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+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.20\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.20  (page no. 588) 
+// Solution
+
+//The upper temperature is given as 120 F and the temperature difference is 
+Ti=120; //Inside temperature //unit:fahrenheit
+To=70; //Outside temperature //unit:fahrenheit
+deltaT=120-70; //unit:fahrenheit //Change in temperature
+//Using figure 11.28, 
+hrdash=1.18; //factor for radiation coefficient //Unit:Btu/(hr*ft^2*F)
+Fe=1; //Emissivity factor to allow for the departure of the surfaces interchanging heat from complete blackness;Fe is a function of the surface emissivities and       configurations
+FA=0.79; //geometric factor to allow for the average solid angle through which one surface "sees" the other
+hr=Fe*FA*hrdash; //The radiation heat-transfer coefficient for the pipe //Unit:Btu/(hr*ft^2*F)
+printf("The radiation heat-transfer coefficient for the pipe is %f Btu/(hr*ft^2*F)\n",hr);
+
+//As a check,Using the results of problem 11.17,
+printf("As a check,using the results of problem 11.17,\n");
+D=3.5/12; //3.5 inch = 3.5/12 feet//Unit:ft //Outside diameter
+L=5; //Length //Unit:ft //From problem 11.10
+A=(%pi*D)*L;  //Area //Unit:ft^2 
+Q=214.5; //heat loss //Unit:Btu/hr
+hr=Q/(A*deltaT); //The radiation heat-transfer coefficient for the pipe //Unit:Btu/(hr*ft^2*F) //Newton's law of cooling
+printf("The radiation heat-transfer coefficient for the pipe is %f Btu/(hr*ft^2*F)\n",hr);