ME 305 Fluid Mechanics I Chapter 2 Fluid Statics These presentations are prepared by Dr. Cüneyt Sert Department of Mechanical Engineering Middle East Technical University nkara, Turkey csert@metu.edu.tr You can get the most recent version of this document from Dr. Sert s web site. Please ask for permission before using them to teach. You are NOT allowed to modify them. 2-1
Fluid Statics Fluids can NOT remain at rest under the presence of shear stress. In other words, fluids at rest can NOT support any shear force. For static fluids we can only talk about normal stress, which is equal to pressure. Determining the pressure variation in a static fluid is the main task here. From high school you already know the basic principle: Pressure increases linearly with depth as you go down in a constant density static fluid. We ll elaborate on this principle. www.cromwell.org.nz/aerial_photos/pages/clyde Dam_jpg.htm www.youtube.com/watch?v=tnm0smha0-i 2-2
Pressure For a fluid at rest, pressure is defined as the normal force acting per unit area exerted on a surface inside in the fluid. It is due to the bombardment of the surface with fluid molecules. 1 Pa = 1 N/m 2 1 bar = 10 5 Pa = 100 kpa 1 atm = 101.325 kpa = 1.01325 bars = 14.7 psi Pound force per square inch tmospheric pressure that we feel is due to the air column sitting on top of us. It is quite high (~10 5 Newton per m 2 or ~10 tones per m 2 ). Famous Magdeburg experiment that demonstrates the power of the atmospheric pressure en.vikipedi.pw/wiki/magdeburg_hemispheres 2-3
Direction Dependency of Pressure in a Static Fluid Exercise : For a fluid at rest, pressure at a point is independent of direction, which is known as Pascal s Law. Find and study its derivation in your fluid mechanics book. This little triangle shows the free surface, here the air - water interface p 3 Dam Point p 1 p 2 p 1 = p 2 = p 3 = In a moving fluid there will be multiple pressure definitions, such as static, dynamic and stagnation. The concept of pressure in a moving fluid is a bit tricky, but we ll come to that later. 2-4
Pressure Variation in a Static Fluid s we dive deep into the sea we feel more pressure in our ears. When we travel to high altitudes atmospheric pressure decreases. Following fluid element in a static fluid is not moving because no net force acts on it. Dam Weight For static fluids Forces = 0 Weight +??? = 0 Net pressure force from the surrounding fluid 2-5
Pressure Variation in a Static Fluid Exercise : In a static fluid with weight being the only body force, derive the following hydrostatic force balance. p + ρ Ԧg = 0 Net pressure force per unit volume Weight per unit volume g = 9.81 It is common to select Ԧg in the negative z direction, i.e. Ԧg = g k In this case the above equation reduces to dp dz + ρg = 0 z dp/dz Pressure changes only in the z direction, not x or y. Remember that ρg = γ (specific weight) g ρg 2-6
Pressure Variation in a Static Fluid (cont d) dp dz = ρg p z =? To evaluate p(z), i.e. to perform the integration, we need to know how ρ and g change with z. ρ z =? g z =? Consider g to be independent of z. t sea level it is 9.807 m/s 2 and at 14 km altitude it is 9.764 m/s 2 (less than 0.5 % change). Depending on the problem, ρ can be constant or can change with z. Let s start with the common and simple case of constant density fluid. If ρg is constant dp dz = ρg p = ρgz + C p + ρgz = C 2-7
Pressure Variation in a Static Fluid (cont d) To get rid of the integration constant we need to have a reference point (elevation) with known pressure. Usually the reference point is a free surface. p =? z p atm H p + ρgz = C t z = H we know that p = p atm C = p atm + ρgh p = p atm + ρg(h z) p atm p =? z h = H z It is easier to use depth, h, instead of z p = p atm + ρgh s we go down in a constant density fluid pressure increases linearly with depth. 2-8
Pressure Variation in a Static Fluid (cont d) Example: How deep in the sea should you dive to feel twice the atmospheric pressure? ρ sea gh = 1 atm = 101325 Pa h = 101325 (1020)(9.81) 10 m Edge of the atmosphere (0 atm) ~100 km Earth s atmosphere is about 100 km thick. This 100 km air column creates 1 atm pressure (Note that the density of air varies considerably over this 100 km s). Sea surface (1 atm) Diver (2 atm) h =? h = ~10 m sea water creates 1 atm pressure. 2-9
If ρ constant: Pressure Variation with Variable Density Exercise : ccording to U.S. Standard atmosphere model (see the next slide), temperature within the first 11 km of the atmosphere drops linearly as T = T 0 + Bz, where the temperature at the ground is T 0 = 288 K (15 ) and the temperature lapse rate is B = 0.0065 K/m. a) Considering that the pressure at the ground level (z = 0) is equal to 101325 Pa and treating air as an ideal gas, obtain the pressure variation as a function of z within the first 11 km of the atmosphere. b) Use the equation derived in part (a) to calculate the atmospheric pressure at the peak of Mount Everest (z = 8848 m). Does your result match with the second figure of the next slide? c) How much percent error would there be in the previous calculation if we assume atmospheric air to be isothermal at - ground temperature of 15 o C - an average temperature of -14 o C? We need a ρ z relation to integrate dp dz = ρg 2-10
Troposphere (11 km) ltitude [km] ltitude [km] U.S. Standard tmosphere Model 60 60 50 50 40 40 30 30 20 20.1 km 20 10 11 km 10 0-56.5 o C -60-40 -20 0 15 Temperature [ o C] 0 0 40 80 120 Pressure [kpa] 2-11
Pressure Variation with Variable Density (cont d) Exercise : In 1960 Trieste sea vessel carried two oceanographers to the deepest point in Earth s oceans, Challenger Deep in the Mariana Trench (10,916 m). Designers of Trieste needed to know the pressure at this depth. They performed two calculations. - First they considered the seawater to be incompressible with a density equal to the value at the ocean surface, which is 1020 kg/m 3. - Then they considered the compressibility of seawater using E v = 2.07 10 9 Pa. Taking the atmospheric pressure at the ocean surface to be 101.3 kpa, calculate the percent error they made in the calculation of pressure at h = 10,916 m when they considered seawater to be incompressible. lso read about James Cameron s 2012 dive at the Challenger Deep. www.deepseachallenge.com http://en.vikipedi.pw/wiki/bathyscaphe_trieste 2-12
bsolute vs Gage Pressure bsolute pressure is measured with respect to complete vacuum, where p = 0. Some pressure measuring devices measure pressure with respect to the ambient pressure, which is usually the atmospheric pressure. This is called gage pressure. Gage pressure is commonly used when we want to get rid of the ambient pressure effect. Example: When your car s manual says that you need to inflate the tires to 30 psi, it is actually trying to say 30 psi gage (psi g). If the local atmospheric pressure is 95 kpa, what is the absolute air pressure inside the tires? bsolute pressure in the tire 101.3 kpa = 30 psi 14.7 psi Gage pressure + 95 kpa mbient pressure = 301 kpa p ambient = 95 kpa Tire pressure is 301 kpa absolute Pressure measuring device measures gage pressure as 30 psi g = 206 kpa g 2-13
Pressure Measuring Devices - Mercury Barometer In 1643 Toricelli demonstrated that atmospheric pressure can be measured using a mercury barometer. Greek word baros means weight. p atm =? B Mercury vapor p B = p mercury vapor 0 C ρ m h Mercury p = p C = p atm p = p B + ρ m gh p atm = ρ m gh For p atm = 101,325 Pa and ρ m = 13595 kg/m 3 mercury rise will be h = 0.76 m. mmhg is another unit used for pressure. It gives the pressure difference across a 1 mm mercury column. 1 mm mercury column Δp = 1 mmhg ρ m gh = 13595 kg m 3 9.81 m s 2 10 3 m = 133.4 Pa 1 atm = 101,325 Pa = 760 mmhg 2-14
Pressure Measuring Devices - neroid Barometer neroid means without fluid. neroid barometer measures absolute pressure. It has a vacuumed chamber with an elastic surface. When pressure is imposed on this surface, it deflects inward. Due to this deflection the needle rotates and shows the pressure. fter proper calibration, a barometer can also be used as an altimeter, to measure altitude. Below a certain altitude, atmospheric pressure decreases 1 millibar for each 8 m of ascent. To read more about the aneroid barometer http://www.bom.gov.au/info/aneroid/aneroid.shtml www.stuffintheair.com www.free-online-private-pilot-ground-school.com 2-15
Pressure Measuring Devices - Bourdon Gage Bourdon gage measures the gage pressure. bent elliptical tube is open and fixed at one end, and closed but free to move at the other end. When pressure is applied to its open end, the tube deflects and the pointer connected to its free end shows the gage pressure (pressure with respect to the ambient pressure outside of the tube). When not pressurized, the pointer shows zero. It can be used for the measurement of liquid and gas pressures upto 100s of MPa. Front www.discoverarmfield.co.uk Back 2-16
Pressure transducers generate an electrical signal as a function of the pressure they are exposed to. They work on many different technologies, such as Piezoresistive Piezoelectric Capacitive Electromagnetic Optical Thermal etc. Pressure Measuring Devices - Pressure Transducer www.lefoo.com They can be used to measure rapid pressure fluctuations in time. Differential types can directly measure pressure differences. 2-17
Manometers Manometers are used to measure pressure differences using liquid columns in tubes. Working principles are any two points at the same elevation in a continuous liquid have the same pressure. pressure increases as ρgh as one goes down in a liquid column. p atm p atm ρ h ρ B h C p = p atm + ρgh p + ρgh = p B = p C = p atm 2-18
Manometers (cont d) h 1 ρ 1 C h 3 h 2 D ρ 3 B ρ h θ B ρ 2 p + ρ 1 gh 1 = p C = p D = p B + ρ 3 gh 3 + ρ 2 gh 2 p B + ρgh sin θ = p Exercise : What s the advantage of using an inclined manometer instead of a vertical one? 2-19
Manometers (cont d) Exercise : The volume rate of flow, Q, through a pipe can be determined by means of a flow nozzle located in the pipe as illustrated below. The nozzle creates a pressure drop, p p B, along the pipe that is related to the flow rate with the equation Q = K p p B, where K is a constant depending on the pipe and nozzle size. The pressure drop is frequently measured with a differential U-tube manometer. a) Determine an equation for p p B in terms of the specific weight of the flowing fluid, the specific weight of the gage fluid, and the various heights indicated. b) For γ 1 = 9.8 kn/m 3, γ 2 = 15.6 kn/m 3, h 1 = 1 m and h 2 = 0.5 m, calculate p p B. Reference: Munson s book 2-20
Hydrostatic Forces cting on Submerged Surfaces Pressure force always acts perpendicular to a surface in a compressive manner. Exercise : Show the variation of pressure force acting on the walls of the following containers. Pay attention to both magnitude and direction. p atm p atm The task is to find the resultant pressure force acting on a submerged surface and point of application of the resultant pressure force. Different techniques can be used such as : 1. Direct Integration Method 2. Pressure Prism Method 3. Force Component Method 2-21
Direct Integration Method This method can be used to calculate the resultant pressure force on any surface (planar or curved). We simply integrate the pressure variation on a surface to get the resultant pressure force F r. Consider the top surface of the following submerged plate lying on the xy plane. p 0 F r =? df h d Pressure distribution y F r = න df = න p d y = න (p 0 + ρgh)d d x 2-22
Direct Integration Method (cont d) F r acts through a point called center of pressure (CP). Coordinates of CP are calculated by equating the moment created by the distributed pressure force along an axis (x or y) to the moment created by F r along the same axis. p 0 F r df F r = න p d x CP F r = න x p d y CP x y y CP y CP F r = න y p d y x CP CP d x Moments created by the resultant pressure force Moments created by the distributed pressure force 2-23
Direct Integration Method (cont d) To use this method correctly you need to - select the coordinate directions and the origin carefully. - select and formulate d carefully. - relate h to coordinate axes carefully. - put integration limits carefully. h = y sin β p 0 β F r = න p d L = න y=0 p 0 + ρgy sin(β) Wdy x CP = W/2 (No need to calculate) y CP = 1 න yp d F r = න (p 0 + ρgh) d = 1 F r න L y=0 y p 0 + ρgy sin(β) Wdy y df y L dy W d = Wdy x 2-24
Direct Integration Method (cont d) There may be cases where analytical integration may seem difficult. You can - use an integration table http://integral-table.com - perform numerical integration calculate the area under the curve roughly https://www.integral-calculator.com - graphical integration plot the curve on a graph paper and count the squares blogs.mathworks.com/cleve 2-25
Exercises for Direct Integration Method Exercise : Rectangular gate of size 2 m 1 m is hinged along B and held by the horizontal force F applied at. Calculate the force F required to keep the gate closed. 4 m 1 m B F =? Side view of the gate 2 m Exercise : Solve the previous problem by considering a triangular gate as shown below. Exercise : Solve the previous problem by considering a semicircular gate as shown below. B F 2 m 1 m B F 2 m 1 m 2-26
Pressure Prism (PP) Method This is an alternative (and sometimes easier) technique to calculate hydrostatic forces acting on submerged planar surfaces (not used for curved surfaces). Consider an imaginary prism with the planar surface of interest being its base and the amount of pressure acting on the surface being its height. p 0 p 0 + ρgh h B df h h Pressure prism y p 0 + ρgh B d p d d PP = p d (Volume of the differential PP) = df 2-27
Pressure Prism (PP) Method df = pd = d PP F r = න df = න d PP = PP Resultant pressure force is equal to the volume of the PP. F r = PP F r passes through the centroid of the pressure prism. 2-28
Pressure Prism (PP) Method (cont d) For ease of calculation, you can divide the PP into sub-prisms. B Pressure prism B = + B F r = PP =? B L =? L F r1 = 1 PP1 B F r2 = PP2 L 2 F r = F r1 + F r2, F r L = F r1 L 1 + F r2 L 2 2-29
Pressure Prism (PP) Method (cont d) Exercise : Solve the problems of slide 2-26 using the PP method. Exercise : The wall shown has a width of 4 m. Determine the total force on the wall due to oil pressure. lso determine the location of the center of pressure from point along the wall. Density of oil is 860 kg/m 3. 1.4 m Oil 45 o Exercise : (Fox s book) s water rises on the left side of the L-shaped gate, it will open automatically. Neglecting the weight of the gate, at what height h above the hinge will this occur? How will the result change (increase, decrease or no change) if the mass of the gate is considered? Water h Hinge Gate 1.5 m 2-30
Pressure Prism (PP) Method (cont d) PP method is easy to use if volume of the pressure prism ( PP ) is easy to calculate. If the surface shape is complicated such that evaluation of PP or its centroid requires integration, then the PP method has no advantage over the direct integration method. Exercise: (Munson s book) The inclined face D of the shown tank is a plane surface containing a gate BC, which is hinged along line BC. If the tank contains water, determine the magnitude of the force that the water exerts on the gate and the moment that it will create around the hinge. Water β B,C C D Hinge y B = 1.2 m BD = 0.6 m B x y = 4 x 2 β = 30 2-31
Pressure Prism (PP) Method (cont d) Exercise : (Fox s book) Gates in the Poe Lock at Michigan, US close a channel, which is 34 m wide and 10 m deep. The geometry of one pair of gates is shown below. Each gate is hinged at the channel wall. When closed, edges of the gates are forced together by the water pressure. Evaluate a) the force exerted by water on gate. b) the reaction forces on the hinges. Hinge Water Gate 15 o 34 m Top View http://www.enerpac.com/infrastructure/fixing-pair-of-loose-hinges 2-32
Forces cting on Submerged Curved Surfaces Force Component Method Consider the curved surface BCD shown below. Direct integration method can be used to calculate F r and its point of application. Or the force component method can be used to calculate F ry and F rz separately. z z B E F r C y, B F r F r y F r z x F D C, D y 2-33
Force Component Method (cont d) z Horizontal component F r y is equal to the force acting on the projected planar surface BEF. B F r Exercise: Prove the above. x F E D C F r y =? y Because BEF is a planar surface, PP method can be used to calculate F r y and its point of action. z Shaded area is the projection of curved surface BCD on the xz plane. B E y F F r y x 2-34
Force Component Method (cont d) z B F r C F r z =? y F r z is the weight of the liquid that will fill the volume between the curved surface and the free surface, i.e. volume BCDEFGH shown below. Exercise: Prove the above. x D F r z acts through the centroid of this volume. F F r z G E H B C D 2-35
Force Component Method (cont d) z Exercise : Calculate the pressure force acting by the liquid acting on the curved surface of the shown quarter cylinder. a) Use direct integration method. b) Use force component method. R R w y x R Exercise : One puzzling detail about the force component method arises when the volume between the curved surface and the free surface is NOT completely filled with liquid. Do you think forces acting by the liquid on the following gates are the same? 2-36
Force Component Method (cont d) Exercise : The following Tainter gate is used to control water flow from a dam. The gate width is 35 m. Determine the magnitude and line of action of the force acting on the gate by the water. What s the advantage of using such a circular gate profile? Tainter gates 20 m 10 m http://www.discover-net.net/~dchs/history/gate_ani.gif http://www.ciltug.com 2-37
Buoyancy Force Consider a body that is fully submerged (could be floating on the surface too) in a static fluid. distributed pressure force acts all around the body. 3D body Using the force component method we can show that the net horizontal pressure force acting on the body is zero. Left and right parts have the same vertical projection. So the horizontal forces acting on them cancel out. 2-38
Buoyancy Force (cont d) But the net vertical pressure force is NOT zero. It is called the buoyancy force. p atm Net vertical force on d is Cylindrical volume d df 1 h 1 d h 2 df = df 2 df 1 = p 2 p 1 d = ρgh 2 ρgh 1 d = ρg h 2 h 1 d df 2 = ρgd Overall vertical force is obtained by integrating the above expression F vertical = F bouyancy = න body ρgd = ρg body Buoyancy force acting on the body is equal to the weight of the fluid displaced by submerging the body into the fluid. This is known as the rchimedes principle. 2-39
Buoyancy Force (cont d) Exercise: Use the force component method to get the following result. F bouyancy = ρg body Exercise: 170 kg granite rock (ρ = 2700 kg/m 3 ) is dropped into a lake. man dives in and tries to lift the rock. Determine how much force he needs to apply to lift it from the bottom of the lake. Do you think he can do it? Exercise: (Fox s book) The cross-sectional shape of a canoe is modeled by the curve y = ax 2, where a = 3.89 m 1, and the coordinates are in meters. ssume the width of the canoe is constant at w = 0.6 m over its entire length L = 5.25 m. Set up a general algebraic expression relating the total mass of the floating canoe and its contents to distance d. Calculate the maximum total mass allowable without swamping the canoe. y w d y = ax 2 x 2-40
Hydrometer Hydrometer uses the principle of buoyancy to measure the density of a liquid. First it is calibrated by dipping it into a liquid of known density, such as water. Marked for water Water mark h Stem Water (ρ w ) Weight = ρ w g 1 Volume inside water (known) nother liquid (ρ) Weight = ρg( 1 h) Cross sectional area of stem (known) ρ = ρ w 1 1 h Movie : Hydrometer We measure h and then calculate the unknown ρ. Stem may be marked so that we can directly read ρ. 2-41
Capillarity When a glass tube is immersed into a liquid, which wets the surface, such as water, adhesive forces between the glass and water exceed cohesive forces in water, and water rises (capillary rise) in the glass tube. This vertical rise continues until the surface tension forces are balanced with the weight of the water column in the tube. For a non-wetting fluid, such as mercury, the force balance results in a different configuration known as capillary drop. Glass tube water mercury 2-42
Capillarity (cont d) Due to surface tension, the meniscus (free surface inside the tube) will be curved and there will be a pressure difference between the two sides of it. p top > p bottom This pressure difference is balanced by the surface tension force. D σ q q p top σ σ p top σ p bot h p top p bottom Meniscus ρ 2-43
Capillarity (cont d) Writing a vertical force balance for the meniscus D σ q q σ Force per length σ(πd) cos(θ) = (p top p bottom ) πd2 4 Length p top p bot h p top Writing the manometer equation for the liquid between the meniscus and the free surface ρ p top = p bottom + ρgh σ p top σ (Note that the pressure change of ambient air over the distance h is negligibly small). p bottom Combining these two equations, capillary rise is given by h = 4 σ cos(θ) ρgd Movie : Capillary Rise 2-44