p0836 - 1166 CHAPTER 8. ONE-DIMENSIONAL COMPRESSIBLE FLOW...

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Unformatted text preview: 1166 CHAPTER 8. ONE-DIMENSIONAL COMPRESSIBLE FLOW 8.36 Chapter 8, Problem 36 Problem: A Pitot-static tube is inserted in the test section of a supersonic wind tunnel with a reservoir pressure of 10 atm. The test-section Mach number is 3 and the working gas is helium (γ = 5/3). Accounting for the detached shock upstream of the tube, what pressure, pt , does the stagnation-pressure tap indicate? p M1 = 3 ............................................................................................................................................................................................................................................................................................................................................................................................................................................................ ..... .. .. .. .. .. .. .. .. .. .. ..... ......................... ...... .................................................... ...... ..................... . . . ........................................................................................................................................................................................................................................................................................................................................... ................................................................. ........................................................................................................................................................................................................................................................ ..... ................ ............................................................................................................................................................................................................................................................................................................................................................................................ .. .... ............... ..... ....................... ..... ..... .................................................................................................................................................................................................................................................. . ..... . .. .. . . .................................................................................................................................................................................................................................................................................................................................................................................................................................................................. . ..... . . . . . . . . . . . . .. ..... ... ........................................................................................................................... .... ........................... ....................... ........................................................................................................................................................................................................................................................................................................................................................................................... .. ... .. .. .. .. .. .. .. . . . . .. . . . .. .......... ...............................................................................................................................t............................................................................................................................................................................................................ ....................................................................................................................................................................................................................................................................................................................................................... .. .. ... ........ ... ........ ... ....... ... ........... ........ ..... . .. ......... ......... ..... . . . . . . . . .. . . .. . . .... . . ... . .. ... . . ... . ....... ..... .............. ..... . ... p Solution: The flow is isentropic on both sides of the shock So the equations for total pressure ahead of and behind the shock are γ−1 2 pt1 = p1 1 + M1 2 γ /(γ −1) γ−1 2 M2 pt2 = p2 1 + 2 , γ /(γ −1) Also, the pressure ratio across the shock and the Mach number after the shock are 2γ p2 2 =1+ M1 − 1 , p1 γ+1 2 M2 = 2 1 + γ −1 M1 2 2 γ M1 − γ −1 2 Now, since M1 = 3 and γ = 5/3 for helium we have 2 M2 1 + 1 32 3 =5 23 1= 11 3 −3 3 Hence, the total pressure ahead of and behind the shock are as follows. 1 pt1 = p1 1 + 32 3 5/2 = 32p1 , pt2 = p2 1 + 13 3 11 Thus we have pt2 1.243p2 p2 = = 0.0388 pt1 32p1 p1 From the normal-shock relation for p2 /p1 quoted above 10/3 2 p2 =1+ 3 − 1 = 11 p1 8/3 So, the total pressure ratio is pt2 = 0.0388(11) = 0.427 pt1 5/2 = 1.243p2 8.36. CHAPTER 8, PROBLEM 36 Finally, since we are given pt1 = 10 atm, there follows pt2 = 0.427(10 atm) = 4.27 atm 1167 ...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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