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clc;
clear;
//Example 3.35
v=2*10^-5 //[m^2/s]
Npr=0.7 //Prandtl number
k=0.03 //[W/m.K]
D=0.25 //Diameter in [m]
L=0.90*D //Characteristic length,let [m]
T1=298 //[K]
T2=403 //[K]
dT=T2-T1 //[K]
Tf=(T1+T2)/2 //[K]
Beta=1/Tf //[K^-1]
A=%pi*(D/2)^2 //Area in[sq m]
g=9.81 //[m/s^2]
//Case 1: Hot surface facing up
Ngr=g*Beta*dT*(L^3)/(v^2) //Grashoff number
Nnu=0.15*((Ngr*Npr)^(1.0/3.0)) //Nusselt number
h=Nnu*k/L //[W/sq m.K]
Q=h*A*dT //[W]
printf("\n Heat transferred when hot surface is facing up is %f W\n",Q);
//Case 2:For hot surface facing down
Nnu=0.27*(Ngr*Npr)^(1.0/4.0); //Grashof Number
h=Nnu*k/L //[W/sqm.K]
Q=h*A*dT //[W]
printf("\n Heat transferred when hot surface is facing down is %f W\n",Q);
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