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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /647/CH10/EX10.16/Example10_16.sce | |
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diff --git a/647/CH10/EX10.16/Example10_16.sce b/647/CH10/EX10.16/Example10_16.sce new file mode 100755 index 000000000..ee0025c9c --- /dev/null +++ b/647/CH10/EX10.16/Example10_16.sce @@ -0,0 +1,94 @@ +clear;
+clc;
+
+// Example: 10.16
+// Page: 429
+
+printf("Example: 10.16 - Page: 429\n\n");
+
+// Solution
+
+// Dew point Pressure
+//*****Data******//
+y1 = 0.10;// [mole fraction of methane in vapour phase]
+y2 = 0.20;// [mole fraction of ethane in vapour phase]
+y3 = 0.70;// [mole fraction of propane in vapour phase]
+T = 10;// [OC]
+//*************//
+
+// Assume P = 690 kPa
+P = 690;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 20.0;
+K2 = 3.25;
+K3 = 0.92;
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+// Since x1 + x2 +x3 < 1, so we assume another value of P = 10135 kPa at 10 OC.
+P = 10135;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 13.20;
+K2 = 2.25;
+K3 = 0.65;
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+// Since x1 + x2 +x3 > 1, so we assume another value of P = 870 kPa at 10 OC.
+P = 870;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 16.0;
+K2 = 2.65;
+K3 = 0.762;
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+// Since x1 + x2 +x3 = 1. Therefore:
+printf("Dew Pressure is %d kPa\n",P);
+printf("Composition of the liquid drop:\n x1 = %.4f\n x2 = %.4f\n x3 = %.4f\n",x1,x2,x3);
+printf("\n");
+
+// Bubble point Pressure
+//*****Data******//
+x1 = 0.10;// [mole fraction of methane in vapour phase]
+x2 = 0.20;// [mole fraction of ethane in vapour phase]
+x3 = 0.70;// [mole fraction of propane in vapour phase]
+T = 10;// [OC]
+//*************//
+
+// Assume P = 2622 kPa
+P = 2622;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 5.60;
+K2 = 1.11;
+K3 = 0.335;
+// From Eqn. 10.83:
+y1 = K1*x1;
+y2 = K2*x2;
+y3 = K3*x3;
+// Since x1 + x2 +x3 > 1, so we assume another value of P = 2760 kPa at 10 OC.
+P = 2760;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 5.25;
+K2 = 1.07;
+K3 = 0.32;
+// From Eqn. 10.83:
+y1 = K1*x1;
+y2 = K2*x2;
+y3 = K3*x3;
+// Since x1 + x2 +x3 < 1, so we assume another value of P = 2656 kPa at 10 OC.
+P = 2656;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 5.49;
+K2 = 1.10;
+K3 = 0.33;
+// From Eqn. 10.83:
+y1 = K1*x1;
+y2 = K2*x2;
+y3 = K3*x3;
+// Since x1 + x2 +x3 = 1. Therefore:
+printf("Bubble Pressure is %d kPa\n",P);
+printf("Composition of the vapour bubble:\n y1 = %.4f\n y2 = %.4f\n y3 = %.4f",y1,y2,y3);
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