Chapter 9 Cmpreible Flw 667 9.57 Air flw frm a tank thrugh a nzzle int the tandard atmphere, a in Fig. P9.57. A nrmal hck tand in the exit f the nzzle, a hwn. Etimate (a) the tank preure; and (b) the ma flw. Slutin: The thrat mut be nic, and the area rati at the hck give the Mach number: Fig. P9.57 4 0.Ma + 0.78Ma A /A* = =.4 =, lve Ma.76 uptream f the hck.8(.76) 0.4 050 Then p /p hck =.46, p = atm, p = 989 Pa.4.46 tank.5 Thu p = p = 989[ + 0.(.76) ] 5900 Pa An. (a) Given that T = 00 C = 7 K and a critical thrat area f 0 cm, we btain m = m max = 0.6847pA* RT = 0.6847(5900)(0.00) 87(7) 0. An. (b) 9.58 Argn (Table A.4) apprache a nrmal hck with V = 700 m/, p = 5 kpa, and T = 50 K. Etimate (a) V, and (b) p. (c) What preure p wuld reult if the ame velcity change V t V were accmplihed ientrpically? Slutin: Fr argn, take k =.67 and R = 08 J/ K. Determine the Mach number uptream f the hck: m 700 a = krt =.67(08)(50) 49 ; Ma = V /a =.0 49 hck p (.67)(.0) 0.67 Then = 4.79, r p = 4.79(5) 599 kpa An. (b) p.67+ 0.67(.0) + m and V /V = 0.47, r V = 0.47(700) = 06 An. (a).67(.0)
670 Slutin Manual Fluid Mechanic, Fifth Editin 9.6 Sea-level tandard air i ucked int a vacuum tank thrugh a nzzle, a in Fig. P9.6. A nrmal hck tand where the nzzle area i cm, a hwn. Etimate (a) the preure in the tank; and (b) the ma flw. Slutin: The flw at the exit ectin ( ) Fig. P9.6 i ubnic (after a hck) therefre mut equal the tank preure. Wrk ur way t and at the hck and thence t in the exit: 050 p0 = 050 Pa, A /A* =.0, thu Ma.97, p = 950 Pa.5 [ + 0.(.) ] p.8(.) 0.4 = = 5.47, p = 5.47(950) 500 Pa p.4 * * * Al cmpute A /A.59, r A =.59 cm Al cmpute p = 050/.59 = 6800 Pa. Finally cmpute A * /A = /.59 =.89, read Ma = 0.7, whence p = 6800/[ + 0.(0.7) ].5 5900 Pa. An. (a). With T = 88 K, the (critical) ma flw = 0.6847p A*/ (RT ) = 0.04 /. An. (b) 9.64 Air in a large tank at 00 C and 50 kpa exhaut t the atmphere thrugh a cnverging nzzle with a 5-cm thrat area. Cmpute the exit ma flw if the atmpheric preure i (a) 00 kpa; (b) 60 kpa; and (c) 0 kpa. Slutin: Chking ccur when p atm < 0.58p tank = 79 kpa. Therefre the firt cae i nt chked, the ecnd tw cae are. Fr the firt cae, with T = 00 C = 7 K, p.5 50 7 (a) = =.5 = ( + 0.Ma e), lve Mae = 0.784, Te = = K pe 00 + 0.(0.784) m and ae =.4(87)() 65, Ve = 0.784(65) = 86 m/, and ρ e = p e/rte =.05 /m, finally: m =.05(0.0005)(86) = 0.50 / An. (a) Bth cae (b) and (c) are chked, with p atm 79 kpa, and the ma flw i maximum and driven by tank cnditin T and p : 0.6847p A* 0.6847(50000)(0.0005) = = = An. (b, c) m m max 0.57 (b, c) (RT ) 87(7)
Chapter 9 Cmpreible Flw 68 f thi hck i 47 K. Calculate (a) the temperature in the large tank; (b) the receiver preure; and (c) the ma flw. Slutin: Firt find the Mach number jut uptream f the hck and the temperature rati: A. cm ( + 0.Ma ) exit = =. =, lve fr Ma =.0 A*.0 cm.78ma T + 47K [ 0.4(.0) ][.8(.0) 0.4] Acr the hck: = = =.95, T = 4K T T (.4) (.0) ( ) T = T = T + 0. Ma = (4 K)[ + 0.(.0) ] = 500 K An. (a) tank Finally, cmpute the preure jut uptream and dwntream f the hck:.5 ( ) p ( ).5 p = p + 0. Ma = (00 kpa)/[ + 0.(.0) ] =.9 kpa.9 kpa p = preceiver = kma k+ = [(.4)(.0) 0.4] k + (.4+ ) = 44 kpa An. (b) 9.86 Air enter a -cm diameter pipe 5 m lng at V = 7 m/, p = 550 kpa, and T = 60 C. The frictin factr i 0.08. Cmpute V, p, T, and p 0 at the end f the pipe. Hw much additinal pipe length wuld caue the exit flw t be nic? Slutin: Firt cmpute the inlet Mach number and then get (fl/d) : m 7 fl a =.4(87)(60 + 7) = 66, Ma = 0.0, read = 4.5, 66 D fr which p/p* = 5.4554, T/T* =.905, V/V* = 0.8, and p/p * =.965 Then (fl/d) = 4.5 (0.08)(5)/(0.0) 5.5, read Ma 0.95 At thi new Ma, read p/p*.68, T/T*.79, V/V* 0.0, p * /p.067. Then V /V* 0.0 m V = V = 7 07 An. (a) V /V* 0.8.68 p = 550 7 kpa 5.455 An. (b).79 T = 0 K.90 An. (c)
684 Slutin Manual Fluid Mechanic, Fifth Editin Nw we need p t get p :.067 = + = 94 kpa.5 p 550[ 0.(0.) ] 566 kpa, p 566.964 The extra ditance we need t chke the exit t nic peed i (fl/d) = 5.5. That i, D 0.0 L = 5.5 = 5.5 9. m f 0.08 An. 9.87 Air enter an adiabatic duct f L/D = 40 at V = 70 m/ and T = 00 K. The flw at the exit i chked. What i the average frictin factr in the duct? Slutin: Nting that Ma exit =.0, cmpute Ma, find fl/d and hence f: Ma V 70 70 = = = = 0.49, a.4(87)(00) 47 L.5 Table B.: read f.5 Then f = 0.09 An. D 40 9.88 Air enter a 5- by 5-cm quare duct at V = 900 m/ and T = 00 K. The frictin factr i 0.0. Fr what length duct will the flw exactly decelerate t Ma =.0? If the duct length i m, will there be a nrmal hck in the duct? If, at what Mach number will it ccur? Slutin: Firt cmpute the inlet Mach number, which i decidedly upernic: V 900 Ma = =.59, a.4(87)(00) 0.05 read (fl/d) 0.45, whence L* Ma= = 0.45. m An. 0.0 [We are taking the hydraulic diameter f the quare duct t be 5 cm.] If the actual duct length = m > L*, then there mut be a nrmal hck in the duct. By trial and errr, we need a ttal dimeninle length (fl/d) = 0.0()/0.05 0.8. The reult i: fl fl Ma =.59, = 0.45, Ma =.4, = 0.45, D D fl hck: Ma = 0.555, = 0.695 D Ttal fl/d = 0.45 0.45 + 0.695 = 0.80 (cle enugh) Ma =.4 An.
686 Slutin Manual Fluid Mechanic, Fifth Editin 9.90 Air, upplied at p 0 = 700 kpa and T 0 = 0 K, flw thrugh a cnverging nzzle int a pipe f.5-cm diameter which exit t a near vacuum. If f = 0.0, what will be the ma flw thrugh the pipe if it length i (a) 0 m, (b) m, and (c) 0 m? Slutin: (a) With n pipe (L = 0), the ma-flw i imply the ientrpic maximum: p π = A* 700000( /4)(0.05) m mmax = 0.6847 = 0.6847 0.764 An. (a) RT 87(0) (b) With a finite length L = m, the flw will chke in the exit plane intead: fl 0.0(.0) Mae =.0, = = 0.88, read Ma (entrance) 0.55 D 0.05 Then T = 0/[ + 0.(0.55) ] = K, a =.4(87)() 54 m/,.5 V = Ma a = 86 m/, p = 700/[ + 0.(0.55) ] = 580 kpa, ρ = p /(RT ) = 6.46 /m Finally, then, m = ρav = (6.46)( π/4)(0.05) (86) 0.590 (% le) An. (b) (c) Repeat part (b) fr a much lnger length, L = 0 m: fl = 0.0(0) = 8.8, Ma = 0.46, T = 6 K, a = 6 m, V = 89 m, D 0.05 al, p = 67 kpa, ρ = 7.7, m = ρav 0.4 (59% le) An. (c) m 9.9 Air flw teadily frm a tank thrugh the pipe in Fig. P9.9. There i a cnverging nzzle n the end. If the ma flw i / and the flw i chked, etimate (a) the Mach number at ectin ; and (b) the preure in the tank. Fig. P9.9
Chapter 9 Cmpreible Flw 687 Slutin: Fr adiabatic flw, T* = cntant = T /. = 7/. = K. The flw chke in the mall exit nzzle, D = 5 cm. Then we etimate Ma frm ientrpic thery: A 6 cm = =.44, read Ma (ubnic) 0.45, fr which fl/d.5, A* 5 cm p /p*.88, p /p*.449, ρ / ρ*.070, T /T* =.5 r T 59 K p π Given m = = ρav = (0.06) (0.45).4(87)(59), 87(59) 4 Slve fr p 640 kpa. Then p* = 640/.88 68 kpa fl fl f L 0.05(9) At ectin, = + =.5 + 5.7, read Ma 0.0 An. (a) D D D 0.06 fr which p /p*.6, r p.6(68) 965 kpa. Auming ientrpic flw in the inlet nzzle, tank.5 p 965[ + 0.(0.0) ] 00 kpa An. (b) 9.9 Mdify Prb. 9.9 a fllw: Let the tank preure be 700 kpa, and let the nzzle be chked. Determine (a) Ma ; and (b) the ma flw. Keep T = 00 C. Slutin: Thi i the revere f Prb. 9.9 and i eaier. The Mach number are the ame, ince they depend nly upn fl/d (which i the ame) and the tw nzzle area rati. If we didn t knw the lutin t Prb. 9.9, we wuld gue Ma, wrk ut Ma and ee if the flw then expand exactly t a nic exit at the ecnd nzzle. Repeat, if neceary, until the prgrein thrugh the pipe and the ecnd nzzle i chked. The reult are: Ma = 0.0, cmpute p = 700/[ + 0.(0.0) ] 658 kpa. In Table B., read 658 p /p*.6, r p* = 8 kpa. Al read fl/d 5.7, ubtract f L/D f.75.6 t find fl/d.5, read Ma 0.45 An. (a) Table B.: A /A*.44.5.44 Then A exit/a* =.0(exactly what we want, nic flw exit). (6/5) G back t ectin r t cmpute m = ρav = ρav.04 / An. (b) 9.9 Air flw adiabatically in a -cm-diameter duct with f = 0.05. At the entrance, V = 950 m/ and T = 50 K. Hw far dwn the tube will (a) the Mach number be.8; and (b) the flw be chked?
698 Slutin Manual Fluid Mechanic, Fifth Editin A heat additin f 504 kj/ i (jut barely) le than maximum, huld nearly chke: q 540000 T 84 T = T + = 05 + 84 K, = = 0.957, Ma 0.78 An. (b) cp 005 T* 880 Finally, withut uing Table B.4, T = 84 /[ + 0.(0.78) ] 75 K An. (c) 9.08 What happen t the inlet flw f Prb. 9.07 if the cmbutin yield 500 kj/ heat additin and p and T remain the ame? Hw much i the ma flw reduced? Slutin: The flw will chke dwn t a lwer ma flw uch that * T = T : 500000 T 05 T = T* = 05 + = 798 K, thu = = 0.7, Ma,new 0.98 005 T* 798 (m/a) = ρv = ρa Ma = ρ a Ma /[+ 0.Ma ] if p, T, ρ are the ame. new mnew 0.98 + 0.(0.0) Then = (abut % le flw). 0.68 An m ld 0.0 + 0.(0.98) 9.09 A jet engine at 7000-m altitude take in 45 / f air and add 550 kj/ in the cmbutin chamber. The chamber cr ectin i 0.5 m, and the air enter the chamber at 80 kpa and 5 C. After cmbutin the air expand thrugh an ientrpic cnverging nzzle t exit at atmpheric preure. Etimate (a) the nzzle thrat diameter, (b) the nzzle exit velcity, and (c) the thrut prduced by the engine. Fig. P9.09 Slutin: At 700-m altitude, p a = 404 Pa, T a = 4.66 K t ue a exit cnditin. p 80000 ρ = = =.00, m = 45 = ρav =.00(0.5)V, V = 90m/ RT 87(78) m
Chapter 9 Cmpreible Flw 699 90 Ma = = 0.7, Table B.4: T * /T 0.9,.4(87)(78) T = 78 + (90) /[(005)] 8 K, T* = 8/0.9 97 K 550000 T 89 Add heat: T = 8 + 89 K, thu = 0.85, read Ma 0.6 005 T* 97 al read p /p*.8, p /p*.54, p = 80(.54/.8) 57 kpa,.5 p = p + 0. Ma = 57[ + 0.(0.6) ] 74 kpa With data nw knwn at ectin, expand ientrpically t the atmphere:.5 p 404 T T = = 0.7 = + 0.Ma, lve Ma = 0.70, = = 0.90, p 57000 T 89.5 e e e e e e e e e e e Slve T 755 K, ρ = p /RT 0.89 /m, a = krt = 55 m/, V = Ma a 85 m/ An. (b) e e e π m = 45 = 0.89(85) D e, lve D e 0.89 m An. (a) 4 Finally, if p = p, frm Prb..68, F = mv = 45(85) 700 N An. (c) e atm thrut e 9.0 Cmpreible pipe flw with heat additin, Sec. 9.8, aume cntant mmentum (p + ρv ) and cntant ma flw but variable tagnatin enthalpy. Such a flw i ften called Rayleigh flw, and a line repreenting all pible prperty change n an temperature-entrpy chart i called a Rayleigh line. Auming air paing thrugh the flw tate p = 548 kpa, T = 588 K, V = 66 m/, and A = m, draw a Rayleigh curve f the flw fr a range f velcitie frm very lw (Ma ) t very high (Ma ). Cmment n the meaning f the maximum-entrpy pint n thi curve. Slutin: Firt evaluate the Mach number and denity at the reference tate: p 548000 V 66 ρ = =.5 ; Ma = = = 0.55 RT 87(588) m krt.4(87)(588) Our baic algebraic equatin are then: Mmentum: p + ρv = 548000 +.5(66) = 778000 (a) Cntinuity: ρv =.5(66), r: ρ = 864/V (b) Entrpy: = 78 ln(t/588) 87 ln(ρ/.5) (c)
70 Slutin Manual Fluid Mechanic, Fifth Editin 0 Then T /T * = = 0.58; Table B.4: read Ma 0.06, read V /V* 0.99 840 S V = 50(0.99) 05 m/ An. (a) Al read p /p*.96, where p* = p /0.58 80 kpa, Hence p.96(80) 5 kpa An. (b) 9. Air enter a cntant-area duct at p = 90 kpa, V = 50 m/, and T = 558 C. It i then cled with negligible frictin until it exit at p = 60 kpa. Etimate (a) V ; (b) T ; and (c) the ttal amunt f cling in kj/. Slutin: We have enugh infrmatin t etimate the inlet Ma and g frm there: m 50 p a =.4(87)(558 + 7) = 578, Ma = 0.90, read =.46, 578 p* 90 p 60 = = = 0.8.46 p* 80 r p* 80.0 kpa, whence.00, read Ma, read T /T* = 0.575, V /V* = 0.87, T /T* 0.49 We have t back ff t ectin t determine the critical (*) value f T, V, T : 558 + 7 Ma = 0.9, T /T* =.045, T* = = 8 K, T = 0.575(8) 466 K An. (b).045 50 al, V /V* = 0.9, V* = = 57 m/, V = 0.87(57) 64 m/ An. (a) 0.9 966 T /T * = 0.99, where T = T+ V /cp = 966 K, T* = = 97 K 0.99 Finally, T = 0.49(97) = 480 K, q cling kj = cp T =.005(966 480) 489 An. (c) 9.4 We have implified thing here by eparating frictin (Sec. 9.7) frm heat additin (Sec. 9.8). Actually, they ften ccur tgether, and their effect mut be evaluated imultaneuly. Shw that, fr flw with frictin and heat tranfer in a cntant-diameter pipe, the cntinuity, mmentum, and energy equatin may be cmbined int the fllwing differential equatin fr Mach-number change: dma + kma dq kma [ + ( k )Ma ] fdx = + Ma Ma ct ( Ma ) D p