Unformatted text preview: A/A* =3.9, we get P o /P b =1.018. Item b is from the limit for part a down to when there is a shock wave in the exit plane of the nozzle for otherwise perfect supersonic expansion, so first do part c. This back pressure that divides c and d is the supersonic branch of the AreaMach number relation. Thus, from IFT at A/A* =7.8, we get P o /P b =94.02 at M e =3.650 (this nearest value in table) and for A/A* =3.9, we get P o /P b =31.59 at M e =2.900. The answer to part b includes flow through from a normal shock at the exit Mach number, and thus, P 2 /P 1 = 9.645 for M e =3.650 and P 2 /P 1 = 15.38 for M e =2.900. Thus, Problem Part Maximum back pressure Minimum back pressure 1 a 12 atm 11.95 atm b 11.95 atm 1.97 atm c 1.97 atm 0.127 atm d 0.127 atm 0 2 a 12 atm 11.8 atm b 11.8 atm 3.66 atm c 3.66 atm 0.380 atm d 0.380 atm 0...
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 Fall '09
 COLLICOTT
 Fluid Dynamics, Mach number, Shock wave, normal shock wave, 12 atmospheres

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