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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 /2090/CH2/EX2.1/Chapter2_example1.sce | |
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Diffstat (limited to '2090/CH2/EX2.1/Chapter2_example1.sce')
| -rwxr-xr-x | 2090/CH2/EX2.1/Chapter2_example1.sce | 38 |
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diff --git a/2090/CH2/EX2.1/Chapter2_example1.sce b/2090/CH2/EX2.1/Chapter2_example1.sce new file mode 100755 index 000000000..5dd63f29c --- /dev/null +++ b/2090/CH2/EX2.1/Chapter2_example1.sce @@ -0,0 +1,38 @@ +clc
+clear
+//Input data
+d=20;//Cylinder bore diameter in cm
+L=25;//Stroke length in cm
+Vc=1570;//The clearance volume in cm^3
+P1=1;//Pressure at the beginning of the compression in bar
+T1=300;//Temperature at the beginning of the compression in K
+T3=1673;//The maximum temperature of the cycle in K
+pi=3.141;//Mathematical constant value of pi
+Cv=0.718;//specific heat at constant volume for air in kJ/kgK
+R=0.287;//Real gas constant in kJ/kgK
+g=1.4;//Isentropic index
+c=500;//Number of cycles per minute
+
+//Calculations
+Vs=(pi/4)*d^2*L;//Swept volume in cm^3
+V1=Vs+Vc;//According to pv diagram Total volume i.e sum of swept and clearance volume in cm^3
+V2=Vc;//Volume according to pv diagram in cm^3
+r=V1/V2;//Compression ratio
+T2=T1*r^(g-1);//In isentropic process, Temperature at point 2 in degree centigrade
+P2=P1*r^g;//In isentropic process, Pressure at point 2 in bar
+P3=P2*(T3/T2);//In constant volume, process Pressure at point 3 in bar
+T4=T3*(1/r)^(g-1);//In isentropic process, Temperature at point 4 in degree centigrade
+P4=P3*(1/r)^(g);//In isentropic process, Pressure at point 4 in bar
+no=(1-(1/r)^(g-1))*100;//Air standard efficiency of otto cycle
+Q1=Cv*(T3-T2);//Heat supplied in kJ/kg
+Q2=Cv*(T4-T1);//Heat rejected in kJ/kg
+W=Q1-Q2;//Work done per unit mass in kJ/kg
+m=[(P1*10^5*V1*10^-6)/(R*T1)]/1000;//The amount of mass in kg
+W1=W*m;//Work done in kJ
+pm=[(W1*10^3)/(Vs*10^-6)]/10^5;//Mean effective pressure in N/m^2
+P=W1*(c/60);//Power developed in kW
+
+//Output
+printf('Temperature at point 2 = %3.1f K \n Pressure at point 2 = %3.3f bar \n Pressure at point 3 = %3.2f bar \n Temperature at point 4 = %3.0f K \n Pressure at point 4 = %3.3f bar \n Air standard efficiency of otto cycle = %3.4f percent \n Work done = %3.2f kJ \n Mean effective pressure = %3.3f bar \n Power developed = %3.1f kW ',T2,P2,P3,T4,P4,no,W1,pm,P)
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