SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 4 SONNTAG BORGNAKKE VAN WYLEN. FUNDAMENTALS of. Thermodynamics. Sixth Edition
|
|
- Elvin Thomas
- 5 years ago
- Views:
Transcription
1 SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 4 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition
2 CHAPTER 4 SUBSECTION PROB NO. Concept-Study Guide Problems 7- Simple processes 3-8 Review Problems 9-30 Polytropic processes 3-34 Multi-step processes, other types of work Rates of work Heat Transfer Rates 4-43 Correspondence table The correspondence between the problem set in this sixth edition versus the problem set in the 5'th edition text. Problems that are new are marked new and the SI number refers to the corresponding SI unit problem. New 5 th Ed. SI New 5 th Ed. SI 7 new 3 new 50 8 new new new new new new New 4 40 new 9 7 new
3 Concept Problems 4.7E The electric company charges the customers per kw-hour. What is that in english units? The unit kw-hour is a rate multiplied with time. For the standard English Eng. units the rate of energy is in Btu/h and the time is in seconds. The integration in Eq.4. becomes kw-hour = 34.4 Btu/h h = 34.4 Btu Conversions are found in Table A. 4.8E Work as F x has units of lbf-ft, what is that in Btu? Conversions are found in Table A. lbf-ft = Btu = 778 Btu 4.9E A work of.5 Btu must be delivered on a rod from a pneumatic piston/cylinder where the air pressure is limited to 75 psia. What diameter cylinder should I have to restrict the rod motion to maximum ft? W = F dx = P dv = PA dx = PA x = P π 4 D x 4W D = πp x = = ft 4.5 Btu π 75 psia ft = lbf-ft π (lbf/ft ) ft
4 4.0E A force of 300 lbf moves a truck with 40 mi/h up a hill. What is the power? Ẇ = F V = 300 lbf 40 (mi/h) = = lbf-ft s =.6 Btu/s lbf-ft s
5 4.E A 00 hp dragster engine drives the car with a speed of 65 mi/h. How much force is between the tires and the road? Power is force times rate of displacement as in Eq.4. Power, rate of work Ẇ = F V = P V. = T ω We need the velocity in ft/s: V = 3600 = ft/s We need power in lbf-ft/s: hp = 550 lbf-ft/s F = Ẇ / V = lbf-ft/s ft/s = 693 lbf
6 4.E A 00 hp dragster engine has a drive shaft rotating at 000 RPM. How much torque is on the shaft? Power is force times rate of displacement as in Eq.4. Power, rate of work Ẇ = F V = P V. = T ω We need to convert the RPM to a value for angular velocity ω ω = RPM π π rad 60 s = s = s We need power in lbf-ft/s: hp = 550 lbf-ft/s 00 hp 550 lbf-ft/s-hp T = Ẇ / ω = rad/s = 35 lbf-ft
7 Simple Processes 4.3E A bulldozer pushes 000 lbm of dirt 300 ft with a force of 400 lbf. It then lifts the dirt 0 ft up to put it in a dump truck. How much work did it do in each situation? W = F dx = F x = 400 lbf 300 ft = lbf-ft = 54 Btu W = F dz = mg dz = mg Z = 000 lbm 3.74 ft/s 0 ft / (3.74 lbm-ft / s -lbf) = lbf-ft =.85 Btu
8 4.4E A steam radiator in a room at 75 F has saturated water vapor at 6 lbf/in. flowing through it, when the inlet and exit valves are closed. What is the pressure and the quality of the water, when it has cooled to 75F? How much work is done? After the valve is closed no flow, constant V and m. T 75 F P : x =, P = 6 lbf/in v = v g = ft 3 /lbm v : T = 75 F, v = v = ft 3 /lbm P = P g = 0.43 lbf/in v = = x ( ) x = W = PdV = 0
9 4.5E A linear spring, F = k s (x x o ), with spring constant k s = 35 lbf/ft, is stretched until it is.5 in. longer. Find the required force and work input. F = k s (x - x 0 ) = 35.5/ = 7.9 lbf W = Fdx = k s (x - x 0 )d(x - x 0 ) = k s (x - x 0 ) = 35 (.5/) = 0.76 lbf ft = Btu
10 4.6E Two hydraulic cylinders maintain a pressure of 75 psia. One has a cross sectional area of 0. ft the other 0.3 ft. To deliver a work of Btu to the piston how large a displacement (V) and piston motion H is needed for each cylinder? Neglect P atm W = F dx = P dv = PA dx = PA H = P V W = Btu = lbf-ft V = W P = lbf-ft lbf/ft = ft3 Both cases the height is H = V/A H = = ft H = = 0.09 ft F F cb
11 4.7E A piston/cylinder has 5 ft of liquid 70 F water on top of the piston (m = 0) with cross-sectional area of ft, see Fig. P.57. Air is let in under the piston that rises and pushes the water out over the top edge. Find the necessary work to push all the water out and plot the process in a P-V diagram. P = P o + ρgh = psia =.8 psia V = H A = 5 = 5 ft 3 lbm/ft 3 ft/s ft (lbm-ft/s -lbf) (in/ft) W = AREA = P dv = ½ (P + P o )(V max -V ) = ½ ( ) psia 5 ft 3 44 (in/ft) = lbf-ft = 49.8 Btu P o P HO P cb Air P0 V Vmax V
12 4.8E A cylinder fitted with a frictionless piston contains 0 lbm of superheated refrigerant R-34a vapor at 00 lbf/in., 300 F. The setup is cooled at constant pressure until the R-34a reaches a quality of 5%. Calculate the work done in the process. Constant pressure process boundary work. State properties from Table F.0 State : Table F.0. v = ft 3 /lbm; State : Table F.0. v = = ft 3 /lbm Interpolated to be at 00 psia, numbers at 0.5 psia could have been used. W = P dv = P (V -V ) = mp (v -v ) = ( ) = Btu P C.P. T C.P. 00 T P = 00 psia cb v v
13 Review Problems 4.9E The gas space above the water in a closed storage tank contains nitrogen at 80 F, 5 lbf/in.. Total tank volume is 50 ft 3 and there is 000 lbm of water at 80 F. An additional 000 lbm water is now forced into the tank. Assuming constant temperature throughout, find the final pressure of the nitrogen and the work done on the nitrogen in this process. Water is compressed liquid, so it is incompressible V H O = mv = = ft3 V N = V tank - V H O = = ft3 V N = V tank - V H O = = 7.85 ft3 N is an ideal gas so P N = P N V N /V N = = lbf/in W = PdV = P V ln V V = ln = Btu
14 4.30E A cylinder having an initial volume of 00 ft 3 contains 0. lbm of water at 00 F. The water is then compressed in an isothermal quasi-equilibrium process until it has a quality of 50%. Calculate the work done in the process assuming water vapor is an ideal gas. State : T, v = V/m = = 500 ft3 /lbm ( > v g ) since P g = 0.95 psia, very low so water is an ideal gas from to. P = P g v g v = V = mv = 0.*350 = 70 ft 3 = lbf/in v 3 = ( ) = 75.0 ft 3 /lbm W = PdV = P V ln V V = ln 00 = Btu W 3 = P =3 m(v 3 - v ) = (75-350) 44 / 778 = -6.6 Btu W 3 = = Btu P C.P. T C.P T 00 3 Psat P v v
15 Polytropic Processes 4.3E Helium gas expands from 0 psia, 600 R and 9 ft 3 to 5 psia in a polytropic process with n =.667. How much work does it give out? Process equation: PV n = constant = P V n = P Vn Solve for the volume at state V = V (P /P ) /n = = ft 3 Work from Eq.4.4 W = P V - P V n = lbf-ft = 43 lbf-ft = 5.43 Btu
16 4.3E Consider a mass going through a polytropic process where pressure is directly proportional to volume (n = ). The process start with P = 0, V = 0 and ends with P = 90 lbf/in., V = 0.4 ft 3.The physical setup could be as in Problem.. Find the boundary work done by the mass. 90 P W = PdV = AREA = (P + P )(V - V ) 0 0 W 0.4 V = (P + 0)( V - 0) = P V = = 59 ft lbf = 3.33 Btu
17 4.33E The piston/cylinder shown in Fig. P4.48 contains carbon dioxide at 50 lbf/in., 00 F with a volume of 5 ft 3. Mass is added at such a rate that the gas compresses according to the relation PV. = constant to a final temperature of 350 F. Determine the work done during the process. From Eq. 4.4 for PV n = const ( n =/ ) W = PdV = P V - P V - n W = mr(t - T ) - n W = ( ) -. Assuming ideal gas, PV = mrt, But mr = P V T = = Btu/R = Btu
18 4.34E Find the work for Problem 3.56E. State : Table F.9 P = 74.6 lbf/in, v = 0.94 ft 3 /lbm Process: Pv = C = P v = P v w = Pdv = C v - dv = C ln v v v State : P = 30 lbf/in P ; v = = / 30 =.76 ft 3 /lbm P w = P v ln v v = P v ln P P = ln = 6845 ft lbf/lbm =.65 Btu/lbm P T v v Notice T is not constant. It is not an ideal gas in this range.
19 Multi-step Processes, Other Types of Work 4.35E Consider a two-part process with an expansion from 3 to 6 ft 3 at a constant pressure of 0 lbf/in. followed by an expansion from 6 to ft 3 with a linearly rising pressure from 0 lbf/in. ending at 40 lbf/in.. Show the process in a P-V diagram and find the boundary work. By knowing the pressure versus volume variation the work is found P 3 W 3 = W + W 3 3 = PdV + PdV = P (V V ) 3 6 W = 0 44 (6-3) + (0 + 40)( - 6) 44 = = ft lbf. = ( / 778) = 44.4 Btu V + (P + P 3 )(V 3 -V )
20 4.36E A piston/cylinder has lbm of R-34a at state with 00 F, 90 lbf/in., and is then brought to saturated vapor, state, by cooling while the piston is locked with a pin. Now the piston is balanced with an additional constant force and the pin is removed. The cooling continues to a state 3 where the R-34a is saturated liquid. Show the processes in a P-V diagram and find the work in each of the two steps, to and to 3. Solution : C.V. R-34a This is a control mass. Properties from table F.0. and 0. State : (T,P) => v = ft 3 /lbm State given by fixed volume and x =.0 State : v = v = v g => W = T = = 55.3 F P = ( ) 0.59 = psia State 3 reached at constant P (F = constant) state 3: P 3 = P and v 3 = v f = ( ) 0.59 = 0.08 ft 3 /lbm W 3 = W + W 3 = 0 + W 3 = P dv = P(V 3 -V ) = mp(v 3 -v ) = ( ) = Btu P 3 cb V
21 4.37E A cylinder containing lbm of ammonia has an externally loaded piston. Initially the ammonia is at 80 lbf/in., 360 F and is now cooled to saturated vapor at 05 F, and then further cooled to 65 F, at which point the quality is 50%. Find the total work for the process, assuming a piecewise linear variation of P versus V. P State : (T, P) Table F.8. v =.767 ft 3 /lbm F State : (T, x) Table F.8. sat. vap. 9 P = 9 psia, v =.3 ft 3 /lbm 05 F State 3: (T, x) P 8 3 = 8 psia, 3 65 F v 3 = ( )/ = W 3 = PdV ( P + P v )m(v - v ) + ( P + P 3 )m(v 3 - v ) = [( )( ) + ( )( )] = -45. Btu
22 4.38E A -ft-long steel rod with a 0.5-in. diameter is stretched in a tensile test. What is the required work to obtain a relative strain of 0.%? The modulus of elasticity of steel is lbf/in.. - W = AEL 0 (e), A = π 4 (0.5) = π 6 in - W = ( π 6 ) (0-3 ) =.94 ft lbf
23 Rates of Work 4.39E An escalator raises a 00 lbm bucket of sand 30 ft in minute. Determine the total amount of work done and the instantaneous rate of work during the process. W = Fdx = F dx = F x = = 6000 ft lbf = (6000/778) Btu = 7.7 Btu. W = W / t = 7.7 / 60 = 0.9 Btu/s
24 4.40E A piston/cylinder of diameter 0 inches moves a piston with a velocity of 8 ft/s. The instantaneous pressure is 00 psia. What is the volume displacement rate, the force and the transmitted power? Rate of work is force times rate of displacement. The force is pressure times area. F = PA = P π D /4 = 00 lbf/in (π/4) 0 in = 7854 lbf. W = FV = 7854 lbf 8 ft s = 4 37 lbf-ft/s = 8.7 Btu/s V P
25 Heat Transfer Rates 4.4E The sun shines on a 500 ft road surface so it is at 5 F. Below the inch thick asphalt, average conductivity of Btu/h ft F, is a layer of compacted rubbles at a temperature of 60 F. Find the rate of heat transfer to the rubbles.. Q = k A T x = = 735 Βtu/h 5 60 /
26 4.4E A water-heater is covered up with insulation boards over a total surface area of 30 ft. The inside board surface is at 75 F and the outside surface is at 70 F and the board material has a conductivity of 0.05 Btu/h ft F. How thick a board should it be to limit the heat transfer loss to 70 Btu/h? Steady state conduction through a single layer board.. Q cond = k A T x = k Α Τ/ Q. x x = (75-70) / 70 = 0.9 ft =.6 in
27 4.43E The black grille on the back of a refrigerator has a surface temperature of 95 F with a total surface area of 0 ft. Heat transfer to the room air at 70 F takes place with an average convective heat transfer coefficient of 3 Btu/h ft R. How much energy can be removed during 5 minutes of operation?. Q = hα T; Q =. Q t = ha T t Q = 3 (Btu/h ft R) 0 ft (95 70) F (5/60) h = 87.5 Btu
Introduction to Engineering thermodynamics 2 nd Edition, Sonntag and Borgnakke. Solution manual
Introduction to Engineering thermodynamics nd Edition, Sonntag and Borgnakke Solution manual Chapter 4 Claus Borgnakke The picture is a false color thermal image of the space shuttle s main engine. The
More informationSOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 5 SONNTAG BORGNAKKE VAN WYLEN. FUNDAMENTALS of. Thermodynamics. Sixth Edition
SOLUTION MANUAL ENGLISH UNIT ROBLEMS CHATER 5 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition CHATER 5 SUBSECTION ROB NO. Concept-Study Guide roblems 39-44 Kinetic and otential
More informationSOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 3 SONNTAG BORGNAKKE VAN WYLEN. FUNDAMENTALS of. Thermodynamics. Sixth Edition
SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 3 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition CHAPTER 3 SUBSECTION PROB NO. Concept-Study Guide Problems 128-132 Phase diagrams
More informationKNOWN: Data are provided for a closed system undergoing a process involving work, heat transfer, change in elevation, and change in velocity.
Problem 44 A closed system of mass of 10 kg undergoes a process during which there is energy transfer by work from the system of 0147 kj per kg, an elevation decrease of 50 m, and an increase in velocity
More informationCHAPTER 2: THE NATURE OF ENERGY
Principles of Engineering Thermodynamics st Edition Reisel Solutions Manual Full Download: http://testbanklive.com/download/principles-of-engineering-thermodynamics-st-edition-reisel-solutions-manual/
More informationChapter 3 First Law of Thermodynamics and Energy Equation
Fundamentals of Thermodynamics Chapter 3 First Law of Thermodynamics and Energy Equation Prof. Siyoung Jeong Thermodynamics I MEE0-0 Spring 04 Thermal Engineering Lab. 3. The energy equation Thermal Engineering
More information+ m B1 = 1. u A1. u B1. - m B1 = V A. /v A = , u B1 + V B. = 5.5 kg => = V tot. Table B.1.
5.6 A rigid tank is divided into two rooms by a membrane, both containing water, shown in Fig. P5.6. Room A is at 200 kpa, v = 0.5 m3/kg, VA = m3, and room B contains 3.5 kg at 0.5 MPa, 400 C. The membrane
More informationThermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am - 10:00am & Monday 8:00am - 9:00am MEP 261 Class ZA
Thermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am - 10:00am & Monday 8:00am - 9:00am MEP 261 Class ZA Dr. Walid A. Aissa Associate Professor, Mech. Engg. Dept. Faculty of Engineering
More informationChapter 5: The First Law of Thermodynamics: Closed Systems
Chapter 5: The First Law of Thermodynamics: Closed Systems The first law of thermodynamics can be simply stated as follows: during an interaction between a system and its surroundings, the amount of energy
More informationFirst Law of Thermodynamics
First Law of Thermodynamics During an interaction between a system and its surroundings, the amount of energy gained by the system must be exactly equal to the amount of energy lost by the surroundings.
More informationfirst law of ThermodyNamics
first law of ThermodyNamics First law of thermodynamics - Principle of conservation of energy - Energy can be neither created nor destroyed Basic statement When any closed system is taken through a cycle,
More information20 m neon m propane. g 20. Problems with solutions:
Problems with solutions:. A -m tank is filled with a gas at room temperature 0 C and pressure 00 Kpa. How much mass is there if the gas is a) Air b) Neon, or c) Propane? Given: T7K; P00KPa; M air 9; M
More informationSUMMARY AND CONCEPTS FOR. Introduction to Engineering Thermodynamics by Sonntag and Borgnakke First edition. Wiley 2001
SUMMARY AND CONCEPTS FOR Introduction to Engineering Thermodynamics by Sonntag and Borgnakke First edition Wiley 2001 This is a collection of end of chapter summaries and main concepts and concept problems
More informationCHAPTER 8 ENTROPY. Blank
CHAPER 8 ENROPY Blank SONNAG/BORGNAKKE SUDY PROBLEM 8-8. A heat engine efficiency from the inequality of Clausius Consider an actual heat engine with efficiency of η working between reservoirs at and L.
More informationDishwasher. Heater. Homework Solutions ME Thermodynamics I Spring HW-1 (25 points)
HW-1 (25 points) (a) Given: 1 for writing given, find, EFD, etc., Schematic of a household piping system Find: Identify system and location on the system boundary where the system interacts with the environment
More informationFirst Law of Thermodynamics Closed Systems
First Law of Thermodynamics Closed Systems Content The First Law of Thermodynamics Energy Balance Energy Change of a System Mechanisms of Energy Transfer First Law of Thermodynamics in Closed Systems Moving
More informationChapter 4. Energy Analysis of Closed Systems
Chapter 4 Energy Analysis of Closed Systems The first law of thermodynamics is an expression of the conservation of energy principle. Energy can cross the boundaries of a closed system in the form of heat
More informationEng Thermodynamics I conservation of mass; 2. conservation of energy (1st Law of Thermodynamics); and 3. the 2nd Law of Thermodynamics.
Eng3901 - Thermodynamics I 1 1 Introduction 1.1 Thermodynamics Thermodynamics is the study of the relationships between heat transfer, work interactions, kinetic and potential energies, and the properties
More informationME 200 Final Exam December 12, :00 a.m. to 10:00 a.m.
CIRCLE YOUR LECTURE BELOW: First Name Last Name 7:30 a.m. 8:30 a.m. 10:30 a.m. 1:30 p.m. 3:30 p.m. Mongia Abraham Sojka Bae Naik ME 200 Final Exam December 12, 2011 8:00 a.m. to 10:00 a.m. INSTRUCTIONS
More informationRelationships between WORK, HEAT, and ENERGY. Consider a force, F, acting on a block sliding on a frictionless surface. x 2
Relationships between WORK, HEAT, and ENERGY Consider a force, F, acting on a block sliding on a frictionless surface x x M F x Frictionless surface M dv v dt M dv dt v F F F ; v mass velocity in x direction
More informationME 201 Thermodynamics
ME 0 Thermodynamics Solutions First Law Practice Problems. Consider a balloon that has been blown up inside a building and has been allowed to come to equilibrium with the inside temperature of 5 C and
More informationMechanical Engineering Department Third year class Gas Dynamics Tables Lecturer: Dr.Naseer Al-Janabi
Mechanical Engineering Department Third year class Gas Dynamics Tables Lecturer: Dr.Naseer Al-Janabi Ref. Fundamentals of Gas Dynamics, 2e - R. Zucker, O. Biblarz Appendixes A. B. C. D. E. F. G. H. I.
More informationChapter One Reviews of Thermodynamics Update on 2013/9/13
Chapter One Reviews of Thermodynamics Update on 2013/9/13 (1.1). Thermodynamic system An isolated system is a system that exchanges neither mass nor energy with its environment. An insulated rigid tank
More informationThe First Law of Thermodynamics. By: Yidnekachew Messele
The First Law of Thermodynamics By: Yidnekachew Messele It is the law that relates the various forms of energies for system of different types. It is simply the expression of the conservation of energy
More informationDepartment of Mechanical Engineering Indian Institute of Technology New Delhi II Semester MEL 140 ENGINEERING THERMODYNAMICS
PROBLEM SET 1: Review of Basics Problem 1: Define Work. Explain how the force is generated in an automobile. Problem 2: Define and classify Energy and explain the relation between a body and energy. Problem
More informationChapter 2: Energy and the 1 st Law of Thermodynamics. The Study of Energy in Closed Systems
Chapter 2: Energy and the 1 st Law of Thermodynamics The Study of Energy in Closed Systems Topics 2.1 Mechanical Concepts of Energy 2.2 Broadening Understanding of Work 2.3 Broadening Understanding of
More informationChapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn
Chapter 7 Entropy by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics: An Engineering Approach, 5th ed.,
More informationNon-Newtonian fluids is the fluids in which shear stress is not directly proportional to deformation rate, such as toothpaste,
CHAPTER1: Basic Definitions, Zeroth, First, and Second Laws of Thermodynamics 1.1. Definitions What does thermodynamic mean? It is a Greeks word which means a motion of the heat. Water is a liquid substance
More informationReadings for this homework assignment and upcoming lectures
Homework #3 (group) Tuesday, February 13 by 4:00 pm 5290 exercises (individual) Thursday, February 15 by 4:00 pm extra credit (individual) Thursday, February 15 by 4:00 pm Readings for this homework assignment
More informationRelationships between WORK, HEAT, and ENERGY. Consider a force, F, acting on a block sliding on a frictionless surface
Introduction to Thermodynamics, Lecture 3-5 Prof. G. Ciccarelli (0) Relationships between WORK, HEAT, and ENERGY Consider a force, F, acting on a block sliding on a frictionless surface x x M F x FRICTIONLESS
More information1.4 Perform the following unit conversions: (b) (c) s. g s. lb min. (d) (e) in. ft s. m 55 h. (f) ft s. km h. (g)
1.4 Perform the following unit conversions: 0.05 ft 1 in. (a) 1L 61in. 1L 1ft (b) 1kJ 650 J 10 J 1Btu 1.0551kJ 0.616 Btu (c) 41 Btu/h 0.15 kw 1kW 1h 600 s 778.17 ft lbf 1Btu ft lbf 99.596 s (d) g 78 s
More informationMAE 11. Homework 8: Solutions 11/30/2018
MAE 11 Homework 8: Solutions 11/30/2018 MAE 11 Fall 2018 HW #8 Due: Friday, November 30 (beginning of class at 12:00p) Requirements:: Include T s diagram for all cycles. Also include p v diagrams for Ch
More informationOutline. Example. Solution. Property evaluation examples Specific heat Internal energy, enthalpy, and specific heats of solids and liquids Examples
Outline Property ealuation examples Specific heat Internal energy, enthalpy, and specific heats of solids and liquids s A piston-cylinder deice initially contains 0.5m of saturated water apor at 00kPa.
More informationES201 - Examination III Richards, North, Berry Fall November 2000 NAME BOX NUMBER
ES201 - Examination III Richards, North, Berry Fall 2000-2001 2 November 2000 NAME BOX NUMBER Problem 1 Problem 2 ( 30 ) ( 30 ) Problem 3 ( 40 ) Total ( 100 ) INSTRUCTIONS Closed book/notes exam. (Unit
More informationThermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am - 10:00am & Monday 8:00am - 9:00am MEP 261 Class ZA
Thermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am - 10:00am & Monday 8:00am - 9:00am MEP 261 Class ZA Dr. Walid A. Aissa Associate Professor, Mech. Engg. Dept. Faculty of Engineering
More informationChapter 1 Introduction
Fundamentals of Thermodynamics Chapter 1 Introduction Prof. Siyoung Jeong Thermodynamics I MEE2022-01 Thermodynamics : Science of energy and entropy - Science of heat and work and properties related to
More informationME 2322 Thermodynamics I PRE-LECTURE Lesson 10 Complete the items below Name:
Lesson 10 1. (5 pt) If P > P sat (T), the phase is a subcooled liquid. 2. (5 pt) if P < P sat (T), the phase is superheated vapor. 3. (5 pt) if T > T sat (P), the phase is superheated vapor. 4. (5 pt)
More informationChapter 5. Mass and Energy Analysis of Control Volumes. by Asst. Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn
Chapter 5 Mass and Energy Analysis of Control Volumes by Asst. Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics:
More information12/21/2014 7:39 PM. Chapter 2. Energy and the 1st Law of Thermodynamics. Dr. Mohammad Suliman Abuhaiba, PE
Chapter 2 Energy and the 1st Law of Thermodynamics 1 2 Homework Assignment # 2 Problems: 1, 7, 14, 20, 30, 36, 42, 49, 56 Design and open end problem: 2.1D Due Monday 22/12/2014 3 Work and Kinetic Energy
More informationME Thermodynamics I
HW-03 (25 points) i) Given: for writing Given, Find, Basic equations Rigid tank containing nitrogen gas in two sections initially separated by a membrane. Find: Initial density (kg/m3) of nitrogen gas
More information- Apply closed system energy balances, observe sign convention for work and heat transfer.
CHAPTER : ENERGY AND THE FIRST LAW OF THERMODYNAMICS Objectives: - In this chapter we discuss energy and develop equations for applying the principle of conservation of energy. Learning Outcomes: - Demonstrate
More information3. First Law of Thermodynamics and Energy Equation
3. First Law of Thermodynamics and Energy Equation 3. The First Law of Thermodynamics for a ontrol Mass Undergoing a ycle The first law for a control mass undergoing a cycle can be written as Q W Q net(cycle)
More informationTHERMODYNAMICS, FLUID AND PLANT PROCESSES. The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial.
THERMODYNAMICS, FLUID AND PLANT PROCESSES The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial. THERMODYNAMICS TUTORIAL 2 THERMODYNAMIC PRINCIPLES SAE
More informationECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 12 June 2006
ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER 1 June 006 Midterm Examination R. Culham This is a hour, closed-book examination. You are permitted to use one 8.5 in. 11 in. crib sheet (one side
More informationEVALUATING PROPERTIES FOR A PURE SUBSTANCES. By Ertanto Vetra
EVALUATING PROPERTIES FOR A PURE SUBSTANCES 1 By Ertanto Vetra Outlines - TV, PV, PT, PVT Diagram - Property Tables - Introduction to Enthalpy - Reference State & Reference Values - Ideal Gas Equation
More informationCourse: TDEC202 (Energy II) dflwww.ece.drexel.edu/tdec
Course: TDEC202 (Energy II) Thermodynamics: An Engineering Approach Course Director/Lecturer: Dr. Michael Carchidi Course Website URL dflwww.ece.drexel.edu/tdec 1 Course Textbook Cengel, Yunus A. and Michael
More informationChapter 1: Basic Definitions, Terminologies and Concepts
Chapter : Basic Definitions, Terminologies and Concepts ---------------------------------------. UThermodynamics:U It is a basic science that deals with: -. Energy transformation from one form to another..
More informationME Thermodynamics I = = = 98.3% 1
HW-08 (25 points) i) : a) 1 Since ν f < ν < ν g we conclude the state is a Saturated Liquid-Vapor Mixture (SLVM) 1, from the saturation tables we obtain p 3.6154 bar. 1 Calculating the quality, x: x ν
More informationME Thermodynamics I
Homework - Week 01 HW-01 (25 points) Given: 5 Schematic of the solar cell/solar panel Find: 5 Identify the system and the heat/work interactions associated with it. Show the direction of the interactions.
More informationDr Ali Jawarneh. Hashemite University
Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University Examine the moving boundary work or P d work commonly encountered in reciprocating devices such as automotive engines and compressors.
More informationThermal Energy Final Exam Fall 2002
16.050 Thermal Energy Final Exam Fall 2002 Do all eight problems. All problems count the same. 1. A system undergoes a reversible cycle while exchanging heat with three thermal reservoirs, as shown below.
More informationECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 13 June 2007
ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER 13 June 2007 Midterm Examination R. Culham This is a 2 hour, open-book examination. You are permitted to use: course text book calculator There are
More informationEngineering Thermodynamics
David Ng Summer 2017 Contents 1 July 5, 2017 3 1.1 Thermodynamics................................ 3 2 July 7, 2017 3 2.1 Properties.................................... 3 3 July 10, 2017 4 3.1 Systems.....................................
More informationI. (20%) Answer the following True (T) or False (F). If false, explain why for full credit.
I. (20%) Answer the following True (T) or False (F). If false, explain why for full credit. Both the Kelvin and Fahrenheit scales are absolute temperature scales. Specific volume, v, is an intensive property,
More informationMAE 110A. Homework 3: Solutions 10/20/2017
MAE 110A Homework 3: Solutions 10/20/2017 3.10: For H 2O, determine the specified property at the indicated state. Locate the state on a sketch of the T-v diagram. Given a) T 140 C, v 0.5 m 3 kg b) p 30MPa,
More informationIntroduction to Engineering thermodynamics 2 nd Edition, Sonntag and Borgnakke. Solution manual
Introduction to Engineering thermodynamics 2 nd Edition, Sonntag and Borgnakke Solution manual Chapter 6 Claus Borgnakke The picture is a false color thermal image of the space shuttle s main engine. The
More informationWeek 5. Energy Analysis of Closed Systems. GENESYS Laboratory
Week 5. Energy Analysis of Closed Systems Objectives 1. Examine the moving boundary work or PdV work commonly encountered in reciprocating devices such as automotive engines and compressors 2. Identify
More informationChapter 5. Mass and Energy Analysis of Control Volumes
Chapter 5 Mass and Energy Analysis of Control Volumes Conservation Principles for Control volumes The conservation of mass and the conservation of energy principles for open systems (or control volumes)
More informationCommon Terms, Definitions and Conversion Factors
1 Common Terms, Definitions and Conversion Factors 1. Force: A force is a push or pull upon an object resulting from the object s interaction with another object. It is defined as Where F = m a F = Force
More informationTwo mark questions and answers UNIT I BASIC CONCEPT AND FIRST LAW SVCET
Two mark questions and answers UNIT I BASIC CONCEPT AND FIRST LAW 1. What do you understand by pure substance? A pure substance is defined as one that is homogeneous and invariable in chemical composition
More informationTo receive full credit all work must be clearly provided. Please use units in all answers.
Exam is Open Textbook, Open Class Notes, Computers can be used (Computer limited to class notes, lectures, homework, book material, calculator, conversion utilities, etc. No searching for similar problems
More information1. (10) Calorically perfect ideal air at 300 K, 100 kpa, 1000 m/s, is brought to rest isentropically. Determine its final temperature.
AME 5053 Intermediate Thermodynamics Examination Prof J M Powers 30 September 0 0 Calorically perfect ideal air at 300 K, 00 kpa, 000 m/s, is brought to rest isentropically Determine its final temperature
More informationFUNDAMENTALS of Thermodynamics
SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 15 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition CONTENT SUBSECTION PROB NO. Correspondence table Concept-Study Guide Problems 1-20 Equilibrium
More informationME 300 Thermodynamics II
ME 300 Thermodynamics II Prof. S. H. Frankel Fall 2006 ME 300 Thermodynamics II 1 Week 1 Introduction/Motivation Review Unsteady analysis NEW! ME 300 Thermodynamics II 2 Today s Outline Introductions/motivations
More information5/6/ :41 PM. Chapter 6. Using Entropy. Dr. Mohammad Abuhaiba, PE
Chapter 6 Using Entropy 1 2 Chapter Objective Means are introduced for analyzing systems from the 2 nd law perspective as they undergo processes that are not necessarily cycles. Objective: introduce entropy
More informationCHAPTER 17 WORK, HEAT, & FIRST LAW OF THERMODYNAMICS
CHAPTER 17 WORK, HEAT, and the FIRST LAW OF THERMODYNAMICS In this chapter, we will examine various thermal properties of matter, as well as several mechanisms by which energy can be transferred to and
More informationCHAPTER INTRODUCTION AND BASIC PRINCIPLES. (Tutorial). Determine if the following properties of the system are intensive or extensive properties: Property Intensive Extensive Volume Density Conductivity
More informationCHAPTER 2 ENERGY INTERACTION (HEAT AND WORK)
CHATER ENERGY INTERACTION (HEAT AND WORK) Energy can cross the boundary of a closed system in two ways: Heat and Work. WORK The work is done by a force as it acts upon a body moving in direction of force.
More information(1)5. Which of the following equations is always valid for a fixed mass system undergoing an irreversible or reversible process:
Last Name First Name ME 300 Engineering Thermodynamics Exam #2 Spring 2008 March 28, 2008 Form A Note : (i) (ii) (iii) (iv) Closed book, closed notes; one 8.5 x 11 sheet allowed. 60 points total; 60 minutes;
More informationThermodynamics I Chapter 2 Properties of Pure Substances
Thermodynamics I Chapter 2 Properties of Pure Substances Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi Malaysia Properties of Pure Substances (Motivation) To quantify the changes
More informationPROBLEM 6.3. Using the appropriate table, determine the indicated property. In each case, locate the state on sketches of the T-v and T-s diagrams.
PROBLEM 63 Using the appropriate table, determine the indicated property In each case, locate the state on sketches of the -v and -s diagrams (a) water at p = 040 bar, h = 147714 kj/kg K Find s, in kj/kg
More informationTDEC202 - Energy II (Recitation #3)
TDEC202 - Energy II (Recitation #3) Summer Term, 2007 M. Carchidi Solution to Problem 3-029 We are given that the substance is R-134a and metric units so that Tables A-11, A-12 and A-13 in Appendix 1 will
More informationChapter 2. Energy and the First Law of Thermodynamics
Chapter 2 Energy and the First Law of Thermodynamics Closed System Energy Balance Energy is an extensive property that includes the kinetic and gravitational potential energy of engineering mechanics.
More informationAnswers to questions in each section should be tied together and handed in separately.
EGT0 ENGINEERING TRIPOS PART IA Wednesday 4 June 014 9 to 1 Paper 1 MECHANICAL ENGINEERING Answer all questions. The approximate number of marks allocated to each part of a question is indicated in the
More informationCHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
Thermodynamics: An Engineering Approach 8th Edition in SI Units Yunus A. Çengel, Michael A. Boles McGraw-Hill, 2015 CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES Lecture slides by Dr. Fawzi Elfghi
More informationChapter 7. Entropy: A Measure of Disorder
Chapter 7 Entropy: A Measure of Disorder Entropy and the Clausius Inequality The second law of thermodynamics leads to the definition of a new property called entropy, a quantitative measure of microscopic
More informationMAE 320 HW 7B. 1e. For an isolated system, please circle the parameter which will change with time. (a) Total energy;
MAE 320 HW 7B his comprehensive homework is due Monday, December 5 th, 206. Each problem is worth the points indicated. Copying of the solution from another is not acceptable. Multi-choice, multi-answer
More informationGAS. Outline. Experiments. Device for in-class thought experiments to prove 1 st law. First law of thermodynamics Closed systems (no mass flow)
Outline First law of thermodynamics Closed systems (no mass flow) Device for in-class thought experiments to prove 1 st law Rubber stops GAS Features: Quasi-equlibrium expansion/compression Constant volume
More information(Heat capacity c is also called specific heat) this means that the heat capacity number c for water is 1 calorie/gram-k.
Lecture 23: Ideal Gas Law and The First Law of Thermodynamics 1 (REVIEW) Chapter 17: Heat Transfer Origin of the calorie unit A few hundred years ago when people were investigating heat and temperature
More informationLecture 44: Review Thermodynamics I
ME 00 Thermodynamics I Lecture 44: Review Thermodynamics I Yong Li Shanghai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Chuan Road Shanghai, 0040, P. R. China Email : liyo@sjtu.edu.cn
More informationNAME: AME Thermodynamics Examination 1: SOLUTION Prof. J. M. Powers 12 February 2010
NAME: AME 0 Thermodynamics Examination : SOLUTION Prof. J. M. Powers February 00 The advantageous use of Steam-power is unquestionably a modern discovery. And yet as much as two thousand years ago the
More informationPrinciples of Energy Conversion Part 6. Review of Engineering Thermodynamic
Principles of Energy Conversion Part 6. Review of Engineering Thermodynamic February 7, 2018 12 From Atmospheric Engines to Thermodynamics 3 12.1 Work from Steam.................................. 4 12.1.1
More informationME 354 THERMODYNAMICS 2 MIDTERM EXAMINATION. Instructor: R. Culham. Name: Student ID Number: Instructions
ME 354 THERMODYNAMICS 2 MIDTERM EXAMINATION February 14, 2011 5:30 pm - 7:30 pm Instructor: R. Culham Name: Student ID Number: Instructions 1. This is a 2 hour, closed-book examination. 2. Answer all questions
More informationEnergy and Energy Balances
Energy and Energy Balances help us account for the total energy required for a process to run Minimizing wasted energy is crucial in Energy, like mass, is. This is the Components of Total Energy energy
More informationChapter 2 Solutions. 2.1 (D) Pressure and temperature are dependent during phase change and independent when in a single phase.
Chater Solutions.1 (D) Pressure and temerature are deendent during hase change and indeendent when in a single hase.. (B) Sublimation is the direct conversion of a solid to a gas. To observe this rocess,
More informationENERGY TRANSFER BY WORK: Electrical Work: When N Coulombs of electrical charge move through a potential difference V
Weight, W = mg Where m=mass, g=gravitational acceleration ENERGY TRANSFER BY WOR: Sign convention: Work done on a system = (+) Work done by a system = (-) Density, ρ = m V kg m 3 Where m=mass, V =Volume
More informationME Thermodynamics I. Lecture Notes and Example Problems
ME 227.3 Thermodynamics I Lecture Notes and Example Problems James D. Bugg September 2018 Department of Mechanical Engineering Introduction Part I: Lecture Notes This part contains handout versions of
More informationCHAPTER. The First Law of Thermodynamics: Closed Systems
CHAPTER 3 The First Law of Thermodynamics: Closed Systems Closed system Energy can cross the boundary of a closed system in two forms: Heat and work FIGURE 3-1 Specifying the directions of heat and work.
More informationKNOWN: Pressure, temperature, and velocity of steam entering a 1.6-cm-diameter pipe.
4.3 Steam enters a.6-cm-diameter pipe at 80 bar and 600 o C with a velocity of 50 m/s. Determine the mass flow rate, in kg/s. KNOWN: Pressure, temperature, and velocity of steam entering a.6-cm-diameter
More informationENERGY ANALYSIS OF CLOSED SYSTEMS
Thermodynamics: An Engineering Approach Seventh Edition in SI Units Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011 Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS Mehmet Kanoglu University of Gaziantep
More informationECE309 THERMODYNAMICS & HEAT TRANSFER MIDTERM EXAMINATION. Instructor: R. Culham. Name: Student ID Number:
ECE309 THERMODYNAMICS & HEAT TRANSFER MIDTERM EXAMINATION June 19, 2015 2:30 pm - 4:30 pm Instructor: R. Culham Name: Student ID Number: Instructions 1. This is a 2 hour, closed-book examination. 2. Permitted
More informationFundamentals of Thermodynamics SI Version
Solution Manual Chapter 2 Fundamentals of hermodynamics SI Version Borgnakke Sonntag 8e Updated July 2013 2 Borgnakke and Sonntag CONEN CHAPER 2 SUBSECION PROB NO. Concept problems 1-15 Phase diagrams,
More informationDownloaded from
Chapter 12 (Thermodynamics) Multiple Choice Questions Single Correct Answer Type Q1. An ideal gas undergoes four different processes from the same initial state (figure). Four processes are adiabatic,
More informationChapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS
Thermodynamics: An Engineering Approach Seventh Edition in SI Units Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011 Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS Copyright The McGraw-Hill Companies,
More informationQ1. A) 46 m/s B) 21 m/s C) 17 m/s D) 52 m/s E) 82 m/s. Ans: v = ( ( 9 8) ( 98)
Coordinator: Dr. Kunwar S. Wednesday, May 24, 207 Page: Q. A hot-air balloon is ascending (going up) at the rate of 4 m/s and when the balloon is 98 m above the ground a package is dropped from it, vertically
More informationChapter 1: INTRODUCTION AND BASIC CONCEPTS. Thermodynamics = Greek words : therme(heat) + dynamis(force or power)
Chapter 1: INTRODUCTION AND BASIC CONCEPTS 1.1 Basic concepts and definitions Thermodynamics = Greek words : therme(heat) + dynamis(force or power) Note that, force x displacement = work; power = work/time
More informationChemical Engineering 141: Thermodynamics Spring 2012
Chemical Engineering 141: Thermodynamics Spring 2012 Thursday, February 23, 2011 Midterm I 80 minutes 115 points total Use of phone devices is not permitted Return your equation sheet with the exam 1.
More informationHEAT- I Part - A C D A B. Te m p. Heat input
e m p HE- I Part -. solid material is supplied with heat at a constant rate. he temperature of the material is changing with heat input as shown in the graph. Study the graph carefully and answer the following
More informationObjectives. Power in Translational Systems 298 CHAPTER 6 POWER
Objectives Explain the relationship between power and work. Explain the relationship between power, force, and speed for an object in translational motion. Calculate a device s efficiency in terms of the
More informationProblem 1.6 Make a guess at the order of magnitude of the mass (e.g., 0.01, 0.1, 1.0, 10, 100, or 1000 lbm or kg) of standard air that is in a room 10
Problem 1.6 Make a guess at the order of magnitude of the mass (e.g., 0.01, 0.1, 1.0, 10, 100, or 1000 lbm or kg) of standard air that is in a room 10 ft by 10 ft by 8 ft, and then compute this mass in
More information