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+
+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