diff options
Diffstat (limited to '530')
86 files changed, 3286 insertions, 0 deletions
diff --git a/530/CH1/EX1.1/example_1_1.sce b/530/CH1/EX1.1/example_1_1.sce new file mode 100755 index 000000000..1696faa88 --- /dev/null +++ b/530/CH1/EX1.1/example_1_1.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 1
+// Introduction
+
+
+// Example 1.1
+// Page 5
+// Given that the viscosity of water at 100 degree Celsius is 28.8 * 10^-6 kgf s/m^2 in MKS system , express this value in SI system.
+printf("Example 1.1, Page 5 \n \n")
+
+// Solution:
+
+//at 100 degree Celsius
+v1=28.8 * 10^-6; // [kgf s/m^2]
+v2=28.8 * 10^-6 * 9.8; // [N s/m^2]
+printf("Viscosity of water at 100 degree celsius in the SI system is %e N.s/m^-2 (or kg/m s)",v2)
\ No newline at end of file diff --git a/530/CH1/EX1.2/example_1_2.sce b/530/CH1/EX1.2/example_1_2.sce new file mode 100755 index 000000000..4818bcfd0 --- /dev/null +++ b/530/CH1/EX1.2/example_1_2.sce @@ -0,0 +1,38 @@ +clear;
+clc;
+// Textbook of Heat Transfer(4th Edition)) , S P Sukhatme
+// Chapter 1 - Introduction
+
+//Example 1.2
+// Page 14
+printf("Example 1.2, Page 14 \n \n")
+//Solution:
+i=950; // radiation flux [W/m^2]
+A=1.5; // area [m^2]
+T_i=61; // inlet temperature
+T_o=69; // outlet temperature
+mdot=1.5; // [kg/min] , mass flow rate
+Mdot=1.5/60; // [kg/sec]
+Q_conductn=50; //[W]
+t=0.95; // transmissivity
+a=0.97;// absoptivity
+// from appendix table A.1 at 65 degree C
+C_p= 4183 ; // [J/kg K]
+// Using Equation 1.4.15 , assuming that the flow through the tubes is steady and one dimensional.
+// in this case (dW/dt)_shaft = 0
+// assuming (dW/dt)_shear is negligible
+// eqn(1.4.15) reduces to
+q=Mdot*C_p*(T_o-T_i);
+
+// let 'n' be thermal efficiency
+n=q/(i*A);
+n_percent=n*100;
+
+
+// equation 1.4.13 yields dQ/dt = 0
+Q_re_radiated=(i*A*t*a)-Q_conductn-q; // [W]
+
+
+printf("Useful heat gain rate is %f W \n",q);
+printf("Thermal efficiency is %e i.e. %f per cent \n",n,n_percent);
+printf("The rate at which energy is lost by re-radiation and convection is %f W",Q_re_radiated)
\ No newline at end of file diff --git a/530/CH1/EX1.3/example_1_3.sce b/530/CH1/EX1.3/example_1_3.sce new file mode 100755 index 000000000..d8d73c9ef --- /dev/null +++ b/530/CH1/EX1.3/example_1_3.sce @@ -0,0 +1,42 @@ +clear;
+clc;
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 1
+// Introduction
+
+
+//Example 1.3
+// Page 16
+printf("Example 1.3, Page 16\n\n");
+
+//Solution:
+// Given
+v_i=10; // [m/s]
+q=1000; // [W]
+d_i=0.04; // [m]
+d_o=0.06; // [m]
+
+// From appendix table A.2
+rho1=0.946; // [kg/m^3] at 100 degree C
+C_p=1009; // [J/kg K]
+
+mdot=rho1*(%pi/4)*(d_i^2)*v_i; // [kg/s]
+
+
+// In this case (dW/dt)_shaft=0 and (z_o - z_i)=0
+// From eqn 1.4.15 , q=mdot*(h_o-h_i)
+// Let dh = (h_o-h_i)
+dh=q/mdot; // [J/kg]
+// Let T_o be the outlet temperature
+T_o=dh/C_p+100;
+
+rho2=0.773; // [kg/m^3] at T_o = 183.4 degree C
+// From eqn 1.4.6
+v_o=mdot/(rho2*(%pi/4)*(d_o)^2); // [m/s]
+
+dKE_kg=(v_o^2-v_i^2)/2; // [J/kg]
+
+
+printf("Exit Temperature is %f degree C \n",T_o);
+printf("Exit velocity is %f m/s \n",v_o);
+printf("Change in Kinetic Energy per kg = %f J/kg",dKE_kg);
\ No newline at end of file diff --git a/530/CH2/EX2.1/example_2_1.sce b/530/CH2/EX2.1/example_2_1.sce new file mode 100755 index 000000000..4e7337d58 --- /dev/null +++ b/530/CH2/EX2.1/example_2_1.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.1
+// Page 27
+printf("Example 2.1, Page 27 \n\n")
+
+d_i=0.02; // [m] inner radius
+d_o=0.04; // [m] outer radius
+r_i=d_i/2; // [m] inner radius
+r_o=d_o/2; // [m] outer radius
+k=0.58; // [w/m K] thermal conductivity of tube material
+t_i=70; //[degree C]
+t_o=100; // [degree C]
+l=1; // [m] per unit length
+// using equation 2.1.5
+q=l*2*(%pi)*k*(t_i-t_o)/log(r_o/r_i);
+printf("Heat flow per unit length is %f W/m",q);
diff --git a/530/CH2/EX2.10.i/example_2_10i.sce b/530/CH2/EX2.10.i/example_2_10i.sce new file mode 100755 index 000000000..f5c667080 --- /dev/null +++ b/530/CH2/EX2.10.i/example_2_10i.sce @@ -0,0 +1,48 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.10(i)
+// Page 58
+printf("Example 2.10(i), Page 58 \n\n")
+
+// Centre of the slab
+// Given data
+b = 0.005 ; // [m]
+t = 5*60; // time, [sec]
+Th = 200 ; // [C]
+Tw = 20 ; // [C]
+h = 150 ; // [W/m^2 K]
+rho = 2200 ; //[kg/m^3]
+Cp = 1050 ; // [J/kg K]
+k = 0.4 ; // [W/m K]
+// Using charts in fig 2.18 and 2.19 and eqn 2.7.19 and 2.7.20
+
+theta = Th - Tw;
+Biot_no = h*b/k;
+a = k/(rho*Cp); // alpha
+Fourier_no = a*t/b^2;
+
+// From fig 2.18, ratio = theta_x_b0/theta_o
+ratio_b0 = 0.12;
+// From fig 2.18, ratio = theta_x_b1/theta_o
+ratio_b1 = 0.48;
+
+// Therefore
+theta_x_b0 = theta*ratio_b0; // [C]
+T_x_b0 = theta_x_b0 + Tw ; // [C]
+theta_x_b1 = theta*ratio_b1; // [C]
+T_x_b1 = theta_x_b1 + Tw ; // [C]
+
+// From Table 2.2 for Bi = 1.875
+lambda_1_b = 1.0498;
+x = 2*sin(lambda_1_b)/[lambda_1_b+(sin(lambda_1_b))*(cos(lambda_1_b))];
+
+// From eqn 2.7.20
+theta_x_b0 = theta*x*(exp((-lambda_1_b^2)*Fourier_no));
+T_x_b0 = theta_x_b0 + Tw;
+printf("Temperature at b=0 is %f degree C\n",T_x_b0);
+
diff --git a/530/CH2/EX2.10.ii/example_2_10ii.sce b/530/CH2/EX2.10.ii/example_2_10ii.sce new file mode 100755 index 000000000..4e0f6ab39 --- /dev/null +++ b/530/CH2/EX2.10.ii/example_2_10ii.sce @@ -0,0 +1,46 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.10(ii)
+// Page 58
+printf("Example 2.10(ii), Page 58 \n\n")
+
+// (ii) Surface of the slab
+
+b = 0.005 ; // [m]
+t = 5*60; // time, [sec]
+Th = 200 ; // [C]
+Tw = 20 ; // [C]
+h = 150 ; // [W/m^2 K]
+rho = 2200 ; //[kg/m^3]
+Cp = 1050 ; // [J/kg K]
+k = 0.4 ; // [W/m K]
+// Using charts in fig 2.18 and 2.19 and eqn 2.7.19 and 2.7.20
+theta = Th - Tw;
+Biot_no = h*b/k;
+a = k/(rho*Cp); // alpha
+Fourier_no = a*t/b^2;
+
+// From fig 2.18, ratio = theta_x_b0/theta_o
+ratio_b0 = 0.12;
+// From fig 2.18, ratio = theta_x_b1/theta_o
+ratio_b1 = 0.48;
+
+// Therefore
+theta_x_b0 = theta*ratio_b0; // [C]
+T_x_b0 = theta_x_b0 + Tw ; // [C]
+theta_x_b1 = theta*ratio_b1; // [C]
+T_x_b1 = theta_x_b1 + Tw ; // [C]
+
+// From Table 2.2 for Bi = 1.875
+lambda_1_b = 1.0498;
+x = 2*sin(lambda_1_b)/[lambda_1_b+(sin(lambda_1_b))*(cos(lambda_1_b))];
+
+// From 2.7.19
+theta_x_b1 = theta_x_b0*(cos(lambda_1_b*1));
+T_x_b1 = theta_x_b1 + Tw;
+printf("Temperature at b=1 is %f degree C\n",T_x_b1);
diff --git a/530/CH2/EX2.11.a/example_2_11a.sce b/530/CH2/EX2.11.a/example_2_11a.sce new file mode 100755 index 000000000..7e5e45c7b --- /dev/null +++ b/530/CH2/EX2.11.a/example_2_11a.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.11(a)
+// Page 65
+printf("Example 2.11(a), Page 65 \n\n")
+
+D = 0.05 ; // [m]
+To = 450 ; // [degree C]
+Tf = 90 ; // [degree C]
+T = 150 ; // [degree c]
+h = 115 ; // [W/m^2 K]
+rho = 8000 ; // [kg/m^3]
+Cp = 0.42*1000 ; // [J/kg K]
+k = 46 ; // [W/m K]
+R = D/2;
+
+// (a)
+// From eqn 2.7.3 for a sphere
+t1 = rho*Cp*R/(3*h)*log((To-Tf)/(T-Tf)); // [sec]
+t1_min = t1/60 ; // [min]
+printf("Time taken by the centre of the ball to reach 150 degree C if internal gradients are neglected is %f seconds i.e. %f minutes \n",t1,t1_min);
diff --git a/530/CH2/EX2.11.b/example_2_11b.sce b/530/CH2/EX2.11.b/example_2_11b.sce new file mode 100755 index 000000000..63b601fb8 --- /dev/null +++ b/530/CH2/EX2.11.b/example_2_11b.sce @@ -0,0 +1,43 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.11(b)
+// Page 65
+printf("Example 2.11(b), Page 65 \n\n")
+
+D = 0.05 ; // [m]
+To = 450 ; // [degree C]
+Tf = 90 ; // [degree C]
+T = 150 ; // [degree c]
+h = 115 ; // [W/m^2 K]
+rho = 8000 ; // [kg/m^3]
+Cp = 0.42*1000 ; // [J/kg K]
+k = 46 ; // [W/m K]
+R = D/2;
+
+// (b)
+// let ratio = theta_R_0/theta_o
+ratio = (T-Tf)/(To - Tf);
+Bi = h*R/k;
+// From Table 2.5
+lambda_1_R = 0.430;
+x = 2*[sin(lambda_1_R) - lambda_1_R*cos(lambda_1_R)]/[lambda_1_R - sin(lambda_1_R)*cos(lambda_1_R)];
+
+// Substituting in equattion 2.7.29, we have an equation in variable y(=at/R^2)
+// Solving
+function[eqn] = parameter(y)
+eqn = ratio - x*exp(-(lambda_1_R^2)*(y));
+funcprot(0);
+endfunction
+
+y = 5; // (initial guess, assumed value for fsolve function)
+Y = fsolve(y,parameter);
+
+a = k/(Cp*rho); // alpha
+t2 = Y*(R^2)/(a); // [sec]
+t2_min = t2/60; // [min]
+printf("Time taken by the centre of the ball to reach 150 degree C if internal temperature gradients are not neglected is %f seconds i.e. %f minutes",t2,t2_min);
diff --git a/530/CH2/EX2.12/example_2_12.sce b/530/CH2/EX2.12/example_2_12.sce new file mode 100755 index 000000000..9d0579278 --- /dev/null +++ b/530/CH2/EX2.12/example_2_12.sce @@ -0,0 +1,50 @@ +clear ;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.12
+// Page 67
+printf("Example 2.12, Page 67 \n\n")
+
+a = 0.12 ; // [m]
+
+T = 400 ; // [C]
+To = 25 ; //[C]
+t = 100/60 ; // [hour]
+h = 10 ; // [W/m^2 K]
+k = 1.0 ; // [W/m K]
+alpha = 3.33*10^-3 ; // [m^2/h]
+// using fig 2.18 and eqn 2.7.20
+
+x1 = h*a/k ;
+x2 = k/(h*a);
+x3 = alpha*t/a^2;
+
+// Let ratio_x = theta/theta_o for x direction, from fig 2.18
+ratio_x = 0.82 ;
+
+// Similarly, for y direction
+ratio_y = 0.41;
+
+// Similarly, for z direction
+ratio_z = 0.30;
+
+// Therefore
+total_ratio = ratio_x*ratio_y*ratio_z ;
+
+T_centre = To + total_ratio*(T-To) ; // [degree C]
+printf("Temperature at the centre of the brick = %f degree C \n\n",T_centre);
+
+// Alternatively
+printf("Alternatively, obtaining Biot number and values of lambda_1_b and using eqn 2.7.20, we get \n")
+
+ratio_x = 1.1310*exp(-(0.9036^2)*0.385);
+ratio_y = 1.0701*exp(-(0.6533^2)*2.220);
+ratio_z = 1.0580*exp(-(0.5932^2)*3.469);
+ratio = ratio_x*ratio_y*ratio_z;
+
+T_centre = To + total_ratio*(T-To) ; // [degree C]
+printf("Temperature at the centre of the brick = %f degree C \n",T_centre);
\ No newline at end of file diff --git a/530/CH2/EX2.13.a/example_2_13a.sce b/530/CH2/EX2.13.a/example_2_13a.sce new file mode 100755 index 000000000..c34632f62 --- /dev/null +++ b/530/CH2/EX2.13.a/example_2_13a.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.13(a)
+// Page 73
+printf("Example 2.13(a), Page 73 \n\n")
+
+D = 0.003 ; // [m]
+L = 0.03 ; // [m]
+h = 10 ; // [W/m^2]
+Tf = 20 ; // [C]
+T1 = 120 ; // [C]
+
+// (a) Copper fin
+k = 350 ; // [W/m K]
+
+// For a circular cross section
+m = [4*h/(k*D)]^(1/2);
+mL = m*0.03 ;
+// T at x = L
+T = Tf + (T1-Tf)/cosh(m*L);
+printf("mL = %f \n",mL);
+printf("Temperature at the tip of fin made of copper is %f degree C \n",T);
diff --git a/530/CH2/EX2.13.b/example_2_13b.sce b/530/CH2/EX2.13.b/example_2_13b.sce new file mode 100755 index 000000000..45e73b79d --- /dev/null +++ b/530/CH2/EX2.13.b/example_2_13b.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.13(b)
+// Page 73
+printf("Example 2.13(b), Page 73 \n\n")
+
+D = 0.003 ; // [m]
+L = 0.03 ; // [m]
+h = 10 ; // [W/m^2]
+Tf = 20 ; // [C]
+T1 = 120 ; // [C]
+
+
+// (b) Stainless steel fin
+k = 15 ; // [W/m K]
+
+// For a circular cross section
+m = [4*h/(k*D)]^(1/2);
+mL = m*0.03 ;
+// T at x = L
+T = Tf + (T1-Tf)/cosh(m*L);
+printf("mL = %f \n",mL);
+printf("Temperature at the tip of fin made of steel is %f degree C \n",T);
diff --git a/530/CH2/EX2.13.c/example_2_13c.sce b/530/CH2/EX2.13.c/example_2_13c.sce new file mode 100755 index 000000000..cc7978cfe --- /dev/null +++ b/530/CH2/EX2.13.c/example_2_13c.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.13(c)
+// Page 73
+printf("Example 2.13(c), Page 73 \n\n")
+
+D = 0.003 ; // [m]
+L = 0.03 ; // [m]
+h = 10 ; // [W/m^2]
+Tf = 20 ; // [C]
+T1 = 120 ; // [C]
+
+// (c) Teflon fin
+k = 0.35 ; // [W/m K]
+
+// For a circular cross section
+m = [4*h/(k*D)]^(1/2);
+mL = m*0.03 ;
+// T at x = L
+T = Tf + (T1-Tf)/cosh(m*L);
+printf("mL = %f \n",mL);
+printf("Temperature at the tip of fin made of teflon is %f degree C \n",T);
diff --git a/530/CH2/EX2.14/example_2_14.sce b/530/CH2/EX2.14/example_2_14.sce new file mode 100755 index 000000000..63af66f42 --- /dev/null +++ b/530/CH2/EX2.14/example_2_14.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.14
+// Page 74
+printf("Example 2.14, Page 74 \n\n")
+
+L = 0.02 ; // [m]
+t = 0.002 ; // [m]
+b = 0.2 ; // [m]
+theta1 = 200 ; // [C]
+h = 15 ; // [W/m^2 K]
+k = 45 ; // [W/m K]
+
+Bi = h*(t/2)/k ;
+
+// We have
+P = 2*(b+t); // [m]
+A = b*t ; // [m^2]
+// Therefore
+mL = ([(h*P)/(A*k)]^(1/2))*L;
+
+// From equation 2.8.6, fin effectiveness n
+n = tanh(mL)/mL;
+printf("Fin Effectiveness = %f \n",n);
+
+q_loss = n*h*40.4*2*10^-4*200; // [W]
+printf("Heat loss rate from fin surface = %f W",q_loss);
\ No newline at end of file diff --git a/530/CH2/EX2.15/example_2_15.sce b/530/CH2/EX2.15/example_2_15.sce new file mode 100755 index 000000000..cb0df11c0 --- /dev/null +++ b/530/CH2/EX2.15/example_2_15.sce @@ -0,0 +1,38 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.1
+// Page 74
+printf("Example 2.15, Page 74 \n\n")
+
+
+// Find Decrease in thermal Resistance
+// Find Increase in heat transfer rate
+
+h = 15 ; //[W/m^2.K]
+k = 300; //[W/m.K]
+T = 200; //[C]
+Tsurr = 30; //[C]
+d = .01; //[m]
+L = .1; //[m]
+A = .5*.5 //[m^2]
+n = 100 //Number of Pins
+
+Bi = h*d/2/k; //Biot Number
+//Value of Biot Number is much less than .1
+//Thus using equation 2.8.6
+mL = (h*4/k/d)^.5*L;
+zi = tanh(mL)/mL;
+Res1 = 1/h/A; // Thermal resistance without fins, [K/W]
+Res2 = 1/(h*(A - n*%pi/4*d^2 + zi*(n*%pi*d*L)));// Thermal resistance with fins,[K/W]
+
+delRes = Res1-Res2; //[K/W]
+// Increase in heat transfer rate
+q = (T-Tsurr)/Res2 - (T-Tsurr)/Res1; //[W]
+
+printf("\n\n Decrease in thermal resistane at surface %.4f K/W.\n Increase in heat transfer rate %.1f W",delRes,q)
+//END
\ No newline at end of file diff --git a/530/CH2/EX2.16/example_2_16.sce b/530/CH2/EX2.16/example_2_16.sce new file mode 100755 index 000000000..efa74023f --- /dev/null +++ b/530/CH2/EX2.16/example_2_16.sce @@ -0,0 +1,15 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.16
+// Page 75
+printf("Example 2.16, Page 75 \n\n")
+
+//Theoretical Problem
+
+printf('\n\n This is a Theoretical Problem, does not involve any mathematical computation.');
+//END
\ No newline at end of file diff --git a/530/CH2/EX2.2/example_2_2.sce b/530/CH2/EX2.2/example_2_2.sce new file mode 100755 index 000000000..a600dd258 --- /dev/null +++ b/530/CH2/EX2.2/example_2_2.sce @@ -0,0 +1,35 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.2
+// Page 31
+printf("Example 2.2, Page 31 \n\n")
+
+d_i=0.02; // [m] inner radius
+d_o=0.04; // [m] outer radius
+r_i=d_i/2; // [m] inner radius
+r_o=d_o/2; // [m] outer radius
+k=0.58; // [w/m K] thermal conductivity of tube material
+t_i=70; //[degree C]
+t_o=100; // [degree C]
+l=1; // [m] per unit length
+
+// thermal resistance of tube per unit length
+R_th_tube=(log(r_o/r_i))/(2*%pi*k*l); // [K/W]
+
+//from table 1.3 , heat transfer co-efficient for condensing steam may be taken as
+h=5000; // [W/m^2 K]
+// thermal resistance of condensing steam per unit length
+R_th_cond=1/(%pi*d_o*l*h);
+
+// since R_th_tube is much less than R_th_cond , we can assume outer surface to be at 100 degree C
+//hence heat flow rate per unit meter is
+q=l*2*(%pi)*k*(t_i-100)/log(r_o/r_i);
+
+printf("Thermal resistance of tube per unit length is %f K/W\n",R_th_tube);
+printf("Thermal resistance of condensing steam per unit length is %f K/W\n",R_th_cond);
+printf("Heat flow per unit length is %f W/m",q);
\ No newline at end of file diff --git a/530/CH2/EX2.3/example_2_3.sce b/530/CH2/EX2.3/example_2_3.sce new file mode 100755 index 000000000..23f936f80 --- /dev/null +++ b/530/CH2/EX2.3/example_2_3.sce @@ -0,0 +1,37 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.3
+// Page 31
+printf("Example 2.3, Page 31 \n\n")
+
+h_w=140; // heat transfer coefficient on water side, [W/m^2 K]
+h_o=150; // heat transfer coefficient on oil side, [W/m^2 K]
+k=30; // thermal conductivity [W/m K]
+r_o=0.01; // inner diameter of GI pipe on inside
+r_i=0.008; // outer diameter GI pipe on inside
+l=1; // [m] , per unit length
+
+// Thermal resistance of inner GI pipe
+R_inner_GI=log((r_o/r_i))/(2*%pi*k*l);
+
+
+// Thermal resistance on the oil side per unit length
+R_oilside=1/(h_o*%pi*2*r_i*l);
+
+
+// Thermal resistance on cold water side per unit length
+R_waterside=1/(h_w*%pi*2*r_o*l);
+
+
+// we see thermal resistance of inner GI pipe contributes less than 0.5 percent to the total resistance
+
+
+printf("Thermal resistance of inner GI pipe = %f K/W \n",R_inner_GI);
+printf("Thermal resistance on the oil side per unit length = %f K/W \n",R_oilside);
+printf("Thermal resistance on cold water side per unit length = %f K/W \n",R_waterside);
+printf("So, Engineer in-charge has made a bad decision");
\ No newline at end of file diff --git a/530/CH2/EX2.4/example_2_4.sci b/530/CH2/EX2.4/example_2_4.sci new file mode 100755 index 000000000..29f619d7f --- /dev/null +++ b/530/CH2/EX2.4/example_2_4.sci @@ -0,0 +1,38 @@ +clear all;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.4
+// Page 32
+printf("Example 2.4, Page 32 \n\n")
+
+Ti = 300; //Internal temp of hot gas in degree Celsius
+OD = 0.1; //Outer diameter of long metal pipe in meters
+ID = 0.04; //Internal diamtere of long metal pipe in meters
+ki = 0.052; //thermal conductivity of mineral wood in W/mK
+To = 50; //Outer surface temperature in degree celsius
+hi = 29; //heat transfer coefficient in the inner side in W/m^2 K
+ho = 12; //heat transfer coefficient in the outer pipe W/m^2 K
+
+//Determination of thickness of insulation
+function[f] = thickness(r)
+ f = r*(10.344 + 271.15*log(r*(0.05)^-1))-11.75
+ funcprot(0);
+endfunction
+r = 0.082;
+while 1
+ rnew = r - thickness(r)/diff(thickness(r));
+ if rnew == r then
+ r3 = rnew;
+ break;
+ end
+ r = rnew;
+end
+t = r3 - OD/2;
+printf("\n Thickness of insulation = %f cm",t*100);
+//Heat loss per unit length
+q = 600*(22/7)*r3;
+printf("\n Heat loss per unit length = %.1f W/m",q);
diff --git a/530/CH2/EX2.5/example_2_5.sce b/530/CH2/EX2.5/example_2_5.sce new file mode 100755 index 000000000..4a868e4b2 --- /dev/null +++ b/530/CH2/EX2.5/example_2_5.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.5
+// Page 34
+printf("Example 2.5, Page 34 \n\n")
+
+Ti = 90; //Temp on inner side in degree celsius
+To = 30; //Temp on outer side in degree celsius
+hi = 500; //heat transfer coeffcient in W/m^2 K
+ho = 10; //heat transfer coeffcient in W/m^2 K
+ID = 0.016; //Internal diameter in meters
+t = [0 0.5 1 2 3 4 5]; //Insulation thickness in cm
+OD = 0.02; //Outer diameter in meters
+r3 = OD/2 + t/100; //radius after insulation in meters
+
+i=1;
+printf("\n Insulation thickness(cm) r3(m) heat loss rate per meter(W/m)");
+while i<=7
+ ql(i) = [2*(%pi)*(ID/2)*(Ti-To)]/[(1/hi)+(0.008/0.2)*log(r3(i)/0.01) + (0.008/r3(i))*(1/ho)];
+printf("\n %.1f %.3f %.1f",t(i),r3(i),ql(i));
+ i = i+1;
+end
+plot(t,ql);
+xtitle("","Insulation thickness(cm)","Heat loss rate per unit length,W/m");
+printf("\n The maxima in the curve is at r_3 = 0.02 m");
diff --git a/530/CH2/EX2.6/example_2_6.sce b/530/CH2/EX2.6/example_2_6.sce new file mode 100755 index 000000000..228e84c14 --- /dev/null +++ b/530/CH2/EX2.6/example_2_6.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.6
+// Page 36
+printf("Example 2.6, Page 34 \n\n")
+
+h_natural = 10;//heat transfer coefficient for natural convection in W/m^2 K
+h_forced = 50; //heat transfer coefficient for forced convection in W/m^2 K
+//for asbestos
+k1 = 0.2; //thermal conductivity in W/m K
+//for mineral wool
+k2 = 0.05; //thermal conductivity in W/m K
+printf("\n critical radius of insulation in cm");
+printf("\n h = 10 h = 50");
+printf("\n Asbestos %.1f %.1f",k1*100/h_natural,k1*100/h_forced);
+printf("\n Mineral wool %.1f %.1f",k2*100/h_natural,k2*100/h_forced);
diff --git a/530/CH2/EX2.7/example_2_7.sce b/530/CH2/EX2.7/example_2_7.sce new file mode 100755 index 000000000..8c51f8567 --- /dev/null +++ b/530/CH2/EX2.7/example_2_7.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.7
+// Page 43
+printf("Example 2.7, Page 43 \n\n")
+
+H = 5 ; // Height, [m]
+L = 10 ; // Length, [m]
+t = 1 ; // thickness, [m]
+b = t/2;
+k = 1.05 ; // [W/m K]
+q = 58 ; // [W/m^3]
+T = 35 ; // [C]
+h = 11.6 ; // Heat transfer coefficient, [W/m^2 K]
+// Substituting the values in equation 2.5.6
+T_max = T + q*b*(b/(2*k)+1/h);
+printf("Maximum Temperature = %f degree C",T_max);
\ No newline at end of file diff --git a/530/CH2/EX2.8/example_2_8.sce b/530/CH2/EX2.8/example_2_8.sce new file mode 100755 index 000000000..515e08d91 --- /dev/null +++ b/530/CH2/EX2.8/example_2_8.sce @@ -0,0 +1,39 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.8
+// Page 47
+printf("Example 2.8, Page 47 \n\n")
+
+// The bar will have two dimensional variation in temperature
+// the differential equation is subject to boundary conditions
+x1 = 0; // [cm]
+Tx1 = 30; // [C]
+x2 = 5; // [cm]
+Tx2 = 30; // [C]
+y1 = 0; // [cm]
+Ty1 = 30; // [C]
+y2 = 10; // [cm]
+Ty2 = 130; // [C]
+// substituting theta = T-30 and using eqn 2.6.11
+// putting x = 2.5cm and y = 5cm in infinite summation series
+
+
+n = 1;
+x1 = (1-cos(%pi*n))/(sinh(2*%pi*n))*sin(n^%pi/2)*sinh(n*%pi);
+
+n = 3;
+x3 = (1-cos(%pi*n))/(sinh(2*%pi*n))*sin(n^%pi/2)*sinh(n*%pi);
+
+n = 5;
+x5 = (1-cos(%pi*n))/(sinh(2*%pi*n))*sin(n^%pi/2)*sinh(n*%pi);
+
+x = x1+x3+x5;
+
+T = x*100+30;
+printf("Steady state temperature = %f C",T);
+
diff --git a/530/CH2/EX2.9/example_2_9.sci b/530/CH2/EX2.9/example_2_9.sci new file mode 100755 index 000000000..3bb790282 --- /dev/null +++ b/530/CH2/EX2.9/example_2_9.sci @@ -0,0 +1,38 @@ +clear all;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 2
+// Heat Conduction in Solids
+
+// Example 2.9
+// Page 51
+printf("Example 2.9, Page 51 \n\n")
+
+k = 330; //thermal conductivity in W/m K
+a = 95*10^(-6); //thermal expansion coefficient
+R = 0.01; //radius in meters
+To = 77; //temperature in kelvins
+Tf = 273+50; //temperature in kelvins
+theta1 = To - Tf;
+T = 273+10; //temperature in kelvins
+theta = T - Tf;
+h = 20; //heat transfer coefficient in W/m^2 K
+printf("\n Theta1 = %d K",theta1);
+printf("\n Theta = %d K ",theta);
+printf("\n v/A = %.3f m",R/2);
+printf("\n k/a = %.4f*10^(6) J/m^3 K",(k/a)*10^(-6));
+
+time = (k/a)*(R/2)/h*log(theta1/theta);
+
+printf("\n Time taken by the rod to heat up = %.1f secs",time);
+Bi = h*R/k;
+printf("\n Biot number Bi = %.2f*10^(-4) ",Bi*10^4);
+printf("\n Since Biot number is much less than 0.1,therefore assumption that internal temperature gradients are negligible is a good one");
+
+
+
+
+
+
+
diff --git a/530/CH3/EX3.1/example_3_1.sce b/530/CH3/EX3.1/example_3_1.sce new file mode 100755 index 000000000..dacbeeb32 --- /dev/null +++ b/530/CH3/EX3.1/example_3_1.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.1
+// Page 114
+printf("Example 3.1, Page 114 \n\n");
+
+T = 5779 ; // [Temperature,in Kelvin]
+// From Wein's law, eqn 3.2.8
+lambda_m = 0.00290/T ; // [m]
+// Substituting this value in plank's law, we get
+e = 2*(%pi)*0.596*(10^-16)/(((0.5018*10^-6)^5)*(exp(0.014387/0.00290)-1)) ; //[W/m^2 m]
+
+e_bl_max= e / 10^6 ;
+
+printf("Value of emissivity on sun surface is %f W/m^2 um \n",e_bl_max); //[W/m^2 um]
+
+e_earth = e_bl_max*((0.695*10^6)/(1.496*10^8))^2 ;
+
+printf("The value of emmissivity on earths surface is %f W/m^2 um", e_earth)
\ No newline at end of file diff --git a/530/CH3/EX3.10/example_3_10.sce b/530/CH3/EX3.10/example_3_10.sce new file mode 100755 index 000000000..011b264cf --- /dev/null +++ b/530/CH3/EX3.10/example_3_10.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.10
+// Page 138
+printf("Example 3.10, Page 138 \n\n")
+
+sigma = 5.670*10^-8 ;
+T1 = 473 ; // [K]
+T2 = 373 ; // [K]
+A1 = 1*2 ; // area, [m^2]
+X = 0.25;
+Y = 0.5 ;
+// From eqn 3.7.4
+F12 = (2/(%pi*X*Y))*[log((((1+X^2)*(1+Y^2))/(1+X^2+Y^2))^(1/2)) + Y*((1+X^2)^(1/2))*atan(Y/((1+X^2)^(1/2))) + X*((1+Y^2)^(1/2))*atan(X/((1+Y^2)^(1/2))) - Y*atan(Y) - X*atan(X) ];
+
+
+q1 = sigma*A1*(T1^4-T2^4)*[(1-F12^2)/(2-2*F12)];
+
+printf("Net radiative heat transfer from the surface = %f W \n",q1);
\ No newline at end of file diff --git a/530/CH3/EX3.11/example_3_11.sci b/530/CH3/EX3.11/example_3_11.sci new file mode 100755 index 000000000..ba9f102c8 --- /dev/null +++ b/530/CH3/EX3.11/example_3_11.sci @@ -0,0 +1,40 @@ +clear all;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.11
+// Page 141
+printf("Example 3.11, Page 141 \n\n")
+
+// All modes of heat transfer are involved
+// let steady state heat flux flowing through the composite slab be (q/a)
+h1 = 20; //[W/m^2 K]
+w1 = 0.2; //[m]
+k1 = 1; //[W/m K]
+e1 = 0.5; //emmisivity at surfce 1
+e2 = 0.4; //emmisivity at surfce 2
+w2 = 0.3; //[m]
+k2 = 0.5; //[W/m K]
+h2 = 10; //[W/m^2 K]
+T1 = 473; //[Kelvin]
+T2 = 273+40; //[Kelvin]
+stefan_cnst = 5.67e-08; //[W/m^2 K^4]
+
+// For resistances 1 and 2
+function[f]=temperature(T)
+ f(1) = (T1-T(1))/(1/h1 + w1/k1) - (T(2) - T2)/(w2/k2 + 1/h2);
+ f(2) = stefan_cnst*(T(1)^4 - T(2)^4)/(1/e1 + 1/e2 -1) - (T(2) - T2)/(w2/k2 + 1/h2);
+ funcprot(0);
+endfunction
+
+T = [10 10]; // assumed initial values for fsolve function
+y = fsolve(T,temperature);
+
+printf("\n Steady state heat flux q/A = %.1f W/m^2",(T1-y(1))/(1/h1 + w1/k1));
+
+
+
+
diff --git a/530/CH3/EX3.12/example_3_12.sce b/530/CH3/EX3.12/example_3_12.sce new file mode 100755 index 000000000..e051d8c66 --- /dev/null +++ b/530/CH3/EX3.12/example_3_12.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.12
+// Page 145
+printf("Example 3.12, Page 145 \n\n")
+
+D = 0.02 ; // [m]
+T1 = 1000+273 ; // [K]
+T2 = 27+273 ; // [K]
+s = 5.670*10^-8 ; // stefans constant
+// Assuming the opening is closed by an imaginary surface at temperature T1
+// Using equation 3.10.3 , we get
+q = s*1*%pi*((D/2)^2)*(T1^4-T2^4); // [W]
+
+printf("Rate at which heat is lost by radiation = %f W",q);
\ No newline at end of file diff --git a/530/CH3/EX3.13/example_3_13.sce b/530/CH3/EX3.13/example_3_13.sce new file mode 100755 index 000000000..24e6809fb --- /dev/null +++ b/530/CH3/EX3.13/example_3_13.sce @@ -0,0 +1,33 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.13
+// Page 146
+printf("Example 3.13, Page 146 \n\n")
+
+D = 0.32 ; // [m]
+D_s = 0.36 ; // [m]
+e = 0.02 ; // emissivity
+l = 201 ; // [kJ/kg]
+rho = 800 ; // [kg/m^3]
+s = 5.670*10^-8 ;
+
+T2 = 303 ; // [K]
+T1 = 77 ; // [K]
+
+// From equation 3.10.1
+q1 = s*4*%pi*((D/2)^2)*(T1^4-T2^4)/[1/e+((D/D_s)^2)*(1/e-1)]; // [W]
+
+evap = abs(q1)*3600*24/(l*1000); // [kg/day]
+mass = 4/3*%pi*((D/2)^3)*rho;
+boiloff = evap/mass*100 ; // percent
+
+T_drop = (abs(q1))/(4*%pi*((D/2)^2))*(1/100); // [C]
+
+printf("Rate at which nitrogen evaporates = %f kg/day \n",evap)
+printf("Boil-off rate = %f percent \n",boiloff);
+printf("Temperature drop between liquid Nitrogen and inner surface = %f C",T_drop);
\ No newline at end of file diff --git a/530/CH3/EX3.14/example_3_14.sce b/530/CH3/EX3.14/example_3_14.sce new file mode 100755 index 000000000..96ed5ccdf --- /dev/null +++ b/530/CH3/EX3.14/example_3_14.sce @@ -0,0 +1,33 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.14
+// Page 147
+printf("Example 3.14, Page 147 \n\n")
+
+D = 1 ; // [m]
+r = 6250 ; // [km]
+D_surf = 300 ; // [km]
+s = 5.670*10^-8;
+e = 0.3 ;
+Tc = -18+273 ; // [K]
+T_surf = 27+273 ; // [K]
+
+// Rate of emissino of radiant energy from the two faces of satellite disc
+r_emission = 2*e*%pi*((D/2)^2)*s*Tc^4; // [W]
+
+// A2*F21 = A1*F12
+sina = (r/(r+D_surf));
+F12 = sina^2;
+
+// Rate at which the satellite receives and absorbs energy coming from earth
+r_receive = e*s*(%pi*((D/2)^2))*F12*T_surf^4; // [W]
+
+r_loss = r_emission - r_receive; // [W]
+
+printf("Net Rate at which energy is leaving the satellite = %f W",r_loss);
+
diff --git a/530/CH3/EX3.15/example_3_15.sce b/530/CH3/EX3.15/example_3_15.sce new file mode 100755 index 000000000..38e7311ab --- /dev/null +++ b/530/CH3/EX3.15/example_3_15.sce @@ -0,0 +1,47 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.15
+// Page 151
+printf("Example 3.15, Page 151 \n\n")
+
+// From example 3.10
+F12 = 0.0363;
+F11 = 0;
+F13 = 1-F11-F12;
+// Similarly
+F21 = 0.0363;
+F22 = 0;
+F23 = 0.9637;
+
+// Now, F31 = A1/A3*F13
+F31 = 2/24*F13;
+// Therefore
+F32 = F31;
+F33 = 1-F31-F32;
+
+// Substituting into equation 3.11.6, 3.11.7, 3.11.8, we have f(1),f(2),f(3)
+
+function[f]=flux(B)
+ f(1)= B(1) - 0.4*0.0363*B(2) - 0.4*0.9637*B(3) - 0.6*(473^4)*(5.670*10^-8);
+ f(2)= -0.4*0.0363*B(1) + B(2) - 0.4*0.9637*B(3) - 0.6*(5.670*10^-8)*(373^4);
+ f(3)= 0.0803*B(1) + 0.0803*B(2) - 0.1606*B(3);
+ funcprot(0);
+endfunction
+
+B = [0 0 0];
+y = fsolve(B,flux);
+printf("\n B1 = %.1f W/m^2",y(1));
+printf("\n B2 = %.1f W/m^2",y(2));
+printf("\n B3 = %.1f W/m^2 \n",y(3));
+
+// Therefore
+H1 = 0.0363*y(2) + 0.9637*y(3) ; // [W/m^2]
+// and
+q1 = 2*(y(1) - H1) ; // [W]
+
+printf("Net radiative heat transfer = %f W",q1);
diff --git a/530/CH3/EX3.2/example_3_2.sce b/530/CH3/EX3.2/example_3_2.sce new file mode 100755 index 000000000..95f099ec6 --- /dev/null +++ b/530/CH3/EX3.2/example_3_2.sce @@ -0,0 +1,43 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.2
+// Page 115
+printf("Example 3.2, Page 115 \n\n")
+
+//Heat emission
+Stefan_constt = 5.67*10^(-8); //(W/m^2.K^4)
+T = 1500; //temperature is in kelvins
+eb = (Stefan_constt)*(T^(4)); //energy radiated by blackbody
+//emission in 0.3um to 1um
+e = 0.9; //emissivity
+lamda1 = 1; //wavelength is in um
+lamda2 = 0.3; //wavelength is in um
+D0_1=0.5*(0.01972+0.00779); //From table 3.1 page- 114
+D0_2=0; //From table 3.1 page- 114
+q = e*(D0_1-D0_2)*Stefan_constt*T^(4);//in W/m^2
+printf("\n wavelength*temp = %d um K",1*1500);
+printf("\n wavelength*temp at 0.3um = %d um K",0.3*1500);
+printf("\n\n Required heat flux, q = %d W/m^2",q);
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/530/CH3/EX3.3/example_3_3.sce b/530/CH3/EX3.3/example_3_3.sce new file mode 100755 index 000000000..4d494575f --- /dev/null +++ b/530/CH3/EX3.3/example_3_3.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.3
+// Page 119
+printf("Example 3.3, Page 119 \n\n")
+
+
+a0_2=1; //absorptivity
+a2_4=1; //absorptivity
+a4_6=0.5; //absorptivity
+a6_8=0.5; //absorptivity
+a8_=0; //absorptivity
+H0_2=0; //Irradiation in W/m^2 um
+H2_4=750; //Irradiation in W/m^2 um
+H4_6=750; //Irradiation in W/m^2 um
+H6_8=750; //Irradiation in W/m^2 um
+H8_=750; //Irradiation in W/m^2 um
+Absorbed_radiant_flux=1*0*(2-0)+1*750*(4-2)+0.5*750*(8-4)+0;
+H = 750*(8-2); //Incident flux
+a = Absorbed_radiant_flux/H;
+p = 1-a; //Since the surface is opaque
+printf("\n Absorbed radiant flux = %d W/m^2",Absorbed_radiant_flux);
+printf("\n Incident flux = %d W/m^2",H);
+printf("\n Absorptivity = %.3f",a);
+printf("\n Since the surface is opaque reflectivity = %.3f",p);
diff --git a/530/CH3/EX3.4.a/example_3_4a.sce b/530/CH3/EX3.4.a/example_3_4a.sce new file mode 100755 index 000000000..7a7a90857 --- /dev/null +++ b/530/CH3/EX3.4.a/example_3_4a.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.4(a)
+// Page 123
+printf("Example 3.4(a), Page 123 \n\n")
+
+
+e = 0.08; //emissivity
+T = 800; //temperature, [K]
+
+Stefan_constt = 5.67*10^(-8); //[W/m^2.K^4]
+// From Stefan Boltzmann law, equation 3.2.10
+q = e*Stefan_constt*T^4; //[W/m^2]
+printf("\n Energy emitted = %.1f W/m^2",q);
+
+// (a)
+// Therefore
+in = (q/(%pi));
+printf("\n Energy emitted normal to the surface = %.1f W/m^2 sr",in);
diff --git a/530/CH3/EX3.4.b/example_3_4b.sce b/530/CH3/EX3.4.b/example_3_4b.sce new file mode 100755 index 000000000..cba9e4da2 --- /dev/null +++ b/530/CH3/EX3.4.b/example_3_4b.sce @@ -0,0 +1,29 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.4(b)
+// Page 123
+printf("Example 3.4(b), Page 123 \n\n")
+
+
+e = 0.08; //emissivity
+T = 800; //temperature, [K]
+
+Stefan_constt = 5.67*10^(-8); //[W/m^2.K^4]
+// From Stefan Boltzmann law, equation 3.2.10
+q = e*Stefan_constt*T^4; //[W/m^2]
+in = (q/(%pi));
+
+// (b)
+// Radiant flux emitted in the cone 0 <= pzi <= 50 degree, 0 <= theta <= 2*pi
+q_cone=2*(%pi)*in*(-cos(100*(%pi/180))+cos(0))/4;
+
+printf("\n Radiant flux emitted in the cone =%.1f W/m^2",q_cone);
+
+Ratio = q_cone/q;
+printf("\n Ratio = %.3f",Ratio);
+
diff --git a/530/CH3/EX3.5/example_3_5.sce b/530/CH3/EX3.5/example_3_5.sce new file mode 100755 index 000000000..8d40922d7 --- /dev/null +++ b/530/CH3/EX3.5/example_3_5.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.5
+// Page 124
+printf("Example 3.5, Page 124 \n\n")
+
+l1 = 0.5 ; // wavelength, [um]
+l2 = 1.5 ; // wavelength, [um]
+l3 = 2.5 ; // wavelength, [um]
+l4 = 3.5 ; // wavelength, [um]
+H1 = 2500 ; // [W/m^2 um]
+H2 = 4000 ; // [W/m^2 um]
+H3 = 2500 ; // [W/m^2 um]
+
+// Since the irridiation is diffuse, the spectral intensity is given by eqn 3.4.14 and 3.4.8
+// Integrating i_lambda over the directions of the specified solid angle and using fig 3.12
+
+
+flux = 3/4*[H1*(l2-l1)+H2*(l3-l2)+H3*(l4-l3)];
+printf("Rate at which radiation is incident on the surface = %f W/m^2",flux);
diff --git a/530/CH3/EX3.6/example_3_6.sce b/530/CH3/EX3.6/example_3_6.sce new file mode 100755 index 000000000..1c235469e --- /dev/null +++ b/530/CH3/EX3.6/example_3_6.sce @@ -0,0 +1,33 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.6
+// Page 132
+printf("Example 3.6, Page 132 \n\n")
+
+// This is a theoretical problem with no numerical data
+printf("This is a theoretical problem with no numerical data \n");
+
+// Considering an elementary ring dA2 of width dr at an arbitary radius r, we have
+// r = h*tanB1
+// dA2 = 2*%pi*r*dr
+// dA2 = 2*%pi*(h^2)*tan(B1)*sec^2(B1)*dB1
+// B2 = B1, since surfaces ate parallel, and
+// L = h/cos(B1)
+// Substituting in eqn 3.6.7
+// F12 = sin^2(a)
+
+
+printf("Considering an elementary ring dA2 of width dr at an arbitary radius r, we have \n");
+printf("r = h*tanB1 \n");
+printf("dA2 = 2*pi*r*dr \n");
+printf("dA2 = 2*pi*(h^2)*tan(B1)*sec^2(B1)*dB1 \n");
+printf("B2 = B1, since surfaces ate parallel, and \n");
+printf("L = h/cos(B1) \n");
+printf("Substituting in eqn 3.6.7 \n");
+printf("F12 = sin^2(a) \n");
+
diff --git a/530/CH3/EX3.7/example_3_7.sce b/530/CH3/EX3.7/example_3_7.sce new file mode 100755 index 000000000..f69b48086 --- /dev/null +++ b/530/CH3/EX3.7/example_3_7.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.7
+// Page 134
+printf("Example 3.7, Page 134 \n\n")
+
+// This is a theoretical problem with no numerical data
+printf("This is a theoretical problem with no numerical data \n");
+
+
+// Considering an elementary circular ring on the surface of the sphere's surface at any arbitary anglr B,
+// we have B1 = B, B2 = 0, L = R and dA_2 = 2*%pi*(R^2)*(sin(B))dB
+// Therefore, from equation 3.6.7
+// F12 = sin^2(a)
+
+printf("Considering an elementary circular ring on the surface of the sphere surface at any arbitary anglr B \n");
+printf("we have B1 = B, B2 = 0, L = R and dA_2 = 2*pi*(R^2)*(sin(B))dB \n");
+printf("Therefore, from equation 3.6.7 \n");
+printf("F12 = sin^2(a)");
\ No newline at end of file diff --git a/530/CH3/EX3.8/example_3_8.sce b/530/CH3/EX3.8/example_3_8.sce new file mode 100755 index 000000000..08c154f7e --- /dev/null +++ b/530/CH3/EX3.8/example_3_8.sce @@ -0,0 +1,33 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.8
+// Page 135
+printf("Example 3.8, Page 135 \n\n")
+
+// From eqn 3.7.5 or fig 3.19
+F65 = 0.22;
+F64 = 0.16;
+F35 = 0.32;
+F34 = 0.27;
+A1 = 3; // [m^2]
+A3 = 3; // [m^2]
+A6 = 6; // [m^2]
+
+// Using additive and reciprocal relations
+// We have F12 = F16 - F13
+
+F61 = F65 - F64 ;
+F31 = F35 - F34 ;
+
+F16 = A6/A1*F61 ;
+F13 = A3/A1*F31 ;
+
+F12 = F16 - F13;
+
+printf("F_1-2 = %f",F12);
+
diff --git a/530/CH3/EX3.9/example_3_9.sce b/530/CH3/EX3.9/example_3_9.sce new file mode 100755 index 000000000..f4173c7b6 --- /dev/null +++ b/530/CH3/EX3.9/example_3_9.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 3
+// Thermal Radiation
+
+// Example 3.9
+// Page 136
+printf("Example 3.9, Page 136 \n\n")
+
+// This is a theoretical problem, does not involve any numerical computation
+printf("This is a theoretical problem, does not involve any numerical computation \n");
+// Denoting area of conical surface by A1
+// Considering an imaginary flat surface A2 closing the conical cavity
+
+F22 = 0 ; // Flat surface
+
+// from eqn 3.7.2 , we have F11 + F12 = 1 and F22 + F21 = 1
+F21 = 1 - F22 ;
+
+// F12 = A2/A1*F21 ;
+// F11 = 1 - F12 ;
+// F11 = 1 - sin(a)
diff --git a/530/CH4/EX4.1/example_4_1.sce b/530/CH4/EX4.1/example_4_1.sce new file mode 100755 index 000000000..a345eb91c --- /dev/null +++ b/530/CH4/EX4.1/example_4_1.sce @@ -0,0 +1,33 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 4
+// Principles of Fluid Flow
+
+// Example 4.1
+// Page 172
+printf("Example 4.1, Page 172 \n\n");
+
+L = 3 ; // Length, [m]
+D = 0.01 ; // ID, [m]
+V = 0.2 ; // Average Velocity, [m/s]
+
+// From Table A.1 at 10 degree C
+rho=999.7 ; // [kg/m^3]
+v=1.306 * 10^-6 ; // [m^2/s]
+
+Re_D=0.2*0.01/(1.306*10^-6) ;
+
+// this value is less than the transition Reynolds number 2300.
+// Hence flow is laminar. From eqn 4.4.19
+f = 16/Re_D;
+
+// from eqn 4.4.17
+delta_p = 4*f*(L/D)*(rho*V^2)/2;
+
+// since flow is laminar
+V_max = 2*V;
+
+printf("Pressure drop is %f Pa \n",delta_p);
+printf(" Maximum velocity is %f m/s",V_max);
diff --git a/530/CH4/EX4.2.a/example_4_2a.sce b/530/CH4/EX4.2.a/example_4_2a.sce new file mode 100755 index 000000000..c8adcf54a --- /dev/null +++ b/530/CH4/EX4.2.a/example_4_2a.sce @@ -0,0 +1,43 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 4
+// Principles of Fluid Flow
+
+// Example 4.2(a)
+// Page 180
+printf("Example 4.2(a), Page 180 \n\n")
+
+L = 3 ; //[m]
+D = 0.01 ; //[m]
+V = 0.2 ; //[m/s]
+
+// (a)
+printf("(a) If the temperature of water is increased to 80 degree C \n");
+
+
+// Properties of water at 80 degree C
+rho = 971.8 ; // [kg/m^3]
+v = 0.365 * 10^-6 ; // [m^2/s]
+
+Re_D = D*V/v;
+
+// flow is turbilent, so from eqn 4.6.4a
+
+f=0.079*(Re_D)^(-0.25);
+delta_p = (4*f*L*rho*V^2)/(D*2); // [Pa]
+printf("Pressure drop is %f Pa \n",delta_p);
+
+// from eqn 4.4.16
+
+// x = (T_w/p)^0.5 = ((f/2)^0.5)*V ;
+x = ((f/2)^0.5)*V ;
+y_plus = 0.005*x/(0.365*10^-6);
+
+// from eqn 4.6.1c & 4.6.2
+
+V_max = x*(2.5* log(y_plus) + 5.5) ; // [m/s]
+ratio = V_max/V;
+printf("V_max = %f m/s \n",V_max);
+printf("V_max/V_bar = %f \n\n",ratio);
diff --git a/530/CH4/EX4.2.b/example_4_2b.sce b/530/CH4/EX4.2.b/example_4_2b.sce new file mode 100755 index 000000000..99e444ab3 --- /dev/null +++ b/530/CH4/EX4.2.b/example_4_2b.sce @@ -0,0 +1,43 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 4
+// Principles of Fluid Flow
+
+// Example 4.2(b)
+// Page 180
+printf("Example 4.2(b), Page 180 \n\n")
+
+L = 3 ; //[m]
+D = 0.01 ; //[m]
+V = 0.2 ; //[m/s]
+
+// (b)
+
+V1=0.7;
+v1 = 1.306 * 10^-6 ; // [m^2/s]
+
+printf("(b) If the velocity is increased to 0.7 \n");
+// if velocity of water is 0.7 m/s
+V1=0.7; // [m/s]
+Re_D1=V1*D/(1.306*10^-6);
+printf("Reynolds no is %f \n",Re_D1);
+
+// flow is again turbulent
+f1 = 0.079*(Re_D1)^(-0.25);
+
+delta_p1 = (4*f1*L*999.7*0.7^2)/(0.01*2); // [Pa]
+printf("Pressure drop is %f Pa \n",delta_p1);
+
+// x1 = (T_w/p)^0.5 = ((f1/2)^0.5)*V ;
+x1 = ((f1/2)^0.5)*V1 ;
+
+y1_plus = 0.005*x1/(v1);
+printf("y+ at centre line = %f \n",y1_plus);
+
+V_max1 = x1*(2.5* log(y1_plus) + 5.5) ; // [m/s]
+printf("V_max is %f m/s \n",V_max1);
+
+ratio1 = V_max1/V1;
+printf("Vmax/Vbar = %f ",ratio1);
diff --git a/530/CH4/EX4.3/example_4_3.sce b/530/CH4/EX4.3/example_4_3.sce new file mode 100755 index 000000000..c40e7ade9 --- /dev/null +++ b/530/CH4/EX4.3/example_4_3.sce @@ -0,0 +1,49 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 4
+// Principles of Fluid Flow
+
+// Example 4.3
+// Page 181
+printf("Example 4.3, Page 181 \n\n")
+P = 80 * 10^3 ; // [Pa]
+L = 10 ; // [m]
+V_bar = 1.9 ; // [m/s]
+l = 0.25 ; // [m]
+b = 0.15 ; // [m]
+
+// Fully developed flow
+
+// From Table A.2, for air at ! atm pressure and 25 degree C
+rho = 1.185 ; // [kg/m^3]
+mew = 18.35 * 10^-6 ; // [kg/m s]
+
+// At 80 kPa and 25 degree C
+rho1 = rho*(80/101.3) ; // [kg/m^3]
+
+// For given duct r=(b/a)
+r = b/l;
+
+D_e = (4*l/2*b/2)/(l/2 + b/2); // [m]
+
+// From eqn 4.6.7
+
+D_l = [2/3 + 11/24*0.6*(2-0.6)]*D_e ; // [m]
+
+// Reynolds no based on D_l
+
+Re = rho1*D_l*V_bar/mew;
+printf("Reynolds no = %f \n",Re);
+
+f = 0.079*(Re^-0.25) ;
+printf("f = %f \n",f);
+
+// From eqn 4.4.17
+
+delta_P = 4*f*(L/D_l)*(rho1*(V_bar^2)/2);
+printf("Pressure drop = %f Pa \n",delta_P);
+
+power = delta_P*(V_bar*l*b)
+printf("Power required = %f W",power);
\ No newline at end of file diff --git a/530/CH4/EX4.4/example_4_4.sce b/530/CH4/EX4.4/example_4_4.sce new file mode 100755 index 000000000..bd8bd8f62 --- /dev/null +++ b/530/CH4/EX4.4/example_4_4.sce @@ -0,0 +1,38 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 4
+// Principles of Fluid Flow
+
+// Example 4.4
+// Page 189
+printf("Example 4.4, Page 189 \n\n")
+
+l = 2 ; // [m]
+b = 1 ; // [m]
+V = 1 ; // [m/s]
+
+// From table A.2
+
+rho = 1.060 ; // [kg/m^3]
+v = 18.97 * 10^-6 ; // [m^2/s]
+
+// At x = 1.5m
+x = 1.5 ; // [m]
+Re = V*x/v; // Reynolds number
+
+// From eqn 4.8.12
+
+d = 5*x/(Re^(1/2))*1000 ; // [mm]
+printf("Thickness of Boundary layer at x = 1.5 is %f mm \n",d)
+
+Re_l = V*l/v;
+
+// From eqn 4.8.19 and 4.8.16
+
+c_f = 1.328*Re_l^-(1/2); // drag coefficient
+printf("Drag Coefficient c_f = %f \n",c_f);
+
+F_d = 0.00409*(1/2)*rho*(2*l*b)*1^2;
+printf("Drag Force F_D = %f N",F_d);
\ No newline at end of file diff --git a/530/CH4/EX4.5/example_4_5.sce b/530/CH4/EX4.5/example_4_5.sce new file mode 100755 index 000000000..569ff6cdd --- /dev/null +++ b/530/CH4/EX4.5/example_4_5.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 4
+// Principles of Fluid Flow
+
+// Example 4.5
+// Page 195
+printf("Example 4.5, Page 195 \n\n");
+
+l = 2 ; // [m]
+v = 4 ; // [m/s]
+
+// From Table A.2
+
+mew = 18.1*10^-6; // [N s/m^2]
+rho = 1.205*1.5; // [kg/m^3]
+
+Re_l = rho*v*l/mew;
+// Boundary layer is partly laminar and partly turbulent, we shall use eqn 4.10.4
+Cf = 0.074*(7.989*10^5)^(-0.2) - 1050/Re_l ;
+printf("Drag coefficieent is %f \n",Cf)
+
+D_f= Cf*1/2*rho*l*v^2;
+printf("Drag force per meter width = %f N \n",D_f);
+
+//from eqn 4.10.1
+
+x = 3*10^5 * (18.1*10^-6)/(1.808*4);
+printf("Value of x_c is %f m",x);
\ No newline at end of file diff --git a/530/CH5/EX5.1.a/example_5_1a.sce b/530/CH5/EX5.1.a/example_5_1a.sce new file mode 100755 index 000000000..b350271a6 --- /dev/null +++ b/530/CH5/EX5.1.a/example_5_1a.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 5
+// Heat Transfer by Forced Convection
+
+
+// Example 5.1(a)
+// Page 209
+printf("Example 5.1(a) \n\n")
+
+D = 0.015 ; // [m]
+Q = 0.05 ; // [m^3/h]
+H = 1000 ; // [W/m^2]
+T_b = 40 ; // [degree C]
+
+// From table A.1, properties at 40 degree C
+k = 0.634 ; // [W/m K]
+v = 0.659*10^-6 ; // [m^2/s]
+
+V_bar = 4*Q/((%pi)*D^2);
+
+Re_D = V_bar*D/v;
+
+// Therefore, Laminar Flow, from eqn 5.2.8
+
+h = 4.364*k/D; // [W/m^2 K]
+
+printf("(a) Local heat transfer coefficient is %f W/m^2 K \n",h);
\ No newline at end of file diff --git a/530/CH5/EX5.1.b/example_5_1b.sce b/530/CH5/EX5.1.b/example_5_1b.sce new file mode 100755 index 000000000..fc3d6371a --- /dev/null +++ b/530/CH5/EX5.1.b/example_5_1b.sce @@ -0,0 +1,34 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 5
+// Heat Transfer by Forced Convection
+
+
+// Example 5.1(b)
+// Page 209
+printf("Example 5.1(b) \n\n")
+
+D = 0.015 ; // [m]
+Q = 0.05 ; // [m^3/h]
+H = 1000 ; // [W/m^2]
+T_b = 40 ; // [degree C]
+
+// From table A.1, properties at 40 degree C
+k = 0.634 ; // [W/m K]
+v = 0.659*10^-6 ; // [m^2/s]
+
+V_bar = 4*Q/((%pi)*D^2);
+
+Re_D = V_bar*D/v;
+
+// Therefore, Laminar Flow, from eqn 5.2.8
+
+h = 4.364*k/D;
+
+// From the definition of h in eqn 5.2.3, the local wal to bulk mean temperature difference is given by
+
+T_w = H/h + T_b;
+
+printf("(b) Wall Temperature Tw = %f degree C",T_w);
\ No newline at end of file diff --git a/530/CH5/EX5.2/example_5_2.sce b/530/CH5/EX5.2/example_5_2.sce new file mode 100755 index 000000000..35587a189 --- /dev/null +++ b/530/CH5/EX5.2/example_5_2.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 5
+// Heat Transfer by Forced Convection
+
+
+// Example 5.2
+// Page 213
+printf("Example 5.2,Page 213 \n\n")
+
+// From eqn 5.2.12 and 4.4.20
+// Let r = Lth/Le
+// r = 0.04305*Pr/0.0575;
+
+function[T]=r(Pr)
+ T = 0.04305*Pr/0.0575
+endfunction
+
+// For Pr = 0.01
+r1 = r(0.01);
+// For Pr = 0.1
+r2 = r(1);
+// For Pr = 100
+r3 = r(100);
+
+printf("Lth/Le at Pr = 0.01 is %f \n",r1);
+printf("Lth/Le at Pr = 1 is %f \n",r2);
+printf("Lth/Le at Pr = 100 is %f",r3);
\ No newline at end of file diff --git a/530/CH5/EX5.3.i/example_5_3i.sce b/530/CH5/EX5.3.i/example_5_3i.sce new file mode 100755 index 000000000..e45eeb248 --- /dev/null +++ b/530/CH5/EX5.3.i/example_5_3i.sce @@ -0,0 +1,42 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 5
+// Heat Transfer by Forced Convection
+
+
+// Example 5.3(i)
+// Page 215
+printf("Example 5.3(i), Page 215 \n\n")
+
+D = 0.015 ; // [m]
+V = 1 ; // [m/s]
+Tw = 90 ; // [degree C]
+Tmi = 50 ; // [degree C]
+Tmo = 65 ; // [degree C]
+
+// (i)
+
+// From Table A.1
+k = 0.656 ; // [W/m K]
+rho = 984.4 ; // [kg/m^3]
+v = 0.497 * 10^-6 ; // [m^2/s]
+Cp = 4178 ; // [J/kg K]
+Pr = 3.12 ;
+rho_in = 988.1 ; // [kg/m^3]
+
+m_dot = %pi*(D^2)*rho_in*V/4 ; // [kg/s]
+
+Re = 4*m_dot/(%pi*D*rho*v) ;
+
+// Using eqn 5.3.2 and 4.6.4a
+f = 0.079*(Re)^-0.25 ;
+
+Nu = (f/2)*(Re-1000)*Pr/[1+12.7*(f/2)^(1/2)*((Pr^(2/3))-1)];
+h = Nu*k/D; // [W/m^2 K]
+
+// From the energy equation, extracting the value of L
+L = m_dot*Cp*(Tmo-Tmi)*[log((Tw-Tmi)/(Tw-Tmo))]/[((Tw-Tmi)-(Tw-Tmo))*h*D*%pi]; // [m]
+
+printf("The length of tube if the exit water temperature is 65 degree C = %f m\n",L);
diff --git a/530/CH5/EX5.3.ii/example_5_3ii.sce b/530/CH5/EX5.3.ii/example_5_3ii.sce new file mode 100755 index 000000000..712bd041e --- /dev/null +++ b/530/CH5/EX5.3.ii/example_5_3ii.sce @@ -0,0 +1,71 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 5
+// Heat Transfer by Forced Convection
+
+
+// Example 5.3(i)
+// Page 215
+printf("Example 5.3(ii), Page 215 \n\n")
+
+D = 0.015 ; // [m]
+V = 1 ; // [m/s]
+Tw = 90 ; // [degree C]
+Tmi = 50 ; // [degree C]
+Tmo = 65 ; // [degree C]
+
+// From Table A.1
+k = 0.656 ; // [W/m K]
+rho = 984.4 ; // [kg/m^3]
+v = 0.497 * 10^-6 ; // [m^2/s]
+Cp = 4178 ; // [J/kg K]
+Pr = 3.12 ;
+rho_in = 988.1 ; // [kg/m^3]
+
+m_dot = %pi*(D^2)*rho_in*V/4 ; // [kg/s]
+
+Re = 4*m_dot/(%pi*D*rho*v) ;
+
+// Using eqn 5.3.2 and 4.6.4a
+f = 0.079*(Re)^-0.25 ;
+
+Nu = (f/2)*(Re-1000)*Pr/[1+12.7*(f/2)^(1/2)*((Pr^(2/3))-1)];
+h = Nu*k/D; // [W/m^2 K]
+
+// From the energy equation, extracting the value of L
+L = m_dot*Cp*(Tmo-Tmi)*[log((Tw-Tmi)/(Tw-Tmo))]/[((Tw-Tmi)-(Tw-Tmo))*h*D*%pi]; // [m]
+
+// (ii)
+printf("\nTrial and error method \n");
+
+// Trial 1
+printf("Trial 1\n");
+printf("Assumed value of Tmo = 70 degree C\n");
+T_mo = 70 ; // [degree C]
+T_b = 60 ; // [degree C]
+
+k1 = 0.659 ; // [W/m K]
+rho1 = 983.2 ; // [kg/m^3]
+v1 = 0.478 * 10^-6 ; // [m^2/s]
+Cp1 = 4179 ; // [J/kg K]
+Pr1 = 2.98 ;
+
+Re1 = 4*m_dot/(%pi*D*rho1*v1);
+
+// From Blasius eqn (4.6.4a), we get
+f1 = 0.005928;
+
+// Substituting this value into the Gnielinski Eqn
+Nu_d = 154.97;
+h = Nu_d*k1/D ; // [W/m^2 K]
+
+// from eqn 5.3.3, we get
+Tmo1 = 73.4 ; // [degree C]
+printf("Value of Tmo obtained = 73.4 degree C\n");
+
+// Trial 2
+printf("Trial 2\n");
+printf("Assume Tmo = 73.4 degree C\n");
+printf("Value of Tmo obtained = 73.6 degree C which is in reasonably close agreement with assumed value.\n")
diff --git a/530/CH5/EX5.4/example_5_4.sce b/530/CH5/EX5.4/example_5_4.sce new file mode 100755 index 000000000..0feb272ec --- /dev/null +++ b/530/CH5/EX5.4/example_5_4.sce @@ -0,0 +1,33 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 5
+// Heat Transfer by Forced Convection
+
+
+// Example 5.4
+// Page 219
+printf("Example 5.4, Page 219 \n\n")
+
+D_i = 0.05 ; // [m]
+m = 300 ; // [kg/min]
+m1 = m/60 ; // [kg/sec]
+rho = 846.7 ; // [kg/m^3]
+k = 68.34 ; // [W/m K]
+c = 1274; // [J/kg K]
+v = 0.2937*10^-6 ; // [m^2/s]
+Pr = 0.00468 ;
+
+Re_D = 4*m1/(%pi*D_i*rho*v);
+
+// Assuming both temperature and velocity profile are fully developed over the length of tube
+// using eqn 5.3.6
+Nu_D = 6.3 + 0.0167*(Re_D^0.85)*(Pr^0.93);
+
+h = Nu_D*k/D_i;
+
+// Equating the heat transferred through the wall of the tube to the change of enthalpy pf sodium
+L = 300/60*1274*(500-400)/(h*%pi*D_i*30)
+
+printf("Length of tube over which the temperature rise occurs = %f m",L)
\ No newline at end of file diff --git a/530/CH5/EX5.5/example_5_5.sce b/530/CH5/EX5.5/example_5_5.sce new file mode 100755 index 000000000..d4485f0b0 --- /dev/null +++ b/530/CH5/EX5.5/example_5_5.sce @@ -0,0 +1,52 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 5
+// Heat Transfer by Forced Convection
+
+
+// Example 5.5
+// Page 231
+printf("Example 5.5, Page 231 \n")
+
+V = 15 ; // [m/s]
+s=0.2 ; // [m]
+T_m = (20+60)/2; // [degree C]
+// Properties at mean temp = 40 degree C
+v = 16.96*10^-6; // [m^2/s]
+rho = 1.128 ; // [kg/m^3]
+k = 0.0276; // [W/m K]
+Pr = 0.699;
+A=s^2;
+Re_L = V*0.2/v;
+// This is less than 3*10^5, hence the boundary layer may be assumed to be laminar over the entire length.
+// from eqn 4.8.19
+
+Cf = 1.328/(Re_L)^0.5
+Fd = 2*Cf*1/2*rho*A*V^2;
+
+// From eqn 5.5.10
+Nu_l = 0.664*(Pr^(1/3))*(Re_L^(1/2));
+
+h = Nu_l*k/s;
+// Therefore rate of heat transfer q is
+q = 2*A*h*(60-20);// [W]
+
+// With a turbulent boundary layer from leading edge, the drag coefficient is given by eqn 4.10.4
+Cf1 = 0.074*(Re_L)^(-0.2);
+Fd1 = 2*Cf1*1/2*rho*A*V^2; // [N]
+
+// from eqn 5.8.3 with C1 = 0
+Nu_l1 = 0.0366*(0.699^(1/3))*(Re_L^(0.8));
+
+h1 = Nu_l1*k/s; // [W/m^2 K]
+q1 = 2*A*h1*(60-20);
+
+printf("For Laminar Boundary Layer \n");
+printf("Rate of Heat transfer = %f W\n",q);
+printf("Drag force = %f N \n \n",Fd)
+
+printf("For Turbulent Boundary Layer from the leading edge \n");
+printf("Rate of Heat transfer = %f W\n",q1);
+printf("Drag force = %f N\n",Fd1)
diff --git a/530/CH5/EX5.6.i/example_5_6i.sce b/530/CH5/EX5.6.i/example_5_6i.sce new file mode 100755 index 000000000..e69b7dba9 --- /dev/null +++ b/530/CH5/EX5.6.i/example_5_6i.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 5
+// Heat Transfer by Forced Convection
+
+
+// Example 5.6(i)
+// Page 235
+printf("Example 5.6(i), Page 235 \n\n")
+
+D = 0.075 ; // [m]
+V = 1.2 ; // [m/s]
+T_air = 20 ; // [degree C]
+T_surface = 100 ; // [degree C]
+T_m = (T_air+T_surface)/2;
+
+v = 18.97*10^-6 ; // [m^2/s]
+k = 0.0290 ; // [W/m K]
+Pr = 0.696 ;
+
+Re_D = V*D/v;
+
+Nu = 0.3 + [(0.62*(Re_D^(1/2))*(Pr^(1/3)))/[(1+((0.4/Pr)^(2/3)))^(1/4)]]*([1+((Re_D/282000)^(5/8))]^(4/5)) ;
+
+h = Nu*k/D ; // [W/m^2 K]
+
+flux = h*(T_surface - T_air); // [W/m^2]
+q = flux*%pi*D*1; // [W/m]
+
+printf("Heat transfer rate per unit length = %f W/m\n",q);
diff --git a/530/CH5/EX5.6.ii/example_5_6ii.sce b/530/CH5/EX5.6.ii/example_5_6ii.sce new file mode 100755 index 000000000..a1982a875 --- /dev/null +++ b/530/CH5/EX5.6.ii/example_5_6ii.sce @@ -0,0 +1,49 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 5
+// Heat Transfer by Forced Convection
+
+
+// Example 5.6(ii)
+// Page 235
+printf("Example 5.6(ii), Page 235 \n\n")
+
+D = 0.075 ; // [m]
+V = 1.2 ; // [m/s]
+T_air = 20 ; // [degree C]
+T_surface = 100 ; // [degree C]
+T_m = (T_air+T_surface)/2;
+
+v = 18.97*10^-6 ; // [m^2/s]
+k = 0.0290 ; // [W/m K]
+Pr = 0.696 ;
+
+Re_D = V*D/v;
+Nu = 0.3 + [(0.62*(Re_D^0.5)*(Pr^(1/3)))/[(1+((0.4/Pr)^(2/3)))^(1/4)]]*[1+(Re_D/282000)^(5/8)]^(5/8) ;
+h = Nu*k/D ; // [W/m^2 K]
+flux = h*(T_surface - T_air); // [W/m^2]
+
+// (ii) Using Trial and error method
+T_avg = 1500/flux*(T_surface - T_air);
+
+T_assumd = 130 ; // [degree C]
+Tm= 75 ; // [degree C]
+
+v1 = 20.56*10^-6 ; // [m^2/s]
+k1 = 0.0301 ; // [W/m K]
+Pr1 = 0.693 ;
+
+Re_D1 = V*D/v1;
+
+
+// Using eqn 5.9.8
+Nu1 = 33.99;
+h = Nu1*k1/D;
+// Therefore
+T_diff = 1500/h; // [degree C]
+T_avg_calc = 129.9 ; // [degree C]
+printf("Assumed average wall temperature = %f degree C\n",T_assumd);
+printf("Calculated average wall Temperature = %f degree C\n",T_avg_calc);
+printf("Hence,Average wall Temperature = %f degree C",T_avg_calc);
\ No newline at end of file diff --git a/530/CH5/EX5.7.i/example_5_7i.sce b/530/CH5/EX5.7.i/example_5_7i.sce new file mode 100755 index 000000000..c846e620b --- /dev/null +++ b/530/CH5/EX5.7.i/example_5_7i.sce @@ -0,0 +1,41 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 5
+// Heat Transfer by Forced Convection
+
+
+// Example 5.7(i)
+// Page 241
+printf("Example 5.7(i), Page 241 \n \n");
+
+// Given data
+D = 0.0125 ; // [m]
+ST = 1.5*D ;
+SL = 1.5*D ;
+V_inf = 2 ; // [m/s]
+
+N = 5;
+Tw = 70; // [degree C]
+Tmi = 30; // [degree C]
+L = 1; // [m]
+// Properties of air at 30 degree C
+rho = 1.165 ; // [kg/m^3]
+v = 16.00 *10^-6 ; // [m^2/s]
+Cp = 1.005 ; // [kJ/kg K]
+k = 0.0267 ; // [W/m K]
+Pr = 0.701;
+
+// From eqn 5.10.2
+Vmax = ST/(SL-D)*V_inf ; // [m/s]
+Re = Vmax*D/v ;
+
+// From fig 5.15
+f = 0.37/4;
+// Also, tube arrangement is square
+X = 1;
+// From eqn 5.10.6
+delta_P = 4*f*N*X*(rho*Vmax^2)/2 ; // [N/m^2]
+
+printf("(i) Pressure drop of air across the bank is %f N/m^2 \n",delta_P);
\ No newline at end of file diff --git a/530/CH5/EX5.7.ii/example_5_7ii.sce b/530/CH5/EX5.7.ii/example_5_7ii.sce new file mode 100755 index 000000000..9ac81bce3 --- /dev/null +++ b/530/CH5/EX5.7.ii/example_5_7ii.sce @@ -0,0 +1,65 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 5
+// Heat Transfer by Forced Convection
+
+
+// Example 5.7(ii)
+// Page 241
+printf("Example 5.7(ii), Page 241 \n \n");
+
+D = 0.0125 ; // [m]
+ST = 1.5*D ;
+SL = 1.5*D ;
+V_inf = 2 ; // [m/s]
+N = 5;
+Tw = 70; // [degree C]
+Tmi = 30; // [degree C]
+L = 1; // [m]
+
+rho = 1.165 ; // [kg/m^3]
+v = 16.00 *10^-6 ; // [m^2/s]
+Cp = 1.005*1000 ; // [J/kg K]
+k = 0.0267 ; // [W/m K]
+Pr = 0.701;
+
+// From eqn 5.10.2
+Vmax = ST/(SL-D)*V_inf ; // [m/s]
+Re = Vmax*D/v ;
+
+// From fig 5.15
+f = 0.37/4;
+// Also, tube arrangement is square
+X = 1;
+// From eqn 5.10.6
+delta_P = 4*f*N*X*(rho*Vmax^2)/2 ; // [N/m^2]
+
+// At 70 degree C
+Pr1 = 0.694 ;
+// From table 5.4 and 5.5
+
+C1 = 0.27;
+m = 0.63;
+C2 = 0.93;
+
+// Substituting in Eqn 5.10.5
+Nu = C1*C2*(Re^m)*(Pr^0.36)*(Pr/Pr1)^(1/4);
+h = Nu*k/D; // [W/m^2 K]
+
+// For 1 m long tube
+m_dot = rho*(10*1.5*D*1)*2; // [kg/s]
+
+// Substituting m_dot in 5.3.4 and solving, we get
+function[f]=temp(Tmo)
+ f(1) = h*(%pi*D*L)*50*[(Tw-Tmi)-(Tw-Tmo(1))]/[log((Tw-Tmi)/(Tw-Tmo(1)))]-m_dot*Cp*(Tmo(1)-Tmi) ;
+ // h*(%pi*D*L)*N*((Tw-Tmi)-(Tw-Tmo))/log[(Tw-Tmi)/(Tw-Tmo)] - m_dot*Cp*(Tmo - Tmi);
+ funcprot(0);
+endfunction
+
+Tmo = 40; // Initial assumed value for fsolve function
+y = fsolve(Tmo,temp);
+printf("Tmo = %f \n",y);
+
+printf("(ii) Exit temperature of air = %f degree C \n",y);
\ No newline at end of file diff --git a/530/CH5/EX5.7.iii/example_5_7iii.sce b/530/CH5/EX5.7.iii/example_5_7iii.sce new file mode 100755 index 000000000..c21befb76 --- /dev/null +++ b/530/CH5/EX5.7.iii/example_5_7iii.sce @@ -0,0 +1,67 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 5
+// Heat Transfer by Forced Convection
+
+
+// Example 5.7(iii)
+// Page 241
+printf("Example 5.7(iii), Page 241 \n \n");
+
+D = 0.0125 ; // [m]
+ST = 1.5*D ;
+SL = 1.5*D ;
+V_inf = 2 ; // [m/s]
+N = 5;
+Tw = 70; // [degree C]
+Tmi = 30; // [degree C]
+L = 1; // [m]
+
+rho = 1.165 ; // [kg/m^3]
+v = 16.00 *10^-6 ; // [m^2/s]
+Cp = 1.005*1000 ; // [J/kg K]
+k = 0.0267 ; // [W/m K]
+Pr = 0.701;
+
+// From eqn 5.10.2
+Vmax = ST/(SL-D)*V_inf ; // [m/s]
+Re = Vmax*D/v ;
+
+// From fig 5.15
+f = 0.37/4;
+// Also, tube arrangement is square
+X = 1;
+// From eqn 5.10.6
+delta_P = 4*f*N*X*(rho*Vmax^2)/2 ; // [N/m^2]
+
+// At 70 degree C
+Pr1 = 0.694 ;
+// From table 5.4 and 5.5
+
+C1 = 0.27;
+m = 0.63;
+C2 = 0.93;
+
+// Substituting in Eqn 5.10.5
+Nu = C1*C2*(Re^m)*(Pr^0.36)*(Pr/Pr1)^(1/4);
+h = Nu*k/D; // [W/m^2 K]
+
+// For 1 m long tube
+m_dot = rho*(10*1.5*D*1)*2; // [kg/s]
+
+// Substituting m_dot in 5.3.4 and solving, we get
+function[f]=temp(Tmo)
+ f(1) = h*(%pi*D*L)*50*[(Tw-Tmi)-(Tw-Tmo(1))]/[log((Tw-Tmi)/(Tw-Tmo(1)))]-m_dot*Cp*(Tmo(1)-Tmi) ;
+ // h*(%pi*D*L)*N*((Tw-Tmi)-(Tw-Tmo))/log[(Tw-Tmi)/(Tw-Tmo)] - m_dot*Cp*(Tmo - Tmi);
+ funcprot(0);
+endfunction
+
+Tmo = 40; // Initial assumed value for fsolve function
+y = fsolve(Tmo,temp);
+
+// Heat transfer rate q
+q = h*(%pi*D*L)*50*((Tw-Tmi)-(Tw-y))/(log((Tw-Tmi)/(Tw-y)));
+
+printf("(iii) Heat transfer rate per unit length to air = %f W",q);
\ No newline at end of file diff --git a/530/CH6/EX6.1/example_6_1.sce b/530/CH6/EX6.1/example_6_1.sce new file mode 100755 index 000000000..2f50868ca --- /dev/null +++ b/530/CH6/EX6.1/example_6_1.sce @@ -0,0 +1,54 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 6
+// Heat Transfer by Natural Convection
+
+
+// Example 6.1
+// Page 258
+printf("Example 6.1, Page 258 \n \n");
+
+H = 0.5 ; // [m]
+T_h = 100; // [degree C]
+T_l = 40; // [degree C]
+
+v = 20.02*10^-6 ; // [m/s]
+Pr = 0.694;
+k = 0.0297; // [W/m K]
+
+T = (T_h+T_l)/2 + 273 ; // [K]
+printf("Mean film temperature = %f K \n",T);
+B = 1/T;
+
+Gr = 9.81*B*((T_h-T_l)*H^3)/(v^2);
+Ra = Gr*Pr;
+
+// (a)
+// Exact analysis - Equation 6.2.17
+disp("(a)");
+printf("Exact analysis\n");
+Nu_a = 0.64*(Gr^(1/4))*(Pr^0.5)*((0.861+Pr)^(-1/4));
+printf("Nu_L = %f \n",Nu_a);
+
+// (b)
+// Integral method - Equation 6.2.29
+disp("(b)");
+printf("Integral method \n");
+Nu_b = 0.68*(Gr^(1/4))*(Pr^0.5)*((0.952+Pr)^(-1/4));
+printf("Nu_L = %f \n",Nu_b);
+
+// (c)
+// McAdams correlation - Equation 6.2.30
+disp("(c)");
+printf("McAdams correlation \n");
+Nu_c = 0.59*(Ra)^(1/4);
+printf("Nu_L = %f \n",Nu_c);
+
+// (d)
+// Churchill and Chu correlation - Equation 6.2.31
+disp("(d)")
+printf("Churchill and Chu correlation\n");
+Nu_d = 0.68 + 0.670*(Ra^(1/4))/[1+(0.492/Pr)^(9/16)]^(4/9);
+printf("Nu_L = %f \n",Nu_d);
\ No newline at end of file diff --git a/530/CH6/EX6.2/example_6_2.sce b/530/CH6/EX6.2/example_6_2.sce new file mode 100755 index 000000000..b3c62df32 --- /dev/null +++ b/530/CH6/EX6.2/example_6_2.sce @@ -0,0 +1,41 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 6
+// Heat Transfer by Natural Convection
+
+
+// Example 6.2
+// Page 259
+printf("Example 6.2, Page 259 \n \n");
+
+Tm = 150 ; // [degree C]
+// From table A.2
+v = 28.95*10^-6 ; // [m^2/s]
+Pr = 0.683;
+k = 0.0357 ; // [W/m K]
+
+B = 1/(273+Tm); // [K^-1]
+
+// from eqn 6.2.30
+printf("Equation 6.2.30 \n h = k/L*0.59*[9.81*B*(Tw-Tinf)*(L^3)*0.683/(v^2)]^(1/4)\n")
+// h = k/L*0.59*[9.81*B*(Tw-Tinf)*(L^3)*0.683/(v^2)]^(1/4);
+// simplifying we get
+// h = 1.38*[(Tw-Tinf)/L]^(1/4)
+printf("Reduces to h = 1.38*[(Tw-Tinf)/L]^(1/4) \n")
+
+
+// From eqn 6.2.33
+// h*L/k = 0.10*[9.81*B*(Tw-Tinf)*(L^3)*0.683/(v^2)]^(1/3);
+printf("Equation 6.2.33 \n h*L/k = 0.10*[9.81*B*(Tw-Tinf)*(L^3)*0.683/(v^2)]^(1/3) \n");
+// simplifying
+// h = 0.95*(Tw-Tinf)^1/3
+printf("Reduces to h = 0.95*(Tw-Tinf)^1/3 \n");
+
+printf("where h is expressed in W/m^2 K, (Tw-Tinf) in C and L in metres \n");
+
+
+
+
+
diff --git a/530/CH6/EX6.3/example_6_3.sce b/530/CH6/EX6.3/example_6_3.sce new file mode 100755 index 000000000..fb624c2dd --- /dev/null +++ b/530/CH6/EX6.3/example_6_3.sce @@ -0,0 +1,51 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 6
+// Heat Transfer by Natural Convection
+
+
+// Example 6.3
+// Page 260
+printf("Example 6.3, Page 260 \n \n");
+
+s = 0.2 ; // [m]
+d = 0.005 ; // [m]
+rho = 7900 ; // [kg/m^3]
+Cp = 460 ; // [J/kg K]
+
+T_air = 20 ; // [C]
+// For 430 C to 330 C
+T_avg = 380 ; // [C]
+Tm = (T_avg + T_air)/2 ; // [C]
+
+
+v = 34.85*10^-6 ; // [m^2/s]
+Pr = 0.680 ;
+k = 0.0393 ; // [W/m K]
+
+Re = 9.81*1/(273+Tm)*(T_avg-T_air)*(s^3)/(v^2)*Pr;
+
+// From eqn 6.2.31
+Nu = 0.68 + 0.670*(Re^(1/4))/[1+(0.492/Pr)^(4/9)]^(4/9);
+
+h = Nu*k/s; // [W/m^2 K]
+t1 = rho*s*s*d*Cp/((s^2)*2*h)*log((430-T_air)/(330-T_air)); // [s]
+printf("Time required for the plate to cool from 430 C to 330 C is %f s\n",t1);
+
+// for 330 to 230
+h2 = 7.348 ; // [W/m^2 K]
+t2 = rho*s*s*d*Cp/((s^2)*2*h2)*log((330-T_air)/(230-T_air)); // [s]
+printf("Time required for the plate to cool from 330 C to 230 C is %f s\n",t2);
+
+// for 230 to 130
+h3 = 6.780; // [W/m^2 K]
+t3 = rho*s*s*d*Cp/((s^2)*2*h3)*log((230-T_air)/(130-T_air)); // [s]
+printf("Time required for the plate to cool from 230 C to 130 C is %f s\n",t3);
+
+// Total time
+
+time = t1+t2+t3;
+minute = time/60;
+printf("Hence, time required for the plate to cool from 430 C to 130 C \n = %f s\n = %f min",time,minute);
diff --git a/530/CH6/EX6.4/example_6_4.sce b/530/CH6/EX6.4/example_6_4.sce new file mode 100755 index 000000000..6ecb437b0 --- /dev/null +++ b/530/CH6/EX6.4/example_6_4.sce @@ -0,0 +1,61 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 6
+// Heat Transfer by Natural Convection
+
+
+// Example 6.4
+// Page 264
+printf("Example 6.4, Page 264 \n \n");
+
+D = 0.006 ; // [m]
+e = 0.1 ;
+Ti = 800 ; // [C]
+Ta = 1000 ; // [C]
+// Rate at which heat gained = net radiant heat, gives h*(Ta-800) = 1306.0 ; // [W/m^2]
+
+// Using trial and error method
+// Trial 1
+printf("Trial 1 \n");
+// Let Ta = 1000 degree C
+printf("Let Ta = 10000 C \n");
+
+Tm = (Ta+Ti)/2;
+// From table A.2
+v = 155.1*10^-6 ; // [m^2/s]
+k = 0.0763 ; // [W/m K]
+Pr = 0.717 ;
+
+Gr = 9.81*1/1173*(200*D^3)/(v^2);
+Ra = Gr*Pr ;
+
+// From eqn 6.3.2
+Nu = 0.36 + 0.518*(Ra^(1/4))/[1+(0.559/Pr)^(9/16)]^(4/9);
+h = Nu*k/D;
+x = h*(Ta-Ti); // [W/m^2]
+printf("Value of h(Ta-800) = %f W/m^2, which is much larger than the required value of 1306 W/m^2 \n",x);
+
+// Trial 2
+printf("\nTrial 2 \n");
+// Let Ta = 900
+printf("Let Ta = 900 C \n");
+Ra2 = 6.42 ;
+Nu2 = 0.9841 ;
+h2 = 12.15 ;
+x2 = h2*(900-800);
+printf("Value of h(Ta-800) = %f W/m^2, which is a little less than the required value of 1306 W/m^2 \n",x2);
+
+// Trial 3
+printf("\nTrial 3 \n");
+// Let Ta = 910
+printf("Let Ta = 910 C \n");
+Ra3 = 6.93 ;
+Nu3 = 0.9963 ;
+h3 = 12.33 ;
+x3 = h3*(910-800);
+printf("Value of h(Ta-800) = %f W/m^2 \nThis value is little more than the required value of 1306 W/m^2 \n",x3);
+// Interpolation
+T = 900 + (910-900)*(1306-x2)/(x3-x2);
+printf("\nThe correct value of Ta obtained by interpolation is %f C",T);
diff --git a/530/CH6/EX6.5/example_6_5.sce b/530/CH6/EX6.5/example_6_5.sce new file mode 100755 index 000000000..0c4952e64 --- /dev/null +++ b/530/CH6/EX6.5/example_6_5.sce @@ -0,0 +1,40 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 6
+// Heat Transfer by Natural Convection
+
+
+// Example 6.5
+// Page 269
+printf("Example 6.5, Page 269 \n \n");
+
+T_p = 75 ; // Temperature of absorber plate , degree C
+T_c = 55 ; // Temperature of glass cover , degree C
+L = 0.025 ; // [m]
+
+H = 2 ; // [m]
+Y = 70 ; // degree
+
+a = 19/180*%pi ; // [Radians]
+
+r = H/L ;
+
+T_avg = (T_p+T_c)/2+273 ; // [K]
+// Properties at 65 degree C
+k = 0.0294 ; // [W/m K]
+v = 19.50*10^-6 ; // [m^2/s]
+Pr = 0.695 ;
+
+Ra = 9.81*(1/T_avg)*(T_p-T_c)*(L^3)/(v^2)*Pr*cos(a);
+
+// From eqn 6.4.3
+Nu = 0.229*(Ra)^0.252;
+
+h = Nu*k/L ; // [W/m^2 K]
+
+Rate = h*2*1*(T_p-T_c); // [W]
+
+printf("Heat transfer rate = %f W",Rate);
+
diff --git a/530/CH6/EX6.6/example_6_6.sce b/530/CH6/EX6.6/example_6_6.sce new file mode 100755 index 000000000..ff28ef8c4 --- /dev/null +++ b/530/CH6/EX6.6/example_6_6.sce @@ -0,0 +1,35 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 6
+// Heat Transfer by Natural Convection
+
+
+// Example 6.6
+// Page 270
+printf("Example 6.6, Page 270 \n \n");
+
+T_air = 30 ; // [C]
+D = 0.04 ; // [m]
+T_s = 70 ; // surface temperature, [C]
+V = 0.3 ; // [m/s]
+
+Tm = (T_air + T_s)/2 ; // [C]
+// Properties at Tm
+v = 17.95*10^-6 ; // [m^2/s]
+Pr = 0.698 ;
+k = 0.0283 ; // [W/m K]
+
+Gr = 9.81*1/323*(T_s-T_air)*(D^3)/v^2;
+Re = V*D/v ;
+X = Gr/Re^2 ;
+printf("Since Gr/Re^2 = %f is > 0.2, we have a combined convection situation. \n\n",X);
+
+// From Eqn 5.9.8
+Nu_forced = 0.3 + 0.62*(Re^0.5)*(Pr^(1/3))/[[1+(0.4/Pr)^(2/3)]^(1/4)]*[1+(Re/282000)^(5/8)]^(4/5);
+
+// Substituting in Eqn 6.5.1
+Nu = Nu_forced*[1+6.275*(X)^(7/4)]^(1/7);
+h = Nu*(k/D);
+printf("The Average heat transfer coefficient = %f W/m^2 K",h);
\ No newline at end of file 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 diff --git a/530/CH8/EX8.1/example_8_1.sce b/530/CH8/EX8.1/example_8_1.sce new file mode 100755 index 000000000..2032008c0 --- /dev/null +++ b/530/CH8/EX8.1/example_8_1.sce @@ -0,0 +1,44 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 8
+// Condensation and Boiling
+
+
+// Example 8.1
+// Page 318
+printf("Example 8.1, Page 318 \n \n");
+Ts = 80 ; // [C]
+Tw = 70 ; // [C]
+L = 1 ; // [m]
+g = 9.8 ; // [m/s^2]
+
+// Assuming condensate film is laminar and Re < 30
+Tm = (Ts + Tw)/2 ;
+// From table A.1
+rho = 978.8 ; // [kg/m^3]
+k = 0.672 ; // [W/m K]
+u = 381 *10^-6 ; // [kg/m s]
+v = u/rho ;
+// At 80 C,
+lambda = 2309 ; // [kJ/kg]
+// Substituting in eqn 8.3.9, we get
+h = 0.943*[(lambda*1000*(rho^2)*g*(k^3))/((Ts-Tw)*u*L)]^(1/4); // [W/m^2 K]
+
+rate = h*L*(Ts-Tw)/(lambda*1000); // [kg/m s]
+Re = 4*rate/u;
+printf("Assuming condensate film is laminar and Re < 30 \n");
+printf("h = %f W/m^2 K\n",h);
+printf("Re_L = %f \n",Re);
+printf("Initial assumption was wrong, Now considering the effect of ripples, we get\n");
+
+// Substituting h = Re*(lambda*1000)*u/(4*L*(Ts-Tw)), in eqn 8.3.12
+Re = [[[4*L*(Ts-Tw)*k/(lambda*1000*u)*(g/(v^2))^(1/3)]+5.2]/1.08]^(1/1.22);
+// From eqn 8.3.12
+h = [Re/(1.08*(Re^1.22)-5.2)]*k*((g/v^2)^(1/3)); // [W/m^2 K]
+m = h*L*10/(lambda*1000); // rate of condensation , [kg/m s]
+
+printf("Re = %f \n",Re);
+printf("Heat Transfer Cofficient = %f W/m^2 K \n",h);
+printf("Rate of condensation = %f kg/m s",m);
\ No newline at end of file diff --git a/530/CH8/EX8.2/example_8_2.sce b/530/CH8/EX8.2/example_8_2.sce new file mode 100755 index 000000000..77775b641 --- /dev/null +++ b/530/CH8/EX8.2/example_8_2.sce @@ -0,0 +1,34 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 8
+// Condensation and Boiling
+
+
+// Example 8.2
+// Page 321
+printf("Example 8.2, Page 321 \n \n");
+
+Ts = 262 ; // [K]
+D = 0.022 ; // [m]
+Tw = 258 ; // [K]
+
+Tm = (Ts+Tw)/2;
+// Properties at Tm
+rho = 1324 ; // [kg/m^3]
+k = 0.1008 ; // [W/m K]
+v = 1.90*10^-7 // [m^2/s];
+lambda = 215.1*10^3 ; // [J/kg]
+g = 9.81 ; // [m/s^2]
+u = v*rho ; // Viscosity
+
+// From eqn 8.4.1
+h = 0.725*[lambda*(rho^2)*g*(k^3)/((Ts-Tw)*u*D)]^(1/4);
+
+rate = h*%pi*D*(Ts-Tw) /lambda ; // [kg/s m]
+Re = 4*rate/u ;
+
+printf("Heat transfer coefficient = %f W/m^2 K\n",h);
+printf("Condensation rate per unit length = %f kg/s m \n",rate);
+printf("Film Reynolds number = %f \n",Re);
diff --git a/530/CH8/EX8.3/example_8_3.sce b/530/CH8/EX8.3/example_8_3.sce new file mode 100755 index 000000000..3e8594cc8 --- /dev/null +++ b/530/CH8/EX8.3/example_8_3.sce @@ -0,0 +1,68 @@ +clear;
+clc;
+
+// A TeTwtbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 8
+// Condensation and Boiling
+
+
+// ETwample 8.3
+// Page 322
+printf("Example 8.3, Page 322 \n \n");
+
+m = 25/60 ; // [kg/sec]
+ID = 0.025 ; // [m]
+OD = 0.029 ; // [m]
+Tci = 30 ; // [C]
+Tce = 70 ; // [C]
+g = 9.8 ; // [m/s^2]
+
+Ts = 100 ; // [C]
+// Assuming 5.3.2 is valid, properties at 50 C
+// Properties at Tm
+rho = 988.1 ; // [kg/m^3]
+k = 0.648 ; // [W/m K]
+v = 0.556*10^-6 // [m^2/s];
+Pr = 3.54 ;
+Re = 4*m/(%pi*ID*rho*v);
+// From eqn 4.6.4a
+f = 0.005635;
+// From eqn 5.3.2
+Nu = 198.39 ;
+h = Nu*k/ID ;
+
+// Assuming average wall temperature = 90 C
+Tw = 90 ; // [C]
+Tm = (Tw+Ts)/2;
+// Properties at Tm
+// Properties at Tm
+rho = 961.9 ; // [kg/m^3]
+k = 0.682 ; // [W/m K]
+u = 298.6*10^-6 ; // [kg/m s]
+lambda = 2257*10^3 ; // [J/kg]
+
+h = 0.725*[lambda*(rho^2)*g*(k^3)/((Ts-Tw)*u*OD)]^(1/4);
+// Equating the heat flow from the condensing steam to the tube wall, to the heat flow from the tube wall to the flowing water.
+// Solving the simplified equation
+function[f] =temp(Tw)
+ f=(100-Tw)^(3/4)-8.3096/[log((Tw-Tci)/(Tw-Tce))];
+ funcprot(0);
+endfunction
+
+T=fsolve(Tw,temp);
+printf("Temperature obtained from trial and error = %f C \n",T);
+
+// Therefore
+hc = 21338.77/(100-T)^(1/4); // [W/m^2 K]
+printf("h_c = %f W/m^2 K \n",hc);
+
+// Now, equating the heat flowing from the condensing steam to the tube wall to the heat gained by the water, we have
+function[g] =lngth(l)
+ g=hc*(%pi*OD*l)*(100-T)-m*4174*(Tce-Tci);
+ funcprot(0);
+endfunction
+
+l = 0; // (initial guess, assumed value for fsolve function)
+L = fsolve(l,lngth);
+printf("\nLength of the tube = %f m \n",L);
+
diff --git a/530/CH8/EX8.4/example_8_4.sci b/530/CH8/EX8.4/example_8_4.sci new file mode 100755 index 000000000..4651e6edd --- /dev/null +++ b/530/CH8/EX8.4/example_8_4.sci @@ -0,0 +1,129 @@ +clear;
+clc;
+
+//Properties at (Tw+Ts)/2 = 100.5 degree celsius
+deltaT1 = 1; //in degree celsius
+p1 = 7.55e-4; //[K^(-1) p1 is coefficient of cubical expansion
+v1 = 0.294e-6; //[m^2/sec] viscosity at 100.5 degree celsius
+k1 = 0.683; //[W/m-k]thermal conductivity
+Pr1 = 1.74; //Prandtl number
+g = 9.81; //acceleration due to gravity
+L = 0.14e-2; //diameter in meters
+//Properties at (Tw+Ts)/2 =102.5
+deltaT2 = 5; //in degree celsius
+p2 = 7.66e-4; //[K^(-1) p1 is coefficient of cubical expansion
+v2 = 0.289e-6; //[m^2/sec] viscosity at 102.5 degree celsius
+k2 = 0.684; //[W/m-k]thermal conductivity
+Pr2 = 1.71; //Prandtl number
+//Properties at (Tw+Ts)/2 =105
+deltaT3 = 10; //in degree celsius
+p3 = 7.80e-4; //[K^(-1) p1 is coefficient of cubical expansion
+v3 = 0.284e-6; //[m^2/sec] viscosity at 105 degree celsius
+k3 = 0.684; //[W/m-k]thermal conductivity
+Pr3 = 1.68; //Prandtl number
+
+function[Ra]=Rayleigh_no(p,deltaT,v,Pr)
+ Ra = [(p*g*deltaT*L^3)/(v^2)]*Pr;
+ funcprot(0);
+endfunction
+
+function[q] = flux(k,deltaT,Rai,v)
+ q=(k/L)*(deltaT)*{0.36+(0.518*Rai^(1/4))/[1+(0.559/v)^(9/16)]^(4/9)};
+ funcprot(0);
+endfunction
+
+Ra = Rayleigh_no(p1,deltaT1,v1,Pr1);
+q1 = flux(k1,deltaT1,Ra,Pr1);
+printf("\n q/A = %.1f W/m^2 at (Tw-Ts)=1",q1);
+Ra = Rayleigh_no(p2,deltaT2,v2,Pr2);
+q2 = flux(k2,deltaT2,Ra,Pr2);
+printf("\n q/A = %.1f W/m^2 at (Tw-Ts)=5",q2);
+Ra = Rayleigh_no(p3,deltaT3,v3,Pr3);
+q3 = flux(k3,deltaT3,Ra,Pr3);
+printf("\n q/A = %.1f W/m^2 at (Tw-Ts)=10",q3);
+
+//At 100 degree celsius
+Cpl = 4.220; //[kJ/kg]
+lamda = 2257; //[kJ/kg]
+ul = 282.4e-6; //viscosity is in kg/m-sec
+sigma = 589e-4; //Surface tension is in N/m
+pl = 958.4; //density in kg/m^3
+pv =0.598; //density of vapour in kg/m^3
+deltap = pl-pv;
+Prl = 1.75; //Prandtl no. of liquid
+Ksf = 0.013;
+function[q1]=heat_flux(deltaT)
+ q1=141.32*deltaT^3;
+ funcprot(0);
+endfunction
+
+printf("\n q/A at deltaT = 5 degree celsius = %.1f W/m^2",heat_flux(5));
+printf("\nq/A at deltaT = 10 degree celsius = %.1f W/m^2",heat_flux(10));
+printf("\n q/A at deltaT =20 degree celsius = %.1f W/m^2",heat_flux(20));
+//qi = [heat_flux(5),heat_flux(10),heat_flux(20)];
+q = [q1 q2 q3];
+i=1;
+while i<=10
+ T(i)=i;
+ ql(i) = heat_flux(i);
+ i=i+1;
+end
+plot2d([1 5 10],q);
+plot2d(T,ql);
+xtitle("Boiling curve","(Tw - Ts)degree celsius","Heat flux,(q/A)W/m^2");
+L1 = (L/2)*[g*(pl-pv)/sigma]^(1/2);
+printf("\n Peak heat flux L = %.3f ",L1);
+f_L = 0.89+2.27*exp(-3.44*L1^(0.5));
+printf("\n f(l) = %.4f",f_L);
+q2 = f_L*{(%pi/24)*lamda*10^(3)*pv^(0.5)*[sigma*g*(pl-pv)]^(0.25)};
+printf("\n q/A = %.3e W/m^2",q2);
+
+Tn = poly([0],'Tn');
+Tn1 = roots(141.32*Tn^3 - q2);
+printf("\n Tw-Ts = %.1f degree celsius",Tn1(3));
+
+
+
+printf("\n\n Minimum heat flux");
+q3 = 0.09*lamda*10^3*pv*[sigma*g*(pl-pv)/(pl+pv)^(2)]^(0.25);
+printf("\n q/A = %d W/m^2",q3);
+printf("\n\n Stable film boiling");
+Ts1 = 140; //surface temperature in degree celsius
+Ts2 = 200; //surface temperature in degree celsius
+Ts3 = 600; //surface temperature in degree celsius
+Twm1 = (140+100)/2; //Mean film temperature
+//properties of steam at 120 degree celsius and 1.013 bar
+kv = 0.02558; //thermal conductivity in W/mK
+pv1 = 0.5654; //vapor density in kg/m^3
+uv=13.185*10^(-6); //viscosity of vapour in kg/m sec
+lamda1 = (2716.1-419.1)*10^(3);//Latent heat of fusion in J/kg
+hc = 0.62*[(kv^3)*pv*(pl-pv)*g*lamda1/(L*uv*(140-100))]^(0.25);
+printf("\n hc = %.2f W/m^2",hc);
+qrad = 5.67*10^(-8)*(413^4 - 373^4)/[(1/0.9)+1-1];
+printf("\n q/A due to radiation = %.2f W/m^2",qrad);
+hr = qrad/(413-373);
+printf("\n hr = %.2f W/m^2 K ",hr);
+
+printf("\n Since hr<hc ");
+printf("\n The total heat transfer coefficient ");
+h = hc + 0.75*hr;
+printf(" h = %.2f W/m^2 K",h);
+printf("\n Total heat flux = %.3f W/m^2 K",h*(140-100));
+
+hc_200 = 0.62*[(kv^3)*pv*(pl-pv)*g*lamda1/(L*uv*(200-100))]^(0.25);
+qrad1 = 5.67*10^(-8)*(473^4 - 373^4)/[(1/0.9)+1-1];
+hr_200 = qrad1/(200-100);
+printf("\n\n hc = %.2f W/m^2",hc_200);
+printf("\n hr = %.2f W/m^2 K",hr_200);
+printf("\n q/A due to radiation = %.2f W/m^2",qrad1);
+h_200 = hc_200 +0.75*hr_200;
+printf("\n Total heat flux = %d W/m^2",h_200*100);
+hc_600 = 0.62*[(kv^3)*pv*(pl-pv)*g*lamda1/(L*uv*(600-100))]^(0.25);
+qrad2 = 5.67*10^(-8)*(873^4 - 373^4)/[(1/0.9)+1-1];
+hr_600 = qrad1/(600-100)
+printf("\n\n hc = %.2f W/m^2",hc_600);
+printf("\n hr = %.2f W/m^2 K",hr_600);
+printf("\n q/A due to radiation = %.2f W/m^2",qrad2);
+
+
+
diff --git a/530/CH8/EX8.5/example_8_5.sce b/530/CH8/EX8.5/example_8_5.sce new file mode 100755 index 000000000..0874299a5 --- /dev/null +++ b/530/CH8/EX8.5/example_8_5.sce @@ -0,0 +1,45 @@ +clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 8
+// Condensation and Boiling
+
+
+// Example 8.5
+// Page 337
+printf("Example 8.5, Page 337 \n \n");
+
+D = 0.02 ; // [m]
+l = 0.15 ; // [m]
+T = 500+273 ; // [K]
+Tc = -196+273 ; // [K]
+e = 0.4;
+s = 5.670*10^-8;
+// Film boiling will occur, hence eqn 8.7.9 is applicable
+Tm = (T+Tc)/2;
+
+// Properties
+k = 0.0349 ; // [W/m K]
+rho = 0.80 ; // [kg/m^3]
+u = 23*10^-6 ; // [kg/m s]
+
+Cp_avg = 1.048 ; // [kJ/kg J]
+rho_liq = 800 ; // [kg/m^3]
+latent = 201*10^3 ; // [J/kg]
+
+lambda = [latent + Cp_avg*(Tm-Tc)*1000]; // [J/kg]
+h_c = 0.62*[((k^3)*rho*799.2*9.81*lambda)/(D*u*(T-Tc))]^(1/4); // [W/m^2 K]
+
+// Taking the emissivity of liquid surface to be unity and using equation 3.9.1, the exchange of radiant heat flux
+flux = s*(T^4-Tc^4)/(1/e+1/1-1); // [W/m^2]
+h_r = flux/(T-Tc);
+
+// Since h_r < h_c, total heat transfer coefficient is determined from eqn 8.7.11
+h = h_c+3/4*h_r; // [W/m^2 K]
+
+flux_i = h*(T-Tc);
+Rate = flux_i*%pi*D*l; // [W]
+
+printf("Initial heat flux = %f W/m^2 \n",flux_i);
+printf("Initial heat transfer rate = %f W",Rate);
\ No newline at end of file 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 --- /dev/null +++ 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 |