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+clc
+h1 = 3230.9 // Enthalpy at state 1 in kJ/kg
+s1 = 6.9212 // Entropy at state 1 in kJ/kgK
+s2 = s1 // Isentropic process
+s3 = s1 // Isentropic process
+h2 = 2796  // Enthalpy at state 2 in kJ/kg
+sf = 0.6493 // ENtropy of fluid onkJ/kgK
+sfg = 7.5009 // Entropy change due to vaporization
+x3 = (s3-sf)/sfg // steam quality
+h3 = 191.83 + x3*2392.8 // Enthalpy at state 3
+h4 = 191.83 // Enthalpy at state 4 in kJ/kg
+h5 = h4 // Isenthalpic process
+h6 = 640.23 // Enthalpy at state 6 in kJ/kg
+h7 = h6 // Isenthalpic process
+m = (h6-h5)/(h2-h5) // regenerative mass
+Wt = (h1-h2)+(1-m)*(h2-h3) // turbine work
+Q1 = h1-h6 // Heat addition
+n_cycle = 100*Wt/Q1 // Cycle efficiency
+sr = 3600/Wt // Steam rate
+s7 = 1.8607  // Entropy at state 7 in kJ/kgK
+s4 = 0.6493 // Entropy at state 4 in kJ/kgK 
+Tm = (h1-h7)/(s1-s7) // Mean temperature of heat addition with regeneration
+Tm1 = (h1-h4)/(s1-s4) // Mean temperature of heat addition without regeneration
+dT = Tm-Tm1 // Change in temperature
+Wt_ = h1-h3 // Turbine work
+sr_ = 3600/Wt_ // Steam rate
+dsr = sr-sr_// Change in steam rate
+n_cycle_ = 100*(h1-h3)/(h1-h4) // Cycle effciency
+dn = n_cycle-n_cycle_// Change in efficiency
+printf("\n Example 12.5\n")
+printf("\n Efficiency of the cycle is %f percent",n_cycle)
+
+printf("\n Steam rate of the cycle is %f kg/kW h",sr)//The answer provided in the textbook is wrong
+
+printf("\n Increase in temperature due to regeneration is %f degree centigrade",dT)
+printf("\n Increase in steam rate due to regeneration is %f kg/kW h",dsr)//The answer provided in the textbook is wrong
+
+printf("\n Increase in Efficiency of the cycle due to regeneration is %f percent",dn)
+
+//The answers vary due to round off error