solhw6 - MAE135 SOLUTIONS TO HW 6 Spring 2010 PROBLEM 1(10...

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Unformatted text preview: MAE135 SOLUTIONS TO HW 6 Spring 2010 PROBLEM 1 (10 points) In the first three cases the outflow is subsonic, therefore the exit pressure naturally matches the ambient pressure, p e = p a . (a) This is an isentropic flow which reaches M = 1 at the throat, then decelerates to subsonic speed in the diverging nozzle. Using the subsonic branch of the Mach number-area relation, we find A e A * = 4 . → M e = 0 . 15 , p /p e = 1 . 014 Since p e = p a , the nozzle pressure ratio is p /p a = p /p e = 1 . 028. So, ideally, in a nozzle with large area ratio it takes very small effort to achieve sonic flow at the throat. (b) The nozzle pressure ratio can be broken down as follows: p p a = p 01 p e = p 01 p 1 p 1 p 2 p 2 p e Using the supersonic branch of the Mach number-area relation, for A 1 /A * = 2 . 0, we get that the shock Mach number is M 1 = 2 . 2 and p 01 /p 1 =10.7. From the shock tables, for M 1 = 2 . 2, p 2 /p 1 = 5 . 48 and M 2 = 0 . 547. The flow from “2” (immediately past the shock) to “e” (nozzle exit) is subsonic and isentropic. It is supplied by a total pressure p 02 and we can imagine that it came from a sonic throat A ** such that A 2 /A ** corresponds to M 2 = 0 . 547. The Mach number-area relation gives A 2 /A ** =1.27. To find the conditions at “e”, we write...
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solhw6 - MAE135 SOLUTIONS TO HW 6 Spring 2010 PROBLEM 1(10...

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