ME 201 Thermodynamics


 Tamsyn Walters
 3 years ago
 Views:
Transcription
1 ME 0 Thermodynamics Second Law Practice Problems. Ideally, which fluid can do more work: air at 600 psia and 600 F or steam at 600 psia and 600 F The maximum work a substance can do is given by its availablity. We will assume that we have a closed system so that ψ = u  u o  T o (s  s o ) We take the dead state to be at STP or 5 C and 00 kpa or 76.4 F and 4.7 psia. Then using the appropriate table we have ψ air = (537) ln = Btu / lb m and = (537) ψ steam = Btu / lb So the steam can do more work m ( ). A heat pump provides 30,000 Btu/hr to maintain a dwelling at 68 F on a day when the outside temperature is 35 F. The power input to the pump is hp. If electricity costs 8 cents per kilowatthour, compare the actual operating cost per day with the minimum theoretical operating cost per day.
2 We sketch our device interactions Dwelling High Temperature Heat Reservoir at T H Q H Heat Pump W net Q L Outside Low Temperature Heat Reservoir at T L The cost is given by Cost = (0.08)W net For the actual cost we have (Cost) act = (0.08)()( kw / hp)(4 hr / day) = $.43 To calculate the minimum cost we will allow the heat pump to operate as a Carnot cycle, so that COP Carnot = = = 6 T 495 L TH 58 Then the minimum possible power input is ( & Q& W ) = H = 30,000 net = 875 Btu / hr min COPCarnot 6 = kw
3 and the minimum cost is (Cost) = (0.08)(0.5495)(4 hr / day) min = $ A cylinder/piston system contains water at 00 kpa, 00 C with a volume of 0 liters. The piston is moved slowly, compressing the water to a pressure of 800 kpa. The process is polytropic with a polytropic exponent of. Assuming that the room temperature is 0 C, show that this process does not violate the second law. To determine if this violate the nd law we will want to calculate the entropy change of the universe and compare it to zero. We have ( S ) = m(s  s ) +  Q sys universe Tsurr We now work this as a first law problem Working Fluid: Water(compressible) System: Closed System Process: Polytropic with n=.0 State State T = 00 C T = 4.7 C P = 00 kpa P = 800 kpa u = kj/kg u = 655.5kJ/kg V = 0.00 m 3 V = m 3 v =.0803 m 3 /kg v = m 3 /kg s = kj/(kg K) s = 6.88 kj/(kg K) phase: sup.vap. phase: sup.vap. italicized values from tables, bold values are calculated Initial State: Fixed Final State: Unknown W sh = 0 Q =???? W bnd =???? We begin by calculating the mass m = V = 0.00 = kg v
4 To fix the final state we use the polytropic relationship P V V = n / n (00)(0.00) / = P = m The specific volume at state is then v = V m = = kg / m which gives us superheated vapor. The boundary work can be shown to be W = P V ln V = (00)(0.00)ln bnd V 0.00 = kj We use the first law to determine the heat transfer Q sys = m(u  u ) + W = (0.085)( ) + (5.55) Then = kj ( S ) =  Q sys surrounds T surr and for the system ( ) =  (5.53) 93 = kj / K S = m(s  s ) = (0.085)( ) system = kj / K So that ( S ) = = kj / K universe Since this is greater than zero the second law is not violated. 4. When a man returns to his wellsealed house on a summer day, he finds that the house is at 3 C. He turns on the air conditioner which cools the entire house to 0 C in 5 minutes. If the COP of the heat pump system is.5, determine the power drawn by the heat pump. Assume the entire mass within the house is equivalent to 800 kg of air. 4
5 We begin by sketching our device interactions House High Temperature Heat Reservoir at T H Q H Heat Pump W net Q L AC System Low Temperature Heat Reservoir at T L By definition we have COP = Q & H W& net So if the required heat transfer can be determined the power can be determined. From a first law analysis on the house, we can write &Q = m u  u = (800) t (5)(60) = kw and Q & =  Q & H = 7.65 kw Then the power required is & W & = Q H COP = 7.65 net = 3.06 kw.5 5
6 5. An innovative way of power generation involves the utilization of geothermal energy, the energy of hot water that exists naturally underground (hot springs), as the heat source. If a supply of hot water at 40 C is discovered at a location where the environmental temperature is 0 C, determine the maximum thermal efficiency a geothermal plant built at that location can have. If the power output of the plant is to be 5 MW, what is the minimum mass flow rate of hot water needed? We begin by sketching our device interactions Geothermal Source High Temperature Heat Reservoir at T H Q H Heat Engine W net Q L Environment Low Temperature Heat Reservoir at T L The maximum thermal efficiency will occur when the heat engine operates as a Carnot cycle, ηth = η L Carnot =  T (0 + 73) =  = 0.9 T H ( ) The minimum mass flow rate of hot water corresponds to the maximum thermal efficiency or (& W& Q ) = net = 5000 H = 7,08 kw min η 0. 9 Carnot 6
7 Performing a first law analysis on the hot water stream we have Q = m& ( hout  h in) For the minimum flow rate we will assume that the hot water is cooled down to the environment temperature, then Q 7,08 m & = = cp( Tout Tin ) ( )( 0 40) = 34. kg / s 6. Air enters an adiabatic nonideal nozzle at 9 m/s, 300 K, and 0 kpa and exits at 00 m/s and 00 kpa. Determine the irreversibility and the reversible work on a per mass basis. We first solve this as a first law problem Working Fluid: Air(ideal gas) System: Control Volume System Process: Nozzle State State T = 300K T = 95 C P = 0 kpa P = 00 kpa h = kj/kg h = kj/kg φ =.7003 kj/(kg K) φ =.6855 kj/(kg K) r r v = 9 m/s v = 00 m/s italicized values from tables, bold values are calculated Initial State: Fixed Final State:??? W sh = 0 Q = 0 We use the first law to fix the final state h + v r = h + v r Then solving for h h = h + v r v r = (9) ( 00) 3 (0 ) = kj / kg which allows us to determine the temperature and φ. Then the reversible work is 7
8 w = h  h  T   R ln P rev HR φ φ P = (98) )ln 0 ( 00 = 5.7 kj / kg Since the actual work is zero the irreversibility is i = w rev = 5.7 kj / kg 7. Determine if a tray of ice cubes could remain frozen when placed in a food freezer having a COP of 9, operating in a room where the temperature is 3 C. We begin by sketching our device interactions Surroundings High Temperature Heat Reservoir at T H Q H Refrigerator W net Q L Freezer Compartment Low Temperature Heat Reservoir at T L 8
9 Assuming that the refrigerator operates on the Carnot cycle, we have COP = COP Carnot = TH TL Solving for T L T T = H = 305 L = 74.5 K + + COP 9 and since this is greater than 0 C the ice cubes will not remain frozen. 8. Air is compressed in a closed system from a state where the pressure is 00 kpa and the temperature is 7 C to a final state at 500 kpa and 77 C. Can this process occur adiabatically? If yes, determine the work per mass. If no, determine the direction of the heat transfer. To determine if the process can occur, we must calculate ( S ) = m(s  s ) +  Q sys universe Tsurr and compare it to zero. Since the process is adiabatic ( s ) = s  s =   R ln P universe φ φ P Going to the air tables we find ( s ) = s  s = )ln 500 universe ( 00 = kj / kg Since this is less than zero, the process cannot be adiabatic. To make s universe greater than zero will require Q sys to be negative, so that the direction of heat transfer is out of the system. 9. The pressure of water is increased by the use of a pump from 4 to 40 psia. A rise in the water temperature from 60 F to 60. F is observed. Determine the irreversibility, the second law efficiency, and the isentropic efficiency of the pump. 9
10 We first solve this as a first law problem Working Fluid: Water (incompressible) System: Control Volume System Process: Pump State State s (ideal) State a (actual T = 60 F T s = T a = 60. F P = 4 psia P = 40 psia P = 40 psia bold values are calculated Initial State: Fixed Final State: fixed W sh =???? Q = 0 To calculate the irreversibility, we use i = T (s  s )  q HR = (537)c ln T p,avg T  0 = (537)(.004)ln = Btu / lb m To determine the second law efficiency we need both the actual work and the ideal work. Starting with the actual work we have w act = h  h + q = c p,avg (T  T ) + v avg (P  P )  0 = (.004)(6060.) + ( )(440) / ( psia ft / Btu) = Btu / lb m The reversible work is given by w rev = i + w act = (0.068) + (0.774) = Btu / lbm which allows us to determine the second law efficiency as η rev II = w = ( ) = 0.55 w act (0.774) 3 0
11 To determine the isentropic efficiency, we must first calculate the ideal work. Recognizing that in a isentropic process, the water will not change temperature, we can write w = v (P  P ) ideal avg 3 = ( )(440) / ( psia ft / Btu) = Btu / lb m Then our isentropic efficiency is η ideal s = w = (0.077) w act (0.774) = Carbon dioxide undergoes an isothermal reversible process from 50 kpa and 300 C to 500 kpa. Determine the heat transfer per mass by using the first law and evaluating the boundary work from Pdv. Compare this to the heat transfer per mass calculated from the entropy change and the second law. We first solve this as a first law problem Working Fluid: CO (ideal gas) System: Closed System Process: Isothermal, Reversible State State T = 300C = 573K T = T = 573K P = 50 kpa P = 500 kpa u = kj/kg u = kj/kg φ = kj/(kg K) φ = kj/(kg K) v = m 3 /kg v = 0.65 m 3 /kg italicized values are from ideal gas relationships Initial State: Fixed Final State: fixed W sh = 0 Q =??? W bnd =????
12 We first go to the CO tables and get our properties. Our boundary work for an ideal gas undergoing an isothermal process is w = RT ln v bnd v = (0.889)(573)ln = Btu / lb m Using the first law our heat transfer is q = u  u + w bnd = ( ) = Btu / lb m From the second law we have q = T (s  s ) = T   R ln P φ φ P = (573) )ln 500 ( 50 = Btu / lb m So the two calculations for heat transfer agree.
20 m neon m propane 20
Problems with solutions:. A m 3 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: T73K; P00KPa; M air 9;
More information Know basic of refrigeration  Able to analyze the efficiency of refrigeration system 
Refrigeration cycle Objectives  Know basic of refrigeration  Able to analyze the efficiency of refrigeration system  contents Ideal VaporCompression Refrigeration Cycle Actual VaporCompression Refrigeration
More informationChapter 11. Refrigeration Cycles
Chapter 11 Refrigeration Cycles The vapor compression refrigeration cycle is a common method for transferring heat from a low temperature space to a high temperature space. The figures below show the objectives
More informationThe Second Law of Thermodynamics
Objectives MAE 320  Chapter 6 The Second Law of Thermodynamics The content and the pictures are from the text book: Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering Approach, McGrawHill,
More information6 18 A steam power plant receives heat from a furnace at a rate of 280 GJ/h. Heat losses to the surrounding air from the steam as it passes through
Thermo 1 (MEP 261) Thermodynamics An Engineering Approach Yunus A. Cengel & Michael A. Boles 7 th Edition, McGrawHill Companies, ISBN9780073529325, 2008 Sheet 6:Chapter 6 6 17 A 600MW steam power
More informationThe Second Law of Thermodynamics
The Second aw of Thermodynamics The 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
More informationThe final numerical answer given is correct but the math shown does not give that answer.
Note added to Homework set 7: The solution to Problem 16 has an error in it. The specific heat of water is listed as c 1 J/g K but should be c 4.186 J/g K The final numerical answer given is correct but
More informationSecond Law of Thermodynamics Alternative Statements
Second Law of Thermodynamics Alternative Statements There is no simple statement that captures all aspects of the second law. Several alternative formulations of the second law are found in the technical
More informationFUNDAMENTALS OF ENGINEERING THERMODYNAMICS
FUNDAMENTALS OF ENGINEERING THERMODYNAMICS System: Quantity of matter (constant mass) or region in space (constant volume) chosen for study. Closed system: Can exchange energy but not mass; mass is constant
More informationUNIT 2 REFRIGERATION CYCLE
UNIT 2 REFRIGERATION CYCLE Refrigeration Cycle Structure 2. Introduction Objectives 2.2 Vapour Compression Cycle 2.2. Simple Vapour Compression Refrigeration Cycle 2.2.2 Theoretical Vapour Compression
More information1. Why are the back work ratios relatively high in gas turbine engines? 2. What 4 processes make up the simple ideal Brayton cycle?
1. Why are the back work ratios relatively high in gas turbine engines? 2. What 4 processes make up the simple ideal Brayton cycle? 3. For fixed maximum and minimum temperatures, what are the effect of
More informationSOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 9 SONNTAG BORGNAKKE VAN WYLEN. FUNDAMENTALS of. Thermodynamics. Sixth Edition
SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 9 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition CONTENT SUBSECTION PROB NO. Correspondence table ConceptStudy Guide Problems 20 Steady
More informationTHE UNIVERSITY OF TRINIDAD & TOBAGO
THE UNIVERSITY OF TRINIDAD & TOBAGO FINAL ASSESSMENT/EXAMINATIONS JANUARY/APRIL 2013 Course Code and Title: Programme: THRM3001 Engineering Thermodynamics II Bachelor of Applied Sciences Date and Time:
More informationPractice Problems on Conservation of Energy. heat loss of 50,000 kj/hr. house maintained at 22 C
COE_10 A passive solar house that is losing heat to the outdoors at an average rate of 50,000 kj/hr is maintained at 22 C at all times during a winter night for 10 hr. The house is to be heated by 50 glass
More informationChapter 11, Problem 18.
Chapter 11, Problem 18. Refrigerant134a enters the compressor of a refrigerator as superheated vapor at 0.14 MPa and 10 C at a rate of 0.12 kg/s, and it leaves at 0.7 MPa and 50 C. The refrigerant is
More informationChapter 5 The Second Law of Thermodynamics
Chapter 5 he Second aw of hermodynamics he second law of thermodynamics states that processes occur in a certain direction, not in just any direction. Physical processes in nature can proceed toward equilibrium
More informationEsystem = 0 = Ein Eout
AGENDA: I. Introduction to Thermodynamics II. First Law Efficiency III. Second Law Efficiency IV. Property Diagrams and Power Cycles V. Additional Material, Terms, and Variables VI. Practice Problems I.
More informationThermodynamics  Example Problems Problems and Solutions
Thermodynamics  Example Problems Problems and Solutions 1 Examining a Power Plant Consider a power plant. At point 1 the working gas has a temperature of T = 25 C. The pressure is 1bar and the mass flow
More informationPhysics 2326  Assignment No: 3 Chapter 22 Heat Engines, Entropy and Second Law of Thermodynamics
Serway/Jewett: PSE 8e Problems Set Ch. 221 Physics 2326  Assignment No: 3 Chapter 22 Heat Engines, Entropy and Second Law of Thermodynamics Objective Questions 1. A steam turbine operates at a boiler
More informationESO 201A/202. End Sem Exam 120 Marks 3 h 19 Nov Roll No. Name Section
ESO 201A/202 End Sem Exam 120 Marks 3 h 19 Nov 2014 Roll No. Name Section Answer Questions 1 and 2 on the Question Paper itself. Answer Question 3, 4 and 5 on the Answer Booklet provided. At the end of
More informationES7A Thermodynamics HW 1: 230, 32, 52, 75, 121, 125; 318, 24, 29, 88 Spring 2003 Page 1 of 6
Spring 2003 Page 1 of 6 230 Steam Tables Given: Property table for H 2 O Find: Complete the table. T ( C) P (kpa) h (kj/kg) x phase description a) 120.23 200 2046.03 0.7 saturated mixture b) 140 361.3
More informationUNITIII PROPERTIES OF PURE SUBSTANCE AND STEAM POWER CYCLE
UNITIII PROPERTIES OF PURE SUBSTANCE AND STEAM POWER CYCLE Pure Substance A Pure substance is defined as a homogeneous material, which retains its chemical composition even though there may be a change
More informationThe First Law of Thermodynamics
Thermodynamics The First Law of Thermodynamics Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic) Reversible and Irreversible Processes Heat Engines Refrigerators and Heat Pumps The Carnot
More informationPhysics 5D  Nov 18, 2013
Physics 5D  Nov 18, 2013 30 Midterm Scores B } Number of Scores 25 20 15 10 5 F D C } A A A + 0 059.9 6064.9 6569.9 7074.9 7579.9 8084.9 Percent Range (%) The two problems with the fewest correct
More information1.5. Which thermodynamic property is introduced using the zeroth law of thermodynamics?
CHAPTER 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 6 Energy Equation for a Control Volume
Chapter 6 Energy Equation for a Control Volume Conservation of Mass and the Control Volume Closed systems: The mass of the system remain constant during a process. Control volumes: Mass can cross the boundaries,
More informationTHE SECOND LAW OF THERMODYNAMICS
1 THE SECOND LAW OF THERMODYNAMICS The FIRST LAW is a statement of the fact that ENERGY (a useful concept) is conserved. It says nothing about the WAY, or even WHETHER one form of energy can be converted
More informationChapter 6: Using Entropy
Chapter 6: Using Entropy Photo courtesy of U.S. Military Academy photo archives. Combining the 1 st and the nd Laws of Thermodynamics ENGINEERING CONTEXT Up to this point, our study of the second law has
More informationChapter 15: Thermodynamics
Chapter 15: Thermodynamics The First Law of Thermodynamics Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic) Reversible and Irreversible Processes Heat Engines Refrigerators and Heat
More informationCHAPTER 7 THE SECOND LAW OF THERMODYNAMICS. Blank
CHAPTER 7 THE SECOND LAW OF THERMODYNAMICS Blank SONNTAG/BORGNAKKE STUDY PROBLEM 71 7.1 A car engine and its fuel consumption A car engine produces 136 hp on the output shaft with a thermal efficiency
More informationENGINEERING COUNCIL CERTIFICATE LEVEL THERMODYNAMIC, FLUID AND PROCESS ENGINEERING C106 TUTORIAL 5 STEAM POWER CYCLES
ENGINEERING COUNCIL CERTIFICATE LEVEL THERMODYNAMIC, FLUID AND PROCESS ENGINEERING C106 TUTORIAL 5 STEAM POWER CYCLES When you have completed this tutorial you should be able to do the following. Explain
More information(b) 1. Look up c p for air in Table A.6. c p = 1004 J/kg K 2. Use equation (1) and given and looked up values to find s 2 s 1.
Problem 1 Given: Air cooled where: T 1 = 858K, P 1 = P = 4.5 MPa gage, T = 15 o C = 88K Find: (a) Show process on a Ts diagram (b) Calculate change in specific entropy if air is an ideal gas (c) Evaluate
More informationAPPLIED THERMODYNAMICS TUTORIAL 1 REVISION OF ISENTROPIC EFFICIENCY ADVANCED STEAM CYCLES
APPLIED THERMODYNAMICS TUTORIAL 1 REVISION OF ISENTROPIC EFFICIENCY ADVANCED STEAM CYCLES INTRODUCTION This tutorial is designed for students wishing to extend their knowledge of thermodynamics to a more
More informationSheet 8:Chapter C It is less than the thermal efficiency of a Carnot cycle.
Thermo 1 (MEP 261) Thermodynamics An Engineering Approach Yunus A. Cengel & Michael A. Boles 7 th Edition, McGrawHill Companies, ISBN9780073529325, 2008 Sheet 8:Chapter 9 9 2C How does the thermal
More informationReversible & Irreversible Processes
Reversible & Irreversible Processes Example of a Reversible Process: Cylinder must be pulled or pushed slowly enough (quasistatically) that the system remains in thermal equilibrium (isothermal). Change
More informationThermodynamic Systems
Student Academic Learning Services Page 1 of 18 Thermodynamic Systems And Performance Measurement Contents Thermodynamic Systems... 2 Defining the System... 2 The First Law... 2 Thermodynamic Efficiency...
More informationProcess chosen to calculate entropy change for the system
Chapter 6 Exaple 6.33 3.  An insulated tank (V 1.6628 L) is divided into two equal parts by a thin partition. On the left
More informationSheet 5:Chapter 5 5 1C Name four physical quantities that are conserved and two quantities that are not conserved during a process.
Thermo 1 (MEP 261) Thermodynamics An Engineering Approach Yunus A. Cengel & Michael A. Boles 7 th Edition, McGrawHill Companies, ISBN9780073529325, 2008 Sheet 5:Chapter 5 5 1C Name four physical
More informationSupplementary Notes on Entropy and the Second Law of Thermodynamics
ME 4 hermodynamics I Supplementary Notes on Entropy and the Second aw of hermodynamics Reversible Process A reversible process is one which, having taken place, can be reversed without leaving a change
More informationME 6404 THERMAL ENGINEERING. PartB (16Marks questions)
ME 6404 THERMAL ENGINEERING PartB (16Marks questions) 1. Drive and expression for the air standard efficiency of Otto cycle in terms of volume ratio. (16) 2. Drive an expression for the air standard efficiency
More informationEsystem = 0 = Ein Eout
AGENDA: I. Introduction to Thermodynamics II. First Law Efficiency III. Second Law Efficiency IV. Property Diagrams and Power Cycles V. Additional Material, Terms, and Variables VI. Practice Problems I.
More informationEntropy. Objectives. MAE 320  Chapter 7. Definition of Entropy. Definition of Entropy. Definition of Entropy. Definition of Entropy + Δ
MAE 320  Chapter 7 Entropy Objectives Defe a new property called entropy to quantify the secondlaw effects. Establish the crease of entropy prciple. Calculate the entropy changes that take place durg
More informationChapter 11 Heat Engines and the. Thermodynamics
Chapter 11 Heat Engines and the Second Law of Thermodynamics The laws of thermodynamics...are critical to making intelligent choices about energy in today s global economy. What Is a Heat Engine? How do
More informationThe First Law of Thermodynamics: Closed Systems. Heat Transfer
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 gained
More informationCHAPTER 20. Q h = 100/0.2 J = 500 J Q c = J = 400 J
CHAPTER 20 1* Where does the energy come from in an internalcombustion engine? In a steam engine? Internal combustion engine: From the heat of combustion (see Problems 19106 to 19109). Steam engine:
More informationTHERMODYNAMICS NOTES  BOOK 2 OF 2
THERMODYNAMICS & FLUIDS (Thermodynamics level 1\Thermo & Fluids Module Thermo Book 2ContentsDecember 07.doc) UFMEQU201 THERMODYNAMICS NOTES  BOOK 2 OF 2 Students must read through these notes and
More informationApplied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture  10 Steam Power Cycle, Steam Nozzle Good afternoon everybody.
More informationENTROPY AND THE SECOND LAW OF THERMODYNAMICS
Chapter 20: ENTROPY AND THE SECOND LAW OF THERMODYNAMICS 1. In a reversible process the system: A. is always close to equilibrium states B. is close to equilibrium states only at the beginning and end
More informationLesson 5 Review of fundamental principles Thermodynamics : Part II
Lesson 5 Review of fundamental principles Thermodynamics : Part II Version ME, IIT Kharagpur .The specific objectives are to:. State principles of evaluating thermodynamic properties of pure substances
More informationSecond Law of Thermodynamics
Thermodynamics T8 Second Law of Thermodynamics Learning Goal: To understand the implications of the second law of thermodynamics. The second law of thermodynamics explains the direction in which the thermodynamic
More information= T T V V T = V. By using the relation given in the problem, we can write this as: ( P + T ( P/ T)V ) = T
hermodynamics: Examples for chapter 3. 1. Show that C / = 0 for a an ideal gas, b a van der Waals gas and c a gas following P = nr. Assume that the following result nb holds: U = P P Hint: In b and c,
More informationTHERMODYNAMICS TUTORIAL 5 HEAT PUMPS AND REFRIGERATION. On completion of this tutorial you should be able to do the following.
THERMODYNAMICS TUTORIAL 5 HEAT PUMPS AND REFRIGERATION On completion of this tutorial you should be able to do the following. Discuss the merits of different refrigerants. Use thermodynamic tables for
More informationCO 2 41.2 MPa (abs) 20 C
comp_02 A CO 2 cartridge is used to propel a small rocket cart. Compressed CO 2, stored at a pressure of 41.2 MPa (abs) and a temperature of 20 C, is expanded through a smoothly contoured converging nozzle
More informationExpansion and Compression of a Gas
Physics 6B  Winter 2011 Homework 4 Solutions Expansion and Compression of a Gas In an adiabatic process, there is no heat transferred to or from the system i.e. dq = 0. The first law of thermodynamics
More informationLecture 36 (Walker 18.8,18.56,)
Lecture 36 (Walker 18.8,18.56,) Entropy 2 nd Law of Thermodynamics Dec. 11, 2009 Help Session: Today, 3:104:00, TH230 Review Session: Monday, 3:104:00, TH230 Solutions to practice Lecture 36 final on
More informationLesson 10 Vapour Compression Refrigeration Systems
Lesson 0 Vapour Compression Refrigeration Systems Version ME, II Kharagpur he specific objectives of the lesson: his lesson discusses the most commonly used refrigeration system, ie Vapour compression
More informationOUTCOME 4 STEAM AND GAS TURBINE POWER PLANT. TUTORIAL No. 8 STEAM CYCLES
UNIT 61: ENGINEERING THERMODYNAMICS Unit code: D/601/1410 QCF level: 5 Credit value: 15 OUTCOME 4 STEAM AND GAS TURBINE POWER PLANT TUTORIAL No. 8 STEAM CYCLES 4 Understand the operation of steam and gas
More informationHeat Engines, Entropy, and the Second Law of Thermodynamics
2.2 This is the Nearest One Head 669 P U Z Z L E R The purpose of a refrigerator is to keep its contents cool. Beyond the attendant increase in your electricity bill, there is another good reason you should
More informationAPPLIED THERMODYNAMICS. TUTORIAL No.3 GAS TURBINE POWER CYCLES. Revise gas expansions in turbines. Study the Joule cycle with friction.
APPLIED HERMODYNAMICS UORIAL No. GAS URBINE POWER CYCLES In this tutorial you will do the following. Revise gas expansions in turbines. Revise the Joule cycle. Study the Joule cycle with friction. Extend
More informationBasic Thermodynamics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Basic Thermodynamics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture  10 Second Law and Available Energy  I Good morning, I welcome you to this
More informationUNIVERSITY ESSAY QUESTIONS:
UNIT I 1. What is a thermodynamic cycle? 2. What is meant by air standard cycle? 3. Name the various gas power cycles". 4. What are the assumptions made for air standard cycle analysis 5. Mention the various
More informationQUESTIONS THERMODYNAMICS PRACTICE PROBLEMS FOR NONTECHNICAL MAJORS. Thermodynamic Properties
QUESTIONS THERMODYNAMICS PRACTICE PROBLEMS FOR NONTECHNICAL MAJORS Thermodynamic Properties 1. If an object has a weight of 10 lbf on the moon, what would the same object weigh on Jupiter? ft ft ft g
More informationA HEAT ENGINE: PRODUCES WORK FROM HEAT BY WASTING A FRACTION OF HEAT INPUT TO A LOW TEMPERATURE RESERVOIR
RANKINE POWER GENERAION CYCE A EA ENGINE: PRODUCES WORK FROM EA BY WASING A FRACION OF EA INPU O A OW EMPERAURE RESERVOIR o C CARACERISICS s (kj/kgk. Rankine cycle is a eat engine comprised of four internally
More informationUNIT 5 REFRIGERATION SYSTEMS
UNIT REFRIGERATION SYSTEMS Refrigeration Systems Structure. Introduction Objectives. Vapour Compression Systems. Carnot Vapour Compression Systems. Limitations of Carnot Vapour Compression Systems with
More informationC H A P T E R T W O. Fundamentals of Steam Power
35 C H A P T E R T W O Fundamentals of Steam Power 2.1 Introduction Much of the electricity used in the United States is produced in steam power plants. Despite efforts to develop alternative energy converters,
More informationMass and Energy Analysis of Control Volumes
MAE 320Chapter 5 Mass and Energy Analysis of Control Volumes Objectives Develop the conservation of mass principle. Apply the conservation of mass principle to various systems including steady and unsteadyflow
More informationEntropy and the Second Law of Thermodynamics. The Adiabatic Expansion of Gases
Lecture 7 Entropy and the Second Law of Thermodynamics 15/08/07 The Adiabatic Expansion of Gases In an adiabatic process no heat is transferred, Q=0 = C P / C V is assumed to be constant during this process
More information4.1 Introduction. 4.2 Work. Thermodynamics
Thermodynamics 41 Chapter 4 Thermodynamics 4.1 Introduction This chapter focusses on the turbine cycle:thermodynamics and heat engines. The objective is to provide enough understanding of the turbine
More informationProblems of Chapter 2
Section 2.1 Heat and Internal Energy Problems of Chapter 2 1 On his honeymoon James Joule traveled from England to Switzerland. He attempted to verify his idea of the interconvertibility of mechanical
More informationAn analysis of a thermal power plant working on a Rankine cycle: A theoretical investigation
An analysis of a thermal power plant working on a Rankine cycle: A theoretical investigation R K Kapooria Department of Mechanical Engineering, BRCM College of Engineering & Technology, Bahal (Haryana)
More information4.28 A car drives for half an hour at constant speed and uses 30 MJ over a distance of 40 km. What was the traction force to the road and its speed?
4.8 A car drives for half an hour at constant speed and uses 30 MJ over a distance of 40 km. What was the traction force to the road and its speed? We need to relate the work to the force and distance
More informationAC 20112088: ON THE WORK BY ELECTRICITY IN THE FIRST AND SECOND LAWS OF THERMODYNAMICS
AC 20112088: ON THE WORK BY ELECTRICITY IN THE FIRST AND SECOND LAWS OF THERMODYNAMICS Hyun W. Kim, Youngstown State University Hyun W. Kim, Ph.D., P.E. Hyun W. Kim is a professor of mechanical engineering
More informationHeat as Energy Transfer. Heat is energy transferred from one object to another because of a difference in temperature
Unit of heat: calorie (cal) Heat as Energy Transfer Heat is energy transferred from one object to another because of a difference in temperature 1 cal is the amount of heat necessary to raise the temperature
More informationESO201A: Thermodynamics
ESO201A: Thermodynamics Instructor: Sameer Khandekar First Semester: 2015 2016 Lecture #1: Course File Introduction to Thermodynamics, Importance, Definitions Continuum, System: closed, open and isolated,
More informationStirling heat engine Internal combustion engine (Otto cycle) Diesel engine Steam engine (Rankine cycle) Kitchen Refrigerator
Lecture. Real eat Engines and refrigerators (Ch. ) Stirling heat engine Internal combustion engine (Otto cycle) Diesel engine Steam engine (Rankine cycle) Kitchen Refrigerator Carnot Cycle  is not very
More informationThermodynamics Lecture Series
Thermodynamics Lecture Series Pure substances Property tables and Property Diagrams Applied Sciences Education Research Group (ASERG) Faculty of Applied Sciences Universiti Teknologi MARA email: drjjlanita@hotmail.com
More informationAvailability. Second Law Analysis of Systems. Reading Problems 10.1 10.4 10.59, 10.65, 10.66, 10.67 10.69, 10.75, 10.81, 10.
Availability Readg Problems 10.1 10.4 10.59, 10.65, 10.66, 10.67 10.69, 10.75, 10.81, 10.88 Second Law Analysis of Systems AVAILABILITY: the theoretical maximum amount of reversible work that can be obtaed
More informationFinal Exam Review Questions PHY Final Chapters
Final Exam Review Questions PHY 2425  Final Chapters Section: 17 1 Topic: Thermal Equilibrium and Temperature Type: Numerical 12 A temperature of 14ºF is equivalent to A) 10ºC B) 7.77ºC C) 25.5ºC D) 26.7ºC
More informationME 6404 THERMAL ENGINEERING UNIT I. PartA
UNIT I PartA 1. For a given compression ratio and heat addition explain why otto cycle is more efficient than diesel cycle? 2. Explain the effect of pressure ratio on the net output and efficiency of
More informationChapter 22. Heat Engines, Entropy, and the Second Law of Thermodynamics
Chapter 22 Heat Engines, Entropy, and the Second Law of Thermodynamics C HAP T E R O UTLI N E 221 Heat Engines and the Second Law of Thermodynamics 222 Heat Pumps and Refrigerators 223 Reversible and Irreversible
More informationLaws of Thermodynamics
Laws of Thermodynamics Thermodynamics Thermodynamics is the study of the effects of work, heat, and energy on a system Thermodynamics is only concerned with macroscopic (largescale) changes and observations
More informationExergy: the quality of energy N. Woudstra
Exergy: the quality of energy N. Woudstra Introduction Characteristic for our society is a massive consumption of goods and energy. Continuation of this way of life in the long term is only possible if
More informationES7A Thermodynamics HW 5: 562, 81, 96, 134; 729, 40, 42, 67, 71, 106 Spring 2003 Page 1 of 7
ES7A hermodynamic HW 5: 56, 8, 96, 34; 79, 4, 4, 67, 7, 6 Sring 3 Page of 7 56 Heat Pum Given: A heat um i ued to maintain a houe at 3 C. he houe loe heat to the outide at a rate of 6, kj/h, and the
More informationCondensers & Evaporator Chapter 5
Condensers & Evaporator Chapter 5 This raises the condenser temperature and the corresponding pressure thereby reducing the COP. Page 134 of 263 Condensers & Evaporator Chapter 5 OBJECTIVE QUESTIONS (GATE,
More informationPhys 2101 Gabriela González
Phys 2101 Gabriela González If volume is constant ( isochoric process), W=0, and Tp 1 is constant. pv=nrt If temperature is constant ( isothermal process), ΔE int =0, and pv is constant. If pressure is
More informationEntropy and The Second Law of Thermodynamics
The Second Law of Thermodynamics (SL) Entropy and The Second Law of Thermodynamics Explain and manipulate the second law State and illustrate by example the second law of thermodynamics Write both the
More informationPhysics Assignment No 2 Chapter 20 The First Law of Thermodynamics Note: Answers are provided only to numerical problems.
Serway/Jewett: PSE 8e Problems Set Ch. 201 Physics 2326  Assignment No 2 Chapter 20 The First Law of Thermodynamics Note: Answers are provided only to numerical problems. 1. How long would it take a
More informationCHAPTER 25 IDEAL GAS LAWS
EXERCISE 139, Page 303 CHAPTER 5 IDEAL GAS LAWS 1. The pressure of a mass of gas is increased from 150 kpa to 750 kpa at constant temperature. Determine the final volume of the gas, if its initial volume
More informationEngineering design experience of an undergraduate thermodynamics course
World Transactions on Engineering and Technology Education Vol.10, No.1, 2012 2012 WIETE Engineering design experience of an undergraduate thermodynamics course Hosni I. AbuMulaweh & Abdulaziz A. AlArfaj
More informationEfficiency Expressions for Combined Cycle
Sanjay Vijayaraghavan Email: sanv@ufl.edu D. Y. Goswami Email: goswami@ufl.edu Mechanical and Aerospace Engineering Department University of Florida, P.O. Box 116300 Gainesville, Florida 326116300 On
More informationThe First Law of Thermodynamics
The First Law of Thermodynamics (FL) The First Law of Thermodynamics Explain and manipulate the first law Write the integral and differential forms of the first law Describe the physical meaning of each
More informationAn introduction to thermodynamics applied to Organic Rankine Cycles
An introduction to thermodynamics applied to Organic Rankine Cycles By : Sylvain Quoilin PhD Student at the University of Liège November 2008 1 Definition of a few thermodynamic variables 1.1 Main thermodynamics
More informationChapter 7 Vapor and Gas Power Systems
Chapter 7 Vapor and Ga Power Sytem In thi chapter, we will tudy the baic component of common indutrial power and refrigeration ytem. Thee ytem are eentially thermodyanmic cycle in which a working fluid
More informationRefrigeration and Air Conditioning Prof. M. Ramgopal Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Refrigeration and Air Conditioning Prof. M. Ramgopal Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture No. # 18 Worked Out Examples  I Welcome back in this lecture.
More informationEngineering Software P.O. Box 2134, Kensington, MD 20891 Phone: (301) 9199670 Web Site:
Engineering Software P.O. Box 2134, Kensington, MD 20891 Phone: (301) 9199670 EMail: info@engineering4e.com Web Site: http://www.engineering4e.com Brayton Cycle (Gas Turbine) for Propulsion Application
More informationThermodynamics. October 12, 2011
Thermodynamics FE Review Session October 12, 2011 What is Thermodynamics? Application of three governing principles to individual components or systems under consideration» Mass Balance» Energy Balance»
More informationLesson. 11 Vapour Compression Refrigeration Systems: Performance Aspects And Cycle Modifications. Version 1 ME, IIT Kharagpur 1
Lesson Vapour Compression Refrigeration Systems: Performance Aspects And Cycle Modifications Version ME, IIT Kharagpur The objectives of this lecture are to discuss. Performance aspects of SSS cycle and
More informationEnergy : ṁ (h i + ½V i 2 ) = ṁ(h e + ½V e 2 ) Due to high T take h from table A.7.1
6.33 In a jet engine a flow of air at 000 K, 00 kpa and 30 m/s enters a nozzle, as shown in Fig. P6.33, where the air exits at 850 K, 90 kpa. What is the exit velocity assuming no heat loss? C.V. nozzle.
More informationEngineering Problem Solving as Model Building
Engineering Problem Solving as Model Building Part 1. How professors think about problem solving. Part 2. Mech2 and BrainFull Crisis Part 1 How experts think about problem solving When we solve a problem
More informationSustainable Energy Utilization, Refrigeration Technology. A brief history of refrigeration. Björn Palm
Sustainable Energy Utilization, Refrigeration Technology Björn Palm bpalm@energy.kth.se This email message was generated automatically. Please do not send a reply. You are registered for the following
More information