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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 /611/CH4/EX4.13 | |
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-rwxr-xr-x | 611/CH4/EX4.13/Chap4_Ex13_R1.sce | 32 |
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diff --git a/611/CH4/EX4.13/Chap4_Ex13_R1.sce b/611/CH4/EX4.13/Chap4_Ex13_R1.sce new file mode 100755 index 000000000..b2a219109 --- /dev/null +++ b/611/CH4/EX4.13/Chap4_Ex13_R1.sce @@ -0,0 +1,32 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 13,Page 112
+//Title:Final temperature and amount of gas entering the tank
+//================================================================================================================
+clear
+clc
+
+//INPUT
+V=1;//volume of tank in m^3
+T0=300;//initial temperature of ideal gas in K
+P0=0.1;//initial pressure of ideal gas in MPa
+T=500;//temperature of ideal gas in the pipeline in K
+P=3;//pressure of ideal gas in the pipeline in MPa
+R=8.314;//universal gas constant in J/molK
+gaamma=1.4;//ratio of the molar heat capacities at constant pressure and constant volume for ideal gas (no unit)
+
+//CALCULATION
+Pf=3;//final pressure reached in the tank in MPa
+//calculation of final temperature of the gas in the tank in K using Eq.(4.44) (and applying u=Cv*T, h=Cp*T and N=P*V/R*T as the gas is taken to be ideal)
+Tf=(Pf*10^6)/((((Pf*10^6)-(P0*10^6))/(gaamma*T))+((P0*10^6)/T0));
+//calculation of the moles of ideal gas entering into the tank using Eq.(4.44) (and applying u=Cv*T, h=Cp*T and N=P*V/R*T as the gas is taken to be ideal)
+N=(V/R)*(((Pf*10^6)/Tf)-((P0*10^6)/T0));
+
+//OUTPUT
+mprintf('\n The final temperature= %0.1f K\n',Tf);
+mprintf('\n The amount of gas that has entered the tank= %0.2f mol\n',N);
+
+//===============================================END OF PROGRAM===================================================
+
+
+
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