3. Review on Energy Balances

Transcription

1 3. Review n Energy Balances Objectives After cmpleting this chapter, students shuld be able t recall the law f cnservatin f energy recall hw t calculate specific enthalpy recall the meaning f heat f frmatin and hw t use it recall the meaning f heat f cmbustin and hw t utilise it recall hw t calculate heat f reactin frm the heat f frmatin and the heat f cmbustin recall hw t perfrm energy-balance calulatins knw hw t perfrm energy-balance calculatins fr unsteady-state prcesses 1

2 3.1 he Law f the Cnservatin f Energy he law f energy cnservatin is ne f the fundamental laws f nature [1]; it is als knwn as the first law f thermdynamics It simply states that, during any interactin, energy can change frm ne frm t anther, but the ttal amunt f energy remains cnstant; in ther wrds, energy cannt be created and destryed [1] When water flw frm the tp f the waterfall, as illustrated in Figure 3.1, ptential energy ( PE ) is cnverted t kinetic energy ( KE ) 2

3 Figure 3.1 Iguazu Falls at the brder f Argentina and Brazil; an illustratin f the cnversin f ne frm f energy (ptential energy) t anther (kinetic energy) (frm Chemical energy stred in fuel (e.g., gasline r diesel) is transfrmed via cmbustin int mechanical energy t drive a car up the hill, as shwn in Figure 3.2 3

4 Figure 3.2 A car is driving up the hill, an illustratin f the cnversin f ne frm f energy (chemical energy) t anther (mechanical energy) (frm he general equatin fr the cnservatin f energy (r energy-balance equatin) can be written as fllws Energy in + Energy generatin Energy ut Energy cnsumptin = Energy accumulatin (3.1) 4

5 Generally, chemical prcesses are either endthermic (energy is transferred int the system) r exthermic (energy is released frm the system) In prcess design, an energy balance is carried ut t determine energy requirements f the prcess energy releases frm the prcess Additinally, energy balance is als perfrmed in the frm f energy audit t determine the pattern f energy cnsumptin, which leads t the suggestin f energy cnservatin, recvery, r savings 5

6 As students have learned frm ChE hermdynamics curses, fr a clsed system, the energy related t mlecular mtin f the system is called the internal energy ( U ) [1-2] Students have als learned that the internal energy is a prperty f a system, and, as such, is a state functin r pint functin, meaning that the change in internal energy ( D U ) depends nly n the state f the system at the initial and final pints, nt n hw it changes r nt n the way (r path) it ges frm the initial pint t the final pint he change in internal energy ( D U ) is caused by the flw f wrk ( W ) and/r heat ( Q ) int and/r ut f the system, as depicted in Figure 3.3 6

7 Figure 3.3 he change f internal energy ( D U ) caused by heat ( Q ) and wrk ( W ) (frm It is ntewrthy that wrk ( W ) and heat ( Q ) is the frm f energy transfer; thus, they are nt the prperties f the system 7

8 In ther wrds, wrk ( W ) and heat ( Q ) are nt the state r pint functins; they are, n the cntrary, path functins, as the amunt f wrk ( W ) and heat ( Q ) depends n the way (path r prcess) they g frm the initial pint t the final pint he energy-balance equatin fr a clsed system can be written as fllws D U = Q- W (3.2) which, when divided by the mass ( m ) f the system, yields D U Q W = - m m m D u = q - w (3.3) where u is the specific internal energy 8

9 Fr an pen system r a cntrl vlume, the change in energy f the system ( D E ) sys with n chemical reactin, r the energy accumulatin term in Eq. 3.1, can be described as fllws ( ) ( ) U + PE + KE + P V + Q + W in in in in in in in - U + PE + KE + P V -Q - W = DE ut ut ut ut ut ut ut sys (3.4) Nte that PE is the ptential energy = mgl 1 KE is the kinetic energy = ˆ2 2 mv PV is the flw wrk (the wrk caused by the flw f a substance/substances int and ut f the system) At steady state, there is n accumulatin f energy f the system; thus, 9

10 D = 0. E sys Hence, Eq. 3.4 can be re-arranged as fllws ( ) ( ) U + PE + KE + P V + Q + W in in in in in in in = U + PE + KE + P V + Q + W ut ut ut ut ut ut ut (3.5) We have learned that the terms U and PV can be gruped tgether, thus creating the new term called enthalpy ( H ) as fllws H = U + PV (3.6) Additinally, fr almst all systems (but it is NO always the case) we encunter, the terms PE and KE are relatively very small and, thus, can be neglected Accrdingly, Eq. 3.5 is reduced t 10

11 ( ) ( ) U + P V + Q + W in in in in in = U + P V + Q + W ut ut ut ut ut and when cmbined with Eq. 3.6 yields H + Q + W = H + Q + W. in in in ut ut ut which can be re-arranged t be ( ) ( ) H - H = Q - Q + W - W ut in in ut in ut ( ) ( ) H - H = Q -Q - W - W ut in in ut ut in Nte that the term: Q in - Q is the net heat flw int the ut system ( Q ) W ut in - W is the net wrk dne by the (3.7) system ( W ) since when wrk is dne by the system, it reduces the energy (r enthalpy) f the system; accrdingly, it has a negative ( ) sign 11

12 Frm the definitins f Q and W abve, Eq. 3.7 can be re-written as fllws r H - H = Q - W (3.8) ut in H - H = Q- W (3.9) 2 1 where subscripts 1 and 2 represent the initial and final states, respectively Once again, when divided Eq. 3.9 by mass ( m ) f the system, it turns t be h - h = q - w (3.10) 2 1 where h 2 r h 1 is the specific enthalpy 3.2 Calculatin f Specific Enthalpy In rder t perfrm energy-balance calculatins, the value f enthalpy r specific enthalpy, fr either input r utput stream, r fr either initial r final state, must be knwn 12

13 Generally, the value f specific enthalpy available in references r handbks is nt the abslute enthalpy; it is rather a relative enthalpy referenced t the reference temperature, nrmally at 25 C (298 K) r 0 C (273 K), where the specific enthalpy at the reference temperature is nrmally set as zer (0) btain the value f enthalpy at any temperature, the fllwing equatin is emplyed where h ref ( ) ( 0) h - h = h - = h =ò c d p ref 13 ref (3.11) = enthalpy at a given temperature h = enthalpy at the reference temperature c p = specific heat capacity fr a cnstant- pressure prcess

14 Nte that Eq is applicable fr the prcess with n phase change and that the term is cmmnly called a sensible heat ò ref cd p he specific heat capacity at a cnstant pressure ( c ) is a functin f temperature, which can p be expressed mathematically as, e.g., p 2 3 c = a + b + c + d (3.12) p = (3.13) c a b c d e c a b c - p 2 = + + (3.14) in which a, b, c, d, and e are the cnstants whse values vary frm substance t substance Fr the prcess with a phase change (e.g., frm liquid phase t vapur phase), Eq is altered t 14

15 where ( ) 1 P h - h = h = c d +D h + c d c p 1 p P p ref ref ò ò 2 P (3.15) = specific heat capacity at a cnstant pressure fr Phase 1 c = specific heat capacity at a cnstant p 2 pressure fr Phase 2 = phase-transitin temperature P D h P = latent heat f the phase change frm Phase 1 t Phase 2 better understand hw Eq wrks, cnsider the fllwing example 15

16 Example Estimate the specific enthalpy f ethyl alchl (EtOH) at 1 atm and 200 C he data frm Ref. [3] are mlecular weight f ethyl alchl = g/ml the nrmal biling pint f ethyl alchl = 78.5 C latent heat f vaprisatin f ethyl alchl = kj/ml c f ethyl alchl p liquid = kj/ml-k at 25 C = kj/ml-k at 78.5 C c f ethyl alchl is expressed as p gas -3-5 = cp gas where is in C and c in kj/ml-k p gas 16

17 f Frm the given infrmatin, the average value c fr the temperature f C is p liquid c = = kj/ml-k p liquid avg 2 ( ) Hence, the term calculated as fllws P ò ref c d p 1 = ( ) ò ( ) P ò cdin Eq can be p1 ref ( ) d æ kj ö = d ç ml-k ò è ø 298 æ kj ö = é( ) Kù çè ml-k ø êë úû P ò ref c d = p kj/ml 17

18 ò P he term c d p 2 ò c d can be cmputed as fllws p2 P 200 é ù = ò d êë úû 200 ò ( -3 c d = ò ) d p P ( ) + ò d ( ) - ò ( ) + ò d (nte that must be in C) d 18

19 ò P c d p 2 = é ê ë ( ) ù úû é ù 2 + ê 2 ú ë û é ù 3 - ê 3 ú ë û é ù 4 + ê 4 ú 78.5 ë û ò P c d p 2 = é ê - ë ( )( ) é ê ë 2 é ê ë 3 é ê ë 4 ù úû ( ) ( ) ù ú û ù ú û ù ú û ( ) 19

20 ò P c d = p kj/ml Eventually, the specific enthalpy f EtOH at 200 C (with reference t 25 C) is ( - ) = ( ) = = 200 ò +D + p P ò p 1 2 r h h h h c d h c d h = = kj/ml 200 kj ml = kj/g = 1,204 kj/kg g ml 3.3 Heat f Frmatin he standard heat f frmatin ( ) 20 D is defined as the enthalpy change when ne mle f the cmpund is frmed frm its cnstituent elements at the standard state [2] (nrmally, 1 atm and 25 C) H f

21 Fr example, liquid ethyl alchl (C 2 H 5 OH) has the standard heat f frmatin f kj/ml, which means that when the elements C (2 atms), H (6 atms), and O (1 atm) are cmbined t frm C 2 H 5 OH in a liquid state at 1 atm and 25 C, energy in the amunt f per 1 ml f C 2 H 5 OH generated is released int the surrundings he standard heat f frmatin f an element in its natural frm is set t zer (0) It is ntewrthy that the term natural frm means the frm that such element appears in the nature 21

22 Fr example, carbn (C) appears in the nature in slid phase as graphite; thus, slid C (r graphite) has the heat f frmatin f 0 Hwever, xygen (O), nitrgen (N) and hydrgen (H), fr instances, appear in the nature in gas phase as O 2, N 2, and H 2, respectively; hence, the heats f frmatin f gaseus O 2, N 2, and H 2 are 0, but the heats f frmatin f atmic O, N and H are nt 0 (fr example, the heat f frmatin O and H are +249 and +218 kj/ml, respectively) Heats f frmatin at ther temperatures ( D H ) f can be cmputed using the fllwing equatin f f p D H = D H +ò c d (3.16) ref 22

23 Fr example, fr liquid EtOH, kj/ml D H f = hus, the heat f frmatin f EtOH at 78.5 C can be calculated as fllws ( ) æ kj ö æ kj ö D H = d f ç ml ò è ø èç ml-k ø ( ) (see Page 17 fr the details f calculatins) é kj ù D H f = ( ) + ( 6.98 ) = ê mlú ë û kj ml 3.4 Heats f Cmbustin he standard heat f cmbustin ( ) H is the amunt f heat (r energy) release when ne mle f a cmbustible substance cmpletely reacts with O 2 at the standard state (nrmally at 1 atm and 25 C) [2] c 23

24 Fr example, liquid EtOH has the heat f cmbustin f 1,366.9 kj/ml, which means that when 1 ml f liquid EtOH is cmpletely burned with O 2, heat in the amunt f 1,366.9 kj is released (frm the cmbustin system) int the surrundings 3.5 Heat f Reactin Fr a prcess with a chemical reactin, t keep the temperature f the system cnstant r t utilise heat prduced frm the system, heat is t be added r remved frm the system he amunt f heat input r release varies frm reactin t reactin 24

25 he standard heat f reactin ( ) D is the amunt f heat input r release when the chemical reactin is taken place under the standard state [2] (nrmally at 1 atm and 25 C) H r Fr example, H 2 (g) + ½O 2 (g) H 2 O (g); D H r = kj (3.17) H 2 (g) + ½O 2 (g) H 2 O (l); D H r = kj (3.18) C 2 H 6 C 2 H 4 + H 2 ; H r D = kj (3.19) 25

26 Nte, frm Rxns , that the phase (r state) f a prduct (and a reactant) has an effect n the amunt f heat f frmatin, as exemplified by Rxns and 3.18 the heat f reactin can be either f negative ( ) sign (e.g., Rxns and 3.18) r f psitive (+) sign (e.g., Rxn. 3.19) when the heat f reactin is f negative ( ) sign, it means that heat is released frm the system and that the reactin is exthermic when the heat f reactin is f psitive (+) sign, it means that heat is t be added int the system (t make the reactin happen) and that the reactin is endthermic 26

27 Generally, the heat f reactin is described as kj per ne mle f a specified substance in the chemical reactin Fr example, frm Rxn. 3.17, the heat f reactin can be described as kj per 1 ml f H 2 (g) r kj/ml H 2 (g) kj per 1 ml f H 2 O (g) r kj/ml H 2 O (g) D H r = D H r = Hwever, when described per 1 ml f O 2 (g), it must be kj per 1 2 ml f O 2 (g) r D H r = kj 1 ml O 2 g 2 ( ) = kj/ml O 2 (g) 27

28 he standard heat f reactin can be cmputed frm the data f either standard heat f frmatin r standard heat f cmbustin using the fllwing equatins: where Frm the heat f frmatin: n i ( n ) ( n ) å å D H = DH - DH H f i r i f i f i i Prducts Reactants (3.20) = stichimetric cefficient f species i D = heat f frmatin f species i where Frm the heat f cmbustin: ( n ) ( n ) å å D H = DH - DH r i c i c i i Reactants Prducts D = heat f cmbustin f species i H c i (3.21) 28

29 he fllwing example illustrates hw the standard heat f reactin ( ) D can be cmputed frm the data f standard heat f frmatin ( D ) and standard heat f cmbustin ( D ) H f H r H c Example Cyclhexane (C 6 H 12 ) can be prduced frm the reactin between benzene (C 6 H 6 ) and hydrgen (H 2 ) in the fllwing reactin: C 6 H 6 (g) + 3H 2 (g) C 6 H 12 (g) (3.22) Calculate the standard heat f reactin ( D ) frm the data f standard heat f frmatin ( ) D and data f standard heat f cmbustin H f ( D ) H c H r 29

30 he standard heat f frmatin ( ) C 6 H 6 (g) = kj/ml [nte that kj/ml] H 2 (g) = 0 D f H f D H f f C 6 H 6 (l) = C 6 H 12 (g) = kj/ml [nte that kj/ml] D H f f C 6 H 12 (l) = By using Eq. 3.20, the standard heat f reactin f Rxn can be calculated as fllws ( n ) ( n ) å å D H = DH - DH r i f i f i i Prducts Reactants é æ kj ö ù D H = ( 1 ml) rrxn ê ç ml ú ë è øû é æ kj ö ê ç ml ë è ø D H = kj r Rxn ( 1 ml) ( 3 ml)( 0) 30 ù ú û

31 D =- H r kj/ml C H 6 12 Rxn he standard heat f cmbustin ( ) C 6 H 6 (g) = 3,301.5 kj/ml [nte that kj/ml] D f H c D H c f C 6 H 6 (l) = 3,267.6 H 2 (g) = kj/ml C 6 H 12 (g) = 3,953.0 kj/ml [nte that kj/ml] D H c f C 6 H 12 (l) = 3,919.9 By emplying Eq. 3.21, the standard heat f reactin f Rxn can be cmputed as fllws ( n ) ( n ) å å D H = DH - DH r i c i c i i Reactants Prducts 31

32 é æ kj ö æ kj öù D H = ( 1 ml) - 3, ( 3 ml) rrxn ê ç ml ç ml ú ë è ø è øû é æ kj ö ù - ( 1 ml) - 3,953.0 ê ç ml ú ë è øû D H = kj r Rxn D =- H r kj/ml C H 6 12 Rxn Nte that this reactin is exthermic, meaning that heat is released frm the system and that the temperature f the system increases as the reactin prceeds 3.6 Examples f Energy-balance Calculatins In this sectin, examples regarding the energybalance calculatins are illustrated as fllws [3-4] 32

33 Example A pump is used t bring water t a prcess unit at the required flw rate f 50 m 3 /h thrugh the inlet pipe with the inner diameter (i.d.) f 12 cm At the exit, which is 10 m abve the inlet, the diameter f the pipe is reduced t 3 cm he temperatures and pressures at the inlet and the exit are almst identical pump Determine the pwer ( W ) required by the In this kind f prblem, ptential energy ( PE ) and kinetic energy ( KE ) cannt be neglected hus, Eq. 3.5 is t be used fr the energybalance calculatins fr this example 33

34 ( ) ( ) U + PE + KE + P V + Q + W in in in in in in in = U + PE + KE + P V + Q + W ut ut ut ut ut ut ut (3.5) Gruping U and PV tgether and re-arranging the resulting equatin yields ( ) ( ) H + PE + KE + Q + W in in in in in = H + PE + KE + Q + W ut ut ut ut ut ( H - H ) + ( PE - PE ) + ( KE -KE ) ut in ut in ut in = ( Q -Q )-( W -W ) in ut ut in ( ) ( ) ( ) H - H + PE - PE + KE - KE = Q - W ut in ut in ut in (3.23) Since, nrmally, the amunt f heat transferred int and ut f a pump is very small and, thus, can be neglected, Eq is reduced t 34

35 ( ) ( ) ( ) H - H + PE - PE + KE - KE =- W ut in ut in ut in (3.24) Additinally, it is given by the prblem statement that the temperatures and pressures at the inlet and the exit are almst identical his results in the fact that the enthalpy change ( H H H ) D = - is very clse t zer (0) ut in Accrdingly, Eq can be re-written and re-arranged as fllws ( ) ( ) PE - PE + KE - KE =- W ut in ut in ( ) ( ) W = PE - PE + KE - KE in ut in ut æ1 ( ) ˆ2 1 ö ˆ2 W = m gl - m gl + m V m V in in ut ut - ç in in ut ut è2 2 ø (3.25) 35

36 Since the system is at a steady state, and there is n chemical reactin, r m = m = m in in ut m = m = m ut herefre, Eq is can be re-written as fllws æ1 ( ) ˆ2 1 ö ˆ2 W = mgl - mgl + mv mv in ut - ç in ut è2 2 ø 1 W = mg L - L + m Vˆ -Vˆ 2 ( ) ( 2 2 ) in ut in ut (3.26) Frm the prblem statement, L in - L =- 10 m ut (an utlet is higher than an inlet by 10 m) 36

37 It is given that the required vlumetric flw rate ( V ) f water is 50 m 3 /h, which can be cnverted t the mass flw rate ( m ) as fllws æ 3 m öæ kg kg 50, ö 1, 000 = 50, 000 = = 13.9 kg/s 3 ç h m è øèç ø h 3,600 ( 3 r = 1, 000 kg/m ) water ( ) he velcities at the inlet ( ) V ˆ and the exit ut V ˆ can be cmputed using the fllwing equatin V Vˆ = (3.27) A where A is the crss-sectinal area f the pipe in 3 m 50 ˆ h 4, 421 = = 4, 421 m/h = = 1.23 m/s 2 p æ 12 ö 3,600 m 4 ç è100 ø V in 37

38 3 m 50 ˆ h 70,736 = = 70, 736 m/h = = 19.6 m/s 2 p æ 3 ö 3600 m 4ç è100 ø V ut hus, mg L æ kgöæ mö ç ç ç 2 è s øèç s ø é m kg ê s - L =-1, ë ut s J =-1, s ( - L ) = ç13.9 ç9.81 (-10 m) in mg L ut ( ) in [the acceleratin f gravity ( g ) is 9.81 m/s 2 ] mg L ( - L ) =-1, W in ut 1 ( 2 2 ) 1æ kgöéæ ˆ ˆ mö æ mö mv - V = in ut - 2 2ç è s ø çè s ø çè s ê ø ë 2 2 ù ú û 2 2 ù ú û 38

39 é m kg 1 ê ( ˆ2 ˆ2 ) s mv - V =-2,659.4 ë in ut 2 s 2 2 ù ú û 1 ( ˆ2 ˆ2 ) J mv - V =- 2,659.4 =-2,659.4 W in ut 2 s Hence, frm Eq. 3.26, 1 W = mg L - L + m Vˆ -Vˆ in ut in ut 2 W = - 1, W + -2,659.4 W ( ) ( 2 2 ) ( ) ( ) W =- 4, 023 W =-4.02 kw. herefre, the required pwer f the pump is æ1.341 hpö 4.02 kw ç 1 kw = 5.39 hp è ø 4.02 kw = ( ) 39

40 Example Steam enters a cndenser (r a heat exchanger) as a saturated steam at 1 bar and leaves the cndenser as a saturated liquid als at 1 bar A cling water used t cndense the steam enters the cndenser at 20 C and flws ut f the cndenser at 35 C Determine the amunt (in kg) f cling water per 1 kg f steam In this kind f prblem, ptential and kinetic energies are negligible hus, Eq. 3.5 ( ) ( ) U + PE + KE + P V + Q + W in in in in in in in = U + PE + KE + P V + Q + W ut ut ut ut ut ut ut (3.5) is reduced t 40

41 r ( ) ( ) H - H = Q -Q - W - W ut in in ut ut in ut in (3.7) H - H = Q - W (3.8) Hwever, in this prblem, there are 2 inlet streams: saturated steam at 1 bar cling water at 20 C 2 utlet streams: saturated liquid at 1 bar cing water at 35 C Eq. 3.8 can be mdified t ( ) ( ) c c s s c c s s m h + m h - m h + m h = Q -W ut ut ut ut in in in in (3.28) which can be re-arranged t 41

42 ( ) ( ) c c c c s s s s m h - m h + m h - m h = Q -W ut ut in in ut ut in in (3.29) and At steady state, m = m = m c c c ut ut in m = m = m s s s in and Additinally, fr a cndenser, generally, Q = 0 (the net heat transfer = 0) W = 0. (there is n wrk) Accrdingly, Eq can be re-written as ( - ) + ( - ) = 0 m h h m h h c c c s s s ut in ut in which can be re-arranged t 42

43 ( - ) =- ( - ) ( - ) = ( - ) m h h m h h c c c s s s ut in ut in m h h m h h c c c s s s ut in in ut ( h -h ) s s m c in = m h h ut ( - ) s c c ut in (3.30) he data we need t search fr are the enthalpy f the inlet saturated steam at 1 bar ( h ) s in the enthalpy f the utlet saturated liquid water at 1 bar ( h ) s ut the enthalpy f the utlet liquid cling water at 35 C ( h ) c ut the enthalpy f the inlet liquid cling water at 20 C ( h ) c in 43

44 data: Frm a steam table, we btain the fllwing h = enthalpy f the inlet saturated steam s in at 1 bar = 2,675.5 kj/kg h = enthalpy f the utlet saturated liquid water at 1 bar = s ut kj/kg h = enthalpy f the utlet liquid water c ut at 35 C = kj/kg h = enthalpy f the inlet liquid water at c in 20 C = kj/kg Hence, frm Eq. 3.30, é( 2, ) kj/kgù m c ê - ú 36.0 = ë û = m é( ) kj/kgù 1 s ê - ë úû he amunt (mass) f cling water per 1 kg f f steam is 36.0 kg 44

45 Example he dehydrgenatin f ethyl alchl (EtOH) t frm acetaldehyde: C 2 H 5 OH CH 3 CHO + H 2 (3.31) is carried ut in a cntinuus reactr he vapur f ethyl alchl is fed t the reactr at 400 C, and the prducts leave the reactr at 100 C Determine the amunt f heat input r release per 1 ml f ethyl alchl frm the reactr fr a cnversin f 30% Basis: 100 ml f ethyl alchl (C 2 H 5 OH) Frm Rxn. 3.31, with the cnversin f 30%, the amunt f each species at equilibrium is C 2 H 5 OH 70 ml CH 3 CHO 30 ml H 2 30 ml 45

46 Since this is an energy balance fr a chemical reactin, the energy-balance equatin (at steady state) can be written frm Eq. 3.1: Energy in + Energy generatin Energy ut Energy cnsumptin = Energy accumulatin (3.1) as fllws éæn ö ù i ( Q = å H ) ç D + n h - n h r ut ut in in n å å êë çè i ø úû (3.32) Nte that, in Eq. 3.32, the term: éæn ö ù i å ( H ) ç D r n is the cmbined generatin êë çè i ø úû and cnsumptin å and å n h are the utlet and in in inlet enthalpies n h ut ut W (wrk) is assumed t be zer (0) 46

47 calculate the amunt f heat ( Q ) fr this prcess, the fllwing data must be btained: Standard heat f frmatin ( ) D f H f C 2 H 5 OH (g) = kj/ml CH 3 CHO (g) = kj/ml H 2 (g) = 0 Specific heat capacity ( c ) f p C 2 H 5 OH (g): = cp CH 3 CHO (g): = cp H 2 (g): = cp (nte that must be in C) 47

48 éæn ö êç i he term ( DH ) êë calculated using Eq. 3.20: ù ú å êç r n ú in Eq can be çè i ø úû ( n ) ( n ) å å D H = DH - DH r i f i f i i Prducts Reactants (3.20) as fllws é æ kj ö ù D H ( 1 ml) ( 1 ml)( 0) r = - + Rxn ê ç ml ú ë è ø û é æ kj ö ù - ( 1 ml) ê èç ml øú ë û D = kj H r Rxn kj D H r = Rxn ml C H OH 2 5 å hus, éæn ö ù æ30 ml EtOHöæ i kj ö ( H ) r n D = ç 1 ç ml EtOH êç ëè i ø úû è øè ø 48

49 å éæn ö ù i ç D = r n êçè ë i ø úû ( H ) 2, 073 kj he specific enthalpy f each species ( h ) i f the inlet stream (at 400 C) can be cmputed as fllws C 2 H 5 OH: é 400 ù ú ë û ò 25 ( ) ( ) ( ) C H OH ê C H OH C H OH h = h - h = c d in C 2 H 5 OH ( ) CHOH h = 2 5 in 33.8 kj/ml p he specific enthalpy f each species f the utlet stream (at 100 C) can be cmputed as fllws C 2 H 5 OH: é 100 ù ú ë û ò 25 ( ) ( ) ( ) CHOH ê CHOH CHOH h = h - h = c d ut C 2 H 5 OH ( ) CHOH h = 5.31 kj/ml 2 5 ut p 49

50 CH 3 CHO: ( ) é CH CHO ê( CH CHO) ( CH CHO) ut 100 ù ú ë û ò 25 h = h - h = c d CH 3 CHO ( ) CH CHO H 2 : h = 3 ut 4.38 kj/ml é 100 ù ú ò ë û 25 ( ) ( ) ( ) H ê H H h = h - h = c d ( ) H ut H 2 h = 2 ut 2.16 kj/ml p p Hence, the ttal amunt f heat ( Q ) f this prcess can be calculated, using Eq. 3.32: éæn ö ù i ( Q = å H ) ç D + n h - n h r ut ut in in n å å êë çè i ø úû (3.32) as fllws 50

51 Q é æ kj ö ù ( 70 ml) 5.31 ç ml è ø æ kj ö é æ kj öù = é2, 073 kjù ( 30 ml) 4.38 ( 100 ml) 33.9 ëê ûú ç ml ê ç ml ú è ø ë è øû æ kj ö + ( 30 ml) 2.16 ê ç è ml øú ë û Q = kj. Since the basis f calculatin is set as 100 ml f EtOH, the amunt f heat utput (r release) (as Q has a sign) per 1 ml f EtOH is kj Q = 100 ml f ethyl alchl q =-7.49 kj/ml f ethyl alchl. 3.7 Unsteady-state Energy Balances Generally, ur energy-balance calculatins are perfrmed fr steady-state prcesses, in which the accumulatin term is taken as zer 51

52 Hwever, at the start-up r fr a batch prcess, the accumulatin term is nt zer, and, usually, we have t slve fr the time required t reach the desired temperature he fllwing example illustrates hw the energy-balance calculatins fr an unsteady-state prcess are carried ut [2] Example In the batch preparatin f an aqueus slutin, water in the amunt f 4,545 kg is heated frm 25 t 80 C in a heater with the heating area ( A ) f 30 m 2 using a superheated steam at the temperature f 150 C he verall heat transfer cefficient ( U ) fr this heater is 285 W/m 2 -K Estimate the heating time 52

53 Frm the heat transfer curse, the rate f heat transfer frm the heater t water can be cmputed using the fllwing equatin: dq UA ( ) h w dt = - (3.33) where = temperature f a heating surce h = temperature f water w he amunt f heat transfer ( dq ) used t increase the water temperature ( d ) can be calculated using the fllwing equatin: in which dq = mc d (3.34) d = - = - p w w w w final initial f i Cmbing Eq with Eq gives mc d p UA h w dt ( ) = - (3.35) 53

54 If the specific heat capacity ( c ) is assumed t p be cnstant thrughut the heating prcess (which is reasnable, as c p f liquid is rather cnstant), Eq can be re-arranged t d mc UA( ) p h w dt = - dt = mc p UA ( - ) h d Integrating Eq yields t 0 dt = mc p w f ò ò w i w d ( - ) UA h w (3.36) t ( - ) h w mcp =- ln UA ( - ) h w f i (3.37) Substituting given numerical values int Eq results in 54

55 t æ kj ö ( 4,545 kg) 4.18 ç é kg K ( ) Kù è ø ê - ú =- ln ë û æ W ö ( ) ( ) 3 é K ù m êë úû ç è m K ø t =- ( ) æ 3 J ö 4,545 kg ç é kg K ( ) Kù è ø ê - ln ë úû æ J ö é( ) Kù - êë úû s ( m ) 2 m K ç çè ø t = 1,288.8 s = 21.5 min 55

56 References [1] Y.A. Cengel and M.A. Bles, hermdynamics: An Engineering Apprach, 8 th ed., McGraw-Hill, [2] R. Sinnt and G. wler, Chemical Engineering Design, 5 th ed., Elsevier, [3] M.J. Mran, H.N. Shapir, D.D. Bettner, and M.B. Bailey, Principles f Engineering hermdynamics, 7 th ed., Wiley, [4] R.M. Felder and R.W. Russeau, Elementary Principles f Chemical Prcesses, Wiley,