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HW29solution

# HW29solution - A/A =3.9 we get P o/P b =1.018 Item b is...

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AAE 334, Fall 2011 Homework 29 Solutions 1. A converging-diverging nozzle operates with an upstream stagnation pressure of 12 atmospheres. The ratio of exit area to throat area is 7.8. If the gas is calorically perfect air, determine the back pressures at which the nozzle will operate: a. Fully subsonic with or without choked throat b. With a normal shock wave in the diverging section c. Over-expanded, and d. Under-expanded 2. If the exit area in #1 is halved, determine the back pressures at which the nozzle will operate: a. Fully subsonic with or without choked throat b. With a normal shock wave in the diverging section c. Over-expanded, and d. Under-expanded Item a in each case is limited on the low exit-pressure side by flow along the subsonic branch of the Area-Mach number relation. Thus, from IFT at A/A* =7.8, we get P o /P b =1.004 (this nearest value in table) and for
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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 Area-Mach 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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