Differential Balances - Alternate Forms

Transcription

1 CBE 6, Lecy Dfferental Balances - Alternate Forms To recall, dfferental balance equatons epress the basc laws of conseraton of mass, momentum, and energy: Dρ = ρ (conseraton of mass) () ρ D ρ D e = B σ (momentum balance) () = q q''' ( σ ) (conseraton of energy) () Here, the operator D/ s the "materal derate" and s defned by D = t Below we consder how these epressons are modfed when appled to certan common types of flud flows.. Equaton of Contnuty (conseraton of mass). Besde the full equaton (), the only other erson mportant for ths course s for an ncompressble flud, for whch = 0 (ncompressble flud) (4) Equaton (4) has already been encountered earler.. Equaton of Change of Momentum (momentum balance)..). Euler Equatons. Perhaps the smplest type of flud s one for whch no scous stresses est. For such a "frctonless" flow, the stress tensor σ n equaton () s smply σ = pδ (5) where p s the (thermodynamc) pressure n the flud (Asde: note that n deal flow the mechancal and thermodynamc pressures are equal). Equaton 5 states that snce there are no scous stresses n a frctonless flow, the only nd of stress present s the normal, compresse stress due to pressure. The

2 CBE 6, Lecy σ th component of σ s, and usng equaton (5) ths becomes p. Therefore, the th component of equaton () s pδ p = δ = D ρ p = B (6) Wrtten out n full for the component n the CCS system, equaton (6) s p ρ = B t (7) wth analogous epressons for the and components (you should be able to readly wrte these out). Equaton (7) and ts and forms are nown as the Euler Equatons. When ewed n terms of the materal derate notaton, equaton (6) states that, for a frctonless flud, the rate of change of -drecton momentum of a flud element mong wth the flow (left hand sde) equals the -drecton body force (frst term on the rght) plus the -drecton pressure force (second term on the rght) actng on the flud element. Recall: For a flud element mong wth the flow, the boundares of the flud element moe wth the bul elocty so that no conecte transfer of any quantty (mass, momentum, or energy) occurs across the surface of the flud element; therefore, momentum of a flud element mong wth the flow can only be altered by forces eerted on the flud element. When usng the Euler equatons to descrbe flud flow, t s assumed that frcton between dfferent parts of the flud that moe relate to each other s neglgble. It s mportant to realze that ths does not necessarly mply the assumpton of a scosty µ that s close to zero. For eample, t s possble that the elocty gradents n the flud are smply so small so that the relate moton between neghborng parts of the flud s ery slght, and the frcton generated s thus neglgble een though the scosty may be sgnfcant..). Naer-Stoes Equatons. The Naer-Stoes equatons are the dfferental momentum balance wrtten for a Newtonan flud. The dfferental momentum balance s gen by equaton () aboe. To specalze t to a Newtonan flud, the Newtonan epresson (.e. the "consttute relaton") for the stress tensor σ must be substtuted nto equaton (). The stress tensor for a Newtonan flud s gen by (note that here, and throughout ths handout, p s the thermodynamc pressure) σ = pδ λ ( ) δ µ e (8)

3 CBE 6, Lecy We recall that the off-dagonal,.e., components of e are proportonal to rate of shear stran, whle the dagonal components of e, =, are related to the rate of dlatatonal stran. In Cartesan notaton, equaton 8 becomes σ = pδ λ ( ) δ µ ( ) (9) The momentum balance requres an epresson for σ. The th component of σ s therefore, usng equaton 9, σ ; ( σ ) = pδ [λ ( ) δ ] [µ ( ) ] = p [λ ( )] [µ ( ) ] (0) The Kronecer deltas were summed oer n arrng at the last lne of equaton 0. Substtutng equaton 0 nto the momentum balance equaton, we obtan the th component of the Naer-Stoes equatons D ρ = B p [λ ( )] [µ ( )] () The Naer-Stoes equatons are the momentum balance epressed for a Newtonan flud. Settng = n equaton produces the component momentum balance: p ρ = B t [µ ] [µ ] [λ ( [µ )] ] ()

4 CBE 6, Lecy 4 Equaton may loo formdable compared to the orgnal momentum balance equaton. Howeer, all we dd was substtute n the consttute relaton for a Newtonan flud, accordng to equaton 9. For an ncompressble flud, = = 0. Moreoer, f we also assume a constant scosty µ then equaton smplfes to D ρ = B p µ ( ) Snce = = 0 due to ncompressblty, D ρ = B p µ () Equaton holds for ncompressble, constant scosty Newtonan fluds and s one of the most frequently used forms of the dfferental momentum balance. Along the three drectons,, and (correspondng to =,, and ), equaton s ρ p = B µ t ρ p = B µ t ρ p = B µ t (4a) (4b) (4c) Equatons 4 can be summarzed much more compactly n tensor notaton as ρ D = B p µ (ncompressble, constant µ Newtonan fluds) (5) When coordnate systems other than CCS are used, equaton 5 wll not need to be redered snce the basc law of momentum conseraton that t represents must reman ald. Howeer, to epress equaton 5 n another coordnate system, the forms for the operators D/, and approprate to

5 CBE 6, Lecy 5 that coordnate system must be used. Equaton 5 states that for a flud element mong wth an ncompressble, constant scosty Newtonan flow, the rate of change of momentum (left hand sde) equals the sum of the body force (frst term on rght), pressure force (second term on rght), and scous force (thrd term on rght) that act on the flud element.. Conseraton of Energy The dfferental law of energy conseraton, equaton, s De ρ = q q ( σ ) (6) To specalze equaton 6 to a Newtonan flud, equaton 9 for the stress tensor σ needs to be substtuted nto the term (σ ). We wll do ust that after rearrangng 6 nto a more useful form. We note that, usng the defnton of the nner product for tensors, n Cartesan notaton the term (σ ) becomes (can you show ths?) (σ ) = / (σ ) (7).). Mechancal Energy Balance. The Mechancal Energy Equaton s concerned wth netc and potental energes only. The nternal energy s not ncluded. Therefore, ths equaton wll not represent the law of energy conseraton, snce t s not a total energy balance. The Mechancal Energy Equaton can be dered drectly from the rate of change of momentum equaton, equaton, whch f epanded n CCS reads (you should be able to perform ths epanson on your own) D D D ρ δ ρ δ ρ δ = ρgδ ρgδ ρgδ σ σ σ σ σ σ σ σ ( ) δ ( ) δ ( Or, n tensor notaton, σ ) δ (8) D ρ D t =ρg σ (9) For defnteness, we too the body force per unt olume B to be equal to the gratatonal force, B = ρ g. Tang the dot product of equaton (9) wth the elocty results n D ρ = ( ρg σ) (0)

6 CBE 6, Lecy 6 We could proe that D ρ D( ) D( / ) ρ = = ρ () Usng equaton, equaton 0 becomes D( / ) ρ = ρ g ( σ) () Now we mae use of some facts: () gz, the gratatonal potental per unt mass at a pont at heght z, does not change wth tme so that (gz)/t = 0, and () f the gratatonal acceleraton g s constant, then the gratatonal force per unt mass g = - gn Z = -g z = - (gz). Here, n Z = z s a unt ector pontng n the poste z drecton. Addng ρ(gz)/t to the left hand sde and substtutng - (gz) for g, equaton rearranges to D( / ) ( gz) ρ ρ{ t (gz)} = ( σ) Note that t was allowable to add ρ(gz)/t only to the left hand sde snce ths term equals 0. Usng the ( gz) defnton of the materal derate, (gz) = D(gz)/, and the aboe epresson t rearranges to D( / ρ gz) = ( σ) () Equaton s called the Mechancal Energy Equaton. The term "mechancal" s meant to emphasze that the equaton s only concerned wth netc and gratatonal potental energes and does not nclude the nternal energy. That equaton s only a balance on netc and gratatonal potental energes s edent from ts left hand sde. We wll net use equaton to dere a dfferental balance on nternal energy..). Dfferental Balance on Internal Energy. A balance on nternal energy s obtaned by subtractng the balance on mechancal (netc potental) energy (equaton ) from the balance on total (nternal netc potental) energy (equaton 6). We hae De ρ = q q ( σ ) (total energy balance) D( / ρ gz) = ( σ) (mechancal energy balance) Recallng that e = u / gz, subtracton of the mechancal energy equaton from the total energy balance leads to

7 CBE 6, Lecy 7 D( u ρ / gz / gz) = q q ( σ ) - ( σ) (4) or D u ρ = q q ( σ ) - ( σ) (5) D t Equaton 5 has not referred to any partcular coordnate system as yet. Howeer, for the rest of the deraton we wll mae use of Cartesan coordnates. Ths s not necessary, but wll be a bt easer to follow than stayng n the general ector notaton. We recall that the dergence of a ector A s A = A (6) Second, from equaton 7 (σ ) = ( σ ) (7) Thrd, by procedures smlar to those used to dere equaton (7), we could show that ( σ) = σ (8) Usng equatons 6, 7, and 8, equaton 5 rearranges to ρ D u q = D t q ( σ ) σ (9) Accordng to the product rule of dfferentaton ( σ ) = σ σ, thus ρ D u q = D t q σ (0) The heat flu ector q can be separated nto heat flu by conducton q C, whch obeys Fourer's Law q C = - κ T, and heat flu by radaton q r. Wth these modfcatons the nternal energy balance becomes

8 CBE 6, Lecy 8 t u ρ D D = C q r q q σ () We wll now specalze equaton to Newtonan fluds that also follow Fourer's Law, q C = - T/ ) (.e. we are now specalzng to flows of Newtonan fluds that obey Fourer's Law): t u ρ D D = T r q q [{ pδ λ ( ) δ µ ( )} ] () The last term on the rght (the term nsde the [ ] parentheses) becomes p δ λ ( ) δ µ ( ) Summng oer n the two terms nolng the Kronecer delta results n (recall that the Kronecer delta δ equals zero unless = ): p λ ( ) µ ( ) () Usng epresson 6 for the dergence of a ector, we see that the p n equaton s multpled by ( ) and the λ s multpled by ( ). Replacng the last term n equaton wth epresson results n t u ρ D D = T r q q p ( ) λ ( ) µ ( ) (4) The last two terms n (4) are usually combned nto the dsspaton functon Φ, Φ = λ ( ) µ ( ) (5) or, summng oer the and ndces, Φ = λ ( ) µ µ (6)

9 CBE 6, Lecy 9 In equaton 6, the dsspaton functon was epanded by summng oer both and ndces n equaton 5. Usng the dsspaton functon, equaton 4 s wrtten, ρ D u T = D t qr q p ( ) Φ (7) or D u ρ = T qr q p ( ) Φ (8) D t Note that, wthout nong Newtonan flud behaor, the dsspaton functon could hae been more generally stated usng tensor notaton as Φ = σ : e p (8b) where e s the rate of stran tensor from Handout 6, e = (/)( T ). Equaton (8) states that the rate of change of nternal energy of a flud element mong wth the flow (left hand sde) equals the sum of: st term on rght: the rate of heat flow nto the element by conducton nd term on rght: the rate of heat flow nto the element by radaton. If there s no net absorpton or emsson of electromagnetc radaton wthn the flud element, the term q r s zero. rd term on rght: the rate at whch heat s generated nsde the element by an eternally coupled mechansm such as a heatng col or wre, for eample. There s an obous conceptual dffculty wth ths term: how eactly does one ft a heatng col nto an nfntesmal flud element? Therefore, t may be temptng to drop ths term from the dfferental balance. On the other hand, f one apples the dfferental energy balance to the nteror of a pece of electrcal wre, for eample, n whch local heat generaton occurs due to resste dsspaton of electrcal energy, then ths term maes perfect sense. Therefore, we wll eep t. 4th term on rght: the rate at whch pressure forces do compresse wor on the flud element (also referred to as P-V wor). Recall that - s the rate of compresson of the flud element; ths allows the 4th term to be dentfed as the rate at whch compresse wor s done on the flud element by pressure forces. The compresse wor s reersble n that t could be recoered; for nstance, by reepandng the flud element to do wor on ts surroundngs. 5th term on rght: the rate at whch wor performed on the flud element by scous stresses s dsspated to nternal energy by frcton n the flud. Ths term can be dentfed as such by mang a connecton wth the nd Law of Thermodynamcs (see below). In thermodynamc termnology, ths s rreersble rate of wor. Ths s the rate at whch wor s conerted to heat. Snce, accordng to the nd Law of Thermodynamcs, heat cannot be fully conerted bac to wor ths process s rreersble.

10 CBE 6, Lecy 0 From equaton 6, we see that all terms n Φ are dered from the presence of elocty gradents (.e. relate rate of separaton of dfferent parts of the flud) and that Φ must always be greater than zero, Φ > 0. (Can you eplan why there s no dsspaton term n the total energy balance, equaton 6?). The unts of all the terms n equaton 8 are energy/(olume tme). For some materals, such as deal gases and strctly ncompressble lquds and solds, the nternal energy u s only a functon of temperature so that u = u(t). In that case, the constant olume specfc heat capacty ĉ V s defned by so that ĉ V = du/dt (9) du = ĉ V dt (40) Usng equaton 40, the rate of change of the nternal energy of a flud element mong wth the flow, Du/, s equal to Du/ = ĉ V DT/ (deal gases or ncompressble lquds) (4) where DT/ s the rate of temperature change of the flud element. Usng equaton (4), the nternal energy balance (equaton 8) becomes DT ρcˆ V = T q r q p ( ) Φ (deal gases or ncompressble lquds) (4) D t or, f the substantal derate s epanded, T ρcˆ V T = T q r q p ( ) Φ (4) t (deal gases or ncompressble lquds) After much manpulaton, we hae arred at a useful form of the energy balance that wll allow us to calculate temperature felds, a quantty of great engneerng mportance. In ths course we wll mostly be concerned wth ncompressble lquds or solds, or deal gases, so that equatons 4 and 4 apply. Howeer, we mae a few bref comments about the case when these equatons are not good appromatons because the nternal energy depends on other arables besdes temperature. For nstance, for sngle component systems (e.g. pure fluds) that are compressble, u becomes a functon of two arables such as T and the densty ρ. The dependence on ρ s usually epressed n terms of /ρ, the olume of flud per unt mass of flud. /ρ s customarly referred to as the "specfc olume" V. If u = u(t, V), then du = (u/t) V dt (u/v) T dv. By defnton, the derate (u/t) V = ĉ V ; therefore du = ĉ V dt (u/v) T dv. Thus, when u depends on V as well as T, equaton 4 gets a term correspondng to the (u /V) T dependence,

11 CBE 6, Lecy DT u DV ρ cˆ V = T q r q p ( ) Φ (44) V T If needed, the term (u /V) T DV / can be further rearranged usng thermodynamc denttes. For many common stuatons equatons 4 and 4 smplfy nto more user frendly forms. For eample, typcally the rate of heat absorpton or emsson by radaton s neglgble, so that q r 0. Moreoer, when there s no heat generaton n the materal by eternally-coupled processes q = 0, and f the thermal conductty s assumed constant then can be taen outsde the operator. So, for eample, under these crcumstances equaton 4 rearranges to T ρcˆ V T = T p ( ) Φ = T p ( ) Φ (45) t ( constant; q = 0; no radate processes) Furthermore, f the flow s ncompressble so that ρ = constant, the terms n equaton 4 and n the dsspaton functon equaton 6 dsappear. Thus T ρcˆ V T = T Φ (ρ, constant; q = 0; no radate processes) (46) t where Φ s gen by equaton 6 but wthout the term. If, n addton, we are consderng the temperature dstrbuton n a sold materal or a flud at rest, then = 0. When = 0, Φ also equals zero (see equaton 6), and equaton 46 becomes T = T (ρ, constant; = 0; q = 0; no radate processes) (47) t ρcˆ V For strctly ncompressble solds and lquds the densty does not depend on temperature, and the constant olume and constant pressure specfc heat capactes are equal, ĉ V = ĉ P (one could proe ths from thermodynamc relatons and the defntons of ĉ V and ĉ P ). Snce most solds and lquds typcally do not epand ery much wth temperature ths equalty s often an ecellent appromaton, so that ĉ P can be used nstead of ĉ V n the aboe equatons. If ĉ P s used nstead of ĉ V n equaton 47, the combnaton of parameters n front of T would be /ρĉ P. Ths partcular combnaton s referred to as the thermal dffusty α, α = /ρĉ P (48) and 47 may be wrtten as T = α T (ρ, constant; = 0; q = 0; no radate processes; ĉ V = ĉ P ) (49) t

12 CBE 6, Lecy Other smplfcatons of the energy balance, n addton to the eamples aboe, are possble. Of course, n any problem the smplfcatons must reflect the physcal nature and statement of the problem. Remars on the nd Law of Thermodynamcs and the conseraton laws. From thermodynamcs, the entropy change of a system, ds, that results from a process nolng the system must be greater than or equal to dq/t, where dq s the heat transferred from surroundngs to the system and T s the absolute temperature of the system. Therefore, ds dq/t (50) Applyng equaton 50 to a fed, open control olume through whch materal can flow (.e. the control olume s the "system") leads to d dt q ρ da (5) T s dv sρ(n ) da n V A A where s s entropy per mass of materal (specfc entropy), V s the control olume, A s the surface enclosng the control olume, n s the outward unt normal to A, s the flow elocty, and q s the heat flu. Equaton 5 s a macroscopc balance on entropy whch states that the rate of accumulaton of entropy nsde the control olume (left hand sde) must be greater than or equal to the conecton of entropy nto the control olume by flud flow (st term on rght) plus the rate of heat flu nto the control olume dded by T (nd term on rght). The equalty n equaton 5 would apply f all transport processes occurrng wthn the system were reersble. All real processes are rreersble to some etent, so that entropy s generated as a consequence. If we ntroduce the rate of entropy generaton per mass s, to account for ths generaton of entropy, we can conert epresson 5 to an equalty, d dt q ( ρdv (5) T sρdv = sρ n ) da n da s V A A V Applyng the dergence theorem and collectng all terms under a sngle ntegral oer V, ( sρ) q s ρ sρ dv = 0 t T V (5) from whch the dfferental balance on entropy s obtaned ( sρ) q s ρ sρ t T = 0 (54) Alternately, usng the product rule of dfferentaton, we can wrte

13 CBE 6, Lecy ρ s q s ρ s( ρ) ρ s s ρ = 0 t t T The sum of the frst and thrd terms, accordng to the equaton of contnuty (equaton ), s zero. Therefore, we are left wth s q ρ ρ s s ρ = 0 (56) t T (55) Ds ρ D t q = s ρ T (57) Note that equaton 57 s wrtten for a closed system (.e. a flud element mong wth the flow nto whch there s no conecton). From thermodynamcs, we recall that for a closed system du = Tds - pdv where V s the specfc olume. If the aboe epresson s appled to a flud element mong wth the flow, t reads Du Ds DV Ds D(/ ρ) Ds p Dρ Ds ρ = ρt pρ = ρt pρ = ρt = ρt p( ) (58) ρ where equaton was used after the last "=" sgn n equaton 58. Therefore, Ds Du ρ T = ρ p( ) (59) Substtutng equaton 57 for the left hand sde n equaton 59, and equaton 8 for the frst term on the rght, Ts q ρ T = T q q r Φ (60) T Splttng the heat transfer q nto conducte (accordng to Fourer's Law) and radate parts, q = q C q r = - T q r rearranges equaton 60 nto Now, T q r q Φ s ρ = T q r (6) T T T T T T and T T ( T ) = T T = T T = T (6) T T T T T T T

14 CBE 6, Lecy 4 q r q r = q r q r = q r T T T T T T (6) Insertng equatons 6 and 6 nto equaton 6 and smplfyng, ( T ) q r q Φ s ρ = T (64) T T T T Equaton 64 shows that the rate of entropy producton per unt olume, s ρ, results from rreersbltes due to the presence of temperature gradents (st term on rght), generaton of heat by eternally coupled processes (rd term on rght), presence of frctonal dsspaton of mechancal energy to heat due to "rubbng" because of elocty gradents (4th term on rght), and couplng between the temperature and electromagnetc radaton felds (nd term on rght). The most mportant concluson from equaton 64 s that presence of elocty gradents, as captured by the dsspaton functon Φ, and of temperature gradents assocated wth heat conducton, always lead to poste generaton of entropy and hence represent rreersbltes assocated wth the transport of momentum and energy. Remars on "Conducton of momentum." In equaton 46, assumng constant alue of the thermal conductty, heat conducton was represented by the term T. In the Naer-Stoes equatons, equaton 5, under assumpton of constant scosty µ there was a term µ. Let's compare the meanng of these two terms. The heat conducton term represents transfer of heat by molecular leel processes. Imagne a control olume - how s heat conducted from one sde of the control olume surface to the other by "molecular leel processes"? Molecular leel processes nclude molecular collsons that transfer energy from more agtated (hotter) to less agtated (cooler) molecules. As such collsons wll occur across the control olume surface een f there s no bul conecton, ths transfer s one eample of heat flow by conducton. It s also possble that hotter molecules from one sde of a control olume enter whle colder molecules leae from the other sde. If ths "molecular echange" s such that the local aerage elocty (.e. the aerage elocty of all the molecules crossng the control olume surface at the pont of nterest) s zero normal to the surface, then ths transfer s also not conecte snce there s no bul flow across the surface. Then the transfer s ewed as conducte. The sum total of all conducte processes, whateer ther physcal orgn, s represented by the conducte heat flu q C that follows Fourer's Law, q C = - T. From ths epresson, t s seen that temperature gradents are the "drng force" for heat conducton. If there are no temperature gradents, there s no heat conducton. How s momentum transferred by molecular leel processes? Faster molecules on one sde of a control olume surface can collde wth slower molecules on the other sde, thus transferrng momentum across the surface. Ths momentum transfer could be consdered "conducte" snce no molecules crossed across the control olume surface. It s also possble that faster molecules cross the control olume surface n one drecton whle slower molecules cross n the other drecton such that the net aerage elocty s zero normal to the surface. Agan, snce there s no bul conecton, such transfer of momentum could be termed to be "conducte." Because the term µ n equaton 5 represents such molecular leel processes, t s at tmes referred to as "conducton of momentum." Note that f there are no elocty gradents, there s no conducton of momentum.

15 CBE 6, Lecy 5 Common Boundary Condtons Integraton of dfferental equatons leads to the appearance of ntegraton constants. The dfferental equatons of transport phenomena that we hae been studyng are no ecepton to ths rule. The ntegraton constants are typcally ealuated usng "boundary" or "ntal" condtons. Boundary condtons mae a physcal statement that s beleed to be true and that we want to enforce at some locaton wthn the system (often, the locaton s the system boundary). Intal condtons represent a physcal statement that we want to enforce at some pont n tme (often, the tme s the startng or ntal tme of a process). Some of the more common boundary and ntal condtons encountered when solng the dfferental equatons of transport phenomena are dscussed below. ). Intal condtons. Eamples: (t = 0) = 0 (65) T(t = 0) = T 0 (66) where 0 and T 0 are nown constants or functons of arables other than t. For eample, we could hae T 0 = f(r) f T 0 depends on poston. ). Impenetrable boundary. Ths boundary condton epresses the fact that materal s not allowed to moe across an mpenetrable boundary. Thus, at the boundary, we must hae n = = 0 (67) where s the materal elocty, n s a unt normal to the boundary, and s the component of perpendcular to the boundary. ). The "no-slp" boundary condton. Ths boundary condton assumes that materal on both sdes of a boundary between two physcal phases moes wth the same elocty. For eample, flud rght net to a statonary sold wall s assumed to be statonary, whle flud rght net to a mong sold wall s assumed to moe wth the same elocty as the wall. Smlarly, at the boundary between two mmscble fluds I and II, the elocty of flud I mmedately on one sde of the boundary s assumed to equal that of flud II mmedately on the other sde. Mathematcally, the no-slp condton can be represented as I = II (68) where I and II are the eloctes of materal at the boundary between phases I and II, respectely. 4). Contnuty of stress across a boundary (at planar boundares wth unform condtons). In ths boundary condton, the stress (normal and shear) eerted by materal on one sde of a boundary s assumed to be equal and opposte to that eerted by the materal on the other sde. Ths statement s not always true, and n fact f the boundary s cured (recall: what happens f a surface s cured?), or f

16 CBE 6, Lecy 6 there are aratons n the condtons at the boundary (e.g. due to composton or temperature changes) then surface tenson effects arse whch must also be accounted for. We wll assume that the boundares are planar, and that the condtons at the boundary are the same eerywhere. In such a smplfed scenaro, the mathematcal statement of the estence of equal and opposte stresses on the two sdes of a boundary can be wrtten n - σ - = - n σ (69) where the subscrpt "-" sgnfes quanttes on the "mnus" sde of (.e. below) the boundary whle "" sgnfes quanttes on the "plus" sde of (.e. aboe) the boundary. n - s a unt normal to the boundary pontng n the negate drecton, whle n s a unt normal that ponts n the poste drecton. Snce n - = - n, the aboe equaton reduces to the smple statement σ - = σ (70) Equaton 70 states that the stress tensor on one sde of the boundary must equal that on the other sde. Ths means that both the normal and shear stresses are contnuous (.e. do not change) across the boundary. For the partcular case when the materal on one sde of the boundary s a gas, because of the low scosty of gases t s often assumed that the scous stresses at the boundary are anshngly small. In that case the scous shear stresses, and scous normal stresses (but not pressure), at the boundary are often appromated to equal zero. 5). Contnuty of stress across a boundary (general case). It s mportant to realze that all the aboe boundary condtons hae nherent assumptons. For eample, the contnuty of the stress tensor can be olated by surface tenson effects. The full condton on the stress tensor that also allows for surface curature and surface tenson gradents s n σ n σ s γ ςγn = 0 (70a) where the surface gradent operator s gen by (note that ths operator taes the gradent of a functon wthn a surface) = n ( n S ) (70b) and the surface mean curature s calculated from ς = mean curature = S n (70c) For a surface specfed by z = f(,y) n the CCS, these epressons become f f f f S = H δ δ δ δ H δ δ f f f f (70d)

17 CBE 6, Lecy 7 / = f f H (70e) = f f f f f f f H ς (70f) 6). Temperature Boundary Condtons. A ery common boundary condton s one n whch temperature s specfed at a pont n space. For eample, T ( = 0) = T o (7) where the alue of T o s gen (notng that T o may be a functon, for eample T o (t) ). 7). Condtons on Heat Flu. One eample of a boundary condton on heat flu deals wth heat conducton at an adabatc boundary or wall. By defnton, an adabatc wall does not transmt heat. From Fourer's Law, the conducte heat flu q C normal to the boundary s gen by q C = - T = - z T d d (7) where z was taen to be the coordnate perpendcular to the boundary (Fg. ) and the subscrpt " " ndcates the component of the conducte heat flu that s normal to the boundary. It wll be assumed that the boundary s located at z = 0. Snce the boundary s adabatc, there can be no conducte heat flu perpendcular to t; therefore, q C must equal zero at z = 0. Ths condton mples that Fg. 0 d d = z z T = 0 (adabatc boundary) (7) The subscrpt z = 0 on the temperature derate ndcates that the derate s to be ealuated rght at the boundary;.e. at z = 0. Now magne that we hae a heat-permeable (dathermal) boundary between systems I and II (Fg. ), and that steady state condtons apply. Under steady state, there can be no accumulaton of heat at the boundary. If heat conducton s the only means by whch heat arres and leaes the boundary,

18 CBE 6, Lecy 8 then the rate at whch heat s conducted to the boundary from system I must equal the rate at whch t s conducted away from the boundary nto system II, Fg. dt dt I = ΙΙ dz I, z = 0 dz II, z = 0 dt I and II are heat conducttes n systems I and II, respectely, and dz the temperature gradents n the two systems, measured at the boundary. I, z = 0 dt and dz (60) II, z = 0 are A thrd eample we consder s that of a heterogeneous reacton occurrng at a boundary between a system and an adabatc wall (Fg. ). Under steady state condtons there can be no accumulaton of heat at the boundary; therefore, the rate at whch heat of reacton s generated at the boundary must equal the rate at whch t leaes. Snce the wall s adabatc, no heat can flow nto the wall; thus, heat can only flow nto the bul system. If conducton s the only mechansm of heat transfer away from the surface, then, n = ) H S dt r = dz z = 0 (74) r S s the rate at whch mass of speces s produced by chemcal reactons at the boundary (n unts of mass / (area tme) ), n s the number of speces partcpatng n the reactons, and Ĥ s the specfc enthalpy (unts: energy / mass) of speces. The left hand sde of equaton 74 s the heat generated per area per tme by chemcal reactons, and the rght hand sde s the rate at whch heat flows away from the surface by conducton. Both terms hae unts of energy / (area tme). Fg. 8

19 CBE 6, Lecy 9 The aboe eamples hae noled conducton of heat to or from a boundary as the domnant mechansm of heat transfer. Snce conducton was noled, Fourer's Law was used to calculate the heat flu. Howeer, n many cases conecton may ad n remoal of heat from a boundary. In such cases, t s more customary to model the heat transport usng so-called heat transfer coeffcents h, q = h (T S - T ) (75) wth T S the temperature at the boundary and T the temperature n the bul of the phase to whch the heat s beng transferred. In the case of radate heat transfer, epressons ntroduced n the pror handout could be used to also calculate radate heat flu at a boundary.