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/CH9/EX9.13/9_13.sce | 54 +++++++++++++++++++++++++++++++++++++++++++++++++
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diff --git a/905/CH9/EX9.13/9_13.sce b/905/CH9/EX9.13/9_13.sce
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+clear;
+clc;
+
+// Illustration 9.13
+// Page: 548
+
+printf('Illustration 9.13 -  Page: 548\n\n');
+
+// solution
+//*****Data*****//
+// w-water a-proteins
+T = 293; // [K]
+d = 2; // [diameter of tube, cm]
+dw = 1; // [g/cubic cm]
+uw = 0.01; // [cP]
+Da = 4*10^-7; // [Diffusivity of proteins, square cm/s]
+vo = 1.5*100; // [m/s]
+Qm = 250*10^-3/3600*100; // [water permeance, cm/s.atm]
+cR = 40; // [g/L]
+
+printf('Illustration 9.13(a) -  Page: 549\n\n');
+// Solution(a)
+
+v = 25*10^-3/3600*100; // [cm/s]
+
+Re = d*vo*dw/uw; // [Renoylds number]
+Sc = uw/(dw*Da); // [Schmidt number]
+Sh = 0.0048*Re^0.913*Sc^0.346; // [Sherwood number]
+ks = Sh*Da/d;
+// From equation 9.87
+cS = cR*exp(v/ks); // [g/L]
+
+// From figure 9.12
+pi1 = 2; // [osmotic pressure, atm]
+// For 100% rejection deltapi = pi1 because pi2 = 0
+// Therefore
+deltapi = pi1; // [atm]
+// From equation 9.83
+deltaP = deltapi+(v/Qm);
+printf("The required pressure differential to produce a water transmembrane volume flux of 25 L/square m.h when the membrane is clean is %f atm.\n\n",deltaP);
+
+
+printf('Illustration 9.13(b) -  Page: 549\n\n');
+// Solution(b)
+
+// Membrane permeance is reduced fivefold by fouling
+Qm = Qm/5; // [cm/s.atm]
+// Here deltaP remains same
+// Equations 9.83 and 9.87, and the osmotic pressure data of Figure 9.12 must be solved simultaneously by trial and error to calculate new values for these three variables.
+// The results are
+cS2 = 213; // [g/L]
+deltapi2 = 1.63; // [atm]
+v2 = 6.53*10^-4; // [cm/s]
+printf("The water flux if the applied pressure differential remains the same as calculated in part (a) is %f L/square m.hr.",v2*1000*10^-2*3600);
-- 
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