p0848 - 1180 CHAPTER 8. ONE-DIMENSIONAL COMPRESSIBLE FLOW...

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Unformatted text preview: 1180 CHAPTER 8. ONE-DIMENSIONAL COMPRESSIBLE FLOW 8.48 Chapter 8, Problem 48 Problem: A compressed gas reservoir has a nozzle attached. There is a normal shock downstream of the nozzle throat. The Mach number of the stream emanating from the nozzle is M = 0.48 and the pressure is p = 101 kPa. Also, the gas temperature is Tt = 300 K in the reservoir. Determine the density in the reservoir, ρt , for air and for helium (γ = 5 ). HINT: To simplify your calculations, express Mach numbers in this 3 problem to two significant figures. .................................................................................. ... ......... .. . ... . .... .. . . .. ......................... ....... ............ .................................................................................. ..... ..... .. .. ..... . . ............................... .. ............... .. ................................................................................. ...... . .. .......... ........ .......... ........ ...................................................................... .... ....... ....... .... .... . .... ........................................................ ........................................................................ ..... .. . .. . ........ ... .................................. .............................................................................. . . . .. ..... . .. . .. .................................... ...... ................................................................. . ............................................. . . .. ............................................................... ... . .. . .... . . . ...................... ........ .... . . .... .................................................................... . .. ......................... .......... ....... ............................................................................. ... ... . .. . ....... .. ........................ ............. . .............................................................................. .. .. ........................ ... ................. .... ................................................................... ............. .............. .................. ..... .. . ... . .... ..... .................................................................. ... .............. .................... ............. ....................................................... ............................................... ... . . .. ...... ... .... ... . .......................................................... .. ... ............. ................................ ... . . .............................................................. ... . ........ ............................................ .... .... ... ..... ........................................................... . . ... .. . .. ... ... ...................... .......................... .......................................................................... ...................... .......... . .......... .. . ....................................................................... .. . .............. ............. . ........... .. ........................................................................... .. .. . ... ...... .. .. .... ..... ............... ........ ......... ............................................................... . .............................. ............ . ........................................................... ............ ................................... T = 300 K ρ =? M = 0 48 ........................................................................................................................................................... . ... . . . . ... .... .... ......... . . .... .... ....... .... ... ... . . . ................................................... .................................. . ............................ ................................................................................................................................................. p = 101 kPa Solution: To solve, we note that the temperature in the reservoir is the total temperature, Tt , which is constant throughout the flow. Also, the pressure in the reservoir is the total pressure, pt so that once we find its value we can use the perfect-gas law to find the density of the gas in the reservoir. Our solution procedure will be as follows: 1. Use the normal-shock relations to compute M1 and p2 /p1 for the given M2 ; 2. Use the isentropic-flow relations to find p1 /pt corresponding to M1 , which permits computing pt ; 3. Use the perfect-gas law to determine the gas density in the reservoir. Air: Since γ = 1.4, we can use the normal-shock and isentropic-flow tables. Focusing first on the shock wave, M2 = 0.48 corresponds to a Mach number ahead of the shock M1 = 2.92 The jump in pressure is p2 /p1 = 9.781. Therefore, the static pressure ahead of the shock, p1 , is p2 101 kPa = p1 = = 10.33 kPa p2 /p1 9.781 Now, using the isentropic-flow tables, the ratio of static to total pressure for a Mach number of 2.92 is p1 /pt = 0.03071. Hence, pt = p1 10.33 kPa = = 336 kPa p1 /pt 0.03071 8.48. CHAPTER 8, PROBLEM 48 1181 Finally, the perfect-gas law tells us that in the reservoir ρt = pt RTt Since ρ = ρt in the reservoir and the perfect-gas constant for air is R = 287 J/(kg·K) and a Joule (J) is one N·m, there follows ρ= kg 3.36 · 105 N/m2 = 3.90 3 [287 N · m/(kg · K)] [300 K] m Helium: Since γ = 5/3, we must use the normal-shock and isentropic-flow equations. Focusing first on the shock wave, the symmetry of the Mach number relation with regard to indices permits computing M1 directly, viz., 2 M1 2 2 1 + 1 M2 1 + 1 (γ − 1)M2 1 + 1 (0.48)2 2 3 = = 5 2 1 = 5 3 2 1 = 21.25 1 2 γ M2 − 2 (γ − 1) M2 − 3 (0.48) − 3 3 3 Taking the square root, we conclude that the Mach number ahead of the shock is M1 = 4.61 Thus, the static pressure ahead of the shock is p1 = 1+ p2 = 2 [M1 − 1] 1+ 2γ γ +1 101 kPa 10/3 8/3 [(4.61)2 − 1] = 3.838 kPa Now, using the isentropic-flow relation, the ratio of static to total pressure for a Mach number of 4.61 is given by γ−1 2 M1 pt = p1 1 + 2 γ /(γ −1) 1 = (3.838 kPa) 1 + (4.61)2 3 5/2 = 713 kPa Finally, since the perfect-gas constant for helium is R = 2077 J/(kg·K), we have 7.13 · 105 N/m2 kg ρ= = 1.144 3 [2077 N · m/(kg · K)] [300 K] m ...
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