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LectureLecture

22Fluid Mechanics and Applications

MECN 3110

Inter American University of Puerto RicoProfessor: Dr. Omar E. Meza Castillo

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Pressure Distribution in a Fluid

Chapter 2

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To calculate the pressure distribution in the atmosphere and the oceans.

To know the use of a simple manometer. To calculate forces on submerged flat and

curved surfaces. To calculate the buoyancy on submerged body,

and To understand the behavior of floating bodies.

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Course Objectives

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on Pressure

For a static fluid, the only stress is the normal stress since by definition a fluid subjected to a shear stress must deform and undergo motion. Normal stresses are referred to as pressure p.

For the general case, the stress on a fluid element or at a point is a tensor

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on Pressure

For the general case, the stress on a fluid element or at a point is a tensor

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on Pressure

As already noted, p is a scalar, which can be easily demonstrated by considering the equilibrium of forces on a wedge-shaped fluid element

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on Pressure

p is single valued at a point and independent of direction

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on Pressure

For a fluid at rest, pressure is the same in all directions at this point. But can vary from point to point, e.g. hydrostatic pressure.

For a fluid in motion additional forces arise due to shearing action and we refer to the normal force as a normal stress. The state of stresses in a fluid in motion is dealt with further in Fluid Mechanics.

In the context of thermodynamics, we think of pressure as absolute, with respect to pressure of a complete vacuum (space) which is zero.

In Fluid Mechanics we often use gage pressure and vacuum pressure.

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on Pressure

Absolute PressureForce per unit area exerted by a fluid

Gage PressurePressure aboveatmospheric

Pgage= Pabs - Patm

Vacuum PressurePressure below atmospheric

Pvac=-Pgage= Patm - Pabs

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on Pressure

Common Pressure Units are:Pa (Pascal), mmHg (mm of Mercury), atm

(atmosphere), psi (lbf per square inch) 1 Pa = 1 N/m2 (S.I. Unit)

1 kPa =103 Pa 1 bar = 105 Pa (note the bar is not an SI unit) 1 MPa = 106 Pa

1 atm = 760 mmHg = 101,325 Pa = 14.696 psi

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on Pressure variation with Elevation

Basic Differential Equation: For a static fluid, pressure varies only with elevation within the fluid. This can be shown by consideration of equilibrium of forces on a fluid element.

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on Pressure variation with Elevation

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on Pressure variation with Elevation

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on Hydrostatic Pressure in Liquids

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HYDROSTATIC PRESSURE

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on Variation of Pressure with Depth

If we take point 1 to be at the free surface of a liquid open to the atmosphere, where the pressure is the atmospheric pressure Patm, then the pressure at a depth h from the free surface becomes

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on Hydrostatic Pressure in Liquids

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on Hydrostatic Pressure in Liquids

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on Example

Oil with a specific gravity of 0.80 forms a layer 0.90 m deep in an open tank that is otherwise filled with water. The total depth of water and oil is 3 m. What is the gage pressure at the bottom of the tank?

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on Example

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on Pressure Variation for Compressible Fluid

Basic equation for pressure variation with elevation

Pressure variation equation can be integrated for γ(p,z) known. For example, here we solve for the pressure in the atmosphere assuming ρ(p,T) given from ideal gas law, T(z) known, and g ≠ g(z).

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on Pressure Variation for Compressible Fluid

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on Pressure Variation in the Troposphere

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on Pressure Variation in the Troposphere

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on Pressure Variation in the Stratosphere

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on Manometer

The elevation change of Δz in a fluid at rest corresponds to ΔP/ρg, which suggests that a fluid column can be used to measure pressure differences. A device based on this principle is called a manometer, and it is commonly used to measure small and moderate pressure differences. A manometer mainly consists of a glass or plastic U-tube containing one or more fluids such as mercury, water, alcohol, or oil.

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on Manometer

Many engineering problems and some manometers involve multiple immiscible fluids of different densities stacked on top of each other. Such systems can be analyzed easily by remembering that:

1. The pressure change across a fluid column of height h is ΔP=ρgh

2. Pressure increases downward in a given fluid and decreases upward (i.e., Pbottom>Ptop), and

3. Two points at the same elevation in a continuous fluid at rest are at the same pressure.

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on Manometer

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Fluid Statics

Chapter 2

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Fluid Statics deals with problems associated with fluids at rest.

In fluid statics, there is no relative motion between adjacent fluid layers.

Therefore, there is no shear stress in the fluid trying to deform it.

The only stress in fluid statics is normal stress Normal stress is due to pressure Variation of pressure is due only to the weight

of the fluid → fluid statics is only relevant in presence of gravity fields.

Applications: Floating or submerged bodies, water dams and gates, liquid storage tanks, etc.

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on Hoover Dam

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on Hoover Dam

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Example of elevation head z converted to velocity head V2/2g. We'll discuss this in more detail in Chapter 5 (Bernoulli equation).

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on Hydrostatic Forces on Plane Surfaces

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When a surface is submerged in a fluid, forces develop on the surface due to the fluid. The determination of these forces is important in the design of storage tanks, ships, dams, and other hydraulic structures.

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In each case, the fluid exerts a force on the surface of interest that acts perpendicular to the surface, considering the basic definition of pressure, p=F/A and the corresponding form, F=pA.

We apply these equations directly only when the pressure is uniform over the entire area of interest.

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on Horizontal Flat Surfaces Under Liquids

The pressure in the water at the bottom of the drum is uniform across the entire area because it is a horizontal plane in a fluid at rest.

We can simply use F =pA to calculate the force on the bottom

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on Horizontal Flat Surfaces Under Liquids

If the drum in the previous slide is open to the atmosphere at the top, calculate the force on the bottom.

We must first calculate the pressure at the bottom of the drum and the area of the bottom.

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kN8.253APF

m07.74/m0.34/DA

kPa9.35kPa715.14kPa189.21)gage(P

)m5.1)(s/m81.9(m/kg1000)m4.2)(s/m81.9(m/kg900)gage(P

ghghPP

)m/kg1000(9.0SG

ghghPP

BottomBottomBottom

222Bottom

Bottom

2323Bottom

wateroilAtmBottom

3wateroiloil

wateroilAtmBottom

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on Horizontal Flat Surfaces Under Liquids

Would there be any difference between the force on the bottom of the cylinder drum and that on the bottom of the cone-shaped container.

The force would be the same because the pressure at the bottom is dependent only on the depth and specific weight of the fluid in the container. The total weight of fluid is not the controlling factor.

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on Hydrostatic Forces on Plane Surfaces

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Rectangular Walls:

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on Hydrostatic Forces on Plane Surfaces

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On a plane surface, the hydrostatic forces form a system of parallel forces

For many applications, magnitude and location of application, which is called center of pressure, must be determined.

Atmospheric pressure Patm can be neglected when it acts on both sides of the surface.

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on Hydrostatic Forces on Plane Surfaces

Rectangular Walls:

The total resultant force can be calculated from the equation

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ApF avgR

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where pavg is the average pressure and A is the total area of the wall.

But the average pressure is that at the middle of the wall and can be calculated from the equation

Where h is the total depth of the fluid. Therefore, we have

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2

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hgpavg

A2

hA

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hgFR

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on Hydrostatic Forces on Plane Surfaces

The center of pressure is at the centroid of the pressure distribution triangle, one third of the distance from the bottom of the wall.

The resultant force acts perpendicular to the wall at this point.

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on Problem

In figure, the fluid is gasoline and the total depth is 3.7m. The wall is 12.2m long. Calculate the magnitude of the resultant force on the wall and the location of the center of pressure.

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68.0SGgasoline

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kN0.557F

)m14.45)(2

m7.3)(s/m81.9)(m/kg1000)(68.0(F

m14.45m2.12*m7.3A

SG

A2

hgF

R

223R

2

watergasolinegasoline

gsolineR

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on Hydrostatic Forces on Plane Surfaces

Rectangular Walls Inclined at an angle θ:

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Atmo PP

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Rectangular Walls Inclined at an angle θ:

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Rectangular Walls Inclined at an angle θ:

Horizontal Wall

Therefore, since θ is constant along the inclined plate

Its line of action passes through the center of pressure CP

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AdhpApdF Atm

AdsinypApdF Atm

AsinypF CAtm

ApAhpF CGCGAtm sinyh CCG

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on Hydrostatic Forces on Plane Surfaces

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on Hydrostatic Forces on Plane Surfaces

In most cases the atmospheric pressure is neglected because it acts on both sides of the plate.

The first moment of the area with respect to x axis:

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AhdApdF

AydsinAdsinyApdF

sinAyF C

ApAhF CGCG

AyAyd C

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on Hydrostatic Forces on Plane Surfaces

To find the coordinates (xCP,yCP), the moment of the resultant force must equal the moment of the distributed pressure force.

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AdysinAdsinyAypdFy 22CP

xxCP IsinFy

Ah

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CG

xxCP

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The negative sign shows that (xCP,yCP) is below the centroid at a deeper level.

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AxydsinAdsinxyAxpdFxCP

xyCP IsinFx

Ah

sinI

Ap

Isinx

CG

xy

CG

xyCP

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Centroidal moments of inertia for various cross sections: (a) rectangle, (b) circle, © triangle, and (d) semicircle

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on Buoyancy, Flotation, and Stability

The same principles used to compute hydrostatic forces can be applied to the net force on a completely submerged or floating body. The results are the two laws of buoyancy discovered by Archimedes in the third century B.C.:1. A body immersed in a fluid experiences a vertical

buoyant force equal to the weight of the fluid it displaces.

2. A floating body displaces its own weight in the fluid in which it floats.

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on Buoyancy, Flotation, and Stability

The same principles used to compute hydrostatic forces can be applied to the net force on a completely submerged or floating body. The results are the two laws of buoyancy discovered by Archimedes in the third century B.C.:1. A body immersed in a fluid experiences a vertical

buoyant force equal to the weight of the fluid it displaces.

2. A floating body displaces its own weight in the fluid in which it floats.

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displaceddisplacedB VgVF

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on Buoyancy, Flotation, and Stability

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on Stability

A body in a fluid is considered stable if it will return to its original position after being rotated a small amount about a horizontal axis.

The condition for stability of bodies completely submerged in a fluid is that the center of gravity of the body must be below the center of buoyancy.

The center of buoyancy of a body is at the centroid of the displaced volume of fluid, and it is through this point that the buoyant force acts in a vertical direction.

The weight of the body acts vertically downward through the center of gravity.

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on Stability of Completely Submerged Bodies

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submerged submarine

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on Stability of Completely Submerged Bodies

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Much of the upper structure is filled with light syntactic foam to provide buoyancy. This causes the center of gravity (cg) to be lower than the center of buoyancy (cb), achieving stability.

Previous figure shows the stability of a submerged submarine.

Because their lines of action are now offset, these forces create a righting couple that brings the vehicle back to its original orientation, demonstrating stability.

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on Stability of Floating Bodies

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Δθ

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on Stability of Floating Bodies

In part (a) of the figure, the floating body is at its equilibrium orientation and the center of gravity (cg) is above the center of buoyancy (cb).

A vertical line through these points will be called the vertical axis of the body.

Figure (b) shows that if the body is rotated slightly, the center of buoyancy shifts to a new position because the geometry of the displaced volume has changed.

The buoyant force and the weight now produce a righting couple that tends to return the body to its original orientation.

Thus, the body is stable.

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on Stability of Floating Bodies

The vertical line upward from the new position of the center of buoyancy intersect the vertical axis of the body in the point (mc) called metacenter, which is independent of Δθ for small angles.

If the point (mc) is above (cg) (that is, if the metacentric height is positive), a restoring moment and the position is stable.

If (mc) is below (cg) (negative metacentric height), the body is unstable and will overturn if disturbed.

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on Stability of Floating Bodies

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on Stability of Floating Bodies

The engineer would determine the distance from M (metacenter) to B (center of buoyancy) from the basic shape and design of the floating body and then make the calculation of the Io (moment of inertia) and the submerged volume vsub.

If the metacentric height MG is positive, the body is stable.

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_____

sub

o______

GBv

IMG

sub

o______

v

IMB

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Due, Wednesday, February ??, 2011

Omar E. Meza Castillo Ph.D.

Homework2 WebPage

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