Actual exergy intake to perform the same task

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1 CHAPER : PRINCIPLES OF ENERGY CONSERVAION INRODUCION Energy conservation rinciles are based on thermodynamics If we look into the simle and most direct statement of the first law of thermodynamics, we can is see that energy is conserved Energy can neither be created nor destroyed First law efficiency indicates conversion of energy in a cycle as given by Eq () Outut η = Inut he net outut can be work or heat deending uon the alication In case of a heat engine, the net outut is in the form of work whereas in case of a solar water heating system it can be the useful heat he difference between the net outut and inut is the energy unused It is always the need from the first law of the thermodynamics to reduce the energy waste from the system First law efficiency can be imroved to limited extent with heavy insulation to the system On the other hand, second law of thermodynamics rovides the criterion as to the robability of various rocesses Sadi Carnot established this law in 84 But it got imortance from last two decades (98 onwards) in view of conservation of energy First law dictates that energy in a system is always conserved here is no way to conserve energy by use of first law of thermodynamics However, it is the exergy, which is a consequence of the second law that is never conserved Unlike energy, exergy always decreases So minimization of exergy loss is nothing but the rincile of energy conservation he net exergy outut to the actual exergy inut to the system is the second law efficiency Minimum exergy intake to erform a given task η = = II Actual exergy intake to erform the same task A min A where, A = exergy o imrove a system, we always try to imrove the second law efficiency Exergy is a tool to identify the loss of energy In comlicated systems, exergy loss at different locations can give an estimation of such losses and measures can be taken u to reduce the same QUALIY OF ENERGY Quantative evaluation of energy in a cycle or in a rocess can be done using the first law of thermodynamics he direction of flow of heat or work is known from the second law of thermodynamics However, it is equally imortant to assign the quality to the energy Energy can be broadly classified into () High grade and () Low grade energy High 38

2 grade form of energy are highly organised in nature and conversion of such energy to W W is not dictated by the second law of some other high grade form thermodynamics Conversion of high grade energy to low grade energy is not desirable However, there may be some conversion to low grade energy as work is converted into other useful form his is because of dissiation of heat due to friction (examle: mechanical work Electricity, some losses are there due to the friction in bearing of machineries) hus both the first and the second law of thermodynamics are to be considered for analysis Low grade energy such as heat due to combustion, fission, fusion reactions as well as internal energies are highly random in nature Conversion of such form of energy into high grade energy ( Q W ) is of interest his is due to the high quality of organised form of energy obtained from low quality energy Second law of thermodynamics dictates that conversion of % heat into work is never ossible hat art of low grade energy which is available for conversion is termed as available energy, availability or exergy he art, which according to the second law of thermodynamics, must be rejected is known as unavailable energy Exergy is the tool, which indicates how far the system dearts from equilibrium state he concet of exergy was ut forward by Gibbs in 878 It was further develoed by Rant in 957 Exergy analysis hels in finding the following: It can be used to determine the tye, location and magnitude of energy losses in a system It can be used to find means to reduce losses to make the energy system more efficient At this oint, it is worth mentioning that the environment lays an imortant role in evaluating the exergy (comosite roerty) 3 IMPORANCE OF EXERGY ANALYSIS Let us take an examle In case of coal fired ower lant, the first law indicates that the condenser greatly effects the ower lant efficiency as large amount of heat is transferred to the cooling water without roviding any clue on the real usefulness of this relatively low temerature fluid Also, energy balances do not rovide information about the internal losses such as throttling valve and heat exchanger Second law or exergy balance, however indicates that there is hardly % exergy loss in the condenser with more than 6% in the boiler he contribution in the boiler exergy loss accounts for irreversibilities 39

3 associated with combustion and finite temerature differences Hence, analysis of exergy lays a deterministic role in identification of rocesses and rectifying the comonents 4 AVAILABLE ENERGY REFERRED O A CYCLE Consider a cyclic heat engine as shown in Fig he maximum work outut obtainable AE he minimum energy that has to be from this engine is the available energy rejected to the sink ( as er nd law of thermodynamics) is called the unavailable energy ( UE ) Q E W max = AE Q = UE Fig Available and unavailable energy in a cycle From energy balance Again, Q = AE + UE (3) W = AE = Q UE (4) max For the given source and sink temeratures & resectively η = rev (5) If is given, η will increase with decrease of he lowest ractical temerature of rev heat rejection is the temerature of the surroundings, (6) η = max 4

4 And, W = Q max (7) 5 AVAILABILIY IN A FINIE PROCESS Let x-y be a finite rocess in which heat is sulied reversibly to a heat engine aking an elementary cycle (shaded area in Fig ), let dq be the heat received by the engine reversibly at temerature hen (8) For the whole rocess x-y, dw = dq = dq dq = AE max y Q xy x Available energy, or energy Wmax = W xy Unavailable energy s = (Sy - Sx) Fig Unavailable energy by nd law of thermodynamics dq y y y y dw = max dq dq = dq x x x x (9) Hence, maximum work obtainable is UE = Q W () x y max UE = s s y x () 4

5 6 AVAILABLE ENERGY FROM A FINIE ENERGY SOURCE m at temerature when the temerature of environment is Consider hot gas of mass g Let the gas be cooled at constant ressure to from state and the heat given u by the gas Q is utilized in heating u reversibly a working fluid of mass m wf from state 3 to along the same ath so that the temerature difference between the gas and working fluid at any instant is zero and hence entroy increase of the universe is also zero he working fluid exands reversibly and adiabatically in an engine or turbine from state to doing work W and then rejects heat Q reversibly and isothermally to return to the E initial state 3 to comlete the heat engine cycle Here, wf Q = m C = m C = Area g g wf () m g W = AE E m wf Q 3 UE Q 5 s 4 Fig 3 Available and unavailable energy in finite rocesses m g C m C g wf wf = (3) Now d s m C gas g g = ( negative as < ) (4) 4

6 d s = m C = m C wf wf wf wf wf ln (5) s = s + s = (6) univ gas wf Q = s = mwf C ln = Area wf wf (7) Available energy W = Q Q max m g C m g C g g = = Area --3- ln (8) Hence, exergy of a gas of mass m g at temerature is AE = m g C g ln (9) 7 DEMONSRAION OF QUALIY OF ENERGY BASED ON EXERGY Consider a hot gas of mass m flowing through a ieline (Fig 4) Due to the heat loss to the surroundings, the temerature of the gas decreases continuously from the inlet state ' a ' to the exit state ' b ' hus, the rocess is irreversible However, for sake of simlicity, let us consider the rocess to be reversible and isobaric between the inlet and outlet For an infinitesimal reversible rocess at constant ressure, 43

7 Fig 4 Energy quality degrades along the flow direction or, ds = mc d () hus the sloe d vice versa ds d ds = mc deends on temerature With increase in, d ds () increases and Let, Q unit of heat lost to the surroundings as temerature of gas decreases from '', being average of the two temeratures ' to Heat loss at section - 44

8 ' '' Q = mc = S Exergy lost with this heat loss at temerature is W = Q S = S S = ( ) S At section -, let heat loss be same as in section -, Q ( < ) ' '' Q = mc = S Exergy loss due to this heat loss at temerature is W = Q S = S Considering the exergy loss at temeratures and W = Q S = S S = S W = Q S = S () (3) (4) (5) (6) (7) As >, S < S W > W (8) Interretations: For the same heat loss at two different temeratures, exergy loss is more with higher temerature For instance, exergy loss is more at K than that at 3 K for kj of heat loss he more the temerature, the more is the quality of energy For examle, quality of energy of a gas at K is suerior to that at 4 K, since the gas at K has the caacity of doing more work than the gas at 4 K 7 SOME MORE EXAMPLES OF QUALIY DEGRADAION (A) HROLE PROCESS For frictionless, steady flow rocess of an ideal gas between sections ()-() and ()-() in a ie (Fig 5), 45

9 P = - P Fig 5 hrottled flow through a ie h = h (9) Again, ds = dh vd (3) h = h and v = R vd Rd (3) ds = = or, or, d s s = R = R [ ln ] s s R R = ln = ln (3) (33) = m Rln S gen I = m R ln (34) (35) 46

10 Equation (35) is the exression for the irreversibility or lost work in the system comrising of the sections ()-() and ()-() It is obvious from the exression that quality degrades logarithmically with ressure dro between the sections as well as with the mass flow rate for an initial ressure and surrounding temerature (B) FLOW WIH FRICION Consider a steady and adiabatic flow of an ideal gas through the segment of a ie Alying the first law of thermodynamics between sections ()-() and ()-() (Fig 6) In su latio n = - P P Fig 6 Adiabatic steady flow of an ideal gas in a straight ie Alying ds relationshi or, or, h = h (36) ds = dh vd (37) dh v ds = d vd d ds = = R (38) (39) d S = mds m R m Rln gen = = (4) 47

11 where ln = Hence, lost work or irreversibility is since W lost = m h s h s = m s s = S = m R [ ] gen < and higher order terms are neglected (4) Hence decrease in exergy is roortional to the ressure dro as well as mass flow rate (C) MIXING OF WO FLUID SREAMS wo fluid streams and of an incomressible fluid or ideal gas mixing adiabatically at constant ressure as shown in Fig 7 m m 3 m I n s u la t i o n 3 Mass balance: Fig 7 Mixing of two fluid streams m + m = m = m (say) 3 (4) Let, m x = m + m By first law of thermodynamics, or, m h + m h = ( m + m ) h 3 ( ) xh + x h = h 3 (43) (44) (45) 48

12 Since, From Eq (343) h h( ) = (46) ( ) x + x = 3 (47) where, x ( x) x ( x) τ 3 = + = + τ = (48) (49) By second law of thermodynamics, S = m s m s m s gen gen 3 3 ( ) S = m s x m s x s S gen 3 ( s s ) x( s s ) = + m 3 (5) (5) (5) or, Let, Sgen = C ln xc ln + m 3 S gen 3 = ln mc N =Entroy generation number= S S gen mc x (53) (54) 49

13 3 Now, substituting x ( x) For, x =, N S τ =, N = For, S = + τ from Eq (48) in Eq (54), we get N S ( x) x + τ = ln S x τ S (, ) N = N x τ = m =, the system is with single stream, temerature of both the streams are same Rate of exergy loss due to mixing is, W W = I = S lost = mc gen lost x τ ( x) x + τ ln (55) (56) (57) (58) 8 EXERGY BALANCE FOR CLOSED AND OPEN SYSEM Exergy balance for the closed and oen systems can be used to determine the locations, tyes and magnitude of losses of otential energy resources (fuels) and ways can be found to reduce such losses for making the energy system more efficient 8 CLOSED SYSEM For a closed system (Fig 8), exergy or availability transfer occurs through heat and work interactions No mass is transferred across the system boundary 5

14 Fig 38 A closed system Ist Law of thermodynamics: nd Law of thermodynamics: E E = dq W dq S S S = gen (59) (6) or, dq ( S S ) S = σ gen (6) Subtracting Eq (6) from Eq (59), we get, dq E E ( S S ) = dq W S σ gen (6) Now, Defining availability function as A = E + V S (63) A A = E E + V V S S (64) 5

15 = dq W S + gen ( V V ) σ = dq W ( V V ) S σ gen A A = dq W ( V V ) S gen σ Exergy transfer withwork Change in exergy Exergy transfer with heat Exergy destruction (65) In the form of rate equation, da dv = Q W j I j d τ dτ j Rate of change of exergy Rate of exergy transfer with heat at boundary Rate of exergy transfer as work Rate of exergy loss due to irreversibility (66) where, = instantaneous temerature at the boundary j dv = rate of change of system volume dτ For an isolated system, I > A A = A = I (67) Since,, the only rocesses allowed by the second law of thermodynamics are those for which the exergy of the isolated system decreases In other words, he exergy of an isolated system can never increase (Counterart of entroy rincile, which states that entroy of an isolated system can never decrease) 5

16 8 OPEN SYSEM: EXERGY BALANCE FOR A SEADY FLOW SYSEM Fig 9 An oen system Consider an oen system (Fig 9) Alying the first law of thermodynamics: (68) mv mv H + + mgz + Q = H + + mgz + W Second law of thermodynamics: or, From Eqs (68) and (7), dq S + S S = σ dq ( S S ) + S I = = gen σ gen (69) (7) V V H H ( S S ) + m + mg ( z z ) = dq W I σ or, A A = dq W I σ (7) (7) 53

17 In the form of rate equation at steady state, where, j Q W m ( a a ) I j + = j CV f f CV V V a a = h h s s + + g z z f f (73) (74) For a single stream entering and leaving, the exergy balance is given by Q W I σ m m m + a f a = f (75) 9 OOLS FOR EXERGY ANALYSIS he following are the tools for exergy analysis Value/Grassman/Shankey diagram Pinch oint technology In general, conversion of energy or roduction of chemical substances in a lant requires determination of all the exergy flows that are transferred between distinguished aaratuses or unit oerations of the lant he resulting exergy losses can rovide useful information with regard to the overall erformance of the lant However, it is generally difficult to judge the thermodynamic losses without any reference Evaluation of lant erformance will usually require a comarison of the thermodynamic erformance of secific aaratuses or unit oerations with available data from reviously built lants hus, exergy efficiencies have aeared to be more useful, for large lants or integrated lants It bears no significance to comonents Exergy efficiencies combined with exergy flow diagrams can rovide the thermodynamic erformance of even for a comlex system 9 VALUE DIAGRAM Consider the heat transfer between two fluids in a recuerative tye of heat exchanger emerature versus heat flow ( Q ) between the two fluids has been resented in Fig 54

18 s Q,-diagram Secondary flow ds s,in,out s,out,in Primary flow d dq Q Fig Heat transfer rocess reresented in a Q- diagram his diagram may be more informative if modified as follows: Abscissa: indicates the heat transfer between the two streams Ordinate: temerature is relaced by the term At =, the ordinate is zero and at = α, ordinate value is unity (Fig ) 55

19 [=~] value diagram 3 4 dq= area / secondary flow ' 5' dex = area rimary flow 5 exergy loss of heat transfer: dex loss =area [=] 6 dq Q Fig Heat transfer rocess reresented in a value diagram If it is assumed that an infinitesimal small amount of heat dq is transferred from the secondary flow, the flow that is cooled down in the heat exchanger, the resulting decrease in temerature d s may be neglected For the energy of this amount of heat can be written as (76) dexs = dq In the value diagram (Fig ) the area equals the amount of heat dq, whereas the area equals the exergy of this amount of heat he term indicates which art of the considered heat can be converted into work and can in rincile be converted into work and can be seen as the exergy fraction of this amount of heat he total exergy from the secondary flow van be determined by integrating Eq(76) from the inlet temerature s, i to the outlet temerature s, out s, out (77) o Exs = dq s s, in 56

20 his amount of exergy equals the whole area below the temerature curve in the value diagram Within the heat exchanger the heat dq is transferred to the rimary flow he exergy of the heat sulied to this flow is (78) dexs = dq s In the value diagram, this exergy is reresented by the area he area indicates the exergy lost due to temerature difference necessary to transfer heat from the secondary to the rimary flow he total exergy absorbed by the rimary flow is, out (79) Ex = dq, in where subscrit indicates the rimary flow he absorbed exergy by the rimary flow equals the area below the temerature curve for this flow in the value diagram he total exergy loss due to heat transfer, Exs Ex, is reresented by the area between the two temerature curves hus,the amount of exergy as well as the exergy loss can be easily resented with the hel of value diagram 9 PROCESS INEGRAION: PINCH POIN ECHNOLOGY A method of using the overall thermal balance of a rocess heoretically redicts minimum energy consumtion Predicts the construction costs of the lant using a heat recovery system First ut into ractical use in 984, at Linhoff March Co Pinch oint technology, or rocess integration, is the name given for a technique develoed by Prof Linnhof and co-workers (978) at Leeds university, UK to otimize the heat recovery in large comlex lants with several hot and cold streams of fluids o illustrate the basic rincile take a case of a lant with two hot and two cold streams, as shown in able 3 57

21 able Data for 4 (four) fluid streams Stream No Initial em ( C ) Final em ( C ) Mass flow rate ( kg / s ) S Heat caacity ( kj / kgk ) Heat caacity rate ( kw / K ) Rate of enthaly increase ( kw ) he hot streams can be combined into an equivalent comosite stream as follows: From able, it is clear that both stream and are having common temerature dro between 75 to 75 C For the common rocesses, we go for rocess integration by considering comosite thermal caacity he hot comosite curve will consists of () stream from 5 to75 C with heat caacity kw/k, () between temerature 75 to 75 C, a combined stream of thermal caacity rate ( +4)=6 kw/k (3) between temerature 75 to 65 C, stream with heat caacity rate kw/k Similarly, the cold streams can be combined he two comosite streams are then lotted on a temeratureheat load grah as shown in Fig he inch oint is defined as the oint where temerature difference between the two comosite curves is minimum emerature/ ( C) K Stream 4 kw Combined hot streams and Stream 3 Pinch oint Stream 9 kw Combined cold streams 3 and 4 stream 3 Rate of change of enthaly (kw) Fig emerature verses rate of enthaly change for comosite hot and cold streams he temerature difference at the inch oint deends on the design of the heat exchanger Smaller the temerature difference, the more exensive is the heat 58

22 exchanger A high value of inch oint indicates high thermal losses due to external irreversibility aking as an examle a temerature difference of K, say, then the cold stream (comosite) can be moved from left to right on the diagram horizontally on the diagram, keeing the hot comosite curve fixed, until the temerature difference at inch oint is K It is then seen then the external heating load of 9 kw and external cooling load of 4 kw are required for the system, all other energy changes can be achieved by the heat exchangers between the various streams Required heating and cooling of each stream above and below the inch oint is shown in able able Cooling and heating above and below inch oint Stream Above Pinch Point Below Pinch Point Cooling: Cooling: 5 5 = 6kW 5 65 = kw Cooling: = kw 3 Heating: = 5kW 4 Heating: = 5kW Net external requirement Cooling: = kw Heating: = 8kW Zero 9 kw (Heating) 4 kw (cooling) From able, we can infer that for a inch oint of K, we require external heating load of 9kW and cooling load of 4kW above and below the inch oint resectively he rocess integration above the inch oint is now as follows: Above inch oint, stream and stream 3 exchange 6 kw heat, stream and stream 4 exchange kw Below inch oint, stream and 3 exchanges 8 kw heat Stream and 4 are to be externally heated with 65 kw and 5 kw resectively to meet the deficit /demand of 9kW Similarly, stream and 3 are to be cooled externally with kw and kw heat exchangers resectively to meet the demand of 4 kw Possible rocesses are shown in Fig 3 59

23 Stream 5 C 6 kw External kw 5 C cooling 65 C Stream 8 C 583 C 5 C 8 kw 45 C Stream3 External heating 65 kw 75 C kw 5 C 8 C Stream3 75 C Stream Stream 55 C External heating 49 4 C 5kW 5 C External cooling kw Stream4 Fig 3 Possible lant to heat and cool four fluid streams for a minimum K temerature difference he following rules should be followed in rocess integration Do not transfer heat from one fluid to another across the inch oint No external heating below inch oint 3 No external cooling above the inch oint 4 A heat exchanger should oerate on one side of the inch, either taking a heat suly from below the inch, or rejecting heat to a fluid above the inch 5 A heat um should oerate across the inch from a cold stream below the inch to a hot stream above the inch Summery: Exergy is the maximum work otential of a system Exergy transfer with heat, work and mass 5 For an isolated system exergy always decreases 3 Exergy remains constant in a reversible rocess 3 Anything that generate entroy is resonsible for decrease of exergy 6

Chapter-6: Entropy. 1 Clausius Inequality. 2 Entropy - A Property

Chapter-6: Entropy. 1 Clausius Inequality. 2 Entropy - A Property hater-6: Entroy When the first law of thermodynamics was stated, the existence of roerty, the internal energy, was found. imilarly, econd law also leads to definition of another roerty, known as entroy.