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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /2705/CH5/EX5.10 | |
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initial commit / add all books
Diffstat (limited to '2705/CH5/EX5.10')
-rwxr-xr-x | 2705/CH5/EX5.10/Ex5_10.sce | 27 |
1 files changed, 27 insertions, 0 deletions
diff --git a/2705/CH5/EX5.10/Ex5_10.sce b/2705/CH5/EX5.10/Ex5_10.sce new file mode 100755 index 000000000..b5d45cd32 --- /dev/null +++ b/2705/CH5/EX5.10/Ex5_10.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+disp('Example 5.10');
+
+// aim : To determine
+// the new temperature of the gas
+
+// Given values
+V1 = .015;// original volume,[m^3]
+T1 = 273+285;// original temperature,[K]
+V2 = .09;// final volume,[m^3]
+
+// solution
+// Given gas is following the law,P*V^1.35=constant
+// so process is polytropic with
+n = 1.35; // polytropic index
+
+// hence
+T2 = T1*(V1/V2)^(n-1);// final temperature, [K]
+
+t2 = T2-273;// [C]
+
+mprintf('\n The new temperature of the gas is = %f C \n',t2);
+
+// there is minor error in book's answer
+
+// End
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