1 Eric G. Paterson Department of Mechanical and Nuclear Engineering Pennsylvania State University Spring 2005
2 Reading and Homework Read Chapter 3. Homework Set #2 has been posted. Due date: Friday 21 January.
3 is defined as a normal force exerted by a fluid per unit area. Units of pressure are N/m 2, which is called a pascal (Pa). Since the unit Pa is too small for pressures encountered in practice, kilopascal (1 kpa = 10 3 Pa) and megapascal (1 MPa = 10 6 Pa) are commonly used. Other units include bar, atm, kgf /cm 2, lbf /in 2 =psi.
4 Absolute, gage, and vacuum pressures Actual pressure at a give point is called the absolute pressure. Most pressure-measuring devices are calibrated to read zero in the atmosphere, and therefore indicate gage pressure, P gage = P abs P atm. below atmospheric pressure are called vacuum pressure, P vac = P atm P abs.
5 at a Point at a any point in a fluid is the same in all directions. has a magnitude, but not a specific direction, and thus it is a scalar quantity. Proof on blackboard.
6 Variation of with Depth In the presence of a gravitational field, pressure increases with depth because more fluid rests on deeper layers. To obtain a relation for the variation of pressure with depth, consider rectangular element Force balance in z direction gives F z = ma z = 0 : P 2 x P 1 x ρg x z = 0 Dividing by x and rearranging gives P = P 2 P 1 = ρg z = γ s z
7 Variation of with Depth in a fluid at rest is independent of the shape of the container. is the same at all points on a horizontal plane in a given fluid.
8 Scuba Diving and
9 Pascal s Law applied to a confined fluid increases the pressure throughout by the same amount. In picture, pistons are at same height: P 1 = P 2 F 1 A 1 = F 2 A 2 F 2 F 1 = A 2 A 1. Ratio A 2 A 1 is called ideal mechanical advantage.
10 P 1 = P 2 P 2 = P atm + ρgh An elevation change of z in a fluid at rest corresponds to P/ρg. A device based on this is called a manometer. A manometer consists of a U-tube containing one or more fluids such as mercury, water, alcohol, or oil. Heavy fluids such as mercury are used if large pressure differences are anticipated.
11 Multifluid For multi-fluid systems 1 change across a fluid column of height h is P = ρgh. 2 increases downward, and decreases upward. 3 Two points at the same elevation in a continuous fluid are at the same pressure. 4 can be determined by adding and subtracting ρgh terms. P atm +ρ 1 gh 1 +ρ 2 gh 2 +ρ 3 gh 3 = P 1.
12 Measuring Drops s are well suited to measure pressure drops across valves, pipes, heat exchangers, etc. Relation for pressure drop P 1 P 2 is obtained by starting at point 1 and adding or subtracting ρgh terms until we reach point 2. P 1 +ρ 1 g(a+h) ρ 2 gh ρ 1 ga = P 2 P 1 P 2 = (ρ 2 ρ 1 )gh If fluid in pipe is a gas, ρ 2 ρ 1 and P 1 P 2 = ρ 2 gh
13 P C + ρgh = P atm P atm = ρgh Atmospheric pressure is measured by a deviced called a barometer; thus, atmospheric pressure is often referred to as the barometric pressure. P C can be taken to be zero since there is only Hg vapor above point C, and it is very low relative to P atm. Change in atmospheric pressure due to elevation has many effects: Cooking, Nose bleeds, engine performance, aircraft performance.
14 deals with problems associated with fluids at rest. In fluid statics, there is no relative motion between adjacent fluid layers. refore, there is no shear stress in the fluid trying to deform it. 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..
15 Hoover Dam
16 Hoover Dam
17 Hoover Dam Example of elevation head z converted to velocity head V 2 2g. We ll discuss this in more detail in Chapter 5 (Bernoulli equation).
18 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 P atm can be neglected when it acts on both sides of the surface.
19 Resultant Force magnitude of F R acting on a plane surface of a completely submerged plate in a homogenous fluid is equal to the product of the pressure P C at the centroid of the surface and the area A of the surface.
20 Center of Line of action of resultant force F R does not pass through the centroid of the surface. In general, it lies underneath where the pressure is higher. Vertical location of Center of is determined by equation the moment of the resultant force to the moment of the distributed pressure force. y P = y C + I xx,c y C A I xx,c is tabulated for simple geometries.
21 F R on a curved surface is more involved since it requires integration of the pressure forces that change direction along the surface. Easiest approach: determine horizontal and vertical components F H and F V separately.
22 Horizontal force component on curved surface: F H = F x. Line of action on vertical plane gives y coordinate of center of pressure on curved surface. Vertical force component on curved surface: F V = F y + W, where W is the weight of the liquid in the enclosed block W = ρgv. x coordinate of the center of pressure is a combination of line of action on horizontal plane (centroid of area) and line of action through volume (centroid of volume). Magnitude of force F R = FH 2 + F V 2 Angle of force is α = arctan (F V /F H )
23 Buoyancy is due to the fluid displaced by a body. F B = ρ f gv. Archimedes principal buoyant force acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body, and it actgs upward through the centroid of the displaced volume.
24 For floating bodies, F B = W. Buoyancy force F B is equal only to the displaced volume ρ f gv displaced. Three scenarios possible ρ body < ρ fluid : Floating body ρ body = ρ fluid : Neutrally buoyant ρ body > ρ fluid : Sinking body
25 Example: Galilean rmometer Galileo s thermometer is made of a sealed glass cylinder containing a clear liquid. Suspended in the liquid are a number of weights, which are sealed glass containers with colored liquid for an attractive effect. As the liquid changes temperature it changes density and the suspended weights rise and fall to stay at the position where their density is equal to that of the surrounding liquid. If the weights differ by a very small amount and ordered such that the least dense is at the top and most dense at the bottom they can form a temperature scale.
26 Example: Floating Drydock Auxiliary Floating Dry Dock Resolute (AFDM-10) partially submerged. Submarine underging repair work on board the AFDM-10. Using buoyancy, a submarine with a displacement of 6,000 tons (12,000,000 lbf) can be lifted.
27 Example: Submarine Ballast Submarines use both static and dynamic depth control. Static control uses ballast tanks between the pressure hull and the outer hull. Dynamic control uses the bow and stern planes to generate trim forces.
28 Example: Submarine Ballast Normal surface trim SSN 711 nose down after accident which damaged fore ballast tanks.
29 Example: Submarine Ballast Ballast Control Panel Emergency Blow Switch: fills ballast tanks with compressed air for emergency ascent. Watch the avi of an emergency ascent.
30 of Immersed Bodies Rotational stability of immersed bodies depends upon relative location of center of gravity G and center of buoyancy B. G below B: stable G above B: unstable G coincides with B: neutrally stable.
31 of Floating Bodies If body is bottom heavy (G lower than B), it is always stable. Floating bodies can be stable when G is higher than B due to shift in location of center buoyancy and creation of restoring moment. Measure of stability is the metacentric height GM. If GM > 1, ship is stable.
32 Rigid-Body re are special cases where a body of fluid can undergo rigid-body motion: linear acceleration, and rotation of a cylindrical container. In these cases, no shear is developed. Newton s 2nd law of motion can be used to derive an equation of motion for a fluid that acts as a rigid body P + ρg k = ρ a In Cartesian coordinates, P x = ρa x, P y = ρa y, P z = ρ (g + a z).
33 Linear Acceleration Gasoline Tanker Container is moving on a straight path with a x 0, a y = a z = 0. P x = ρa x, P P y = 0, z = ρg. Total differential of P dp = ρa x dx ρgdz difference between two points P 2 P 1 = ρa x (x 2 x 1 ) ρg (z 2 z 1 ) Find the rise by selecting 2 points on free surface, P 2 = P 1 z s = z s2 z s1 = ax g (x 2 x 1 )
34 Rotation in a Cylindrical Container Container is rotating about the z-axis a r = rω 2, a θ = a z = 0. P r = ρrω 2, P P θ = 0, z = ρg. Total differential of P dp = ρrω 2 dr ρgdz On an isobar, dp = 0 dz isobar dr = rω2 g z isobar = ω2 2g r 2 + C 1 Equation of the free surface ( z s = h 0 ω2 4g R 2 2r 2)
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