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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 /2417/CH6/EX6.19 | |
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initial commit / add all books
Diffstat (limited to '2417/CH6/EX6.19')
-rwxr-xr-x | 2417/CH6/EX6.19/Ex6_19.sce | 16 |
1 files changed, 16 insertions, 0 deletions
diff --git a/2417/CH6/EX6.19/Ex6_19.sce b/2417/CH6/EX6.19/Ex6_19.sce new file mode 100755 index 000000000..e6e8ee493 --- /dev/null +++ b/2417/CH6/EX6.19/Ex6_19.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.19\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.19 (page no. 262)
+// Solution
+
+//from the equation, deltas/cv = (k*log(v2/v1))+ log(p2/p1) //change in entropy
+k=1.4; //k=cp/cv //ratio of specific heats
+//deltas=(1/4)*cv //so,
+// 1/4= (k*log(v2/v1))+ log(p2/p1)
+v2=1/2; //Because,v2=(1/2)*v1 //initial specific volume
+v1=1; //final specific volume
+p2byp1=exp((1/4)-(k*log(v2/v1))); //increase in pressure
+printf("p2/p1=%f\n",p2byp1);
+printf("So,increase in pressure is %f ",p2byp1);
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