Fluid Mechanics. Fluid Statics [3-1] Dr. Mohammad N. Almasri. [3] Fall 2010 Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics
|
|
- Horatio Logan Cook
- 7 years ago
- Views:
Transcription
1 1 Fluid Mechanics Fluid Statics [3-1] Dr. Mohammad N. Almasri
2 Fluid Pressure Fluid pressure is the normal force exerted by the fluid per unit area at some location within the fluid Fluid pressure has the units: N/m 2 2 Pascal 1 Pa = 1 N/m 2 psi pound-force per square inch = 6, Pa Bar 1 bar = 10 5 Pa = 100 Kpa Atmosphere 1 atm = 101,325 Pa = bars 1 kgf/cm 2 = N/m 2 psia (pounds-force per square inch absolute) gauge pressure plus local atmospheric pressure
3 Absolute pressure is the actual pressure at a given position Absolute Pressure However, the absolute pressure is measured relative to absolute zero pressure This is why most pressuremeasuring devices (called gages) read zero in the atmosphere 3
4 4 Gage and Vacuum Pressures When the pressure gages read a pressure value this value would be the difference between the absolute pressure and the local atmospheric pressure This difference is called gage pressure (P gage ) P gage = P abs P atm If Pgage is negative then it is called vacuum pressure P vac = P atm - P abs
5 5 Gage and Vacuum Pressures
6 6 The gage used to measure the air pressure in an automobile tire reads the gage pressure Gage Pressure Therefore, the common reading of 32 psi indicates a pressure of 32 psi above the atmospheric pressure At a location where the atmospheric pressure is 14.3 psi, for example, the absolute pressure in the tire is = 46.3 psi
7 7 Example A vacuum gage connected to a chamber reads 5.8 psi at a location where the atmospheric pressure is 14.5 psi Determine the absolute pressure in the chamber We know that P abs = P atm P vac = = 8.7 psi
8 8 Pressure at a Point Consider a small wedgeshaped fluid element of unit length (into the page) in equilibrium The mean pressures at the three surfaces are P n, P x, and P z, and the force acting on a surface is the product of mean pressure and the surface area
9 9 Pressure at a Point By writing the equation of equilibrium for the x direction, we obtain: For the z direction, we obtain: Now, when we divide this equation by the product ΔlΔycosα and the fluid element shrinks to a point (Δl 0)m the last term disappears. Thus we have: Ultimately, this implies that : Since α is arbitrary and Pn is independent of α, we conclude that the pressure at a point in a static fluid acts with the same magnitude in all directions:
10 10 Variation of Pressure with Depth To obtain a relation for the variation of pressure with depth, consider the element shown in the figure with a length of Δl, ΔA in cross-sectional area, and inclined at angle α with the horizontal A force balance in the vertical z-direction gives: Upon simplification: However, if we let the length approach zero, then Δp/Δl=dp/dl. Note also that sin α = dz/dl. Therefore:
11 Variation of Pressure with Depth This is the basic equation for hydrostatic pressure variation with elevation It implies that pressure changes with the vertical distance but remains constant in the other directions The pressure changes inversely with elevation. If one travels upward in the fluid (positive z direction), the pressure decreases; and if one goes downward (negative z), the pressure increases 11 Of course, a pressure increase is exactly what a diver experiences when descending in a lake or pool
12 Variation of Pressure with Depth If the specific weight of the fluid is uniform then it is constant. Thus the previous equation (dp/dz = -γ) becomes after integration: The sum of the terms p/γ and z is called the piezometric head. This is constant throughout an incompressible static fluid 12 One can easily relate the piezometric head at one point to that of another point in the fluid as in the following:
13 13 Variation of Pressure with Depth Thus, the pressure difference between two points in a constant density fluid is proportional to the vertical distance Δz between the points and the density ρ of the fluid In other words, pressure in a fluid increases linearly with depth as we go down and vice versa
14 14 Variation of Pressure with Depth A pressure difference of 10 psi can be specified in terms of pressure head as 23.1 ft of water (γ = 62.4 lb/ft 3 ), or 518 mm of Hg (γ = 133 kn/m 3 ) As illustrated by the figure, a ft-tall column of water with a crosssectional area of 1 in. 2 weighs 10 lb
15 15 Variation of Pressure with Depth If we take point 1 to be at the free surface of a liquid open to the atmosphere where the pressure is the atmospheric pressure (P atm ) then the pressure at a depth h from the free surface becomes: P = P atm + ρgh or P gage = ρgh
16 Example Determine the pressure in psi at a depth of 20 ft below the free surface of a body of water if γ = 62.4 lb/ft 3 P = γ h = = 1,248 lb/ft 2 P = 1,248/144 = 8.67 psi 16
17 17 Example Determine the pressure at a depth of 9 m in oil of specific gravity 0.75 P = γ h = = 66.1 kpa
18 18 Example What depth of oil of specific gravity 0.75 will produce a pressure of 40 psi? What depth of water?
19 19 Example Find the pressure at the bottom of a tank containing glycerin under pressure
20 20 Example What is the water pressure at a depth of 35 ft in the tank shown?
21 21 Example Oil with a specific gravity of 0.80 forms a layer 0.90 m deep in an open tank that is otherwise filled with water. The total depth of water and oil is 3 m. What is the gage pressure at the bottom of the tank?
22 22 Example Because of a leak in a buried gasoline storage tank, water has seeped in to the depth shown in the figure. The specific gravity of the gasoline is 0.68 Determine the pressure at the gasoline water interface and at the bottom of the tank
23 23 Example
24 24 Example
25 25 Pressure Transmission In a closed system, a pressure change produced at one point in the system will be transmitted throughout the entire system This phenomenon of pressure transmission, along with the ease with which fluids can be moved, has led to the widespread development of hydraulic controls for operating equipment such as hydraulic presses
26 26 Pressure Transmission In the figure the air pressure from a compressor establishes the pressure in the oil system, which in turn acts against the piston in the lift It can be seen that if a pressure of 600 kn/m 2, for example, acts on the 25-cm diameter piston, then a force equal to p A, or 29.5 kn, will be exerted on the piston To handle larger or smaller loads it is necessary only to increase or decrease the pressure
27 27 Example A hydraulic jack has the dimensions shown. If one exerts a force F of 100 N on the handle of the jack, what load, F 2, can the jack support? Neglect lifter weight
28 28 Example
29 29 Example In the figure the areas of the plunger A and cylinder B are 6.00 and 600 in 2, respectively, and the weight of B is 9000 lb. The vessel and the connecting passages are filled with oil of specific gravity What force F is required for equilibrium, neglecting the weight of A?
30 30 Example
31 31 Pressure Transmission The pressure is the same at all points on a horizontal plane in a given fluid Note that the pressures at points A, B, C, D, F. F, and C are the same since they are at the same depth, and they are interconnected by the same static fluid However, the pressures at points H and I are not the same since these two points cannot be interconnected by the same fluid although they are at the same depth
32 32 Pressure Transmission
33 Measurement of Pressure Barometer Absolute atmospheric pressure is measured by a device called barometer and thus the atmospheric pressure is often referred to as the barometric pressure If we immerse the open end of a tube in a liquid that is open to the atmosphere (atmospheric pressure), and if we exhaust air from the tube, the liquid will rise in the tube If the tube is long enough and if we have removed all the air, the only pressure on the surface of the liquid in the tube will be that of its own vapor pressure, and the liquid will have reached its maximum possible height 33
34 34 Measurement of Pressure Barometer From the concepts developed earlier, we see that the pressure at 0 within the tube and at a on the surface of the liquid outside the tube must be the same; that is: P 0 = P a = P atm In addition, we have: P 0 = γy + P vapor or P atm = γy + P vapor If we neglect the vapor pressure, then we would have: P atm = γy
35 35 Measurement of Pressure Barometer The liquid used in barometers is usually mercury because: its density is sufficiently great to enable a reasonably short tube to be used Its vapor pressure is negligibly small at ordinary temperatures Standard sea-level atmospheric pressure in different ways are: in Hg = 760 mm Hg ft of water = m of water
36 36 Example Determine the atmospheric pressure at a location where the barometric reading is 740 mm Hg and the gravitational acceleration is g = 9.81 m/s 2. assume that the temperature of mercury to be 10 C at which the density is 13,570 kg/m 3 We know that: P atm = γy + P vapor Neglecting the vapor pressure we have: P atm = γy = ρgh = 13, = Pa
37 Example What would be the reading on a barometer containing carbon tetrachloride at 68 F at a time when the atmospheric pressure was equivalent to in Hg? For carbon tetrachloride at 68 F we have: ρ = 3.08 slugs/ft 3 and P vapor = 1.9 psia Convert P atm to the units of psia P atm = / = psia Y = (P atm P vapor )/ρg = ( )/( ) = ft of carbon tetrachloride 37
38 38 Measurement of Pressure Piezometer Measuring pressure involves the use of devices that use liquid columns. These measuring devices are called manometers The simplest type of manometer, called a piezometer tube, consists of a vertical tube, open at the top, and attached to the container in which the pressure is desired
39 39 Measurement of Pressure Piezometer Because manometers involve columns of fluids at rest, the fundamental equation describing their use is P = γh + P 0 which gives the pressure at any elevation in terms of the vertical distance h between P and P 0 Application of this equation to the piezometer tube indicates that the pressure P A can be determined by a measurement of h 1 through the relationship P A = γ 1 h 1 where γ 1 is the specific weight of the liquid in the container Since the tube is open at the top, the pressure P 0 can be set equal to zero (we are now using gage pressure for P A ) Because point (1) and point A within the container are at the same elevation then P A = P 1
40 40 Measurement of Pressure U-Tube Manometer Another type of manometer is used and consists of a tube formed into the shape of a U To find the pressure PA in terms of the various column heights, we start at one end of the system and work our way around to the other end, simply utilizing equation: P 1 = γh + P 2
41 41 Measurement of Pressure U-Tube Manometer Figure 1. The liquid is at the same height in each leg and pressure is equal at both ends Figure 2. When positive pressure is applied to one leg, the liquid is forced down in that leg and up in the other. The difference in height, "h," which is the sum of the readings above and below zero, indicates the pressure Figure 3. When a vacuum is applied to one leg, the liquid rises in that leg and falls in the other. The difference in height, "h," indicates the amount of vacuum
42 Measurement of Pressure U-Tube Manometer Thus, we will start at point A and work around to the open end The pressure at points A and (1) are the same As we move from point (1) to (2) the pressure will increase by y 1 h 1 The pressure at point (2) is equal to the pressure at point (3), since the pressures at equal elevations in a continuous mass of fluid at rest must be the same 42
43 43 Measurement of Pressure U-Tube Manometer With the pressure at point (3) specified we now move to the open end where the pressure is zero As we move vertically upward the pressure decreases by an amount y 2 h 2 As such we can write P A + γ 1 h 1 γ 2 h 2 = 0 And the pressure PA can be written as column heights as: P A = γ 2 h 2 γ 1 h 1
44 44 Measurement of Pressure U-Tube Manometer
45 45 Example A closed tank contains compressed air and oil (SG oil = 0.90). A U-tube manometer using mercury (SG Hg = 13.6) is connected to the tank. The column heights are h 1 = 36 in., h 2 = 6 in., and h 3 = 9 in Determine the pressure reading (in psi) of the gage
46 46 Example Following the general procedure of starting at one end of the manometer system and working around to the other, we will start at the air oil interface in the tank and proceed to the open end where the pressure is zero. The pressure at level (1) is: This pressure is equal to the pressure at level (2), as these two points are at the same elevation in a homogeneous fluid at rest. As we move from level (2) to the open end, the pressure must decrease by Y Hg h 3, and at the open end the pressure is zero. Thus, the manometer equation can be expressed as:
47 47 Example Assume that the gage pressure remains at 3.06 psi, but the manometer is altered so that it contains only oil. That is, the mercury is replaced by oil. A simple calculation shows that in this case the vertical oil-filled tube would need to be h 3 = 11.3 ft tall, rather than the original h 3 = 9 in. There is an obvious advantage of using a heavy fluid such as mercury in manometers
48 48 Measurement of Pressure Differential Manometers The U-tube manometer is also widely used to measure the difference in pressure between two containers or two points in a given system Consider a manometer connected between containers A and B as is shown in the figure. The difference in pressure between A and B can be found by again starting at one end and working around to the other end
49 Measurement of Pressure Differential Manometers For example, at A the pressure is P A, which is equal to P 1, and as we move to point (2) the pressure increases by γ 1 h 1 The pressure at P 2 is equal to P 3, and as we move upward to point (4) the pressure decreases by γ 2 h 2 Similarly, as we continue to move upward from point (4) to (5) the pressure decreases by γ 3 h 3 Finally, P 5 = P B, as they are at equal elevations. Thus, 49 P A + γ 1 h 1 γ 2 h 2 γ 2 h 2 = P B P A P B =γ 2 h 2 + γ 2 h 2 γ 1 h 1 The pressure difference
50 50 Example The nozzle creates a pressure drop, P A P B, along the pipe. The pressure drop is measured with a differential U-tube manometer of the type illustrated (a) Determine an equation for P A P B in terms of the specific weight of the flowing fluid, γ 1, the specific weight of the gage fluid, γ 2 and the various heights indicated
51 51 Example If we start at point A and move vertically upward to level (1), the pressure will decrease by γ 1 h 1 and will be equal to the pressure at (2) and at (3). We can now move from (3) to (4) where the pressure has been further reduced by γ 2 h 2 The pressures at levels (4) and (5) are equal, and as we move from (5) to B the pressure will increase by γ 1 (h 1 + h 2 ). Thus, in equation form P A γ 1 h 1 γ 2 h 2 + γ 1 (h 1 + h 2 ) = P B P A P B =h 2 (γ 2 γ 1 ) It is to be noted that the only column height of importance is the differential reading, h 2
52 52 Measurement of Pressure Inclined Manometers To measure small pressure changes, a manometer of the type shown in the figure is frequently used One leg of the manometer is inclined at an angle θ and the differential reading l 2 is measured along the inclined tube. The difference in pressure P A P B can be expressed as: P A + γ 1 h 1 γ 2 l 2 sin θ γ 3 h 3 = P B P A P B = γ 2 l 2 sin θ + γ 3 h 3 γ 1 h 1 where it is to be noted the pressure difference between points (1) and (2) is due to the vertical distance between the points, which can be expressed as l 2 sin θ Thus, for relatively small angles the differential reading along the inclined tube can be made large even for small pressure differences
53 53 Example For the open tank, with piezometers attached on the side, containing two different immiscible liquids as shown in the figure, find: (a) The elevation of the liquid surface in piezometer A (b)the elevation of the liquid surface in piezometer B (c) The total pressure at the bottom of the tank
54 Example (a) Liquid A will simply rise in piezometer A to the same elevation as liquid A in the tank (i.e., to elevation 2 m) (b) Liquid B will rise in piezometer B to elevation 0.3 m (as a result of the pressure exerted by liquid B) plus an additional amount, h A, as a result of the overlying pressure of liquid A, P A P A = γh = ( )(1.7) = kpa h A = p/ γ = 11.98/( ) = m Liquid B will rise in piezometer B to elevation = m (c) Pressure at bottom = ( )(1.7) + ( )(0.3) = 18.9 kpa 54
55 55 Example Determine the pressure at A in psi gage due to the deflection of the mercury, SG = in the U-tube gage shown in the figure Pressure head at B = pressure head at C P A /γ ft water = ft water Solving, P A / γ = 34.0 ft water, and P A = ( )/144 = 14.7 psi
56 Example A manometer is attached to a tank containing three different fluids, as shown in the figure. Find the difference in elevation of the mercury column in the manometer (for instance find the value of y in the figure) 56 pressure at A = pressure at B 30 + ( )(3) + (9.79)(3.00) = ( )(y) Y = m
57 57 Example Oil of specific gravity flows through the nozzle shown in the figure and deflects the mercury in the U-tube gage. Determine the value of h if the pressure at A is 20.0 psi
58 58 Example
59 59 Example For a gage pressure at A of kpa, find the specific gravity of the gage liquid B in the figure
60 60 Example
61 61 Example For a gage reading at A of 2.50 psi, determine: (a) the elevations of the liquids in the open piezometer columns E, F, and G (b) the deflection of the mercury in the U-tube gage i
62 62 Example Since the unit weight of the air (about 0.08 lb/ft 3 ) is very small compared with that of the liquids, the pressure at elevation 49 may be considered to be psi without introducing significant error in the calculations
63 63 Example
64 64 Example
65 65 Example The loss through a device X is to be measured by a differential gage using oil of specific gravity as the gage fluid. The flowing liquid has SG Find the change in pressure head between A and B for the deflection of the oil shown in the figure
66 66 Example
67 67 Example A differential mercury manometer is connected to two pressure taps in an inclined pipe. Water at 50 F is flowing through the pipe. The deflection of mercury in the manometer is 1 inch. Find the change in piezometric pressure and piezometric head between the two points
68 68 Example
69 69 Example
When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid.
Fluid Statics When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid. Consider a small wedge of fluid at rest of size Δx, Δz, Δs
More informationp atmospheric Statics : Pressure Hydrostatic Pressure: linear change in pressure with depth Measure depth, h, from free surface Pressure Head p gh
IVE1400: n Introduction to Fluid Mechanics Statics : Pressure : Statics r P Sleigh: P..Sleigh@leeds.ac.uk r J Noakes:.J.Noakes@leeds.ac.uk January 008 Module web site: www.efm.leeds.ac.uk/ive/fluidslevel1
More informationCHAPTER 3: FORCES AND PRESSURE
CHAPTER 3: FORCES AND PRESSURE 3.1 UNDERSTANDING PRESSURE 1. The pressure acting on a surface is defined as.. force per unit. area on the surface. 2. Pressure, P = F A 3. Unit for pressure is. Nm -2 or
More informationPhysics 1114: Unit 6 Homework: Answers
Physics 1114: Unit 6 Homework: Answers Problem set 1 1. A rod 4.2 m long and 0.50 cm 2 in cross-sectional area is stretched 0.20 cm under a tension of 12,000 N. a) The stress is the Force (1.2 10 4 N)
More informationoil liquid water water liquid Answer, Key Homework 2 David McIntyre 1
Answer, Key Homework 2 David McIntyre 1 This print-out should have 14 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making
More informationa cannonball = (P cannon P atmosphere )A cannon m cannonball a cannonball = (P cannon P atmosphere ) πd 2 a cannonball = 5.00 kg
2.46 A piston/cylinder with a cross-sectional area of 0.01 m 3 has a mass of 100 resting on the stops as shown in the figure. With an outside atmospheric pressure of 100 kpa what should the water pressure
More informationCE 3500 Fluid Mechanics / Fall 2014 / City College of New York
1 Drag Coefficient The force ( F ) of the wind blowing against a building is given by F=C D ρu 2 A/2, where U is the wind speed, ρ is density of the air, A the cross-sectional area of the building, and
More informationChapter 13 - Solutions
= Chapter 13 - Solutions Description: Find the weight of a cylindrical iron rod given its area and length and the density of iron. Part A On a part-time job you are asked to bring a cylindrical iron rod
More informationOUTCOME 1 STATIC FLUID SYSTEMS TUTORIAL 1 - HYDROSTATICS
Unit 41: Fluid Mechanics Unit code: T/601/1445 QCF Level: 4 Credit value: 15 OUTCOME 1 STATIC FLUID SYSTEMS TUTORIAL 1 - HYDROSTATICS 1. Be able to determine the behavioural characteristics and parameters
More informationChapter 3: Pressure and Fluid Statics
Pressure Pressure is defined as a normal force exerted by a fluid per unit area. Units of pressure are N/m 2, which is called a pascal (Pa). Since the unit Pa is too small for pressures encountered in
More informationBarometric Effects on Transducer Data and Groundwater Levels in Monitoring Wells D.A. Wardwell, October 2007
Barometric Effects on Transducer Data and Groundwater Levels in Monitoring Wells D.A. Wardwell, October 2007 Barometric Effects on Transducer Data Barometric Fluctuations can Severely Alter Water Level
More information= 800 kg/m 3 (note that old units cancel out) 4.184 J 1000 g = 4184 J/kg o C
Units and Dimensions Basic properties such as length, mass, time and temperature that can be measured are called dimensions. Any quantity that can be measured has a value and a unit associated with it.
More informationChapter 15. FLUIDS. 15.1. What volume does 0.4 kg of alcohol occupy? What is the weight of this volume? m m 0.4 kg. ρ = = ; ρ = 5.
Chapter 15. FLUIDS Density 15.1. What volume does 0.4 kg of alcohol occupy? What is the weight of this volume? m m 0.4 kg ρ = ; = = ; = 5.06 x 10-4 m ρ 790 kg/m W = D = ρg = 790 kg/m )(9.8 m/s )(5.06 x
More informationCO 2 41.2 MPa (abs) 20 C
comp_02 A CO 2 cartridge is used to propel a small rocket cart. Compressed CO 2, stored at a pressure of 41.2 MPa (abs) and a temperature of 20 C, is expanded through a smoothly contoured converging nozzle
More informationc KEY EQUATIONS c EXERCISES: THINGS ENGINEERS THINK ABOUT c PROBLEMS: DEVELOPING ENGINEERING SKILLS Problems: Developing Engineering Skills 27
Problems: Developing Engineering Skills 27 c KEY EQUATIONS n 5 m/m (1.8) p. 14 Relation between amounts of matter on a mass basis, m, and on a molar basis, n. T(8R) 5 1.8T(K) (1.16) p. 21 Relation between
More information01 The Nature of Fluids
01 The Nature of Fluids WRI 1/17 01 The Nature of Fluids (Water Resources I) Dave Morgan Prepared using Lyx, and the Beamer class in L A TEX 2ε, on September 12, 2007 Recommended Text 01 The Nature of
More information2.016 Hydrodynamics Reading #2. 2.016 Hydrodynamics Prof. A.H. Techet
Pressure effects 2.016 Hydrodynamics Prof. A.H. Techet Fluid forces can arise due to flow stresses (pressure and viscous shear), gravity forces, fluid acceleration, or other body forces. For now, let us
More informationFluid Mechanics: Static s Kinematics Dynamics Fluid
Fluid Mechanics: Fluid mechanics may be defined as that branch of engineering science that deals with the behavior of fluid under the condition of rest and motion Fluid mechanics may be divided into three
More informationGases. Macroscopic Properties. Petrucci, Harwood and Herring: Chapter 6
Gases Petrucci, Harwood and Herring: Chapter 6 CHEM 1000A 3.0 Gases 1 We will be looking at Macroscopic and Microscopic properties: Macroscopic Properties of bulk gases Observable Pressure, volume, mass,
More informationChapter 27 Static Fluids
Chapter 27 Static Fluids 27.1 Introduction... 1 27.2 Density... 1 27.3 Pressure in a Fluid... 2 27.4 Pascal s Law: Pressure as a Function of Depth in a Fluid of Uniform Density in a Uniform Gravitational
More informationTemperature Measure of KE At the same temperature, heavier molecules have less speed Absolute Zero -273 o C 0 K
Temperature Measure of KE At the same temperature, heavier molecules have less speed Absolute Zero -273 o C 0 K Kinetic Molecular Theory of Gases 1. Large number of atoms/molecules in random motion 2.
More informationFLUID FORCES ON CURVED SURFACES; BUOYANCY
FLUID FORCES ON CURVED SURFCES; BUOYNCY The principles applicable to analysis of pressure-induced forces on planar surfaces are directly applicable to curved surfaces. s before, the total force on the
More informationDensity Measurement. Technology: Pressure. Technical Data Sheet 00816-0100-3208 INTRODUCTION. S min =1.0 S max =1.2 CONSTANT LEVEL APPLICATIONS
Technical Data Sheet 00816-0100-3208 Density Measurement Technology: Pressure INTRODUCTION Pressure and differential pressure transmitters are often used to measure the density of a fluid. Both types of
More informationPhysics 181- Summer 2011 - Experiment #8 1 Experiment #8, Measurement of Density and Archimedes' Principle
Physics 181- Summer 2011 - Experiment #8 1 Experiment #8, Measurement of Density and Archimedes' Principle 1 Purpose 1. To determine the density of a fluid, such as water, by measurement of its mass when
More informationMECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS
MECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS This is the second tutorial on bending of beams. You should judge your progress by completing the self assessment exercises.
More informationMercury is poured into a U-tube as in Figure (14.18a). The left arm of the tube has crosssectional
Chapter 14 Fluid Mechanics. Solutions of Selected Problems 14.1 Problem 14.18 (In the text book) Mercury is poured into a U-tube as in Figure (14.18a). The left arm of the tube has crosssectional area
More informationGas Laws. The kinetic theory of matter states that particles which make up all types of matter are in constant motion.
Name Period Gas Laws Kinetic energy is the energy of motion of molecules. Gas state of matter made up of tiny particles (atoms or molecules). Each atom or molecule is very far from other atoms or molecules.
More informationCE 6303 MECHANICS OF FLUIDS L T P C QUESTION BANK PART - A
CE 6303 MECHANICS OF FLUIDS L T P C QUESTION BANK 3 0 0 3 UNIT I FLUID PROPERTIES AND FLUID STATICS PART - A 1. Define fluid and fluid mechanics. 2. Define real and ideal fluids. 3. Define mass density
More informationDensity (r) Chapter 10 Fluids. Pressure 1/13/2015
1/13/015 Density (r) Chapter 10 Fluids r = mass/volume Rho ( r) Greek letter for density Units - kg/m 3 Specific Gravity = Density of substance Density of water (4 o C) Unitless ratio Ex: Lead has a sp.
More informationXI / PHYSICS FLUIDS IN MOTION 11/PA
Viscosity It is the property of a liquid due to which it flows in the form of layers and each layer opposes the motion of its adjacent layer. Cause of viscosity Consider two neighboring liquid layers A
More informationPressure. Pressure. Atmospheric pressure. Conceptual example 1: Blood pressure. Pressure is force per unit area:
Pressure Pressure is force per unit area: F P = A Pressure Te direction of te force exerted on an object by a fluid is toward te object and perpendicular to its surface. At a microscopic level, te force
More informationENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P
ENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P This material is duplicated in the Mechanical Principles module H2 and those
More informationUnderstanding Pressure and Pressure Measurement
Freescale Semiconductor Application Note Rev 1, 05/2005 Understanding Pressure and Pressure Measurement by: David Heeley Sensor Products Division, Phoenix, Arizona INTRODUCTION Fluid systems, pressure
More information9. Hydrostatik I (1.2 1.5)
9. Hydrostatik I (1.2 1.5) Vätsketryck, tryck-densitet-höjd Tryck mot plana ytor Övningstal: H10 och H12 HYDROSTATICS Hydrostatics: Study of fluids (water) at rest No motion no shear stress viscosity non-significant
More informationChapter 5 MASS, BERNOULLI AND ENERGY EQUATIONS
Fluid Mechanics: Fundamentals and Applications, 2nd Edition Yunus A. Cengel, John M. Cimbala McGraw-Hill, 2010 Chapter 5 MASS, BERNOULLI AND ENERGY EQUATIONS Lecture slides by Hasan Hacışevki Copyright
More informationPressure in Fluids. Introduction
Pressure in Fluids Introduction In this laboratory we begin to study another important physical quantity associated with fluids: pressure. For the time being we will concentrate on static pressure: pressure
More informationExperiment 3 Pipe Friction
EML 316L Experiment 3 Pipe Friction Laboratory Manual Mechanical and Materials Engineering Department College of Engineering FLORIDA INTERNATIONAL UNIVERSITY Nomenclature Symbol Description Unit A cross-sectional
More informationForces. Definition Friction Falling Objects Projectiles Newton s Laws of Motion Momentum Universal Forces Fluid Pressure Hydraulics Buoyancy
Forces Definition Friction Falling Objects Projectiles Newton s Laws of Motion Momentum Universal Forces Fluid Pressure Hydraulics Buoyancy Definition of Force Force = a push or pull that causes a change
More informationUnit 1 INTRODUCTION 1.1.Introduction 1.2.Objectives
Structure 1.1.Introduction 1.2.Objectives 1.3.Properties of Fluids 1.4.Viscosity 1.5.Types of Fluids. 1.6.Thermodynamic Properties 1.7.Compressibility 1.8.Surface Tension and Capillarity 1.9.Capillarity
More informationUNDERSTANDING REFRIGERANT TABLES
Refrigeration Service Engineers Society 1666 Rand Road Des Plaines, Illinois 60016 UNDERSTANDING REFRIGERANT TABLES INTRODUCTION A Mollier diagram is a graphical representation of the properties of a refrigerant,
More informationSURFACE TENSION. Definition
SURFACE TENSION Definition In the fall a fisherman s boat is often surrounded by fallen leaves that are lying on the water. The boat floats, because it is partially immersed in the water and the resulting
More informationHomework 9. Problems: 12.31, 12.32, 14.4, 14.21
Homework 9 Problems: 1.31, 1.3, 14.4, 14.1 Problem 1.31 Assume that if the shear stress exceeds about 4 10 N/m steel ruptures. Determine the shearing force necessary (a) to shear a steel bolt 1.00 cm in
More informationThe Versatile Differential Pressure Transmitter. By David Gunn Honeywell Process Solutions
The Versatile Differential Pressure Transmitter By David Gunn Honeywell Process Solutions The Versatile Differential Pressure Transmitter 2 Table of Contents Abstract... 3 Pressure Fundamentals... 3 Applications...
More informationGrade 8 Science Chapter 9 Notes
Grade 8 Science Chapter 9 Notes Force Force - Anything that causes a change in the motion of an object. - usually a push or a pull. - the unit for force is the Newton (N). Balanced Forces - forces that
More informationWrite True or False in the space provided.
CP Physics -- Exam #7 Practice Name: _ Class: Date: Write True or False in the space provided. 1) Pressure at the bottom of a lake depends on the weight density of the lake water and on the volume of the
More informationChapter 4 Atmospheric Pressure and Wind
Chapter 4 Atmospheric Pressure and Wind Understanding Weather and Climate Aguado and Burt Pressure Pressure amount of force exerted per unit of surface area. Pressure always decreases vertically with height
More informationRusty Walker, Corporate Trainer Hill PHOENIX
Refrigeration 101 Rusty Walker, Corporate Trainer Hill PHOENIX Compressor Basic Refrigeration Cycle Evaporator Condenser / Receiver Expansion Device Vapor Compression Cycle Cooling by the removal of heat
More informationTHE KINETIC THEORY OF GASES
Chapter 19: THE KINETIC THEORY OF GASES 1. Evidence that a gas consists mostly of empty space is the fact that: A. the density of a gas becomes much greater when it is liquefied B. gases exert pressure
More informationFor Water to Move a driving force is needed
RECALL FIRST CLASS: Q K Head Difference Area Distance between Heads Q 0.01 cm 0.19 m 6cm 0.75cm 1 liter 86400sec 1.17 liter ~ 1 liter sec 0.63 m 1000cm 3 day day day constant head 0.4 m 0.1 m FINE SAND
More informationLiquid level measurement using hydrostatic pressure and buoyancy
iquid level measurement using hydrostatic pressure and buoyancy This worksheet and all related files are licensed under the Creative Commons Attribution icense, version 1.0. To view a copy of this license,
More informationFLUID MECHANICS. TUTORIAL No.7 FLUID FORCES. When you have completed this tutorial you should be able to. Solve forces due to pressure difference.
FLUID MECHANICS TUTORIAL No.7 FLUID FORCES When you have completed this tutorial you should be able to Solve forces due to pressure difference. Solve problems due to momentum changes. Solve problems involving
More informationDifferential Relations for Fluid Flow. Acceleration field of a fluid. The differential equation of mass conservation
Differential Relations for Fluid Flow In this approach, we apply our four basic conservation laws to an infinitesimally small control volume. The differential approach provides point by point details of
More informationSTRESS AND DEFORMATION ANALYSIS OF LINEAR ELASTIC BARS IN TENSION
Chapter 11 STRESS AND DEFORMATION ANALYSIS OF LINEAR ELASTIC BARS IN TENSION Figure 11.1: In Chapter10, the equilibrium, kinematic and constitutive equations for a general three-dimensional solid deformable
More informationChapter 28 Fluid Dynamics
Chapter 28 Fluid Dynamics 28.1 Ideal Fluids... 1 28.2 Velocity Vector Field... 1 28.3 Mass Continuity Equation... 3 28.4 Bernoulli s Principle... 4 28.5 Worked Examples: Bernoulli s Equation... 7 Example
More informationCopyright 2011 Casa Software Ltd. www.casaxps.com. Centre of Mass
Centre of Mass A central theme in mathematical modelling is that of reducing complex problems to simpler, and hopefully, equivalent problems for which mathematical analysis is possible. The concept of
More informationBuoyant Force and Archimedes Principle
Buoyant Force and Archimedes Principle Predict the behavior of fluids as a result of properties including viscosity and density Demonstrate why objects sink or float Apply Archimedes Principle by measuring
More information5.2. Vaporizers - Types and Usage
5.2. Vaporizers - Types and Usage 5.2.1. General Vaporizers are constructed in numerous designs and operated in many modes. Depending upon the service application the design, construction, inspection,
More informationPump Formulas Imperial and SI Units
Pump Formulas Imperial and Pressure to Head H = head, ft P = pressure, psi H = head, m P = pressure, bar Mass Flow to Volumetric Flow ṁ = mass flow, lbm/h ρ = fluid density, lbm/ft 3 ṁ = mass flow, kg/h
More informationWEEK 1. Engineering Calculations Processes Process Variables
WEEK 1 Engineering Calculations Processes Process Variables 2.1 Units and Dimensions Units and dimensions are important in science and engineering A measured quantity has a numerical value and a unit (ex:
More informationW i f(x i ) x. i=1. f(x i ) x = i=1
Work Force If an object is moving in a straight line with position function s(t), then the force F on the object at time t is the product of the mass of the object times its acceleration. F = m d2 s dt
More informationConcept Questions Archimedes Principle. 8.01t Nov 24, 2004
Concept Questions Archimedes Principle 8.01t Nov 24, 2004 Pascal s Law Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel
More informationValve Sizing. Te chnic al Bulletin. Flow Calculation Principles. Scope. Sizing Valves. Safe Product Selection. www.swagelok.com
www.swagelok.com Valve Sizing Te chnic al Bulletin Scope Valve size often is described by the nominal size of the end connections, but a more important measure is the flow that the valve can provide. And
More informationFluids and Solids: Fundamentals
Fluids and Solids: Fundamentals We normally recognize three states of matter: solid; liquid and gas. However, liquid and gas are both fluids: in contrast to solids they lack the ability to resist deformation.
More informationFloating between two liquids. Laurence Viennot. http://education.epsdivisions.org/muse/
Floating between two liquids Laurence Viennot http://education.epsdivisions.org/muse/ Abstract The hydrostatic equilibrium of a solid floating between two liquids is analysed, first as a classical exercise,
More informationEXPERIMENT 15: Ideal Gas Law: Molecular Weight of a Vapor
EXPERIMENT 15: Ideal Gas Law: Molecular Weight of a Vapor Purpose: In this experiment you will use the ideal gas law to calculate the molecular weight of a volatile liquid compound by measuring the mass,
More informationAP Physics - Chapter 8 Practice Test
AP Physics - Chapter 8 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A single conservative force F x = (6.0x 12) N (x is in m) acts on
More informationAmbient Pressure = and Pressure, collectively
Three in One Pressure: Ambient, Barometric, Atmospheric 1 Ambient Pressure Overhead Transparency Ambient Pressure = and Pressure, collectively When surrounded by air, pressure =? pressure =? pressure The
More informationUnit 24: Applications of Pneumatics and Hydraulics
Unit 24: Applications of Pneumatics and Hydraulics Unit code: J/601/1496 QCF level: 4 Credit value: 15 OUTCOME 2 TUTORIAL 2 HYDRAULIC AND PNEUMATIC CYLINDERS The material needed for outcome 2 is very extensive
More informationName Class Date. F 2 2269 N A 1 88.12 cm 2 A 2 1221 cm 2 Unknown: Step 2: Write the equations for Pascal s principle and pressure, force, and area.
Skills Worksheet Math Skills Pascal s Principle After you study each sample problem and solution, work out the practice problems on a separate sheet of paper. Write your answers in the spaces provided.
More information( ) where W is work, f(x) is force as a function of distance, and x is distance.
Work by Integration 1. Finding the work required to stretch a spring 2. Finding the work required to wind a wire around a drum 3. Finding the work required to pump liquid from a tank 4. Finding the work
More informationPipe Flow-Friction Factor Calculations with Excel
Pipe Flow-Friction Factor Calculations with Excel Course No: C03-022 Credit: 3 PDH Harlan H. Bengtson, PhD, P.E. Continuing Education and Development, Inc. 9 Greyridge Farm Court Stony Point, NY 10980
More informationPipe Sizes For Water Distribution System Design
Appendix D Pipe Sizes For Water Distribution System Design D-. This appendix contains information to help determine pipe sizes when designing a water distribution system. Use Table D- and Tables D- through
More informationPhysics 101 Hour Exam 3 December 1, 2014
Physics 101 Hour Exam 3 December 1, 2014 Last Name: First Name ID Discussion Section: Discussion TA Name: Instructions Turn off your cell phone and put it away. Calculators cannot be shared. Please keep
More informationAn Introduction to Fluid Mechanics
0. Contents of the Course Notes For the First Year Lecture Course: An Introduction to Fluid Mechanics School of Civil Engineering, University of Leeds. CIVE1400 FLUID MECHANICS Dr Andrew Sleigh January
More informationName Date Class STATES OF MATTER. SECTION 13.1 THE NATURE OF GASES (pages 385 389)
13 STATES OF MATTER SECTION 13.1 THE NATURE OF GASES (pages 385 389) This section introduces the kinetic theory and describes how it applies to gases. It defines gas pressure and explains how temperature
More informationBasic Hydraulics and Pneumatics
Basic Hydraulics and Pneumatics Module 1: Introduction to Pneumatics PREPARED BY IAT Curriculum Unit March 2011 Institute of Applied Technology, 2011 ATM 1122 Basic Hydraulics and Pneumatics Module 1:
More informationC B A T 3 T 2 T 1. 1. What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N
Three boxes are connected by massless strings and are resting on a frictionless table. Each box has a mass of 15 kg, and the tension T 1 in the right string is accelerating the boxes to the right at a
More informationThe Gas Laws. Our Atmosphere. Pressure = Units of Pressure. Barometer. Chapter 10
Our Atmosphere The Gas Laws 99% N 2 and O 2 78% N 2 80 70 Nitrogen Chapter 10 21% O 2 1% CO 2 and the Noble Gases 60 50 40 Oxygen 30 20 10 0 Gas Carbon dioxide and Noble Gases Pressure Pressure = Force
More informationEffects of Atmospheric Pressure on Gas Measurement
Effects of Atmospheric Pressure on Gas Measurement Class # 1390.1 March 2012 / White paper by Denis Rutherford Regional Sales Manager Central US Schneider Electric Telemetry & Remote SCADA Solutions 6650
More informationSoil Suction. Total Suction
Soil Suction Total Suction Total soil suction is defined in terms of the free energy or the relative vapor pressure (relative humidity) of the soil moisture. Ψ = v RT ln v w 0ω v u v 0 ( u ) u = partial
More informationThree Methods for Calculating the Buoyant Force Gleue: Physics
Three Methods for Calculating the Buoyant Force Gleue: Physics Name Hr. The Buoyant Force (F b ) is the apparent loss of weight for an object submerged in a fluid. For example if you have an object immersed
More informationPhysics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam
Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to fill your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry
More informationFluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 20 Conservation Equations in Fluid Flow Part VIII Good morning. I welcome you all
More informationCHEMISTRY GAS LAW S WORKSHEET
Boyle s Law Charles Law Guy-Lassac's Law Combined Gas Law For a given mass of gas at constant temperature, the volume of a gas varies inversely with pressure PV = k The volume of a fixed mass of gas is
More informationBuoyancy Problem Set
Buoyancy Problem Set 1) A stone weighs 105 lb in air. When submerged in water, it weighs 67.0 lb. Find the volume and specific gravity of the stone. (Specific gravity of an object: ratio object density
More informationCHAPTER 6 WORK AND ENERGY
CHAPTER 6 WORK AND ENERGY CONCEPTUAL QUESTIONS. REASONING AND SOLUTION The work done by F in moving the box through a displacement s is W = ( F cos 0 ) s= Fs. The work done by F is W = ( F cos θ). s From
More informationAt the skate park on the ramp
At the skate park on the ramp 1 On the ramp When a cart rolls down a ramp, it begins at rest, but starts moving downward upon release covers more distance each second When a cart rolls up a ramp, it rises
More informationLecture 2 PROPERTIES OF FLUID
Lecture 2 PROPERTIES OF FLUID Learning Objectives Upon completion of this chapter, the student should be able to: Define three states of matter: Solid, liquid and gas. Define mass density, specific weight
More informationPrelab Exercises: Hooke's Law and the Behavior of Springs
59 Prelab Exercises: Hooke's Law and the Behavior of Springs Study the description of the experiment that follows and answer the following questions.. (3 marks) Explain why a mass suspended vertically
More informationStructural Axial, Shear and Bending Moments
Structural Axial, Shear and Bending Moments Positive Internal Forces Acting Recall from mechanics of materials that the internal forces P (generic axial), V (shear) and M (moment) represent resultants
More information9. The kinetic energy of the moving object is (1) 5 J (3) 15 J (2) 10 J (4) 50 J
1. If the kinetic energy of an object is 16 joules when its speed is 4.0 meters per second, then the mass of the objects is (1) 0.5 kg (3) 8.0 kg (2) 2.0 kg (4) 19.6 kg Base your answers to questions 9
More informationPractice Problems on Pumps. Answer(s): Q 2 = 1850 gpm H 2 = 41.7 ft W = 24.1 hp. C. Wassgren, Purdue University Page 1 of 16 Last Updated: 2010 Oct 29
_02 A centrifugal with a 12 in. diameter impeller requires a power input of 60 hp when the flowrate is 3200 gpm against a 60 ft head. The impeller is changed to one with a 10 in. diameter. Determine the
More informationEXPERIMENT 3 Analysis of a freely falling body Dependence of speed and position on time Objectives
EXPERIMENT 3 Analysis of a freely falling body Dependence of speed and position on time Objectives to verify how the distance of a freely-falling body varies with time to investigate whether the velocity
More informationExam 4 Review Questions PHY 2425 - Exam 4
Exam 4 Review Questions PHY 2425 - Exam 4 Section: 12 2 Topic: The Center of Gravity Type: Conceptual 8. After a shell explodes at the top of its trajectory, the center of gravity of the fragments has
More informationWORK DONE BY A CONSTANT FORCE
WORK DONE BY A CONSTANT FORCE The definition of work, W, when a constant force (F) is in the direction of displacement (d) is W = Fd SI unit is the Newton-meter (Nm) = Joule, J If you exert a force of
More informationCENTRIFUGAL PUMP OVERVIEW Presented by Matt Prosoli Of Pumps Plus Inc.
CENTRIFUGAL PUMP OVERVIEW Presented by Matt Prosoli Of Pumps Plus Inc. 1 Centrifugal Pump- Definition Centrifugal Pump can be defined as a mechanical device used to transfer liquid of various types. As
More informationSwissmetro travels at high speeds through a tunnel at low pressure. It will therefore undergo friction that can be due to:
I. OBJECTIVE OF THE EXPERIMENT. Swissmetro travels at high speeds through a tunnel at low pressure. It will therefore undergo friction that can be due to: 1) Viscosity of gas (cf. "Viscosity of gas" experiment)
More information