//(9.14) A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1.0 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4, determine the mass flow rate, in kg/s, and the exit Mach number for back pressures of (a) 500 kPa and (b) 784 kPa. //solution //variable initialization Tnot = 360 //in kelvin pnot = 1 //in MPa A2 = .001 //in m^2 k = 1.4 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 p2 = pstar //in MPa printf('the exit mach number for back pressure of 500kpa is: %f',M) T2 = Tnot/(1+((k-1)/2)*(M^2)) //exit temperature in kelvin R = 8.314 //universal gas constant, in SI units M = 28.97 //molar mass of air in grams V2 = sqrt(k*(R/M)*T2*10^3) //exit velocity in m/s mdot = (p2/((R/M)*T2))*A2*V2*10^3 //mass flow rate in kg/s printf('\nthe mass flow rate in kg/s for back pressure of 500kpa is: %f',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 //exit pressure in kpa M2 = {(2/(k-1))*[(pnot*10^3/p2)^((k-1)/k)-1]}^.5 //exit mach number T2 = Tnot/(1+((k-1)/2)*(M2^2)) //exit temperature in kelvin V2 = M2*sqrt(k*(R/M)*10^3*T2) //exit velocity in m/s mdot2 = (p2/((R/M)*T2))*A2*V2 //mass flow rate in kg/s printf('\n\nthe mass flow rate at the exit in kg/s for back pressure of 784kpa is: %f',mdot2) printf('\nthe exit mach number for back pressure of 784 kpa is: %f',M2)