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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 /1775/CH2/EX2.22/Chapter2_Example22.sce | |
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| -rwxr-xr-x | 1775/CH2/EX2.22/Chapter2_Example22.sce | 45 |
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diff --git a/1775/CH2/EX2.22/Chapter2_Example22.sce b/1775/CH2/EX2.22/Chapter2_Example22.sce new file mode 100755 index 000000000..cc6ebb314 --- /dev/null +++ b/1775/CH2/EX2.22/Chapter2_Example22.sce @@ -0,0 +1,45 @@ +//Chapter-2, Illustration 22, Page 81
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+C1=280;//Velocity of aircraft in m/s
+P1=48;//Pressure at point 1 kPa
+T1=260;//Temperature at point 1 in K
+rp=13;//Pressure ratio
+T4=1300;//Temperature at point 4 in K
+Cp=1005;//Specific heat at constant pressure in J/kg
+y=1.4;//Ratio of specific heats
+
+//CALCULATIONS
+x=(y-1)/y;//Ratio
+T2=T1+((C1^2)/(2*Cp));//Temperature at point 2 in K
+P2=P1*((T2/T1)^(1/x));//Pressure at point 2 in kPa
+P3=rp*P2;//Pressure at point 3 in kPa
+P4=P3;//Pressure at point 4 in kPa
+T3=T2*(rp^x);//Temperature at point 3 in K
+T5=T4-T3+T2;//Temperature at point 5 in K
+P5=P4*((T5/T4)^(1/x));//Pressure at point 5 in kPa
+P6=P1;//Pressure at point 6 in kPa
+T6=T5*((P6/P5)^x);//Temperature at point 6 in K
+C6=sqrt(2*Cp*(T5-T6));//Velocity of air at nozzle exit in m/s
+W=(C6-C1)*C1;//Propulsive power in J/kg
+Q=Cp*(T4-T3);//Total heat transfer rate in J/kg
+nP=(W/Q)*100;//Propulsive efficiency
+
+//OUTPUT
+mprintf('Pressure at the turbine exit is %3.1f kPa \n Velocity of exhaust gases are %3.1f m/s \n Propulsive efficiency is %3.1f percent',P5,C6,nP)
+
+
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
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