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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 /1919/CH4/EX4.10/Ex4_10.sce | |
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initial commit / add all books
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| -rwxr-xr-x | 1919/CH4/EX4.10/Ex4_10.sce | 34 |
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diff --git a/1919/CH4/EX4.10/Ex4_10.sce b/1919/CH4/EX4.10/Ex4_10.sce new file mode 100755 index 000000000..68b4b7c81 --- /dev/null +++ b/1919/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,34 @@ +
+// Theory and Problems of Thermodynamics
+// Chapter 4
+// Energy Analysis of Process
+// Example 10
+
+clear ;clc;
+
+//Given data
+V = 2 // volume in m^3
+gam = 1.67 // gamma ideal gas coefficient
+P0 = 0.1 // Initial Pressure in MPa
+T0 = 300 // Initial Temperature in K
+Pf = 3 // Final Pressure in MPa
+T1 = 500 // Final Temperature in K
+R = 8.314 // Gas constant
+
+// Calculation
+// the first law of thermodynamics for a transient flow and after rewriting
+Tf = Pf/(((Pf-P0)/(gam*T1))+(P0/T0)) // Final temperature
+
+Pf = Pf * 1e6 // units conversion from MPa to Pa
+P0 = P0 * 1e6 // units conversion from MPa to Pa
+
+q = (V/R)*(Pf/Tf - P0/T0)
+
+// Output Results
+mprintf('Temperature of helium in the tank = %6.2f K',Tf)
+mprintf('\nQuantity of helium entered into tank = %6.2f mol',q)
+
+
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