// Given:- Tnot = 360.00 // in kelvin pnot = 1.00 // in MPa A2 = 0.001 // in m^2 k = 1.4 // Calculations pstarbypnot = (1+(k-1)/2)**(k/(1-k)) pstar = pstarbypnot*pnot // Part(a) // Since back pressure of 500 kpa is less than critical pressure pstar(528kpa in this case) found above, the nozzle is choked // At the exit M = 1.00 p2 = pstar // in MPa T2 = Tnot/(1+((k-1)/2)*(M**2)) // exit temperature in kelvin R = 8.314 // universal gas constant, in SI units Mwt = 28.97 // molar mass of air in grams V2 = ((k*(R/Mwt)*T2*10**3)**0.5) // exit velocity in m/s mdot = (p2/((R/Mwt)*T2))*A2*V2*10**3 // mass flow rate in kg/s // Results printf( ' The exit mach number for back pressure of 500kpa is: %.2f',M) printf( ' The mass flow rate in kg/s for back pressure of 500kpa is: %.2f',mdot) // Part(b) // Since the back pressure of 784kpa is greater than critical pressure of pstar determined above,the flow throughout the nozzle is subsonic and the exit pressure equals the back pressure, p2 = 784.00 // exit pressure in kpa // Calculations M2 = (((2.00)/(k-1))*(((pnot*10**3)/p2)**((k-1)/k)-1))**0.5 // exit mach number T2 = Tnot/(1+((k-1)/2)*(M2**2)) // exit temperature in kelvin V2 = M2*((k*(R/Mwt)*10**3*T2)**0.5) // exit velocity in m/s mdot2 = (p2/((R/Mwt)*T2))*A2*V2 // mass flow rate in kg/s // Results printf( ' The mass flow rate at the exit for back pressure of 784kpa is: %.2f kg/s.',mdot2) printf( ' The exit mach number for back pressure of 784 kpa is: %.2f',M2)