From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 536/CH12/EX12.4/Example_12_4.sce | 46 ++++++++++++++++++++++++++++++++++++++++
 1 file changed, 46 insertions(+)
 create mode 100755 536/CH12/EX12.4/Example_12_4.sce

(limited to '536/CH12/EX12.4/Example_12_4.sce')

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)
-- 
cgit