Fluid Mechanics and Applications MECN 3110 Inter American University of Puerto Rico Professor: Dr. Omar E. Meza Castillo

Pressure Distribution in a Fluid Chapter 2 Pressure Distribution in a Fluid

To know the use of a simple manometer. Course Objectives 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. Thermal Systems Design Universidad del Turabo

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

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

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

p is single valued at a point and independent of direction Pressure p is single valued at a point and independent of direction

In Fluid Mechanics we often use gage pressure and vacuum 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.

Pvac=-Pgage= Patm - Pabs Pressure Absolute Pressure Force per unit area exerted by a fluid Gage Pressure Pressure above atmospheric Pgage= Pabs - Patm Vacuum Pressure Pressure below atmospheric Pvac=-Pgage= Patm - Pabs

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

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.

Pressure variation with Elevation

Pressure variation with Elevation

Hydrostatic Pressure in Liquids

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

Hydrostatic Pressure in Liquids

Hydrostatic Pressure in Liquids

Application Problems

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?

Example

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).

Pressure Variation for Compressible Fluid

Pressure Variation in the Troposphere

Pressure Variation in the Troposphere

Pressure Variation in the Stratosphere

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.

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: The pressure change across a fluid column of height h is ΔP=ρgh Pressure increases downward in a given fluid and decreases upward (i.e., Pbottom>Ptop), and Two points at the same elevation in a continuous fluid at rest are at the same pressure.

Manometer

Application Problems

Chapter 2 Fluid Statics

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.

Hoover Dam

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

Hydrostatic Forces on Plane Surfaces 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.

Hydrostatic Forces on Plane Surfaces 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.

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

Application Problems

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.

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.

Hydrostatic Forces on Plane Surfaces Rectangular Walls:

Hydrostatic Forces on Plane Surfaces 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.

Hydrostatic Forces on Plane Surfaces Rectangular Walls: The total resultant force can be calculated from the equation

Hydrostatic Forces on Plane Surfaces 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

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.

Application Problems

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.

Problem

Hydrostatic Forces on Plane Surfaces Rectangular Walls Inclined at an angle θ:

Hydrostatic Forces on Plane Surfaces Rectangular Walls Inclined at an angle θ:

Hydrostatic Forces on Plane Surfaces 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

Hydrostatic Forces on Plane Surfaces

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:

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.

Hydrostatic Forces on Plane Surfaces The negative sign shows that (xCP,yCP) is below the centroid at a deeper level.

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

Application Problems

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Problem

Problem

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Problem

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.: 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.

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.: 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.

Buoyancy, Flotation, and Stability

Application Problems

Problem

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.

Stability of Completely Submerged Bodies submerged submarine

Stability of Completely Submerged Bodies 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.

Stability of Floating Bodies Δθ

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.

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.

Stability of Floating Bodies

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.

Application Problems

Problem

Due, Wednesday, February ??, 2011 Homework2  WebPage Due, Wednesday, February ??, 2011 Omar E. Meza Castillo Ph.D.