Course Introduction January 22 – 24, 2008

Today’s Class Course Introduction • First class day items: roll, outline, etc. • Class goals and learning objectives Larry Caretto • Assessment quiz Mechanical Engineering 390 • Discussion of dimensions and units Fluid Mechanics – Physical quantities have dimensions – Several units measure same dimension January 22 and 24, 2008 – Use SI system of units (meter, kilogram, ... – Also use engineering units (feet, pounds ...

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Thursday Class Basic Information • Fluid properties • Larry Caretto, Jacaranda (Engineering) 3333, – Density [email protected], 818.677.6448 – Bulk modulus • Office hours Monday and Wednesday, 4:30 to 5:15 pm; Tuesday and Thursday 2:45 to – Viscosity 3:45 pm; other times by email, phone, drop-in, – Vapor pressure or appointment – Viscosity • http://www.csun.edu/~lcaretto/me390 –Surface Tension • Munson, Young, and Okiisii, Fundamentals of • Start discussion of fluid statics on using Fluid Mechanics (fifth edition), Wiley, 2006. next set of notes

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Email Course Learning Objectives • Campus policy requires students to • Understand the and be able to formu- monitor their CSUN email addresses late and solve problems using basic – These addresses will be used class email fluid properties: density, specific weight, list [email protected] viscosity and mechanical quantities: • Setup your CSUN email account if you pressure, velocity, force and stress have not done so already • solve problems to determine pressures • If desired, forward CSUN email to in static fluids and manometers another address • understand limits of and solve problems with Bernoulli equation

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ME 390 – Fluid Mechanics 1 Course Introduction January 22 – 24, 2008

More Learning Objectives Still More Learning Objectives • understand definition and be able to use • understand the differences between concepts of system and control volume laminar and turbulent flows and be able to determine if a flow is laminar or • use continuity equation to use mass turbulent based on the Reynolds conservation in problem solving number for the flow • solve problems to determine forces in • solve problems in laminar and turbulent moving fluids using control volumes flows in pipes • use dimensionless parameters and apply • be familiar with the basic ideas of the concept of similitude for fluid boundary layers and irrotational flows mechanics experimentation outside these boundary layers 7 8

Learning Objectives Concluded Thermodynamics • solve problems of lift and drag in • Often a prerequisite for fluids, but not external flows presently a prerequisite at CSUN • understand the important variables used to solve problems in open channel and • Students advised to complete ME 370 compressible flows prior to taking ME 390 • solve problems in one of the following • Instead of a 370 review this course will areas (a) compressible flows (b) open use “just-in-time” Thermodynamics channel flows • Cover specific topics as required for course in nature of review

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Class Operation Quizzes • Tuesday: lecture on new material • Twelve during the semester – Review text and notes before class • Based on group work and homework • Thursday: group problem solving – Homework assigned, but not collected or graded • Tuesday: 30-minute quiz at start of • Solutions available on line class followed by new material lecture • Count ten highest quiz grades for final • Starts next week – No makeup quizzes; final quiz grade based – Introduction during first week only on quizzes taken if fewer than ten • First quiz is on Tuesday, February 5 • First few quizzes closed book; remainder will be open book and equation sheet

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ME 390 – Fluid Mechanics 2 Course Introduction January 22 – 24, 2008

Grading See the Course Outline • Quiz grades 45% • Download from course web site • Midterm (March 13) 22% – http://www.csun.edu/~lcaretto/me390 • Contains lecture schedule and homework • Final (May 13) 33% assignments (homework also on web) • Plus/minus grading will be used • Also read information on the following items • Grading criteria in course outline – Class participation and courtesy – Collaboration versus plagiarism: students found • No make-up quizzes or exams cheating receive F grade in course • Students are responsible for changes to outline announced in class

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You cannot Goals for this Course teach people • My goal is to help all students find within themselves sufficient knowledge of fluid anything; you mechanics so that they will all get an A can only help grade in the course • What is your goal for this course? Galileo Galilei them find it within (1564-1642) • What will you do to achieve that goal? themselves.

http://space.about.com/od/astronomyhistory/a/galileoquotes.htm

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How to get your A How to Get your A, Part II • Spend six to ten hours per week outside • Study with fellow students and try to class studying for the course answer each other’s questions • Prepare for lecture and be ready to ask • Do the homework as well as you can questions before reviewing the on-line solutions – Read the assigned reading before class • Contact me by email, telephone or – Download, print, and review the lecture office visits to ask questions presentations before class • Develop a good working relation with • Use these as notes so that you can follow the lecture; write additional notes on these other members of your self-study group presentations 17 18

ME 390 – Fluid Mechanics 3 Course Introduction January 22 – 24, 2008

What I will do to help Preliminary Assessment • Arrive at class a few minutes early to • Designed to help instruction answer any questions you may have • One set of questions on student • Give lectures that stress application of background basics to problem solving • Second set of questions is ungraded quiz • Return quizzes and exams promptly so that you can learn from your errors • Take about 10 minutes for this assessment • Be available for questions by email, • Hand yours in when finished office visits or phone calls – Will call time when most students are done – Send entire class emails as appropriate 19 20

Dimensions and Units Systems of Units • Any physical quantity has a unique • Arbitrary units for fundamental dimension: e.g., mass, length, time, … dimensions, e.g. mass (M), length (L), • Several units may be available for any time (T), and temperature (Θ). dimension • Units for other physical quantities from – Length is measured in meters, feet, miles, the physical relations to quantities with fathoms, furlongs, yards, light-years, etc. fundamental units – You cannot measure length in units with – Velocity dimensions are length/time, L/T the dimension of mass – Acceleration dimensions are length/time2 – Force dimension of (mass)(length)/(time)2

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More Dimensions Still More Dimensions • Pressure = force per unit area • Another energy term 2 = [force] / [length] – Potential energy = mgh = = [(mass) (length) / (time)2] / (length)2 2 -2 -1 (mass)(acceleration)(length) = = (mass) / [(time) (length)] or MT L (mass)(length)2/(time)2 • Common dimensions for energy terms • Power = (energy)/(time) are (mass)(length)2/(time)2 or ML2T-2 = (mass) (length)2 /(time)3 or ML2T-3 – Work = force times distance = (force)(length) • Thermodynamic work is PdV = (mass)(length)2/(time)2 or ML2T-2 – This is like Fdx where P = F/A and dV = – Kinetic energy = mV2/2 Adx (A is area) 2 = (mass)(velocity) 3 = (mass)(length)2/(time)2 or ML2T-2 – PdV dimensions are (length) (force)/(area) which also is (mass)(length)2/(time)2 23 24

ME 390 – Fluid Mechanics 4 Course Introduction January 22 – 24, 2008

SI Units Other Units • Basic definitions for fundamental units • Light intensity and molar units – Mass: kilogram (kg) = international prototype • Units for velocity and acceleration are – Time: second (s) = time for 9 192 631 770 m/s and m/s2 periods of radiation from Cs133 2 – Length: meter (m) = length light travels in • Units for force are kg·m/s 1/299 792 458 of a second – 1 newton (N) = 1 kg·m/s2 – Temperature: kelvin (K) = 1/273.16 of the • Units for energy are kg(m/s)2 = N·m triple point of water – 1 joule (J) = 1 N·m = 1 kg·m2/s2 – Current: ampere (A) defined in terms of electrostatic force

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Still More Units Some Prefixes • Power: (energy)/(time) = joules/second pico, p nano, n micro, μ milli, m – 1 watt (W) = 1 J/s = 1 N·m/s = 1 kg·m2/s3 • Pressure: (force)/(area) = newtons per square meter (1 atm = 101,325 Pa) 10-12 10-9 10-6 10-3 – 1 pascal (Pa) = 1 N/m2 = 1 kg/(m·s2 ) • Note that Sir Isaac Newton has a capital tera, t giga, g mega, M kilo, k N; 1 newton of force does not, unless it is abbreviated as 1 N (true for all units named after individuals) 1012 109 106 103

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Engineering Units Why Use a Pound Force? • Second is the basic unit of time • From the definition of a pound force, the • The foot = 0.3048 m (exactly) is the weight, W = mg, of a pound mass in a basic unit of length standard gravitational field is 1 lbf • Pound is confusing because it can be 2 32.174 ft lb f ⋅ s used to represent two dimensions W = mg = (m lbm ) 2 = m lbf s 32.174lbm ⋅ ft – Mass: pound-mass (lbm = 0.453592 kg) 2 – Force: pound force (lbf = 32.174 lbm·ft/s ) • This is convenient, but the same name • What is SI equivalent for pound force? for two dimensions is confusing and the 1 lbf = 4.4482 N conversion factor is awkward 29 30

ME 390 – Fluid Mechanics 5 Course Introduction January 22 – 24, 2008

Two Engineering Unit Systems More Engineering Units • English engineering units use mass as • foot-pound is work (energy unit) pound mass and force as pound force • British thermal unit (Btu = 778.16 ft-lbf) –1 lb= 32.174 lb ·ft/s2 2 f m • Pressure in lbf/in (psi) – 1 atm = 14.696 2 • British gravitational (BG) system uses psi = (144)(14.696) lbf/ft (psf) slug as the mass unit • Horsepower as power unit –1 lb= 1 slug·ft/s2 6 f – 1 hp·hr = 2,545 Btu = 1.98x10 ft·lbf • Which mass is larger, slug or lbm? – 1 kW·hr = 3,412 Btu What is their conversion factor? • The metric unit, calorie = 1/252 Btu 2 –1 lbf·s /ft = 32.174 lbm = 1 slug • The food Calorie is a kilocalorie 31 32

Calculating Units Calculating Units II

• What is kinetic energy of a 100 lbm • What is kinetic energy of a 3 slug mass mass moving at 10 ft/s moving at 10 ft/s 2 2 2 2 2 -2 •mV/2 = (100 lbm)(10ft/s) /2 = •mV/2 = (3 slugs)(10ft/s) /2=15 slug·ft ·s 5000 lbm·ft2·s-2 • Unit conversion • Unit conversion 2 2 (3 slugs) ⎛10 ft ⎞ lb f ⋅ s 2 lb ⋅ s2 KE = ⎜ ⎟ =150 ft ⋅lb f (100 lbm ) ⎛10 ft ⎞ f 2 ⎝ s ⎠ 1slug ⋅ ft KE = ⎜ ⎟ =165.4 ft ⋅lbf 2 ⎝ s ⎠ 32.174lbm ⋅ ft • Note algebraic cancellation of units • Note algebraic cancellation of units – Especially simple if numeric value is one!

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Units quiz Another Quiz • What is the change in potential energy • Some European engineering when a mass of 20 slugs is raised a calculations use the kilogram-force, distance of 15 ft? defined in the same way as the pound 2 • Do you need more data to answer this force and measure pressure in kgf/cm

question? • What exactly is the definition of a kgf? 2 • What is g? Use 5 ft/s for this problem • How many newtons are in a kgf? 2 • How many pascals are in a kg /cm2? 5 ft lb f ⋅ s f PE = mgh = (20 slug) ()15 ft =1500 ft ⋅lb f s2 1slug ⋅ ft

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ME 390 – Fluid Mechanics 6 Course Introduction January 22 – 24, 2008

Solutions to Another Quiz A Few Other Units

•One kgf is the force required to • Volume is sometimes measured in liters accelerate 1 kg at an acceleration of (or litres), L, where 1 L = 1000 cm3 = standard gravity, g = 9.80665 m/s2 0.001 m3 9.80665 m 1 N ⋅ s2 • Gallons, gal, is another volume 1 kg f =1 kg = 9.80665 N measure; 7.4805 gal = 1 ft3 s2 kg ⋅m • Speed is sometimes measured in miles 2 2 kg f kg f 9.80665 N ⎛100 cm ⎞ Pa ⋅m 1 =1 = 98066.5 Pa per hour, mph; 30 mph = 44 ft/s and 2 2 ⎜ ⎟ cm cm kg f ⎝ m ⎠ 1 N 1 mph = 0.44704 m/s kg • 1 hogshead = 63 gallons 1 f ≈105 Pa ≈1 atm cm2 37 38

Working With Units Temperature Units • Carrying units in the calculation is a • SI unit: absolute temperature in K good approach for correct results • Degrees Celsius, oC = K – 273.15 • If you do not want to do that, here are • Degrees Fahrenheit, oF = 1.8(oC) + 32 some hints for correct unit results • Rankine, R = oF + 459.67 is absolute – In the BG system convert all lengths to feet temperature for Fahrenheit scale 2 time to seconds, and pressures to lbf/ft 2 • T(R) = 1.8 T(K) (psf); 1 lbf = 1 slug·ft/s – In the SI system always use m, Pa and N – What is a ΔT of 25oF in Rankine? 25 R (instead of mm, cm, kPa, kN, etc.; 1 N = 1 – What is 15oC in Rankine? kg·m/s2) •15oC = 288.15 K = 59oF = 518.67 R 39 40

Density and Related Properties Density and Related Properties II • Density, ρ, is mass per unit volume (ρ = • Specific gravity, SG, of a substance: 1/v, where v is specific volume used – ratio of the substance density to the more commonly in thermodynamics) density of a reference substance at a specified temperature – Units for density are (SI) kg/m3, (EE) 3 3 • Reference substance is usually water for liquids lbm/ft , and (BG) slug/ft and air for gases • Specific weight, γ = ρg, typically • Water reference temperature: 4oC (39.4oF) where ρ = 1000 kg/m3 = 1.94 slugs/ft3 tabulated at standard gravity, g = water 9.80665 m/s2 = 32.174 ft/s2, in N/m3 for • The specific gravity of mercury at 68oF 3 is 13.56 (relative to water at 39.4oF). SI and lbf/ft for both EE and BG What is its density at this temperature? 41 4 3 3 42 ρHg = 1.356x10 kg/m = 26.3 slugs/ft

ME 390 – Fluid Mechanics 7 Course Introduction January 22 – 24, 2008

Density and Related Summary States of Matter (Phases) • Density: ρ = mass per unit volume with • Triple point: units of kg/m3 or slugs/ft3 unique point for • Specific weight: γ = ρg with units of each substance N/m3 or lb /ft3 (varies with local g) where solid, f Liquid Critical liquid and vapor • Specific gravity: SG = ρ/ρref = γ/γref Point o 3 coexist – Liquid ρref: water at 4 C with ρ = 1000 kg/m Solid and γ = 9806.65 N/m3 or water at 60oF with Pressure • No liquid-gas ρ = 1.94 slugs/ft3 and γ = 62.4 lb /ft3 f Triple Point Gas transition above o o –Gas ρref: air at 15 C (59 F) with ρ = 1.23 critical point kg/m3 = 0.00238 slugs/ft3 and γ = 62.4 lb /ft3 f Temperature 43 44

Transitions Between Phases Vapor Pressure •Forphase • Pressure exerted by a liquid transitions in equilibrium with a vapor pressure and • Value depends on the nature temperature are of the liquid and temperature Liquid e n li related Melting line Melting • For water the vapor pressure g in • Vapor pressure is o il at 100 C is 101.325 kPa Pressure o B Gas the pressure at • Vapor pressure is pressure

Solid which liquid-vapor at which liquids change to Sublimation curve transition occurs gas at constant temperature Temperature 45 Figure on page 23 Fundamentals of Fluid Mechanics, 5/E by 46 Bruce Munson, Donald Young, and Theodore Okiishi,Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

Volume Change Compressibility • Changing the pressure on a • Isothermal bulk modulus fluid can change its volume ⎛ ∂P ⎞ ⎛ ∂P ⎞ E = − V = ρ⎜ ⎟ – The volume change can be T ⎜ ⎟ ⎜ ⎟ ⎝ ∂V ⎠T ⎝ ∂ρ ⎠T done in different ways • In an isothermal process the • Isentropic bulk modulus temperature is constant – No heat added to fluid • An isentropic (constant entropy) ⎛ ∂P ⎞ ⎛ ∂P ⎞ process is one in which no heat Es = − V ⎜ ⎟ = ρ⎜ ⎟ is added to the fluid and there is ⎝ ∂V ⎠s ⎝ ∂ρ ⎠s no friction; this ideal process is approached for short times •ET ≈ Es for liquids

Figure on page 20Fundamentals of Fluid Mechanics, 5/E by 47 Figure on page 20Fundamentals of Fluid Mechanics, 5/E by 48 Bruce Munson, Donald Young, and Theodore Okiishi,Copyright © Bruce Munson, Donald Young, and Theodore Okiishi,Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. 2005 by John Wiley & Sons, Inc. All rights reserved.

ME 390 – Fluid Mechanics 8 Course Introduction January 22 – 24, 2008

Ideal Gases Ideal Gases II • From chemistry: PV = nRT (V is volume) • Isothermal: P = ρRT – n = m / M is the number of moles ⎛ ∂P ⎞ 1 P • for mass in kg n is in kilogram moles (kmol); for ET = ρ⎜ ⎟ = ρ = ρ = P ⎝ ∂ρ ⎠T RT ρ mass in lbm, n is in pound moles (lbmol) k – R = 8.31447 kJ/kmol·K = 10.7316 psia·ft3 / • Isentropic: P/ρ = C lbmol·R is universal gas constant ⎛ ∂P ⎞ k−1 k –R = R/M is engineering gas constant that is Es = ρ⎜ ⎟ = ρCkρ = kCρ = kP ⎝ ∂ρ ⎠ different for each gas s – Real gases like ideal gases at low pressures • k is heat capacity ratio, a –P = nRT / V = (m/M)RT / V = (m/V)(R/M)T gas property, = 1.4 for air 49 Figure on page 20Fundamentals of Fluid Mechanics, 5/E by 50 P = ρRT Bruce Munson, Donald Young, and Theodore Okiishi,Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

Introduction to Viscosity ∂u τ = μ ∂y τ is shear stress on plate τ is (opposite sign from Velocity gradient, du/dy (∂u/∂y) shear shear stress on fluid) stress on fluid

Figure 1.2 (p. 13) Figure 1.3 (p. 14) (a) Deformation of material placed between two Behavior of a fluid placed between two parallel parallel plates. (b) Forces acting on upper plate. plates (top one moving, bottom stationary.) Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi,Copyright © 2005 by John Wiley & 51 Young, and Theodore Okiishi,Copyright © 2005 by John Wiley & 52 Sons, Inc. All rights reserved. Sons, Inc. All rights reserved.

Viscosity Viscosity II How does μ Newtonian for water Newtonian and Fluids change with non-Newtonian have a linear temperature? Variation of shear variation of stress with rate of shearing strain for shearing stress ∂u with rate of τ = μ several types of ∂y fluids, including shearing strain ∂u – slope is common non- τ = μ viscosity Newtonian fluids. ∂y Figure 1.4 (p. Figure 1.5 (p. 16)

15) Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi,Copyright © 2005 by John Wiley & 53 Young, and Theodore Okiishi,Copyright © 2005 by John Wiley & 54 Sons, Inc. All rights reserved. Sons, Inc. All rights reserved.

ME 390 – Fluid Mechanics 9 Course Introduction January 22 – 24, 2008

What is τ for Velocity Profile? Viscosity Dimensions • What are viscosity dimensions? (τ = μ∂u/∂y) • What are dimensions of other variables? – τ has dimensions of FL-2 = (MLT-2)L-2 = ML-1T-2 – ∂u / ∂y has dimensions of (L/T) / L = T-1 – μ has dimensions of τ / (∂u/∂y) = FL-2 / T-1 – What are μ dimensions in terms of mass? • Dimensions of μ are FTL-2 = ML-1T-1 2 2 – SI units: N·s/m = kg/m·s; BG units: lbf·s/ft = Figure E1.5 (p. 19) 2 slug/ft·s; EE units: lbf·s/ft = 32.174 lbm/ft·s Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald 55 56 Young, and Theodore Okiishi,Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

Viscosity III Surface Tension • Forces generated • Viscosity of at liquid-gas or gases increases liquid-liquid with temperature interfaces • Viscosity of • Surface tension, σ, liquids a fluid property, decreases with Figure 1.6 (p. 17) with dimensions temperature F/L

Figure 1.7 Forces acting on one-half of a liquid drop. Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald 57 Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald 58 Young, and Theodore Okiishi,Copyright © 2005 by John Wiley & Young, and Theodore Okiishi,Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Sons, Inc. All rights reserved.

Surface Tension Effects Surface Tension Effects II

• Vertical force balance: γπR2h = 2πRσcosθ Figure 1.8 Capillary action in small tubes. (a) Rise – Surface tension depends on fluid and of column for a liquid that wets the tube. (b) Free- temperature, wetting angle, θ, depends on fluid body diagram for calculating column height. (c) and surface 2σcosθ h = Depression of column for a nonwetting liquid. γR Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald 59 Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald 60 Young, and Theodore Okiishi,Copyright © 2005 by John Wiley & Young, and Theodore Okiishi,Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Sons, Inc. All rights reserved.

ME 390 – Fluid Mechanics 10 Course Introduction January 22 – 24, 2008

Surface Tension Problem Tube Diameter vs. Capillary Rise • Find the capillary rise for water at 60oF 3 2σcosθ (γ = 62.4 lbf/ft , σ = 0.00503 lbf/ft) in a h = circular tube with a diameter of 0.5 in? γR – For water in clean glass, θ = 0o Water at 20oC 0.00503 lb 144 in2 2 f ()cos0 2σcosθ ft ft 2 h = = = 0.093 in γR 62.4 lb 0.5 f in 0.5 in = 12.7 mm gives h = 0.093 in = 2.36 mm ft3 2

61 Figure E1.8, Fundamentals of Fluid Mechanics, 5/E by Bruce 62 Munson, Donald Young, and Theodore Okiishi,Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

Typical Units Quantity SI units EE units BG units

3 3 3 Density kg/m lbm/ft slug/ft 2 Pressure & kPa = 1 psi = 1 lbf/in = shear stress kN/m2 2 144 psf = 144 lbf/ft Velocity m/s ft/s 2 2 2 Viscosity N·s/m = lbf·s/ft = lbf·s/ft = kg/m·s 32.2 lbm/ft·s slug/ft·s 3 3 Specific N/m lbf/ft weight = ρg Tabulated values at standard gravity

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ME 390 – Fluid Mechanics 11