Modern Compressible Flow: With Historical Perspective 3th Edition

ISBN 13
978-0072424430
ISBN 10
0072424435
Authors
John Anderson
has exam prep Exam Prep  has homework help Homework Help
17 documents found.
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978-0072424430 Chapter 1
1 1 1 CHAPTER 1 = p RT =( . )( ) ()( ) 56 2116 1716 850 = 0.00812 slug/ft3 v = 1 = 123 ft3/slug v = 1 = 123 ft3/slug v = 1 = 123 ft3/slug 1.2 = p RT = ()(
Modern Compressible Flow: With Historical Perspective 3rd Edition
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978-0072424430 Chapter 10
98 CHAPTER 10 10.1 (a) shock = 0.592 rad = 33.9 (b) ps/p = 1.286 ps = 1.286 (1.01 x 105) = 1.3 x 105 N/m2 s/ = 1.196 s = 1.196 (1.23) = 1.47 kg/m3 Ts/T = 1.075 Ts = 1.075 (288)
Modern Compressible Flow: With Historical Perspective 3rd Edition
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978-0072424430 Chapter 11
CHAPTER 11 11.1 The solution is given in the following table and graph of the contour. The point numbers in the table correspond to the numbered points on the graph. Point K- = K+ = = = M = # + - (K-+K+) (K-+K+) (deg)
Modern Compressible Flow: With Historical Perspective 3rd Edition
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978-0072424430 Chapter 12 Part 1
107 CHAPTER 12 12.1 Define the following dimensionless variables: = /o, u = u/ao, T = T/To, p = p/po, x = x/L, t = t/(L/ao), and A = A/A*, = /o, u = u/ao, T = T/To, p = p/po, x = x/L,
Modern Compressible Flow: With Historical Perspective 3rd Edition
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978-0072424430 Chapter 12 Part 2
122 123
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978-0072424430 Chapter 14
137 ANSWERS TO PROBLEMS REFERRED TO IN THE TEXT OF CHAPTER 14 Please note: There are two problems that are described within the text of Chapter 14; these are not specifically itemized at the end of the chapter. In this Solutions Manual, they will be
Modern Compressible Flow: With Historical Perspective 3rd Edition
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Homework Help
978-0072424430 Chapter 15
142 ANSWERS TO PROBLEMS REFERRED TO IN THE TEXT OF CHAPTER 15 Please note: There are two problems that are described within the text of Chapter 15; these are not specifically itemized at the end of the chapter. In this Solutions Manual, they will be
Modern Compressible Flow: With Historical Perspective 3rd Edition
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978-0072424430 Chapter 16
147 CHAPTER 16 16.1 W = j g g N N j j j j ! ( )! ! n W = j [ n gj! - n (gj Nj)! - n Nj!] Employ Sterlings formula. n W = [gj n gj gj (gj Nj) n (gj Nj) n
Modern Compressible Flow: With Historical Perspective 3rd Edition
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978-0072424430 Chapter 16B
163 which is Eq. (13.54c). Finally, multiply Eqs. (Q16) and (Q17), and divide by (Q18). ( )Ne e e QABABA B oAB= ( )N NNeQ QQA BABkT A BABo=/ which is Eq. (16.55), where o = oA + oB - oAB. ( )Ne e
Modern Compressible Flow: With Historical Perspective 3rd Edition
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978-0072424430 Chapter 17
174 CHAPTER 17 17.1 1 u1 = 2 u2 (continuity) or, 2 = 1 u2 (A1) p1 + 1 u12 = p2 + 2 u22 (momentum) 2 = 1 u2 (A1) p1 + 1 u12 = p2 + 2 u22 (momentum)
Modern Compressible Flow: With Historical Perspective 3rd Edition
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978-0072424430 Chapter 2
4 CHAPTER 2 2.1 Consider a two-dimensional body in a flow, as sketched in Figure A. A control volume is drawn around this body, as given in the dashed lines in Figure A. The control volume is bounded by: 1. The upper and lower streamlines far above
Modern Compressible Flow: With Historical Perspective 3rd Edition
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978-0072424430 Chapter 3
11 11 11 CHAPTER 3 3.1 From Table A.1, for M = 0.7; po/p = 1.387 and To/T = 1.098. Hence, po = p p= 0.9 (1.01 x 10-5)(1.387) = 1.26 x 105 N/m2 (or 1.248 atm) To = T TTo = 250 (1.098) =
Modern Compressible Flow: With Historical Perspective 3rd Edition
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978-0072424430 Chapter 4 Part 1
23 CHAPTER 4 4.1 a1 = RT = ( . )(14 1716)(520) a1 = 1118 ft/sec M1 = v1/a1 = 3355/1118 = 3.0 Mn1 = M1 sin = 3.0 sin (35) = 1.72 From Table A.l. P2/p1 = 3.285
Modern Compressible Flow: With Historical Perspective 3rd Edition
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978-0072424430 Chapter 4 Part 2
34 M2 = = = 2.03 M2 = = = 2.03 M2 = = = 2.03 Mn2 sin( ) 06099 37 520 . sin( . ) M2 is the Mach number ahead of the normal shock on the Pitot tube. From
Modern Compressible Flow: With Historical Perspective 3rd Edition
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978-0072424430 Chapter 5
45 CHAPTER 5 5.1 From Table A.1 for Me = 2.4: Ae/A* = 2.403 , po/pe = 14.62 and To/Te = 2.152. Hence: po = ppoe pe = 14.62 (1 atm) = 14.62 atm To = To Te = 2.152 (519R) = 1117R Hence:
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978-0072424430 Chapter 7
62 CHAPTER 7 7.1 e2 e1 = p p 1 2 2 + (v1 v2) e = cv T = RT vRTp=1; Hence, Eq. (1) becomes R1 (T2 T1) = p p RTpRTp1 2 11222+ 211 121122121RTTppRppRTT= + 21 TT21 - 21 = 1 + pp21 - pp12TT21 -
Modern Compressible Flow: With Historical Perspective 3rd Edition
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978-0072424430 Chapter 9
82 2222VVV Therefore, the resulting coefficient of ux on the right-hand-side of Eq. (9.4) reduces to M2 ( + 1) uV. For subsonic and supersonic flow, this term is relatively small (u/V << 1) in comparison to (1 - M2) on the left-hand-side, and the linearized result, Eq. (9.5)
Modern Compressible Flow: With Historical Perspective 3rd Edition

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