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+//scilab 5.4.1

+clear;

+clc;

+printf("\t\t\tProblem Number 11.25\n\n\n");

+// Chapter 11 : Heat Transfer

+// Problem 11.25  (page no. 602) 

+// Solution

+

+//In problem 11.24, A COUNTERFLOW HEAT EXCHANGER is operated in the parallel flow

+//Hot oil enters at 215 F and leaves at 125 F

+//Water enters the unit at 60 F and leaves at 90 F

+//Therefore,From figure 11.35, 

+thetaA=215-60; //the greatest temperature difference between the fluids(at either inlet or outlet) //Unit:fahrenheit

+thetaB=125-90; //the least temperature difference between the fluids(at either inlet or outlet) //Unit:fahrenheit

+deltaTm=(thetaA-thetaB)/log(thetaA/thetaB); //logarithmic mean temperature difference //Unit:fahrenheit 

+//From the oil data,

+m=400*60;  //mass //Unit:lb/sec //1 min=60 sec

+Cp=0.85; //Specific heat of the oil //Unit:Btu/(lb*F)

+deltaT=215-125;  //Change in temperature  //Unit:fahrenheit

+Q=m*Cp*deltaT //The heat transfer //Unit:Btu/hr

+//Q=U*A*deltaTm

+U=40;//The overall coefficient of heat transfer of the unit //Unit:Btu/(hr*ft^2*F)

+A=Q/(U*deltaTm); //Umit:ft^2 //The outside surface area

+printf("The outside surface area required is %f ft^2",A);

+

+

+

+

+