//Chapter-12, Example 12.3, Page 509 //============================================================================= clc clear //INPUT DATA mh=10000;//Mass flow rate of oil in kg/h ch=2095;//Specific heat of oil J/kg.K Thi=80;//Inlet temperature of oil in degree C Tho=50;//Outlet temperature of oil in degree C mc=8000;//Mass flow rate of water in kg/h Tci=25;//Inlet temperature of water in degree C U=300;//Overall heat ransfer coefficient in W/m^2.K cc=4180;//Specific heat of water in J/kg.K //CALCULATIONS Q=(mh*ch*(Thi-Tho));//Heat transfer rate in W Tco=((Q/(mc*cc))+Tci);//Outlet temperature of water in degree C T=(Thi-Tco);//Temperature difference between oil inlet temperature and water outlet temperature in degree C t=(Tho-Tci);//Temperature difference between oil outlet temperature and water inlet temperature in degree C A=(((Q/U)*log(t/T))/(3600*(t-T)));//Area of heat exchanger in m^2 //OUTPUT mprintf('Area of heat exchanger is %3.2f m^2',A) //=================================END OF PROGRAM==============================