PROBLEM SET 3 - SOLUTIONS

Similar documents
1.060 Engineering Mechanics II Spring Problem Set 4

Where does Bernoulli's Equation come from?

HOW TO GET A GOOD GRADE ON THE MME 2273B FLUID MECHANICS 1 EXAM. Common mistakes made on the final exam and how to avoid them

Basic Fluid Mechanics

Part A: 1 pts each, 10 pts total, no partial credit.

VARIED FLOW IN OPEN CHANNELS

Chapter (6) Energy Equation and Its Applications

1.060 Engineering Mechanics II Spring Problem Set 8

6.041SC Probabilistic Systems Analysis and Applied Probability, Fall 2013 Transcript Tutorial:A Random Number of Coin Flips

Pressure Head: Pressure head is the height of a column of water that would exert a unit pressure equal to the pressure of the water.

Hydraulics Prof Dr Arup Kumar Sarma Department of Civil Engineering Indian Institute of Technology, Guwahati

Shell/Integral Balances (SIB)

FACULTY OF CHEMICAL & ENERGY ENGINEERING FLUID MECHANICS LABORATORY TITLE OF EXPERIMENT: MINOR LOSSES IN PIPE (E4)

ACCOUNTING FOR FRICTION IN THE BERNOULLI EQUATION FOR FLOW THROUGH PIPES

CEE 3310 Control Volume Analysis, Oct. 7, D Steady State Head Form of the Energy Equation P. P 2g + z h f + h p h s.

FE Fluids Review March 23, 2012 Steve Burian (Civil & Environmental Engineering)

CEE 481 Midterm Exam 1b, Aut 2008

Lesson 6 Review of fundamentals: Fluid flow

MITOCW 6. Standing Waves Part I

Forces and Motion in One Dimension

MOMENTUM PRINCIPLE. Review: Last time, we derived the Reynolds Transport Theorem: Chapter 6. where B is any extensive property (proportional to mass),

CEE 3310 Control Volume Analysis, Oct. 10, = dt. sys

150A Review Session 2/13/2014 Fluid Statics. Pressure acts in all directions, normal to the surrounding surfaces

Hi, I'm Jocelyn, and we're going to go over Fall 2009, Exam 1, problem number 2.

MAE 224 Notes #4a Elements of Thermodynamics and Fluid Mechanics

Math Week 1 notes

A Question about free-body diagrams

4. Find the average velocities and average accelerations of a particle moving in 1-D given its position at various times.

Advanced Hydraulics Prof. Dr. Suresh A. Kartha Department of Civil Engineering Indian Institute of Technology, Guwahati

Putting the Shot. 3. Now describe the forces that you exerted on the weight. Is it constant? Explain why you think so.

Math Week 1 notes

Fluids. Fluids in Motion or Fluid Dynamics

CVE 372 HYDROMECHANICS EXERCISE PROBLEMS

To factor an expression means to write it as a product of factors instead of a sum of terms. The expression 3x

MITOCW watch?v=poho4pztw78

If a stream of uniform velocity flows into a blunt body, the stream lines take a pattern similar to this: Streamlines around a blunt body

Physics 123 Unit #1 Review

OPEN QUIZ WHEN TOLD AT 7:00 PM

BERNOULLI EQUATION. The motion of a fluid is usually extremely complex.

The Impulse-Momentum Principle

Hydraulics Prof. Dr. Arup Kumar Sarma Department of Civil Engineering Indian Institute of Technology, Guwahati

Review for 3 rd Midterm

Basic Thermodynamics Prof. S.K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

MITOCW 5. Traveling Waves without Damping

Student Exploration: Roller Coaster Physics

For example an empty bucket weighs 2.0kg. After 7 seconds of collecting water the bucket weighs 8.0kg, then:

Bernoulli and Pipe Flow

MITOCW MITRES18_005S10_DerivOfSinXCosX_300k_512kb-mp4

of 8 28/11/ :25

Advanced Hydrology Prof. Dr. Ashu Jain Department of Civil Engineering Indian Institute of Technology, Kanpur. Lecture 6

MITOCW 18. Quiz Review From Optional Problem Set 8

Conservation of Momentum using Control Volumes

Homework 6. Solution 1. r ( V jet sin( θ) + ω r) ( ρ Q r) Vjet

CIVE HYDRAULIC ENGINEERING PART I Pierre Julien Colorado State University

Chapter 6 The Impulse-Momentum Principle


MITOCW big_picture_derivatives_512kb-mp4

Free-Body Diagrams: Introduction

Objectives. Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation

vector H. If O is the point about which moments are desired, the angular moment about O is given:

The Inductive Proof Template

Chapter 5: Mass, Bernoulli, and Energy Equations

3.8 The First Law of Thermodynamics and the Energy Equation

Computational Fluid Dynamics Prof. Dr. Suman Chakraborty Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

for the initial position and velocity. Find formulas for v t and x t. b) Assuming x 0 = 0 and v 0 O 0, show that the maximum value of x t is x max

Chapter Four fluid flow mass, energy, Bernoulli and momentum

Chapter 7 The Energy Equation

LECTURE 9. Hydraulic machines III and EM machines 2002 MIT PSDAM LAB

An overview of the Hydraulics of Water Distribution Networks

Most people said that they understand force and acceleration. GOOD!

MITOCW watch?v=ko0vmalkgj8

Reservoir Oscillations with Through Flow

Hydraulic (Piezometric) Grade Lines (HGL) and

MITOCW MITRES18_005S10_DiffEqnsMotion_300k_512kb-mp4

LAB: FORCE AND MOTION

MITOCW R4. Free Body Diagrams

Duct System Losses Are Total Pressure Losses

Physics 4230 Set 2 Solutions Fall 1998

Momentum. Momentum. Momentum Momentum = mass velocity or Momentum = mass speed (when direction is unimportant) Momentum = mv. Impulse.

Experiment (3): Impact of jet

Basics of fluid flow. Types of flow. Fluid Ideal/Real Compressible/Incompressible

LABORATORY V CONSERVATION OF MOMENTUM

Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

Chapter (3) Water Flow in Pipes

the probability of getting either heads or tails must be 1 (excluding the remote possibility of getting it to land on its edge).

Section 2.6 Solving Linear Inequalities

Which one of the following graphs correctly shows the relationship between potential difference (V) and current (I) for a filament lamp?

Solving Linear Inequalities: Introduction and Formatting (page 1 of 7)

MITOCW free_body_diagrams

V/ t = 0 p/ t = 0 ρ/ t = 0. V/ s = 0 p/ s = 0 ρ/ s = 0

COURSE NUMBER: ME 321 Fluid Mechanics I 3 credit hour. Basic Equations in fluid Dynamics

Forces I. Newtons Laws

Classical Mechanics Lecture 4

Exam #2: Fluid Kinematics and Conservation Laws April 13, 2016, 7:00 p.m. 8:40 p.m. in CE 118

Chapter 11 - Fluids in Motion. Sections 7-9

MITOCW watch?v=ybcsmyk5xmg

MITOCW watch?v=pqkyqu11eta

The First Derivative Test

MITOCW ocw-18_02-f07-lec02_220k

Forces. Video Demos. Graphing HW: October 03, 2016

Transcription:

Comments on Problem Set 4 PROBLEM 1: PROBLEM SET 3 - SOLUTIONS - Be careful with the directions of the other forces when you apply momentum conservation in a control volume. Typically, you draw and calculate the forces exerted by the surroundings on the control volume of fluid. At the end, you need calculate the forces exerted by the fluid on the surroundings (in this case on the pipe), which have the same magnitude as the forces on the fluid but opposite direction. Many groups calculated F x and F y without specifying if those were the forces on the fluid or the forces on the pipe, and some groups got confused because of this. - Onto the same issue, the best way to specify the direction of a force is to show it in a sketch. PROBLEM 2: - Remember that we have seen two ways of applying conservation of energy. First, you can apply Bernoulli between two points along a streamline. This is our good old expression: z 1 + p 1 /(ρg) + v 1 2 /(2g) = z 2 + p 2 /(ρg) + v 2 2 /(2g) + (losses between 1 and 2) where z 1, p 1, and v 1 are the elevation, pressure, and velocity at point 1 (same for point 2). We used this many times in the old times (a.k.a. before test 1) when we assumed inviscid flow and didn t have losses. Now, we are dealing with real flows, in which there are usually losses. Furthermore, these real flows are not usually uniform in the cross-section, and we are not usually interested in determining velocities at specific points of the cross-section. Rather, we are interested in determining the cross-sectional averaged properties of the flow, and thus we work with the cross-sectional average velocity, V. For this reason, we have developed the control volume analysis, which deals with cross-sectional averaged quantities. The conservation of energy between the inflow and the outflow of a Control Volume reads z 1,CG + p 1,CG /(ρg) + V 1 2 /(2g) = z 2,CG + p 2,CG /(ρg) + V 2 2 /(2g) + (losses between 1 and 2) It looks similar to our old good Bernoulli, but it is conceptually different! Now we apply conservation of energy between two sections (not two points), which are the inflow and the outflow of our Control Volume. z 1,CG is the elevation of the center of gravity of section 1, p 1,CG the pressure at the center of gravity of section 1, and V 1 the average velocity in section 1 (not the velocity at a particular point). So, in the future, be explicit about which version of Bernoulli you are applying, and between which and which section (or which and which point). NOTE: To apply conservation principles, the control volume must always be chosen so that the flow is wellbehaved both in the inflow and the outflow. For this reason, pressure is hydrostatic at these sections. Therefore, the piezometric head, z 1,CG + p 1,CG /(ρg), is constant in all the points on the inflow section (and same for the outflow), and you can evaluate the sum of these two terms at any point, not necessarily at the center of gravity. PROBLEM 3: - Notice that the reason why we can neglect the headloss in problems 1 and 3 is because in both cases we have a short transition of a converging flow. Since the transition is short, the friction loss is very small. Since the flow is always converging, there is no minor loss.

Problem #4: A few groups didn't have the formula for head loss quite right we must also take into account the elevation difference from one side of the hydraulic jump to the other. The formula can easily be derived from the equation: H_1 = H_2 + HL_(1-->2). See solutions for details. A handful of groups also answered the last part incorrectly. Conceptually, if we lose energy going from 1 --> 2, we'd have to gain energy if we went from 2 --> 1 (by conservation of energy). This would yield a negative head loss and negative energy dissipation (which implies a head gain and energy gain). This situation is physically impossible, so the flow only goes from 1 to 2. See solution for mathematical solutions. Problem #5: Don't ever assume that two graphs are the same simply by looking at a single plot! They may look the same with the scale you've used, but there still may be large errors. Such is the case in this problem for values of y near the bottom (y = 0) where errors reach 25%. Everyone lost a point or two on this. Also, use Excel or another graphing program to make your plots. Handwritten plots are not acceptable (especially when units and other things are missing), especially for something like this where you have to be very accurate. If you don't know how to make graphs in Excel, ask the TAs. It's an important skill that you'll be able to use throughout your MIT life and beyond. Problem #6: Don't forget to explicitly say which way the force is acting.

- PROBLEM No. 2: