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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/CH5/EX5.36 | |
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initial commit / add all books
Diffstat (limited to '2417/CH5/EX5.36')
-rwxr-xr-x | 2417/CH5/EX5.36/Ex5_36.sce | 13 |
1 files changed, 13 insertions, 0 deletions
diff --git a/2417/CH5/EX5.36/Ex5_36.sce b/2417/CH5/EX5.36/Ex5_36.sce new file mode 100755 index 000000000..38f0ba3e6 --- /dev/null +++ b/2417/CH5/EX5.36/Ex5_36.sce @@ -0,0 +1,13 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.36\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.36 (page no. 219)
+// Solution
+
+//Because the tank volume is 10 ft^3,the final specific volume of the steam is 10 ft^3/lbm.Interpolations in Table A.2 yield a final pressure of 42 psia.The heat added is simply difference in internal energy between the two states.
+u2=1093.0; //internal energy //Btu/lbm
+u1=117.95; //internal energy //Btu/lbm
+q=u2-u1; //heat added //Btu/lbm
+printf("The final pressure is 42 psia and the heat added is %f Btu/lbm\n",q);
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