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+//scilab 5.4.1

+clear;

+clc;

+printf("\t\t\tProblem Number 11.11\n\n\n");

+// Chapter 11 : Heat Transfer

+// Problem 11.11  (page no. 567) 

+// Solution

+

+//From problem 11.9,

+//A bare steel pipe

+r2=3.50; //Outside diameter //Unit:in.

+r1=3.00; //inside diameter //Unit:in.

+Ti=240; //Inside temperature //unit:fahrenheit

+L=5; //Length //Unit:ft

+k1=26; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity

+ans1=(inv(k1)*log(r2/r1));

+

+//Now,in problem 11.11,

+//Mineral wool

+r3=5.50; //inside diameter //Unit:in.

+r2=3.50; //outside diameter //Unit:in.

+To=85; //Outside temperature //unit:fahrenheit

+deltaT=Ti-To; //Change in temperature //unit:fahrenheit

+k2=0.026 //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity

+ans2=(inv(k2)*log(r3/r2));

+

+Q=(2*%pi*L*deltaT)/(ans1+ans2); //The heat loss from the pipe  //unit:Btu/hr

+printf("The heat loss from the pipe is %f Btu/hr",Q);

+

+