Introduction to the Exergy Concept 2.83/2.813 T. G. Gutowski 2 - PowerPoint PPT Presentation

Introduction to the “Exergy” Concept 2.83/2.813 T. G. Gutowski 2

Readings 1. Ch.2 Thermodynamics Revisited (p 7-13), and 2. Ch 6 Exergy, a Convenient Concept (p 59-62), Jakob de Swaan Arons, 2004 3. Ch 6 Thermodynamic Analysis of Resources Used in Mfg Processes, Gutowski & Sekulic, in TDR 2011 4. Appendix, Tables of Standard Chemical Exergy, Szargut, Morris & Steward, 1988 3

Exergy Accounting • Exergy Units: Joules J , rate in Watts, W • Exergy symbols: B, Ex, X, and b, e x , x • Also called “ Availability ” by Keenan, 1941 and “ Available Energy ” , Ω R , by Gyftopoulos and Beretta, 1991 4

Definition of Exergy “ Exergy is the amount of work obtainable when some matter is brought to a state of thermodynamic equilibrium with the common components of the natural surroundings by means of reversible processes, involving interaction only with the above mentioned components of nature ” [Szargut et al 1988]. 5

Exergy System State Maximum work obtainable between System and Reference States. Reference State 6

Exergy System State The minimum work needed to raise System from the reference state to the System State Reference State 7

Aggregate Exergy Accounting Ex out Ex in Ex lost Ex in � Ex out = Ex lost 8

Thermodynamics Overview • Systems • Heat Interactions • Exergy of heat interaction • Entropy and Enthapy • Physical and Chemical Exergy 9

Open System Work interaction W mass in Mass mass out Q Heat interaction 10

Closed System X W mass in X Mass mass out Q 11

Isolated System X X W mass in X Mass mass out X Q 12

The 1st Law U Q W � = � in out dU Q W = � � � in out dQ dW 0 � � � = 13

Heat Interaction Q T H Q W out = Q � Q o Q o � = W out = 1 � Q o Q Q T o 14

Carnot’s “Reversible” Heat Engine Q T out � reversible = f ( T H , T L ) L = Q T in H T L 1 � = � T H 15

Maximum Work Output T max L W Q ( 1 ) = � out in T H • Q in is at T H , let T L be T ref or T o • This gives the “ available energy ” of a heat interaction at T H in reference T o . • Work and Heat are no longer equivalent! 16

Exergy “ Ex ” of Heat Interaction Q = Q (1 � T o E X T ) • Exergy, “ Ex ” is the available energy w.r.t. a reference environment, T 0 , and P 0 … • Ex(work) = W; Ex (heat) = Q(1-T 0 /T) 17

Availability “ The First Law states that in every cyclic process either work is converted into heat or heat is converted into work. In this sense it makes no distinction between work and heat except to indicate a means of measuring each in terms of equivalent units. Once this technique of measurement is established, work and heat become entirely equivalent for all applications of the First Law. ” Keenan, 1941 18

Availability • “ The Second Law, on the other hand, marks the distinction between these two quantities by stating that heat from a single source whose temperature is uniform cannot be completely converted into work in any cyclic process, whereas work from a single source can always be completely converted into heat. ” 19 Keenan, 1941

Availability “ The efforts of the engineer are very largely directed toward the control of processes so as to produce the maximum amount of work, or so as to consume the minimum amount of it. The success of these efforts can be measured by comparing the amount of work produced in a given process with the maximum possible amount of work that could be produced in the course of a change of state identical with that which occurs in the process. ” 20 Keenan, 1941

Energy, E and Exergy, B = Ex E 2 , B 2 E 1 , B 1 Properties for two different states of the system shown by the boxes. This change may come about due to spontaneous changes or due to heat or work interaction, or mass transfer. reversible process B 1 - B 2 = E 1 - E 2 irreversible process B 1 - B 2 > E 1 - E 2 Ref: Gyftopoulos and Beretta 21

Define Entropy S 1 = S o + 1 ( ) � ( B 1 � B o ) E 1 � E o � � � � C R • They show C R = T R = T 0 • Entropy is a Property • Entropy is a measure of something lost Ref: Gyftopoulos and Beretta 22

Entropy Difference S 2 � S 1 = 1 ( ) � ( B 2 � B 1 ) E 2 � E 1 � � � � C R • ∆ S = 0, reversible process • ∆ S > 0, irreversible process Ref: Gyftopoulos and Beretta 23

Example, Heat Interaction Q, T T 0 T E 2 = E 1 +Q B 2 = B 1 + Q(1-T 0 /T) Δ S = (1/T 0 )(Q – Q + Q(T 0 /T)) = Q/T Δ S = Q/T 24

Example, Work Interaction W E 2 = E 1 +W B 2 = B 1 + W Δ S = (W - W) = 0 25

Homeworks 1 & 2 T H Q in 1. Calculate the entropy change for a reversible heat engine,and 2. Calculate the entropy loss for a reversible heat engine. Use the results given in this Presentation. Q out T L 26

Answers for 1 & 2 Consider the process in two stages; 1) you transfer heat in, and 2) You transfer heat and work out. Use the result from Carnot to Show that the change in entropy is zero. This leads to the result that The exergy lost is also zero. � S = S 2 � S 1 = 1 � = Q L � Q H ( ) � ( B 2 � B 1 ) E 2 � E 1 � � � T o T L T H B lost = Q H � Q L + T o � S � W 27

Properties or State Variables • T = temperature • P = pressure intensive variables • V = volume • U = internal energy • E = energy extensive • B = exergy and • H = enthalpy (H = U + PV) intensive variables • S = entropy 28

State Variables 2 dU = 0 � d � � = � � � 2 1 1 dQ d ( U PV ) 0 � + = 0 � T = 29

Enthalpy H=U+PV 2 1 Here the Work done is W = P(V 2 – V 1 ) The First Law can be written as Q = (U+PV) 2 – (U + PV) 1 The quantity in parenthesis is Enthalpy H = U + PV Q in = Δ H The First Law can be written as Constant Pressure Equilibrium Process 30

For Flow System Energy U + P � + 1 2 mV 2 + mgz = H + 1 2 mV 2 + mgz Control Volume F d W boundary = Fd = P ν 31

Open System with H, S W H, S p o , T o H, S p, T Q, T o Consider the Work to bring the system from the reference environment at standard conditions, T o , p o to the state at T, p See Ch 6 de Swaan Arons 32

From EQ 1 & 2(Ch 6), de Swaan Arons � � � � H H Q W 0 � � + = in out out in � Q � � � S S out S 0 � � + = in out generated T o � � � � W H T S T S = � � � + o o generated Steady State Work to bring system from P o , T o to P, T 33

Minimum Work = Exergy � W W rev H T S = = � � � min o � m ( H H ) T ( S S ) = � � � p , T p , T o p , T p , T o o o o B ( H T S ) ( H T S ) = � � � o o o � W rev W min = m = B out � B in � 34

Lost Work & Lost Exergy Recall: � W = � B out � � B in + T o � S generated Let: � W = � W min + W lost then � lost = � B lost = T o � W S generated 35

Exergy also is ⋯ W p, T � W max b w � = p o , T o max � m Q B ( H T S ) ( H T S ) = � � � o o o ⋯ the maximum amount of work that can be obtained from a system in reference to the environment at standard conditions, T o , P o � Standard ref. values T 298 . 2 K , P 101 . 3 kPa = = 0 o 36

Open flow system 37

First Law for a Flow System 2 dU u � � � � � i m h gz � � � = + + � � � � i i i dt 2 � � � � in cv 2 u � � � i m j h gz � � � � + + � � j j 2 � � out � � � � Q Q W W � � � � + � + � in out sh , in sh , out 2 u � � � � � � m H g z Q W 0 EQ 1 � � � + + � + � = � � in out 2 � � 38 one stream steady state

Second Law for a Flow System dS � � � � m S m S � � = � � � i i j j dt � � in out cv � � Q Q � � � k 1 S �� �� + � + generated T T in out � Q � � m S surr S 0 EQ 2 � + + = generated T 39 one stream steady state

From EQ 1 & 2 2 u � � � � � � m H g z Q W 0 � � � + + � + � = � � out 2 � � � Q � � m S S 0 � + + = generated T 0 2 u � � � � � � W m H g z m T S � � = � + + � � o � � � rev 2 � � 40

Physical and Chemical Exergy • B = B ph + B ch • B ph (T=T o , p=p o , µ= µ * ≠ µ ο ) =0 – this is the “ restricted dead state ” • B ch ( µ * = µ ο ) = 0 • when B = B ph + B ch = 0 – this is the “ dead state ” 41

Thank you Jan Szargut 42

Chemical Reaction, at T o , p o � n 1 R 1 + n 2 R 2 → n 3 Π 3 R 1 Π 3 R 2 Q 43

Chemical Reactions stoichiome tric mass balance v R v R .... v v ... + + � � + � + a a b b j j k k exergy " balance" v b v b .... v b v b B + + � � = a R b R j k lost � � a b j k where exergy b is given in kJ/mole 44