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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /2417/CH8/EX8.1/Ex8_1.sce | |
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-rwxr-xr-x | 2417/CH8/EX8.1/Ex8_1.sce | 27 |
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diff --git a/2417/CH8/EX8.1/Ex8_1.sce b/2417/CH8/EX8.1/Ex8_1.sce new file mode 100755 index 000000000..991fdefef --- /dev/null +++ b/2417/CH8/EX8.1/Ex8_1.sce @@ -0,0 +1,27 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.1\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.1 (page no. 380)
+// Solution
+
+//From the Steam Tables or Mollier chart in Appendix 3,we find that
+hf=340.49; //Unit:kJ/kg //at 50kPa //enthalpy
+h1=hf; //at 50kPa //hf=enthalpy of saturated liquid //Unit:kJ/kg
+h4=3230.9; //Unit:kJ/kg //enthalpy
+h5=2407.4; //Unit:kJ/kg ////enthalpy
+//Here,point 5 is in the wet steam region.
+printf("Solution for (a)\n");
+//Neglecting pump work (h2=h1) gives
+nR=(h4-h5)/(h4-h1); //Thermal efficiency of the cycle
+printf("The thermal efficiency of the cycle is %f percentage\n\n",nR*100);
+
+printf("Solution for (b)\n");
+p2=3000; //Unit:kPa //Upper pressure
+p1=50; //Unit:kPa //Lower pressure
+vf=0.001030; //Specific volume of saturated liquid //m^3/kg
+Pumpwork=(p2-p1)*vf; //Unit:kJ/kg //pump work
+//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);
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