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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 |
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diff --git a/611/CH5/EX5.19/Chap5_Ex19_R1.sce b/611/CH5/EX5.19/Chap5_Ex19_R1.sce new file mode 100755 index 000000000..2c72ed36f --- /dev/null +++ b/611/CH5/EX5.19/Chap5_Ex19_R1.sce @@ -0,0 +1,29 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 19,Page 186
+//Title: Power consumed by the compressor
+//================================================================================================================
+clear
+clc
+
+//INPUT
+Ti=25;//temperature of air taken in by the adiabatic air compressor in degree celsius
+Pi=0.1;//pressure of air taken in by the adiabatic air compressor in MPa
+Pe=1;//discharge pressure of air in MPa
+n_c=0.8;//isentropic efficiency of the compressor (no unit)
+gaamma=1.4;//ratio of molar specific heat capacities (no unit)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+Ti=Ti+273.15;//conversion of temperature in K
+Te=Ti*(((Pe*10^6)/(Pi*10^6))^((gaamma-1)/gaamma));//calculation of the discharge temperature of air using Eq.(4.35) in K (for reversible and adiabatic compression)
+W_s=(((R*gaamma)/(gaamma-1))*(Te-Ti))*10^-3;//calculation of the power consumed by the isentropic compressor using Eq.(5.69) in kW
+Ws=W_s/n_c;//calculation of the power consumed by an actual compressor per mole of air using Eq.(5.68)in kW
+Te_actual=((Ws*10^3*(gaamma-1))/(R*gaamma))+Ti;//calculation of the exit temperature of air in K
+
+//OUTPUT
+mprintf("\n The exit temperature of air=%0.2f K\n",Te_actual);
+mprintf("\n The power consumed by the compressor =%f kW/mol\n",Ws);
+
+
+//===============================================END OF PROGRAM===================================================
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