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+clc;

+warning("off");

+printf("\n\n example11.5 - pg521");

+// given

+Nre=50000;

+d=0.04;  //[m] - diameter of pipe

+// physical properties of water

+T1=293.15;  //[K]

+T2=303.15;  //[K]

+T3=313.15;  //[K]

+p1=999;  //[kg/m^3] - density of water at temperature T1

+p2=996.0;  //[kg/m^3] - density of water at temperature T2

+p3=992.1;  //[kg/m^3] - density of water at temperature T3

+mu1=1.001;  //[cP] - viscosity of water at temperature T1

+mu2=0.800;  //[cP] - viscosity of water at temperature T2

+mu3=0.654;  //[cP] - viscosity of water at temperature T3

+k1=0.63;  //[W/m*K] - thermal conductivity of water at temperature T1

+k2=0.618;  //[W/m*K] - thermal conductivity of water at temperature T2

+k3=0.632;  //[W/m*K] - thermal conductivity of water at temperature T3

+cp1=4182;  //[J/kg*K] - heat capacity of water at temperature T1

+cp2=4178;  //[J/kg*K] - heat capacity of water at temperature T2

+cp3=4179;  //[J/kg*K] - heat capacity of water at temperature T3

+Npr1=6.94;  // prandtl no. at temperature T1

+Npr2=5.41;  // prandtl no. at temperature T2

+Npr3=4.32;  // prandtl no. at temperature T3

+// (a)  Dittus -Boelter-this correction evalutes all properties at the mean bulk temperature,which is T1

+kmb=0.603

+h=(kmb/d)*0.023*((Nre)^(0.8))*((Npr1)^0.4);

+printf("\n\n (a) Dittus -Boelter\n the heat transfer coefficient is \n h = %f W/m^2*K = %f Btu/ft^2*h^-1*degF",h,h*0.17611);

+// (b) Seider Tate-this correlation evaluates all the properties save muw at the mean bulk temperature 

+h=(kmb/d)*(0.027)*((Nre)^0.8)*((Npr1)^(1/3))*((mu1/mu3)^0.14);

+printf("\n\n (b) Seider Tate\n the heat transfer coefficient is \n h = %f W/m^2*K = %f Btu/ft^2*h^-1*degF",h,h*0.17611);

+// (c) Sleicher-Rouse equation

+a=0.88-(0.24/(4+Npr3));

+b=(1/3)+0.5*exp((-0.6)*Npr3);

+Nref=Nre*(mu1/mu2)*(p2/p1);

+Nnu=5+0.015*((Nref)^a)*((Npr3)^b);

+h=Nnu*(kmb/d);

+printf("\n\n (c) Sleicher-Rouse equation\n the heat transfer coefficient is \n h = %f W/m^2*K = %f Btu/ft^2*h^-1*degF",h,h*0.17611);

+// (d) Colbum Analogy- the j factor for heat transfer is calculated

+jh=0.023*((Nref)^(-0.2));

+Nst=jh*((Npr2)^(-2/3));

+U=(Nre*mu1*10^-3)/(d*p1);

+h=Nst*(p1*cp1*U);

+printf("\n\n (d) Colbum Analogy\n the heat transfer coefficient is \n h = %f W/m^2*K = %f Btu/ft^2*h^-1*degF",h,h*0.17611);

+// (e) Friend-Metzner

+f=0.005227;

+Nnu=((Nre)*(Npr1)*(f/2)*((mu1/mu3)^0.14))/(1.20+((11.8)*((f/2)^(1/2))*(Npr1-1)*((Npr1)^(-1/3))));

+h=Nnu*(kmb/d);

+printf("\n\n (e) Friend-Metzner\n the heat transfer coefficient is \n h = %f W/m^2*K = %f Btu/ft^2*h^-1*degF",h,h*0.17611);

+// (f) Numerical analysis

+Nnu=320;

+h=Nnu*(kmb/d);

+printf("\n\n (f) Numerical analysis\n the heat transfer coefficient is \n h = %f W/m^2*K = %f Btu/ft^2*h^-1*degF",h,h*0.17611);

+

+

+

+