Fluid Mechanics PDF

CHAPTER 4 Fluid Mechanics Dr. Ranjini Rajammal  Chapter Outline: Pressure and its application Archimedes’s Principle Pascal’s Principle Bernoulli’s Principle States of Matter Before we begin to understand the nature of a fluid we must understand the basic nature of all states o...

CHAPTER 4 Fluid Mechanics Dr. Ranjini Rajammal  Chapter Outline: Pressure and its application Archimedes’s Principle Pascal’s Principle Bernoulli’s Principle States of Matter Before we begin to understand the nature of a fluid we must understand the basic nature of all states of matter: The 3 primary states of matter: ① solid - definite shape and volume. ② liquid -takes the shape of its container yet has a definite volume. ③ gas - takes the shape and volume of its container. What is a Fluid? By definition, a fluid is any material that is unable to withstand a shear stress. Unlike an elastic solid which responds to a shear stress with a recoverable deformation, a fluid responds with an irrecoverable flow. Examples of fluids include gases and liquids. Why fluids are useful in physics? Typically, liquids are considered to be incompressible. That is once you place a liquid in a sealed container you can DO WORK on the FLUID as if it were a rigid object. The PRESSURE you apply is transmitted throughout the liquid and over the entire length of the fluid itself. Fluid Properties - Density The mass density  of any substance is the mass of the substance divided by the volume it occupies: m unit: kg/m3  mass can be found as m = V V and weight as mg = Vg The density of water is 1000kg/m3 Fluid Properties - Pressure Any fluid can exert a force perpendicular to its surface on the walls of its container. This force is described in terms of the pressure it exerts, or force per unit area: Other units of pressure: N/m2 = Pascal psi (pounds per square inch) atmosphere The same force applied over a smaller area results in greater pressure – think of poking a balloon with your finger and then with a needle. Problem Example A waterbed is 2.0 m on a side and 30.0 cm deep. a) Find its weight if the density of water is 1000 kg/m3. b) Find the pressure the waterbed exerts on the floor. Assume that the entire lower surface of the bed makes contact with the floor. Atmospheric Pressure Atmospheric pressure is due to the weight of the atmosphere above us. N 1 2 = 1 Pascal (Pa) m  Fluid Properties - Hydrostatic Pressure If a Fluid (such as a liquid) is at REST, the pressure it exerts is called HYDROSTATIC PRESSURE Two important points A fluid will exert a pressure in all directions A fluid will exert a pressure perpendicular to any surface it compacts Notice that the arrows on TOP of the objects are smaller than at the BOTTOM. This is because pressure is greatly affected by the DEPTH of the object. Since the bottom of each object is deeper than the top the pressure is greater at the bottom. Pressure vs. Depth Fatm Suppose we had an object submerged in water with the top part touching the atmosphere. If we were to draw an FBD for this object we would have three forces 1. The weight of the object Fwater mg 2. The force of the atmosphere pressing down 3. The force of the water pressing up Fwater= Fatm + mg Pressure vs. Depth Fatm But now suppose the “object” is just a cylindrical container holding water. This water also has weight and the same equation applies. Fwater mg Finally, make the container imaginary. The water is still there and its weight. This weight is that affects the pressure below it. Fwater= Fatm + mg Pressure vs. Depth But pressure is force per unit area. So express the force of the fluids in terms of pressure. F P Fwater  Fatm  mg A PA  Po A  mg Note: The initial pressure in this m   m  V case is atmospheric V pressure, which is a PA  Po A  Vg CONSTANT. V  Ah Po = 1 x 105 N/m2 PA  Po A  Ahg P  Po  gh A closer look at Pressure vs. Depth P  Po  gh Depth below surface ABSOLUTE PRESSURE Initial Pressure – May or The weight of the May NOT be atmospheric volume of fluid. pressure P  gh Gauge Pressure = CHANGE in pressure or the DIFFERENCE in the initial and absolute pressure Example Calculate the absolute pressure at an ocean depth of 1000 m. Assume that the density of water is 1000 kg/m3 and that Po= 1.01 x 105 Pa (N/m2).. Pascal’s Principle “states that when pressure is applied to an enclosed fluid, the pressure is transmitted equally to every point in the fluid and to every point on the walls of the container.” If you take a liquid and place it in a system that is CLOSED (plumbing or a car’s brake line), the PRESSURE is the same everywhere in the system. The fact that the pressure is the same in a closed system has MANY applications. This is called PASCAL’S PRINCIPLE Pascal’s Principle Another Example - Brakes In the case of a car's brake pads, a small initial force applied by you on the brake pedal is transferred via a brake line, which has a small cylindrical area. The brake fluid then enters a chamber with more AREA allowing a LARGE FORCE to be applied on the brake shoes, which in turn slow the car down. P1  P2 Fbrake pedal Fbrake pad / shoe  Abrake pedal Abrake pad / shoe Another Example – Hydraulic Lift In a car lift in a service station, compressed air exerts a force on a small piston of circular cross section having a radius of 5.0 cm. This pressure is transmitted by an incompressible liquid to a second piston of radius 15 cm. What force must the compressed air exert in order to lift a car weighing 13300 N? P1  P2 F1 F2  A1 A2 A1 F1  F2 A2  0.052 F1  13300  0.152 F1  1478 N Buoyancy When an object is immersed in a fluid, such as a liquid, it is buoyed UPWARD by a force called the BUOYANT FORCE. When the object is placed in fluid is DISPLACES a certain amount of fluid. If the object is completely submerged, the VOLUME of the OBJECT is EQUAL to the VOLUME of FLUID it displaces. Archimedes' Principle " An object is buoyed up by a force equal to the weight of the fluid displaced." In the figure, we see that the difference between the weight in AIR and the weight in WATER is 3 lbs. This is the buoyant force that acts upward to cancel out part of the force. If you were to weight the water displaced it also would weigh 3 lbs. Archimedes' Principle Buoyant Force Weight (Force) of displaced fluid FB = (mg)Fluid m = r V FB = ( r Vg)Fluid Weight (Force) in terms of density of the fluid. Vobject = VFluid Problem Example A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weight in air to be 7.84 N. She then weighs the crown while it is immersed in water (density of water is 1000 kg/m3) and now the scale reads 6.86 N. Is the crown made of pure gold if the density of gold is 19.3 x 103 kg/m3? Fobject (air ) - Fobject (water ) = Fbuoyant 7.84 - 6.86 = FB = 0.98 N NO! This is NOT gold as 8000