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+//clear//

+clear;

+clc;

+

+//Example 18.6

+//Given

+xF = 0.40;

+P = 1; //[atm]

+D = 5800; //[kg/h]

+R = 3.5;

+LbyV = R/(1+R);

+//Solution

+//Physical properties of methanol

+M = 32;

+Tnb = 65; //[C]

+rho_v = M*273/(22.4*338); //[kg/^3]

+rho_l_0 = 810; //[kg/m^3], At 0C, from Perry, Chemical Engineers' Handbook 

+rho_l_20 = 792; //[kg/m^3], At 20C, from Perry, Chemical Engineers' Handbook 

+rho_l = 750; //[kg/m^3], At 65C

+sigma = 19; //[dyn/cm], from Lange's Handbook of Chemistry

+//(a)

+//Vapor velocity and column diameter

+//Using Fig. 18.28, the abscissa is

+abscissa = LbyV*(rho_v/rho_l)^(1/2);

+//for 18-in. plate spacing

+Kv = 0.29;

+//Allowable vapor velocity

+uc = Kv*((rho_l-rho_v)/rho_v)^(1/2)*(sigma/20)^(0.2); //[ft/s]

+//Vapor flow rate

+V = D*(R+1)/(3600*rho_v); //[m^3/s]

+//Cross setional area of the column

+Bubbling_area = V/2.23; //[m^2]

+//If the bubble area is 0.7 of the total column area

+Column_area = Bubbling_area/0.7; //[m^2]

+//Column diameter

+Dc = sqrt(4*Column_area/%pi); //[m]

+disp('respectively','m',Dc,'and','ft/s',uc,'the allowable velocity and colmn diameter are')

+

+//(b)

+//Pressure drop:

+//Area of one unit of three holes on a trangular 3/4-in. pitch is

+//1/2*3/4*(3/4*sqrt(3/2)) in.^2. The hole area in this section (half a hole)is

+//1/2*%pi/4*(1/4)^2 in.^2. Thus the hole area is %pi/128*64/9*sqrt(3), or 10.08 percent

+//of the bubbling area.

+//Vapor velocity through holes:

+uo = 2.23/0.1008; //[m/s]

+//Using Eq.(18.58),

+//From Fig. 18.27

+Co = 0.73;

+hd = 51.0*uo^2*rho_v/(Co^2*rho_l); //[mm methanol]

+//Head of liquid on plate:

+//Weir height

+hw = 2*25.4; //[mm]

+//Height of the liquid above weir:

+//Assuming the downcomer area is 15 percent of the column 

+//area on each side of th column. From Perry, the chord

+//length for sucha segmental downcomer is 1.62 times the radius

+//of the colmn, so

+Lw = 1.62*2.23/2; //[m]

+//Liqiud Flow rate:

+qL = D*(R+1)/(rho_l*60); //[m^3/min]

+//From Eq.(18.60)

+how = 43.4*(qL/Lw)^(2/3) //[mm]

+//From Eq.(18.59), with 

+beeta = 0.6;

+hI = beeta*(hw+how); //[mm]

+//Total height of liquid, from Eq.(18.62)

+hT = hd+hI; //[mm]

+disp('mm methanol',hT,'pressure drop per plate is')

+

+//(c)

+//Froth height in th downcomer :

+//Using Eq.(18.62).,Estimating 

+hf_L = 10; //[mm methanol]

+//Then,

+Zc = (2*hI)+hd+hf_L; //[mm]

+//from Eq.(18.63)

+Z = Zc/0.5; //[mm]

+disp('mm methanol',Z,'Froth height in the downcomer is')