ENZO ZANCHINI Università di Bologna +39 051 20 93295 AVAILABILITY FUNCTIONS AND THERMODYNAMIC EFFICIENCY 3. Availability functions.

Slides:



Advertisements
Similar presentations
Chapter 2 Introduction to Heat Transfer
Advertisements

PTT 201/4 THERMODYNAMICS SEM 1 (2012/2013) 1. light Energy can exist in numerous forms: Thermal Mechanical Kinetic Potential Electric Magnetic Chemical.
Chapter 4 Mass and Energy Analysis of Control Volumes (Open Systems)
Lecture# 9 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
Advanced Thermodynamics Note 6 Applications of Thermodynamics to Flow Processes Lecturer: 郭修伯.
Lesson 4 THERMODYNAMIC SYSTEMS AND PROCESSES DESCRIBE the following types of thermodynamic systems: – Isolated system – Closed system – Open system DEFINE.
Exergy: A Measure of Work Potential Study Guide in PowerPoint
Gas Dynamics ESA 341 Chapter 1 Dr Kamarul Arifin B. Ahmad PPK Aeroangkasa.
ES 202 Fluid and Thermal Systems Lecture 7: Mechanical Energy Balance (12/16/2002)
Shaft Power Cycles Ideal cycles Assumptions:
EGR 334 Thermodynamics Chapter 6: Sections 11-13
Chapter 7 ENTROPY Mehmet Kanoglu
PTT 201/4 THERMODYNAMIC SEM 1 (2012/2013). Objectives Apply the second law of thermodynamics to processes. Define a new property called entropy to quantify.
Chapter II Isentropic Flow
BASIC EQUATIONS IN INTEGRAL FORM FOR A CONTROL VOLUME – CH 4
5. MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
Calculating Entropy Change
Energy Balance Equation
AME 513 Principles of Combustion Lecture 7 Conservation equations.
The First Law of Thermodynamics
Analysis of A Disturbance in A Gas Flow P M V Subbarao Associate Professor Mechanical Engineering Department I I T Delhi Search for More Physics through.
Control Volume Analysis Using Energy
Chapter 6 Using Entropy.
Lecture slides by Mehmet Kanoglu
Entropy Rate Balance for Closed Systems
Conservation of mass If we imagine a volume of fluid in a basin, we can make a statement about the change in mass that might occur if we add or remove.
CHAPTER 5: Mass and Energy Analysis of Control Volumes
Last Time Where did all these equations come from?
Chapter 4 Control Volume Analysis Using Energy. Learning Outcomes ►Distinguish between steady-state and transient analysis, ►Distinguishing between mass.
Reynolds Transport Theorem We need to relate time derivative of a property of a system to rate of change of that property within a certain region (C.V.)
Chapter 7 Energy and Energy Balance By : Mrs. Norazian Mohamed Noor
Chapter 4: Entropy: an Additional Balance Equation.
Chapter 10 Vapor and Combined Power Cycles Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 7th edition by Yunus.
Dr. Jason Roney Mechanical and Aerospace Engineering
Entropy, the Second and Third Law of Thermodynamics By Doba Jackson, Ph.D. Associate Professor of Chemistry and Biochemistry Huntingdon College.
ME 200 L23: Clausius Inequality and Control Volume Example Problems Kim See’s Office ME Gatewood Wing Room 2172 Please check your HW Grades on Blackboard.
MME 2009 Metallurgical Thermodynamics
Abj1 1.System, Surroundings, and Their Interaction 2.Classification of Systems: Identified Volume, Identified Mass, and Isolated System Questions of Interest:
Entropy Rate Balance for Closed Systems
Second Law of Thermodynamics Alternative Statements
Control Volume Analysis Using Energy
Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
The Second Law of Thermodynamics Entropy and Work Chapter 7c.
Gestão de Sistemas Energéticos 2015/2016 Exergy Analysis Prof. Tânia Sousa
1 12. Thermodynamics The essential concepts here are heat and temperature. Heat is a form of energy transferred between a system and an environment due.
Chapter 7. Application of Thermodynamics to Flow Processes
Chapter 3: Conservation of Energy. Important Notation 2.
Chapter 5 Part 2 Mass and Energy Analysis of Control Volumes Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 8th edition.
ENZO ZANCHINI Università di Bologna AVAILABILITY FUNCTIONS AND THERMODYNAMIC EFFICIENCY 2. Conditions for mutual.
1 Chapter 5 Mass and Energy Analysis of Control Volumes.
Chapter: 07 ENTROPY.
First Law of Thermodynamics applied to Flow processes
Chapter 8 Exergy: A Measure of Work Potential Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 8th edition by Yunus.
Objectives Develop the conservation of mass principle.
Chapter: 06 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES.
Mass and Energy Analysis of Control Volumes
Chapter 8 Exergy: A Measure of Work Potential Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 5th edition by Yunus.
Chapter 6 Energy and Energy Balance
Chapter 5 The First Law of Thermodynamics for Opened Systems
Chapter 7 Entropy: A Measure of Disorder
Fluid kinematics Chapter 3
Thermodynamics: An Engineering Approach Yunus A. Cengel, Michael A
An Engineering Approach
Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
Mass and Energy Analysis of Control Volumes (Open Systems)
Fundamentals of Thermal-Fluid Sciences
Chapter Seven: Entropy
Chapter 8 EXERGY: A MEASURE OF WORK POTENTIAL
Ch. 4 The first law of thermodynamics: Control Volume
Chapter 6: Entropy First law: Energy conservation concept.
Presentation transcript:

ENZO ZANCHINI Università di Bologna AVAILABILITY FUNCTIONS AND THERMODYNAMIC EFFICIENCY 3. Availability functions for closed systems and flow availability Bologna - September

AVAILABILITY FUNCTIONS We call availability function for a system A a property of A whose decrease gives the highest work (or, the highest shaft work) which can be obtained in a process A 1 → A 2 of A which fulfils prefixed conditions. For simplicity, we will consider only simple systems or unions of simple systems ADIABATIC AVAILABILITY We call adiabatic availability of a system A a property of A whose value gives, for every initial state A 1, the maximum work obtainable by a weight process for A starting from A 1, such that the region of space occupied by A in the final state coincides with that occupied in the initial state A 1. By Theorem 1, the adiabatic availability is zero for every initial stable equilibrium state. 2

3 E S A1A1 AsAs Adiabatic availability of A 1 V

We call available energy of a system AR, where A is a closed system and R is a thermal reservoir, the property Ω = E - T 0 S, where E and S are the energy and the entropy of A, and T 0 is the temperature of the reservoir R. AVAILABLE ENERGY Given the states A 1 and A 2 of a system A and a thermal reservoir R, the highest work obtainable by a weight process for AR from A 1 to A 2 is Ω 1 – Ω 2, and is obtained only by any reversible process. THEOREM 3.1 4

PROOF Let us consider the energy and the entropy balance for AR; then, we combine algebrically the two balance equations; X is a system which performs a cycle. Energy balance (3.1) Entropy balance (3.2) (3.3) Summing (3.1) and(3.3): (3.4) The Gibbs equation for R gives From (3.4): (3.5) W A X R A 1 → A 2 T0T0 5

PROBLEM A closed system A composed of a mass m = 100 kg of liquid water, is kept at constant volume (the pressure is p = bar at the temperature T = 58.5 °C). In the initial state, the temperature is T 1 e =e363 K (  90 °C). The external environment can be considered as a thermal reservoir R with temperature T 0 = 300 K (  27 °C). Find the maximum work that can be obtained by a weight process for AR (adiabatic availability of AR, in the initial state). A X R W T0T0 H 2 O, m = 100 kg V = cost To obtain the maximum work, one must bring A in mutual stable equilibrium with R, namely at the temperature T 0. SOLUTION By Theorem 3.1: 6

Since the volume is constant: Then: v = kJ/(kg K) c v evaluated at pressure bar and temperature 58.5 °C 7

KEENAN’S AVAILABILITY FUNCTION,  Let A and B be two closed simple systems, such that B crosses only stable equilibrium states with the same values of temperature T 0 and pressure p 0, and the volume of AB is fixed. We call Keenan’s availability function  of AB the property  = E – T 0 S + p 0 V, where E, S and V are the energy, the entropy and the volume of A. Comment. System B has a behaviour very similar to that of a thermal reservoir, but its volume can change, with changes that are opposite to those of A. A B T 0, p 0 8

THEOREM 3.2 For given states A 1 and A 2, the maximum work obtainable in a weight process for AB from A 1 to A 2, with fixed volume of AB, is  1 –  2, and is obtained only by any reversible process. PROOF A B T 0, p 0 X W A1→A2A1→A2 Energy balance Entropy balance  T 0 Sum (3.6) 9

Gibbs equation for B: (3.7) By substituting (3.8) in (3.6) one obtains: (3.9) (3.8) From (3.7): 10

PROBLEM SOLUTION p 1, T 0 p 0, T 0 Since air has the same composition as the environment, the system can be considered as closed. The availability function, per mol, is 11 A student decides to build a rocket which works by compressed air. The mass of the rocket, without the air, is m = 0.5 kg. The vessel which contains the compressed air has a volume V = 0.5 dm 3. The compressed air has the same temperature as the environment air, T 0 = 300 K. The compressor compresses the air from the environment pressure p 0 = 1 bar to p 1 = 10 bar. Which is the theoretical maximum high which could be reached by the rocket in the absence of the viscous drag of the environment air?

T 1 = T 0 = 300 K p 1 = 10 bar p 0 = 1 bar V 1 = 0.5 dm 3 12

13

We call control volume a fixed region of space crossed by a fluid in motion (portion of a duct). The fluid is considered as the union of infinite infinitely small simple systems, called fluid elements, which move in a uniform gravity field g. The specific energy (per unit mass) of a fluid element is The energy of the fluid that, at an instant , is contained within the control volume, which occupies the region of space V, is given by (3.10) (3.11) (internal energy + potential energy + kinetic energy) CONTROL VOLUME 14

Let C be a control volume, with inlet section 1 and outlet section 2, crossed by a fluid in motion. We call shaft work performed by the fluid, in the time interval [  1,  2 ], the difference between the total work performed and the work performed against the external fluid (upstream and downstream). (3.12) ENERGY BALANCE FOR A CONTROL VOLUME By neglecting the work against viscous forces: 15 To introduce in C the infinitesimal volume dV 1 of fluid, pressure p 1 performs on C the work p 1 dV 1 (work per unit mass p 1 v 1 ); to expel from C the infinitesimal volume dV 2 of fluid, system C performs the work p 2 dV 2 against the pressure p 2 of the external fluid (work per unit mass p 2 v 2 ). 1 2 V C QQ  W sh dV1dV1 dV2dV2 p1p1 p2p2 dV 1 = v 1 dm dV 2 = v 2 dm

16 The energy balance equation per unit time is 1 2 V C (3.13) “Specific energy” of fluid at inlet: e 1 +p 1 v 1, of fluid at outlet e 2 +p 2 v 2. This is not a rigorous proof, but a simple way to get the result

If C is in a steady state, the left hand side of Eq. (3.13) is zero; moreover (3.14) From Eq. (3.13) with vanishing left hand side, by employing Eq. (3.14), one gets (3.16) = heat received by every kg of fluid which crosses C; = = shaft work performed by every kg of fluid which crosses C. (3.13) (3.15) CONTROL VOLUME IN STEADY STATE By dividing both sides of Eq. (3.15) by, one obtains 17

(3.16) Mean specific energy e in a cross section: fluid elements have the mean velocity W and are placed at the high z of the centre of the cross section (3.17) (3.18) (3.19) By introducing the specific enthalpy 18

1 2 R0R0 T0T0 C  C IN STEADY STATE ENTROPY BALANCE FOR A CONTROL VOLUME (3.20) (3.21) 19

Let C be a control volume in a stationary state, crossed by a fluid in motion. The states 1 and 2 of the fluid in sections 1 and 2 are fixed, and C can exchange heat with only one thermal reservoir R 0, with temperatureT 0 (which represents the external environment). 1 2 R0R0 T0T0 C TEOREM 3.3: FLOW AVAILABILITY The maximum shaft work that can be obtained from every kg of fluid which goes from section 1 to section 2 of C is, where  = h – T 0 s is called flow availability, and is obtained only by any reversible process. 20

Energy balance (by neglecting the changes in kinetic and potential energy): (3.22) Entropy balance:   (3.23) By substituting Eq. (3.23) in Eq. (3.22) one obtains: (3.24) PROOF 21

GIBBS FREE ENERGY AS A SPECIAL CASE OF FLOW AVAILABILITY 1 2 R T 0 = T C T T If in sections 1 and 2 the fluid has the same temperature T, which coincides with the temperature T 0 of R, one has : Since,, if T = T 0 one has  = g (3.25)Per kg, or mol: For a mass m or n moles:(3.26) 22