From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

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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')
-- 
cgit