1. NON-NEWTONIAN FLUIDS
Fluids that do not follow the linear law of newton’s law of viscosity are called non-Newtonian fluids. For the non linear curves, the slope at any point is called the apparent viscosity.

2. DILATANT
This fluid is shear thickening, increasing its resistance with increasing strain rate.
Examples : Quick sand, which stiffens up if one thrashes out.

3. PSEUDOPLASTIC
A shear thinning fluid , which is less resistant at higher strain rates. A very strong thinning is called plastic.
Examples: Polymer solutions, paper pulp in water, latex paint, blood plasma, syrup and molasses.
The classic case is paint, which is thick when poured but thin when brushed at a high strain rate.

4. BINGHAM PLASTIC
The limiting case of a plastic substance is one that required a finite yield stress before it begins to flow.
Examples: Clay suspensions, drilling mud, toothpaste , mayonnaise, chocolate and mustard.
The classic case is ketchup, which will not come out of the bottle until you stress it by shaking.

6. CHAPTER 2 PRESSURE DISTRIBUTION IN FLUIDS

7. Introduction to Fluid Statics
Fluid at rest:
– No shear stresses
– Only normal forces due to pressure
Normal forces are important:
– Overturning of concrete dams
– Bursting of pressure vessels
– Breaking of lock gates on canals

8. Introduction to Fluid Statics (Cont’d)
For design: compute magnitude and location of normal forces
Development of instruments that measure pressure
Development of systems that transfer pressure, e.g.,
– automobile breaks
– hoist

9. Introduction to Fluid Statics (Cont’d)
Average pressure intensity p = force per unit area
Let:
F = total normal pressure force on a finite area A
dF = normal force on an infinitesimal area dA
The local pressure on the infinitesimal area is
p = 𝑑𝐹 𝑑𝐴
If pressure is uniform, p = F/A
● BG units: psi (=lb/in2) or psf (=lb/ft2)
● SI units: Pa (=N/m2), kPa (=kN/m2)
● Metric: bar, millibar; 1 mb = 100 Pa

10. Isotropy of Pressure
Small wedge of fluid at rest of Δx by Δz by Δs and depth b into the paper
px, pz and pn ; Element is assumed small, so pressure is constant on each face.
Δs sin θ= Δz
Δs cos θ= Δx

11. Isotropy of Pressure (cont’d)
Δz = 0
Two important principles of hydrostatics:
There is no pressure change in the horizontal direction
There is a vertical change in pressure proportional to the density, gravity, and depth change
pressure p at a point in a static fluid is independent of orientation

12. Pressure Force on a Fluid Element
The total net force vector on the element due to pressure is :
f is net force per unit element volume
Thus it is not the pressure but the pressure gradient causing a net force which must be balanced by gravity or acceleration or some other effect in the fluid

13. Equilibrium of a Fluid Element
By Newton’s law , the sum of “per unit volume forces” equals the mass per unit volume (density , ρ) times the acceleration a of the fluid element.

14. Absolute and Gage Pressures
Pressure measured:
Relative to absolute zero (perfect vacuum): absolute
Relative to atmospheric pressure: gage
If p < patm , we call it a vacuum, its gage value = how much below atmospheric
Absolute pressure values are all positive
Gage pressures:
Positive: if above atmospheric
Negative: if below atmospheric
Relationship:
Pabs = Patm + Pgage

15. Absolute and Gage Pressures (Cont’d)

16. Absolute and Gage Pressures (Cont’d)
Atmospheric pressure is also called barometric pressure
Atmospheric pressures varies:
with elevation
with changes in meteorological conditions
Use absolute pressure for most problems involving gases and vapor (thermodynamics)
Use gage pressure for most problems related to liquids

17. NOTE : In lecture 1 , we have used γ symbol for specific gravity , but from now on as according to Fluid Mechanics book by Frank M white , the symbol gamma (γ) will be used for specific weight and S.G. for specific gravity.

18. Hydrostatic Pressure Distributions
Hydrostatics Condition ; a = 0 and fvisc = 0
The fluid in hydrostatic equilibrium will align its constant pressure surfaces everywhere to the local gravity vector. The maximum pressure increase will be in the direction of gravity that is “down”

19. Hydrostatic Pressure Distributions (cont’d)
The local gravity vector for the small scale problem is g = -g k
𝝏𝒑 𝝏𝒙 =𝟎 ; 𝝏𝒑 𝝏𝒚 =𝟎 ; 𝝏𝒑 𝝏𝒛 =−ρ𝒈=−γ
p is independent of x and y. So hydrostatic condition reduces to
𝒅𝒑 𝒅𝒛 =−𝜸
𝑝 2 − 𝑝 1 =- 1 2 γ 𝑑𝑧

20. Hydrostatic Pressure Distributions (cont’d)
Pressure in a continuously distributed uniform static fluid varies only with vertical distance and is independent of the shape of the container. The pressure is the same at all points on a given horizontal plane in the fluid. The pressure increases with depth in the fluid.

21. Hydrostatic Pressure Distributions (cont’d)

22. Hydrostatic Pressure Distributions (cont’d)

23. Example
Newfound Lake, a freshwater lake near Bristol, New Hampshire, has a maximum depth of 60 m, and the mean atmospheric pressure is 91 kPa. Estimate the absolute pressure in kPa at this maximum depth.

24. Mercury Barometer Measures atmospheric pressure
A tube is filled with mercury and inverted while submerged in a reservoir.
Causes near vacuum in the closed upper end because mercury has an extremely small vapor pressure at room temperatures (0.16 Pa at 20oC).
Since atmospheric pressure forces a mercury column to rise a distance h into the tube, the upper mercury surface is at zero pressure.

25. Mercury Barometer (cont’d)
Substituting
p1 = 0 at z1 = h ;
p2 = pa at z2=0
At sea-level standard
Pa = 101,350 Pa
γM = 133,100 N/m3
The barometric height is
h = 101,350/133,100
= m or 761 mm

26. Application to Manometery
Static column of one or more liquids or gases can be used to measure pressure differences between two points. Such a device is called a manometer

27. Application to Manometery (cont’d)

28. Application to Manometery (cont’d)
Professor John Foss of Michigan State University
Keep adding on PRESSURE INCREMENTS as you move down through the layered fluid

29. Simple Open Manometer

30. Simple Open Manometer (cont’d)
Jump Across :
The physical reason that we can “jump across” at section 1 is that a continuous length of the same fluid connects these two equal elevations.
Any two points at the same elevation in a continuous mass of the same static fluid will be at the same pressure
This idea of jumping across to equal pressures facilitates multiple-fluid problems.

31. Example

32. Example

33. Example