4 files changed, 144 insertions, 0 deletions
diff --git a/536/CH12/EX12.1/Example_12_1.sce b/536/CH12/EX12.1/Example_12_1.sce
new file mode 100755
index 000000000..3cb711638
--- /dev/null
+++ b/536/CH12/EX12.1/Example_12_1.sce
@@ -0,0 +1,33 @@
+clc;
+clear;
+
+printf("\n Example 12.1\n");
+
+d=250e-3; //internal diameter of pipe
+u=15; //Velocity of air through the pipe
+y1=50e-3; //First point where velocity is to be found out
+y2=5e-3; //Second point where velocity is to be found out
+rho_air=1.10; //Density of air
+Meu_air=20e-6; //Viscosity of air
+
+Re=d*u*rho_air/Meu_air;
+//Hence, from Figure 3.7: X=R/(rho*u^2)=0.0018
+X=0.0018;
+u_s=u/0.817;
+u_star=u*X^0.5;
+
+//At 50 mm from the wall:
+y1_r=2*y1/d;// y/r
+//Hence, from equation 12.34:
+u_x1=u_s+2.5*u_star*log(y1_r);
+printf("\n The fluid velocity at 50 mm from the wall = %.1f m/s",u_x1);
+
+//At 5 mm from the wall:
+y2_r=2*y2/d;// y/r
+//Hence, from equation 12.34:
+u_x2=u_s+2.5*u_star*log(y2_r);
+printf("\n The fluid velocity at 5 mm from the wall = %.1f m/s",u_x2);
+
+//The thickness of the laminar sub-layer is given by equation 12.54:
+del_b=5*d/(Re*X^0.5);
+printf("\n The thickness of the laminar sub-layer = %.3f mm",del_b*1e3);
+\ No newline at end of file
diff --git a/536/CH12/EX12.2/Example_12_2.sce b/536/CH12/EX12.2/Example_12_2.sce
new file mode 100755
index 000000000..f0523fc01
--- /dev/null
+++ b/536/CH12/EX12.2/Example_12_2.sce
@@ -0,0 +1,19 @@
+clc;
+clear;
+
+printf("\n Example 12.2\n");
+
+u=10; //Velocity of air
+T=330; //Temperature of air
+d=25e-3; //Inner diameter of pipe
+T_p=415; //Temperature at which the pipe is maintained
+DP_l=80; //Drop of static pressure along the pipe per unit length
+
+//From equations 12.98 and 3.18:
+//we get h=0.05*Cp
+//The heat taken up per unit time by the air dQ=0.0052*Cp*dT......(i)
+//The heat transferred through the pipe wall is also given by: = 0.039*Cp*(415-T)......(ii)
+//Equating (i) & (ii)
+//On integrating we get
+T_0=415-(85/exp(0.45))
+printf("\n The required air Temperature = %d K",T_0);
diff --git a/536/CH12/EX12.3/Example_12_3.sce b/536/CH12/EX12.3/Example_12_3.sce
new file mode 100755
index 000000000..58a18d17d
--- /dev/null
+++ b/536/CH12/EX12.3/Example_12_3.sce
@@ -0,0 +1,46 @@
+clc;
+clear;
+
+printf("\n Example 12.3\n");
+
+u=3.5; //Velocity of water
+d=25e-3; //Diameter of the pipe
+l=6; //Length of the pipe
+T1=300; //Temperature at enterance
+T2=330; //Temperature at exit
+rho=1000; //density of water at 310 K
+Meu=0.7e-3; //Viscosity of water at 310 K
+//Taking the fluid properties at 310 K and assuming that fully developed flow exists
+Cp=4.18e3; //heat capapcity
+k=0.65; //Thermal conductivity
+
+Re=d*u*rho/Meu;
+Pr=Cp*Meu/k;
+
+printf("\n (a) Reynolds analogy");
+h1=0.032*(Re^-0.25)*Cp*rho*u;//....Equation 12.139
+printf("\n h = %.2f kW/m^2 K",h1*1e-3);
+// on solving we get final equation as
+theta_dash1=330-10^(log10(30)-(0.0654*h1*1e-3/2.303));
+printf("\n The outlet temperature = %.1f K",theta_dash1)
+
+printf("\n\n (b) Taylor Prandtl Equation");
+h2=0.032*(Re^-0.25)*(1+2*Re^(-1/8)*(Pr-1))^-1*Cp*rho*u;
+printf("\n h = %.2f kW/m^2 K",h2*1e-3);
+// on solving we get final equation as
+theta_dash2=330-10^(log10(30)-(0.0654*h2*1e-3/2.303));//....Equation 12.140
+printf("\n The outlet temperature = %.1f K",theta_dash2)
+
+printf("\n\n (c) Universal velocity profile equation");
+h3=0.032*(Re^-0.25)*(1+0.82*Re^(-1/8)*((Pr-1)+log(0.83*Pr+0.17)))^-1*Cp*rho*u;//...equation 12.141
+printf("\n h = %.2f kW/m^2 K",h3*1e-3);
+// on solving we get final equation as
+theta_dash3=330-10^(log10(30)-(0.0654*h3*1e-3/2.303));
+printf("\n The outlet temperature = %.1f K",theta_dash3)
+
+printf("\n\n (d) Nu=0.023*Re^0.8*Pr^0.33");
+h4=k/d*0.023*Re^0.8*Pr^0.33;
+printf("\n h = %.2f kW/m^2 K",h4*1e-3);
+// on solving we get final equation as
+theta_dash4=330-10^(log10(30)-(0.0654*h4*1e-3/2.303));
+printf("\n The outlet temperature = %.1f K",theta_dash4)
diff --git a/536/CH12/EX12.4/Example_12_4.sce b/536/CH12/EX12.4/Example_12_4.sce
new file mode 100755
index 000000000..0c4aeacfb
--- /dev/null
+++ b/536/CH12/EX12.4/Example_12_4.sce
@@ -0,0 +1,46 @@
+clc;
+clear;
+
+printf("\n Example 12.4\n");
+
+u=3.5; //Velocity of air
+d=25e-3; //Diameter of the pipe
+l=6; //Length of the pipe
+T1=290; //Temperature at enterance
+T2=350; //Temperature at exit
+rho=29/22.4*273/310; //density of air at 310 K
+Meu=0.018e-3; //Viscosity of air at 310 K
+//Taking the physical properties at 310 K and assuming that fully developed flow exists
+Cp=1.003e3; //heat capapcity
+k=0.024; //Thermal conductivity
+
+Re=d*u*rho/Meu;
+Pr=Cp*Meu/k;
+
+printf("\n (a) Reynolds analogy");
+h1=0.032*(Re^-0.25)*Cp*rho*u;//....Equation 12.139
+printf("\n h = %.2f W/m^2 K",h1);
+// on solving we get final equation as
+theta_dash1=350-10^(log10(60)-(239.88*h1*1e-3/2.303));
+printf("\n The outlet temperature = %.1f K",theta_dash1)
+
+printf("\n\n (b) Taylor Prandtl Equation");
+h2=0.032*(Re^-0.25)*(1+2*Re^(-1/8)*(Pr-1))^-1*Cp*rho*u;
+printf("\n h = %.2f W/m^2 K",h2);
+// on solving we get final equation as
+theta_dash2=350-10^(log10(60)-(239.88*h2*1e-3/2.303));//....Equation 12.140
+printf("\n The outlet temperature = %.1f K",theta_dash2)
+
+printf("\n\n (c) Universal velocity profile equation");
+h3=0.032*(Re^-0.25)*(1+0.82*Re^(-1/8)*((Pr-1)+log(0.83*Pr+0.17)))^-1*Cp*rho*u;//...equation 12.141
+printf("\n h = %.2f W/m^2 K",h3);
+// on solving we get final equation as
+theta_dash3=350-10^(log10(60)-(239.88*h3*1e-3/2.303));
+printf("\n The outlet temperature = %.1f K",theta_dash3)
+
+printf("\n\n (d) Nu=0.023*Re^0.8*Pr^0.33");
+h4=k/d*0.023*Re^0.8*Pr^0.33;
+printf("\n h = %.2f W/m^2 K",h4);
+// on solving we get final equation as
+theta_dash4=350-10^(log10(60)-(239.88*h4*1e-3/2.303));
+printf("\n The outlet temperature = %.1f K",theta_dash4)