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+//scilab 5.4.1

+clear;

+clc;

+printf("\t\t\tProblem Number 10.4\n\n\n");

+// Chapter 10 : Refrigeration

+// Problem 10.4  (page no. 506) 

+// Solution

+

+COP=4.5; //Coefficient of performance //From problem 10.1

+HPperTOR=4.717/COP; //Horsepower per ton of refrigeration //Unit:hp/ton

+Qremoved=1000; //Unit:Btu/min //From problem 10.1

+//1000 Btu/min /200 Btu/min ton = 5 tons of refrigeration

+HPrequired=HPperTOR*5; //The horsepower required //unit:hp

+printf("The horsepower required is %f hp\n",HPrequired);

+//In problem 10.1, 77.2 Btu/min was required

+printf("The power required is %f hp\n",77.2*778*inv(33000)); //1 Btu=778 ft*lbf //1 min*hp = 33000 ft*lbf

+//The ratio of the power required in each problem is the same as the inverse ratio of the COP value

+//Therefore,

+printf("The power required is %f hp\n",(COP/12.95)*HPrequired); //COP(in problem 10.1)=12.95

+printf("This checks our results")