1. Chapter 8 Introduction and Basic Fluid Properties

2. Fluid Mechanics
Fluid Mechanics is concerned with the behavior of fluids at rest and in
motion
Distinction between solids and fluids:
-A solid is “hard” and not easily deformed.
-A fluid is “soft” and deforms easily.
-Fluid is a substance that alters its shape in response to any force
however small, that tends to flow or to conform to the outline of its
container, and that includes gases and liquids and mixtures of
solids and liquids capable of flow.
-A fluid is defined as a substance that deforms continuously when
acted on by a shearing stress of any magnitude.
- A shearing stress is created whenever a tangential force acts on a surface.

4. Fluid and Solid When a constant shear force is applied:
Solid deforms or bends (ie can resists a shear stress)
Fluid continuously deforms (can not resists shear stress)

5. Dimensions and Units
In fluid mechanics we must describe various fluid characteristics in terms of certain basic quantities such as length, time and mass
A dimension is the measure by which a physical variable is expressed qualitatively, i.e. length is a dimension associated with distance, width, height, displacement.
Basic dimensions: Length, L
(or primary quantities) Time, t
Mass, M
Temperature, θ
We can derive any secondary quantity from the primary quantities i.e. Force = (mass) x (acceleration) : F = M L t-2
A unit is a particular way of attaching a number to the qualitative dimension: Systems of units can vary from country to country, but dimensions do not

6. Dr Mustafa Nasser

7. Example BG and SI Units 1/2
A tank of water having a total mass of 36 kg rests on the floor of an elevator. Determine the force (in newtons) that the tank exerts on the floor when the elevator is accelerating upward at 7 ft/sec2.
Solution
A free-body diagram of the tank is shown in Figure where is the weight of the tank and water (w) is the reaction of the floor on the tank. Application of Newton’s second law of motion to this body gives :
SF = ma or Ff = ma +w ……1

8. Example BG and SI Units 2/2
where we have taken upward as the positive direction. Since Eq. 1 can be written as :
Ff = m (g + a) …… 2
Before substituting any number into Eq. 2 we must decide on a system of units, and then be sure all of the data are expressed in these units. Since we want Ff in newtons we will use SI units so that
Ff = 36 kg [9.81 m/ s2 +(7 ft /s2) ( m/ft)] = kg . m /s2
Since1 N = 1 kg . m/ s2 it follows that
Ff = 430 N (downward on floor) (Ans)
The direction is downward since the force shown on the free-body diagram is the force of the floor on the tank so that the force the tank exerts on the floor is equal in magnitude but opposite in direction.

9. Density and Specific Volume
The density of a fluid, designated by the Greek symbol  (rho), is defined as its mass per unit volume (ie =m/V).
Density is used to characterize the mass of a fluid system.
In the BG system  has units of slug/ft3 and in SI the units are kg/m3.
The value of density can vary widely between different fluids, but for liquids, variations in pressure and temperature generally have only a small effect on the value of density.
The specific volume, is the volume per unit mass that is, ν=1/ρ

10. Specific Weight
The specific weight of a fluid, designated by the Greek symbol  (gamma), is defined as its weight per unit volume (with unit N/m3).
Under conditions of standard gravity (g= m/ s2
or = 9.81 m/ s2 ).
water at 60ºF has a specific weight of 9.80kN/m3. The density of water is 998kg/m3.

11. Specific Weight Weight per unit volume (e.g., @ 20 oC, 1 atm)
gwater = (998 kg/m3)(9.807 m/s2)
= 9790 N/m3
[= 62.4 lbf/ft3]
gair = (1.205 kg/m3)(9.807 m/s2)
= 11.8 N/m3
[= lbf/ft3]

12. Specific Gravity  H2O , 4oC= 999kg/m3≈1000kg/m3 . For liquid
The specific gravity of a fluid, designated as SG, is defined as the ratio of the density of the fluid to the density of WATER (for liquids) or to the density of AIR (for gases) at some specified temperature.
For liquid
 H2O , 4oC= 999kg/m3≈1000kg/m3 .

13. Simple Flows Flow between a fixed and a moving plate
Fluid in contact with the plate has the same velocity as the plate
u = x-direction component of velocity
u=U
Moving plate
Fixed plate y x U
u=0 b
Fluid

14. Simple Flows
Flow through a long, straight pipe Fluid in contact with the pipe wall has the same velocity as the wall
u = x-direction component of velocity r x R U
Fluid

15. Fluidity of Fluid 1/3
How to describe the “fluidity” of the fluid?
The bottom plate is rigid fixed, but the upper plate is free to move.
If a solid, such as steel, were placed between the two plates and loaded with the force F, the top plate would be displaced through some small distance, a.
The vertical line AB would be rotated through the small angle, , to the new position AB'. F
F =τA
where τ is shearing
stress
Acting opposite to F
A is area F

16. Fluidity of Fluid 2/3
What happens if the solid is replaced with a fluid such as water?
When the force F is applied to the upper plate, it will move continuously with a velocity U.
The fluid “sticks” to the solid boundaries and is referred to as the NON-SLIP conditions.
The fluid between the two plates moves with velocity u=u(y) that would be assumed to vary linearly, u=Uy/b. In such case, the velocity gradient is
du / dy = U / b. F

17. Fluidity of Fluid 3/3 tan (δβ)=δβ = δa / b
In a small time increment, δt, an imaginary vertical line AB would rotate through an angle, δβ , so that
tan (δβ)=δβ = δa / b
Since δa = U δt it follows that
δβ= U δt / b
Defining the rate of shearing strain, as
If the shearing stress is increased by F, the rate of shearing strain is increased in direct proportion, F
The common fluids such as water, oil, gasoline, and air, the shearing stress and rate of shearing strain can be related with a relationship

18. Shear in Different Fluids
Shear-stress relations for different types of fluids
Newtonian fluids: linear relationship
Slope of line (coefficient of proportionality) is “viscosity”

19. Viscosity Newton’s Law of Viscosity Viscosity Units Water (@ 20oC)
m = 1x10-3 N-s/m2
Air 20oC)
m = 1.8x10-5 N-s/m2
Kinematic viscosity

20. Kinematic Viscosity Defining kinematic viscosity
The dimensions of kinematic viscosity are L2/t.
The units of kinematic viscosity in BG system are ft2/s and SI system are m2/s.
In the CGS system, the kinematic viscosity has the units is called a stoke, abbreviated St.
It is sometimes expressed in terms of centiStokes (cSt).
1 St = 1 cm2·s−1 = 10−4 m2·s−1.
1 cSt = 1 mm2·s−1 = 10−6m2·s−1.
Water at 20 °C has a kinematic viscosity of about 1 cSt.

21. Example Newtonian Fluid Shear Stress 1/3

22. Example Newtonian Fluid Shear Stress 2/3

23. Example Newtonian Fluid Shear Stress 3/3

24. Dimension and Unit of Viscosity
The dimension of μ : Ft/L2 (FT/L2) or M/Lt.
The unit of μ:
In BG : lbf . s/ft2 or slug/(ft.s)
In SI : kg/(m . s) or N . s/m2 or Pa . s
In the Absolute Metric: poise=1g/(cm . s)
The primary parameter correlating the viscous
behavior of all Newtonian fluids is the dimensionless Reynolds number (Re)
ρuD uD
Re= =
μ υ
u is the average velocity, D is the diameter and υ is the Kinematic viscosity

25. Generally, the first thing a fluids engineer should do is estimate the Reynolds number range of the flow under study.
Very low Re indicates viscous creeping motion, where inertia effects are negligible.
Moderate Re implies a smoothly varying laminar flow.
High Re probably spells turbulent flow, which is slowly varying in the time-mean but has superimposed strong random high-frequency fluctuations..
For a given value of u and D in a flow, these fluids exhibit a spread of four orders of magnitude in the Reynolds number

26. Example Viscosity and Dimensionless Quantities

27. Example Viscosity and Dimensionless Quantities

28. Fluid Statics Basic Principles: Fluid is at rest : no shear forces
Pressure is the only force acting
What are the forces acting on the block?
Air pressure on the surface - neglect
Weight of the water above the block
Pressure only a function of depth
The word STATIC fluid = non-moving fluid.

29. Section 1: Pressure

30. Pascal’s Laws Pascals’ laws:
Pressure acts uniformly in all directions on a small volume (point) of a fluid
In a fluid confined by solid boundaries, pressure acts perpendicular to the boundary – it is a normal force.

31. Direction of fluid pressure on boundaries
Furnace duct
Pipe or tube
Heat exchanger
Pressure is due to a Normal Force
(acting perpendicular to
the surface)
It is also called a Surface Force
Dam

32. What are the z-direction forces?
Let Pz and Pz+Dz denote the pressures at the base and top of the cube, where the elevations are z and z+Dz respectively. z y x

33. Pressure distribution for a fluid at rest
Force at base of cube: Pz A=Pz (Dx Dy)
Force at top of cube: Pz+Dz A= Pz+Dz (Dx Dy)
Force due to gravity: m g=r V g = r (Dx Dy Dz) g
A force balance in the z direction
For an infinitesimal element (Dz0) 

34. Variation of pressure with elevation

35. Equality of pressure at the same level in a static fluid

36. Equality of pressure at the same level in a continuous fluid

37. Incompressible fluid (Liquid)
Liquids are incompressible i.e. their density is assumed to be constant
When we have a liquid with a free surface,
the pressure P at any depth below the free
surface is:
where Po is the pressure at the free surface (ie Po=Patm) and h = zfree surface - z
By using gauge pressures we can simply write:

38. Example: Pressure-Depth Relationship
Because of a leak in a buried gasoline storage tank, water has seeped in to the depth shown in figure. If the specific gravity of the gasoline is SG=0.68. Determine the pressure at the gasoline-water interface and at the bottom of the tank. Express the pressure in units of lb/ft2, lb/in2, and pressure head in feet of water.

39. Solution 1/2
Since we are dealing with liquid at rest, the pressure distribution will be hydrostatic, and therefore the pressure variation can be found from the equation: p = γ h + po
With po corresponding to the pressure at the free surface of the gasoline, then the pressure at the interface is
If we measure the pressure relative to atmosphere pressure (gage pressure), it follows that po =0, and therefore

40. Solution 2/2
We can now apply the same relationship to determine the pressure at the tank bottom; that is,

41. Buoyancy
The same principle used to compute hydrostatic forces on surfaces can be applied to the net pressure force on a completely submerged or floating body. The results are the laws of buoyancy discovered by Archimedes:
A body immersed in a fluid experiences a vertical buoyant force equal to the weight of the fluid it displaces
A floating body displaces its own weight in the fluid in which it floats
Free liquid surface F1 h1
The upper surface of the body is subjected to a smaller force than the lower surface
A net force is acting upwards H h2 F2

42. Buoyancy where ρ = the fluid density
The net force due to pressure in the vertical direction is:
FB = F2- F1 = (Pbottom - Ptop) (DxDy)
The pressure difference is:
Pbottom – Ptop = r g (h2-h1) = r g H As
FB = r g H (DxDy)
Thus the buoyant force is:
FB = r g V
where ρ = the fluid density
If the fluid density is greater than the average density of the object, the object floats. If less, the object sinks

43. Archimedes Principle Archimedes Principle FB = weight displaced fluid
Line of action passes through the centroid of displaced volume

44. Example: Buoyant Force on a Submerged Object
A spherical buoy has a diameter of 1.5 m, weighs 8.50kN, and is anchored to the seafloor with a cable as is shown in Figure a. Although the buoy normally floats on the surface, at certain times the water depth increases so that the buoy is completely immersed as illustrated. For this condition what is the tension of the cable?

45. Solution1/2
We first draw a free-body diagram of the buoy as is shown earlier, where FB is the buoyant force acting on the buoy, W is the weight of the buoy, and T is the tension in the cable. For Equilibrium it follows that
As FB= ρgV
and for seawater with γ= 10.1 kN/m3 and V = πd3/6 then p
The tension in the cable can now be calculated as

46. Solution2/2
Note that we replaced the effect of the hydrostatic pressure force on the body by the buoyant force, FB. Another correct free-body diagram of the buoy is shown earlier. The net effect of the pressure forces on the surface of the buoy is equal to the upward force of magnitude, FB (the buoyant force). Do not include both the buoyant force and the hydrostatic pressure effect in your calculations－use one or the other.

47. Surface Tension 1/3
At the interface between a liquid and a gas, or between two immiscible liquids, forces develop in the liquid surface which cause the surface to behave as if it were a “skin” or “membrane” stretched over the fluid mass.
Although such a skin is not actually present, this conceptual analogy allows us to explain several commonly observed phenomena.

48. Surface tension 2/3
Force at interface between liquid and solid or liquid and gas.
s is the surface tension coefficient.

49. Surface Tension 3/3
Surface tension: the intensity of the molecular attraction per unit length along any line in the surface and is designated by the Greek symbol  (sigma).
Where pi is the internal pressure and pe is the external pressure

50. capillary
A common phenomena associated with (caused by) surface tension is the rise or fall of a liquid in a capillary tube.
Effect of capillary action in small tubes. (a) Rise of column for a liquid that wets the tube. (b) Free-body diagram for calculating column height. (c) Depression of column for a nonwetting liquid.

51. Generally σ decreases with liquid temperature and is equal to
zero at the critical point. Values of σ for water are given in Figure
bellow. σ

52. Example : Surface Tension