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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/CH5/EX5.22/Example5_22.sce | |
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Diffstat (limited to '647/CH5/EX5.22/Example5_22.sce')
| -rwxr-xr-x | 647/CH5/EX5.22/Example5_22.sce | 33 |
1 files changed, 33 insertions, 0 deletions
diff --git a/647/CH5/EX5.22/Example5_22.sce b/647/CH5/EX5.22/Example5_22.sce new file mode 100755 index 000000000..87d682b7e --- /dev/null +++ b/647/CH5/EX5.22/Example5_22.sce @@ -0,0 +1,33 @@ +clear;
+clc;
+
+// Example: 5.22
+// Page: 174
+
+printf("Example: 5.22 - Page: 174\n\n");
+
+// Solution
+
+//*****Data*****//
+T1 = 400;// [K]
+P1 = 300;// [kPa]
+V1 = 1;// [cubic m]
+V2 =2;// [cubic m]
+R = 8.314;// [kJ/kmol K]
+//**************//
+
+// Since the system is well insulated, there is no scope of transferring heat between system & surrounding.
+deltaQ = 0;// [kJ]
+deltaW = 0;// [kJ]
+// By first law of thermodynamics:
+deltaU =deltaQ - deltaW;// [kJ]
+// As the internal energy of the gas depends only on temperature,
+deltaT = 0;// [K]
+T2 = T1 + deltaT;// [K]
+P2 = (P1*V1/T1)*(T2/V2);// [kPa]
+n = P1*V1/(R*T1);// [kmol]
+deltaS_system = n*R*log(P1/P2);// [kJ/K]
+// Since process is adiabatic:
+deltaS_surrounding = 0;// [kJ/K]
+deltaS = deltaS_system + deltaS_surrounding;// [kJ/K]
+printf("Change in Entropy of the gas is %.4f kJ/K",deltaS);
\ No newline at end of file |