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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/EX3.10/Example3_10.sce | |
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| -rwxr-xr-x | 647/CH3/EX3.10/Example3_10.sce | 42 |
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diff --git a/647/CH3/EX3.10/Example3_10.sce b/647/CH3/EX3.10/Example3_10.sce new file mode 100755 index 000000000..7f8a27f88 --- /dev/null +++ b/647/CH3/EX3.10/Example3_10.sce @@ -0,0 +1,42 @@ +clear;
+clc;
+
+// Example: 3.10
+// Page: 103
+
+printf("Example: 3.10 - Page: 103\n\n");
+
+// Solution
+
+//*****Data*****//
+beeta = 1.487*10^(-3);// [1/OC]
+alpha = 62*10^(-6);// [1/bar]
+V1 = 1.287;// [cubic cm /g]
+//************//
+
+// Solution (a)
+// The value of derivative (dP/dT) at constant V:
+// dV/V = beeta*dT - alpha*dP
+// dV = 0
+// dP/dT = beeta/alpha
+// Value = dP/dT
+Value = beeta/alpha;// [bar/OC]
+printf("Value of derivative is %.2f bar/OC\n",Value);
+
+// Solution (b)
+P1 = 1;// [bar]
+T1 = 20;// [OC]
+T2 = 30;// [OC]
+// Applying the same equation:
+P2 = P1 +(beeta/alpha)*(T2 - T1);// [bar]
+printf("The pressure generated by heating at constant Volume is %.2f Pa\n",P2);
+
+// Solution (c)
+T2 = 0;// [OC]
+T1 = 20;// [OC]
+P2 = 10;// [bar]
+P1 = 1;// [bar]
+// The change in Volume can be obtained as:
+V2 = V1*exp((beeta*(T2 - T1)) - alpha*(P2 - P1));// [cubic cm/g]
+deltaV = V2 - V1;// [cubic cm/g]
+printf("The change in Volume is %.3f cubic cm/g\n",deltaV);
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