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 --- 905/CH7/EX7.5/7_5.sce | 54 +++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 54 insertions(+) create mode 100755 905/CH7/EX7.5/7_5.sce (limited to '905/CH7/EX7.5/7_5.sce') diff --git a/905/CH7/EX7.5/7_5.sce b/905/CH7/EX7.5/7_5.sce new file mode 100755 index 000000000..ae269489b --- /dev/null +++ b/905/CH7/EX7.5/7_5.sce @@ -0,0 +1,54 @@ +clear; +clc; + +// Illustration 7.5 +// Page: 444 + +printf('Illustration 7.5 - Page: 444\n\n'); + +// solution +//*****Data*****// +// C-nicotine A-water B-kerosene +// F-feed R-raffinate S-solvent +F = 1000; // [feed rate, kg/h] +xAF = 0.99; // [fraction of water in feed] +// Because the solutions are dilute therefore +xCF = 0.01; // [fraction of nicotene in feed, kg nicotene/kg water] +xCR = 0.001; // [fraction of nicotene in raffinate, kg nicotene/kg water ] +m = 0.926; // [kg water/kg kerosene] +//*****// + +printf('Illustration 7.5(a) - Page: 444\n\n'); +// Solution(a) + +yCS = 0; // [kg nicotene/kg water] + +// Because, in this case, both the equilibrium and operating lines are // straight,if the minimum solvent flow rate Bmin is used, the concentration // of the exiting extract, yCmax, will be in equilibrium with xCF. Therefore +yCmax = m*xCF; // [kg nicotene/kg kerosene] + +A = F*xAF; // [kg water/h] +// From equation 7.17 +Bmin = A*(xCF-xCR)/(yCmax-yCS); // [kg kerosene/h] +printf("The minimum amount of solvent which can be used is %f kg kerosene/h.\n\n",Bmin); + +printf('Illustration 7.5(b) - Page: 444\n\n'); +// Solution(b) + +B = 1.2*Bmin; // [kg kerosene/h] +EF = m*B/A; +Nt = log((xCF-yCS/m)/(xCR-yCS/m)*(1-1/EF)+1/EF)/log(EF); + +printf("The number of theoretical stages if the solvent rate used is 20 percent above the minimum is %f .\n\n",Nt); + +printf('Illustration 7.5(c) - Page: 444\n'); +// Solution(c) + +Eme = 0.6; // [Murphree stage efficiency] +// from equation 7.20 +Eo = log(1+Eme*(EF-1))/log(EF); // [overall efficiency] +Nr = Nt/Eo; // [number of real stages] +disp(Nr); +// The nearest integer to number of real stages is 11 +// Therefore +Nr = 11; +printf("The number of real stages required is %f.\n\n",Nr); \ No newline at end of file -- cgit