initial commit / add all books

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+
+// 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)