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diff --git a/1808/CH4/EX4.4/Chapter4_Example4.sce b/1808/CH4/EX4.4/Chapter4_Example4.sce
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+clc
+clear
+//INPUT DATA
+pb=100;//Saturated vapour in bar
+pc=0.1;//Saturated liquid in bar
+two=35;//Cooling water exit temperature in degree C
+twi=20;//Cooling water entry temperature in degree C
+S1=5.6198;//Entropy in kJ/kg.K
+S3=0.649;//Entropy in kJ/kg.K
+S5=8.15;//Entropy in kJ/kg.K
+h1=2727.7;//Entropy in kJ/kg
+h2=1778.3;//Entropy in kJ/kg
+h3=191.8;//Enthalpy in kJ/kg
+h4=201.79;//Enthalpy in kJ/kg
+h5=2584.7;//Enthalpy in kJ/kg
+x2=0.63;//Quality of steam
+V3=0.001;//Specific volume in m^3/kg
+Cpw=4.18;//specific heat of water in kJ/kgk
+nt=0.8;//Turbine efficiency in percentage
+np=0.9;//Pump efficiency in percentage
+
+
+//CALCULATIONS
+h21=h1-nt*(h1-h2);//Entropy in kJ/kg
+h41=((h4-h3)/np)+h3;//Entropy in kJ/kg
+nRA=((h1-h21)-(h41-h3))/(h1-h4)*100;//Actual thermal efficiency
+m=pb*1000/((h1-h21)-(h41-h3));//Mass flow rate of steam
+mx=(m*3600);//Mass flow rate in kg/hr
+QS1=m*(h1-h41);//Rate of heat transfer into working medium in MW
+QR1=m*(h21-h3);//Rate of heat transfer from the condenser in MW
+mw1=(mx*(h21-h3))/((Cpw)*(two-twi))/10^7;//mass flow rate of water in the condenser in kg/s
+RwA=((h1-h21)-(h41-h3))/(h1-h21);//work ratio
+ 
+ //OUTPUT
+printf('(i) The Actual Thermal efficiency is %3.2f percent \n(ii)The mass flow rate of steam is %3.2f kJ/s \n(iii) The rate of heat transfer into working medium is %3.1f kJ/s \n(iv)The rate of heat transfer from condenser is %3.2f kJ/s\n(v)mass flow rate of water in condenser is %3.3f *10^7 kg/s \n(vi) The work ratio is %f ',nRA,m,QS1,QR1,mw1,RwA)