Application of the Momentum Equation

Slides:



Advertisements
Similar presentations
A Level Mechanics Bodies in Equilibrium. Since the method of resolving forces can be applied to any of these problems, we’ll use it in the following examples.
Advertisements

Fluid mechanics. Pressure at depth In a cylindrical column of water, as in any cylinder the volume is the height x cross sectional area The density of.
Pharos University ME 259 Fluid Mechanics for Electrical Students Revision for Mid-Term Exam Dr. A. Shibl.
Experiment (3) Impact of jet
Conservation of Linear Momentum.
Review Chap. 5 Applying Newton’s laws
Flow Over Notches and Weirs
Forces applied at an Angle & Inclined Planes
WEDGES AND FRICTIONAL FORCES ON FLAT BELTS
 12-ME-65 Farrukh Zaheer  12-ME-62 Saqib Ali  12-ME-81 M.Umer Nissar  12-ME-80 Waqar Ahmed  12-ME-64 Ashfaq Younis.
Force due to a jet hitting an inclined plane
Forces on Submerged Surfaces in Static Fluids
Forces in Two Dimensions
Forces on Submerged surfaces—plane surfaces Problem consider a plane surface of area A Draw an y and x axis passing through the centroid x y Place surface.
Water Pressure and Pressure Forces
Negative if into control volume Positive if out of control volume In simple unidirectional flow casesIn general Unit normal pointing out from control volume.
Water Pressure and Pressure Force (Revision)
Hinge Statics ? Surface Forces.
Static Equilibrium And Elasticity (Keseimbangan Statik dan Kekenyalan)
Static Surface Forces hinge 8 m water ? 4 m.
Monroe L. Weber-Shirk S chool of Civil and Environmental Engineeringhinge ? Statics Surface Forces 
Forces Acting on a Control Volume Body forces: Act through the entire body of the control volume: gravity, electric, and magnetic forces. Surface forces:
Chapter 4: Forces Newton’s Third Law You cannot touch
1 Example of Groundwater Primer - Yours will be fluid mechanics primer – see homework assignment sheet
Hinge Statics ? Surface Forces.
Fluid mechanics 3.1 – key points
Assignment No. 1 [Grup 8] Figure below shows a portion of a hydraulic circuit. The pressure point B must be 200 psig when the volume flow rate is 60 gal/min.
NEWTON’S FIRST LAW Definition- An object at rest will remain at rest, or if it is moving, it will continue to move with constant velocity, unless acted.
Chapter 6: Momentum Analysis of Flow Systems
Momentum. NEWTON’S LAWS Newton’s laws are relations between motions of bodies and the forces acting on them. –First law: a body at rest remains at rest,
CHAPTER 6 MOMENTUM PRINCIPLE Dr. Ercan Kahya Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson Copyright © 2005 by John Wiley & Sons, Inc.
Example Water at 95°C is flowing at a rate of 2.0 ft3/s through a 60° bend, in which there is a contraction from 4 to 3 inches internal diameter. Compute.
SURVIVAL MODE Quiz 3 –
Force on Floating bodies:
Chapter 4 Changes in Motion Objectives
Unit: IV-Fluid Dynamic
Problems Chapter 4,5.
PTT 204/3 APPLIED FLUID MECHANICS SEM 2 (2012/2013)
Part 2a: Newton and His Laws
Physics. Simple Harmonic Motion - 3 Session Session Objectives.
Pressure & it’s Measurement. Pressure & it’s Measurement  Pressure & Measurement -Pascal’s law -Piezo-meter & Manometer -Atmospheric - Absolute - Gauge.
Water Pressure and Pressure Force (Revision) The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Hydraulics - ECIV 3322.
Applications & Examples of Newton’s Laws. Forces are VECTORS!! Newton’s 2 nd Law: ∑F = ma ∑F = VECTOR SUM of all forces on mass m  Need VECTOR addition.
Chapter 4: Forces. Newton’s Second Law of Motion Example 4-2: Force to accelerate a fast car. Estimate the net force needed to accelerate (a) a 1000-kg.
Friction Friction Problem Situations. Friction Friction F f is a force that resists motion Friction involves objects in contact with each other. Friction.
Chapter 8 - Friction Sections Friction w Frictional forces resist movement of a rigid body over a rough surface. w It is assumed that the.
Forces on Inclined Planes Unit 3, Presentation 3.
A positive point charge is moving directly toward point P. The magnetic field that the point charge produces at point P Q points from the charge.
Elementary Mechanics of Fluids CE 319 F Daene McKinney Momentum Equation.
Objects on Inclined Planes I guess you could call them ramps if you want.
Statika Fluida Section 3. Fluid Dynamics Objectives Introduce concepts necessary to analyse fluids in motion Identify differences between Steady/unsteady.
Physics 215 – Fall 2014Lecture Welcome back to Physics 215 Today’s agenda: More on free-body diagrams and force components Applying Newton’s laws.
Physics Section 4.4 Describe various types of forces Weight is a measure of the gravitational force exerted on an object. It depends upon the objects.
An 7.3-kg object rests on the floor of an elevator which is accelerating downward at a rate of 1.0 m/s 2. What is the magnitude of the force the object.
Transport process In molecular transport processes in general we are concerned with the transfer or movement of a given property or entire by molecular.
Physics Section 4.4 Describe various types of forces Weight is a measure of the gravitational force exerted on an object. It depends upon the objects.
Section 5.3 Section 5.3 Force and Motion in Two Dimensions In this section you will: Section ●Determine the force that produces equilibrium when.
Today: (Ch. 3) Tomorrow: (Ch. 4) Forces and Motion in Two and Three Dimensions Equilibrium and Examples Projectile Motion.
A Presentation on Continuity, Momentum and Energy Equation Enrolment No
Momentum Equation and its Applications
1 Rotational Dynamics The Action of Forces and Torques on Rigid Objects Chapter 9 Lesson 2 (a) Translation (b) Combined translation and rotation.
IMPACT OF JETS PREPARED BY KIRIT S DAYMA ( )
NEWTON'S LAWS OF MOTION Philosophiae Naturalis Principia Mathematica (1686)
Chapter 6: Momentum Analysis of Flow Systems
Pimpri Chinchwad polytechnics
Physics 111: Mechanics Lecture 5
GLOBAL CONSERVATION EQUATIONS
Control volume approach (검사체적 방법)
1. A ball of mass m is suspended from two strings of unequal length as shown above. The magnitudes of the tensions T1 and T2 in the strings must satisfy.
Pascals Law Pascal’s law states that: “The pressure in a confined fluid is transmitted equally to the whole surface of its container”
Presentation transcript:

Application of the Momentum Equation We will consider the following examples: 1. Force due to the flow of fluid round a pipe bend. 2. Force on a nozzle at the outlet of a pipe. 3. Impact of a jet on an inclined plane surface. 4. Force due to flow round a curved vane.

The force due the flow around a pipe bend Consider a pipe bend with a constant cross section lying in the horizontal plane and turning through an angle of θ

Because the fluid changes direction, a force (very large in the case of water supply pipes,) will act in the bend. If the bend is not fixed it will move and eventually break at the joints. We need to know how much force a support (thrust block) must withstand. Step in Analysis: 1. Draw a control volume 2. Decide on co-ordinate axis system 3. Calculate the total force (rate of change of momentum) 4. Calculate the pressure force 5. Calculate the body force 6. Calculate the resultant force

The control volume is draw in the above figure, with faces at the inlet and outlet of the bend and encompassing the pipe walls. It is convenient to choose the co-ordinate axis so that one is pointing in the direction of the inlet velocity. In the above figure the x-axis points in the direction of the inlet velocity.

Calculate the total force: In the x-direction: In the y-direction:

Calculate the pressure force F p = pressure force at 1 - pressure force at 2 Fpx = p1 A1 cos 0 – p2 A2 cos θ = p1 A1 – p2 A2 cos θ Fpy = p1 A1 sin 0 – p2 A2 sin θ = – p2 A2 sin θ Calculate the body force There are no body forces in the x or y directions. The only body force is that exerted by gravity (which acts into the paper in this example - a direction we do not need to consider). Fbx = Fby = 0

Calculate the resultant force FRx = Ftx - Fpx - Fbx FRy = Fty - Fpy - Fby

And The force on the bend is the same magnitude but in the opposite direction R = - Fresultant

Force on a pipe nozzle Force on the nozzle at the outlet of a pipe. Because the fluid is contracted at the nozzle forces are induced in the nozzle. Anything holding the nozzle (e.g. a fireman) must be strong enough to withstand these forces.

The same analysis will be followed as before. Step in Analysis: 1. Draw a control volume 2. Decide on co-ordinate axis system 3. Calculate the total force 4. Calculate the pressure force 5. Calculate the body force 6. Calculate the resultant force

1 & 2 Control volume and Co-ordinate axis are shown in the figure above. 3 Calculate the total force Using the continuity, Q = A u

4 Calculate the pressure force Fp = Fpx = pressure force at 1 - pressure force at 2 Friction is neglected for short distance and the nozzle is horizontal i.e. z1 and z2 are equal and p2 is atmospheric i.e. 0 and with continuity

5 Calculate the body force The only body force is the weight due to gravity in the y-direction - but we need not consider this as the only forces we are considering are in the x-direction. 6 Calculate the resultant force

So the fireman must be able to resist the force of