Hydrostatics. ENGR 5961 Fluid Mechanics I: Dr. Y.S. Muzychka


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1 1 Hydrostatics
2 2 Introduction In Fluid Mechanics hydrostatics considers fluids at rest: typically fluid pressure on stationary bodies and surfaces, pressure measurements, buoyancy and flotation, and fluid masses under rigid body motion, to name a few. We will consider a simple element of stationary fluid and examine the how the pressure, which is a scalar field, varies throughout the element. A scalar field unlike a vector field, only has a magnitude associated with it at any point, and not direction. Pressure at any point is the same in all directions.
3 3 Pressure Distribution p = p(x, y,z)
4 4 Pressure Gradient Pressure Distribution If we examine the rate of change of pressure through the element in any direction we find (using a Taylor series approximation): p x = p p x +dx = p + p x dx p x p x +dx = p p + p x dx = p x dx Similiarly, we find: p x p x +dx = p p + p x dx = p x dx p y p y +dy = p p + p y dy = p y dy p z p z+dz = p p + p z dz = p z dz ** We can also develop the Taylor expansion using a central Coordinate as given in the figure. For simplicity we choose point O to be located at the lower left corner. We get the same result.
5 5 Pressure Distribution Considering force balances across each element displacement, we obtain: df s = p x dx dydzˆ i p y dy dxdzˆ j p z dz dxdyk ˆ df s = p i x ˆ p y ˆ j p k ˆ dxdydz = p dxdydz z The body force is the weight per unit volume: d F g = ρ g dxdydz
6 6 Pressure Distribution For a stationary fluid, the force balance is strictly between pressure forces and body forces (weight): d F = d F s + d F g = 0 The force balance after defining the change in pressure across a differential length is: p i x ˆ + p y ˆ j + p k ˆ dxdydz + ρ z g dxdydz = 0 Body Force Pressure p + ρ g = 0
7 7 Pressure Distribution This is a vector equation, which must be expanded into its three components: p x + ρg x = 0 p y + ρg y = 0 p z + ρg z = 0 x direction y direction z direction If our coordinates are gravity oriented (positive z axis upwards), then: p x = 0 p y = 0 p z = ρg x direction y direction z direction
8 8 Pressure Distribution We can extend this to fluids in rigid body motion using Newton s law F = ma, to obtain: p x + ρg x = ρa x p y + ρg y = ρa y p z + ρg z = ρa z x direction y direction z direction This form allows us to determine the effect of acceleration on fluid masses held in containers which are accelerating or decelerating.
9 9 Standard Atmosphere The pressure in the atmosphere varies with elevation. We can model the first 11 km (troposphere) using a simple linear relationship for the change in temperature and use ideal gas law.
10 10 Standard Atmosphere Defining the following equations: dp dz + ρg = 0 p = ρrt T(z) = T o βz where R = J/kgK (gas constant for air), β = K/m (lapse rate in the first 11 km), and T o = K (15 C) we obtain: p(z) = p a 1 βz T o g Rβ =101.3( z) where p a = kpa is the sea level (z = 0) pressure of the atmosphere.
11 11 Standard Atmosphere Sea level conditions of the Standard Atmosphere
12 12 Example  5 Calculate the absolute pressure you would experience if you were at an altitude of 10 km and a depth of 10 km below the ocean surface. Place the final results in terms the number of atmospheres for each fluid.
13 Manometers Variation of pressure within a fluid is frequently used for making pressure measurements: 13 We desire a simple methodology to predict the pressure difference in these devices. Modern devices use strain gages attached to membranes.
14 14 Manometers Types of Manometers Piezotube UTube (for vacuum pressures) UTube differential (for pressure difference) Inclined (for small pressure difference) Manometers allow for accurate pressure measurement provided the manometer fluid s density is accurately known. Typical fluids include: water (S.G. = 1), oil (S.G ), mercury (S.G. = 13.55).
15 15 Manometers The basic equation is: dp dz = ρg Separating variables and integrating we find: dp = ρg dz More simply put, we define: p p o p p o = ρg(z z o ) = ρg(z o z) z z o p p o = ρgh h = z o z h > 0 h < 0 down up
16 16 Manometers Reference scale for manometry problems:
17 Manometers We can use this simple approach to define the pressure change (increases) as we move downwards through the fluid and (decreases) as we move upwards. We also know, since the fluid is stationary that pressure along any horizontal line is constant. At any interface, we jump across. 17 p = p o + ρgh p A + ρ H 2 Og(10") ρ Hg g(3") + ρ Oilg(4") ρ Hgg(5") ρ H 2 Og(8") = p B Jump Across Jump Across Jump Across
18 18 Example  6 Consider the system we used on the last slide: Assuming that points A and B are connected to a pipe with flowing water in it, determine the actual pressure difference between A and B. Use the properties of fluids from your text.
19 19 Example  7 A mixture of crude oil (SG = 0.87) and water is pumped into a settling tank that is 10 m high and vented to the atmosphere. After sometime a layer of oil 2 m thick forms at the top of the tank with 7.5 m of water below. Find: The pressure at the bottom of the tank Mean density of the combined fluids Pressure gradient in the system
20 20 Forces on Submerged Surfaces Consider the following system:
21 21 Forces on Submerged Surfaces The pressure and force distribution is as follows:
22 22 Forces on Submerged Surfaces We wish to resolve the pressure distribution and obtain the following information: Resultant force (F R ) Location of the Centroid of the surface (CofG) Location of the Center of Pressure (CofP) This will allow us to make appropriate calculations for forces and moments required in systems. We will also require some information about moments of area. Later we will consider curved surfaces and develop an alternate approach for these systems.
23 23 Forces on Submerged Surfaces The force on the plate is obtained by integrating the pressure force over the surface: F R = We also know we can write the pressure at any point as: A pda p = p atm + ρgh where Thus: F R = h = y sin(θ) [ p atm + ρgy sinθ ] da A F R = p atm A + ρgsinθ yda A
24 24 Forces on Submerged Surfaces The integral is the first moment of area about the xaxis, which may be simplified according to: A yda = y C A Where y C is y coordinate of the location of the centroid of the surface. Thus we may finally write: F R = p atm A + ρgsinθy C A = ( p atm + ρgh C ) Absolute Pressure at Centroid Frequently, we do not need need to consider the absolute pressure, as the net force is due to gage pressure. A
25 25 Forces on Submerged Surfaces Since the pressure varies with depth, the resultant force does not act the centroid of the surface. To find the location of the resultant force, we need to consider moments about the xaxis: y P F R = y( p atm + ρgy sinθ)da = p atm y + ρgy 2 sinθ A y P F R = p atm A A ( )da yda + ρgsinθ y 2 da A First Moment Second Moment A A yda = y C A y 2 da = I xx = I xx,c + Ay C 2 Combining gives: y P F R = ( p atm + ρgh C )y C A + ρgsin(θ)i xx,c
26 26 Forces on Submerged Surfaces Simplifying and solving for y P we get: y P F R = F R y C + ρgsin(θ)i xx,c y P = y C + ρgsin(θ)i xx,c F R I xx,c is the second moment about the centroid. It is tabulated for many shapes. If we are dealing with gage pressure, which we frequently do, the above formula simplifies to give: y P = y C + I xx,c Ay C since for gage pressure F R = ρgy c A. y P is the center of pressure.
27 27 Forces on Submerged Surfaces For surfaces which are not symmetric with respect to the yaxis, we must also locate the C of P in the x direction. The analysis is similar and leads to the following: or x P = x C + ρgsin(θ)i xy,c F R x P = x C + I xy,c Ay C Here I xy,c is product of inertia about the centroid. For symmetric surfaces, this parameter is equal to zero, for nonsymmetric surfaces it is not zero!
28 28 Forces on Submerged Surfaces Summary: Resultant force is the product of pressure at the C of G and the area of the surface. It acts a the C of P which is below the C of G. The location of the C of P requires the second moment about the centroid. If the plate is not symmetric with respect to the yaxis, we need to locate the C of P in the xdirection as well. Frequently we use gage pressure in the resultant force. Taking the coordinate system at the C of G rather than the free surface we have: F R = ρgh c A = p C A y P = ρgsinθ I xx,c p C A x P = ρgsinθ I xy,c p C A
29 29 Example  8 Find the force on the submerged gate shown in the figure (we will sketch in class). The gate is 5 ft into the page. The specific weight of water in BG units is 64.2 lbf/ft 3. Compute the force at the wall and at the reaction forces at the hinge.
30 30 Curved Surfaces Forces on curved surfaces are also important, but we utilize a different approach, than in plane surfaces, since integration is more complex.
31 31 Curved Surfaces The horizontal force component is calculated for a virtual vertical surface using: F H = ρgh c A = p C A The vertical force component is found to be the weight of the fluid over the surface: F V = ρghda z = ρgdv = ρgv A z V The line of action of the vertical weight is through the centroid of the volume of fluid above the curved surface. This can be difficult to find but not impossible.
32 32 Example  9 Calculate the hydrostatic force on the curved surface of the dam (sketched in class). The dam level is 24 ft and is described by the parabolic curve: y = y 0 where x 0 = 10 ft and y 0 = 24 ft. Assume the specific weight of water is in BG units is 64.2 lbf/ft 3. x x 0 2
33 33 Buoyancy and Stability Floating bodies or suspended bodies in a fluid are another common application of fluid statics. The buoyant force is due to the hydrostatic pressure distribution around a submerged or partially submerged body: Archimedes principle: F B = ρgv displaced
34 34 Buoyancy and Stability Proof is simple (referring to the figure on the last slide): but so df z = (ρgh 2 ρgh 1 )da = ρg( h 2 h 1 )da F B = ( h 2 h 1 )da = dv df z = ρgdv = ρgv displaced that is the buoyant force equals the weight of fluid displaced. F B > F G F B = F G F B < F G Body Floats Neutrally Buoyant Body Sinks
35 35 Example  10 It is desired to make a floatation device from steel in the shape of a sphere. The device is to have an outer diameter of 30 cm. How thick should the vessel wall be to assure floatation in fresh water. What happens if the sphere is then placed in oil S.G. = 0.8 and seawater S.G ? Assume that the steel has a density of 7854 kg/m 3. What about the contents inside, if any?
36 36 Example  11 A helium weather balloon is designed to carry a payload of 250 [kg] (carriage, instruments, balloon fabric etc.). The balloon has a nearly spherical shape with a diameter of 12 [m] and is filled with helium gas having a density ρ = 1.66 x 101 [kg/m 3 ]. The air temperature of the standard atmosphere varies linearly with altitude up to 11 km. If the balloon is constructed of a material which prevents significant heat transfer from the inside of the balloon to the atmospheric air outside, i.e. assume constant temperature inside the balloon. Determine the altitude to which the balloon will rise, i.e find z. Hint: consider the air as an ideal gas.
37 37 Buoyancy and Stability Stability of floating bodies is quite important. The relative locations of the center of gravity of a body and its center of buoyancy determine stability.
38 38 Buoyancy and Stability In general, the buoyant force and gravitational force of a body in equilibrium are inline. Keep in mind that the buoyant force is located at the center of buoyancy which is the centroid of the submerged portion of the body. Its location changes as a body is displaced, the center of mass of the body does not move. As a result, if the body is displaced in such a manner that they are no longer in line, a moment arises which may restore the body to its position (stable) or overturn the body (unstable).
39 39 Buoyancy and Stability
40 40 Buoyancy and Stability In general, we can define the Metacentric height of a floating body as: MG = I o V submerged GB MG GB I o V submerged Metacentric Height Separation of CofG and CofB Moment of Inertia of Waterline Volume of Submerged Portion If we find that: MG > 0 MG < 0 Stable Unstable The analysis leading to the above equation is based on the assumption of a small displacement occuring.
41 41 Example  12 A floating rectangular barge has a displacement in water of H, with a total height of 2H, a width 2L, and a length B (into the page B >> 2L). Assuming (in this case) that the barge has a weight such that its CofG is always at the waterline, determine the ratio of L/H which ensures stability for small displacements.
42 42 Example  13 A 4 m wood log having a 15 cm diameter is to be weighted at one end in an attempt to provide vertical stability. The specific weight of the wood is 0.5 that of water. A 20 kg cylindrical lead weight sleeve of length 30 cm is added to one end. Assuming that the weight has negligible volume as compared with the volume of the log, will the log float vertical or will it find another stable orientation, i.e. find the metacentric height. Hint, you will need to find the CofG of the weighted log using a simple moment (pivot) analysis.
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