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clc;
clear;
//Example 2.51
//Given
k=75 //Thermal conductivity [W/(m.K)]
T_water=363 //[K]
T_air=303 //[K]
dT=T_water-T_air //delta T
h1=150 // for water[W/(sq m.K)]
h2=15 //for air [W/(sq m.K)]
W=0.5 //Width of wall[m]
L=0.025 //[m]
Area=W^2 //Base Area [sq m]
t=1 //[mm]
t=t/1000 //[m]
pitch=10 //[mm]
pitch=pitch/1000 //[m]
N=W/pitch //[No of fins]
//Calculations
A=N*W*t //Total cross-sectional area of fins in [sq m]
Ab=Area-A //[sq m]
Af=2*W*L //Surface area of fins [sq m]
//CASE 1: HEAT TRANSFER WITHOUT FINS
A1=Area //[sq m]
A2=A1 //[sq m]
Q=dT/(1/(h1*A1)+1/(h2*A2)); //[W]
printf("\nWithout fins,Q=%f W\n",Q);
//CASE 2: Fins on the water side
P=2*(t+W);
A=0.5*10^-3;
m=sqrt(h1*P/(k*A))
nfw=tanh(m*L)/(m*L) //Effeciency on water side
Aew=Ab+nfw*Af*N //Effective area on the water side [sq m]
Q=dT/(1/(h1*Aew)+1/(h2*A2)); //[W]
printf("\n With fins on water side,Q=%f W \n",Q);
//CASE 3: FINS ON THE AIR SIDE
m=sqrt(h2*P/(k*A))
nf_air=tanh(m*L)/(m*L) //Effeciency
Aea=Ab+nf_air*Af*N //Effective area on air side
Q=dT/(1/(h1*A1)+1/(h2*Aea)); //[W]
printf("\n With Fins on Air side,Q=%f W \n",Q)
//BOTH SIDE:
Q=dT/(1/(h1*Aew)+1/(h2*Aea)); //[W]
printf("\n With Fins on both side,Q=%f W \n",Q);
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