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Diffstat (limited to '2873/CH5/EX5.15/Ex5_15.sce')
-rwxr-xr-x | 2873/CH5/EX5.15/Ex5_15.sce | 20 |
1 files changed, 20 insertions, 0 deletions
diff --git a/2873/CH5/EX5.15/Ex5_15.sce b/2873/CH5/EX5.15/Ex5_15.sce new file mode 100755 index 000000000..ddab98c28 --- /dev/null +++ b/2873/CH5/EX5.15/Ex5_15.sce @@ -0,0 +1,20 @@ +// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Engineering Thermodynamics by Onkar Singh Chapter 5 Example 15")
+p1=0.5;//initial pressure of air in Mpa
+T1=400;//initial temperature of air in K
+p2=0.3;//final pressure of air in Mpa
+T2=350;//initial temperature of air in K
+R=0.287;//gas constant in KJ/kg K
+Cp=1.004;//specific heat at constant pressure in KJ/kg K
+disp("let the two points be given as states 1 and 2,")
+disp("let us assume flow to be from 1 to 2")
+disp("so entropy change(deltaS1_2)=s1-s2=Cp*log(T1/T2)-R*log(p1/p2)in KJ/kg K")
+deltaS1_2=Cp*log(T1/T2)-R*log(p1/p2)
+disp("deltaS1_2=s1-s2=0.01254 KJ/kg K")
+disp("it means s2 > s1 hence the assumption that flow is from 1 to 2 is correct as from second law of thermodynamics the entropy increases in a process i.e s2 is greater than or equal to s1")
+disp("hence flow occurs from 1 to 2 i.e from 0.5 MPa,400K to 0.3 Mpa & 350 K")
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