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+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.23\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.23  (page no. 358) 
+// Solution
+
+//The cooling tower
+//From the Steam tables,
+//For water:
+h100F=68.05;  //Btu/lbm //enthalpy at 100 F
+h70F=38.09;   //Btu/lbm  //enthalpy at 70 F
+//For air:
+h=20.4; //Unit:Btu/lb //at inlet,total heat/lb dry air
+w=38.2; //Unit:grains/lb  //at inlet,moisture pickup/lb dry air (at 60F D.B. and 50% R.H.)
+H=52.1; //Unit:Btu/lb //at outlet,total heat/lb dry air
+W=194.0; //Unit:grains/lb //at outlet,moisture pickup/lb dry air (at 90F D.B. and 90% R.H.)
+
+//Per pound of dry air,the heat interchange is H-h Btu per pound of dry air.
+//Per pound of dry air,the moisture increase is (W-w)/7000 lb per pound of dry air.
+//From the equation, ma*(H-h) = 200000*h100F - mwout*h70F             //ma=mass of air  mwout=mass of cooled water 
+//and  ma*((W-w)/7000) = 200000 - mwout
+//Solving the latter equation for mwout,we have  mwout=200000-(ma*((W-w)/7000))
+//Substituting this into the heat balance yields,
+// ma*(H-h) = 200000*h100F - 200000*h70F + ma*h70F*((W-w)/7000)
+//Solving gives us,
+ma=(200000*(h100F-h70F))/((H-h)-(h70F*((W-w)/7000))); //The amount of air required per hour //Unit:lbm/hr of dry air
+printf("The amount of air required per hour is %f lbm/hr of dry air\n",ma);
+printf("The amount of water lost per hour due to evaporation is %f lbm/hr\n",ma*((W-w)/7000));
+//note that the water evaporated is slightly over 2% of the incoming water,and this is the makeup that has to be furnished to the tower.
+//answer are slightly differ because of value of (W-w)/7000 is given 0.0233 instead of 0.0225