1 files changed, 28 insertions, 0 deletions
diff --git a/2417/CH8/EX8.2/Ex8_2.sce b/2417/CH8/EX8.2/Ex8_2.sce
new file mode 100755
index 000000000..a9b1049dc
--- /dev/null
+++ b/2417/CH8/EX8.2/Ex8_2.sce
@@ -0,0 +1,28 @@
+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.2\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.2  (page no. 381) 
+// Solution
+
+//Using the computer disk to obtain the neccesary properties
+printf("Solution for (a)\n");
+//For the conditions given in problem8.1,the properties are found to be
+hf=340.49; //Unit:kJ/kg //at 50kPa //enthalpy
+h1=hf; //at 50kPa //hf=enthalpy of saturated liquid
+h2=h1; //Enthalpy //Unit:kJ/kg
+h4=3230.9; //Unit:kJ/kg //enthalpy
+h5=2407.4; //Unit:kJ/kg //enthalpy
+//Neglecting pump work 
+nR=(h4-h5)/(h4-h2); //Thermal efficiency of the cycle
+printf("The thermal efficiency of the cycle is %f percentage\n\n",nR*100);
+
+printf("Solution for (b)\n");
+//For the pump work,we do not need the approximation,because the computerized tables give us the necessary values directly.
+//Assuming that the condensate leaving the condenser is saturated liquid gives us an enthalpy of 340.54 kJ/kg and an entropy of 1.0912 kJ/kg*K for an isentropic compression, the final cond-ition is the boiler pressure of 3Mpa and an entropy of 1.0912 kJ/kg*K. For these values,the program yields an enthalpy of 343.59 kJ/kg*K.The isentropic pump work is equal to 
+Pumpwork=343.59-340.54; //Unit:kJ/kg //pumpwork
+//The efficiency of the cycle including pump work is
+nR=((h4-h5)-Pumpwork)/((h4-h1)-Pumpwork); //Thermal efficiency of the cycle
+printf("The thermal efficiency of the cycle including pump work is %f percentage\n\n",nR*100);
+//Final results in this problem agree with the result in problem8.1