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

---
 1394/CH10/EX10.4.1/Ex10_4_1.sce | 50 +++++++++++++++++++++++++++++++++++++++++
 1 file changed, 50 insertions(+)
 create mode 100755 1394/CH10/EX10.4.1/Ex10_4_1.sce

(limited to '1394/CH10/EX10.4.1')

diff --git a/1394/CH10/EX10.4.1/Ex10_4_1.sce b/1394/CH10/EX10.4.1/Ex10_4_1.sce
new file mode 100755
index 000000000..df473897a
--- /dev/null
+++ b/1394/CH10/EX10.4.1/Ex10_4_1.sce
@@ -0,0 +1,50 @@
+
+
+clc
+//initialization of variables
+y1in = 0.37 // mole fraction of Ammonia in gas mixture entering
+y2in =0.16 // mole fraction of nitrogen in gas mixture entering
+y3in = 0.47 // mole fraction of hydrogen in gas mixture entering
+x1out = 0.23 // mole fraction of Ammonia in liquid coming out
+y1out = 0.01 // mole fraction of ammonia in gas coming out
+G0 = 1.20 // Gas glow entering in m^3/sec
+Mu = 1.787*0.01*0.3048/2.23 // liquid viscousity in american units
+dl = 62.4 // Density of  liquid in lb/ft^3
+KG = 0.032 // Overall m.t.c in gas phase in  gas side m/sec
+a = 105 // surface area in m^2/m^3
+gc = 32.2 // acceleration due to gravity in ft/sec^2
+dg = 0.0326 // Density of gas in lb/ft^3
+//Molecular weights of Ammonia , N2 , H2
+M1 = 17
+M2 = 28
+M3 = 2
+Nu = 1 // Viscousity 
+//Calculations
+AG0 = (y2in+y3in)*G0/22.4 // Total flow of non absorbed gases in kgmol/sec
+ANH3 = y1in*G0/22.4- (y1out*AG0)/(1-y1out) // Ammonia absorbed kgmol/sec
+AL0 = ((1-x1out)/x1out)*ANH3 // the desired water flow in kgmol/sec
+avg1 = 11.7 // Average mol wt of gas
+avg2 = 17.8 // avg mol wt of liquid
+TFG = avg1*AG0/(y2in+y3in)//Total flow of gas in kg/sec
+TFL = avg2*AL0/(1-x1out)//total flow of liquid in kg/sec
+F = 45 // Packing factor
+GFF =  1.3*((dl-dg)^0.5)/((F^0.5)*(Nu^0.05))// Flux we require in lb/ft^2-sec
+GFF1 = GFF*0.45/(0.3^2) // in kg/m^2-sec (answer wrong in textbook)
+Area = TFG/GFF1 // Area of the cross section of tower
+dia = (sqrt(4*Area/%pi)) // diameter in metres
+HTU = (22.4*AG0/%pi*dia^2)/(KG*a*4)
+NTU = 5555
+l = HTU*NTU // Length of the tower
+//Results
+printf("The flow of pure water into the top of the tower %.4f kgmol/sec",AL0)
+printf("\n The diameter of the tower is %.1f m",dia)
+printf("\n The length of the tower is %.f m",l)
+
+
+
+
+
+
+
+
+
-- 
cgit