Chate 2 Cmbustn and Themchemsty 뷔르거작허풍선이남작의모험 B.C.Ch 1 ety Relatns 1. Extensve and Intensve ety - Extensve ety : V(m 3 ), U(J), H(J)=U+V - Intensve ety : v(m 3 /kg), u(j/kg), h(j/kg) v-(m 3 /kml), u-(j/kml), h-(j/kml) 2. Equatn State V = N R u T R u :unvesal gas cnstant (=8315J/kml-K) V = m R T v = RT B.C.Ch 2
3. Calc Equatns State u = u(t, v) h = h(t,) du = ( u/ T) v dt + ( u/ v) T dv dh = ( h/ T) dt + ( h/ ) T d whee, cnstant-vlume secc heat : c v = ( u/ T) v cnstant-essue secc heat : c = ( h/ T) F deal gas, ( u/ v) T dv=, ( h/ ) T d= Integatn, T e T u(t) u e = c v dt h(t) h e = c dt B.C.Ch 3 F bth eal and deal gas, c v c = (T) Mla cnst.-essue secc heats B.C.Ch 4
4. Ideal-gas Mxtues mle actn seces, X X = N /(N 1 +N 2 + N 3 +. + N + ) = N /N tt smlaly, the mass actn seces, Y Y = m /(m 1 +m 2 + m 3 +.+ m + )=m /m tt =X MW / MW mx ΣX = 1, ΣY = 1 Mxtue mlecula weght, MW mx = ΣX MW = 1/{ΣY / MW } atal essue the th seces, = Σ, = X B.C.Ch 5 - 이상기체혼합물에서한성분의엔트로피계산시그성분의분압이용 T 등온 S(T,) T =1 atm 이상기체-표준상태 (T, ) S S(T, ) = S (T, ) R uln KJ/(Kml K) F th seces Ideal-gas mxtue, S (T, ) y m = S (T, ) R uln KJ/(Kml K) Q = y m S = Rln : atal essue the th seces y : mle actn the th seces S m : mxtue ttal essue B.C.Ch 6
5. Latent heat vazatn h g = at a gven tem., n a cnstant-essue, the heat equed t cmletely vaze a unt mass lqud h g (T,) = h va (T,) h lqud (T,) Latent heat vazatn = Enthaly vazatn *Heatng Value(HV) : (1) Lwe HV, LHV Wate aeas as a gas n the ducts. (2) Hghe HV, HHV Wate aeas as a lqud n the ducts. HHV = HLV + (N H2O h H2O ) kj/kg uel B.C.Ch 7 1 st Law Analyss Reactng Systems 1. 화학 Enegy 변화의표현 1 Steady-lw Systems Enthaly = h + h ) Sensble enthaly eactve at 25C,1atm (KJ/Kml) Ke, e 무시 Enthaly at 25C, 1atm h h h 표준상태에서 enthaly matn 주어진상태에서 Sensble enthaly 표준상태 sensble enthaly (KJ/S) 에너지보존법칙은 Q W = n n (KJ/S) n : the mlal lw the duct n : the mlal lw the eactant B.C.Ch 8
연소에서는 mle 로표현 Q W = N N (KJ/Kml) N : 단위연료 ml당 duct의수 N : 단위연료ml당eactant의수 즉 Q W = H d H eact H H = N + h h d ) = N + h h eact ) (KJ/Kml uel) (KJ/Kml uel) (KJ/Kml uel) 특수한반응의경우 h c ( 연소 enthaly) 가주어진다면 Q W = h c + N h ) N h ) 정상유동과정은일 (wk) 을동반하지않는다. (KJ/Kml uel) B.C.Ch 9 2 Clsed System Enegy 보존식 U d : 생성물의내부에너지 Q W = U d U eact (KJ/kml U uel) eact : 반응물의내부에너지 내부에너지로표현 : 생성내부에너지를 u 사용해야하지만, 이것을피하기위해 enthaly의정의를이용. u = h v ( u + u u ) = Q W = N + h h ) - v + h h - v) N + h h - v) (KJ) v는고체, 액체에서무시가능. 기체의경우는이상기체로가정 (R u T) B.C.Ch 1
Reactant and duct mxtues Stchmetc : 연료를완전연소시키기위한최소공기량 CxHy + a(o 2 +3.76N 2 ) = x CO 2 + (y/2)h 2 O + 3.76 a N 2 a= x + y/4 ( by O balance : 2a = 2x +y/2) The stchmetc a-uel at (A/F) stc = (m a /m uel ) = 4.76a/1 (MW a /MW uel ) - Excess a at, λ = (A/F) / (A/F) stc - The equvalence at, φ = (A/F) stc / (A/F) = 1/ λ B.C.Ch 11 Enthaly Fmatn & Enthaly Cmbustn Chemcal enegy 핵에너지 mlecula Latent enegy atm Sensble enegy - 지금까지는화학에너지 ( 분자구조 ), 잠재에너지및감지에너지만고려핵에너지 ( 원자구조 ) 는불고려 - 화학반응 : 원자들을묶은분자로만드는화학결합이깨어지고새로운분자가생성 화학반응시화학에너지동반 - 표준상태 : 25, 1 atm ( 모든물질에동일기준상태제시 ) 윗첨자 으로나타냄, 예 ) h 연소공학에서는표준상태를 m 3 N 으로표현 B.C.Ch 12
Steady State Flw Q net = -393 MJ/kml 1kml C 1kml O 2 (at 25C, 1atm) eactants Cmbustn chambe CO 2 At 25C, 1atm duct C + O 2 = CO 2 발열반응 (exthemc eactn) : 반응중화학에너지가열의형태로방출 Q W = H d - H eact = -393 MJ/kml 열역학제1법칙에의거 Q net + Σ N h = Σ Ne he R Q net + H = H Q net = H - H Sensble Enegy Bken Chemcal Bnd 분자분자분자 B.C.Ch 13 * Enthaly eactn h : 완전반응에대하여생성물의엔탈피와반응물의엔탈피차 * Enthaly cmbustn hc : hc = H H 표준상태 (25C,1atm) 에서 Δh c = H H = Σ Ne )e - Σ N ) duct eact. * Enthaly matn h : 주어진상태에서화학조성으로인한물질의엔탈피 H, hc 는연소상태에따라변화근원적상태량필요 kml kj/kml B.C.Ch 14
Adabatc Flame Temeatues Insulatn Fuel A Cmbustn Chambe ducts T max Q W = Q = n n Q=, W = H d= H eact ( n = n ) ( N = N ) (KJ/s) B.C.Ch 15 Adabatc Flame Temeatues(T ad ) h(kj/kg) T eactants h = h T ad ducts T Fm 1 st law themdynamcs U (T, ) = U (T ad, ) H=U+V, U=H-V H H V( ) = V=Σn R u T = N R u T V =Σn R u T ad = N R u T ad H H R u (N T N T ad ) = N = m mx /MW N = m mx /MW H H R u (T m mx /MW T ad m mx /Mw ) = h h R u (T /MW T ad /MW ) = B.C.Ch 16
1 연료 1kg 이완전연소 LH V = (kj/kg) Cmbustn Sensble Heat H l 2 연소가스 G w (Kg/Kg uel) 3 연소가스평균정압비열 C KJ/(KgK) c = ( h/ T) 4 연소전온도, T 5 Adabatc lame Tem. = T ab G C (T T ) = H w m ab Hl T ab = + T G C w m l T ab 22K CH 4 -A 1. Fuel/A at B.C.Ch 17 Calculatn T ad. H d= H eact ( N = N ) (1) 반응식 : CxHy + a(o 2 +3.76N 2 ) = xco 2 + (y/2)h 2 O (+bco) + 3.76 a N 2 a= x + y/4 (2) ( N = N ) (3) Reactants 의물성치 물질 CH 4 (methane) C 3 H 8 (ane) C 8 H 18 (ctane) -249,95 - O 2 8682 N 2 8669 H 2 O(gas) -241,82 994 CO 2-393,52 9364 CO h (kj/kml) -74,85-118,91-11,53 h 298K (kj/kml) - - 8669 B.C.Ch 18
Calculatn T ad. (m=1.) (4) (x kml CO 2 )[-393,52 + hc2-9364 (kj/kml CO2)] + (y/2 kml H2O)[-241,82 + hh2o - 994 (kj/kml H 2 O)] + (3.76 a kml N 2 )[ + hn2-8669 (kj/kml N2)] = (1 kml uel, h) [-249,95 kj/kml C8H18] (5) x hc2 + y/2 hh2 + 3.76a hn2 5,896, = -249,95 kj x hc2 + y/2 hh2 + 3.76a hn2 = 5,646, kj = D 미지수가3개 (cl) 지만실제미지수는온도하나, 왜, 이상기체엔탈피는온도만의함수이므로 (6) 첫번째 D/(x+y/2+3.76a) = 88,2 kj/kml, 이엔탈피값은, 생성물 N2의경우 265K, H2O의경우 21K, CO2의경우 18K에해당 (Aendx A 참조 ) 생성물중질소비중이높으므로질소값 (265K) 과유사한값을단열화염온도로가정 약 24K 가정 (7) 24K에서각생성물의엔탈피 c2, hh2, hn2) 를찾음. (8) (7) 엔탈피 x 생성물의몰수 = D? (9) 아니면가정한단열온도값을조정하여맞을때까지반복계산 ( 보간법으로추정 ) * FOTRAN, C+, BASIC 등으로B.C.Ch 수치계산가능 19 T ad Octane (C 8 H 18 ) 3, Adabatc Flame Tem. (K) 2,5 2, 1,5 1, 2,393 2,157 C8H18 CH4 C3H8 942 5 - - 1. 2. 3. 4. 5. B.C.Ch A at [ ] 2
Chemcal Equlbum 2 nd Law Cnsdeatns Q S d Seact = δ + S T gen.sys. (KJ/K) Enty Change Enty tanse wth heat Enty geneatn wthn system Q 단열, Q= Reactant S eact Reactn Chambe ΔS sys ducts S d 과정중전체엔트로피는엔트로피형성식을계와외부비가역성이발생하는주위를포함하는확장계로표현 보통어떤 cess에서 enty의생성은 S S + S gen = sys su (KJ/K) B.C.Ch 21 반응 System 온도 T R 인저장소와 Q R 의열량을교환 ; 밀폐반응 S sys = S S 는각성분의Enty에대한공통기준이필요 d S eact sys S su = S d Seact S N S = N Q = T R (KJ/K) - 따라서, 모든물질의 enty에대한공통기준을정해야한다. Thd law themdynamcs 절대온도 K에서순수결정물질의 enty는 이다. -기준- 절대엔트로피 반응물과생성물에서가스상태성분을이상기체로가정하여 enty 를결정할경우, enty 는이상기체에서도온도와압력의함수. R B.C.Ch 22
- 이상기체혼합물에서한성분의엔트로피계산시그성분의분압이용 Q = y m T 등온 S(T,) T =1 atm 이상기체-표준상태 (T, ) S S(T, ) = S (T, ) R uln KJ/(Kml K) S = Rln S Ideal gas mxtue 의 성분으로나타내면, S (T, ) y m = S (T, ) R uln KJ/(Kml K) Q = y m : 분압 y : 성분의 mle 분율 m : 혼합물의전체압력 B.C.Ch 23 2nd Law Analyss Reactng Systems Avalablty Exegy Revesble WORK 계수행에서할수있는최대일 eactants ducts W u : Useul wk Wu I : Ievesblty I State Enty 생성으로부터비가역성을계산 I = W W = T S (KJ) ev u gen T : 주위의전개온도 B.C.Ch 24
정상유동연소과정에서 W ev = 2 v m + + gz TS ) 2 me 2 ve e + + gze 2 T S ) e (KW/s ) 여기서 h h + h) Wev = N + h h T S) N + h h T S) (kw) B.C.Ch 25 Chemcal Equlbum 2 nd -law cnsdeatns Cmlete cmbustn eactn CO + 1/2O 2 CO 2 [CO + 1/2O 2 ] [(1-α)CO 2 + αco + α /2O 2 ] The enty ducts, Themal dsscatn 2CO 2 2CO + O 2 S mx (T,) =ΣN s (T, ) = (1-α)S CO 2 + αs CO + α /2S O 2 S(T, ) = S (T, ) R ln u B.C.Ch 26
at α=.5, S max, The enty change ntenal t the system ds The cndtn equlbum( 더이상화학조성변화는없다 ) (ds) U,V,m = B.C.Ch 27 Gbbs unctn Fxed enegy system( 질량과체적이일정 ) 에서는 2 nd law가 chemcal equlbum을위해유용 Real mxtue(t,, st.) 에사용은불가 Gbbs unctn(g) 를 enty 대신사용 ; 열역학의물성을중요시 -일정질량으로구성된단순압축성시스템이일정온도, 압력을가지고있다면 1st law : dq-dv=du 2nd law : ds dq/t du+dv-tds -Gbbs unctn G = H TS -양변미분 (dg) T, = dh TdS SdT= (du+dv+vd)-tds-sdt = du + dv - TdS 2 nd law can be exessed as B.C.Ch (dg) T,,m 28
Gbbs unctn (dg) T,,m - 일정온도, 압력하에서화학반응은 Gbbs unctn 을감소시키는방향으로진행 - 화학반응은 Gbbs unctn 이최소가될때 equlbum dg mx = G dg< dg>(2 nd law 위배 ) dg= 1% 반응물평형조성 1% 생성물 Enty? B.C.Ch 29 Gbbs unctn deal gases Gbbs unctn matn Gbbs unctn mxtue deal gases Equlbum cndtn B.C.Ch 3
geneal system aa + bb + ee + F +. equlbum 화학종몰수 (dn) = 이론반응계수 (k)xml 수 k 공통 생략 B.C.Ch 31 Standad-state Gbbs unctn change K :equlbum cnstant altenately Statement chemcal equlbum at cnt. T and ΔG T > : 반응물쪽으로진행 ln K < (K <1) ΔG T < :? B.C.Ch 32
ncle Le Chatele Themal dsscatn 1. At any xed T, nceasng the essue suesses the dsscatn. 2. At xed, nceasng the tem. mtes the dsscatn. B.C.Ch 33 Equlbum ducts Cmbustn cmlete-cmbustn T max =2278K, Φ=1.5(ch) H 2 O max at Φ =1.15 - 연소열과비열이동시에감소 -Φ>1.5 연소열이비열보다빠르게감소 At Φ =1., by themal dsscatn -esence O 2, CO, H 2 Maj duct seces C 3 H 8 -a cmbustn at 1 atm B.C.Ch 34
Equlbum ducts Cmbustn - O, H adcals - OH, NO adcals - CO s mn seces n lean - O2 s mn seces n ch lean ch Mn seces dstbutns C 3 H 8 -a cmbustn at 1 atm B.C.Ch 35 Ret 1 1. Slve Examle 2.8 t 2.1 ( 첨부 gam 실행 ) 2. Revew questn(.69); dd N. 3.blems(Chate 2); 1, 14, 27, 38, 47 B.C.Ch 36