p0816 - p = p t and T = T t . Because = 1 . 4 for both air...

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8.16. CHAPTER 8, PROBLEM 16 1137 8.16 Chapter 8, Problem 16 Problem: Gas flows with velocity U = 3000 ft/sec, pressure p =20psiandtemperature T =500 o F. We want to determine the pressure and temperature the gas would reach if it were brought to rest isentropically. Do your computations for air and for hydrogen. Solution: First, note that the Mach number, M ,is M = U γ RT Also, the absolute temperature is T =459 . 67 o R+500 o F=959 . 67 o R . So, for the given conditions, M air = 3000 ft / sec ± 1 . 4 [1716 ft · lb(slug · o R)] [959 . 67 o R] =1 . 98 M H 2 = 3000 ft / sec ± 1 . 4 [24677 ft · lb(slug · o R)] [959 . 67 o R] =0 . 52 When the gas is brought to rest isentropically,
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Unformatted text preview: p = p t and T = T t . Because = 1 . 4 for both air and hydrogen, we can use the isentropic-flow tables, from which we find M = 1 . 98 = p p t = 0 . 1318 , T T t = 0 . 5605 M = 0 . 52 = p p t = 0 . 8317 , T T t = 0 . 9487 Therefore, when the gas is air , p t = p p/p t = 20 psi . 1318 = 152 psi T t = T T/T t = 959 . 67 o R . 5605 = 1712 . 17 o R = 1252 o F Similarly, when the gas is hydrogen , p t = p p/p t = 20 psi . 8317 = 24 psi T t = T T/T t = 959 . 67 o R . 9487 = 1011 . 56 o R = 552 o F...
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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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