Archives: Solution Manual
978-0134610672 Chapter 6 Part 2
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 6 Part 1
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 5 Part 2
208 5–21. The power transmission cable weighs 10 lb>ft . If the resultant horizontal force on tower BD is required to be zero, determine the sag h of cable BC. AB hC D 300 ft 10 ft 200 ft SOLUTION […]
978-0134610672 Chapter 5 Part 1
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 4 Part 4
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 4 Part 3
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 4 Part 2
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 4 Part 1
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 3 Part 3
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 3 Part 2
9 2 Ans. FCE =FEG =FEF =0 FDE =2.00 k 1C2 FFG =3.00 k 1C2 FAG =2.00 k 1C2 FCD =4.375 k 1C2 FAD =2.625 k 1T2 FCF =4.375 k 1C2 FBF =0.625 k 1T2 FAB =0.625 k 1T2 Equations […]
978-0134610672 Chapter 3 Part 1
7 2 © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or […]
978-0134610672 Chapter 2 Part 3
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 2 Part 2
4 6 © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or […]
978-0134610672 Chapter 2 Part 1
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 1 Part 2
1 4 © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or […]
978-0134610672 Chapter 1 Part 1
1 © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by […]
978-0131206687 Chapter 15
C Ch ha ap pt te er r F Fi if ft te ee en n M ME EA AS SU UR RE EM ME EN NT TS S I IN N C CO OM MP PR RE ES SS […]
978-0131206687 Chapter 14 Part 3
325 . Point Area A/A* M (Exact Solution) M (MOC) % Error 4 A4 10.7188 4.00 4.0000 11 1.0637A4 12.3051 4.1558 4.1557 -0.0014 18 1.1334A4 14.1642 4.3172 4.3207 0.0799 (c) Minitial = 2.0, total wedge angle of 24˚ and γ […]
978-0131206687 Chapter 14 Part 2
310 0 x ua au t 10 01 = ∂ ∂ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + ∂ ∂ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ww where the dependent column vector is w = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ […]
978-0131206687 Chapter 14 Part 1
290 Chapter Fourteen C CH HA AR RA AC CT TE ER RI IS ST TI IC CS S Problem 1. – Use the Method of Indeterminate Derivatives to obtain equations of the characteristics for the following equation in the […]
978-0131206687 Chapter 13
272 Chapter Thirteen L LI IN NE EA AR RI IZ ZE ED D F FL LO OW WS S Problem 1. – The lift coefficient versus angle of attack for an airfoil, as measured in a low-speed wind tunnel, […]
978-0131206687 Chapter 12 Part 2
uvu y = Φ vvv y =Φ Now use the chain rule and write () ( ) ( ) 1 u y yu x xuu u 1vuxyuuxx xxx =Φφ+Φφ= ∂ ∂ ∂ φ∂ + ∂ ∂ ∂ φ∂ = ∂ […]
978-0131206687 Chapter 12 Part 1
250 Chapter Twelve E EX XA AC CT T S SO OL LU UT TI IO ON NS S Problem 1. – The velocity potential equation can be written in a variety of forms. For example, Taylor and Maccoll, Ref. […]
978-0131206687 Chapter 11
Chapter Eleven E EQ QU UA AT TI IO ON NS S O OF F M MO OT TI IO ON N F FO OR R M MU UL LT TI ID DI IM ME EN NS SI IO ON […]
978-0131206687 Chapter 10 Part 2
() Kkg/J5.1004 14.1 2874.1 1 R cp⋅= − = −γ γ = ()( ) ( ) s/m8990.4004002874.1RTa 11 ==γ= 087304.0 35 V M 1 1=== 8990.400 a 1 At this Mach number we find from the Rayleigh relations that 03587.0 […]
978-0131206687 Chapter 10 Part 1
207 Chapter Ten F FL LO OW W W WI IT TH H H HE EA AT T A AD DD DI IT TI IO ON N O OR R H HE EA AT T L LO OS SS S […]
978-0131206687 Chapter 9 Part 2
(a) For A2/A1 = A2/A* = 2.9, M2 = 2.6015. Therefore, (fLmax/D)2 = 0.45288. Now fL/D = (0.02)(20)/(1) = 0.4. Hence, L < Lmax so the flow cannot reach Me = 1. To compute the exit Mach number we have […]
978-0131206687 Chapter 9 Part 1
Chapter Nine F FL LO OW W W WI IT TH H F FR RI IC CT TI IO ON N Problem 1. – Draw the T-s diagram for the adiabatic flow of a gas with γ = 1.4 in […]
978-0131206687 Chapter 8
149 Chapter Eight A AP PP PL LI IC CA AT TI IO ON NS S I IN NV VO OL LV VI IN NG G S SH HO OC CK KS S A AN ND D E EX XP […]
978-0131206687 Chapter 7
Chapter Seven P PR RA AN ND DT TL L– –M ME EY YE ER R F FL LO OW W Problem 1. – Use a trigonometric development to demonstrate that for an expansion flow around a convex corner, Vn2 […]
978-0131206687 Chapter 6 Part 2
118 ()()()() 8350.09948.09959.09966.08457.0 p p p p p p p p p p 1o 2o 2o 3o 3o 4o 4o 5o 1o 5o === Four oblique shocks: °=δ 6 9972.0 p 4o 5o = Normal Shock M 5 = 1.5184, […]
978-0131206687 Chapter 6 Part 1
106 C Ch ha ap pt te er r S Si ix x O OB BL LI IQ QU UE E S SH HO OC CK K W WA AV VE ES S Problem 1. – Uniform airflow (γ = […]
978-0131206687 Chapter 5 Part 2
() ()() 1 2 113 1 1 1 p 3= γ = −γ−−γ = ⎟ ⎟ ⎠ ⎜ ⎜ ⎝ +γ − ⎟ ⎟ ⎠ ⎜ ⎜ ⎝ +γ = () ()() 2 11 1 1 1 1 1 13 […]
978-0131206687 Chapter 5 Part 1
C Ch ha ap pt te er r F Fi iv ve e M MO OV VI IN NG G N NO OR RM MA AL L S SH HO OC CK K W WA AV VE ES S Problem […]
978-0131206687 Chapter 4 Part 2
69 () () b2 b1 2 1 b2 1b 2 1 2 1 1o 2o bM1b 1 b1bM M p p −+ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ −+ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ −+ […]
978-0131206687 Chapter 4 Part 1
56 C Ch ha ap pt te er r F Fo ou ur r S ST TA AT TI IO ON NA AR RY Y N NO OR RM MA AL L S SH HO OC CK K W WA […]
978-0131206687 Chapter 3 Part 2
Problem 16. – Steam is to be expanded to Mach 2.0 in a converging-diverging nozzle from an inlet velocity of 100 m/s. The inlet area is 50 cm2; inlet static temperature is 500 K. Assuming isentropic flow, determine the throat […]
978-0131206687 Chapter 3 Part 1
32 C Ch ha ap pt te er r T Th hr re ee e I IS SE EN NT TR RO OP PI IC C F FL LO OW W O OF F A A P PE ER RF […]
978-0131206687 Chapter 2
C Ch ha ap pt te er r T Tw wo o W WA AV VE E P PR RO OP PA AG GA AT TI IO ON N I IN N C CO OM MP PR RE ES SS […]
978-0131206687 Chapter 1
C Ch ha ap pt te er r O On ne e B BA AS SI IC C E EQ QU UA AT TI IO ON NS S O OF F C CO OM MP PR RE ES SS SI […]
978-0073529363 Chapter 12
Chapter 12 l2-l Air@300K k-0.a262 Vertical plate GrPr – 104 ; Nu : trllk : 54.2( TL)t t4 Fig l2-8 CrrPr – 104 Nu :6 : lOe Nu:100 l2-2, Openended problem r2-3 d : 15 mm T. : 800 K […]
978-0073529363 Chapter 11
Chapter 11 11-l 1t-3 T* =25oC u- = 1.5 m/sec .b= 30 cm h-*to.urlt)Re LttzPrtt3 hD= E [:rlt” L’ ‘ L’J “Ir pfcp[6r) Ig)’ t6r,/ ‘[ofJ -l’098 Prl/3-0-884 OF Tw OC hfr xl0{ cw Tf OK v xl0{ kcp Pr […]
978-0073529363 Chapter 10 Part 2
Chapter l0 10-65 ho =0.8 kg/sec To, =30″C Toz =’loc Tr, =3oC U = 55 + h* =0.75 kg/se c Co= (0.8X1005) = 804 m’ .oc s) = 1.67 5 air flow by i \TO .919– LT’=26″C – 1.7 A […]
978-0073529363 Chapter 10 Part 1
Chapter 10 10-1 Pipe nearly constant temperature Tn =82oC L = 30oC € = 0.g h — |s2(12 -3o)t’o = 7.27 L tG.2 q = mrcrLTn = (2)(4180X90 – 60) = 2.508 x 105 W ^4 40- 50 Lr^=;eI=44.81oC q=UALTn […]
978-0073529363 Chapter 9
Chapter 9 9-1 {4- \r/3 \kr”prF) = o’oo77Re ro’o ( ,, )r/r’8 [ortp,” ) i* _ (0.00 77fr o’6 fit 3 k rzt zlaee – Tw)ft3 @ y2 ils t e N.U d3 = l4PrkL(r, -r)1t’ o ‘1″ – L […]
978-0073529363 Chapter 8 Part 4
Chapter I C) t^tl 8-130 zv * 150 Tf= = 85oQ = 358 K Pr-0.7 Re= (25X0.5) 21.58 x 10-6 r/ = 2I 58 x 10-6 – 5 .7g2x 105 k – 0.030b Nu4 = 0., * (0.62’t(s.tg LmL’.t. ,er”oooJ […]
978-0073529363 Chapter 8 Part 3
Chapter I Outer Surface: coolins _ 91,051- (}f<0l _ r.238 +e3.55)rc(0.6×1323 _643)__98,124 Wlm 8-E1 Eur=37,194 Eh=14,513 9=g.tSZ | +=0.3142 LL + =e.548 +=2.122 f- =42.44 etAt ez\, 4ryses – -” 8-82 Assume large number of pins so that array behaves […]
978-0073529363 Chapter 8 Part 2
Chapter I 8-45 (5.669 x lo-8xgzg)4(+.,g” tOa \ -26OoC = 533 K t3 = 1.0 Fn: o-96 €2 = 0.5 A2-A3=n(1.25)z Ar = n(2,5)(7.5) : 58.9 ,*2 =4.91 cm2 4z= q3 = +Fzt:(0.9 q( +*)=0.08 ,^r ‘ ‘\58.9) – 13,456 […]
978-0073529363 Chapter 8 Part 1
8-1 (a) T – 800’C = 1073 K br = (aXIAn)= 4292 Chapter I L{ =(0.2X1073) – 214.6 Euo-h, = oEUo-f, = 0.53131 E6 – (5.66,0 x to-811t otl)a = 7.515 x lOa w I ^’ 4trans = (0.53131X0.9X7.515 x […]
978-0073529363 Chapter 7 Part 3
Chap|€ir 7 7_99 Tf = 35O K v = 20.76x l0{ /c = 0.03003 pr = 0.7 4.=0.04 ro=0.05 6=0.01 Gr5 pr – (s’s)(#)l+00 – :00×0’01)3(0’7) (20’76* ro 6F— = 4548 lt = p.zzt)(4548)o.226 = 1.53 k ft” = (0.03003X1.53) […]