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### S08_chap1_web

Course: ECE 309, Fall 2009
School: W. Alabama
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Word Count: 776

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Basic Concepts of Thermodynamics Reading 2-1 2-8 Problems 2-44, 2-48, 2-78, 2-100 Thermal Sciences The thermal sciences involve the storage, transfer and conversion of energy. We will study the basic laws and principles that govern the three disciplines of thermal sciences, namely, thermodynamics, heat transfer and uid mechanics. Thermodynamics Conservation of mass Conservation of energy Second law of...

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Basic Concepts of Thermodynamics Reading 2-1 2-8 Problems 2-44, 2-48, 2-78, 2-100 Thermal Sciences The thermal sciences involve the storage, transfer and conversion of energy. We will study the basic laws and principles that govern the three disciplines of thermal sciences, namely, thermodynamics, heat transfer and uid mechanics. Thermodynamics Conservation of mass Conservation of energy Second law of thermodynamics Properties Heat Transfer Conduction Convection Radiation Conjugate yn am ics He at mo d a Tr ns Thermal Systems Engineering Th er fer Fluids Mechanics Fluid Mechanics Fluid statics Conservation of momentum Mechanical energy equation Modeling 1 Thermodynamic Systems Isolated Boundary Surroundings W ork He at System Surroundings - everything that interacts with the system System Boundary (real or imaginary fixed or deformable) System - may be as simple as a melting ice cube - or as complex as a nuclear power plant SYSTEM: any speci ed collection of matter under study. thermodynamic systems are classi ed as either open or closed Closed System: composed of a control (or xed) mass where heat and work can cross the boundary but no mass crosses the boundary. Open System: composed of a control volume (or region in space) where heat, work, and mass can cross the boundary or the control surface WORK &amp; HEAT TRANSFER: work and heat transfer are NOT properties they are the forms that energy takes to cross the system boundary 2 <a href="/keyword/thermodynamic-properties/" >thermodynamic properties</a> of Systems Basic De nitions Thermodynamic Property: Any observable or measurable characteristic of a system. Any mathematical combination of the measurable characteristics of a system Intensive Properties Pressure Temperature P T Extensive Properties N/m2 or P a K Volume Mass V m3 m kg Other Properties (Intensive and Extensive) V /m = v = 1/v = m/V E/m = e U/m = u H/m = h = u + P v S/m = s Cp, Cv R = Cp Cv m3 /kg kg/m3 J/kg J/kg J/kg J/kg K J/kg K J/kg K Speci c VolumeA Density Total EnergyB Internal EnergyB EnthalpyB EntropyB Speci c Heat Ideal Gas Constant Notes: A The term speci c denotes that the property is independent of mass. There are two exceptions: 1) speci c gravity (or relative density) is given where: s = / H2 O and H2 O is 1000 kg/m3 . 2) speci c weight is weight per unit volume (i.e. W/V = mg/V = g = ) B The given internal energy, enthalpy and entropy are speci c properties. The term speci c is dropped. 3 Pressure P ressure = F orce Area ; N m2 Pa in uids, this is pressure (normal component of force per unit area) in solids, this is stress 1 kP a = 103 P a 1 bar = 105 P a : : 1 M P a = 106 P a 1 atm = 101, 325 P a Pressure gauge pressure Patm vacuum pressure absolute vacuum pressure ABSOLUTE ATMOSPHERIC PRESSURE absolute pressure Manometer a device that measures pressure using a column of liquid 4 the uid pressure at any point, 1 in a manometer is given as P1 = Patm + gh Bourdon Tube a device that measures pressure using mechanical deformation Pressure Transducers devices that use piezoelectrics to measure pressure Barometer device that measures atmospheric pressure 5 Temperature temperature is a pointer for the direction of energy transfer as heat TA &gt; TB TA Q TB TA &lt; TB TA Q TB It follows that temperature is a property two systems have in common when they are in thermal equilibrium with each other Thermal Equilibrium: TA = TB and Q=0 0th Law of Thermodynamics: if system C is in thermal equilibrium with system A, and also with system B, then TA = TB = TC two bodies are in thermal equilibrium if they have the same temperature, even if they are not in contact with one another State and Equilibrium the state of a system is its condition as described by a set of relevant energy related properties. thermodynamic equilibrium refers to a condition of equilibrium with...

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W. Alabama - ECE - 309
Chapter 1: Basic Concepts of ThermodynamicsEvery science has its own unique vocabulary associated with it. Precise definition of basic concepts forms a sound foundation for development of a science and prevents possible misunderstandings. Careful st
W. Alabama - ECE - 309
ECE309 INTRODUCTION TO THERMODYNAMICS &amp; HEAT TRANSFER 10 August 2005 Final Examination R. Culham &amp; M. Bahrami This is a 2 - 1/2 hour, closed-book examination. You are permitted to use one 8.5 in. 11 in. crib sheet (both sides), Conversion Factors
W. Alabama - ECE - 309
ECE309 INTRODUCTION TO THERMODYNAMICS &amp; HEAT TRANSFER 3 August 2004 Final Examination R. Culham This is a 3 hour, closed-book examination. You are permitted to use one 8.5 in. 11 in. crib sheet (both sides), Conversion Factors (inside cover of tex
W. Alabama - ECE - 309
Properties of Pure SubstancesReading 4-1 4-7 Problems 4-31, 4-43, 4-58, 4-61, 4-64, 4-79, 4-80, 4-88Phases of Pure Substances a pure substance may exist in different phases, where a phase is considered to be a physically uniform form of the matt
W. Alabama - ECE - 309
Conduction Heat TransferReading 10-1 10-6 11-1 11-2 Problems 10-38, 10-48, 10-57, 10-70, 10-71, 10-78, 10-92 10-117, 10-121, 10-153 11-14, 11-19, 11-39, 11-45, 11-53, 11-91Fourier Law of Heat Conductionx=L x+Dx x x=0 Qx A x Qx+Dxinsulatedg
W. Alabama - ECE - 309
Example: Transient Conduction An aluminum plate made of Al 2024-T6 with a thickness of 0.15 m is initially at a temperature of 300 K. It is then place in a furnace at 800 K with a convection coefcient of 500 W/m2 K. Find: i) ii) the time (s) for the
W. Alabama - ECE - 309
ECE-309 Heat Transfer &amp; Thermodynamics Spring 2005 Tutorial # 12 Radiation heat transfer-exchanged between surfaces Problem 1: A furnace cavity, which is in the form of a cylinder of 75-mm diameter and 150-mm length, is open at one end to large surro
W. Alabama - ECE - 309
ECE309Introduction to Thermodynamics and Heat TransferTutorial 1Basic Concepts of Thermodynamics 1-51 The basic barometer can be used as an altitude-measuring device inairplanes. The ground control reports a barometric reading of 753mmHg while
Allan Hancock College - DODAB - 2002259
THE LEGISLATIVE ASSEMBLY FOR THE AUSTRALIAN CAPITAL TERRITORY DRUGS OF DEPENDENCE AMENDMENT BILL 2002 A BILL TO AMEND THE DRUGS OF DEPENDENCE ACT 1989 EXPLANATORY MEMORANDUM Circ
W. Alabama - ECE - 309
ECE309 Introduction to Thermodynamics and Heat TransferSpring 2005Tutorial #3First Law of Thermodynamics: Closed SystemsProblem 3-73 A 0.3-m3 tank contains oxygen initially at 100kPa and 27C. A paddlewheel within the tank is rotated until th
W. Alabama - ECE - 309
Thermodynamics and Heat transfer ECE 309 Tutorial 5# The Second law of Thermodynamics Problem 1 A gas turbine has an efficiency of 17 percent and develops apower output of 6000 kW . Determine the fuel consumption rate of this gas turbine, in L / min
W. Alabama - ECE - 309
ECE 309 Heat Transfer &amp; Thermodynamics Spring 2005 Tutorial # 8 Transient Heat Conduction _ Problem 1: A long aluminium cylinder 50 mm in diameter [k=215 W/(mC), = 2700 kg/m3 , Cp = 0.9 kJ/(kgC), and = 8.4 10-5 m2/s] and initially at 200C is sudde
Allan Hancock College - DODAB - 2002259
THE LEGISLATIVE ASSEMBLY FOR THE AUSTRALIAN CAPITAL TERRITORYDRUGS OF DEPENDENCE AMENDMENT BILL 2002A BILL TO AMEND THE DRUGS OF DEPENDENCE ACT 1989EXPLANATORY MEMORANDUMCirculated by authority of the Minister for Health Jon Stanhope MLA 2002
W. Alabama - ECE - 309
ECE 309 Introduction to Thermodynamics and Heat Transfer Spring 2005Tutorial # 2 Properties of Pure SubstancesProblem 1 Complete the following table for substance water: P [kPa] 500 500 1400 T [oC] 20 200 300 v [m3/kg] 0.20 0.8 xa. b. c. d.Sol
W. Alabama - ECE - 309
ECE309 Introduction to Thermodynamics and Heat TransferSpring 2005Tutorial # 11Radiation Heat TransferProblem 1 Two parallel disks of diameter D=0.6m separated by L=0.4m are locateddirectly on top of each other. Both disks are black, and are
W. Alabama - ECE - 309
ECE309 Introduction to Thermodynamics and Heat TransferSpring 2005Tutorial # 7Steady State ConductionProblem 1Consider a naked person standing in a room at 20C with an exposed surface area of 1.7m2. The deep body temperature of the human body
W. Alabama - ECE - 309
ECE 309 Introduction to Thermodynamics and Heat Transfer Spring 2005Tutorial # 6 Entropy Problem 1 Air is compressed steadily by a 5-kW compressor from 100 kPa and 17oC to 600 kPa and 167oC at a rate of 1.6 kg/min. During this process, some heat tra
W. Alabama - ECE - 309
ECE 309 Introduction to Thermodynamics and Heat Transfer Spring 2005Tutorial # 10 Natural ConvectionProblem Consider a 15 cm 20 cm printed circuit board (PCB) that has electronic components on one side. The board is placed in a room at 20oC. The
W. Alabama - ECE - 309
Thermodynamics and Heat Transfer ECE 309 Tutorial # 4 First Law of Thermodynamics: Control Volumes _ Problem 1: Air enters an adiabatic nozzle steadily at 300 kPa, 200C, and 30 m/s and leaves at 100 kPa and 180 m/s. The inlet area of the nozzle is 80
W. Alabama - ECE - 309
Forced ConvectionECE309 Introduction to Thermodynamics and Heat TransferTutorial 9QuestionA truck, pulling a refrigerated trailer, travels at 105 km/hr. Solar radiation heats the roof of the truck, with the net radiatvie ux being4 qrad = 375
W. Alabama - ECE - 309
Chapter 10: Steady Heat ConductionQuestion 10-38 10-48 10-57 10-70 10-71 10-78 10-92 10-117 10-121 10-153 0.8%, 99.2%,Answers 29.9 W/m C (b) 6.4 C (b) 406 W (b) 17.9 C (b) 16,730 kg(a) 142.4 W,(a) 6.588 C/W, (a) 15,957 W, (a) 64,600 W, 59.
W. Alabama - ECE - 309
Chapter 15: Radiation Heat TransferQuestion 15-27 15-33 15-50 15-58 15-77 15-79 15-86 15-106 15-107Answers (a) 2.184 105 kW, 0.575, 0.119, 32.6 kW/m2 0.118 (b) 55.8 kW [2( s2 + D2 s]/(D) 543 K 747 K 1245 W, 631 K, 0.44, 725 W 751.6 W, 54.4 kW
W. Alabama - ECE - 309
Chapter 11: Transient Heat ConductionQuestion 11-14 11-19 11-39 11-45 11-53 10-91 27.8 s 4 min 445 C, 46.2 min 11.2 C, 482 s,Answers448 C2.7 C,17.2 kJ513 s (chart)
W. Alabama - ECE - 309
Chapter 5: The First Law: Closed SystemsQuestion 3-40 3-49 5-9 5-10 5-25 5-29 5-35 5-42 5-61 (a) 98.1 kW, 4.578 kW, 22.16 kJ, 364.6 K, 37.18 kJ,Answers (b) 188.1 kW, (c) -21.9 kW3.29 m2 36.79 kJ, 1.64 kJ -34.86 kJ, -6.97 kJ, -4.65 kJ 151.8 C
W. Alabama - ECE - 309
ECE309 INTRODUCTION TO THERMODYNAMICS &amp; HEAT TRANSFER MIDTERM EXAMINATION June 4, 2008 5:30 pm - 6:45 pm Instructor: R. CulhamInstructions This is a 75 minute examination. The only aides allowed are the course textbook, Introduction to Thermodynam
W. Alabama - ECE - 309
ECE309 INTRODUCTION TO THERMODYNAMICS &amp; HEAT TRANSFER MIDTERM EXAMINATION June 4, 2008 5:30 pm - 6:45 pm Instructor: R. CulhamInstructions This is a 75 minute examination. The only aides allowed are the course textbook, Introduction to Thermodynam
W. Alabama - ECE - 309
Convection Heat TransferReading 12-1 12-8 13-1 13-6 14-1 14-4 Problems 12-41, 12-46, 12-53, 12-57, 12-76, 12-81 13-39, 13-47, 13-59 14-24, 14-29, 14-47, 14-60Introduction convection heat transfer is the transport mechanism made possible throug
W. Alabama - ECE - 309
Chapter 5: The Second Law of ThermodynamicsThe second law of thermodynamics asserts that processes occur in a certain direction and that the energy has quality as well as quantity. The first law places no restriction on the direction of a process, a
W. Alabama - ECE - 309
Chapter 12: Radiation Heat TransferRadiation differs from Conduction and Convection heat t transfer mechanisms, in the sense that it does not require the presence of a material medium to occur. Energy transfer by radiation occurs at the speed of lig
W. Alabama - ECE - 309
Chapter 3: The First Law of Thermodynamics: Closed SystemsThe first law of thermodynamics can be simply stated as follows: during an interaction between a system and its surroundings, the amount of energy gained by the system must be exactly equal t
W. Alabama - ECE - 309
Chapter 8: Steady Heat ConductionIn thermodynamics, we considered the amount of heat transfer as a system undergoes a process from one equilibrium state to another. Thermodynamics gives no indication of how long the process takes. In heat transfer,
W. Alabama - ECE - 309
ECE309 INTRODUCTION TO THERMODYNAMICS &amp; HEAT TRANSFER MIDTERM EXAMINATION #2 July 2, 2008 5:30 pm - 6:45 pm Instructor: R. CulhamInstructions This is a 75 minute examination. The only aides allowed are the course textbook, Introduction to Thermody
W. Alabama - ECE - 309
ECE309 INTRODUCTION TO THERMODYNAMICS &amp; HEAT TRANSFER 16 June 2004 Midterm Examination R. Culham This is a 90 minute, closed-book examination. You are permitted to use one 8.5 in. 11 in. crib sheet (one side only), Conversion Factors (inside cover
W. Alabama - ECE - 309
ECE309 INTRODUCTION TO THERMODYNAMICS &amp; HEAT TRANSFER 20 June 2005 Midterm Examination R. Culham &amp; M. Bahrami This is a 90 minute, closed-book examination. You are permitted to use one 8.5 in. 11 in. crib sheet (one side only), Conversion Factors
W. Alabama - ECE - 309
Radiation Heat TransferReading 21-1 21-6 22-1 21-5 Problems 21-21, 21-24, 21-41, 21-61, 21-69 22-11, 22-17, 22-26, 22-36, 22-71, 22-72IntroductionIt should be readily apparent that radiation heat transfer calculations required several additiona
W. Alabama - MATH - 239
Faculty of MathematicsUniversity of WaterlooThe Second Year Tutorial Centre - MC 4067Students Helping StudentsIn the Fall Term 2007, the second year Tutorial Centre will provide a tutorial service for the courses MATH 235, MATH 237 and MATH 239
W. Alabama - MATH - 239
Math 239 Assignment 1DUE: NOON Friday 21 September 2007 in the drop boxes opposite the Math Tutorial Centre MC 4067 (except St. Jeromes section, drop box in St. Jeromes). 1. Give two proofs, one combinatorial and one using the formula for binomial c
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#11.5University of WaterlooSTAT 231 W. H. CherryFigure 11.5. OBSERV TIONAL PLANS: Physical Activity and Health AEM9005: Cambridge Report, February 24, 1990, page 7ATheres proof walking reduces risk of deathFITNESS AND YOUIts hard sometime
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#10.4University of WaterlooSTAT 231 W. H. CherryFigure 10.4. EXPERIMENTAL PLANS: Cigarette Advertising and Tobacco ConsumptionThe article reprinted below is of interest because it illustrates how limited blocking and replicating and the absen
W. Alabama - STAT - 231
Head injury (Ch. 10) University of Waterloo EM0204: The Globe and Mail, January 21, 2002, page A5 STAT 231 - W. H. CherryHead injuries linked to brain diseases _Some sports elevate risk of Alzheimers, U.S. scientists warn BY ANDRE PICARDPUBLIC H
W. Alabama - STAT - 231
#10.3University of WaterlooSTAT 231 W. H. CherryFigure 10.3. EXPERIMENTAL PLANS: Oat Bran and Serum Cholesterol LevelsEM9001: The Globe and Mail, January 18, 1990, page A12Boston researchers chew up oat bran as cholesterol ghterReuter __
W. Alabama - STAT - 231
#6.1University of WaterlooSTAT 231 - W. H. Cher ryFigure 6.1. MEASURING: Investigating Coin WeightsThe problem of getting good measurements and nding ways to describe them concisely has been our chief concern in the preceding chapters. Only in
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CO 466/666: Continuous Optimization Winter 2007 Problem Set 8 S. Vavasis Handed out: 2007-Mar-27. Due: 2007-Apr-3, in lecture. 1. (Textbook exercise 17.9) Consider the box-constrained optimization problem min{f (x) : l x u} where l Rn and u Rn ar
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UNIVERSITY OF WATERLOO WATERLOO ONTARIOTEST 1 C&amp;O 22010:00 a.m.- 11:20am Instructor: P. Haxell INSTRUCTIONS 1. Fill in the following: January 31, 2006Name:I.D.#:2. Non-programmable calculators only are allowed. 3. A complete paper has 6 page
W. Alabama - CO - 466
CO 466/666: Continuous Optimization Winter 2008 Problem Set 4 S. Vavasis Handed out: 2008-Feb-11. Due: 2008-Feb-25 in lecture.1. Consider the trust region subproblem in which the linear term is absent: min m(p) = pT Bp/2 s.t. p . Give a simple cha
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Math 117 A Fall 2001Final ExamName: ID Number:Part A[3] [4] 1. 2. Given f (x ) = Sin 1(e x ), find f (x ) . If g( x ) = cosh(ln (x ) , evaluate g (1) DO NOT leave your answer in terms of hyperbolic trig functions. [3] [4] 3. 4. Evaluate, if po
W. Alabama - ACTS - 331
Actsc 331 - Life Contingencies 1 Time: Class 2:30-3:20 MWF Tutorial 4:30-5:20 W Room: RCH 112 Instructor: Gord Willmot Office: MC 6026B Extension: 6594 Office hours: 3:30-4:20 MW Teaching Assistants: Seoh Oh, MC 6129, X-3812, s2oh@uwaterloo.ca Xuemin
W. Alabama - ACTS - 331
ACTSC 331 Fall 2004ASSIGNMENT 5 DUE: 10.00 am Thursday 18 November 20041. Anne is aged 80 and Jane is aged 85. Both lives survival functions follow de Moivres law with = 100 and the lives are independent of each other. Y is the present value ran
W. Alabama - ACTS - 331
Life Contingencies 1 - ACTSC 331, Winter 2003 Course Review Notes of Chapter 8The propsepective formula for benefit reserves for general fully discrete insurances on (x): hV = j=0 bh+j+1 v j+1 j px+h qx+h+j -j=0h+j v j j px+h ,where bj+1 is
W. Alabama - ACTS - 331
ACTSC 331, SPRING 2003 ASSIGNMENT #1Due in class on Thursday, June 5, 2003. Note: Solutions to Assignment #1 will be posted on the course website after the due date.1. On the basis of De Moivres law with lx = 105x and the interest rate of 7%, calc
W. Alabama - ACTS - 331
ACTSC 331 WINTER 2004ASSIGNMENT 8 DUE: Not to be handed in 1. The gure below shows a model of employment/unemployment in which mortality and migration are ignored. 01 x+t Employed 0 10 x+t (a) Write down expressions for t p00 and t p11 x x (b) Sho
W. Alabama - ACTS - 331
SOLUTIONS TO ASSIGNMENT #2, ACTSC 331, FALL 2003 1. (a) t V = 0 1t Ext 0s v s s px ds tt 0 bsv s s px x (s) ds , t 0. 0 t[2 marks] (b) It follows from 0 V = 0 that 0 = intot 0(bs v s s px x (s) s v s s px ) ds. Then, splitting
W. Alabama - ACTS - 331
Review Notes for Chapter 8The prospective formula for benet reserves for general fully discrete insurances on (x): hV =j=0bh+j+1 v j+1 j px+h qx+h+j j=0h+j v j j px+h ,where bj+1 is the death benet in the j + 1-th policy year if death is
W. Alabama - ACTS - 331
Solutions to Exercises for Chapter 1010.2 (a) We are given t p50 = e (b) fT (t) = (c) fJ (j) =2 j=1 50 0 ( )t 3 ds 0 50s= ( 50t )3 . Thus, fT,J (t, j) = t p50 50 (t) = t t)2 dt = j/3.( )(j)j(50t)2 . 503fT,J (t, j) =fT,J (t, j)dt =3(
W. Alabama - ACTS - 331
University of WaterlooACTSC 331 Past Exam Questions1. You are given the following extract from a select life table with 4-year select period. A select individual aged 41 purchased a fully discrete 3-year term insurance with a sum insured of \$200,0
W. Alabama - ACTS - 331
An introduction to multiple state models for life contingenciesMary R Hardy March 9, 20041IntroductionIn all the material of Actuarial Mathematics up to Chapter 9 we are concerned with only two states for a life alive or dead and we valued b
W. Alabama - ACTS - 331
ACTSC 331 WINTER 2004ASSIGNMENT 4 DUE: Friday 6 February 20041. An insurer issues a 5 year fully discrete term insurance policy with sum insured 100,000 to a life age 40. The interest rate used in the premium calculation is 6% per year, mortality
W. Alabama - ACTS - 331
SOLUTIONS TO ASSIGNMENT #1A 1. (a) By the EP, b A40 = 230 a40 = 230 1- 40 . We have = ln1.06 and A40 = 0v t t p40 40 (t) dt =60 0e-t dt = 0.2773594. 60 [5 marks] [5 marks]Hence, b = 10, 284.20. (b) The benefit reserve = b A60 - 23
W. Alabama - ACTS - 331
Solutions to Assignment #31. We are given that = 1/20 = 0.05. (a) The expected time is t p50:60 E[T (50 : 60)] = =0 0dt =0sT (50) (t) + sT (60) (t) sT (50) (t) sT (60) (t) et dt(e0.03t + e0.02t e0.05t )e0.05t dt = 16.7857. [5 ma
East Los Angeles College - ES - 368
SOLAR ENERGY FOR COOLING AND REFRIGERATIONDr. R.E. Critoph and Mr. K. Thompson Engineering Department, University of Warwick, Coventry CV4 7AL, UKABSTRACT Solar refrigeration may have applications in both developed and developing countries. Applic
W. Alabama - ACTS - 331