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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /536/CH12/EX12.3/Example_12_3.sce | |
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Diffstat (limited to '536/CH12/EX12.3/Example_12_3.sce')
-rwxr-xr-x | 536/CH12/EX12.3/Example_12_3.sce | 46 |
1 files changed, 46 insertions, 0 deletions
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)
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