p0884

# p0884 -

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Unformatted text preview: 8.84. CHAPTER 8, PROBLEM 84 1231 8.84 Chapter 8, Problem 84 Problem: A normal shock stands ahead of a blunt body in the test section of a supersonic wind tunnel. Conditions ahead of the shock are M1 = 3 and p1 = 0.4 psi. A nozzle has been attached downstream of the test section and the flow becomes supersonic after passing through the nozzle. If the fluid is air, what is the pressure, p∗ , 2 at the nozzle’s throat? ................................................................ ................................................................................................................................................................................................ ..... .... .... ............. .... ............. .... .... .. .................................................................................................................................................................................................... .... ........... .. . ...... .............. .. . ................................ . ........ ........................................................ . ................. . .... ...................................................................... . ................................................................................................. .. ....... . ......... . ............................................................. ............. ....... .. M1 M >1 ............ ......................................................................................................................................................... ....................... ....................................................................................................................................................................................................... ........................................................................................................... .... . ... ............ ....................... ...... ........................................................ ...... ........... .. ....................................................................................................................................................................... ..................................................................................... ................................................................................................................................................................................................. ........................ . ............................................................. ...... ..................... ........................................................................................................................................................................................... ... ...... ............................................................................................................................................................................................. ................... . ............................................. ................................ ................................................................ ....... ................................ .... . .. .. .. ... .. ... .. ... . . . . . . . . .. . .. . ...................................................................................................................................................................................................... .......................................................................................................................................................................................................... . ................................................................................. .. .... .. ... .. ... .. .. .. .. .. . .. ... .. ... .. ... .. ... .......................................................................................... .. ... .. ... .. .. .. .. ..................... .. ... .. ... .. ... .. ... .. ......................................................................................................................................................................................................................... . ................................................................................. .. .. .. .. .. .. .. ... .. ... .. .................... .. .. .. .. .. .. .. .. Solution: Since we are dealing with air, which has γ = 1.4, we use the normal-shock tables. The given value of the Mach number ahead of the shock is M1 = 3, and reference to the normal-shock tables tells us that behind the shock p2 M2 = 0.4752 and = 10.333 p1 Behind the shock, the flow is isentropic. Hence, we use the isentropic-flow tables to determine p2 /pt . Linear interpolation gives M2 p2 /pt 0.4600 0.8650 0.4752 0.8567 0.4800 0.8541 Therefore, the ratio of static pressure to total pressure behind the shock is p2 = 0.8567 pt Finally, because the flow again becomes supersonic after passing through the nozzle, we know that sonic conditions exist at the nozzle. Again appealing to the isentropic-flow ∗ tables, this time with M2 = 1, we find p∗ 2 = 0.5283 pt We now have sufficient information to solve for p∗ , viz., 2 p∗ 2 p∗ /pt p2 0.5283 (10.333)(0.4 psi) = 2.55 psi =2 p1 = p2 /pt p1 0.8567 ...
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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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