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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+clc;
+clear;
+x1=0.5;//m
+x2=0.3;//m
+Acritical=0.1;//m^2
+//at x1, Max1 is found as
+Max1=2.8;
+//and px/p0,x is found as
+pratio1=0.04;
+//For this value of Max, py/px is found as
+prat1=9;
+pfraction1=prat1*pratio1;//where pfraction=py/p0,x = pIII/p0,x
+//at x2, Max2 is found as
+Max2=2.14;
+//for this value of Max2, the following are found
+prat2=5.2;
+prat22=0.66;//where prat22=p0,y/p0,x
+May=0.56;
+//for this valur of May, Ay/Acritical is found as
+Aratio=1.24;
+Arat=(Acritical+(x1^2))/(Acritical+(x2^2));//where Aratio=A2/Ay
+Afraction=Aratio*Arat;//where Afraction=A2/Acritical
+A2=Acritical+(x1^2);//m^2
+Acritical1=A2/Afraction;//where Acritical1 critical area for the isentropic flow downstream of the shock
+//with the value of Afraction, the following are found
+Ma2=0.26;
+pfraction=0.95;//where pfraction=p2/p0,y
+//hence,
+pfrac=pfraction*prat22;//where pfrac=p2/p0,x
+disp(pfraction1,"The ratio of back pressure to inlet stagnation pressure that will result in a normal shock at the exit of the duct=")
+disp(pfrac,"The value of back pressure to inlet stagnation pressure required to position the shock at (x=0.3 m)=")