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+clc;
+clear;
+//Example 5.15
+
+//CASE I:
+Cp=4*10^3;           //[J/kg.K]
+t1=295;              //[K]
+t2=375;              //[K]
+sp=1.1;              //Specific gravity of liquid 
+v1=1.75*10^-4;        //Flow of liquid in [m^3/s]
+rho=sp*1000         //[kg/m^3]
+m_dot=v1*rho         //[kg/s]
+Q=m_dot*Cp*(t2-t1)  //[W]
+T=395;               //[K]
+dT1=T-t1            //[K]
+dT2=T-t2            //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2) //[K]
+U1A=Q/dTlm          //[W/K]
+
+//CASE-II
+v2=3.25*10^-4                   //Flow in [m^3/s]
+T2=370                          //[K]
+m_dot=v2*rho          //[kg/s]
+Q=m_dot*Cp*(T2-t1)                //[W]
+dT1=T-t1                          //[K]
+dT2=T-T2                          //[K]
+dTlm=(dT1-dT2)/log(dT1/dT2)       //[K]
+U2A=Q/dTlm                        //[W/K]
+//since u is propn to v
+//hi =C*v^0.8
+
+U2_by_U1=U2A/U1A                
+
+ho=3400                         //Heat transfer coeff for condensing steam in [W/sq m.K]
+C=poly(0,"C")
+//Let C=1 and v=v1
+//C=1;
+v=v1;                           //=1.75*10^-4 m^3/s
+hi=C*v^0.8
+U1=1/(1/ho+1/hi)                 //
+
+//When v=v2
+v=v2;
+hi=C*v^0.8
+U2=1/(1/ho+1/hi)               //
+
+//Since U2=1.6U1
+//On solving we get:
+C=142497
+v=v1
+hi=C*v^0.8
+U1=1/(1/ho+1/hi)                 //
+A=U1A/U1                //Heat transfer area in [sq m]
+printf("\n Overall heat transfer coefficient is %f W/sq m.K and\n\nHeat transfer area is %f sq m",U1,A);
+