initial commit / add all books

5 files changed, 151 insertions, 0 deletions

diff --git a/581/CH8/EX8.1/Example8_1.sce b/581/CH8/EX8.1/Example8_1.sce
new file mode 100755
index 000000000..72cf1d2ea
--- /dev/null
+++ b/581/CH8/EX8.1/Example8_1.sce
@@ -0,0 +1,27 @@
+

+clear;

+clc;

+

+printf("\t Example 8.1\n");

+

+T1=313;     //fluid temp.,K

+T2=287;     //air temp.,K

+H=0.4;      //height of sides,m

+Pr=0.711;   //prandtl no.

+

+

+ b=1/T2;    // b=1/v*d(R*T/p)/dt=1/To   characterisation constant of thermal expansion of solid, K^-1

+ RaL=9.8*b*(T1-T2)*H^3/((1.566*10^-5)*(2.203*10^-5));    //Rayleigh no.

+ 

+ Nu=0.678*RaL^(0.25)*(Pr/(0.952+Pr))^(1/4);        // nusselt no.

+ h=Nu*0.02614/H     // average heat transfer coefficient, W/m^2/K

+ 

+ q=h*(T1-T2)        // average heat transfer,W/m^2

+ c=3.936*((0.952+Pr)/Pr^2)^(1/4)*(1/(RaL/Pr)^0.25);   //boundary layer thickness.,m

+ printf("\t  average heat transfer coefficient is : %.2f W/m^2/K\n",h);

+ printf("\t average heat transfer is : %.1f W/m^2\n",q);

+ printf("\t boundary layer thickness is : %.3f m\n",c);

+ 

+ printf("\t thus the BL thickness at the end of the plate is only 4 percent of the height, or 1.72 cm thick.this is thicker thsan typical forced convection BL but it is still reasonably thin.")

+  

+  //end
\ No newline at end of file
diff --git a/581/CH8/EX8.3/Example8_3.sce b/581/CH8/EX8.3/Example8_3.sce
new file mode 100755
index 000000000..6afb5398a
--- /dev/null
+++ b/581/CH8/EX8.3/Example8_3.sce
@@ -0,0 +1,26 @@
+

+clear;

+clc;

+

+printf("\t Example 8.3\n");

+

+T1=323;     //wall temp.,K

+T2=293;     //air temp.,K

+H=0.3;      //height of wall, m

+v1=16.45*10^-6;   // molecular diffusivity, m^2/s

+b=1/T2;    // b=1/v*d(R*T/p)/dt=1/To   characterisation constant of thermal expansion of solid, K^-1

+v2=2.318*10^-5;    //molecular diffusivity, m^2/s

+Pr=0.71;   //prandtl no.

+

+ Ral=9.8*b*(T1-T2)*H^3/((1.566*10^-5)*(2.203*10^-5));    // Rayleigh no.

+ Nu=0.678*Ral^(0.25)*(Pr/(0.952+Pr))^(1/4);        // nusselt no.

+ h=Nu*0.0267/H     // average heat transfer coefficient, W/m^2/K

+ 

+ Nu1=0.68+0.67*((Ral)^(1/4)/(1+(0.492/Pr)^(9/16))^(4/9));    //churchill correlation 

+ 

+ h1=Nu1*(0.0267/0.3)-.11;     //average heat transfer coefficient, W/m^2/K

+ 

+ 

+printf("\t correlation average heat transfer coefficient is :%.2f W/m^2/K\n",h1)

+printf("\t the prediction is therefore within 5 percent of corelation .we should use the latter result in preference to the theoritical one, although the difference is slight.")

+  //end
\ No newline at end of file
diff --git a/581/CH8/EX8.4/Example8_4.sce b/581/CH8/EX8.4/Example8_4.sce
new file mode 100755
index 000000000..33dd6ed59
--- /dev/null
+++ b/581/CH8/EX8.4/Example8_4.sce
@@ -0,0 +1,33 @@
+
+clear;
+clc;
+
+printf("\t Example 8.4\n");
+
+T1=400;     //hot oil temp.,K
+D=0.005;          //diameter of line carrying oil, m
+T2=300;    //temp. of air around the tube,K
+Tav=350;    //average BI temp.,K
+
+//we evaluate properties at this temp. and write g as ge*(g-level), where ge is g at the earth surface and the g-level is the fraction of ge in the space vehicle.
+b=1/T2;    // b=1/v*d(R*T/p)/dt=1/To   characterisation constant of thermal expansion of solid, K^-1
+
+v1=2.062*10^-5;   // molecular diffusivity, m^2/s
+v2=2.92*10^-5;    //molecular diffusivity, m^2/s
+Pr=0.706;   //prandtl no.
+
+g=[10^-6 10^-5 10^-4 10^-2];
+i=1;
+while(i<5)
+Ral=(9.8*b*((T1-T2))*(D^(3))/(v1*v2))*g(i);    // Rayleigh no.
+Nu(i)=(0.6+0.387*(Ral/(1+(0.559/Pr)^(9/16))^(16/9))^(1/6))^2;
+//Nu(i)=(0.6+0.387*((Ral)/(1+(0.559/Pr)^(9/16))^(16/9))^1/6)^2;    //churchill correlation. 
+printf("\t Nusselt no. are : %.3f\n",Nu(i)); 
+h(i)=Nu(i)*0.0297/D;    // convective heat transfer coefficient,W/(m^2*K)
+printf("\t convective heat transfer coefficient are : %.2fW/(m^2*K)\n",h(i));
+Q(i)=%pi*D*h(i)*(T1-T2);    //heat transfer,W/m
+printf("\t heat transfer is :%.2fW/m of tube\n",Q(i));
+i=i+1;
+end 
+
+//end
\ No newline at end of file
diff --git a/581/CH8/EX8.5/Example8_5.sce b/581/CH8/EX8.5/Example8_5.sce
new file mode 100755
index 000000000..94f4e9997
--- /dev/null
+++ b/581/CH8/EX8.5/Example8_5.sce
@@ -0,0 +1,34 @@
+

+clear ;

+clc;

+

+printf("\t Example 8.5\n");

+

+T2=300;     //air temp.,K

+P=15;       //delivered power,W

+D=0.17;       //diameter of heater,m

+v1=1.566*10^-5;   // molecular diffusivity, m^2/s

+b=1/T2;    // b=1/v*d(R*T/p)/dt=1/To   characterisation constant of thermal expansion of solid, K^-1

+Pr=0.71;   //prandtl no.

+v2=2.203*10^-5;    //molecular diffusivity, m^2/s

+v3=3.231*10^-5;    //molecular diffusivity at a b except at 365 K., m^2/s

+v4=2.277*10^-5;    //molecular diffusivity at a b except at 365 K., m^2/s

+k1=0.02614;        //thermal conductivity

+k2=0.0314;        //thermal conductivity

+

+//we have no formula for this situation, so the problem calls for some guesswork.following the lead of churchill and chau, we replace RaD with RaD1/NuD in eq.   

+//(NuD)^(6/5)=0.82*(RaD1)^(1/5)*Pr^0.034

+

+delT=1.18*P/(3.14*D^(2)/4)*(D/k1)/((9.8*b*661*D^(4)/(0.02164*v1*v2))^(1/6)*Pr^(0.028));

+

+//in the preceding computation, all the properties were evaluated at T2.mow we must return the calculation,reevaluating all properties except b at 365 K.

+

+delTc=1.18*661*(D/k2)/((9.8*b*661*D^(4)/(k2*v3*v4))^(1/6)*(0.99));

+

+TS=T2+delTc;

+TS1=TS-271.54

+

+printf("\t average surface temp. is :%.0f K\n",TS1);

+

+printf("\t that is rather hot.obviously, the cooling process is quite ineffective in this case.")

+//end

diff --git a/581/CH8/EX8.6/Example8_6.sce b/581/CH8/EX8.6/Example8_6.sce
new file mode 100755
index 000000000..52a9415fe
--- /dev/null
+++ b/581/CH8/EX8.6/Example8_6.sce
@@ -0,0 +1,31 @@
+

+clear;

+clc;

+

+printf("\t Example 8.6\n");

+

+T2=363;       // temp. of strip,K

+T1=373;       //saturated temp.,K

+H=0.3;        // height of strip,m

+Pr=1.86;   //prandtl no.

+Hfg=2257;    //latent heat. kj/kg

+ja=4.211*10/Hfg;   //jakob no.

+a1=961.9;       //density of water,kg/m^3

+a2=0.6;         //density of air,kg/m^3

+k=0.677;        //thermal conductivity,W/(m*K)

+

+Hfg1=Hfg*(1+(0.683-0.228/Pr)*ja);    //corrected latent heat,kj/kg

+

+delta=(4*k*(T1-T2)*(2.99*10^(-4))*0.3/(a1*(a1-a2)*9.806*Hfg1*1000))^(0.25)*1000;

+

+Nul=4/3*H/delta;    //average nusselt no.

+q=Nul*k*(T1-T2)/H;   // heat flow on an area about half the size of a desktop,W/m^2

+Q=q*H;   //overall heat transfer per meter,kW/m

+

+m=Q/(Hfg1);     //mass rate of condensation per meter,kg/(m*s)

+

+printf("\t overall heat transfer per meter is :%.1f kW/m^2\n",Q);

+printf("\t  film thickness at the bottom is :%.3f mm\n",delta);

+printf("\t mass rate of condensation per meter. is : %.4f kg/(m*s)\n",m);

+

+//end
\ No newline at end of file