Fluid Mechanics (Phy101) - Mansoura University - PDF

Chapter 3 Physics Department Faculty of Science Mansoura University Dr. Ehab Salim First year student Phy101 Matter is usually said to exist in three phases: Solid Liquid Gas...

Chapter 3 Physics Department Faculty of Science Mansoura University Dr. Ehab Salim First year student Phy101 Matter is usually said to exist in three phases: Solid Liquid Gas Have the characteristic of being able to flow are hard bodies that A liquid flows A gas also flows resist and takes the into a container shape of and spreads out deformation whatever until it occupies container in the entire which it is volume of the placed containr. Fluid is defined as any substance that can flow, flow, and hence liquids and gases are both considered to be fluids. fluids.  Properties of fluid 1. Density is defined as the ratio of the mass of a fluid to its volume. 2. Specific volume the volume of a fluid occupied by a unit mass or volume per unit mass of a fluid. 3. Viscosity of liquid is defined as the property of a fluid which offers resistance to the movement of one layer of fluid over another adjacent layer of fluid. 4. Fluid is treated as a continuous isotropic substance ‫ﻻﺗﻌﺘﻤﺪ ﺧﻮﺍﺻﻬﺎ ﻋﻠﻰ ﺍﻻﺗﺠﺎﻩ‬ 4. Fluids do not sustain shearing stresses or tensile stresses. ‫ﻻ ﺗﺤﺎﻓﻆ ﺍﻟﺴﻮﺍﺋﻞ ﻋﻠﻰ ﺇﺟﻬﺎﺩ ﺍﻟﻘﺺ ﺃﻭ ﺇﺟﻬﺎﺩ ﺍﻟﺸﺪ‬  Types of Fluids Ideal fluid is a fluid that has several properties including the fact that it is : ‫ﺍﻟﻤﺎﺋﻊ ﺍﻟﻤﺜﺎﻟﻰ ﻫﻮ‬ 1. a non-existent ‫ﻣﺎﺋﻊ ﺍﻓﺘﺮﺍﺿﻰ ﻏﻴﺮ‬ ‫ﻣﻮﺟﻮﺩ ﻳﻔﺘﺮﺽ ﺍﻧﻪ‬ 2. Incompressible – the density is constant ‫ﻟﻴﺲ ﻟﻪ ﻟﺰﻭﺟﺔ‬ 3. Irrotational – the flow is smooth, no turbulence ‫ﻭﻏﻴﺮ ﻗﺎﺑﻞ‬ ‫ﻟﻼﻧﻀﻐﺎﻁ‬ 4. Nonviscous –(Inviscid) fluid has no internal friction ( η = 0) Real fluid Fluid that have viscosity (η (η > 0) and their motion known as viscous flow. All the fluids in actual practice are real fluids. A fluid with compressibility but without viscosity is occasionally discriminated and called a perfect fluid Pressure is a scalar quantity because it is proportional to the magnitude of the force. The units of pressure: Example 3.1 The mattress of a water bed is 2.00 m long by 2.00 m wide and 30.0 cm deep. (A) Find the weight of the water in the mattress. (B) Find the pressure exerted by the water bed on the floor when the bed rests in its normal position. Assume the entire lower surface of the bed contacts the floor. Absolute pressure and gauge pressure There are two methods used to express the pressure: Absolute pressure : uses absolute zero as its zero point (the perfect vacuum). gauge pressure uses atmospheric pressure as its zero point. Due to varying atmospheric pressure, gauge pressure measurement is not precise, while absolute pressure is always definite gauge pressure = absolute pressure- atmospheric pressure, gauge pressure = P - P0 In gauge pressure, a pressure under 1 atmospheric pressure is expressed as a negative pressure.  Variation of Pressure with Depth Select a parcel of the liquid (uniform density ρ) contained within an imaginary block of cross-sectional area A at depth h. The pressure exerted by the liquid on the bottom face of the parcel is P = PA, and the pressure on the top face is P0 = P0 A. The mass of liquid in the parcel is M = ρ V = ρ A h; therefore; Mg = ρ A h g. Because the parcel is at rest, so that the net force acting on it must be zero. The pressure is the same at all points having the same depth, independent of the shape of the container. 1 pascal = 1 newton per square metre, 1 N/m2  Measurement of pressure Aim: barometer used to measure atmospheric pressure. Construction: A long tube closed at one end is filled with mercury and then inverted into a dish of mercury. The closed end of the tube is nearly a vacuum, Idea: pressure at (A), due to the column of mercury equal pressure at (B), due to the atmosphere. Therefore, P0 = ρHg g h, where ρHg is the density of the mercury and h is the height of the mercury column. Conclusion: As atmospheric pressure varies, the height of the mercury column varies, so the height can be calibrated to measure atmospheric pressure. P0 = 1 atm = 1.013 x105 Pa: one atmosphere of pressure is defined to be the pressure equivalent of a column of mercury that is exactly 0.760 m in height at 0oC. 2: Manometer Aim: A device which measures the fluid pressure by the height of a liquid column. Construction: U-tube filled with a liquid with density ρ, one branch connected with a bulb enclosed gas of unknown pressure p and the other branch exposed to the atmospheric pressure po Idea: Pressure of the gas p can be obtained by measuring the height of liquid h coming upwards into a manometer as shown in the figure. P0 is the atmospheric pressure, ρ is the density; Note: when the pressure p is large, this is inconvenient because h is too high. So, a U-tube manometer, containing a high-density liquid such as mercury is used. Pascal’s law A change in the pressure applied to a fluid is transmitted undiminished to every point of the fluid and to the walls of the container. the hydraulic press is an application of Pascal’s law. A force of magnitude F1 is applied to a small piston of surface area A1. The pressure is transmitted through an incompressible liquid to a larger piston of surface area A2. P = F1 / A1 = F2 / A2 , we can say that force F2 greater than F1 by A2 / A1  Without any change in the volume of the enclosed liquid, as the piston moves downward through a displacement ∆x1 the liquid pushed up the right piston through a displacement ∆x2. That is, A1 ∆x1 = A2 ∆x2; therefore, P = F2 / F1 = A2 / A1 = ∆x1 /∆x2 F1∆x1 = F2∆x2 ; is the work done by the force on its respective piston. Therefore, the work done by F1 equals the work done by F2, energy conservation. Example 3.2 In a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section of radius 5.00 cm. This pressure is transmitted by a liquid to a piston that has a radius of 15.0 cm. (A) What force must the compressed air exert to lift a car weighing 13300 N? (B) What air pressure produces this force? Solution: This pressure is approximately twice atmospheric pressure. 3.4. Buoyant Forces and Archimedes’s Principle When an object is placed in a fluid, the fluid exerts an upward force comes from the pressure exerted on the object by the fluid; called the buoyant force. Archimedes’ principle the magnitude of the buoyant force on an object equals the weight of the fluid displaced by the object.  the pressure Pbot at the bottom of the cube is greater than the pressure Ptop at the top by an amount ρfluid g h, where h is the height of the cube and ρfluid is the density of the fluid. The upward force = Pbot A, where A is the area of the bottom face. The pressure at the downward force = Ptop A. The resultant of these two forces is the buoyant force  Vdisp = Ah is the volume of the fluid displaced by the cube. ρfluid V disp is equal to the mass of fluid displaced by the object, B = Mg, where Mg is the weight of the fluid displaced by the cube. Case 1: Totally Submerged Object If an object (Mass M, Volume Vobj and density ρobj) totally submerged in fluid; the volume Vdisp of the displaced fluid = the volume Vobj of the object; so, the upward buoyant force is B = ρfluid g Vobj,. If the object has a mass M and density ρobj, its weight is equal to Fg = Mg = ρobj gVobj, and the net force on the object is B – Fg = Mg = (ρfluid - ρobj) g Vobj. if ρobj > ρfluid, Fg > B and the object sinks (Fig. b) If ρobj < ρfluid, Fg < B and the object accelerates upward(Fig. a). If the density of the object ρobj = ρfluid, Fg = B, and the object remains in equilbrium. Example 3.3 Wood floats. A block of oak wood 5.00 cm high, 5.00 cm wide, and 10.0 cm long is placed into a tub of water. The density of the wood is 7.20 × 102 kg/m3. How far will the block of wood sink before it floats? Solution: Wwood = (mwood) (g) = (ρwood ) (Vwood ) (g ) = (7.2×102 kg/m3) (0.05x0.05x0.1 m3) (9.80 m/s2) = 1.76 N The buoyant force = the weight of the water displaced Wwater Thus, the block sinks to a depth of 3.59 cm. At this point the buoyant force becomes equal to the weight of the wooden block and the wooden block floats. Example 3.4 Iron sinks. Repeat example 3.3 for a block of iron of the same dimensions. (The density of iron is 7860 kg/m3.) Solution: The weight of the iron block is given by But the block is only 10 cm high. Hence, the buoyant force is not great enough to lift an iron block of this size, and the iron block sinks to the bottom. Another way to look at this problem calculate the buoyant force on this piece of iron. The buoyant force on the iron, the upward buoyant force is B = ρfluid g Vobj, Thus, the net force upward on a block of iron of this size is 2.45 N. But the block weighs 19.3 N. Hence, Wiron > B  Example 3.5 An iron boats. The iron block of example 3.4 is cut into 16 slices, each 5.00 cm by 10.0 cm by 5/16 cm. They are now welded together to form a box 20.0 cm wide by 10.0 cm long by 10.0 cm high. Will this iron body now float, or will it sink? (The density of iron is 7860 kg/m3.) Solution: the weight of the box is also equal to 19.3 N. Because the iron box is 10 cm high, it sinks to a depth of 9. 48 cm and it then floats.