SOLUTION Applied Aerodynamics (UG & PG) Compressible Flow Assignment 1 SOLUTION Due: 5pm, Friday October 23, 2020 (a) Air flows in a pipe of 50 mm internal diameter. The air enters at supersonic speed at a temperature of 300 K and a flow rate of 1.70 m3/s. The friction coefficient can be assumed constant at Cr = 0.005 ( f= 0.005 in the Anderson text book notation). (i) State any assumptions used to solve this problem. (ii) Find the Mach number of the flow at the pipe entrance. [1 mark] [5 marks] [7 marks] (ii) What is the length of the pipe if the Mach number at exit is 1.5? (b) Consider a convergent-divergent nozzle with an exit-to-throat area ratio of 3.5. The working fluid is air, with k = 1.4. The inlet flow is from a large air storage vessel with a total pressure of 2 atm. The nozzle discharges into ambient air with a static pressure of 1 atm. For this pressure ratio, a normal shock wave stands somewhere inside the divergent section of the nozzle. Using the "direct" or mass-flow-rate matching method, find the exit Mach number of the nozzle. [7 marks] Notes: 1. If using tables, it is acceptable to use the nearest value. 2. For air R = 287 J/kg.K and k = 1.4. 3. Use the problem-solving protocol, state all assumptions, and show all working. 4. Your submission must have a signed cover sheet attached.
1 Compressible Flow Assignment 1. a) Given: 50 mm pipe with frictional air flow Find : Mach number at entrance and the length of the pipe if Mexit = 1.5. Schematic Diagram +Data: 6=0.005 D = 0.05m T1 = 300 K Q= 1.70 m/s 1. L1-2 2 L* M=1 Note: This diagram * is the key to solving this problem . Assume: (1) Steady, workless, frictional, adiabatic flow of an ideal gas (Fanno flow). Solution : (ii) Find M1. We are given Q, = 1.70 m3/s Q = AV1 So V1 = Q/A = 1.70 $ (0-05)2= 865.8 m/s. Speed of sound C=1 KRT =V14×287×300=347.19m/s M1= V16 = 865.8/347.19 = 2.494 =2.5. (11) Find Li-2 if M2 = 1.5. Fram diagram (1-2 = Lx - Lz Multiply by 44: 4 Cc(1-2 = 44Li D D D Table E3 gives MI = 2.5 = 4 Cfli = 0.4320 M2 = 1.5 => 49t= = 0.1361 D D 4 CL2
2 Continued: 80 4CL1-2 = 0.4320 - 0.1361 = 0.2959 D D 0.05×0.2959 .º C1-2 = 46 ×0.2959 = 4 × 0.005 = 0.7398 m 2 0.74 m. b) Use the direct method to find the exit Mach number of a nozzle. The nozzle is specified as follows: Pressure Ratio Pef = 1 = 0.5 (from given data) Poi N Ael Avea Ratio = 3.5 (given) From the mass flow rate matching equation: Pe Ate - Pe Ac _ (Pe)/Ael /Pel Pop At Po, A+ 00 e )(AF) = (*) ( * ) =?×3.5 = 1.75. Mass flow rate matching equation: