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Fluids At ordinary temperature, matter exists in one of three states (5 if you include plasma and Bose-Einstein condensate). Solid - has a shape and forms a surface Liquid - has no shape but forms a surface Gas - has no shape and forms no surface What do we mean by “fluids”? Fluids are “substances that flow”…. “substances that take the shape of the container” Atoms and molecules are free to move. No long range correlation between positions. 1x10^-9 K

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**Fluids What parameters do we use to describe fluids? Density units :**

kg/m3 = 10-3 g/cm3 If any substance is more dense than water, it will sink r(water) = x 103 kg/m = g/cm3 r(ice) = x 103 kg/m = g/cm3 r(air) = kg/m = 1.29 x 10-3 g/cm3 r(Hg) = 13.6 x103 kg/m = 13.6 g/cm3

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**Fluids What parameters do we use to describe fluids? Pressure**

units : 1 N/m2 = 1 Pa (Pascal) 1 bar = 105 Pa 1 mbar = 102 Pa 1 torr = Pa 1 atm = x105 Pa = 1013 mbar = 760 Torr = 14.7 lb/ in2 (=PSI) Any force exerted by a fluid is perpendicular to a surface of contact, and is proportional to the area of that surface. Force (a vector) in a fluid can be expressed in terms of hydrostatic pressure (a scalar) as: A n

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Practice An 80 kg metal cylinder, 2.0 m long and with each end having an area of 25 cm2. It stands vertically on one of the ends. What pressure does the cylinder exert on the floor?

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Solution An 80 kg metal cylinder, 2.0 m long and with each end having an area of 25 cm2. It stands vertically on one of the ends. What pressure does the cylinder exert on the floor?

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Practice A waterbed is 2m on a side and 30cm deep. Find a) Its weight. b) The pressure the waterbed exerts on the floor. Waterbed

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**A waterbed is 2m on a side and 30cm deep. Find a) Its weight. **

Solution A waterbed is 2m on a side and 30cm deep. Find a) Its weight. b) The pressure the waterbed exerts on the floor. Water density is kg/m3 Let’s rearrange for Mass, M Now, we write weight in terms of density , volume and g

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**The force is the weight of the bed.**

Solution A waterbed is 2m on a side and 30cm deep. Find a) Its weight. b) The pressure the waterbed exerts on the floor. The force is the weight of the bed. Surface area of contact between the floor and the bed. Weight is 1.18 x N

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Specific Gravity An old-fashioned but still very common way of expressing the density of a substance is to relate it to the density of water. The ratio of the density of any substance to the density of water is known as the specific gravity of the substance. If the specific gravity is less than 1, it will float in water, if it is greater than 1, it will sink.

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Act 3 A cork has a volume of 5 cm3 and weighs 0.03N. What is the specific gravity of cork?

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Act 4 Solution A cork has a volume of 5 cm3 and weighs 0.03N. What is the specific gravity of cork? First let’s find the mass Now for the density Finally the specific gravity

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**Pressure vs. Depth Incompressible Fluids (liquids)**

The Bulk Modulus of a substance measures the substance’s resistance to uniform compression. It is defined as the pressure increase needed to cause a relative decrease in volume. When the pressure is much less than the bulk modulus of the fluid, we treat the density as a constant independent of pressure: LIQUID: incompressible (density almost constant) GAS: compressible (density depends a lot on pressure) For an incompressible fluid, the density IS the same everywhere, but the pressure is NOT Bulk Modulus Change in pressure

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**Pressure vs. Depth For a fluid in an open container:**

pressure same at a given depth independent of the container fluid level is the same everywhere in a connected container (assuming no surface forces) Why is this so? Why, in equilibrium, does the pressure below the surface depend only on depth? Imagine a tube that would connect two regions at the same depth. If the pressures were different, fluid would flow in the tube! However, if fluid did flow, then the system was NOT in equilibrium, since no equilibrium system will spontaneously leave equilibrium. The horizontal forces acting at the left and right side of a cylinder at specific depth are PAL and PAR. Since the cylinder is in equilibrium PAL= PAR Therefore The pressure, P at either side is the same.

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ACT 5 I r1 r2 dI 1. What happens with two different fluids?? Consider a U tube containing liquids of density r1 and r2 as shown: Compare the densities of the liquids: A) r1 < r2 B) r1 = r2 C) r1 > r2

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**ACT 5 Solution I p C) r1 > r2**

At the depth of the interface, the pressures in each side must be equal. Since there’s more liquid above this depth on the left side, that liquid must be less dense! I r1 r2 dI p C) r1 > r2

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**Pressure Measurements: Barometer**

Invented by Torricelli A long closed tube is filled with mercury and inverted in a dish of mercury The closed end is nearly a vacuum Measures atmospheric pressure as One 1 atm = m (of Hg)

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**Pressure vs. Depth Incompressible Fluids (liquids)**

Due to gravity, the pressure depends on depth in a fluid Consider an imaginary fluid volume (a cube, each face having area A) The sum of all the forces on this volume must be ZERO as it is in equilibrium. There are three vertical forces: The weight (mg) The upward force from the pressure on the bottom surface (F2) The downward force from the pressure on the top surface (F1) Since the sum of these forces is ZERO, we have: F2-F1=mg Same fluid)

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**Pressure vs. Depth Incompressible Fluids (liquids)**

Now since Force=pressure× Area Same fluid Now since Mass=Density× Volume Volume is area x height The pressure at the bottom of a vertical column is equal to the pressure at the top of the column (Po , Usually atmospheric) , plus the pressure due to all the water in the column of height , h. (density x gravity x height).

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GAUGE Pressure One must be careful when we are talking about pressure. Do we want to include atmospheric pressure in our calculations? If we want the pressure over and above atmospheric pressure, we have P0=0. This is called the Gauge Pressure. If we need to include atmospheric pressure we use: P0=1x105N/m2 = 100 kPa Most pressure gauges register the pressure over and above atmospheric pressure. For example a tire gauge registers 220 kPa, the actual pressure within the tire is 220 kPa +100 kPa = 320 kPa

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**Gauge Pressure in a Tank Filled with Gasoline and Water**

What is the pressure at point A? At point B? At point A: 10m 3 m At point B:

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**Pascal’s Principle So far we have discovered (using Newton’s Laws):**

Pressure depends on depth: DP = r g Dy Pascal’s Principle addresses how a change in pressure is transmitted through a fluid. Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel. Pascal’s Principle explains the working of hydraulic lifts i.e., the application of a small force at one place can result in the creation of a large force in another. Will this “hydraulic lever” violate conservation of energy? No

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**Free Energy? Where did this extra force come from?**

Pascal’s Principle Consider the system shown: A downward force F1 is applied to the piston of area A1. This force is transmitted through the liquid to create an upward force F2. Pascal’s Principle says that increased pressure from F1 , [P=(F1/A1)] , is transmitted throughout the liquid. Yes, but what gives? Free Energy? Where did this extra force come from? Since A2 is larger than A1, we have a force multiplier.

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Pascal’s Principle Let’s check the if the work done by F1 equal the work done by F2 Displaced volumes are the same, so… Therefore energy is conserved With a hydraulic lever, a given force applied over a given distance can be transformed to a greater force applied over a smaller distance.

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ACT In the hydraulic system shown below in the diagram, the 200kg cylinder has a cross sectional area of 100 cm2. The cylinder on the right has a cross sectional area of 10.0 cm2 Determine the amount of weight (F) you must apply on the right side of the system to hold the system in equilibrium (Stop the left side from moving down). If the left cylinder is pushed down 5.0cm, determine the distance F will move. Determine the mechanical advantage of the system

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**Apply Pascal’s Principle for F1**

ACT Solutions In the hydraulic system shown below in the diagram, the 200kg cylinder has a cross sectional area of 100 cm2. The cylinder on the right has a cross sectional area of 10.0 cm2 Determine the weight (F) required to hold the system in equilibrium If the left cylinder is pushed down 5.0cm, determine the distance the right cylinder will move. Determine the mechanical advantage of the system Apply Pascal’s Principle for F1

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ACT 7 Solutions In the hydraulic system shown below in the diagram, the 200kg cylinder has a cross sectional area of 100 cm2. The cylinder on the right has a cross sectional area of 10.0 cm2 Determine the weight (F) required to hold the system in equilibrium If the left cylinder is pushed down 5.0cm, determine the distance the right cylinder will move.. Determine the mechanical advantage of the system The volume of fluid displaced by the 100 cm2 cylinder equals the change in the volume of fluid in the right hand cylinder or

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ACT 7 Solutions In the hydraulic system shown below in the diagram, the 200kg cylinder has a cross sectional area of 100 cm2. The cylinder on the right has a cross sectional area of 10.0 cm2 Determine the weight (F) required to hold the system in equilibrium If the left cylinder is pushed down 5.0cm, determine the distance F will move. Determine the mechanical advantage of the system Determine the Ideal Mechanical Advantage (IMA) of the system. IMA=(output Force/ Input Force)

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**Archimedes’ Principle**

Suppose we weigh an object in air (1) and in water (2). How do these weights compare? W1 < W2 W1 = W2 W1 > W2 Why? Since the pressure at the bottom of the object is greater than that at the top of the object, the water exerts a net upward force, the buoyant force, on the object. W2? W1

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**Archimedes’ Principle**

W2? W1 The buoyant force is equal to the difference in the pressures times the area. Therefore, the buoyant force is equal to the weight of the fluid displaced.

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**Archimedes’ Principle**

A 60 kg rock lies at the bottom of a pool. Its volume is 3.0 x 104 cm3. What is its apparent weight? The buoyant force on the rock due to the water is equal to the weight of 3.0 x 104 cm3 = 3.0 x 10-2 m3 of water: The weight of the rock is: This is as if the rock had a mass of 31kg

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Sink or Float? The buoyant force is equal to the weight of the liquid that is displaced. If the buoyant force of a fully submerged object is larger than the weight of the object, it will float; otherwise it will sink. F mg B y We can calculate how much of a floating object will be submerged in the liquid: Object is in equilibrium

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**Sink of Float? Object is in equilibrium F mg y B**

The Tip of The Iceberg: What fraction of an iceberg is submerged? If the Density of ice is 917 kg/m3 and the density of water is 1024 kg/m3

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Archimedes Archimedes is said to have discovered his principle in his bath while thinking how he might determine whether the king’s new crown was pure gold or fake. Gold has a specific gravity of 19.3, somewhat higher than most metals, but a determination of specific gravity or density is not easy with a irregularly shaped object. Archimedes realised that if the crown is weighed in air and weighed in water, the density can de determined.

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Archimedes A 14.7 kg crown has an apparent weight of 13.4 kg when submerged in water. Is it gold? The apparent weight of the submerged crown, w`, equals its actual weight, w, minus the buoyant force, FB. We can find the specific gravity : This corresponds to a density of 11,300 kg/m3 (that of lead) Alternate solution

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**Let’s find the volume of the displaced water (this is also the volume of the crown)**

We can now determine the density of the crown

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Practice A lead weight is fastened to a large styrofoam block and the combination floats on water with the water level with the top of the Styrofoam block as shown. If you turn the Styrofoam + Pb upside down, what happens? styrofoam Pb A) It sinks C) B) D) styrofoam Pb

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**Solution A) It sinks C) B) D)**

styrofoam Pb A) It sinks C) B) D) styrofoam Pb If the object floats right-side up, then it also must float upside-down. It displaces the same amount of water in both cases However, when it is upside-down, the Pb displaces some water. Therefore the styrofoam must displace less water than it did when it was right-side up (when the Pb displaced no water).

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**Weight of ship = Buoyant force =Weight of displaced water**

Question Which weighs more: 1. A large bathtub filled to the brim with water. 2. A large bathtub filled to the brim with water with a battle- ship floating in it. 3. They will weigh the same. Tub of water Tub of water + ship Overflowed water Weight of ship = Buoyant force =Weight of displaced water 15

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**Act 8 Must be same! FB = Water g Vdisplaced FB = Weight of ice**

Suppose you float a large ice-cube in a glass of water, and after you place the ice in the glass the level of the water is at the very brim. When the ice melts, the level of the water in the glass will: 1. Go up, causing the water to spill out of the glass. 2. Go down. 3. Stay the same. Must be same! FB = Water g Vdisplaced FB = Weight of ice = ice g Vice W g Vmelted_ice

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Example Problems At what depth is the water pressure two atmospheres? (It is one atmosphere at the surface. 1.01105 Pa) What is the pressure at the bottom of the deepest oceanic trench (about 104 meters)? Solution: For d = 104 m: This assumes that water is incompressible. If water were compressible, would the pressure at the bottom of the ocean be greater or smaller than the result of this calculation?

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**We are displacing this much water**

Example Problems (2) Have you ever tried to submerge a beach ball (r = 50 cm) in a swimming pool? It’s difficult. How big a downward force must you exert to get it completely underwater? Solution: We are displacing this much water I’m ignoring the weight of the beach ball. The force is the weight of a 523 kg object.

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More Fun With Buoyancy Cup I Cup II Two cups are filled to the same level with water. One of the two cups has plastic balls floating in it. Which cup weighs more? Archimedes principle tells us that the cups weigh the same. Each plastic ball displaces an amount of water that is exactly equal to its own weight.

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Still More Fun! A plastic ball floats in a cup of water with half of its volume submerged. Oil (roil < rball <rwater) is slowly added to the container until it just covers the ball. Relative to the water level, the ball moves up. Why? water oil For oil to cover the ball, the ball must have “displaced” some water. Therefore, the buoyant force on the ball increases. Therefore, the ball moves up (relative to the water). Note that we assume the buoyant force of the air on the ball is negligible (it is!); the buoyant force of the oil is not.

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Archimedes Summary Archimedes Principle states that the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object. If the weight of the water displaced is less than the weight of the object (buoyant force is less than the weight of object) , the object will sink. Otherwise the object will float, with the weight of the water displaced equal to the weight of the object.

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Fluids in Motion Up to now we have described fluids in terms of their static properties: Density r Pressure P To describe fluid motion, we need something that can describe flow: Velocity v There are different kinds of fluid flow of varying complexity non-steady / steady compressible / incompressible rotational / irrotational viscous / ideal

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**Types of Fluid Flow Laminar flow**

Each particle of the fluid follows a smooth path The paths of the different particles never cross each other The path taken by the particles is called a streamline Turbulent flow An irregular flow characterized by small whirlpool like regions Turbulent flow occurs when the particles go above some critical speed

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**Types of Fluid Flow Laminar flow**

Each particle of the fluid follows a smooth path The paths of the different particles never cross each other The path taken by the particles is called a streamline Turbulent flow An irregular flow characterized by small whirlpool like regions Turbulent flow occurs when the particles go above some critical speed

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**Onset of Turbulent Flow**

The SeaWifS satellite image of a von Karman vortex around Guadalupe Island, August 20, 1999

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Ideal Fluids Fluid dynamics is very complicated in general (turbulence, vortices, etc.) Consider the simplest case first: the Ideal Fluid No “viscosity” - no flow resistance (no internal friction) Incompressible - density constant in space and time Simplest situation: consider ideal fluid moving with steady flow - velocity at each point in the flow is constant in time In this case, fluid moves on streamlines streamline

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**Ideal Fluids Streamlines do not meet or cross**

Velocity vector is tangent to streamline Volume of fluid follows a tube of flow bounded by streamlines Streamline density is proportional to velocity Flow obeys continuity equation Volume flow rate Q = A·v is constant along flow tube. Follows from mass conservation if flow is incompressible. A1v1 = A2v2

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**The Equation of Continuity**

Q: Have you ever used your thumb to control the water flowing from the end of a hose? A: When the end of a hose is partially closed off, thus reducing its cross-sectional area, the fluid velocity increases. This kind of fluid behavior is described by the equation of continuity.

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Continuity A four-lane highway merges down to a two-lane highway. The officer in the police car observes 8 cars passing every second, at 30 mph. How many cars does the officer on the motorcycle observe passing every second? A) B) C) 16 How fast must the cars in the two-lane section be going? A) 15 mph B) 30 mph C) 60 mph New slide Fall 2005 not in handouts. All cars stay on the road… Must pass motorcycle too Must go faster, else pile up!

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Exercise A housing contractor saves some money by reducing the size of a pipe from 1cm diameter to 1/2cm diameter at some point in your house. v1 v1/2 Assuming the water moving in the pipe is an ideal fluid, relative to its speed in the 1cm diameter pipe, how fast is the water going in the ½ cm pipe? (A) 2 v1 (B) 4 v1 (C) 1/2 v1 (D) 1/4 v1

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**Continuity of Fluid Flow**

Fluid In= Fluid Out Watch “plug” of fluid moving through the narrow part of the tube (A1) Time for “plug” to pass point t = x1 / v1 Mass of fluid in “plug” m1 = Vol1 = A1 x1 or m1 = A1v1 t Watch “plug” of fluid moving through the wide part of the tube (A2) Time for “plug” to pass point t = x2 / v2 Mass of fluid in “plug” m2 = Vol2 =A2 x2 or m2 = A2v2t Continuity Equation says m1 = m2 fluid isn’t appearing or disappearing A1 v1 = A2 v2

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Faucet Preflight A stream of water gets narrower as it falls from a faucet (try it & see). Explain this phenomenon using the equation of continuity A1 A2 V1 V2 “Since the area inside the faucet is large, the velocity of the water will be low. Due to the equation of continuity, the area of the water has to be lower due to the velocity increasing outside of the faucet. ” “Velocity and area are inversely proportional. Velocity increases, so area decreases.”

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Fluid Flow Concepts Continuity: A1 v1 = A2 v2 Mass flow rate: Av (kg/s) Volume flow rate: Av (m3/s) i.e., mass flow rate the same everywhere e.g., flow of river Since we have an ideal fluid (incompressible), then we can simply use: 24

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Pressure, Flow and Work Continuity Equation says fluid speeds up going to smaller opening, slows down going to larger opening But, if the fluid is moving faster, then it has more kinetic energy. Therefore something did work on the fluid to accelerate it (apply a force) Acceleration due to change in pressure. P1 > P2 Smaller tube has faster water and LOWER pressure Change in pressure does work! W = F1Dx1 –F2Dx2= P1A1Dx1 - P2A2Dx2 = (P1 – P2)V In essence, Bernoulli’s Principle states that where the velocity of a fluid is high, the pressure is low and where the velocity is low, the pressure is high.

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Pressure ACT What will happen when I “blow” air between the two plates? A) Move Apart B) Come Together C) Nothing Several Demos here include cans on straws.

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Bernoulli’s Equation For steady flow, the speed, pressure, and elevation of an incompressible and nonviscous fluid are related by an equation discovered by Daniel Bernoulli (1700–1782). It is the application of the conservation of energy. P is the pressure of the fluid at the specified spot. y is the vertical height at the specified spot, v is the velocity at the specified spot.

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**Pressure, Flow and Work Work done = Gain in KE + Gain in PE or**

Work is done by the difference in pressure Fluid to left gives pressure. This is the work done to move the coloured fluid from left to right or

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**Bernoulli’s Equation Kinetic Energy per Volume**

The Fluid is pushed from Point 1 to Point 2 Kinetic Energy per Volume Potential energy per unit Volume External Pressure + + This sum has the same value at all points along a streamline.

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Question The container below is filled to a height h with liquid of density ρ. The atmospheric pressure is Patm, determine the pressure of the fluid at each labelled point . Compare the pressure at point 3 with that at point 5 with A3=1/2A5 h 1 2 4 5 3 (Cross sectional area are the same, v1=v2=v3, height are the same, h1=h2=h3) Let’s see how

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Question The container below is filled to a height h with liquid of density ρ. The atmospheric pressure is Patm, determine the pressure of the fluid at each labelled point . Compare the pressure at point 3 with that at point 5 with A3=1/2A5 h 1 2 4 5 3 (Cross sectional area are the same, v1=v2=v3, height are the same, h1=h2=h3)

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**Bernoulli ACT Through which hole will the water come out fastest?**

P1+rgy1 + ½ rv12 = P2+rgy2 + ½rv22 A Note: Outside all three holes the external pressure is the same: P=1 Atm (the ρgy term accounts for the pressure of water in column) B y2 y1 C y2 > y1 v1 > v2 Smaller y gives larger v. Hole C is fastest

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Exercise A housing contractor saves some money by reducing the size of a pipe from 1cm diameter to 1/2cm diameter at some point in your house. v1 v1/2 (A) 2 v1 (B) 4 v1 (C) 1/2 v1 (D) 1/4 v1 For equal volumes in equal times then ½ the diameter implies ¼ the area so the water has to flow four times as fast. But if the water is moving four times as fast the it has 16 times as much kinetic energy. Something must be doing work on the water (the pressure drops at the neck and we recast the work as W=P DV = (F/A) (ADx) = F Dx )

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Practice A large bucket full of water has two drains. One is a hole in the side of the bucket at the bottom, and the other is a pipe coming out of the bucket near the top, which bent is downward such that the bottom of this pipe is even with the other hole, like in the picture below: Though which drain is the water spraying out with the highest speed? 1. The hole 2. The pipe 3. Same Speed is determined by pressure difference where water meets the atmosphere. Relevant height is where the hole is. Both are exiting at the same height!

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Example A garden hose w/ inner diameter 2 cm, carries water at 2.0 m/s. To spray your friend, you place your thumb over the nozzle giving an effective opening diameter of 0.5 cm. What is the speed of the water exiting the hose? What is the pressure difference between inside the hose and outside? Continuity Equation A1 v1 = A2 v2 v2 = v1 ( A1/A2) = v1 ( πr12 / πr22) = 2 m/s 16 = 32 m/s Lift on house if v=30 m/s. Density of air = 1.27 kg / m^3 Bernoulli Equation P1+gy1 + ½ v12 = P2+gy2 + ½ v22 P1 – P2 = ½ (v22 – v12) = ½ (1000 kg/m3) (1024 m2/s2 – 4 m2/s2 ) = 5.1 105 Pa

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**Bernoulli’s Principle**

A housing contractor saves some money by reducing the size of a pipe from 1cm diameter to 1/2cm diameter at some point in your house. v1 v1/2 1) What is the pressure in the 1/2cm pipe relative to the 1cm pipe? (A) smaller (B) same (C) larger

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**Applications of Bernoulli's Equation**

The tarpaulin that covers the cargo is flat when the truck is stationary but bulges outward when the truck is moving.

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Household Plumbing In a household plumbing system, a vent is necessary to equalize the pressures at points A and B, thus preventing the trap from being emptied. An empty trap allows sewer gas to enter the house.

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Curveball Pitch

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**Applications of Fluid Dynamics**

Streamline flow around a moving airplane wing Lift is the upward force on the wing from the air Drag is the resistance The lift depends on the speed of the airplane, the area of the wing, its curvature, and the angle between the wing and the horizontal higher velocity lower pressure lower velocity higher pressure Note: density of flow lines reflects velocity, not density. We are assuming an incompressible fluid.

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**Applications of Fluid Dynamics**

The constriction in the Subclavian artery causes the blood in the region to speed up and thus produces low pressure. The blood moving UP the LVA is then pushed DOWN instead of down causing a lack of blood flow to the brain. This condition is called TIA (transient ischemic attack) or “Subclavian Steal Syndrome.

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Flash Files Pressure Buoyancy Fluid Flow Bernoulli’s Equation

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Summary These are the formulas that you need to know:

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**Archimedes’ Principle:**

Pascal’s Principle: Continuity Equation: Av is flow rate Bernoulli’s Equation:

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Fluids Mechanics Carlos Silva December 2 nd 2009.

Fluids Mechanics Carlos Silva December 2 nd 2009.

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