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+//clear//

+clear;

+clc;

+

+//Example 6.1

+//Given

+gama = 1.4;

+M = 29;

+R = 82.0568*10^-3; //[atm-m^3/Kg mol-K]

+Nma = 0.8;

+gc = 1; //[ft-lb/lbf-s^2]

+//At Entrance 

+p0 = 20; //[atm]

+T0 = 555.6; //[K]

+

+//(a)

+// Using Eq.(6.28)

+//Pressure at throat

+pt  = (1/(1+((gama-1)/2)*Nma^2)^(1/(1-1/gama)))*p0 //[atm]

+//From Eq.(6.10)

+rho0 = (p0*M)/(R*T0); //[kg/m^3]

+// Using Eq.(6.10) and Eq.(6.26), the velocity in the throat

+ut = sqrt((2*gama*gc*R*T0)/(M*(gama-1))*(1-(pt/p0)^(1-1/gama))); // [m^3-am/kg]^0.5

+//In terms of [m/s], Using Appendix 2, 1 atm = 1.01325*10^ [N/m^2]

+ut = ut*sqrt(1.01325*10^5) //[m/s]

+//Using Eq.(6.23), density at throat 

+rho_t  = rho0*(pt/p0)^(1/gama) //[kg/m^3]

+//The mass velocity at the throat,

+Gt = ut*rho_t //[kg/m^2-s]

+//Using Eq.(6.24), The temperature at throat 

+Tt = T0*(pt/p0)^(1-1/gama) // [K]

+

+//(b)

+// From Eq.(6.29)

+pstar = ((2/(gama+1))^(1/(1-1/gama)))*p0 //[atm]

+//From Eq.(6.24) and (6.29)

+Tstar = T0*(pstar/p0)^(1-1/gama) //[K]

+//From Eq.(6.23)

+rho_star = rho0*(pstar/p0)^(1/gama) //[Kg/m^3]

+//From Eq.(6.30)

+G_star = sqrt(2*gama*gc*rho0*p0*101.325*10^3/(gama-1))*(pstar/p0)^(1/gama)*sqrt(1-(pstar/p0)^(1-1/gama)) //[Kg-m^2/s]

+u_star = G_star/rho_star //[m/s]

+

+//(c) 

+// By continuity, G inversely proportional to S, the mass velocity at dischage is

+G_r  = G_star/2 // [Kg/m^3-s]

+//Using Eq.(6.30)

+// Let x = pr/p0

+err = 1;

+eps = 10^-3;

+x = rand(1,1);

+

+while(err>eps)  

+  xnew  = ((0.1294)/sqrt(1-x^(1-1/1.4)))^1.4;

+  err = x-xnew;

+  x=xnew;

+end

+

+//Using Eq.(6.27)

+//The Mach Number at dischage is

+Nmr = sqrt((2/(gama-1))*(1/x^(1-1/gama)-1))