Archives

1 CHAPTER 1   = p RT =( . )( ) ()( ) 56 2116 1716 850 = 0.00812 slug/ft3 v = 1 = 123 ft3/slug  1.2  = p RT = ()( . ) ()( ) 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 = […]

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 […]

107 CHAPTER 12 12.1 Define the following dimensionless variables: where the subscript zero denotes reservoir conditions. The governing equations are Eqs. (12.5) – (12.9). Replacing p by RT in Eq. (12.6) and e by cvT in Eq. (12.7), and noting […]

122 123 124 125 126 127 128 129 130 131 (b) 12.3 The Newtonian pressure coefficient is independent of Mach number. Cp = 2 sin2 c where c = 15. Hence, Cp = 2(0.2588)2 = 0.134. In Problem 10.3, it […]

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 […]

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 […]

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 Sterling’s formula. n […]

163 which is Eq. (13.54c). Finally, multiply Eqs. (Q16) and (Q17), and divide by (Q18). ( ) N e e e Q AB AB A B oAB = − − −    ( ) N N NeQ Q […]

174 CHAPTER 17 17.1 1 u1 = 2 u2 (continuity) or, 2 = 1 u 2 (A1) p1 + 1 u12 = p2 + 2 u22 (momentum) u 1 or, p2 = p1 = 1 u12 – 2 u22 (A2) […]

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: Consider […]

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) […]

23 CHAPTER 4 4.1 a1 =  RT = ( . )(14 1716)(520) Hence: p2 = 3.285 (2000) = 6570 lb/ft2 T2 = 1.473 (520) = 766R Mt1 = M1 cos  = 3 cos (35) = 2.457 w1 = […]

34 Mn2 sin( )   − 06099 37 520 . sin( . )− M2 is the Mach number ahead of the normal shock on the Pitot tube. From Table A.2 for M2 = 2.03: p p o o 3 […]

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 = p p o e pe = 14.62 (1 atm) = 14.62 atm To = T […]

62 CHAPTER 7 7.1 e2 – e1 = p p 1 2 2 + (v1 – v2) e = cv T = RT vRT p  −= 1; Hence, Eq. (1) becomes R  −1 (T2 – T1) = p […]

82      2 2 2 2 V V V Therefore, the resulting coefficient of   u x on the right-hand-side of Eq. (9.4) reduces to M2 ( + 1) u V . For subsonic and […]