//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))