initial commit / add all books

author: priyanka 2015-06-24 15:03:17 +0530
committer: priyanka 2015-06-24 15:03:17 +0530
commit: b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch)
tree: ab291cffc65280e58ac82470ba63fbcca7805165 /530/CH9
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12 files changed, 374 insertions, 0 deletions
diff --git a/530/CH9/EX9.1/example_9_1.sce b/530/CH9/EX9.1/example_9_1.sce
new file mode 100755
index 000000000..bb6a96c8a
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+++ b/530/CH9/EX9.1/example_9_1.sce
@@ -0,0 +1,22 @@
+clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 9
+// Mass Transfer
+
+
+// Example 9.1
+// Page 349
+printf("Example 9.1, Page 349 \n \n");
+
+w_a = 0.76 ;
+w_b = 0.24 ;
+m_a = 28 ; // [kg/kg mole]
+m_b = 32 ; // [kg/kg mole]
+
+x_a = (w_a/m_a)/(w_a/m_a+w_b/m_b);
+x_b = (w_b/m_b)/(w_a/m_a+w_b/m_b);
+printf("The molar fractions are given by \n");
+printf("x_a = %f\n",x_a);
+printf("x_b = %f",x_b);
diff --git a/530/CH9/EX9.2/example_9_2.sce b/530/CH9/EX9.2/example_9_2.sce
new file mode 100755
index 000000000..995b860e1
--- /dev/null
+++ b/530/CH9/EX9.2/example_9_2.sce
@@ -0,0 +1,17 @@
+clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 9
+// Mass Transfer
+
+
+// Example 9.2
+// Page 350
+printf("Example 9.2, Page 350 \n \n");
+
+// From Table 9.1 at 1 atm and 25 C
+Dab = 0.62*10^-5 ; // [m^2/s]
+// Therefore at 2 atm and 50 C
+Dab2 = Dab*(1/2)*(323/298)^1.5 ;
+printf("Dab at 2 atm & 50 C = %e m^2/s",Dab2);
+\ No newline at end of file
diff --git a/530/CH9/EX9.3.a/example_9_3a.sce b/530/CH9/EX9.3.a/example_9_3a.sce
new file mode 100755
index 000000000..aa453a1a5
--- /dev/null
+++ b/530/CH9/EX9.3.a/example_9_3a.sce
@@ -0,0 +1,24 @@
+clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 9
+// Mass Transfer
+
+
+// Example 9.3(a)
+// Page 352
+printf("Example 9.3(a), Page 352 \n \n");
+
+t = 0.04 ; // [m]
+A = 2 ; // [m^2]
+rho1 = 0.10 ;
+rho2 = 0.01 ;
+D_400 = 1.6*10^-11 ; // at 400K [m^2/s]
+
+// Mass Diffusion in solid solution, assuming Ficks law is valid & steady state and one dimensional diffusion
+
+// Subtituting the values in eqn 9.3.3 , At 400 K
+
+m_400 = A*D_400*(rho1-rho2)/t; // [kg/s]
+printf("Rate of diffusion of Hydrogen at 400 K = %e kg/s \n",m_400);
+\ No newline at end of file
diff --git a/530/CH9/EX9.3.b/example_9_3b.sce b/530/CH9/EX9.3.b/example_9_3b.sce
new file mode 100755
index 000000000..551d0a404
--- /dev/null
+++ b/530/CH9/EX9.3.b/example_9_3b.sce
@@ -0,0 +1,26 @@
+clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 9
+// Mass Transfer
+
+
+// Example 9.3(b)
+// Page 352
+printf("Example 9.3(b), Page 352 \n \n");
+
+t = 0.04 ; // [m]
+A = 2 ; // [m^2]
+rho1 = 0.10 ;
+rho2 = 0.01 ;
+D_1200 = 3.5*10^-8 ; // at 1200k [m^2/s]
+
+// Mass Diffusion in solid solution, assuming Ficks law is valid & steady state and one dimensional diffusion
+
+// At 1200 K
+// From eqn 9.3.3
+
+m_1200 = A*D_1200*(rho1-rho2)/t ;
+printf("(b) Rate of diffusion of Hydrogen at 1200 K = %e kg/s \n",m_1200);
+
diff --git a/530/CH9/EX9.4.a/example_9_4a.sce b/530/CH9/EX9.4.a/example_9_4a.sce
new file mode 100755
index 000000000..5f5a3ff5d
--- /dev/null
+++ b/530/CH9/EX9.4.a/example_9_4a.sce
@@ -0,0 +1,28 @@
+clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 9
+// Mass Transfer
+
+
+// Example 9.4(a)
+// Page 356
+printf("Example 9.4(a), Page 356 \n \n");
+
+L = 1 ; // [m]
+D = 0.005 ; // [m]
+Pa1 = 1 ; // [atm]
+Pa2 = 0 ; 
+R = 8314 ;
+T = 298 ; // [K]
+
+// Assuming Equimolal counter diffusion
+// From Table 9.1
+Dab = 2.80*10^-5 ; // [m^2/s]
+// Substituing in eqn 9.4.12
+Na = -[Dab/(R*T)*(Pa2-Pa1)*(1.014*10^5)/L]*(%pi*(D/2)^2);
+R_NH3 = Na*17 ; // [kg/s]
+
+printf("Na = -Nb = %e (kg mole)/m^2 s\n",Na);
+printf("Rate at which ammonia is lost through the tube = %e kg/s \n",R_NH3);
+\ No newline at end of file
diff --git a/530/CH9/EX9.4.b/example_9_4b.sce b/530/CH9/EX9.4.b/example_9_4b.sce
new file mode 100755
index 000000000..05be33bbd
--- /dev/null
+++ b/530/CH9/EX9.4.b/example_9_4b.sce
@@ -0,0 +1,31 @@
+clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 9
+// Mass Transfer
+
+
+// Example 9.4(b)
+// Page 356
+printf("Example 9.4(b), Page 356 \n \n");
+
+L = 1 ; // [m]
+D = 0.005 ; // [m]
+Pa1 = 1 ; // [atm]
+Pa2 = 0 ; 
+R = 8314 ;
+T = 298 ; // [K]
+
+// Since the tank is large and the pressure and temperature at the two ends of the same tube are same, we are assuming Equimolal counter diffusion
+// From Table 9.1
+Dab = 2.80*10^-5 ; // [m^2/s]
+// Substituing in eqn 9.4.12
+Na = -[Dab/(R*T)*(Pa2-Pa1)*(1.014*10^5)/L]*(%pi*(D/2)^2);
+
+// Since equimolal counter diffusion is taking place
+Nb = - Na ;
+// therefore rate at which air enters the tank
+R_air = abs(Nb)*29 ; // [kg/s]
+
+printf("Rate at which air enters the tank = %e kg/s",R_air);
+\ No newline at end of file
diff --git a/530/CH9/EX9.5/example_9_5.sce b/530/CH9/EX9.5/example_9_5.sce
new file mode 100755
index 000000000..03b7756c9
--- /dev/null
+++ b/530/CH9/EX9.5/example_9_5.sce
@@ -0,0 +1,32 @@
+clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 9
+// Mass Transfer
+
+
+// Example 9.5
+// Page 359
+printf("Example 9.5, Page 359 \n \n");
+
+// Evaporation of water, one dimensional
+T_w = 20+273 ; // [K]
+D = 0.04 ; // [m]
+h = 0.20 ; // [m]
+h_w = 0.03 ; // [m]
+
+P = 1.014*10^5; // [Pa]
+R = 8314 ; // [J/kg mole K]
+P_sat = 0.02339 ; // [bar]
+x_a1 = P_sat/1.014 ; // mole fraction at liq-vap interface
+x_a2 = 0 ; // mole fraction at open top
+c = P/(R*T_w);
+// From Table 9.2
+Dab = 2.422*10^-5 ; // [m^2/s]
+
+// Substituting above values in eqn 9.4.18
+flux = 0.041626*Dab/0.17*log((1-0)/(1-x_a1)); // [kg mole/m^2 s]
+rate = flux*18*(%pi/4)*(D^2);
+
+printf("Rate of evaporation of water = %e kg/s",rate);
+\ No newline at end of file
diff --git a/530/CH9/EX9.6/example_9_6.sce b/530/CH9/EX9.6/example_9_6.sce
new file mode 100755
index 000000000..849b2624c
--- /dev/null
+++ b/530/CH9/EX9.6/example_9_6.sce
@@ -0,0 +1,33 @@
+clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 9
+// Mass Transfer
+
+
+// Example 9.6
+// Page 364
+printf("Example 9.6, Page 364 \n \n");
+
+l = 1; // length, [m]
+w = 0.25; // width, [m]
+T = 293 ; // Temperature, [K]
+rho_infinity = 0; // [kg/m^3]
+R = 8314; // [J/ kg K]
+
+// From Table A.2
+v = 15.06*10^-6; // [m^2/s]
+// From Table 9.2
+Dab = 2.4224*10^-5; // [m^2/s]
+Re = 2.5/v;
+Sc = v/Dab;
+// Since Re > 3*10^5, we may assume laminar boundary layer
+Sh = 0.664*Sc^(1/3)*Re^(1/2); // Sherwood number
+h = Sh*Dab;
+
+p_aw = 2339; // Saturation pressure of water at 20 degree C. [N/m^2]
+rho_aw = p_aw/(R/18*T); // [kg/m^3]
+rho_a_inf = 0 ; // since air in the free stream is dry
+m_h = h*(2*l*w)*(rho_aw-rho_infinity);
+printf("Rate of evaporation from plate = %e kg/s",m_h);
+\ No newline at end of file
diff --git a/530/CH9/EX9.7.a/example_9_7a.sce b/530/CH9/EX9.7.a/example_9_7a.sce
new file mode 100755
index 000000000..683fea477
--- /dev/null
+++ b/530/CH9/EX9.7.a/example_9_7a.sce
@@ -0,0 +1,37 @@
+clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 9
+// Mass Transfer
+
+
+// Example 9.7(a)
+// Page 366
+printf("Example 9.7(a), Page 366 \n \n");
+
+D = 0.04 ; // [m]
+V = 1.9 ; // [m/s]
+
+// (a) Colburn anology and Gnielinski equation
+// 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);
+// Substituting this value into Gnielinski equation 5.3.2
+Nu = [(f/2)*(Re-1000)*Pr]/[1+12.7*((f/2)^(1/2))*((Pr^(2/3))-1)];
+h = Nu*k/D;
+h_m = h/r; // [m/s]
+
+printf("h_m using Colburn anology and Gnielinski equation = %f \n",h_m);
+\ No newline at end of file
diff --git a/530/CH9/EX9.7.b/example_9_7b.sce b/530/CH9/EX9.7.b/example_9_7b.sce
new file mode 100755
index 000000000..4bbc20885
--- /dev/null
+++ b/530/CH9/EX9.7.b/example_9_7b.sce
@@ -0,0 +1,36 @@
+clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 9
+// Mass Transfer
+
+
+// Example 9.7(b)
+// Page 366
+printf("Example 9.7(b), Page 366 \n \n");
+
+D = 0.04 ; // [m]
+V = 1.9 ; // [m/s]
+
+// (b) mess transfer correlation equivalent to the Gleilinski equation
+
+// 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;
+
+// From Blasius equation 4.6.4a
+f = 0.079*Re^(-0.25);
+
+// Substituting in eqn 9.6.7
+Sh_D = [(f/2)*(Re-1000)*Sc]/[1+12.7*((f/2))*((Sc^(2/3))-1)];
+h_m1 = Sh_D*Dab/D;
+
+printf("(b) h_m = %f \n",h_m1);
+\ No newline at end of file
diff --git a/530/CH9/EX9.7.c/example_9_7c.sce b/530/CH9/EX9.7.c/example_9_7c.sce
new file mode 100755
index 000000000..ee0e75c4f
--- /dev/null
+++ b/530/CH9/EX9.7.c/example_9_7c.sce
@@ -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
diff --git a/530/CH9/EX9.8/example_9_8.sce b/530/CH9/EX9.8/example_9_8.sce
new file mode 100755
index 000000000..72c96259a
--- /dev/null
+++ b/530/CH9/EX9.8/example_9_8.sce
@@ -0,0 +1,45 @@
+clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 9
+// Mass Transfer
+
+
+// Example 9.8
+// Page 369
+printf("Example 9.8, Page 369 \n \n");
+
+V = 0.5 ; // [m/s]
+T_h = 30 ; // [C]
+T_c = 26 ; // [C]
+Tm = (T_h+T_c)/2;
+// From table A.2
+rho = 1.173 ; // [kg/m^3]
+Cp = 1005 ; // [J/kg K]
+k = 0.02654 ; // [W/m K]
+
+alpha = k/(rho*Cp); // [m^2/s]
+
+// From Table 9.2 at 301 K
+Dab = 2.5584*10^-5 ; // [m^2/s]
+lambda = 2439.2*10^3 ; // [J/kg]
+
+// Substituting in equation 9.7.5
+// let difference = rho_aw-rho_a infinity
+difference = rho*Cp*((alpha/Dab)^(2/3))*(T_h-T_c)/lambda;
+
+// From steam table
+Psat = 3363;
+rho_aw = Psat/(8314/18*299);
+rho_inf = rho_aw - difference;
+x = rho_inf/rho; // mole fraction of water vapour in air stream
+
+PP = rho_inf*8314/18*303; // Partial pressure of water vapour in air stream
+// From steam table partial pressure of water vapour at 30 C
+PP_30 = 4246 ; // [N/m^2]
+
+rel_H = PP/PP_30;
+percent = rel_H*100;
+
+printf("Relative humidity = %f i.e. %f percent ",rel_H,percent);
+\ No newline at end of file
author	priyanka	2015-06-24 15:03:17 +0530
committer	priyanka	2015-06-24 15:03:17 +0530
commit	b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch)
tree	ab291cffc65280e58ac82470ba63fbcca7805165 /530/CH9
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