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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);