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+

+clc;

+clear;

+//Example 6.4

+Cpf=3.98        //Specific heat of feed in kJ/(kg.K)

+lambda_s=2202    //Latent heat of conds of heat at 0.2MPa in [kJ/kg]

+lambda=2383     //Latent heat of vaporisation of water aty 323 [kJ/kg

+ic=0.1          //Initial concentration of soilds in [%]

+fc=0.5          //Final concentration

+m_dot=30000     //Feed to evaporator in [kg/h]

+mdash_dot=ic* m_dot/fc  //Mass flow rate of thick liquor in [kg/h]

+mv_dot=m_dot-mdash_dot      //Water evaporated in [kg/h]

+

+//Case 1: Feed at 293K

+mf_dot=30000        //[kg/h]

+mv_dot=24000        //[kg/h]

+Cpf=3.98        //[kJ/(kg.K)]

+Ts=393      //Saturation temperature of steam in [K]

+T=323       //Boiling point of solution [K]

+lambda_s=2202       //Latent heat of condensation [kJ/kg]

+lambda=2383     //Latent heat of vaporisation[kJ/kg]

+Tf=293          //Feed temperature

+//Enthalpy balance over the evaporator:

+ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s       //Steam consumption[kg/h]

+eco=(mv_dot/ms_dot)         //Steam economy

+printf("\nWhen Feed introduced at 293 K ,Steam economy is %f\n",eco);

+dT=Ts-T                     //[K]

+U=2900          //[W/sq m.K]

+Q=ms_dot*lambda_s           //Heat load =Rate of heat transfer in [kJ/h]

+Q=Q*1000/3600               //[J/s]

+A=Q/(U*dT)              //Heat transfer area required [sq m]

+printf("\n ANSWER-(i)\n\n At 293 K,Heat transfer area required is %f sq m\n",A);

+

+//Case2: Feed at 308K

+Tf=308      //[Feed temperature][K]

+ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s           //Steam consumption in [kg/h]

+eco=mv_dot/ms_dot               //Economy of evaporator

+printf("\n ANSWER-(ii)\n\n When T=308 K \nEconomy of evaporator is %f\n",eco);

+Q=ms_dot*lambda_s               //[kJ/h]

+Q=Q*1000/3600                   //[J/s]

+A=Q/(U*dT)                      //Heat transfer area required [sq m]

+printf('\nANSWER-(iii) \n When T=308 K,\nHeat transfer Area required is %f sq m\n",A);