) stress force per unit area) that tends to change its shape.
Distinction between solid and fluid?
Solid: can resist an applied shear by deforming but will not continuously deform.
Fluid: deforms continuously under applied shear.

3. WHAT IS A FLUID?
Stress is defined as force per unit area and is determined by dividing the force by the area upon which it acts. The normal component of the force acting on a surface per unit area is called the normal stress, and tangential component of a force acting on a surface per unit area is called shear stress. In a fluid at rest, the normal stress is called pressure.
A fluid at rest is at a state of zero shear stress. When walls are removed or a liquid container is tiled, a shear develops as the liquid moves to re-establish a horizontal free surface.

4. WHAT IS A FLUID?
Liquid
Group of molecules can move relative to each other, but the volume remains relatively constant because of strong cohesive forces between the molecules. As a result, a liquid takes the shape of the container it is in, and it forms a free surface in a larger container in a gravitational field.
Gas
Expands until it encounters the walls of the container and fills the entire available space because cohesive forces are very small.
Gases cannot form a free surface.
Compressible.
Density not constant with pressure
Gas and vapor are often used as synonymous words

5. Properties of Fluids
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.

6. Density And Specific Gravity
The density of a substance is the quantity of matter contained in a unit
volume of the substance. It can be expressed in three different ways.
Mass Density
Mass Density, p, (rho), is defined as the mass of substance per unit volume.
Units: Kilograms per cubic metre, kg / m3 (or kgm-3 )
Typical values:
Water = 1000 kgm-3 , Mercury = kgm-3 Air = 1.23 kgm-3 , Paraffin Oil = 800 kgm-3 .(at pressure =1.013 x 10-5 N m-2 and Temperature = K.)
Specific volume
Is defined as v = 1/r = V/m.
For a gas, density depends on temperature and pressure.
The density gases is proportional to pressure and inversely proportional to temperature.
Unit is m3/kg

7. Specific gravity, or relative density
Specific weight
Specific Weight , (sometimes (gamma), and sometimes known as specific gravity) is defined as the weight per unit volume. or
The force exerted by gravity, g, upon a unit volume of the substance.
The Relationship between g and can be determined by Newton’s 2nd Law, since
weight per unit volume = mass per unit volume x g
= pg
Units: Newton’s per cubic metre, N/m3(or Nm-3 )
Typical values:
Water =9814 N m-3 , Mercury = N m-3 , Air =12.07 N m-3 , Paraffin Oil =7851 N m-3
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=p/pH20 at 4°C. SG is a dimensionless quantity.

8. EXAMPLE 1
Calculate the weight of a reservoir of oil if it has a mass of 825 kg.
Solution

9. EXAMPLE 2 5.6m3 of oil weighs 46 800 N. Find its mass density, , and
relative density, g.
Solution
Weight = mg
Mass m = / 9.81 = kg
Mass density, = Mass / volume = / 5.6 = 852 kg/m3
Relative density

10. EXAMPLE 3
The density of an oil is 850 kg/m3. Find its relative density and Kinematic viscosity if the dynamic viscosity is 5 x 10-3 kg/ms.
Solution
oil = 850 kg/m3
water = 1000 kg/m3
= 850 / 1000 = 0.85
Dynamic viscosity = m = 5 x 10-3 kg/ms
Kinematic viscosity = = /

11. 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 or
where P is the absolute pressure, v is the specific volume, T is the thermodynamic (absolute) temperature, and R is the constant. The value of R for several substance is given in Table A-1.
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-4 in textbook.

12. EXAMPLE 4
Determine the density, specific gravity, and mass of the air in a room whose dimensions are 4 m x 5 m x 6 m at 100 kPa and 25°C.
The density of air is determined from the ideal-gas relation to be
Then the specific gravity of air becomes

13. Finally, the volume and the mass of air in the room are

14. Viscosity
Viscosity is a property that represents the internal resistance of a fluid to motion or the “fluidity”.
To obtain a relation for viscosity, consider a fluid layer between two very large parallel plates separated by a distance ℓ. A constant parallel force F is applied to the upper plate while the lower plate is held fixed. The fluid in contact with the upper plate sticks to the plate surface and moves with it at the same velocity, and shear stress acting on this fluid layer is

15. Using the no-slip condition, u(0) = 0 and u(ℓ) = V, the velocity profile and gradient are
Shear stress for “Newtonian Fluids” obey the linear relationship given by Newton’s law of viscosity,
is the “coefficient of dynamic viscosity” – see next slide.
Newtonian fluids -Fluids for which the rate of deformation is proportional to the shear stress.
The shear force acting on a Newtonian fluid layer is
where A is the contact area between the plate and the fluid

16. Variation of shear stress with the rate of deformation for Newtonian and non-Newtonian fluids (the slope of a curve at a point is the apparent viscosity of the fluid at that point).
The rate of deformation (velocity gradient) of a Newtonian fluid is proportional to shear stress, and the constant of proportionality is the viscosity.

17. Coefficient of Dynamic Viscosity (absolute viscosity)
The Coefficient of Dynamic Viscosity, , is defined as the shear force, per unit area, (or shear stress, ),required to drag one layer of fluid with unit velocity past another layer a unit distance away.
Units: Newton seconds per square metre, N sm-2 or Kilograms per meter per second, kgm-1 s-1 .
(Although note that is often expressed in Poise, P, where 10 P = 1 kgm-1 s-1 .)
Typical values:
Water =1.14 x10-3 kgm-1 s-1 , Air =1.78 x10-5 kgm-1 s-1 , Mercury =1.552 kgm-1 s-1, Paraffin Oil =1.9 kgm-1 s-1 .

18. Coefficient of Kinematic Viscosity
Kinematic Viscosity, , is defined as the ratio of dynamic viscosity to mass density.
Units: square metres per second, m2 s-1
(Although note that n is often expressed in Stokes, St, where 104 St = 1 m2 s-1 .)
Typical values:
Water =1.14 x 10-6 m2 s-1 , Air =1.46 x 10-5 m2 s-1 , Mercury =1.145 x 10-4 m2 s-1 , Paraffin Oil =2.375 x 10-3 m2 s-1 .

19. More About Viscosity
The viscosity of liquids decreases and the viscosity of gases increases with temperature. (dynamic & kinematics viscosity)

20. Example 5
If the viscosity of a liquid is 0.01 pa.s, find its viscosity in centipoises.
Solution
1 pa.s = 10 poise
1 pa.s = 1000 centipoise
Therefore

21. Example 6
The density of an oil is 850 kg/m3. Find its relative density and Kinematic viscosity if the dynamic viscosity is 5 x 10-3 kg/ms.
Solution
oil = 850 kg/m3 r water = 1000 kg/m3
SGoil = 850 / 1000 = 0.85
Dynamic viscosity = m = 5 x 10-3 kg/ms
Kinematic viscosity = n = m / r

22. Example 7
The velocity distribution of a viscous liquid (dynamic viscosity = 0.9 Ns/m2) flowing over a fixed plate is given by u = 0.68y - y2 (u is velocity in m/s and y is the distance from the plate in m). What are the shear stresses at the plate surface and at y=0.34m?
Solution
At the plate face y = 0m,

23. Calculate the shear stress at the plate face
At y = 0.34m,
As the velocity gradient is zero at y=0.34 then the shear stress must also be zero.

24. Example 8
In a fluid the velocity measured at a distance of 75mm from the boundary is 1.125m/s. The fluid has absolute viscosity Pa s and relative density What is the velocity gradient and shear stress at the boundary assuming a linear velocity distribution.
Solution
= Pa s
= 0.913

25. Example 9
For a parallel plate arrangement of the type shown in slide 14, it is found that when the distance between plates is 2 mm, a shearing stress of 150 Pa develops at the upper plate when it is pulled at a velocity of 1 m/s. Determine the viscosity of the fluid between the plates.
Answer:

26. Example 10
A 25-mm-diameter shaft is pulled through a cylindrical bearing as shown in above. The lubricant that fills the 0.3-mm gap between the shaft and bearing is an oil having a kinematic viscosity of 0.8 x 10^-4 m^2/s and specific gravity of Determine the force P required to pull the shaft at a velocity of 3 m/s. Assume the velocity distribution in the gap is linear.

27. Answer:

28. 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.

29. MEASUREMENT OF VISCOSITY

30. Example 11
The space between two 40-cm-long concentric cylinders is filled with glycerin (viscosity = N.s/m^2). The inner cylinder has a radius of 11 cm and the gap width between cylinder is 0.15 cm. Determine the torque and power required to rotate the inner cylinder at 300 rev/min. The outer cylinder is fixed. Assume the velocity distribution in the gap to be linear.

31. Surface Tension Examples
Water droplets from rain hang from branches or leaves of trees (Picture)

32. WHAT IS SURFACE TENSION
The cohesive forces between molecules down into a liquid are shared with all neighboring atoms. Those on the surface have no neighboring atoms above, and exhibit stronger attractive forces upon their nearest neighbors on the surface. This enhancement of the intermolecular attractive forces at the surface is called surface tension.

33. Capillary Effect
Capillary effect: The rise or fall of a liquid in a small-diameter tube inserted into the liquid.
Capillaries: Such narrow tubes or confined flow channels.
The capillary effect is partially responsible for the rise of water to the top of tall trees.
Meniscus: The curved free surface of a liquid in a capillary tube.
The strength of the capillary effect is quantified by the contact (or wetting) angle, defined as the angle that the tangent to the liquid surface makes with the solid surface at the point of contact.
The meniscus of colored water in a 4-mm-inner-diameter glass tube. Note that the edge of the meniscus meets the wall of the capillary tube at a very small contact angle.
The contact angle for wetting and nonwetting fluids.

34. A liquid is said to wet the surface when Ø<90° and not to wet the surface when Ø>90°. In atmospheric air, the contact angle of water (and most other organic liquids) with glass is nearly zero, Ø≈0°.
When the attractive forces are between unlike molecules, they are said to be adhesive forces. The adhesive forces between water molecules and the walls of a glass tube are stronger than the cohesive forces (attraction between like molecules) lead to an upward turning meniscus at the walls of the vessel and contribute to capillary action.

35. Vapor Pressure
The pressure at which a liquid will boil is called its vapor pressure. This pressure is a function of temperature (vapor pressure increases with temperature). In this context we usually think about the temperature at which boiling occurs.
For example, water boils at 100C at sea-level atmospheric pressure (1 atm abs). However, in terms of vapor pressure, we can say that by increasing the temperature of water at sea level to 100C, we increase the vapor pressure to the point at which it is equal to the atmospheric pressure (1 atm abs), so that boiling occurs.
It is easy to visualize that boiling can also occur in water at temperatures much below 100°C if the pressure in the water is reduced to its vapor pressure. For example, the vapor pressure of water at 10°C is 0.01 atm.
Therefore, if the pressure within water at that temperature is reduced to that value, the water boils. Such boiling often occurs in flowing liquids, such as on the suction side of a pump or tip regions of impellers. When such boiling does occur in the flowing liquids, vapor bubbles start growing in local regions of very low pressure (high velocity) and then collapse in regions of high downstream pressure (low viscosity). This phenomenon is called as cavitation. Cavitations must be avoided (or at least minimized) in most flow systems since it reduces performances, generates annoying vibrations and noise, and causes damage to equipment.