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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+clc
+//page 19
+printf("\t example 2.5 \n");
+printf("\t approximate values are mentioned in the book \n");
+t1=150; // assume temperature of outer surface of rockwool,F
+ta=70; // temperature of surrounding air,F
+ha=2.23; // surface coefficient,Btu/(hr)*(ft^2)*(F)
+q=(3.14)*(300-70)/(((2.3/(2*0.033))*log10(3.375/2.375))+(1/((2.23)*(3.375/12)))); // using formula for heat loss,Btu/(hr)*(lin ft), calculation mistake
+printf("\t heat loss for linear foot is : %.1f Btu/(hr)*(lin ft) \n",q);
+printf("\t Check between ts and t1, since delt/R = deltc/Rc \n");
+t1=300-(((104.8)*((2.3)*(log10(3.375/2.375))))/((2)*(3.14)*(.033))); // using eq 2.31,F
+printf("\t t1 is : %.1f F \n",t1);
+t1=125; // assume temperature of outer surface of rockwool,F
+ha=2.10; // surface coefficient,Btu/(hr)*(ft^2)*(F)
+q=((3.14)*(300-70))/(((2.3/(2*0.033))*log10(3.375/2.375))+(1/((2.10)*(3.375/12)))); // using formula for heat loss,Btu/(hr)*(lin ft)
+printf("\t heat loss for linear foot is : %.1f Btu/(hr)*(lin ft) \n",q);
+printf("\t Check between ts and t1, since delt/R = deltc/Rc \n");
+t1=300-(((103)*((2.3)*(log10(3.375/2.375))))/((2)*(3.14)*(.033))); // using eq 2.31,F
+printf("\t t1 is : %.1f F \n",t1);
+// end