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authorpriyanka2015-06-24 15:03:17 +0530
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+clc
+clear
+// solution
+// initialization of variables
+// Please refer to fig of question 6.1 for quantities
+effi1=0.357//efficiency from example 6.3
+P2=2*1000 //higher pressure converted in in kPa
+P1=10 // lower pressure in kPa
+rho=1000 // density of water in Kg/m^3
+T2=600// max temperature of cycle in degree celsius
+h1=192 // enthalpy at state 1 in kJ/kg
+h3=3690 // enthalpy at state 3 in kJ/kg, 600*C and 2MPa pressure
+h4=2442 // enthalpy from example 6.3
+h6=505 // specific enthalpy @ 200 kPa from steam table
+h7=h6 // isenthalpic process
+s3=7.702// entropy at state 3 in kJ/kg.K, 600*C and 2MPa pressure
+
+h2=h1 // isenthalpic process
+s5=s3 // isentropic process
+h5=(s3-7.509)*(2971-2870)/(7.709-7.509)+2870 // interpolationg from steam table 2 200 kPa using s5=s3= 7.702 kJ/kg.
+
+m6=1 // let mass of steam =1 Kg
+m5=(h6-h2)*(m6)/(h5-h2)
+m2=m6-m5 // conservation of mass
+
+wt=h3-h5+(h5-h4)*m2 // work done by turbine
+qb=h3-h7 // heat given to bolier
+effi2=(wt)/qb // efficiency of power cycle
+printf(" The Efficiency is %.3f or %.1f %% \n",effi2,effi2*100)
+%increase=((effi2-effi1)/effi1)*100
+printf(" The %% increase in Efficiency is %.2f \n",%increase)
+// The anwer is different in textbook as there the intermediate values are approximated while in scilab the calculations are precise