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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 […]
978-0131206687 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 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 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 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 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 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 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 14 Part 2
310 0 x ua au t 10 01 = ∂ ∂ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + ∂ ∂ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ww where the dependent column vector is w = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ […]
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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 […]