Chapter 2. Fluid Statics Fluid statics is concerned with the balance of forces which stabilize fluids at rest. In the case of a liquid, as the pressure largely changes according to its height, it is necessary to take its depth into account. Furthermore, even in the case of relative rest (e.g. the case where the fluid is stable relative to its vessel even when the vessel is rotating at high speed), the fluid can be regarded as being at rest if the fluid movement is observed in terms of coordinates fixed upon the vessel. Pressure When a uniform pressure acts on a flat plate of area A and a force F pushes the plate, then the pressure p is : p = F/A When the pressure is not uniform, the pressure acting on the minute area A is expressed by the following equation: F df Characteristics of pressure The pressure has the following three characteristics. 1. The pressure of a fluid always acts perpendicular to the wall in contact with the fluid. 2. The values of the pressure acting at any point in a fluid at rest are equal regardless of its direction. Imagine a minute triangular prism of unit width in a fluid at rest as shown in Fig. 1 Let the pressure acting on the small surfaces da 1, da 2, and da be p 1, p 2 and p respectively. The following equations are obtained from the balance of forces in the horizontal and vertical directions: 1 Fig.1
The weight of the triangle pillar is doubly infinitesimal, so it is omitted. From geometry, the following equations are obtained: Therefore, the following relation is obtained: Pressure of fluid at rest In general, in a fluid at rest the pressure varies according to the depth. Consider a minute column in the fluid as shown in Fig. 2. Assume that the sectional area is da and the pressure acting upward on the bottom surface is p and the pressure acting downward on the upper surface (dz above the bottom surface) is p + (dp/dz)dz. Then, from the balance of forces acting on the column, the following equation is obtained: Since p is constant for liquid, the following equation ensues: Fig.2 When the base point is set at zo below the upper surface of liquid as shown in Fig. 3 and p o is the pressure acting on that surface, then p = po when z = zo, so : Thus it is found that the pressure inside a liquid increases in proportion to the depth. Fig.3 2
Hydrostatic Pressure in Gases: Gases are compressible, with density nearly proportional to pressure. Thus density must be considered as a variable in the above equation : Separate the variables and integrate between points 1 and 2: The integral over z requires an assumption about the temperature variation T(z). One common approximation is the isothermal atmosphere, where T=To : In the spatial cases the gases become incompressible fluids (ρ=constant) when the pressure and temperature at absolute values so that the above integral become : Ln(p2/p1)=-k( z2 z1 ) where k = g/(rt)=1.139 10-4 m -1, for z 2 z 1 = 1 m and at 27 o C p2/p1 = 0.9989, if p 1 = 100 kn/m 2 then p 2 = 99.989 kn/m 2 P=0 It can be neglected the head pressure of the gases. Manometry :The technique of pressure measurement by employing the hydrostatic law i.e., dp/dz=ρg is known as manometry and the devises used for measurement are called manometers 1-single tube manometers : the tube is attached to the point at it is lower and left open to the atmosphere at the upper end. Since the column of liquid in the tube is at rest, the gage pressure at the lower end of the tube must be balanced by the hydrostatic pressure due to the column of liquid in it : p = po + ρ g h h P h cosα α 3
This manometer may serve to measure the pressure above the atmospheric pressure For a small gage pressure, it is better to use an inclined single tube. 2- U-tube Manometers : Consists of two tubes joined at one end to form a U- shape tube. This manometer can be employed to measure the pressure above and below atmospheric in any fluid, liquid or gas. p h h Liquid under pressure P=ρgh Liquid under vacuum P=-ρgh h' h Gas under pressure P=ρmgh' Gas under vacuum P=-ρmgh Generally, the pressure gages divided to: 1- Tube gage: i- Piezometer (small positive), ii- Manometer (simple (small positive and negative), micro, differential and invert of differential (large positive and negative))), 2-Mechaniical gage (Bourdon tube), c-type, spiral, twist and helical. 4
Bourdon gage 3-Piezoelectric transducers, also called solid state pressure transducers, work on the principle that an electric potential is generated in a crystalline substance when it is subjected to mechanical pressure. 4-5
For a fluid under a very high pressure, consider the U-tube, multi-fluid manometer as shown in below figure.if we first label all inter mediate points between A & a, the total pressure difference can be expressed in terms of a series of intermediate terms as following : PA+ ρ1g(za Z1 ) - ρ2g (Z2 Z1 ) - ρag ( Za Z2 ) = Pa PA-Pa= - ρ1g(za Z1 )+ ρ2g (Z2 Z1 )+ ρag ( Za Z2 ) For a multiple-fluid manometer connected to the two chambers A and B the difference in pressure between two chambers A and B. PA+ ρ1g (ZA Z1 ) - ρ2g (Z2 Z1 ) + ρ3g (Z2 Z3 ) - ρ4g (Z B Z3 ) = PB PA- PB = - ρ1g (ZA Z1 ) + ρ2g (Z2 Z1 ) - ρ3g (Z2 Z3 ) + ρ4g (Z B Z3 ) Manometer Example: Given the indicated manometer, determine the gage pressure at A. ρg w = 9810 N/m 3 ρg A = 0.83*9810 = 6807 N/m 3 6
ρg air = 11.8 N/m 3 With the indicated points labeled on the manometer, we can write P A + ρ 1 g(z A Z 1 ) - ρ 2 g (Z 2 Z 1 ) - ρ a g ( Z a Z 2 ) = P a P A -P a = - ρ 1 g(z A Z 1 )+ ρ 2 g (Z 2 Z 1 ) + ρ a g ( Z a Z 2 ) Neglect the contribution due to the air column. Substituting values, we obtained: P A -P a = - (6807 0.1) + (9810 0.18) = 1085.1 N/m 2 Buoyancy Fluid pressure acts all over the wetted surface of a body floating in a fluid, and the resultant pressure acts in a vertical upward direction. This force is called buoyancy. The buoyancy of air is small compared with the gravitational force of the immersed body, so it is normally ignored. Suppose that a cube is located in a liquid of density ρ as shown in below Figure. For the vertical direction, where the atmospheric pressure is Po, the force F1 acting on the upper surface A is expressed by the following equation: F1= ( Po + ρgh1 )A The force F2 acting on the lower surface is F2= ( Po + ρgh2)a So, when the volume of the body in the liquid is V, the resultant force F from the pressure acting on the whole surface of the body is F = F2 F1 = ρg ( h2 h1)a F= ρgvb 7
Pressure distribution in rigid body motion We will now consider an extension of our static fluid analysis to the case of rigid body motion, where the entire fluid mass moves and accelerates uniformly (as a rigid body). The container of fluid shown below is accelerated uniformly up and to the right as shown. Fist law of motion a m =ƩFx a x dm = 1 dm =ρdx dy ax = - (1/ρ) a y dm = - g dm 1 ay = - g (1/ρ) dp = + dy P = -ρ ax x - ρ( g + ay ) y + C tanθ = - ax / (g + ay ) 8
Example: The tank of liquid in the right figure accelerates to the right with the fluid in rigid-body motion. Compute A(-100,-15) y x a x in m/s 2. Find the gage pressure at point A if the fluid is glycerin at 20 o C. Solution: (a) The slope of the liquid gives us the acceleration: tan a x g a y at x=0 and y=0 then C=P o =0 Rotational motion Let us study the height of the water surface in the case where a cylindrical vessel filled with liquid is rotating at constant angular velocity ω. The movement at constant angular velocity like this is sometimes called gyrostatics, where the liquid surface poses a concave free surface. Then let us take cylindrical coordinates (r,θ,z). Consider a minute element of mass m on the equipressure plane.the forces acting on it are mg due to the gravitational acceleration g in the vertical direction and mrω 2 due to the centripetal acceleration rw 2 in the horizontal direction. 9
If z = h o at r = 0, c = h o, and the following equation is obtained : The free surface is now a rotating parabolic surface. Example : A 16-cm-diameter open cylinder 27 cm high is full of water. Find the central rigid-body rotation rate for which (a) one third of the water will spill out; and (b) the bottom center of the can will be exposed. ω Solution: (a) One-third will spill out if the resulting parabolic surface is 18 cm deep: (at z=27cm, r=8cm then z-h o =27-9=18cm) ω 2 R 2 ω 2 (0.08) 2 ω= 23.5 rad /s b) The bottom is barely exposed if the parabolic surface is 27 cm deep: (at z=27cm, r=8cm then z-h o =27-0=27cm) ω 2 (0.08) 2 ω= 28.8 rad /s 11
Problems 1. What is the water pressure on the sea bottom at a depth of 6500 m? The specific gravity of sea water is assumed to be 1.03. 2. Obtain the pressure p at point A in Fig.1 (a), (b) and (c). Fig.1 3. Obtain the pressure difference p1 - p2 in Fig.2 (a) and (b) as shown blow. Fig.2 4. An iceberg of specific gravity 0.92 is floating on the sea with a specific gravity of 1.025. If the volume of the iceberg above the water level is 100 m 3, what is the total volume of the iceberg? 5. A cylindrical vessel of radius ro filled with water to height h is rotated around the central axis, and the difference in height of water level is h'. What is the rotational angular velocity? Furthermore, assuming ro = 10 cm and h = 18 cm, obtain ω when h' = 10 cm and also the number of revolutions per minute n when the cylinder bottom begins to appear. 11
Answer 1. 6.57 107 pa 2. (a) P = Po + PgH, (b) P = Po PgH, ( c) P = Po + ρ ' gh ' - ρgh 4. 976 m 3 5-12