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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/CH3/EX3.3/Example3_3.sce | |
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Diffstat (limited to '647/CH3/EX3.3/Example3_3.sce')
| -rwxr-xr-x | 647/CH3/EX3.3/Example3_3.sce | 28 |
1 files changed, 28 insertions, 0 deletions
diff --git a/647/CH3/EX3.3/Example3_3.sce b/647/CH3/EX3.3/Example3_3.sce new file mode 100755 index 000000000..f7d8a16ca --- /dev/null +++ b/647/CH3/EX3.3/Example3_3.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+// Example: 3.3
+// Page: 89
+
+printf("Example: 3.3 - Page: 89\n\n");
+
+// Solution
+
+//*****Data*****//
+V1 = 6;// [cubic m]
+P1 = 500;// [kPa]
+R = 0.287;// [kJ/kg K]
+//*************//
+
+// Applying the charectarstic equation to the gas initially:
+// P1*V1 = m1*R*T1.......................................(i)
+// Applying the charectarstic equation to the gas which was left in the vessel after one-fifth of the gas has been removed:
+// P2*V2 = m2*R*T2.......................................(ii)
+// V2 = V1;
+// T2 = T1;
+// m2 = (4/5)*m1;
+// Eqn (ii) becomes:
+// P2*V1 = (4/5)*m1*R*T1..................................(iii)
+// Dividing eqn (i) by eqn (iii), we get:
+P2 = (4/5)*P1;// [kPa]
+printf("The pressure of the remaining air is %d kPa\n",P2);
\ No newline at end of file |