当前位置:高等教育资讯网  >  中国高校课件下载中心  >  大学文库  >  浏览文档

《化工原理公式》(全)

资源类别:文库,文档格式:DOC,文档大小:1.38MB,文档页数:39,团购合买
点击下载完整版文档(DOC)

T'=T"=T (2-1) ==F〓 (2-2) fi=fi=f fi=fi fi /y fi /x, P a'y P=x P (2-8) y,P=yixi (2-10) y Yi xi K1=y2/x1 Vi/y, K K K y 1n=1 RT V-lnz (2-15) dP RT So L On,/Pm (2-16) RT P Vm -bm To vm(vm +b a,=0.42748R272/P b,=0.08664RT/P 27R272/64P

T = T = T = (2-1) P = P = P = (2-2) ˆ ˆ ˆ i i i f f f    = = = (2-3) L i V i f ~ f ~ = (2-4) f / y P ~ ˆ i v i i V  = (2-5) f / x P ~ ˆ i L i i L  = (2-6) l i OL i ˆ f i xf  = (2-7) ˆ i yiP ˆ i xiP V L  = (2-8) oL i i i i i V y P x f ˆ  =  (2-9) 2 i 1 i f ~ f ~ = (2-10) 2 i 2 i 1 i 1 i  x =  x (2-11) (2-12) j i i j i j ij K K x x y y  = = (2-13) V L i i i i i   ˆ ˆ x y K = = (2-14) t m t v i i d ln 1 ln j V Z V RT n P RT ˆ T ,V ,n  −            −             =    (2-15) dP 1 ln j 0 i t i           −             =  P RT n V RT ˆ T ,P,n P  (2-16) ( ) m m m m m m T v v b a V b RT P + − − = 0.5 ai R Tc,i Pc,i 2 2 = 0.42748 bi = 0.08664 RTc,i Pc,i ai 27R Tc,i 64Pc,i 2 2 = i i i K = y x

b=RT。/8P bn=∑y SRK方程 RT v(v+b) a1=aci·aa(r) an,=0.42748R2T2/P =p+m-(/r m1=0.48508+1.55171m-0.1561302 y b=∑y RT a b 0 Zm=PV/RT=V/V, -b)-a/RTV, 2√aa InI z RTt K Vi Yi ④"P 当x;→1时 ni=f/xf

bi = RTc,i 8Pc,i ( ) 2 m =  i ai a y m =  ibi b y SRK 方程: v(v b) a v b RT P + − − = ai ac,i  i(T ) = • c,i c,i i ac, R T P 2 2 = 0.42748 ( )   ( )  2 0.5 i T = 1+ mi 1- T Tc,i i mi i 2 = 0.48508 +1.55171 − 0.15613 ( ) 2 =  i ai a y =  ibi b y  + − = 0      − + P ab V P a V P RT Vt b t t 3 2 ( ) Zm = PVt RT =Vt Vt − b − a RTVt t i m t i i RTV aa ln Z V b b ˆ ln 2 V b 1- t  −              − −  = ˆ P f x y K V i oL i i i i i   = = (2-17) 当 xi →1 时,  i →1 (2-18) L i i L i i f ˆ f x ˆ  = (2-19)

f 1 RT 2-20) P RT dP|=indi In P =Pas·er(p-P)yR」 (2-21) 当x1→0时,y;*→1 f=H=1im左 fi =H x (n, RTlny) (2-26) RTI In A A21x2 A hy=x2[42+2(421-A12kx] hy2=x2[41+2(42-A1)x2 X Iny,=l 4k 4=exr(n-2) x, g In y G Gxk

i i f P  v - dP 1 ln P 0 i i L        = P RT P RT f (2-20) ( ) s i s i L s i i P P i P i L i P P ln RT v P P dP ln P RT dP v P RT v P f ln s i s i − − = +              + −      = −    RT 0 1 f P expv (P P ) RT  S S i L i i S i L i =  • − (2-21) 当 xi → 0 时,  i →1 (2-22) i L i 0 oL i lim i x f ~ f H x → =  (2-23) i L f i Hx ˆ = (T, P 一定, xi → 0 ) (2-24) ˆ f i xiH L i =   (2-25) ( ) = = c i 1 i ln i n RT  E G (2-26) i ln j i RT n G T ,P,n Ei =           (2-27) 2 1 2 1 2 1 2 2 1 2 2 21 2 1 2 1 1 2 1 1 1         + =         + = A x A x A ln A x A x A ln   ( )   ( )  21 12 21 2 2 12 21 12 1 2 1 2 ln 1 = x2 A + 2 A − A x ln = x A + 2 A − A x    −       = k j j j k i j j k x k x ln i x     j 1-ln exp ( ) RT  v v L ij ii i L j ij = −  −            = + −       k k kj l j l j l l i j k k ki j i j k k ki j j i j i j i G x G x G x x G G x G x ln      ij = (gij − gij) RT ( ) ij ij ij G = exp − 

gii-g B C +-+一+ (2-28) 1+Bp+cpa d=2∑y;B Ing y: B:-lnz (2-31) B niy3ij =1+ (2-33) yi P P (2-34) y T yiyi Pq eXp RT>>VP-p K (2-37) K=YiPi 2-38) 1P13x1 y

gij − gij = = + + 2 + B 1 v C RT v Pv Z (2-28) = = 1+ BP + CP 2 + RT Pv Z (2-29) RT P ˆ B        =  − = C j 1 lni 2 y iB ij (2-30) = = − C j 1 i i ij ln 2 ln ˆ y B Z v  (2-31) = = = c i 1 c i 1 B yi yj Bij (2-32) RT v Pv Z B = = 1+ (2-33)   = =  c i 1 i ci c i 1 i ci 2 y T y P T P (2-34) ( )       − = = RT S i L i V i S i S i i i i i exp V P p ˆ P P x y K    (2-35) ( ) exp 1 S i L i        − RT V P p ( ) S P pi RT v L  i − K P P S i = i (2-36) i S i i x p p y = (2-37) p p K S i i i  = (2-38) p p x y i S i i i  = (2-39)

P K pi ①'P K;=2/f (2-41) (2-42) K yPs「r-p) exp (2-43 ④P 1=K1x1(i Vi 1 (2-46) K: =f(P,T,x, y) InK: =A-B /(T+Ci) (2-48) lnK1=A1-B1/(T+18-0.197) ∑Kx1 f()=∑K 设T→由PK图查K→∑K,x1-|f(m)≤e →结束 调整T

( )       − = RT S i L i V i S i S i i exp V P p ˆ P P K   (2-40) V i L i i K = f f (2-41) L i i L i f x f ~ = (2-42) ( )       − = RT S i L i V i S i S i i i exp V P p P P K    (2-43) yi = Ki xi (i =1,2, c) (2-44) = = c i 1 yi 1 (2-45) = = c i 1 xi 1 (2-46) Ki = f(P,T,x, y) (2-47) lnKi = Ai − Bi (T +Ci ) (2-48) lnKi = Ai − Bi (T +18 −0.19Tb ) (2-49) = = c i 1 Ki xi 1 (2-50) ( ) -1 0 c i 1  i i = f T = K x = (2-51) 设 T ⎯给定P ⎯→ 由 P-T-K 图查 Ki → = c i 1 i i K x → ︳ f(T) ︱≤ε ⎯⎯Y→    i x T →结束 N 调整 T

y K·a K ∑a1 (K+1) K K;x;) G(1T=1n∑K1x1=0 f(P)=∑K1x1-1=0 Pixi (2-57) P=∑rPx1 (2-58) ∑(y1/K)=1.0 (2-59) f()=∑(y/K)-1.0=0 f(P)=∑(y/K,)-1.0=0 (2-61) 设T一给→由PK图查K→∑(y/K)→1f)1≤e→}一结束

i K i i y = K • • x (2-52)  = i i K 1 x K  (2-53) = + = C i 1 (K) i i (K) (K 1) K K ( K x ) K K ) (2-54) = = = C i 1 G(1 T) ln K ix i 0 (2-55) ( ) -1 0 c i 1  i i = f P = K x = (2-56) i c i 1 S 泡 i P P x = = (2-57) i c i 1 S 泡 i P P x  i = =  (2-58)  y Ki .0 = = c i 1 ( i ) 1 (2-59) ( ) ( )-1. 0 c i 1  i = f T = y Ki 0 = (2-60) ( ) ( )-1. 0 c i 1  i = f P = y Ki 0 = (2-61) 设 T ⎯给定P ⎯→ 由 P-T-K 图查 Ki → = c i 1 ( yi Ki)→ ︳ f(T) ︱≤ε ⎯⎯Y→    i x T →结束

调整T 1+∑(y:/K) -1+∑(/ OK B,yi K: T()+C F1=Lx1+吃1i=1,2, (2-63) F=L+y FHF+O=VH+LH, (2-65) L+VK Y=V/F (2-67) 1+y(K1-1) 2 1.0 1+y(K:-1) 11 1.0 1+y(K1-1 f()=(K:-1)=1-=0 (2-71) 1+y(K1-1)

N 调整 T ( ) ( )             + − + = +           − + = + + 2 i i i i i i 2 i i ( 1) ( ) 1 ( i ) ( ) 1 ( ) K T C B y y K T T K K y K T T K K K i K i y (2-62) Fzi = Lxi +Vyi i =1,2,c (2-63) F = L +V (2-64) FHF +Q =VHV + LHL (2-65) i i i L VK Fz x + = i i i F V VK Fz x − + = (2-66)  =V F ( K ) z x 1 i 1 i i + − =  (2-67) ( K ) K z y 1 i 1 i i i + − =  (2-68) 1.0 1 1 c i 1 i i = + − = ( K ) z  (2-69) 1.0 1 1 c i 1 i i i = + − = ( K ) K z  (2-70) 0 1 1 ( 1) ( ) c i 1 i i i = + − − = = ( K ) K z f   (2-71)

df(y())dy (K;-1)2 +y(k) (2-73) dp (K H1=∑yH2(T,P) (2-74) H x: H (T, P) (2-75) (2-76) Tp-TB K.(r(g+1) (7 G(T)=VH+LHL-FHF G(T=H+(1-y (2-78) G(T)) +L-=vCPr LCPl (2-79) △T=d·△T计算 f(T)= (2-80) 1+y(K1-1)

( ) ( ) ( ) ( )     d d - k k ( 1) ( ) f K K f = + (2-72) ( ) ( ) ( )   = + − − = − c i 1 2 i k i 2 i k 1 ( 1) ( 1) d d K K z   f  (2-73) = = c i 1 HV yiHvi(T ,P) (2-74) = = c i 1 HL xiHLi(T ,P) (2-75) D B B T T T T − −  = (2-76) ( ) ( ) ( ) ( ) ( ) ( ) K K K 1 ref ref 1 d f T K T K T +  = + ( ) G T =VHV + LHL − FHF ( ) ( ) G T =HV + 1− HL − HF (2-77) ( ) ( ) ( ) ( ) ( ) ( ) K (K ) K K K dG T dT G T T = T − +1 (2-78) ( ) ( ) ( ) PV PL V L K K VC LC dT dH L dT dH V dT dG T = + = + (2-79) ︱ (K ) (K) T =T +1 ︱≤ε △T=d·△T 计算 = + − − = c i 1 i i i 1 1 ( 1) ( ) ( K ) K z f T  (2-80)

G()=+(1-y)H-H U(K+1) (2-82) (2-83) N=N-N f=c-I+2 Ny=f+1=(-1+2)=c+2 N"=∑N+N-N (3-2) (3-3a) ax (3-3b) L AD (3-5) =a,·a N1·C (3-7) aA=1·a,·a N lga (3-10

( ) ( ) G =HV + 1− HL − HF (2-81) ( ) (K ) V L K F L H H H H         − − = +1  (2-82) ( ) ( ) ( ) ( ) K K K  = + d − + 计 1 (2-83) Ni = Nv − Nc (3-1) f = c − + 2 Nv = f +1= (c −1+ 2) = c + 2 =  + − i N N N N u r c e i u i (3-2) ( )  = + − m 1 i i i,D m R x    (3-3a)  = − 1- q i i i,F    x (3-3b) 1 D 1 1         =         =         B A B A B A x x x x y y  (3-4) A,2 A,1 DxA,D V2 y = L1 x − (3-5) 1         =         B A B A x x y y 2 (3-6) B W A N B A x x x x         = • • •         1  2  3 N-1  D (3-7)   1 1 2 3 N-1 N N       AB = • • • AB m lg lg                          = B W A B A x x x x N D (3-8) 3   D  W  F 平均 = • • (3-9)   D  W 平均 = • (3-10)

N Iga 台(kr (3-12) (-ax0) 9xD·9h,W N (3-15) lg +54.4X)(x- Y=l-exp (3-16) 1+117.2XX Y=0.75-0.75X0566 (3-17) D Ig =N lga KI pir, K2 p212 n|=4(2-x1)+x(2-2x12-42) (3-21) +x,[41、-A2+2x(A1-A2)-x,(42-A2)-C(x2-x1 2|=410-x)-2x)+x(4,-) (3-22) A2(1-x,1-2x)+x,(4-42,) P A2(1-2x1)

 平均 lg lg m                   = A w B d w d N (3-11) ( ) Nm  i,r        =        r r i i w d w d (3-12) ( ) LK LK 1 LK d f w f LK,D LK ; = −LK,D = • (3-13) ( ) H HK 1 H HK d f w f HK HK,W K ; = − K,W = • (3-14) ( )( ) N W LK - K LK,D K, LK,D HK,W H H m lg 1 1 lg            − − • = (3-15) ( )( ) ( )       + + − = − X X X X Y 11 117.2 1 54.4 1 exp 1 (3-16) 0.5668 Y = 0.75 − 0.75X (3-17) 0.206 2 HK,D LK,W LK,F HK,F               •                 = D W x x z z N N S R (3-18) m AB B,W B,D A,W A,D lg lg N lg x x x x − = (3-19) 2 s 2 1 s 1 2 1    p p K K = = (3-20) ( ) ( )( )  ( ) ( ) ( ) s 1 s s 2 1 s 1 s 2 2 s s 2 2 1 2 1 2 1 2 2 1 1 2 2 1 2 1 2 ln 2 x A A x A A x A A C x x A x x x x x A A s s + − + − − − − − = − + − −           (3-21) ( )( ) ( ) s 2s s = A − x − x  + x A − A         1 s 1s 2 1 ln 12 1 1 2   (3-22) ( )( ) ( ) s 2s T 3 s 2 s 1 s A x x x A A P P a +  − −  +  −          ln = ln 12 1 1 2 1 s 1 s (3-23) ( ) 1 2 1 ln = A − 2x         12 1   (3-24)

点击下载完整版文档(DOC)VIP每日下载上限内不扣除下载券和下载次数;
按次数下载不扣除下载券;
24小时内重复下载只扣除一次;
顺序:VIP每日次数-->可用次数-->下载券;
共39页,可试读13页,点击继续阅读 ↓↓
相关文档

关于我们|帮助中心|下载说明|相关软件|意见反馈|联系我们

Copyright © 2008-现在 cucdc.com 高等教育资讯网 版权所有