initial commit / add all books

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diff --git a/530/CH7/EX7.1/example_7_1.sce b/530/CH7/EX7.1/example_7_1.sce
new file mode 100755
index 000000000..9deb9d28d
--- /dev/null
+++ b/530/CH7/EX7.1/example_7_1.sce
@@ -0,0 +1,20 @@
+clear;

+clc;

+

+// A Textbook on HEAT TRANSFER by S P SUKHATME

+// Chapter 7

+// Heat Exchangers

+

+

+// Example 7.1

+// Page 285

+printf("Example 7.1, Page 285 \n \n");

+

+h = 2000 ; // [W/m^2 K]

+// From Table 7.1

+U_f = 0.0001 ; // fouling factor, m^2K/W

+h_f = 1/[1/h+U_f];

+printf("Heat transfer coefficient including the effect of foulung = %f W/m^2 K \n",h_f);

+

+p = (h-h_f)/h*100;

+printf("Percentage reduction = %f  \n",p);
\ No newline at end of file
diff --git a/530/CH7/EX7.2/example_7_2.sce b/530/CH7/EX7.2/example_7_2.sce
new file mode 100755
index 000000000..1bc9bc1b1
--- /dev/null
+++ b/530/CH7/EX7.2/example_7_2.sce
@@ -0,0 +1,48 @@
+clear;

+clc;

+

+// A Textbook on HEAT TRANSFER by S P SUKHATME

+// Chapter 7

+// Heat Exchangers

+

+

+// Example 7.2

+// Page 294

+printf("Example 7.2, Page 294 \n \n");

+

+m = 1000 ; // [kg/h]

+Thi = 50 ; // [C]

+The = 40 ; // [C]

+Tci = 35 ; // [C]

+Tce = 40 ; // [C]

+U = 1000 ; // OHTC, W/m^2 K

+

+// Using Eqn 7.5.25

+q = m/3600*4174*(Thi-The) ; // [W]

+

+delta_T = ((Thi-Tce)-(The-Tci))/log((Thi-Tce)/(The-Tci)); // [C]

+printf("delta T = %f \n\n",delta_T);

+

+// T1 = Th and T2 = Tc

+R = (Thi-The)/(Tce-Tci) ;

+S = (Tce-Tci)/(Thi-Tci) ;

+// From fig 7.15,

+F =0.91 ;

+

+printf("Taking T1 = Th and T2 = Tc \n")

+printf("R = %f, S = %f \n",R,S);

+printf("Hence, F = %f \n \n",F);

+

+// Alternatively, taking T1 = Tc and T2 = Th

+R = (Tci-Tce)/(The-Thi);

+S = (The-Thi)/(Tci-Thi);

+

+// Again from fig 7.15,

+F =0.91 ;

+

+printf("Taking T1 = Tc and T2 = Th \n")

+printf("R = %f, S = %f \n",R,S);

+printf("Hence, F = %f \n",F);

+

+A = q/(U*F*delta_T);

+printf("\nArea = %f m^2",A);

diff --git a/530/CH7/EX7.3/example_7_3.sce b/530/CH7/EX7.3/example_7_3.sce
new file mode 100755
index 000000000..617dd51de
--- /dev/null
+++ b/530/CH7/EX7.3/example_7_3.sce
@@ -0,0 +1,29 @@
+clear;

+clc;

+

+// A Textbook on HEAT TRANSFER by S P SUKHATME

+// Chapter 7

+// Heat Exchangers

+

+

+// Example 7.3

+// Page 295

+printf("Example 7.3, Page 295 \n \n");

+

+// Because of change of phase , Thi = The

+Thi = 100 ; // [C], Saturated steam

+The = 100 ; // [C], Condensed steam

+Tci = 30 ; // [C], Cooling water inlet

+Tce = 70 ; // [C], cooling water outlet

+

+R = (Thi-The)/(Tce-Tci) ;

+S = (Tce-Tci)/(Thi-Tci) ;

+

+// From fig 7.16

+F = 1;

+

+// For counter flow arrangement

+Tm_counter = ((Thi-Tce)-(The-Tci))/log((Thi-Tce)/(The-Tci)); // [C]

+// Therefore

+Tm = F*Tm_counter ;

+printf("Mean Temperaature Difference = %f C",Tm)

diff --git a/530/CH7/EX7.4.a/example_7_4a.sce b/530/CH7/EX7.4.a/example_7_4a.sce
new file mode 100755
index 000000000..c4880ad6c
--- /dev/null
+++ b/530/CH7/EX7.4.a/example_7_4a.sce
@@ -0,0 +1,37 @@
+clear;

+clc;

+

+// A Textbook on HEAT TRANSFER by S P SUKHATME

+// Chapter 7

+// Heat Exchangers

+

+

+// Example 7.4(a)

+// Page 302

+printf("Example 7.4(a), Page 302 \n \n");

+

+// (a)

+printf("(a) \n");

+// Using Mean Temperature Difference approach

+m_hot = 10 ; // [kg/min]

+m_cold = 25 ; // [kg/min]

+hh = 1600 ; // [W/m^2 K], Heat transfer coefficient on hot side

+hc = 1600 ; // [W/m^2 K], Heat transfer coefficient on cold side

+

+Thi = 70 ; // [C]

+Tci = 25 ; // [C]

+The = 50 ; // [C]

+

+// Heat Transfer Rate, q

+q = m_hot/60*4179*(Thi-The); // [W]

+

+// Heat gained by cold water = heat lost by the hot water

+Tce = 25 + q*1/(m_cold/60*4174); // [C]

+

+// Using equation 7.5.13

+Tm = ((Thi-Tci)-(The-Tce))/log((Thi-Tci)/(The-Tce)); // [C]

+printf("Mean Temperature Difference = %f C \n",Tm);

+

+U = 1/(1/hh + 1/hc); // [W/m^2 K]

+A = q/(U*Tm); // Area, [m^2]

+printf("Area of Heat Exchanger = %f m^2 \n",A);

diff --git a/530/CH7/EX7.4.b/example_7_4b.sce b/530/CH7/EX7.4.b/example_7_4b.sce
new file mode 100755
index 000000000..40873abec
--- /dev/null
+++ b/530/CH7/EX7.4.b/example_7_4b.sce
@@ -0,0 +1,54 @@
+clear;

+clc;

+

+// A Textbook on HEAT TRANSFER by S P SUKHATME

+// Chapter 7

+// Heat Exchangers

+

+

+// Example 7.4(b)

+// Page 302

+printf("Example 7.4(b), Page 302 \n \n");

+

+// Using Mean Temperature Difference approach

+m_hot = 10 ; // [kg/min]

+m_cold = 25 ; // [kg/min]

+hh = 1600 ; // [W/m^2 K], Heat transfer coefficient on hot side

+hc = 1600 ; // [W/m^2 K], Heat transfer coefficient on cold side

+

+Thi = 70 ; // [C]

+Tci = 25 ; // [C]

+The = 50 ; // [C]

+

+// Heat Transfer Rate, q

+q = m_hot/60*4179*(Thi-The); // [W]

+

+// Heat gained by cold water = heat lost by the hot water

+Tce = 25 + q*1/(m_cold/60*4174); // [C]

+

+// Using equation 7.5.13

+Tm = ((Thi-Tci)-(The-Tce))/log((Thi-Tci)/(The-Tce)); // [C]

+U = 1/(1/hh + 1/hc); // [W/m^2 K]

+A = q/(U*Tm); // Area, [m^2]

+

+m_hot = 20 ; // [kg/min]

+// Flow rate on hot side i.e. 'hh' is doubled

+hh = 1600*2^0.8 ; // [W/m^2 K]

+U = 1/(1/hh + 1/hc); // [W/m^2 K]

+m_hC_ph =  m_hot/60*4179 ; // [W/K]

+m_cC_pc = m_cold/60*4174 ; // [W/K]

+// Therefore

+C = m_hC_ph/m_cC_pc ;

+NTU = U*A/m_hC_ph ;

+printf("NTU = %f \n",NTU);

+

+// From equation 7.6.8

+e = [1 - exp(-(1+C)*NTU)]/(1+C) ;

+

+// Therefore (Thi - The)/(Thi - Tci) = e , we get

+The = Thi - e*(Thi - Tci); // [C]

+

+// Equating the heat lost by water to heat gained by cold water , we get

+Tce = Tci + [m_hC_ph*(Thi-The)]/m_cC_pc;

+printf("Exit temperature of cold and hot stream are %f C and %f C respectively.",Tce,The);

+

diff --git a/530/CH7/EX7.5/example_7_5.sce b/530/CH7/EX7.5/example_7_5.sce
new file mode 100755
index 000000000..a493c9186
--- /dev/null
+++ b/530/CH7/EX7.5/example_7_5.sce
@@ -0,0 +1,93 @@
+clear;

+clc;

+

+// A Textbook on HEAT TRANSFER by S P SUKHATME

+// Chapter 7

+// Heat Exchangers

+

+

+// Example 7.5

+// Page 304

+printf("Example 7.5, Page 304 \n \n");

+

+mc = 2000 ; // [kg/h]

+Tce = 40 ; // [C]

+Tci = 15 ; // [C]

+Thi = 80 ; // [C]

+U = 50 ; // OHTC, [W/m^2 K]

+A = 10 ; // Area, [m^2]

+

+// Using effective NTU method

+// Assuming m_c*C_pc = (m*C_p)s

+NTU = U*A/(mc*1005/3600);

+e = (Tce-Tci)/(Thi-Tci);

+// From fig 7.23, no value of C is found corresponding to the above values, hence assumption was wrong.

+// So, m_h*C_ph must be equal to (m*C_p)s, proceeding by trail and error method

+

+

+printf("m_h(kg/h      NTU      C        e        T_he(C)       T_he(C) (Heat Balance)");

+ 

+mh = rand(1:5);

+NTU = rand(1:5);

+The = rand(1:5);

+The2 = rand(1:5);

+

+mh(1) = 200

+NTU(1) = U*A/(mh(1)*1.161);

+//Corresponding Values of C and e from fig 7.23

+C = .416;

+e = .78;

+//From Equation 7.6.2 Page 297

+The(1) = Thi - e*(Thi-Tci)

+//From Heat Balance

+The2(1) = Thi -  mc*1005/3600*(Tce-Tci)/(mh(1)*1.161);

+printf("\n\n   %i     %.3f      %.3f      %.3f      %.2f       %.2f",mh(1),NTU(1),C,e,The(1),The2(1));

+

+mh(2) = 250

+NTU(2) = U*A/(mh(2)*1.161);

+//Corresponding Values of C and e from fig 7.23

+C = .520;

+e = .69;

+//From Equation 7.6.2 Page 297

+The(2) = Thi - e*(Thi-Tci)

+//From Heat Balance

+The2(2) = Thi -  mc*1005/3600*(Tce-Tci)/(mh(2)*1.161);

+printf("\n\n   %i     %.3f      %.3f      %.3f      %.2f       %.2f",mh(2),NTU(2),C,e,The(2),The2(2));

+

+mh(3) = 300

+NTU(3) = U*A/(mh(3)*1.161);

+//Corresponding Values of C and e from fig 7.23

+C = .624;

+e = .625;

+//From Equation 7.6.2 Page 297

+The(3) = Thi - e*(Thi-Tci)

+//From Heat Balance

+The2(3) = Thi -  mc*1005/3600*(Tce-Tci)/(mh(3)*1.161);

+printf("\n\n   %i     %.3f      %.3f      %.3f      %.2f       %.2f",mh(3),NTU(3),C,e,The(3),The2(3));

+

+mh(4) = 350

+NTU(4) = U*A/(mh(4)*1.161);

+//Corresponding Values of C and e from fig 7.23

+C = .728;

+e = .57;

+//From Equation 7.6.2 Page 297

+The(4) = Thi - e*(Thi-Tci)

+//From Heat Balance

+The2(4) = Thi -  mc*1005/3600*(Tce-Tci)/(mh(4)*1.161);

+printf("\n\n   %i     %.3f      %.3f      %.3f      %.2f       %.2f",mh(4),NTU(4),C,e,The(4),The2(4));

+

+mh(5) = 400

+NTU(5) = U*A/(mh(5)*1.161);

+//Corresponding Values of C and e from fig 7.23

+C = .832;

+e = .51;

+//From Equation 7.6.2 Page 297

+The(5) = Thi - e*(Thi-Tci)

+//From Heat Balance

+The2(5) = Thi -  mc*1005/3600*(Tce-Tci)/(mh(5)*1.161);

+printf("\n\n   %i     %.3f      %.3f      %.3f      %.2f       %.2f",mh(5),NTU(5),C,e,The(5),The2(5));

+

+clf();

+plot(mh,The,mh,The2,[295 295 200],[0 39.2 39.2])

+xtitle('The vs mh','mh (kg/hr)','The (C)');

+printf("\n\n From the plot, value of mh = 295 kg/hr and correspondingly The = 39.2 C")
\ No newline at end of file