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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")