P = ρ{ g a } + µ 2 V II. FLUID STATICS

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1 II. FLUID STATICS From a force analysis on a triangular fluid element at rest, the following three concepts are easily developed: For a continuous, hydrostatic, shear free fluid: 1. Pressure is constant along a horizontal plane, 2. Pressure at a point is independent of orientation, 3. Pressure change in any direction is proportional to the fluid density, local g, and vertical change in depth. These concepts are key to the solution of prolems in fluid statics and lead to the following: 1. Two points at the same depth in a static fluid have the same pressure. 2. The orientation of a surface has no earing on the pressure at a point in a static fluid. 3. Vertical depth is a key dimension in determining pressure change in a static fluid. If we were to conduct a more general force analysis on a fluid in motion with constant density and viscosity, we would otain the following: P = ρ{ g a } + µ 2 V Thus the pressure change in fluid in general depends on: effects of fluid statics (ρ g), Ch. II inertial effects (ρ a), Ch. III viscous effects ( µ 2 V ) Chs VI & VII Note: For prolems involving the effects of oth (1) fluid statics and (2) inertial effects, it is the net g " a " acceleration vector that controls oth the magnitude and direction of the pressure gradient. II-1

2 This equation can e simplified for a fluid at rest (ie., no inertial or viscous effects) to yield: p = ρ g p x = 0; p y = 0; p z = dp dz = ρ g P 2 P 1 = ρgdz 2 1 It is appropriate at this point to review the differences etween asolute, gage, and vacuum pressure. These differences are illustrated in Fig. 2.3 (shown elow) and are ased on the following definitions: Patm asolute pressure - Pressure measured relative to asolute zero. gage pressure - Pressure > Patm measured relative to Patm vacuum pressure - Pressure < Patm measured relative to Patm Patm - local asolute pressure due to the local atmosphere only. standard Patm at sea level = 1atm = kpa = 2116 lf/ft 2 P (kpa) Pressure aove atmospheric e.g P = 120 kpa asolute or P = 30 kpa gage P (relative to Patm) = Pgage = 30 kpa gage 0 P(relative to Patm) = P(vac) = 40 kpa P(as) = 50 kpa Local atmospheric pressure e.g at 1000 m Patm 90 kpa = 0 Pa gage = 0 Pa vacuum Pressure elow atmospheric e.g. P = 50 kpa as Asolute zero, perfect vacuum P(as) = 0, P(vac) = 90kPa, P(gage) = - 90 kpa Fig. 2.3 Illustration of asolute, gage, and vacuum pressure II-2

3 Hydrostatic Pressure in Liquids For liquids and incompressile fluids, the previous integral expression for P2 - P1 integrates to Note: P 1 P 2 = -ρg (Z 2 Z 1 ) z Free surface Pressure = Pa P 2 Z 2 Z 1 is positive for Z 2 aove Z 1. ut P 2 P 1 is negative for Z 2 aove Z 1. We can now define a new fluid parameter: γ = ρg specific weight of the fluid x Z 2 h = Z - Z 2 1 Z 1 P 1 y With this, the previous equation ecomes (for an incompressile, static fluid) P 2 P 1 = - γ (Z 2 Z 1 ) The most common application of this result is that of manometry. Consider the U-tue, multifluid manometer shown on the right. If we first lael all intermediate points etween A & a, we can write for the overall pressure change II-3

4 P A - P a = (P A - P 1) + (P 1 - P 2 ) + (P 2 - P a ) This equation was otained y adding and sutracting each intermediate pressure. The total pressure difference is now expressed in terms of a series of intermediate pressure differences. Sustituting the previous result for static pressure difference, we otain P A - P B = - ρ g(z A - Z 1) ρ g (Z 1 Z 2 ) ρ g (Z 2 - Z B ) Again note: Z positive up and Z A > Z 1, Z 1 < Z 2, Z 2 < Z a. In general, follow the following steps when analyzing manometry prolems: 1. On the manometer schematic, lael points on each end of the manometer and at each intermediate point where there is a fluid-fluid interface, e.g. A B. 2. Express the overall pressure difference in terms of appropriate intermediate pressure differences. P A - P B = (P A - P 1 ) + (P 1 - P 2 ) + (P 2 - P B ) 3. Express each intermediate pressure difference in terms of an appropriate product of specific weight (or ρ g) * elevations change ( and watch the signs). P A - P B = -γ (z A - z 1 ) + -γ (z 1 - z 2 ) + -γ (z 2 - z B ) When developing a solution for manometer prolems, take care to: 1. Include all pressure changes. 2. Use correct Z and γ with each fluid. 3. Use correct signs with Z. If pressure difference is expressed as P A P 1, the elevation change should e written as Z A Z Watch units. II-4

5 Manometer Example: Given the indicated manometer, determine the gage pressure at A. Pa = kpa. The fluid at A is Meriam red oil no. 3. ρg w = 9790 N/m 3 A S.G. = cm H 0 2 a 2 18 cm ρg A = S.G.*ρg w = 0.83*9790 N/m ρg A = 8126 N/m 3 ρg air = 11.8 N/m 3 With the indicated points laeled on the manometer, we can write P A - P a = P A (gage) = (P A - P 1 ) + (P 1 P 2 ) + (P 2 - P a ) Sustituting the manometer expression for a static fluid, we otain P A (gage) = - ρg A (z A - z 1) ρg w (z 1 z 2 ) ρg a (z 2 - z a ) Neglect the contriution due to the air column. Sustituting values, we otain P A (gage) = N/m 3 * 0.10 m 9790 N/m 3 * = N/m 2 Ans Note why: (z A - z 1 ) = 0.10 m and (z 1 z 2 ) = m, & we did not use P a Review the text examples for manometry. See Tale 2.1 for values of specific weight, γ, in oth B.G. and S.I. units. II-5

6 Hydrostatic Pressure in Gases Since gases are compressile, density is a non-constant variale in the previous expression for dp/dz. Assuming the gas is an ideal gas, we can write dp dz P 2 = ρg = RT g or dp = ln P 2 = g 2 P P 1 R For an isothermal atmosphere with T = T o, this integrates to P 2 = P 1 exp gz 2 z 1 RT o 1 ( ) 1 dz T Up to an altitude of approximately 36,000ft (l l,000 m), the mean atmospheric temperature decreases nearly linearly and can e represented y T T o - B z where B is the lapse rate The following values are assumed to apply for air from sea level to 36,000 ft: To = R = K (15 C) B = R/ft = K/m Sustituting the linear temperature variation into the previous equation and integrating we otain P = P a 1 Bz T o g /( RB) where g RB = 5.26 ( for air ) and R = 287 m 2 /(s 2 K) Review example 2.2 in the text. II-6

7 Hydrostatic Forces on Plane Surfaces Consider a plane surface of aritrary shape and orientation, sumerged in a static fluid as shown: If P represents the local pressure at any point on the surface and h the depth of fluid aove any point on the surface, from asic physics we can easily show that the net hydrostatic force on a plane surface is given y (see text for development): F = PdA A = P cg A Thus, asic physics says that the hydrostatic force is a distriuted load equal to the integral of the local pressure force over the area. This is equivalent to the following: The hydrostatic force on one side of a plane surface sumerged in a static fluid equals the product of the fluid pressure at the centroid of the surface times the surface area in contact with the fluid. Also: Since pressure acts normal to a surface, the direction of the resultant force will always e normal to the surface. Note: In most cases, since it is the net hydrostatic force that is desired and the contriution of atmospheric pressure Pa will act on oth sides of a surface, the result of atmospheric pressure Pa will cancel and the net force is otained y F = ρ gh cg A F = P cg A Pcg is now the gage pressure at the centroid of the area in contact with the fluid. II-7

8 Therefore, to otain the net hydrostatic force F on a plane surface: 1. Determine depth of centroid hcg for the area in contact with the fluid. 2. Determine the (gage) pressure at the centroid Pcg. 3. Calculate F = PcgA. The following page shows the centroid, and other geometric properties of several common areas. It is noted that care must e taken when dealing with layered fluids. The procedure essentially requires that the force on the part of the plane area in each individual layer of fluid must e determined separately for each layer using the steps listed aove. We must now determine the effective point of application of F. This is commonly called the center of pressure - cp of the hydrostatic force. Note: This is not necessarily the same as the c.g. Define an x y coordinate system whose origin is at the centroid, c.g, of the area. The location of the resultant force is determined y integrating the moment of the distriuted fluid load on the surface aout each axis and equating this to the moment of the resultant force aout that axis. Therefore, for the moment aout the x axis: Fy cp = A ypda Applying a procedure similar to that used previously to determine the resultant force, and using the definition (see text for detailed development), we otain Y cp = ρgsinθ I xx P cg A 0 where: I xx is defined as the Moment of Inertia, or the 2 nd moment of the area Therefore, the resultant force will always act at a distance ycp elow the centroid of the surface ( except for the special case of sin θ = 0 ). II-8

9 PROPERTIES OF PLANE SECTIONS Geometry Centroid Moment of Inertia I x x Product of Inertia I x y Area y x L L 2, 2 L L y x 2R π R 4 0,0 0 π R 2 4 L y x 3, L 3 L L 2 72 L 2 a y x R 0,a = 4R 3π R 4 π π 0 π R 2 2 y x s L a = L 3 L 3 36 ( 2s)L L y x R a = 4R 3 π π π R π R 4 π R 2 4 y 1 x h a ) = h ( ( + 1 ) h 3 ( ( + 1 ) ) 0 ( + 1 ) h 2 Fluid Specific Weight Air 1f/ft N /m Seawater 3 1f /ft N /m ,050 Oil ,996 Glycerin ,360 Water Ethyl ,790 7,733 Mercury Caron ,100 15,570 II-9

10 Proceeding in a similar manner for the x location, and defining I xy = product of inertia, we otain X cp = ρgsinθ I xy P cg A where Xcp can e either positive or negative since Ixy can e either positive or negative. Note: For areas with a vertical plane of symmetry through the centroid, i.e. the y - axis (e.g. squares, circles, isosceles triangles, etc.), the center of pressure is located directly elow the centroid along the plane of symmetry, i.e., X cp = 0. Key Points: The values X cp and Y cp are oth measured with respect to the centroid of the area in contact with the fluid. X cp and Y cp are oth measured in the inclined plane of the area; i.e., Y cp is not necessarily a vertical dimension, unless θ = 90 o. Special Case: For most prolems where (1) we have a single, homogeneous fluid (i.e. not applicale to layers of multiple fluids) and (2) the surface pressure is atmospheric, the fluid specific weight γ cancels in the equation for Y cp and X cp and we have the following simplified expressions: F = ρ gh cg A Y cp = I xx sinθ h cg A X cp = I xy sinθ h cg A However, for prolems where we have either (1) multiple fluid layers, or (2) a container with surface pressurization > P atm, these simplifications do not occur and the original, asic expressions for F, Y cp, and X cp must e used; i.e., take II-10

11 care to use the approximate expressions only for cases where they apply. The asic equations always work. II-11

12 Summary: 1. The resultant force is determined from the product of the pressure at the centroid of the surface times the area in contact with the fluid. 2. The centroid is used to determine the magnitude of the force. This is not the location of the resultant force. 3. The location of the resultant force will e at the center of pressure which will e at a location Ycp elow the centroid and Xcp as specified previously. 4. Xcp = 0 for areas with a vertical plane of symmetry through the c.g. Example 2.5 Given: Gate, 5 ft wide Hinged at B Holds seawater as shown Find: a. Net hydrostatic force on gate 15 Seawater 64 lf/ft 3 h c.g. 9 A. Horizontal force at wall - A c. Hinge reactions - B B θ c.g. 6 8 a. By geometry: θ = tan -1 (6/8) = o Neglect P atm Since the plate is rectangular, h cg = 9 ft + 3ft = 12 ft A = 10 x 5 = 50 ft 2 P cg = γ h cg = 64 lf/ft 3 * 12 ft = 768 lf/ft 2 F p = P cg A = 768 lf/ft 2 * 50 ft 2 = 38,400 lf II-12

13 II-13

14 . Horizontal Reaction at A Must first find the location, c.p., for F p y cp = ρ gsinθ I xx P cg A = I xx sinθ h cg A For a rectangular wall: I xx = h 3 /12 I xx = 5 * 10 3 /12 = 417 ft 4 Note: The relevant area is a rectangle, not a triangle. Bx Bz θ Fw y c.p. c.g. c.p. 8 ft θ P 6 ft Note: Do not overlook the hinged reactions at B. y cp = 417 ft = ft elow c.g. 12 ft 50 ft x cp = 0 due to symmetry M B =0 P ( ) 38,400 6 P = 0 P = 29,330 lf Bx Bz θ Fw y c.p. c.g. c.p. θ 6 ft 8 ft II-14

15 c. F x = 0, B x + Fsinθ P = 0 B x + 38,400*0.6-29,330 = 0 B x = 6290 lf F z = 0, B z Fcosθ = 0 B z = 38,400 * 0.8 = 30,720 lf Note: Show the direction of all forces in final answers. Summary: To find net hydrostatic force on a plane surface: 1. Find area in contact with fluid. 2. Locate centroid of that area. 3. Find hydrostatic pressure P cg at centroid, typically = γ h cg (generally neglect P atm ). 4. Find force F = P cg A. 5. The location will not e at the c.g., ut at a distance y cp elow the centroid. y cp is in the plane of the area. II-15

16 Plane Surfaces in Layered Fluids For plane surfaces in layered fluids, the part of the surface in each fluid layer must essentially e worked as a separate prolem. That is, for each layer: 1.) Identify the area of the plate in contact with each layer, 2.) Locate the c.g. for the part of the plate in each layer and the pressure at the c.g., and 3. Calculate the force on each layered element using F 1 = Pc.g 1 A 1. Repeat for each layer. Use the usual procedure for finding the location of the force for each layer. Review all text examples for forces on plane surfaces. II-16

17 Forces on Curved Surfaces Since this class of surface is curved, the direction of the force is different at each location on the surface. Therefore, we will evaluate the x and y components of net hydrostatic force separately. Consider curved surface, a-. Force alances in x & y directions yield Fh = F H Fv = Wair + W 1 + W 2 From this force alance, the asic rules for determining the horizontal and vertical component of forces on a curved surface in a static fluid can e summarized as follows: Horizontal Component, Fh The horizontal component of force on a curved surface equals the force on the plane area formed y the projection of the curved surface onto a vertical plane normal to the component. The horizontal force will act through the c.p. (not the centroid) of the projected area. h cg F h y cp a Projected vertical plane Curved cp surface a II-17

18 Therefore, to determine the horizontal component of force on a curved surface in a hydrostatic fluid: 1. Project the curved surface into the appropriate vertical plane. 2. Perform all further calculations on the vertical plane. 3. Determine the location of the centroid - c.g. of the vertical plane. 4. Determine the depth of the centroid - h cg of the vertical plane. 5. Determine the pressure - Pcg = ρ g h cg at the centroid of the vertical plane. 6. Calculate Fh = P cg A, where A is the area of the projection of the curved surface into the vertical plane, ie. the area of the vertical plane. 7. The location of Fh is through the center of pressure of the vertical plane, not the centroid. Get the picture? All elements of the analysis are performed with the vertical plane. The original curved surface is important only as it is used to define the projected vertical plane. Vertical Component - Fv The vertical component of force on a curved surface equals the weight of the effective column of fluid necessary to cause the pressure on the surface. The use of the words effective column of fluid is important in that there may not always actually e fluid directly aove the surface. (See graphic that follows.) This effective column of fluid is specified y identifying the column of fluid that would e required to cause the pressure at each location on the surface. Thus, to identify the effective volume - Veff: 1. Identify the curved surface in contact with the fluid. 2. Identify the pressure at each point on the curved surface. 3. Identify the height of fluid required to develop the pressure. II-18

19 4. These collective heights comine to form V eff. V eff P P a P Fluid aove the surface a V eff P P P fluid No fluid actually aove surface These two examples show two typical cases where this concept is used to determine V eff. The vertical force acts vertically through the centroid (center of mass) of the effective column of fluid. The vertical direction will e the direction of the vertical components of the pressure forces. Therefore, to determine the vertical component of force on a curved surface in a hydrostatic fluid: 1. Identify the effective column of fluid necessary to cause the fluid pressure on the surface. 2. Determine the volume of the effective column of fluid. 3. Calculate the weight of the effective column of fluid - Fv = ρgv eff. 4. The location of Fv is through the centroid of V eff. Finding the Location of the Centroid A second prolem associated with the topic of curved surfaces is that of finding the location of the centroid of V eff. Recall: II-19

20 Centroid = the location where a point area, volume, or mass can e place to yield the same first moment of the distriuted area, volume, or mass, e.g. x cg V 1 = V1 xdv This principle can also e used to determine the location of the centroid of complex geometries. For example: If V eff = V 1 + V 2 then x cg V eff = x 1 V 1 + x 2 V 2 V 1 V 2 a or for the second geometry V T = V 1 + V eff V eff x T V T = x 1 V 1 + x cg V eff a V 1 fluid Note: In the figures shown aove, each of the x values would e specified relative to a vertical axis through since the cg of the quarter circle is most easily specified relative to this axis. II-20

21 Example: Gate AB holds ack 15 ft of water. Neglecting the weight of the gate, determine the magnitude (per unit width) and location of the hydrostatic forces on the gate and the resisting moment aout B. 15 ft Water F H A F V Width - W B a. Horizontal component γ = ρg = 62.4 lf/ft 3 Rule: Project the curved surface into the vertical plane. Locate the centroid of the projected area. Find the pressure at the centroid of the vertical projection. F = P cg A p Note: All calculations are done with the projected area. The curved surface is not used at all in the analysis. P cg h cg a A B The curved surface projects onto plane a - and results in a rectangle, (not a quarter circle) 15 ft x W. For this rectangle: h cg = 7.5, P cg = γh cg = 62.4 lf/ft 3 * 7.5 ft = 468 lf/ft 2 F h = P cg A = 468 lf/ft 2 * 15 ft*w= 7020 W lf per ft of width Location: I xx = h 3 /12 = W * 15 3 /12 = W ft 4 y cp = I xx sinθ h cg A = W ft4 sin90 o 7.5 ft15w ft 2 = 2.5 ft The location is 2.5 ft elow the c.g. or 10 ft elow the surface, 5 ft aove the ottom. II-21

22 . Vertical force: A C c.g. Rule: F v equals the weight of the effective column of fluid aove the curved surface (shown y the dashed diagonal lines). F v B Q: What is the effective volume of fluid aove the surface? What volume of fluid would result in the actual pressure distriution on the curved surface? Vol = Vol A - B - C V rec = V qc + V ABC, V ABC = V rec - V qc V ABC = V eff = 15 2 W - π 15 2 /4*W = W ft 3 F v = ρg V eff = 62.4 lf/ft 3 * ft 3 = 3013 lf per ft of width Note: F v is directed upward even though the effective volume is aove the surface. c. What is the location? A C Rule: F v will act through the centroid of the effective volume causing the force. F v c.g. B We need the centroid of volume A-B-C. How do we otain this centroid? Use the concept which is the asis of the centroid, the first moment of an area. Since: A rec = A qc + A ABC M rec = M qc + M ABC M ABC = M rec - M qc II-22

23 Note: We are taking moments aout the left side of the figure, ie., point. WHY? (The c.g. of the quarter circle is known to e at 4 R/ 3 π w.r.t..) x cg A = x rec A rec - x qc A qc x cg { π*15 2 /4} = 7.5* {4*15/3/π}* π*15 2 /4 x cg = ft { distance to rt. of to the centroid } Q: Do we need a y location? Why? d. Calculate the moment aout B needed for equilirium. A c.g. C M B =0 clockwise positive. F H M B + 5F h + ( 15 x v )F v = 0 F v M B B M B W + ( ) 3013W = 0 M B + 35,100W +10,093.6W = 0 M B = 45,194W ft lf per unit width Why is the answer negative? (What did we assume for an initial direction of M B?) The hydrostatic forces will tend to roll the surface clockwise relative to B, thus a resisting moment that is counterclockwise is needed for static equilirium. II-23

24 Always review your answer (all aspects: magnitude, direction, units, etc.) to determine if it makes sense relative to physically what you understand aout the prolem. Begin to think like an engineer. II-24

25 Buoyancy An important extension of the procedure for vertical forces on curved surfaces is that of the concept of uoyancy. The asic principle was discovered y Archimedes. It can e easily shown that (see text for detailed development) the uoyant force F is given y: P atm F = ρ g V where V is the volume of the fluid displaced y the sumerged ody and ρ g is the specific weight of the fluid displaced. V c.g. F Thus, the uoyant force equals the weight of the fluid displaced, which is equal to the product of the specific weight times the volume of fluid displaced. The location of the uoyant force is through a vertical line of action, directed upward, which acts through the centroid of the volume of fluid displaced. Review all text examples and material on uoyancy. II-25

26 Pressure Distriution in Rigid Body Motion All of the prolems considered to this point were for static fluids. We will now consider an extension of our static fluid analysis to the case of rigid ody motion, where the entire fluid mass moves and accelerates uniformly (as a rigid ody). The container of fluid shown elow is accelerated uniformly up and to the right as shown. From a previous analysis, the general equation governing fluid motion is P = ρ( g a ) + µ 2 V For rigid ody motion, there is no velocity gradient in the fluid, therefore µ 2 V = 0 The simplified equation can now e written as P = ρ( g a ) = ρg where G = g a the net acceleration vector acting on the fluid. II-26

27 This result is similar to the equation for the variation of pressure in a hydrostatic fluid. However, in the case of rigid ody motion: * P = f {fluid density & the net acceleration vector- G = g a } * P acts in the vector direction of G = g a. * Lines of constant pressure are perpendicular to G. The new orientation of the free surface will also e perpendicular to G. The equations governing the analysis for this class of prolems are most easily developed from an acceleration diagram. Acceleration diagram: For the indicated geometry: Free surface θ a a z θ = tan 1 a x g + a z dp ds = ρg where G = a 2 x + (g + a z ) 2 { } 1 2 -a G θ g s a x P 2 P 1 and P 2 P 1 = ρg(s 2 s 1 ) Note: P 2 P 1 ρ gz ( 2 z 1 ) and s 2 s 1 is not a vertical dimension Note: s is the depth to a given point perpendicular to the free surface or its extension. s is aligned with G. II-27

28 In analyzing typical prolems with rigid ody motion: 1. Draw the acceleration diagram taking care to correctly indicate a, g, and θ, the inclination angle of the free surface. 2. Using the previously developed equations, solve for G and θ. 3. If required, use geometry to determine s 2 s 1 (the perpendicular distance from the free surface to a given point) and then the pressure at that point relative to the surface using P 2 P 1 = ρ G (s 2 s 1 ). Key Point: Do not use ρg to calculate P 2 P 1, use ρ G. Example 2.12 Given: A coffee mug, 6 cm x 6 cm square, 10 cm deep, contains 7 cm of coffee. The mug is accelerated to the right with a x = 7 m/s 2. Assuming rigid ody motion and ρ c = 1010 kg/m 3, Determine: a. Will the coffee spill?. P g at a &. z θ a x 6 cm 7 cm 10 cm c. F net on left wall. a. First draw schematic showing the original orientation and final orientation of the free surface. ρ c = 1010 kg/m 3 a x = 7m/s 2 a z = 0 g = m/s 2 a We now have a new free surface at an angle θ where -a θ = tan 1 a x g + a z θ θ = tan = 35.5 z = 3 tan 35.5 = 2.14 cm G g a II-28

29 h max = = 9.14 cm < 10 cm Coffee will not spill.. Pressure at a &. P a = ρ G s a G = {a 2 x + g 2 } 1/2 = { } 1/2 z θ a x 6 cm G = m/s 2 10 cm s a = {7 + z} cos θ s a = 9.14 cm cos 35.5 = 7.44 cm θ s a 7 cm P a = 1010 kg/m 3 *12.05m/s 2 * m P a = 906 (kg m/s 2 )/m 2 = 906 Pa Note: P ρ gy G g a a Q: How would you find the pressure at, P? c. What is the force on the left wall? We have a plane surface, what is the rule? Find cg, P cg, F = P cg. A Vertical depth to cg is: z cg = 9.14/2 = 4.57 cm s cg = 4.57 cos 35.5 = 3.72 cm z θ cg a x 6 cm θ s cg 7 cm 10 cm P cg = ρ G s cg P cg = 1010 kg/m 3 *12.05 m/s 2 * m P cg = N/m 2 F = P cg A = N/m 2 *0.0914*0.06m 2 a F = 2.48 N II-29

30 What is the direction? Horizontal, perpendicular to the wall; i.e. Pressure always acts normal to a surface. Q: How would you find the force on the right wall? II-30

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