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### HW30solution

Course: AAE 334, Fall 2010
School: Purdue
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Word Count: 475

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334, AAE Fall 2011 Homework 30 SOLUTIONS Due Friday, December 2, beginning of class. Air flows through a converging diverging nozzle in a steady flow. The air entering the nozzle has stagnation conditions po 150psi and To 500R . 1. The flow upstream of the throat is subsonic. At a certain axial position in the converging section of the nozzle, the Mach number is 0.3 and the cross-sectional area is 1 square inch....

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334, AAE Fall 2011 Homework 30 SOLUTIONS Due Friday, December 2, beginning of class. Air flows through a converging diverging nozzle in a steady flow. The air entering the nozzle has stagnation conditions po 150psi and To 500R . 1. The flow upstream of the throat is subsonic. At a certain axial position in the converging section of the nozzle, the Mach number is 0.3 and the cross-sectional area is 1 square inch. Find the mass flow and momentum flow through this cross section. 2. The throat is choked and at a cross section downstream of the throat, the area is 1.6 times the throat area. Find the mass flow and momentum flow through this cross section for: a. Subsonic flow b. Supersonic flow Preliminaries: Mass flow is m uA and momentum flow is L u 2 A . 12in p 150lb in2 or Stagnation density is o o 1 o 1 o RTo 1716 ft lb slug R 500 R ft 150144slug o 0.025slug ft 3 1716500 ft 3 Area: 1 square inch = 0.0069 square feet. 2 To solve for state 1, accelerate isentropically from stagnation conditions to M 0.3 , o 1 1 M 2 1 1 2 0.025slug ft 3 1 0.2 0.32 1 / 2 2.5 0.024slug ft 3 We need velocity too. Velocity is Mach number times speed of sound: 1 2 u Ma M RT M RTo 1 M 2 1 u 326 ft s Now compute: 0.3 1.4 1716 ft lb slug1 o R1500 o R 0.99 m uA 0.024slug ft 3 326 ft s1 0.0069 ft 2 or m 0.054slug s1 2 L u2 A 0.024slug ft 3 326 ft s1 0.0069 ft 2 or L 17.6 slug ft s2 To solve for state 2, At area ratio of 1.6, M=1.94. Throat area is found the from M=0.3 point, A 0.0069 ft 2 0.0034 ft 2 . The area in the diverging section is then 2.04 Adiv 0.0034 ft 2 1.6 0.0055ft 2 . But the throat is choked, so for mass flow use the formula you derived in an earlier homework: p A* 2 m o To R 1 the throat is choked. 1 / 1 12in 150lb in2 0.0034 ft 2 2 ft 1.4 2.4 or 1716 ft lb slug1 o R1 500 o R 6 2 m 0.054slug s1 as above. Note that this is the mass flow answer for both 2a and 2b because Momentum flow at the cross section in the diverging section (label with subscript div ) is: 2a: subsonic: find the velocity after finding that at A 1.6 you have M div 0.4 : A or udiv 432 ft s1 . udiv Mdiv RTdiv Mdiv Now density: 1 1 2 RTo 1 Mdiv 2 1 / 2 2.5 1 2 1 or div 0.024slug ft 3 div o 1 M div 0.026slug ft 3 1 0.2 0.42 2 2 2 Ldiv divudiv Adiv 0.024slug ft 3 432 ft s1 0.0055 ft 2 or Ldiv 2.5 slug ft s2 which is lower than in part 1. 2b. Supersonic: Mdiv 1.94 so we find udiv Mdiv RTdiv Mdiv Now density: 1 1 2 RTo 1 Mdiv 2 1 / 2 or udiv 1610 ft s1 . 2.5 1 2 1 or div 0.0064slug ft 3 div o 1 M div 0.026slug ft 3 1 0.2 1.942 2 2 2 Ldiv divudiv Adiv 0.0064slug ft 3 1610 ft s1 0.0055 ft 2 or Ldiv 91slug ft s2 which is over 5 times that in in part 1 and over 36 times that in 2a. This shows that while mass flow is constant, the momentum flux, which is a big part of thrust, is much greater in supersonic diverging section flow than for subsonic flow.
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