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+clc
+Qr = 500 // Heat release in kW
+Tr = 2000 // Fuel burning temperature in K
+T0 = 300 // Ambient temperature in K
+// Part (a)
+printf("\n Example 8.12")
+Qa = 480 // Energy absorption by furnace in kW
+Ta = 1000 // Furnace temperature in K
+n1a = (Qa/Qr) // first law efficiency
+n2a = n1a*(1-(T0/Ta))/(1-(T0/Tr)) //second law efficiency
+
+//The answers vary due to round off error
+printf(" \n\n PART (A)")
+printf("\n The first law efficiency is %d percent" ,n1a*100)
+printf("\n The second law efficiency is %d percent",n2a*100)
+
+// Part (b)
+Qb = 450 // Energy absorption in steam generation in kW
+Tb = 500// steam generation temperature in K
+n1b = (Qb/Qr)// first law efficiency
+n2b = n1b*(1-(T0/Tb))/(1-(T0/Tr))//second law efficiency
+printf(" \n\n PART (B)")
+printf("\n The first law efficiency is %d percent" ,n1b*100)
+printf("\n The second law efficiency is %f percent",n2b*100)
+// Part (c)
+Qc = 300 // Energy absorption in chemical process in kW
+Tc = 320 // chemical process temperature in K
+n1c = (Qc/Qr) // first law efficiency
+n2c = n1c*(1-(T0/Tc))/(1-(T0/Tr))//second law efficiency
+printf(" \n\n PART (C)")
+printf("\n The first law efficiency is %d percent",n1c*100)
+printf("\n The second law efficiency is %f percent" ,n2c*100)
+// Part (d)
+Qd = 450
+n1d = (Qd/Qr)
+n2a_= n1d*(1-(T0/Ta))/(1-(T0/Tr))
+n2b_= n1d*(1-(T0/Tb))/(1-(T0/Tr))
+n2c_= n1d*(1-(T0/Tc))/(1-(T0/Tr))
+printf(" \n\n PART (D)")
+printf("\n The First law efficiency for all the three cases would remain same and here is %d percent",n1d*100) //The answer provided in the textbook is wrong
+
+printf("\n The Second law efficiency of part (a) is %f percent",n2a_*100)
+
+printf("\n The Second law efficiency of part (b) is %f percent",n2b_*100)
+
+printf("\n The Second law efficiency of part (c) is %f percent",n2c_*100)
+