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

---
 530/CH9/EX9.7.c/example_9_7c.sce | 43 ++++++++++++++++++++++++++++++++++++++++
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 create mode 100755 530/CH9/EX9.7.c/example_9_7c.sce

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diff --git a/530/CH9/EX9.7.c/example_9_7c.sce b/530/CH9/EX9.7.c/example_9_7c.sce
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@@ -0,0 +1,43 @@
+clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 9
+// Mass Transfer
+
+
+// Example 9.7(c)
+// Page 366
+printf("Example 9.7(c), Page 366 \n \n");
+
+D = 0.04 ; // [m]
+V = 1.9 ; // [m/s]
+
+// (c) To show that mass flux of water is very small compared to the mass flux of air flowing in the pipe
+// Properties of air at 27 degree C
+v = 15.718*10^-6 ; // [m^2/s]
+rho = 1.177 ; // [kg/m^3]
+Pr = 0.7015 ;
+Cp = 1005 ; // [J/kg K]
+k = 0.02646 ; // [W/m K]
+// From Table 9.2
+Dab = 2.54 * 10^-5 ; // [m^2/s]
+Sc = v/Dab ;
+Re = V*D/v;
+// The flow is turbulent and eqn 9.6.5 may be applied
+// let r = h/h_m 
+r = rho*Cp*((Sc/Pr)^(2/3));
+// From Blasius equation 4.6.4a
+f = 0.079*Re^(-0.25);
+
+// From steam table
+rho_aw = 1/38.77 ; // [kg/m^3]
+// let X = (m_a/A)_max
+X = f*rho_aw; // [kg/m^2 s]
+
+// let Y = mass flux of air in pipe = (m/A)
+Y = rho*V ; // [kg/m^2 s]
+ratio = X/Y ;
+percent = ratio*100;
+
+printf("(c) (m_a/A)_max/(m_a/A) = %f percent Thus, mass flux of water is very small compared to the mass flux of air flowing in the pipe. ",percent );
\ No newline at end of file
-- 
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