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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 /479/CH3/EX3.10/Example_3_10.sce | |
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| -rwxr-xr-x | 479/CH3/EX3.10/Example_3_10.sce | 26 |
1 files changed, 26 insertions, 0 deletions
diff --git a/479/CH3/EX3.10/Example_3_10.sce b/479/CH3/EX3.10/Example_3_10.sce new file mode 100755 index 000000000..c998d9d00 --- /dev/null +++ b/479/CH3/EX3.10/Example_3_10.sce @@ -0,0 +1,26 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.10
+clear;
+clc;
+
+//Given
+V = 0.3;//Volume of the tank in m^3
+P1 = 1;//Initial pressure of the tank in atm
+P2 = 0;//Final pressure of the tank in atm
+T = 298;//Temperature of the tank in K
+t = 10;//evacuation time in min
+
+//delN=(V/(R*T)*delP)..(a) change in moles as V and T are constant
+//delW=delN*R*T*lnP..(b)pump work required
+//From (a)&(b),delW=V*delP*lnP
+
+//To calculate the pump work required
+//On doing integration of dW we will get
+
+W = V*(P1-P2);//pump work done in J/sec
+W1=(W*(1.033*10^4))/(75*600);
+mprintf('The pump work required is %f hp',W1);
+//end
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