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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 /2705/CH5/EX5.9/Ex5_9.sce | |
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initial commit / add all books
Diffstat (limited to '2705/CH5/EX5.9/Ex5_9.sce')
| -rwxr-xr-x | 2705/CH5/EX5.9/Ex5_9.sce | 22 |
1 files changed, 22 insertions, 0 deletions
diff --git a/2705/CH5/EX5.9/Ex5_9.sce b/2705/CH5/EX5.9/Ex5_9.sce new file mode 100755 index 000000000..016e3d2ee --- /dev/null +++ b/2705/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+disp('Example 5.9');
+
+// aim : To determine
+// the new pressure of the gas
+
+// Given values
+P1 = 300;// original pressure,[kN/m^2]
+T1 = 273+25;// original temperature,[K]
+T2 = 273+180;// final temperature,[K]
+
+// solution
+// since gas compressing according to the law,P*V^1.4=constant
+// so,for polytropic process,T1/T2=(P1/P2)^((n-1)/n),here n=1.4
+
+// hence
+P2 = P1*(T2/T1)^((1.4)/(1.4-1));// [kN/m^2]
+
+mprintf('\n The new pressure of the gas is = %f kN/m^2\n',P2);
+
+// End
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