Ch.10 Stage and Continuous Gas-Liquid Separation Process bsorption(1 bsorption(1 1
Introduction V 1, 1 V 2, 2 L 0, x 0 L 1, x 1 설계목표 1.Plate tower : 액상의흐름으로기상을 2 1 으로의농도변화에필요한단수 - 도식적으로 2.Packed tower : 액상의흐름으로기상을 2 1 으로의농도변화에필요한탑높이 간단한설계법, 이동단위수 bsorption(1 2
10.2 Equlibrium Relations between Phases Phase rule 주어진조건에서두상사이의평형이상률에의해제한 f=c-p+2 한 solute의평형에영향을끼치는주요변수들은온도, 압력, 조성 ex CO 2 -air-water sstem에서 f=3-2+2=3 온도, 압력이결정되면임의로결정할수있는변수가하나남는다. 액상에서 CO 2 의몰분율을 x 라두면기상에서몰분율 나분압p 는자동적으로결정 bsorption(1 3
기-액평형데이터 Henr s Law p =Hx (10.2-2 H : Henr 상수 [atm/mole fraction] cf. p =p o x : Raoult s Law 양변을 total P로나누면 =H x (10.2-3 p V x L bsorption(1 4
. p Hx 에서공기중산소의분압은 0.21atm 0.21 4.3810 x 4 x -6 4.810 n n n -6 n n 1mole 이면 n 4.810 mole B B -6 4.810 mole O2 32g 1mole H2O 용존산소= -6 (1-4.810 mole H O 1mole O 18g 2 2 8.5310 4 go 100gwater 2 bsorption(1 5
10.3 Single and Multiple Equilibrium Contact Stages. Single stage Equilibrium Contact V 1 V 2 L 0 L 1 상이한두상이서로밀접하게접촉하고나서분리된두상사이에평형상태에도달 bsorption(1 6
일반적인경우 :solute B:air C:water total material balance L 0 + V 2 = L 1 + V 1 = M (10.3-1 가정 : 세개의성분,B,C가모든흐름에존재 와 C 에대한 solute material balance L 0 x 0 + V 2 2 = L 1 x 1 + V 1 1 = M x M (10.3-2 L 0 x C0 + V 2 C2 = L 1 x C1 + V 1 C1 = M x CM (10.3-3 x +x B +x C =1 이기때문에 B 에대한방정식불필요 known value : (in V 2, 2, C2, L 0, x 0, x C0 (target 1 unknown value : L 1, V 1, x 1, x C1, C1 5개방정식수 : (10.3-1~(10.3-3 3개평형관계식 : x 1 vs 1, x C1 vs C1 2개 bsorption(1 7
간략화한경우 가정 1.air(B 는 water phase 에 insoluble 2.water(C 는 gas phase 로증발하지않는다. V 1, 1 L 0, x 0 +B gas +C liq V 2, 2 L 1, x 1 known value L, x, V, L V L V Lx V Lx V C L x L x L (1 x 0 0 2 2 unknown value L, x, V, 총괄성분성분평형 1 1 1 1 0 2 1 1 0 0 2 2 1 1 1 1 0 C0 1 C1 1 1 H x 1 1 bsorption(1 8
간략화한경우의해결법 가정 1.air 는 water phase 에 insoluble V 2.water 는 gas phase 로증발하지않는다. L V 1 (V 1 L 0 (L x 0 +B gas +C liq V 2 (V 2 V : inert air 의 mole flow rate L 1 (L x 1 L : inert water 의 mole flow rate L L L [ L L (1 x ] 0 0 0 0 L L x ( L L x L 0 0 0 0 0 0 x 1 x 0 0 L bsorption(1 9
성분에대한 material balance는 L V L V 0 2 1 1 inert base로고쳐쓰면 x 0 2 x 1 ( ( ( 1 L V L V( (10.3 4 1x 1 1x 1 평형관계는 0 2 1 1 H x (10.3 5 1 1 1. (10.3 5 를 (10.3 4 에대입하여 x 을계산 2. x 을(10.3 5 에대입하여 을계산 1 L V 3. L1, V1 1x 1 1 1 1 1 bsorption(1 10
. V 1 300kgmole water/hr=l 0 x O =0 V 2 =100kgmole/hr, 2 =0.2 L 1. bsorption(1 11
inert air flow: V V (1 100 (1 0.2 80 kgmole / hr 2 2 2 CO 에대한 material balance x 0 2 x 1 ( ( ( 1 L V L V( 1x 1 1x 1 0 2 1 1 4 from 3-18 H 0.14210 atm / mole fraction H 4 H 0.142 10 mole fraction gas / mole fractionliq P 0 0.2 x1 1 300( 80( 300( 80( 10 10.2 1x 1 x 0.14210 x 20 300( 80( 4 1 1 4 1 x1 1 0.14210 x1 4 x1 1.4110, 1 0.2 1 1 L 300 L 300[ kgmole/ hr] 1 4 1 x1 1 1.4110 V 80 V1 100[ kgmole/ hr] 1 10.2 1 bsorption(1 12
10.3C Countercurrent Multiple-Contact Stages 전 stage에대한총괄물질수지 L V L V M 0 N1 N 1 성분수지 (, B, C중한성분에대한 Lx V L x V Mx 0 0 N1 N1 N N 1 1 M known values 1. Input condition ; V,, L, x 2. target value ; 1 N1 N1 0 0 unknown value ; x, L, V N N 1 bsorption(1 13
1~ n단까지의총괄물질수지 L V L V 0 n1 n 1 1~ n단까지의성분수지 Lx V Lx V 0 0 n1 n1 n n 1 1 n1 L V x V Lx n 1 1 0 0 n n1 Vn1 " operating line( 조작선 " Vstream에서의농도 n1 n 과 L stream에서의농도 x 과연관짓는방정식 bsorption(1 14
두개의 stream L 과 V 가상호 immiscible 하면단지 만이전달조작선의기울기 (L n /V n+1 는단마다흐름이다르기때문에곡선형태 N+1 에서 1 으로의농도감소처럼주어진분리를얻기위한이상단수 (ideal stage 의결정은도식적으로수행 bsorption(1 15
bsorption(1 16
T=300K, P=101.325kPa, =2.53x V 1 중아세톤 0.3 0.1=0.03 kgmol/hr V 1 =29.7+0.03=29.73 x 0 =0 L 0 =90 kgmole H 2 O/hr N+1 =0.01 V N+! =30 kgmole/hr 아세톤 : 30 0.01=0.3 kgmol/hr 공기 : 30 (1-0.01=29.7kgmol/hr L N 중아세톤 : 0.3 0.9=0.27kgmol/hr L N =90+0.27=90.27 bsorption(1 17
Solution 조작선의기울기 L n /V n+1 : 90/29.73~90.27/30 3 (x 0, 1 =(0, 1 =(0,0.00101 1 =0.03/(29.7+0.03=0.00101 (x N, N+1 =(x N,0.01=(0.003,0.01 x N =0.27/(90+0.27=0.003 N=5.2 이론단수가필요 bsorption(1 18
* 대류물질전달계수 Thin viscous 표면에근접한영역 sublaer 분자확산 큰농도강하 buffer laer 분자확산 + 난류확산 turbulent core laer 난류확산 매우적은농도강하 bsorption(1 19
c 등몰상호확산에서물질전달계수 dc N( DB M x( N NB 에서 dz N N B ( D N ( c c k ( c c k B M 1 2 c 1 2 z2 z1 ( DB M z z 2 1 기체 : N k ( c c k ( p p k ( c 1 2 G 1 2 1 2 bsorption(1 20
2 1 c 1 2 일방확산의물질전달계수 dc c N( DB M ( N NB 에서 dz c ( DB M NB0 and kc z z N c c c ( D c z z B M BM BM N k ( c c 1 2 2 1 2 1 ( DB M 1 ( c1 c2 z z 기체 : N k ( c c k ( p p k ( c 1 2 G 1 2 1 2 bsorption(1 21
10.4 Mass Transfer between Phases Mass transfer of solute from one fluid phase b convection and then through a second fluid phase b convection 계면 (interface 에서는평형에대한저항이없으므로 i 와 x i 는평형 i =f(x i bsorption(1 23
10.4C Mass Transfer Using Film Mass-Transfer Coefficients and Interface Concentrations 1. 등몰상호확산 : diffusion from gas to liquid B : equimolar counterdiffusion from liquid to gas N k ( k ( x x G i x i L k (10.4 3 k x x x G i L i k 두개의막계수 kx과 k을안다면기울기 으로 k PM을그려서계면에서의조성을결정 bsorption(1 24 x
2. 일방확산 N k ( k ( x x k G i x i L k kx kx (1 (1 x (1 (1 x im im im im (1 i (1 G ln[(1 /(1 ] i L G (1 xl (1 xi ln[(1 x /(1 x ] i k k N ( ( x x G i i L (1 im (1 im x k /(1 x k (10.4 9 k k x x x im x G i /(1 im L i bsorption(1 25
식 (10.4-9 를이용하여계면조성을구하는방법 step.1 (1- im 과 (1-x im 을 1로가정식 (10.4-9 을이용기울기와 i, x i 를구한다 step.2 i 와 x i 값을이용새로운 i 와 x i 값을구하기위해새로운기울기를계산 step.3 계면조성이변하지않을때까지빈복 (3 회정도면충분 bsorption(1 26
bsorption(1 27
x L =0.1 G =0.38 wetted-wall tower T=298K, P=1.013 10 5 Pa k =1.465 10-3 kgmol /m 2 s mole frac k x =1.967 10-3 kgmol /m 2 s mole frac 계면농도 i, x i 와 N 를계산하여라 bsorption(1 28
P(x L, G =P(0.1,0.38 계면농도를구하기위한과정 step.1 (1- im 과 (1-x im 을 1 로가정식 (10.4-9 을이용기울기와 i, x i 를구한다 k /(1 x 1.967 10 1.342 M(0.247,0.183 k 3 x im 3 /(1 im 1.46510 bsorption(1 29
step.2 i 와 x i 값을이용새로운 i 와 x i 값을구하기위해새로운기울기를계산 (1 i (1 G (10.183 (10.38 (1 im 0.715 ln[(1 i /(1 G ] ln[(1 0.183/(1 0.38] (1 xl (1 xi (10.1 (10.247 (1 x im 0.825 ln[(1 xl /(1 xi ] ln[(1 0.1/(1 0.247] 3 k /(1 x 1.967 10 /0.825 1.163 M 3 1(0.257, 0.197 k 1.46510 /0.715 x im /(1 im bsorption(1 30
step.3 계면조성이변하지않을때까지빈복 (1 i (1 G (1 0.197 (1 0.38 (1 im 0.709 ln[(1 /(1 ] ln[(1 0.197/(1 0.38] (1 xl (1 xi (10.1 (10.257 (1 x im 0.820 ln[(1 x /(1 x ] ln[(1 0.1/(1 0.257] k k M 1 3 /(1 x 1.967 10 /0.820 3 1.46510 /0.709 x im /(1 im i L (0.257, 0.197 을취한다. G i 1.160 이기울기는 -1.163 과거의같으므로계면의농도는 bsorption(1 31
k N ( ( x x kx G i i L (1 im (1 x im 3 1.46510 (0.38 0.197 0.709 0.820 3 1.967 10 (0.257 0.1 kgmole m s 4 2 3.78 10 [ / ] bsorption(1 32
10.4D Overall Mass-Transfer Coefficients and Driving Forces N =K ( G -* N =K x (x* -x L x* * 등몰상호확산과희석용액에서의확산 N k ( k ( x x G i x i L ( ( * * G G i i * i m 이므로 x i x L * G ( G i m( xi xl N N m N K k k x 1 1 m (10.4 15 K k k x bsorption(1 33
유사한방법으로 * * x x ( x x ( x x m" x L i i L G * x i i * G i x xl ( xi xl m" N N N K m k k x " x 1 1 1 (10.4 18 K k m" k x x bsorption(1 34
일방확산 K N ( ( x x * Kx * G L (1 * M (1 x * M k ( ( (1 (1 kx G i xi xl im x im ( ( * * G G i i ( m( x x G i i L 1 1 m (10.4 24 K /(1 k /(1 k /(1 x * M im x im K x 1 /(1 x (1 (1 x 1 1 (10.4 25 * M m" k /(1 im kx /(1 x im * (1 (1 G * M * G ln[(1 /(1 ] * (1 xl (1 x * M * xl x ln[(1 /(1 ] bsorption(1 35
bsorption(1 36
G =0.38 x L =0.1 wetted-wall tower T=298K, P=1.013X10 5 Pa k =1.465 10-3 kgmol /m 2 s mole frac k x =1.967 10-3 kgmol /m 2 s mole frac K, N, 기체와액체막에서퍼센트저항을계산하여라 bsorption(1 37
Solution (0.257,0.197 0.052= (0.1,0.052 1 1 m K /(1- k /(1- k /(1- x * M im x im * i - 0.197-0.052 m 0.923 x - x 0.257-0.1 i (1- L * (1- -(1- G * M * G ln[(1 - /(1 - ] (1-0.052-(1-0.38 0.773 ln[(1-0.052/(1-0.38] 에서 bsorption(1 38
K 1 1 0.923 /0.773 1.46510 /0.709 1.967 10 /0.820-3 -3 484( 기체막저항 384.8( 액체막저항 868.8( 총저항 K 8.9010 [ kgmol / m smole frac] 4 2 484 기체막저항퍼센트 100 55.7% 868.8 384.8 액체막저항퍼센트 100 44.3% 868.8 K 4 * 8.9010 N ( G (0.38 0.052 (1 0.773 * M 4 2 3.7810 [ / ] kgmol m s bsorption(1 39
ssignment #3 1. 10-2-2 2. 10-3-1. 3. 10-4-2 ( kakaka, x, kakaka, x, bsorption(1 40
bsorption(1 41