Chapter 3. Solution

Transcription

1 Chapter 3 Eample In an eperimental stud of the absorption of ammonia b water in a wetted-wall column, the value of K G was found to be kmol/m 2 s kpa. t one point in the column, the composition of the gas and liquid phases were 8.0 and mole % NH 3, respectivel. The temperature was 300 K and the total pressure was 1 atm. Eight-five percent of the total resistance to mass transfer was found to be in the gas phase. t 300 K, ammonia-water solution follow Henr s law up to 5 mole % ammonia in the liquid, with m = 1.64 when the total pressure is 1 atm. Calculate the individual film coefficients and the interfacial concentrations. Solution N = K G (p p * ) = K G P( * ) = K ( * ) K = K G P = ( )(101.3) = kmol/m 2 s Since 1 K resistance: = 1 k + m k for gas phase resistance that accounts for 85% of the total 1/ k 1/ K m / k 1/ K = 0.85 k = K /0.85 = /0.85 = kmol/m 2 s = 0.15 k = mk /0.15 = /0.15 = kmol/m 2 s * = m = = N = K ( * ) = ( ) = kmol/m 2 s N = k ( i ) i = N k = = i = m i i = i m = = Benitez, J. Principle and Modern pplications of Mass Transfer Operations, Wile, 2009, p

2 Eample wetted-wall absorption tower is fed with water as the wall liquid and an ammonia-air miture as the central-core gas. t a particular point in the tower, the ammonia concentration in the bulk gas is 0.6 mole fraction, that in the bulk liquid is 0.12 mole fraction. The temperature is 300 K and the pressure is 1 atm. Ignoring the vaporization of water, calculate the local ammonia mass-transfer flu. Data: k = 3.5/(1 ) lm mol/m 2 s, and k = 2.0/(1 ) lm mol/m 2 s; equilibrium relation: i = 10.5 i [ i (5.765 i 1)]. Solution The molar flu of ammonia () is given b i N = k (1 i ) (1 ) i ln 1 i = k (1 ) (1 i ) ln 1 i k ln 1 i = k ln 1 i 1 1 i = 1 i k / k i = 1 (1 i ) 1 i k / k 0.88 = i 1.75 (E-1) i and i are the solutions of Eq. (E-1) and the equilibrium relation: i = 10.5 i [ i (5.765 i 1)] (E-2) The following Matlab codes solve Eqs. (E-1) and (E-2) and plot out these equations. The intersection of the two curves from these equations also provides the solution. v=fminsearch('f3d2d3',[ ]); i=v(1);i=v(2); fprintf('i = %8.3f, i = %8.3f\n',i,i) =0:0.01:0.28; 1=1-0.4*(0.88./(1-)).^1.75; 2=10.5*.*( *.*(5.765*-1)); plot(,1,,2) label('_');label('_') grid on function ff=f3d2d3(v) i=v(1);i=v(2); f1=i-1+0.4*(0.88/(1-i))^1.75; f2=i-10.5*i*( *i*(5.765*i-1)); 2 Benitez, J. Principle and Modern pplications of Mass Transfer Operations, Wile, 2009, p

3 ff=f1*f1+f2*f2; >> e3d2d3 i = 0.231, i = The interfacial concentrations are: i = and i = The molar flu of is then N = k ln 1 i = 2 ln = 4.7 mol/m2 s Eample miture of methanol (substance 1, the more volatile) and water (substance 2) is being distilled in a packed tower. t a point along the tower where the temperature is 360 K, the methanol content of the bulk of the gas phase is 36 mol%; that of the bulk of the liquid phase is 20 mol%. ssume equimolar counter diffusion with mass transfer coefficients k = kmol/m 2 s and k = kmol/m 2 s, estimate the local flu of methanol from the liquid to the gas phase. Solve the problem b plotting equilibrium curve at 360 K. The activit coefficients for this sstem are given b: ln γ 1 = ln( ) V2 ; 12 = V ep a12 1 RT 3 Benitez, J. Principle and Modern pplications of Mass Transfer Operations, Wile, 2009, p

4 12 21 ln γ 2 = ln( ) V1 ; 21 = V ep a21 2 RT Recommended values of the parameter for this sstem are: V 1 = cm 3 /mol, V 2 = cm 3 /mol, a 12 = cal/mol, and a 21 = cal/mol. The vapor pressure for methanol and water can be determined from the following equations: ln P 1 vap = T 33 ln P 2 vap = T 47 Solution The equilibrium curve at 360 K is obtained from the following steps: 1) Choose a value of mole fraction in the liquid phase, 1, between 0.1 and 0.25; 2) Evaluate the activit coefficients; 3) Evaluate the vapor pressures 4) Evaluate the partial pressures: p 1 = γ 1 1 P 1 vap, and p 2 = γ 2 2 P 2 vap 5) The mole fraction in the vapor phase, 1, is then p 1 /( p 1 + p 2 ) The local flu is obtained from the following epressions: N 1 = k ( i ) = k ( i ) k i = + k ( i) = (0.2 i) (E-1) The above equation is a straight line that can be plotted on the graph with the equilibrium curve. The intersection of flu equation (E-1) and the equilibrium curve provides the interfacial compositions i and i. The Matlab codes, listed in Table E-1, plots the equilibrium curve and flu equation (E-1). From the closed up view of the graph (Figure E-2), the interfacial compositions are i = and i = The local flu is then N 1 = k ( i ) = ( ) = kmol/m 2 s 3-22

5 Table E-1 Matlab codes to plot equilibrium curve and flu equation R=1.987;T=360;v1=40.73;v2=18.07;a12=107.38;a21=469.55; RT=R*T; d12=v2*ep(-a12/(rt))/v1;d21=v1*ep(-a21/(rt))/v2; 1=0.1:0.01:0.25;2=1-1; con=d12./(1+2*d12)-d21./(2+1*d21); gam1=ep(-log(1+2*d12)+2.*con); gam2=ep(-log(2+1*d21)-1.*con); pv1=ep( /(t-33)); pv2=ep( /(t-47)); p1=pv1*gam1.*1;p2=pv2*gam2.*2; 1=p1./(p1+p2); plot(1,1) =0.15; =0.36+(0.2-)*.0149/.0017; plot(1,1,[ 0.2],[ 0.36],'--') label('_');label('_') grid on legend('equilibrium','flu equation') Figure E-1 Graphical solution 3-23

6 Figure E-2 Graphical solution (closed-up view) The interfacial compositions i and i can be obtained directl from the following Matlab codes without graphing: function ff=f3d2d4(v) 1=v(1);1=v(2);2=1-1; R=1.987;T=360;v1=40.73;v2=18.07;a12=107.38;a21=469.55; RT=R*T; d12=v2*ep(-a12/(rt))/v1;d21=v1*ep(-a21/(rt))/v2; con=d12/(1+2*d12)-d21/(2+1*d21); gam1=ep(-log(1+2*d12)+2*con); gam2=ep(-log(2+1*d21)-1*con); pv1=ep( /(t-33)); pv2=ep( /(t-47)); p1=pv1*gam1*1;p2=pv2*gam2*2; f1=1-p1/(p1+p2); f2= *(0.2-1)/0.0017; ff=f1*f1+f2*f2; >> fminsearch('f3d2d4',[ ]) ans =