Physics 5D - Nov 18, 2013

Size: px
Start display at page:

Transcription

1 Physics 5D - Nov 18, Midterm Scores B } Number of Scores F D C } A- A A Percent Range (%) The two problems with the fewest correct answers were #10 and # What would be the greatest effect on the ideal gas law if there is a slight attractive force between the molecules? A) At low densities, the pressure would be less than that predicted by the ideal gas law. This is why the first factor in the van der Waals equation is (P + a/v 2 ) Text Text Text Text Score: (out of 155)

2 18-5 Van der Waals Equation of State We assume that some fraction b of the volume per mole is unavailable due to the finite size of the molecules. We also expect that the pressure will be reduced by a factor proportional to the square of the density, due to interactions between the molecules. This gives the Van der Waals equation of state; the constants a and b are found experimentally for each gas: Text Text Text Text From Lecture 2

3 Problem 24 was one that I discussed in Lecture 5: An athlete is sitting unclothed in a locker room whose dark walls are at a temperature of 15 C. Calculate his net rate of heat loss by radiation, assuming a skin temperature of 34 C and ε = Take the surface area of the body not in contact with the chair to be 1.5 m 2. Answer: Let T1 = 34 = 307K, T2 = 15 = 288K (T1 4 - T2 4 ), where = (0.70)(5.67x10-8 W/m 2 K 4 )(1.5 m 2 ) (T1 4 - T2 4 ) = 120 W You have to include the radiation from the walls! Text Text Text Text

4 Physics 5D - Nov 18, 2013 The 2nd Law of Thermodynamics The second law of thermodynamics is a statement about which processes occur and which do not. This is the Clausius statement of the second law: Heat can flow spontaneously from a hot object to a cold object; it will not flow spontaneously from a cold object to a hot object. TCOLD YES! NO! THOT

5 20-2 Heat Engines No heat engine can have an efficiency of 100%. This is the Kelvin-Planck statement of the second law of thermodynamics: No device is possible whose sole effect is to transform a given amount of heat completely into work. Today s lecture will introduce a third statement of the second law, in terms of entropy.

6 20-3 Reversible Processes Classical thermodynamics only considers systems in equilibrium states. The equilibrium state of an ideal gas can be specified by giving T, V, and P; then PV/RT=n. A reversible process is one carried out so slowly that every intermediate state is in equilibrium. (Any process carried out rapidly will cause nonuniformities of temperature and pressure within the gas, so we cannot know the exact path in a P-V diagram.) Text Text Text Text

7 20-3 Reversible and Irreversible Processes; the Carnot Cycle The Carnot cycle allows us to calculate the maximum possible efficiency of a heat engine. Each leg of the cycle is reversible. The Carnot cycle consists of: Isothermal expansion ab Adiabatic expansion bc Isothermal compression cd Adiabatic compression da Adiabat Isotherm Isotherm Adiabat

8 The Carnot cycle consists of: Isothermal expansion ab Adiabatic expansion bc Isotherm Isothermal compression cd Adiabatic compression da Adiabat Isotherm Adiabat We will show that for an ideal gas Carnot cycle the efficiency because for a Carnot cycle

9 Proof that for a Carnot cycle with an ideal gas In the isothermal processes and For the adiabatic processes and for an ideal gas Dividing, Dividing these eqns, or

10 Proof that for a Carnot cycle with an ideal gas In the isothermal processes and We just found that Substituting into the expressions above for QH, QL we see that Thus for a Carnot cycle, the efficiency

11 We see that for an ideal reversible Carnot cycle, the efficiency can be written in terms of the temperatures: From this we see that 100% efficiency can be achieved only if the cold reservoir is at absolute zero, which is impossible. Thus we derive the Kelvin-Planck statement of the 2nd Law: No device is possible whose sole effect is to transform a given amount of heat completely into work. Real engines have some frictional losses; the best achieve 60 80% of the Carnot value of efficiency.

12 20-3 Reversible and Irreversible Processes; the Carnot Engine Example 20-2: A phony claim? An engine manufacturer makes the following claims: An engine s heat input per second is 9.0 kj at 435 K. The heat output per second is 4.0 kj at 285 K. Can you believe these claims?

13 20-3 Reversible and Irreversible Processes; the Carnot Engine Example 20-2: A phony claim? An engine manufacturer makes the following claims: An engine s heat input per second is 9.0 kj at 435 K. The heat output per second is 4.0 kj at 285 K. Can you believe these claims? Answer: The claims imply an efficiency of This is greater than the maximum possible e so the claim is not credible.

14 20-4 Refrigerators, Air Conditioners, and Heat Pumps These appliances are essentially heat engines operating in reverse. By doing work, heat is extracted from the cold reservoir and exhausted to the hot reservoir.

15 20-4 Refrigerators, Air Conditioners, and Heat Pumps Refrigerator performance is measured by the coefficient of performance (COP): What we want Substituting: What we pay for For an ideal Carnot refrigerator QH/QL = TH/TL, so

16 For an ideal refrigerator, Since TH - TL > 0, a perfect refrigerator (W = 0) is impossible. This is the Clausius statement of the 2nd Law, mentioned earlier: Heat can flow spontaneously from a hot object to a cold object; it will not flow spontaneously from a cold object to a hot object. TCOLD YES! NO! THOT

17 20-4 Refrigerators, Air Conditioners, and Heat Pumps Example 20-4: Making ice. A freezer has a COP of 3.8 and uses 200 W of power. How long would it take this otherwise empty freezer to freeze an ice-cube tray that contains 600 g of water at 0 C? (L = 333 kj/kg)

18 20-4 Refrigerators, Air Conditioners, and Heat Pumps Example 20-4: Making ice. A freezer has a COP of 3.8 and uses 200 W of power. How long would it take this otherwise empty freezer to freeze an ice-cube tray that contains 600 g of water at 0 C? (L = 333 kj/kg) Answer: COP = QL/W = 3.8, so QL/s = 3.8 W/s = 3.8 (200 W) = 760 W = 0.76 kj/s The latent heat of the ice-water phase transition is L = 333 kj/kg, so for 0.6 kg the needed QL = 0.6 (333 kj) = 200 kj. The time required is t = 200 kj/0.76 kj/s = 260 s = 4.4 minutes.

19 20-4 Refrigerators, Air Conditioners, and Heat Pumps A heat pump can heat a house in the winter: The heat pump coefficient of performance is defined as As usual, it s what we want divided by what we pay for.

20 Example 20-5: Heat pump. A heat pump has a coefficient of performance of 3.0 and is rated to do work at 1500 W. (a) How much heat can it add to a room per second? (b) If the heat pump were turned around to act as an air conditioner in the summer, what would you expect its coefficient of performance to be, assuming all else stays the same?

21 Example 20-5: Heat pump. A heat pump has a coefficient of performance of 3.0 and its power requirement is 1500 W. (a) How much heat can it add to a room per second? Answer: QH/s= (COP)W/s = 4500 W. (b) If the heat pump were turned around to act as an air conditioner in the summer, what would you expect its coefficient of performance to be, assuming all else stays the same? Answer: Its refrigerator COP = QL/W where QL = QH - W = 3000 J/s. Thus its air conditioning COP = 3000/1500 = 2.0.

22 20-5 Entropy Definition of the change in entropy S when an amount of heat Q is added: if the process is reversible and the temperature is constant. Otherwise, I propose to name the quantity S the entropy of the system, after the Greek word [τροπη trope], the transformation. I have deliberately chosen the word entropy to be as similar as possible to the word energy: the two quantities to be named by these words are so closely related in physical significance that a certain similarity in their names appears to be appropriate. Clausius probably chose "S" in honor of Sadi Carnot, whose 1824 paper had inspired Clausius. Text Rudolf Text Clausius Text Text Sadi Carnot

23 Which has higher entropy a kilogram of ice at 0 ºC or a kilogram of water at 0 ºC? A. Ice B. Water C. No difference

24 Which has higher entropy a kilogram of ice at 0 ºC or a kilogram of water at 0 ºC? A. Ice 1 kg of water has ΔS = Lfusion/T B. Water = 333 kj / 273K = 1.22 kj/k higher entropy than a kg of ice C. No difference

25 20-5 Entropy Any reversible cycle can be written as a succession of Carnot cycles; therefore, what is true for a Carnot cycle is true of all reversible cycles.

26 20-5 Entropy Since for any Carnot cycle Q H /Q L = T H /T L so Q H /T H Q L /T L = 0. Thus, if we approximate any reversible cycle as an infinite sum of Carnot cycles, we see that the integral of dq/t around any closed path is zero. This means that entropy is a state variable, like internal energy. The change in its value depends only on the initial and final states. (Note the analogy with the path integral argument that shows the potential energy change depends only on the initial and final states, not the path.)

27 Change of Entropy An ideal gas of n moles undergoes the reversible process from a to b shown at right, where Ta = Tb. Calculate the change of entropy. Text Text Text Text Text Text Text Text

28 Text Text Text Text Change of Entropy An ideal gas of n moles undergoes the reversible process from a to b shown at right, where Ta = Tb. Calculate the change of entropy ΔS = Sb - Sa. Text Text Text Text Answer: Entropy is a state variable, so ΔS is the same for any path. Since Ta = Tb consider an isothermal process connecting a and b, so that Q = W = nrt ln(vb/va). Then ΔS = Q/T = W/T = nr ln(vb/va).

29 Analogy between Entropy and Volume Suppose that we follow various reversible paths in a P-T diagram. We plot T vs. P, T vs. V, etc., and look for an invariant. We plot 1/P vs. Wrev and find that the area under the curves is always the same! In differential form this is or knew. Entropy Text TextEducation, Text CopyrightText 2009 Pearson Inc. as we already is like this, but hard to visualize.

30 20-6 Entropy and the Second Law of Thermodynamics TL Q TH ΔS = Q/TL - Q/TH > 0 The total entropy always increases when heat flows from a warmer object to a colder one in an isolated two-body system. The heat transferred is the same, and the cooler object is at a lower average temperature than the warmer one, so the entropy gained by the cooler one is always more than the entropy lost by the warmer one.

31 20-6 Entropy and the Second Law of Thermodynamics The fact that after every interaction the entropy of the system plus the environment never decreases is another way of putting the 2nd Law: The entropy of an isolated system never decreases. It either stays constant (reversible processes) or increases (irreversible processes).

32 20-6 Entropy and the Second Law of Thermodynamics The fact that after every interaction the entropy of the system plus the environment never decreases is another way of putting the 2nd Law: The entropy of an isolated system never decreases. It either stays constant (reversible processes) or increases (irreversible processes). ΔS = Q/TL - Q/TH > 0 for TL Q TH shows that Heat can flow spontaneously from a hot object to a cold object; it will not flow spontaneously from a cold object to a hot object. (Clausius 2nd Law)

33 20-6 Entropy and the Second Law of Thermodynamics Example 20-D: Heat transfer. Suppose that 1 kg of ice is in contact with a heat reservoir infinitesimally warmer than 0. Given that Lice-water = 333 kj/kg, calculate a. ΔSice b. ΔSheat reservoir c. ΔStotal

34 20-6 Entropy and the Second Law of Thermodynamics Example 20-D: Heat transfer. Suppose that 1 kg of ice is in contact with a heat reservoir infinitesimally warmer than 0. Given that Lice-water = 333 kj/kg, calculate a. ΔSice = kj / 273 K = 1.22 kj/k b. ΔSheat reservoir c. ΔStotal

35 20-6 Entropy and the Second Law of Thermodynamics Exercise 20-D: Heat transfer. Suppose that 1 kg of ice is in contact with a heat reservoir infinitesimally warmer than 0. Given that Lice-water = 333 kj/kg, calculate a. ΔSice = kj / 273 K = 1.22 kj/k b. ΔSheat reservoir = 1.22 kj/k c. ΔStotal = 0 This is a good example of a reversible transformation.

The 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

Supplementary 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

The 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, McGraw-Hill,

The 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

ME 201 Thermodynamics

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.

FUNDAMENTALS 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

CHAPTER 7 THE SECOND LAW OF THERMODYNAMICS. Blank

CHAPTER 7 THE SECOND LAW OF THERMODYNAMICS Blank SONNTAG/BORGNAKKE STUDY PROBLEM 7-1 7.1 A car engine and its fuel consumption A car engine produces 136 hp on the output shaft with a thermal efficiency

The 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

Phys222 W11 Quiz 1: Chapters 19-21 Keys. Name:

Name:. In order for two objects to have the same temperature, they must a. be in thermal equilibrium.

a) Use the following equation from the lecture notes: = ( 8.314 J K 1 mol 1) ( ) 10 L

hermodynamics: Examples for chapter 4. 1. One mole of nitrogen gas is allowed to expand from 0.5 to 10 L reversible and isothermal process at 300 K. Calculate the change in molar entropy using a the ideal

We will try to get familiar with a heat pump, and try to determine its performance coefficient under different circumstances.

C4. Heat Pump I. OBJECTIVE OF THE EXPERIMENT We will try to get familiar with a heat pump, and try to determine its performance coefficient under different circumstances. II. INTRODUCTION II.1. Thermodynamic

UNIT 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

Chapter 8 Maxwell relations and measurable properties

Chapter 8 Maxwell relations and measurable properties 8.1 Maxwell relations Other thermodynamic potentials emerging from Legendre transforms allow us to switch independent variables and give rise to alternate

Statistical Mechanics, Kinetic Theory Ideal Gas. 8.01t Nov 22, 2004

Statistical Mechanics, Kinetic Theory Ideal Gas 8.01t Nov 22, 2004 Statistical Mechanics and Thermodynamics Thermodynamics Old & Fundamental Understanding of Heat (I.e. Steam) Engines Part of Physics Einstein

Exergy: 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

Problem Set 1 3.20 MIT Professor Gerbrand Ceder Fall 2003

LEVEL 1 PROBLEMS Problem Set 1 3.0 MIT Professor Gerbrand Ceder Fall 003 Problem 1.1 The internal energy per kg for a certain gas is given by U = 0. 17 T + C where U is in kj/kg, T is in Kelvin, and C

Chapter 2 Classical Thermodynamics: The Second Law

Chapter 2 Classical hermodynamics: he Second Law 2.1 Heat engines and refrigerators 2.2 he second law of thermodynamics 2.3 Carnot cycles and Carnot engines 2.4* he thermodynamic temperature scale 2.5

Chapter 10 Temperature and Heat

Chapter 10 Temperature and Heat What are temperature and heat? Are they the same? What causes heat? What Is Temperature? How do we measure temperature? What are we actually measuring? Temperature and Its

Answer, Key Homework 6 David McIntyre 1

Answer, Key Homework 6 David McIntyre 1 This print-out should have 0 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making

Thermodynamics AP Physics B. Multiple Choice Questions

Thermodynamics AP Physics B Name Multiple Choice Questions 1. What is the name of the following statement: When two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium

6 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, McGraw-Hill Companies, ISBN-978-0-07-352932-5, 2008 Sheet 6:Chapter 6 6 17 A 600-MW steam power

THE KINETIC THEORY OF GASES

Chapter 19: THE KINETIC THEORY OF GASES 1. Evidence that a gas consists mostly of empty space is the fact that: A. the density of a gas becomes much greater when it is liquefied B. gases exert pressure

Lesson. 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

Problem Set 3 Solutions

Chemistry 360 Dr Jean M Standard Problem Set 3 Solutions 1 (a) One mole of an ideal gas at 98 K is expanded reversibly and isothermally from 10 L to 10 L Determine the amount of work in Joules We start

Chapter 18 Temperature, Heat, and the First Law of Thermodynamics. Problems: 8, 11, 13, 17, 21, 27, 29, 37, 39, 41, 47, 51, 57

Chapter 18 Temperature, Heat, and the First Law of Thermodynamics Problems: 8, 11, 13, 17, 21, 27, 29, 37, 39, 41, 47, 51, 57 Thermodynamics study and application of thermal energy temperature quantity

HEAT UNIT 1.1 KINETIC THEORY OF GASES. 1.1.1 Introduction. 1.1.2 Postulates of Kinetic Theory of Gases

UNIT HEAT. KINETIC THEORY OF GASES.. Introduction Molecules have a diameter of the order of Å and the distance between them in a gas is 0 Å while the interaction distance in solids is very small. R. Clausius

Final Exam CHM 3410, Dr. Mebel, Fall 2005

Final Exam CHM 3410, Dr. Mebel, Fall 2005 1. At -31.2 C, pure propane and n-butane have vapor pressures of 1200 and 200 Torr, respectively. (a) Calculate the mole fraction of propane in the liquid mixture

) and mass of each particle is m. We make an extremely small

Umeå Universitet, Fysik Vitaly Bychkov Prov i fysik, Thermodynamics, --6, kl 9.-5. Hjälpmedel: Students may use any book including the textbook Thermal physics. Present your solutions in details: it will

Temperature. Temperature

Chapter 8 Temperature Temperature a number that corresponds to the warmth or coldness of an object measured by a thermometer is a per-particle property no upper limit definite limit on lower end Temperature

Introduction to the Ideal Gas Law

Course PHYSICS260 Assignment 5 Consider ten grams of nitrogen gas at an initial pressure of 6.0 atm and at room temperature. It undergoes an isobaric expansion resulting in a quadrupling of its volume.

Diesel Cycle Analysis

Engineering Software P.O. Box 1180, Germantown, MD 20875 Phone: (301) 540-3605 FAX: (301) 540-3605 E-Mail: info@engineering-4e.com Web Site: http://www.engineering-4e.com Diesel Cycle Analysis Diesel Cycle

REFRIGERATION (& HEAT PUMPS)

REFRIGERATION (& HEAT PUMPS) Refrigeration is the 'artificial' extraction of heat from a substance in order to lower its temperature to below that of its surroundings Primarily, heat is extracted from

APPLIED 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

Chapter 1 Classical Thermodynamics: The First Law. 1.2 The first law of thermodynamics. 1.3 Real and ideal gases: a review

Chapter 1 Classical Thermodynamics: The First Law 1.1 Introduction 1.2 The first law of thermodynamics 1.3 Real and ideal gases: a review 1.4 First law for cycles 1.5 Reversible processes 1.6 Work 1.7

Chapter 6 The first law and reversibility

Chapter 6 The first law and reversibility 6.1 The first law for processes in closed systems We have discussed the properties of equilibrium states and the relationship between the thermodynamic parameters

Define the notations you are using properly. Present your arguments in details. Good luck!

Umeå Universitet, Fysik Vitaly Bychkov Prov i fysik, Thermodynamics, 0-0-4, kl 9.00-5.00 jälpmedel: Students may use any book(s) including the textbook Thermal physics. Minor notes in the books are also

Technical Thermodynamics

Technical Thermodynamics Chapter 2: Basic ideas and some definitions Prof. Dr.-Ing. habil. Egon Hassel University of Rostock, Germany Faculty of Mechanical Engineering and Ship Building Institute of Technical

The first law: transformation of energy into heat and work. Chemical reactions can be used to provide heat and for doing work.

The first law: transformation of energy into heat and work Chemical reactions can be used to provide heat and for doing work. Compare fuel value of different compounds. What drives these reactions to proceed

Thermodynamics of Mixing

Thermodynamics of Mixing Dependence of Gibbs energy on mixture composition is G = n A µ A + n B µ B and at constant T and p, systems tend towards a lower Gibbs energy The simplest example of mixing: What

The First Law of Thermodynamics

The First aw of Thermodynamics Q and W are process (path)-dependent. (Q W) = E int is independent of the process. E int = E int,f E int,i = Q W (first law) Q: + heat into the system; heat lost from the

Chapter 17: Change of Phase

Chapter 17: Change of Phase Conceptual Physics, 10e (Hewitt) 3) Evaporation is a cooling process and condensation is A) a warming process. B) a cooling process also. C) neither a warming nor cooling process.

I. A mechanical device shakes a ball-spring system vertically at its natural frequency. The ball is attached to a string, sending a harmonic wave in the positive x-direction. +x a) The ball, of mass M,

18 Q0 a speed of 45.0 m/s away from a moving car. If the car is 8 Q0 moving towards the ambulance with a speed of 15.0 m/s, what Q0 frequency does a

First Major T-042 1 A transverse sinusoidal wave is traveling on a string with a 17 speed of 300 m/s. If the wave has a frequency of 100 Hz, what 9 is the phase difference between two particles on the

High Speed Aerodynamics Prof. K. P. Sinhamahapatra Department of Aerospace Engineering Indian Institute of Technology, Kharagpur

High Speed Aerodynamics Prof. K. P. Sinhamahapatra Department of Aerospace Engineering Indian Institute of Technology, Kharagpur Module No. # 01 Lecture No. # 06 One-dimensional Gas Dynamics (Contd.) We

Heat and Work. First Law of Thermodynamics 9.1. Heat is a form of energy. Calorimetry. Work. First Law of Thermodynamics.

Heat and First Law of Thermodynamics 9. Heat Heat and Thermodynamic rocesses Thermodynamics is the science of heat and work Heat is a form of energy Calorimetry Mechanical equivalent of heat Mechanical

Problem Set 4 Solutions

Chemistry 360 Dr Jean M Standard Problem Set 4 Solutions 1 Two moles of an ideal gas are compressed isothermally and reversibly at 98 K from 1 atm to 00 atm Calculate q, w, ΔU, and ΔH For an isothermal

AN INTRODUCTION TO THE CONCEPT OF EXERGY AND ENERGY QUALITY. Truls Gundersen

AN INRODUION O HE ONEP OF EXERGY AND ENERGY QUALIY by ruls Gundersen Department of Energy and Process Engineering Norwegian University of Science and echnology rondheim, Norway Version 4, March 211 ruls

So T decreases. 1.- Does the temperature increase or decrease? For 1 mole of the vdw N2 gas:

1.- One mole of Nitrogen (N2) has been compressed at T0=273 K to the volume V0=1liter. The gas goes through the free expansion process (Q = 0, W = 0), in which the pressure drops down to the atmospheric

Freezing Point Depression: Why Don t Oceans Freeze? Teacher Advanced Version

Freezing Point Depression: Why Don t Oceans Freeze? Teacher Advanced Version Freezing point depression describes the process where the temperature at which a liquid freezes is lowered by adding another

How To Calculate The Performance Of A Refrigerator And Heat Pump

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

Dynamic Process Modeling. Process Dynamics and Control

Dynamic Process Modeling Process Dynamics and Control 1 Description of process dynamics Classes of models What do we need for control? Modeling for control Mechanical Systems Modeling Electrical circuits

An 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)

ES-7A Thermodynamics HW 1: 2-30, 32, 52, 75, 121, 125; 3-18, 24, 29, 88 Spring 2003 Page 1 of 6

Spring 2003 Page 1 of 6 2-30 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

Give all answers in MKS units: energy in Joules, pressure in Pascals, volume in m 3, etc. Only work the number of problems required. Chose wisely.

Chemistry 45/456 0 July, 007 Midterm Examination Professor G. Drobny Universal gas constant=r=8.3j/mole-k=0.08l-atm/mole-k Joule=J= Nt-m=kg-m /s 0J= L-atm. Pa=J/m 3 =N/m. atm=.0x0 5 Pa=.0 bar L=0-3 m 3.

= 1.038 atm. 760 mm Hg. = 0.989 atm. d. 767 torr = 767 mm Hg. = 1.01 atm

Chapter 13 Gases 1. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. Gases have volumes that depend on their conditions, and can be compressed or expanded by

SOLAR COOLING WITH ICE STORAGE

SOLAR COOLING WITH ICE STORAGE Beth Magerman Patrick Phelan Arizona State University 95 N. College Ave Tempe, Arizona, 8581 bmagerma@asu.edu phelan@asu.edu ABSTRACT An investigation is undertaken of a

Kinetic Theory & Ideal Gas

1 of 6 Thermodynamics Summer 2006 Kinetic Theory & Ideal Gas The study of thermodynamics usually starts with the concepts of temperature and heat, and most people feel that the temperature of an object

COURSE TITLE : REFRIGERATION AND AIR CONDITIONING COURSE CODE : 4029 COURSECATEGORY : A PERIODS/WEEK : 5 PERIODS/SEMESTER : 90 CREDITS : 4 OBJECTIVES

COURSE TITLE : REFRIGERATION AND AIR CONDITIONING COURSE CODE : 4029 COURSECATEGORY : A PERIODS/WEEK : 5 PERIODS/SEMESTER : 90 CREDITS : 4 TIME SCHEDULE MODULE TOPICS PERIODS 1 Introduction 22 Principles

Isentropic flow. Wikepedia

Isentropic flow Wikepedia In thermodynamics, an isentropic process or isoentropic process (ισον = "equal" (Greek); εντροπία entropy = "disorder"(greek)) is one in which for purposes of engineering analysis

Gas Laws. vacuum. 760 mm. air pressure. mercury

Gas Laws Some chemical reactions take place in the gas phase and others produce products that are gases. We need a way to measure the quantity of compounds in a given volume of gas and relate that to moles.

Gibbs Free Energy and Chemical Potential. NC State University

Chemistry 433 Lecture 14 Gibbs Free Energy and Chemical Potential NC State University The internal energy expressed in terms of its natural variables We can use the combination of the first and second

Entropy From Wikipedia, the free encyclopedia Entropy is a thermodynamic property that can be used to determine the energy not available for work in a thermodynamic process, such as in energy conversion

(1) The size of a gas particle is negligible as compared to the volume of the container in which the gas is placed.

Gas Laws and Kinetic Molecular Theory The Gas Laws are based on experiments, and they describe how a gas behaves under certain conditions. However, Gas Laws do not attempt to explain the behavior of gases.

15 THERMODYNAMICS. Learning Objectives

CHAPTER 15 THERMODYNAMICS 505 15 THERMODYNAMICS Figure 15.1 A steam engine uses heat transfer to do work. Tourists regularly ride this narrow-gauge steam engine train near the San Juan Skyway in Durango,

Passive Solar Design and Concepts

Passive Solar Design and Concepts Daylighting 1 Passive Solar Heating Good architecture? The judicious use of south glazing coupled with appropriate shading and thermal mass. Summer Winter Passive solar

Module P7.3 Internal energy, heat and energy transfer

F L E X I B L E L E A R N I N G A P P R O A C H T O P H Y S I C S Module P7.3 Internal energy, heat and energy transfer 1 Opening items 1.1 Module introduction 1.2 Fast track questions 1.3 Ready to study?

6-2. A quantum system has the following energy level diagram. Notice that the temperature is indicated

Chapter 6 Concept Tests 6-1. In a gas of hydrogen atoms at room temperature, what is the ratio of atoms in the 1 st excited energy state (n=2) to atoms in the ground state(n=1). (Actually H forms H 2 molecules,

ENGINEERING INFORMATION Hot water and steam service

ENGINEERING INFORMTION Hot water and steam service WHT IS STEM? Like other substances, water can exist in the form of a solid, when we call it ice; as a liquid when we call it water or as a gas when we

Chapter 4 EFFICIENCY OF ENERGY CONVERSION

Chapter 4 EFFICIENCY OF ENERGY CONVERSION The National Energy Strategy reflects a National commitment to greater efficiency in every element of energy production and use. Greater energy efficiency can

Chapter 10: Temperature and Heat

Chapter 10: Temperature and Heat 1. The temperature of a substance is A. proportional to the average kinetic energy of the molecules in a substance. B. equal to the kinetic energy of the fastest moving

Electrochemical Kinetics ( Ref. :Bard and Faulkner, Oldham and Myland, Liebhafsky and Cairns) R f = k f * C A (2) R b = k b * C B (3)

Electrochemical Kinetics ( Ref. :Bard and Faulkner, Oldham and Myland, Liebhafsky and Cairns) 1. Background Consider the reaction given below: A B (1) If k f and k b are the rate constants of the forward

IDEAL AND NON-IDEAL GASES

2/2016 ideal gas 1/8 IDEAL AND NON-IDEAL GASES PURPOSE: To measure how the pressure of a low-density gas varies with temperature, to determine the absolute zero of temperature by making a linear fit to

Practice Test. 4) The planet Earth loses heat mainly by A) conduction. B) convection. C) radiation. D) all of these Answer: C

Practice Test 1) Increase the pressure in a container of oxygen gas while keeping the temperature constant and you increase the A) molecular speed. B) molecular kinetic energy. C) Choice A and choice B

Thermodynamics: Lecture 2

Thermodynamics: Lecture 2 Chris Glosser February 11, 2001 1 OUTLINE I. Heat and Work. (A) Work, Heat and Energy: U = Q + W. (B) Methods of Heat Transport. (C) Infintesimal Work: Exact vs Inexact Differentials

Refrigeration and Airconditioning Prof. M. Ramgopal Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

Refrigeration and Airconditioning Prof. M. Ramgopal Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture No. # 22 Refrigeration System Components: Compressor (Continued)

1/7 2009/11/14 上 午 11:10 Manage this Assignment: Chapter 16 Due: 12:00am on Saturday, July 3, 2010 Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy

Energy savings in commercial refrigeration. Low pressure control

Energy savings in commercial refrigeration equipment : Low pressure control August 2011/White paper by Christophe Borlein AFF and l IIF-IIR member Make the most of your energy Summary Executive summary

Engineering Problem Solving as Model Building

Engineering Problem Solving as Model Building Part 1. How professors think about problem solving. Part 2. Mech2 and Brain-Full Crisis Part 1 How experts think about problem solving When we solve a problem

Condensers & 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,

CFD SIMULATION OF SDHW STORAGE TANK WITH AND WITHOUT HEATER

International Journal of Advancements in Research & Technology, Volume 1, Issue2, July-2012 1 CFD SIMULATION OF SDHW STORAGE TANK WITH AND WITHOUT HEATER ABSTRACT (1) Mr. Mainak Bhaumik M.E. (Thermal Engg.)

Daikin Altherma hybrid heat pump. The natural combination

Daikin Altherma hybrid heat pump The natural combination Daikin Altherma hybrid heat pump, the natural combination Why choose Daikin Altherma hybrid heat pump? What the customer wants: more energy efficient

Thermodynamics - 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

How Ground/Water Source Heat Pumps Work

How Ground/Water Source s Work Steve Kavanaugh, Professor Emeritus of Mechanical Engineering, University of Alabama Ground Source s (a.k.a. Geothermal s) are becoming more common as the costs of energy

Lecture Notes: Gas Laws and Kinetic Molecular Theory (KMT).

CHEM110 Week 9 Notes (Gas Laws) Page 1 of 7 Lecture Notes: Gas Laws and Kinetic Molecular Theory (KMT). Gases Are mostly empty space Occupy containers uniformly and completely Expand infinitely Diffuse

Chapter 4: Transfer of Thermal Energy

Chapter 4: Transfer of Thermal Energy Goals of Period 4 Section 4.1: To define temperature and thermal energy Section 4.2: To discuss three methods of thermal energy transfer. Section 4.3: To describe

Thermodynamics. Thermodynamics 1

Thermodynamics 1 Thermodynamics Some Important Topics First Law of Thermodynamics Internal Energy U ( or E) Enthalpy H Second Law of Thermodynamics Entropy S Third law of Thermodynamics Absolute Entropy

CHAPTER 12. Gases and the Kinetic-Molecular Theory

CHAPTER 12 Gases and the Kinetic-Molecular Theory 1 Gases vs. Liquids & Solids Gases Weak interactions between molecules Molecules move rapidly Fast diffusion rates Low densities Easy to compress Liquids

PERFORMANCE ANALYSIS OF VAPOUR COMPRESSION REFRIGERATION SYSTEM WITH R404A, R407C AND R410A

Int. J. Mech. Eng. & Rob. Res. 213 Jyoti Soni and R C Gupta, 213 Research Paper ISSN 2278 149 www.ijmerr.com Vol. 2, No. 1, January 213 213 IJMERR. All Rights Reserved PERFORMANCE ANALYSIS OF VAPOUR COMPRESSION

Practical Applications of Freezing by Boiling Process

Practical Applications of Freezing by Boiling Process Kenny Gotlieb, Sasha Mitchell and Daniel Walsh Physics Department, Harvard-Westlake School 37 Coldwater Canyon, N. Hollywood, CA 9164 Introduction

Boyle s law - For calculating changes in pressure or volume: P 1 V 1 = P 2 V 2. Charles law - For calculating temperature or volume changes: V 1 T 1

Common Equations Used in Chemistry Equation for density: d= m v Converting F to C: C = ( F - 32) x 5 9 Converting C to F: F = C x 9 5 + 32 Converting C to K: K = ( C + 273.15) n x molar mass of element

Energy Matters Heat. Changes of State

Energy Matters Heat Changes of State Fusion If we supply heat to a lid, such as a piece of copper, the energy supplied is given to the molecules. These start to vibrate more rapidly and with larger vibrations

We will study the temperature-pressure diagram of nitrogen, in particular the triple point.

K4. Triple Point of Nitrogen I. OBJECTIVE OF THE EXPERIMENT We will study the temperature-pressure diagram of nitrogen, in particular the triple point. II. BAKGROUND THOERY States of matter Matter is made

Physical Chemistry Laboratory I CHEM 445 Experiment 6 Vapor Pressure of a Pure Liquid (Revised, 01/09/06)

1 Physical Chemistry Laboratory I CHEM 445 Experiment 6 Vapor Pressure of a Pure Liquid (Revised, 01/09/06) The vapor pressure of a pure liquid is an intensive property of the compound. That is, the vapor

HEAT PUMPS A KEY COMPONENT IN LOW CARBON FUTURE

HEAT PUMPS A KEY COMPONENT IN LOW CARBON FUTURE Satish Joshi Managing Director CONTENTS 1. INTRODUCTION, APPLICATIONS 2. TECHNOLOGY, PROJECTS DONE, COST COMPARISION 3. HEAT PUMPS IN THE RENEWABLES DIRECTIVE,

Availability. 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

Calculating Heat Loss by Mark Crombie, Chromalox

Calculating Heat Loss by Mark Crombie, Chromalox Posted: January 30, 2006 This article deals with the basic principles of heat transfer and the calculations used for pipes and vessels. By understanding

Mohan Chandrasekharan #1

International Journal of Students Research in Technology & Management Exergy Analysis of Vapor Compression Refrigeration System Using R12 and R134a as Refrigerants Mohan Chandrasekharan #1 # Department

Energy, Entropy and Exergy Concepts and Their Roles in Thermal Engineering

Entropy 2001, 3, 116-149 entropy ISSN 1099-4300 www.mdpi.org/entropy/ Energy, Entropy and Exergy Concepts and Their Roles in Thermal Engineering Ibrahim Dincer 1 and Yunus A. Cengel 2 1 Department of Mechanical