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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 /572/CH12/EX12.6/c12_6.sce | |
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initial commit / add all books
Diffstat (limited to '572/CH12/EX12.6/c12_6.sce')
| -rwxr-xr-x | 572/CH12/EX12.6/c12_6.sce | 83 |
1 files changed, 83 insertions, 0 deletions
diff --git a/572/CH12/EX12.6/c12_6.sce b/572/CH12/EX12.6/c12_6.sce new file mode 100755 index 000000000..675a8e27d --- /dev/null +++ b/572/CH12/EX12.6/c12_6.sce @@ -0,0 +1,83 @@ +//(12.6) At steady state, 100 m3/min of dry air at 32C and 1 bar is mixed adiabatically with a stream of oxygen (O2) at 127C and 1 bar to form a mixed stream at 47C and 1 bar. Kinetic and potential energy effects can be ignored. Determine (a) the mass flow rates of the dry air and oxygen, in kg/min, (b) the mole fractions of the dry air and oxygen in the exiting mixture, and (c) the time rate of entropy production, in kJ/K . min
+
+//solution
+
+//variable initialization
+T1 = 32 //temperature of dry air in degree celcius
+p1 = 1 //pressure of dry air in bar
+AV1 = 100 //volume rate of dry air in m^3/min
+T2 = 127 //temperature of oxygen stream in degree celcius
+p2 = 1 //pressure of oxygen stream in bar
+T3 = 47 //temperature of mixed stream in degree celcius
+p3 = 1 //pressure of mixed stream in bar
+
+//part(a)
+Rbar = 8314 //universal gas constant
+Ma = 28.97 //molar mass of air
+Mo = 32 //molar mass of oxygen
+
+va1 = (Rbar/Ma)*(T1+273)/(p1*10^5) //specific volume of air in m^3/kg
+ma1dot = AV1/va1 //mass flow rate of dry air in kg/min
+
+//from table A-22 and A-23
+haT3 = 320.29 //in kj/kg
+haT1 = 305.22 //in kj/kg
+hnotT2 = 11711 //in kj/kmol
+hnotT1 = 9325 //in kj/kmol
+
+modot = ma1dot*(haT3-haT1)/[(1/Mo)*(hnotT2-hnotT1)] //in kg/min
+printf('the mass flow rate of dry air in kg/min is: %f',ma1dot)
+printf('\nthe mass flow rate of oxygen in kg/min is: %f',modot)
+
+//part(b)
+nadot = ma1dot/Ma //molar flow rate of air in kmol/min
+nodot = modot/Mo //molar flow rate of oxygen in kmol/min
+
+ya = nadot/(nadot+nodot) //mole fraction of air
+yo = nodot/(nadot+nodot) //mole fraction of oxygen
+
+printf('\n\nthe mole fraction of dry air in the exiting mixture is: %f',ya)
+printf('\nthe mole fraction of dry oxygen in the exiting mixture is: %f',yo)
+
+//part(c)
+//with the help of tables A-22 and A-23
+sanotT3 = 1.7669 //in kj/kg.K
+sanotT1 = 1.71865 //in kj/kg.K
+sbarT3 = 207.112 //in kj/kmol.K
+sbarT2 = 213.765 //in kj/kmol.K
+
+sigmadot = ma1dot*[sanotT3-sanotT1-(8.314/Ma)*log(ya)] + (modot/Mo)*[sbarT3-sbarT2-8.314*log(yo)]
+printf('\n\nthe time rate of entropy production, in kJ/K . min is: %f',sigmadot)
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