p atmospheric Statics : Pressure Hydrostatic Pressure: linear change in pressure with depth Measure depth, h, from free surface Pressure Head p gh

Size: px
Start display at page:

Download "p atmospheric Statics : Pressure Hydrostatic Pressure: linear change in pressure with depth Measure depth, h, from free surface Pressure Head p gh"

Transcription

1 IVE1400: n Introduction to Fluid Mechanics Statics : Pressure : Statics r P Sleigh: r J January 008 Module web site: Unit 1: Fluid Mechanics asics Flow Pressure Properties of Fluids Fluids vs. Solids Viscosity : Statics Hydrostatic pressure Manometry / Pressure measurement Hydrostatic forces on submerged surfaces Unit 3: ynamics The continuity equation. The ernoulli Equation. pplication of ernoulli equation. The momentum equation. pplication of momentum equation. Unit 4: Effect of the boundary on flow Laminar and turbulent flow oundary layer theory n Intro to imensional analysis Similarity 3 lectures 3 lectures 7 lectures 4 lectures Hydrostatic Pressure: linear change in pressure with depth p Measure depth, h, from free surface bsolute pressure auge pressure p1 g z z1 p gh p absolute g h + p atmospheric p gauge g h p atmospheric IVE1400: Fluid Mechanics Lecture 4 45 IVE1400: Fluid Mechanics Lecture 4 46 Pressure Head Examples of pressure head calculations: The lower limit of any pressure is the pressure in a What is a pressure of 500 knm - in head of water of density, 1000 kgm -3 Use p gh, Pressure measured above a perfect vacuum (zero) is known as pressure h In head of Mercury density kgm -3. Pressure measured relative to atmospheric pressure is known as pressure gauge pressure can be given using height of any fluid. h In head of a fluid with relative density 8.7. remember water ) h p gh This height, h, is referred to as the If pressure is quoted in head, the of the fluid also be given. IVE1400: Fluid Mechanics Lecture 4 47 IVE1400: Fluid Mechanics Lecture 4 48

2 Manometers for Measuring Pressure n Examples of a Piezometer. Manometers use the relationship between and to measure pressure 1. What is the maximum gauge pressure of water that can be measured by a Piezometer of height 1.5m? Piezometer Tube simple open tube attached to the top of a container with liquid at pressure. Liquid rises to a height, h, equal to the pressure in the container. p gh. If the liquid had a relative density of 8.5 what would the maximum measurable gauge pressure? The pressure measured is relative to so it measures pressure Problems with the Piezometer: IVE1400: Fluid Mechanics Lecture 4 49 IVE1400: Fluid Mechanics Lecture 4 50 Equality of Pressure t The Same Level In Static Fluid onsider these two tanks, one much larger than the other, and linked together by a thin tube: rea Fluid density ρ p l, p r, P Q Face L weight, mg Face Horizontal cylindrical element cross sectional area mass density left end pressure p l right end pressure p r For equilibrium the sum of the forces in the x direction is zero. Pressure in the horizontal direction is z L We have shown p l p r For a vertical pressure change we have p l and p r so p p p q z This true for any fluid. Pressure at the two equal levels are. IVE1400: Fluid Mechanics Lecture 4 51 IVE1400: Fluid Mechanics Lecture 4 5

3 n improved manometer: The U -Tube U is connected as shown and filled with manometric fluid. Fluid density ρ U -Tube enables the pressure of both liquids and gases to be measured Important points: 1. U -Tube enables the pressure of both liquids and gases to be measured h 1 h. The manometric fluid density should be measured. man > Manometric fluid density ρ man Using the fact that pressure at two levels are equal: pressure at p For the left hand arm pressure at p For the right hand arm pressure at p p p p Subtract p atmospheric to give gauge pressure p IVE1400: Fluid Mechanics Lecture The two fluids should they must be immiscible. What if the fluid is a gas? changes. The manometric fluid is liquid (usually mercury, oil or water) nd Liquid density is much greater than gas, man >> gh is negligible and pressure is given by p IVE1400: Fluid Mechanics Lecture 4 54 n example of the U-Tube manometer. Using a u-tube manometer to measure gauge pressure of fluid density 700 kg/m 3, and the manometric fluid is mercury, with a relative density of What is the gauge pressure if: a) h 1 0.4m and h 0.9m? b) h and h -0.1m? Pressure ifference Measurement Using a U -Tube Manometer. The U -tube manometer can be connected at both ends to measure pressure difference between these two points Fluid density ρ E h b h h a Manometric fluid density ρ man IVE1400: Fluid Mechanics Lecture 4 55 IVE1400: Fluid Mechanics Lecture 4 56

4 pressure at p p p n example using the u-tube for pressure difference measuring In the figure below two pipes containing the same fluid of density 990 kg/m 3 are connected using a u-tube manometer. a) What is the pressure between the two pipes if the manometer contains fluid of relative density 13.6? Fluid density ρ Fluid density ρ iving the pressure difference h a 1.5m h 0.5m E h b 0.75m p - p gain if the fluid is a gas man >>, then the terms involving can be neglected, Manometric fluid density ρ man 13.6 ρ p - p IVE1400: Fluid Mechanics Lecture 4 57 IVE1400: Fluid Mechanics Lecture 4 58 dvances to the U tube manometer Problem: Two reading are required. Solution: esult: p 1 p diameter z 1 atum line diameter d z If the manometer is measuring the pressure difference of a gas of (p 1 - p ) as shown, we know p 1 - p volume of liquid moved from the left side to the right The fall in level of the left side is z 1 z d / 4 / 4 d z Putting this in the equation, d p1 p g z z d gz 1 If >> d then (d/) is very small so p p 1 IVE1400: Fluid Mechanics Lecture 4 59 IVE1400: Fluid Mechanics Lecture 4 60

5 Inclined manometer Problem: Small changes difficult to see Incline the arm: same height change but bigger movement. p p 1 diameter d z 1 diameter θ Scale eader x z atum line Example of an inclined manometer. n inclined tube manometer consists of a vertical cylinder 35mm diameter. t the bottom of this is connected a tube 5mm in diameter inclined upward at an angle of 15 to the horizontal, the top of this tube is connected to an air duct. The vertical cylinder is open to the air and the manometric fluid has relative density a) etermine the pressure in the air duct if the manometric fluid moved 50mm along the inclined tube. a) What is the error if the movement of the fluid in the vertical cylinder is ignored? The pressure difference is still given by the height change of the manometric fluid. but, p p p gz 1 z p 1 The sensitivity to pressure change can be further by a inclination. IVE1400: Fluid Mechanics Lecture 4 61 IVE1400: Fluid Mechanics Lecture 4 6 hoice Of Manometer Take care when fixing the manometer to vessel urrs cause local pressure variations. Lecture 5: Forces in Static Fluids : Statics isadvantages: response - only really useful for very slowly varying pressures - no use at all for fluctuating pressures; For the U tube manometer measurements must be taken to get the h value. It is often difficult to measure variations in pressure. It cannot be used for pressures unless several manometers are connected in series; For very accurate work the and relationship between temperature and must be known; dvantages of manometers: They are very. No is required - the pressure can be calculated from first principles. From earlier we know that: 1. static fluid can have acting on it.. ny force between the fluid and the boundary must. F n Pressure force normal to the boundary. True also for curved surfaces any imaginary plane in the fluid n element of fluid at rest is in equilibrium: 3. The sum of forces in any direction is 4. The sum of the moments of forces about any point is F n F F 1 1 IVE1400: Fluid Mechanics Lecture 4 63 IVE1400: Fluid Mechanics Lecture 5 64

6 eneral submerged plane Horizontal submerged plane F 1 p 1 δ 1 F n p n δ n F p δ The total or resultant force,, on the plane is the sum of the forces on the small elements i.e. and This force will act through the centre of. For a plane surface all forces acting can be represented by one single force, acting at right-angles to the plane through the centre of. The pressure, p, will be at all points of the surface. The resultant force will be given by urved submerged surface Each elemental force is a different magnitude and in a different direction (but still normal to the surface.). It is, in general, not easy to calculate the resultant force for a curved surface by combining all elemental forces. The sum of all the forces on each element will always be than the sum of the individual forces, p. IVE1400: Fluid Mechanics Lecture 5 65 IVE1400: Fluid Mechanics Lecture 5 66 esultant Force and entre of Pressure on a general plane surface in a liquid. Fluid density ρ esultant Force P z z area δ x S c θ Q area Take pressure as zero at the surface. s O d O elemental area δ x z is known as the 1 st Moment of rea of the plane PQ about the free surface. nd it is known that z is the area of the plane z is the distance to the centre of ( ) Measuring down from the surface, the pressure on an element, depth z, p So force on element F esultant force on plane (assuming and g as constant). In terms of distance from point O z (as z x sin ) The resultant force on a plane Pressure at centre of gravity X rea IVE1400: Fluid Mechanics Lecture 5 67 IVE1400: Fluid Mechanics Lecture 5 68

7 This resultant force acts at right angles through the centre of pressure,, at a depth. How do we find this position?. Sum of moments Moment of about O Equating s the plane is in equilibrium: The moment of will be equal to on all the elements about the same point. It is convenient to take moment about O The force on each elemental area: Force on the moment of this force is: Moment of Force on about O g s sin s gsins, g and are the same for each element, giving the total moment as The position of the centre of along the plane measure from the point O is: s Sc x How do we work out the summation term? This term is known as the Second Moment of rea, I o, of the plane (about an axis through O) nd moment of area about O I o It can be easily calculated for many common shapes. s IVE1400: Fluid Mechanics Lecture 5 69 IVE1400: Fluid Mechanics Lecture 5 70 The position of the centre of pressure along the plane measure from the point O is: nd S c nd Moment of area about a line through O Moment of area about a line through O How do you calculate the nd moment of area? nd moment of area is a geometric property. It can be found from tables - UT only for moments about an axis through its centroid I. and epth to the centre of pressure is We need it for an axis through O Use the parallel axis theorem to give us what we want. The parallel axis theorem can be written I o IVE1400: Fluid Mechanics Lecture 5 71 IVE1400: Fluid Mechanics Lecture 5 7

8 We then get the following equation for the position of the centre of pressure The nd moment of area about a line through the centroid of some common shapes. S c Shape rea nd moment of area, I, about an axis through the centroid ectangle b bd bd 3 h 1 (In the examination the parallel axis theorem and the I will be given) Triangle ircle b h h/3 bd bd Semicircle (4)/(3π) IVE1400: Fluid Mechanics Lecture 5 73 IVE1400: Fluid Mechanics Lecture 5 74 Example 1: etermine the resultant force due to the water acting on the 1m by m rectangular area shown in the diagram below. [ N,.37m from O O P Example : etermine the resultant force due to the water acting on the 1.5m by.0m triangular area shown in the example above. The apex of the triangle is at. [ N,.81m from P] 1.m 1.0m 45.0 m.0 m IVE1400: Fluid Mechanics Lecture 5 75 IVE1400: Fluid Mechanics Lecture 5 76

9 Example 3: Find the moment required to keep this triangular gate closed on a tank which holds water..0m 1.m Lecture 6: Pressure iagrams and Forces on urved Surfaces : Statics Pressure diagrams 1.5m For vertical walls of constant width it is possible to find the resultant force and centre of pressure graphically using a pressure diagram. We know the relationship between pressure and depth: p So we can draw the diagram below: IVE1400: Fluid Mechanics Lecture 5 77 IVE1400: Fluid Mechanics Lecture 6 78 This is know as a pressure diagram. Pressure increases from zero at the surface linearly by p, to a maximum at the base of p. The area of this triangle represents the resultant force on the vertical wall, Units of this are per metre. rea The force acts through the centroid of the pressure diagram. For a triangle the centroid is at its height i.e. the resultant force acts horizontally through the point z. For a vertical plane the depth to the centre of pressure is given by esultant force per unit width IVE1400: Fluid Mechanics Lecture 6 79 IVE1400: Fluid Mechanics Lecture 6 80

10 heck this against the moment method: The same technique can be used with combinations of liquids are held in tanks (e.g. oil floating on water). For example: The resultant force is given by: oil ρ o 0.8m water ρ 1.m and the depth to the centre of pressure by: and by the parallel axis theorem (with width of 1) I I x o ρg0.8 ρg1. Find the position and magnitude of the resultant force on this vertical wall of a tank which has oil floating on water as shown. epth to the centre of pressure IVE1400: Fluid Mechanics Lecture 6 81 IVE1400: Fluid Mechanics Lecture 6 8 Forces on Submerged urved Surfaces E If the surface is curved the resultant force must be found by combining the elemental forces using some vectorial method. F O H alculate the and components. ombine these to obtain the resultant force and direction. (lthough this can be done for all three dimensions we will only look at one vertical plane) v The fluid is at rest in equilibrium. So of fluid such as is also in. In the diagram below liquid is resting on top of a curved base. IVE1400: Fluid Mechanics Lecture 6 83 IVE1400: Fluid Mechanics Lecture 6 84

11 onsider the Horizontal forces The sum of the horizontal forces is zero. The resultant horizontal force of a fluid above a curved surface is: H F No horizontal force on as there are no shear forces in a static fluid Horizontal forces act only on the faces and as shown. H We know 1. The force on a vertical plane must act horizontally (as it acts normal to the plane).. That H must act through the same point. So: H acts horizontally through the of the of the curved surface onto an vertical plane. F, must be equal and opposite to H. is the projection of the curved surface onto a vertical plane. We have seen earlier how to calculate resultant forces and point of action. Hence we can calculate the resultant horizontal force on a curved surface. IVE1400: Fluid Mechanics Lecture 6 85 IVE1400: Fluid Mechanics Lecture 6 86 onsider the Vertical forces The sum of the vertical forces is zero. E esultant force The overall resultant force is found by combining the vertical and horizontal components vectorialy, esultant force v There are no shear force on the vertical edges, so the vertical component can only be due to the of the fluid. So we can say The resultant vertical force of a fluid above a curved surface is: V. It will act vertically down through the centre of gravity of the mass of fluid. nd acts through O at an angle of. The angle the resultant force makes to the horizontal is The position of O is the point of intersection of the horizontal line of action of H and the vertical line of action of V. IVE1400: Fluid Mechanics Lecture 6 87 IVE1400: Fluid Mechanics Lecture 6 88

12 typical example application of this is the determination of the forces on dam walls or curved sluice gates. Find the magnitude and direction of the resultant force of water on a quadrant gate as shown below. 1.0m ate width 3.0m What are the forces if the fluid is below the curved surface? This situation may occur or a curved sluice gate. F O H Water ρ 1000 kg/m 3 v The force calculation is very similar to when the fluid is above. IVE1400: Fluid Mechanics Lecture 6 89 IVE1400: Fluid Mechanics Lecture 6 90 Horizontal force Vertical force F O H The two horizontal on the element are: The horizontal reaction force H The force on the vertical plane. The resultant horizontal force, H acts as shown in the diagram. Thus we can say: The resultant horizontal force of a fluid below a curved surface is: H v What vertical force would keep this in equilibrium? If the region above the curve were all water there would be equilibrium. Hence: the force exerted by this amount of fluid must equal he resultant force. The resultant vertical force of a fluid below a curved surface is: v Weight of the volume of fluid the curved surface. IVE1400: Fluid Mechanics Lecture 6 91 IVE1400: Fluid Mechanics Lecture 6 9

13 The resultant force and direction of application are calculated in the same way as for fluids above the surface: Example 1: m wide wall as shown in the figure below is fitted with a sluice gate at the base. Find the force on the gate. esultant force.5m nd acts through O at an angle of. The angle the resultant force makes to the horizontal is 1m 1m Force on a submerged surface, pressure at centroid rea of surface IVE1400: Fluid Mechanics Lecture 6 93 IVE1400: Fluid Mechanics Lecture 6 94 Exampel : What would be the force if the gate was changed to a circular one arranged as below?.5m 1m Vertical force weight of (real or imaginary) fluid above the surface. v gvolume above surface v v h 1m ( h + v ) 1/ Horizontal force on curved surface force on projection on to a vertical surface v i.e. equivalent to this figure:.5m h 1m h gh h IVE1400: Fluid Mechanics Lecture 6 95 IVE1400: Fluid Mechanics Lecture 6 96

14 Example 3: curved sluice gate which experiences force from fluid below. 1.5m long cylinder lies as shown in the figure, holding back oil of relative density 0.8. If the cylinder has a mass of 50 kg find a) the reaction at b) the reaction at E IVE1400: Fluid Mechanics Lecture 6 97

An 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

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

oil 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

OUTCOME 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

Fluid 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

Mercury 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

FLUID 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

Experiment 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

2.016 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

01 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

Structural 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

9. 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

MECHANICS 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.

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 1 NON-CONCURRENT COPLANAR FORCE SYSTEMS 1. Be able to determine the effects

Copyright 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

XI / 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

ENGINEERING 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

MECHANICAL PRINCIPLES HNC/D MOMENTS OF AREA. Define and calculate 1st. moments of areas. Define and calculate 2nd moments of areas.

MECHANICAL PRINCIPLES HNC/D MOMENTS OF AREA The concepts of first and second moments of area fundamental to several areas of engineering including solid mechanics and fluid mechanics. Students who are

Open channel flow Basic principle

Open channel flow Basic principle INTRODUCTION Flow in rivers, irrigation canals, drainage ditches and aqueducts are some examples for open channel flow. These flows occur with a free surface and the pressure

Stack Contents. Pressure Vessels: 1. A Vertical Cut Plane. Pressure Filled Cylinder

Pressure Vessels: 1 Stack Contents Longitudinal Stress in Cylinders Hoop Stress in Cylinders Hoop Stress in Spheres Vanishingly Small Element Radial Stress End Conditions 1 2 Pressure Filled Cylinder A

1. Fluids Mechanics and Fluid Properties. 1.1 Objectives of this section. 1.2 Fluids

1. Fluids Mechanics and Fluid Properties What is fluid mechanics? As its name suggests it is the branch of applied mechanics concerned with the statics and dynamics of fluids - both liquids and gases.

Practice Problems on Boundary Layers. Answer(s): D = 107 N D = 152 N. C. Wassgren, Purdue University Page 1 of 17 Last Updated: 2010 Nov 22

BL_01 A thin flat plate 55 by 110 cm is immersed in a 6 m/s stream of SAE 10 oil at 20 C. Compute the total skin friction drag if the stream is parallel to (a) the long side and (b) the short side. D =

du u U 0 U dy y b 0 b

BASIC CONCEPTS/DEFINITIONS OF FLUID MECHANICS (by Marios M. Fyrillas) 1. Density (πυκνότητα) Symbol: 3 Units of measure: kg / m Equation: m ( m mass, V volume) V. Pressure (πίεση) Alternative definition:

SURFACE 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

Chapter 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

CE 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

CE 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

Physics 1A Lecture 10C

Physics 1A Lecture 10C "If you neglect to recharge a battery, it dies. And if you run full speed ahead without stopping for water, you lose momentum to finish the race. --Oprah Winfrey Static Equilibrium

FLUID 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

CHAPTER 29 VOLUMES AND SURFACE AREAS OF COMMON SOLIDS

CHAPTER 9 VOLUMES AND SURFACE AREAS OF COMMON EXERCISE 14 Page 9 SOLIDS 1. Change a volume of 1 00 000 cm to cubic metres. 1m = 10 cm or 1cm = 10 6m 6 Hence, 1 00 000 cm = 1 00 000 10 6m = 1. m. Change

Chapter 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

SURFACE AREA AND VOLUME

SURFACE AREA AND VOLUME In this unit, we will learn to find the surface area and volume of the following threedimensional solids:. Prisms. Pyramids 3. Cylinders 4. Cones It is assumed that the reader has

Physics 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)

Fluids 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.

Lecture 24 - Surface tension, viscous flow, thermodynamics

Lecture 24 - Surface tension, viscous flow, thermodynamics Surface tension, surface energy The atoms at the surface of a solid or liquid are not happy. Their bonding is less ideal than the bonding of atoms

Hydrostatic Force on a Submerged Surface

Experiment 3 Hydrostatic Force on a Submerged Surface Purpose The purpose of this experiment is to experimentally locate the center of pressure of a vertical, submerged, plane surface. The experimental

Solution Derivations for Capa #11

Solution Derivations for Capa #11 1) A horizontal circular platform (M = 128.1 kg, r = 3.11 m) rotates about a frictionless vertical axle. A student (m = 68.3 kg) walks slowly from the rim of the platform

Physics 201 Homework 8

Physics 201 Homework 8 Feb 27, 2013 1. A ceiling fan is turned on and a net torque of 1.8 N-m is applied to the blades. 8.2 rad/s 2 The blades have a total moment of inertia of 0.22 kg-m 2. What is the

Differential 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

Physics 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

a 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

Three 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

Chapter 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

Fluid 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

Geometry and Measurement

The student will be able to: Geometry and Measurement 1. Demonstrate an understanding of the principles of geometry and measurement and operations using measurements Use the US system of measurement for

Reflection and Refraction

Equipment Reflection and Refraction Acrylic block set, plane-concave-convex universal mirror, cork board, cork board stand, pins, flashlight, protractor, ruler, mirror worksheet, rectangular block worksheet,

Homework 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

MECHANICS OF SOLIDS - BEAMS TUTORIAL 1 STRESSES IN BEAMS DUE TO BENDING. On completion of this tutorial you should be able to do the following.

MECHANICS OF SOLIDS - BEAMS TUTOIAL 1 STESSES IN BEAMS DUE TO BENDING This is the first tutorial on bending of beams designed for anyone wishing to study it at a fairly advanced level. You should judge

Chapter 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

Experiment (13): Flow channel

Introduction: An open channel is a duct in which the liquid flows with a free surface exposed to atmospheric pressure. Along the length of the duct, the pressure at the surface is therefore constant and

ME 111: Engineering Drawing

ME 111: Engineering Drawing Lecture # 14 (10/10/2011) Development of Surfaces http://www.iitg.ernet.in/arindam.dey/me111.htm http://www.iitg.ernet.in/rkbc/me111.htm http://shilloi.iitg.ernet.in/~psr/ Indian

Chapter 5: Distributed Forces; Centroids and Centers of Gravity

CE297-FA09-Ch5 Page 1 Wednesday, October 07, 2009 12:39 PM Chapter 5: Distributed Forces; Centroids and Centers of Gravity What are distributed forces? Forces that act on a body per unit length, area or

Figure 1.1 Vector A and Vector F

CHAPTER I VECTOR QUANTITIES Quantities are anything which can be measured, and stated with number. Quantities in physics are divided into two types; scalar and vector quantities. Scalar quantities have

EXAMPLE: Water Flow in a Pipe

EXAMPLE: Water Flow in a Pipe P 1 > P 2 Velocity profile is parabolic (we will learn why it is parabolic later, but since friction comes from walls the shape is intuitive) The pressure drops linearly along

Chapter 13 OPEN-CHANNEL FLOW

Fluid Mechanics: Fundamentals and Applications, 2nd Edition Yunus A. Cengel, John M. Cimbala McGraw-Hill, 2010 Lecture slides by Mehmet Kanoglu Copyright The McGraw-Hill Companies, Inc. Permission required

Lecture 17. Last time we saw that the rotational analog of Newton s 2nd Law is

Lecture 17 Rotational Dynamics Rotational Kinetic Energy Stress and Strain and Springs Cutnell+Johnson: 9.4-9.6, 10.1-10.2 Rotational Dynamics (some more) Last time we saw that the rotational analog of

Algebra Geometry Glossary. 90 angle

lgebra Geometry Glossary 1) acute angle an angle less than 90 acute angle 90 angle 2) acute triangle a triangle where all angles are less than 90 3) adjacent angles angles that share a common leg Example:

MECHANICS OF SOLIDS - BEAMS TUTORIAL TUTORIAL 4 - COMPLEMENTARY SHEAR STRESS

MECHANICS OF SOLIDS - BEAMS TUTORIAL TUTORIAL 4 - COMPLEMENTARY SHEAR STRESS This the fourth and final tutorial on bending of beams. You should judge our progress b completing the self assessment exercises.

Applied Fluid Mechanics

Applied Fluid Mechanics Sixth Edition Robert L. Mott University of Dayton PEARSON Prentkv Pearson Education International CHAPTER 1 THE NATURE OF FLUIDS AND THE STUDY OF FLUID MECHANICS 1.1 The Big Picture

Fundamentals of Fluid Mechanics

Sixth Edition. Fundamentals of Fluid Mechanics International Student Version BRUCE R. MUNSON DONALD F. YOUNG Department of Aerospace Engineering and Engineering Mechanics THEODORE H. OKIISHI Department

Chapter 2. Derivation of the Equations of Open Channel Flow. 2.1 General Considerations

Chapter 2. Derivation of the Equations of Open Channel Flow 2.1 General Considerations Of interest is water flowing in a channel with a free surface, which is usually referred to as open channel flow.

Backwater Rise and Drag Characteristics of Bridge Piers under Subcritical

European Water 36: 7-35, 11. 11 E.W. Publications Backwater Rise and Drag Characteristics of Bridge Piers under Subcritical Flow Conditions C.R. Suribabu *, R.M. Sabarish, R. Narasimhan and A.R. Chandhru

PHYS 211 FINAL FALL 2004 Form A

1. Two boys with masses of 40 kg and 60 kg are holding onto either end of a 10 m long massless pole which is initially at rest and floating in still water. They pull themselves along the pole toward each

Lecture L22-2D Rigid Body Dynamics: Work and Energy

J. Peraire, S. Widnall 6.07 Dynamics Fall 008 Version.0 Lecture L - D Rigid Body Dynamics: Work and Energy In this lecture, we will revisit the principle of work and energy introduced in lecture L-3 for

Head Loss in Pipe Flow ME 123: Mechanical Engineering Laboratory II: Fluids

Head Loss in Pipe Flow ME 123: Mechanical Engineering Laboratory II: Fluids Dr. J. M. Meyers Dr. D. G. Fletcher Dr. Y. Dubief 1. Introduction Last lab you investigated flow loss in a pipe due to the roughness

Midterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m

Midterm Solutions I) A bullet of mass m moving at horizontal velocity v strikes and sticks to the rim of a wheel a solid disc) of mass M, radius R, anchored at its center but free to rotate i) Which of

1. A wire carries 15 A. You form the wire into a single-turn circular loop with magnetic field 80 µ T at the loop center. What is the loop radius?

CHAPTER 3 SOURCES O THE MAGNETC ELD 1. A wire carries 15 A. You form the wire into a single-turn circular loop with magnetic field 8 µ T at the loop center. What is the loop radius? Equation 3-3, with

Unit 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

Ch 2 Properties of Fluids - II. Ideal Fluids. Real Fluids. Viscosity (1) Viscosity (3) Viscosity (2)

Ch 2 Properties of Fluids - II Ideal Fluids 1 Prepared for CEE 3500 CEE Fluid Mechanics by Gilberto E. Urroz, August 2005 2 Ideal fluid: a fluid with no friction Also referred to as an inviscid (zero viscosity)

Chapter 6 Work and Energy

Chapter 6 WORK AND ENERGY PREVIEW Work is the scalar product of the force acting on an object and the displacement through which it acts. When work is done on or by a system, the energy of that system

Floating 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,

Buoyant 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

Chapter 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

So if ω 0 increases 3-fold, the stopping angle increases 3 2 = 9-fold.

Name: MULTIPLE CHOICE: Questions 1-11 are 5 points each. 1. A safety device brings the blade of a power mower from an angular speed of ω 1 to rest in 1.00 revolution. At the same constant angular acceleration,

Chapter 4. Forces and Newton s Laws of Motion. continued

Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting

Geometry Notes PERIMETER AND AREA

Perimeter and Area Page 1 of 57 PERIMETER AND AREA Objectives: After completing this section, you should be able to do the following: Calculate the area of given geometric figures. Calculate the perimeter

Pressure 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

C 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

Lecture 5 Hemodynamics. Description of fluid flow. The equation of continuity

1 Lecture 5 Hemodynamics Description of fluid flow Hydrodynamics is the part of physics, which studies the motion of fluids. It is based on the laws of mechanics. Hemodynamics studies the motion of blood

Gases. 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,

MATHEMATICS FOR ENGINEERING BASIC ALGEBRA

MATHEMATICS FOR ENGINEERING BASIC ALGEBRA TUTORIAL 4 AREAS AND VOLUMES This is the one of a series of basic tutorials in mathematics aimed at beginners or anyone wanting to refresh themselves on fundamentals.

CHAPTER 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

Solving Simultaneous Equations and Matrices

Solving Simultaneous Equations and Matrices The following represents a systematic investigation for the steps used to solve two simultaneous linear equations in two unknowns. The motivation for considering

Lecture 07: Work and Kinetic Energy. Physics 2210 Fall Semester 2014

Lecture 07: Work and Kinetic Energy Physics 2210 Fall Semester 2014 Announcements Schedule next few weeks: 9/08 Unit 3 9/10 Unit 4 9/15 Unit 5 (guest lecturer) 9/17 Unit 6 (guest lecturer) 9/22 Unit 7,

STRESS 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

Vector Algebra II: Scalar and Vector Products

Chapter 2 Vector Algebra II: Scalar and Vector Products We saw in the previous chapter how vector quantities may be added and subtracted. In this chapter we consider the products of vectors and define

Solutions to Homework 10

Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x

Module 7 (Lecture 24 to 28) RETAINING WALLS

Module 7 (Lecture 24 to 28) RETAINING WALLS Topics 24.1 INTRODUCTION 24.2 GRAVITY AND CANTILEVER WALLS 24.3 PROPORTIONING RETAINING WALLS 24.4 APPLICATION OF LATERAL EARTH PRESSURE THEORIES TO DESIGN 24.5

Lecture L2 - Degrees of Freedom and Constraints, Rectilinear Motion

S. Widnall 6.07 Dynamics Fall 009 Version.0 Lecture L - Degrees of Freedom and Constraints, Rectilinear Motion Degrees of Freedom Degrees of freedom refers to the number of independent spatial coordinates

PROBLEM SET. Practice Problems for Exam #1. Math 1352, Fall 2004. Oct. 1, 2004 ANSWERS

PROBLEM SET Practice Problems for Exam # Math 352, Fall 24 Oct., 24 ANSWERS i Problem. vlet R be the region bounded by the curves x = y 2 and y = x. A. Find the volume of the solid generated by revolving

1. The volume of the object below is 186 cm 3. Calculate the Length of x. (a) 3.1 cm (b) 2.5 cm (c) 1.75 cm (d) 1.25 cm

Volume and Surface Area On the provincial exam students will need to use the formulas for volume and surface area of geometric solids to solve problems. These problems will not simply ask, Find the volume

Chapter 11 Equilibrium

11.1 The First Condition of Equilibrium The first condition of equilibrium deals with the forces that cause possible translations of a body. The simplest way to define the translational equilibrium of

Urban Hydraulics. 2.1 Basic Fluid Mechanics

Urban Hydraulics Learning objectives: After completing this section, the student should understand basic concepts of fluid flow and how to analyze conduit flows and free surface flows. They should be able

For 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

= 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.

Open Channel Flow. M. Siavashi. School of Mechanical Engineering Iran University of Science and Technology

M. Siavashi School of Mechanical Engineering Iran University of Science and Technology W ebpage: webpages.iust.ac.ir/msiavashi Email: msiavashi@iust.ac.ir Landline: +98 21 77240391 Fall 2013 Introduction