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+//scilab 5.4.1

+clear;

+clc;

+printf("\t\t\tProblem Number 11.21\n\n\n");

+// Chapter 11 : Heat Transfer

+// Problem 11.21  (page no. 589) 

+// Solution

+

+//Because the conditions of illustrative problem 11.15 are the same as for problem 11.19 and 11.20,we can solve this problem in two ways to obtain a check.

+//Thus,adding the results of these problems yields,

+printf("Adding the results of the problems yields,\n")

+Qtotal=206.2+214.5; //Unit:Btu/hr //total heat loss

+printf("The heat loss due to convection is %f Btu/hr\n",Qtotal);

+

+//We can also approach this solution by obtaining radiation and convection heat-transfer co-efficcient.Thus,

+hcombined=0.9+0.94; //Coefficient of heat transfer //Unit:Btu/(hr*ft^2*F)

+D=3.5/12; //3.5 inch = 3.5/12 feet //Unit:ft //Outside diameter

+Ti=120; //Inside temperature //unit:fahrenheit

+To=70; //Outside temperature //unit:fahrenheit

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

+L=5; //Length //Unit:ft //From problem 11.10

+A=(%pi*D)*L;  //Area //Unit:ft^2 

+Qtotal=hcombined*A*deltaT; //Unit:Btu/hr //total heat loss due to convection //Newton's law of cooling

+printf("By obtaining radiation and convection heat-transfer co-efficcient,\n")

+printf("The heat loss due to convection is %f Btu/hr",Qtotal);