R c = g c-1. c pc R t = g t-1. g c. c pt. g t. a 0 = g c R c g c T 0. t r = 1+ g c-1 2. p r = t c êhg c -1L r. c pc T 0.
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1 MAE 113 SS1 2009, HW #4 Solutions 7.1 Develop a set of equations for parametric analysis of a ramjet engine with losses. Calculate the performance of a ramjet with losses over a Mach number range of 1 to 3 for the following input data: p dmax 0.95, T 0 217K, 1.4, c pc kjêhkgÿkl p b 0.94, h b 0.96, g t 1.3, c pt kjêhkgÿkl p n 0.95, P 0 P 9 1, T t4 1800K, h PR kjêkg For analysis, a ramjet can be considered as a turbojet with t c p c 1, and t t p t 1, since a ramjet does not have a compressor or a turbine. Using equations 7-20 a-y, with equations changed for the ramjet written in red, we get R c -1 c pc R t g t-1 g t c pt a 0 R c T 0 V 0 a 0 M 0 t r M 2 0 g p r t c êh -1L r h r 1 form 0 1 h r HM 0-1L 1.35 form 0 > 1 p d p dmax h r t l c ptt t4 c pc T 0 t c 1 h c undefined, since no compressor t f l -t r h PR h b ëic pc T 0 M-t l t t 1 p t 1 h t undefined, since no turbine P t9 P 0 p P 9 P r p d p b p n 9 M 9 2 g t -1 BJP t9 P 9 N Hg t-1lêg t - 1F
2 2 MAE113_solutions_HW4.nb T 9 T 0 t l c pc HP t9 êp 9 L Igt-1Mëgt c pt V 9 a 0 M 9 g t R t T 9 R c T 0 F m 0 a 0BH1+fL V 9 -M a 0 +H1+fL R tt 9 êt 0 0 R c V 9 êa 0 S f Fêm 0 h T a 0 2 AH1+fLHV 9 êa 0 L 2 -M 2 0 E h P 2 fh PR 2 V 0 HFêm 0L a 0 2 AH1+fLHV 9 êa 0 L 2 -M 0 2 E h 0 h P h T H1-P 0 êp 9 L F If we start to plug in numbers, we find R c R t a 0 1.4J286.9 t l kjêhkgÿkl kj kjêhkgÿkl kj J NH1LH217KL 295.2mês kj 1800K kj 217K I'll let Mathematica do the rest of the values over various Mach numbers: -1 g t -1-1 R c c pc ;R t c pt ;a 0 R c 1000 T 0 ;V0@M_D:a 0 M; tr@m_d1+ M 2 ; hr@m_d HM -1L 1.35 ; g t 2 c pt T t4 t l - tr@md P 0 pr@m_d tr@md gcëigc-1m ; pd@m_d p dmax hr@md; t l ; f@m_d: ; Pt9P9@M_D pr@mdpd@mdp b p n ; c pc T 0 h PR h b ëic pc T 0 M-t l P 9 M9@M_D 2 t-1mëg g t -1 JHPt9P9@MDLIg t -1N ; T9T0@M_D ;V9a0@M_D M9@MD HPt9P9@MDL Igt-1Mëgt c pt Fm0@M_D: a 0 H1 + f@mdlv9a0@md -M + H1 + f@mdl R tt9t0@md H1 -P 0 êp 9 L f@md ;S@M_D R c V9a0@MD Fm0@MD ; t l c pc g t R t R c T9T0@MD ;
3 MAE113_solutions_HW4.nb 3 NumberFormBGridB:8"M 0 ", "1.0", "1.25", "1.5", "1.75", "2.0", "2.25", "2.5", "2.75", "3.0"<, 8"τ r ", τr@1d, τr@1.25d, τr@1.5d, τr@1.75d, τr@2d, τr@2.25d, τr@2.5d, τr@2.75d, τr@3d<, 8"π r ", πr@1d, πr@1.25d, πr@1.5d, πr@1.75d, πr@2d, πr@2.25d, πr@2.5d, πr@2.75d, πr@3d<, 8"f", f@1d, f@1.25d, f@1.5d, f@1.75d, f@2d, f@2.25d, f@2.5d, f@2.75d, f@3d<, 8"P t9 êp 9 ", Pt9P9@1D, Pt9P9@1.25D, Pt9P9@1.5D, Pt9P9@1.75D, Pt9P9@2D, Pt9P9@2.25D, Pt9P9@2.5D, Pt9P9@2.75D, Pt9P9@3D<, 8"M 9 ", M9@1D, M9@1.25D, M9@1.5D, M9@1.75D, M9@2D, M9@2.25D, M9@2.5D, M9@2.75D, M9@3D<, 8"V 9 êa 0 ", V9a0@1D, V9a0@1.25D, V9a0@1.5D, V9a0@1.75D, V9a0@2D, V9a0@2.25D, V9a0@2.5D, V9a0@2.75D, V9a0@3D<, :"Fêm 0 H N L", Fm0@1D, Fm0@1.25D, Fm0@1.5D, kgês Fm0@1.75D, Fm0@2D, Fm0@2.25D, Fm0@2.5D, Fm0@2.75D, Fm0@3D>, :"S H mgês L", S@1D, N S@1.25D, S@1.5D, S@1.75D, S@2D, S@2.25D, S@2.5D, S@2.75D, S@3D>>, 8Frame All<F, 4F M τ r π r f P t9 êp M V 9 êa Fêm 0 H N L kgës S H mgës L N
4 4 MAE113_solutions_HW4.nb 7.3 Calculate and compare the performance of turbojet engines with the basic data of Example 7.1 for components with technology level 2 values in Table 6.2 (assume cooled turbine and the same diffuser and nozzle values as in Example 7.1). Comment on the changes in engine performance. M 0 2, T K, 1.4, c pc kjêhkgÿkl,g t 1.3 c pt kjêhkgÿkl,h PR kjêkg, p dmax 0.95, p b 0.92 p n 0.96, e c 0.84, e t 0.83, h b 0.94, h m 0.97, P 0 êp 9 0.5, T t4 1390K, p c 10 The equations we use are the same as in example 7.1 R c -1 c g pc kjêhkgÿkl kj c 1.4 R t g t-1 c g pt kjêhkgÿkl kj t 1.3 a 0 R c T 0 1.4µ286.9 J µ 1µ216.7K 295.0mês f t t 1- V 0 a 0 M 0 H295.0mêsLH2L 590.0mês t r M g p r t c êh -1L r h r HM 0-1L p d p dmax h r t l c ptt t4 c pc T 0 t l -t r t c h PR h b ëic pc T 0 M-t l 1 h m H1+fL H1.239 kjêhkgÿklh1390kl H1.004 kjêhkgÿklh216.7kl gc-1 gcec t c p c H0.84L h c p c gc-1 gc -1 t c H1.8L H2.1884L H kjêkgµ0.94lêh1.004 kjêhkgÿklµ216.7kl t r t l Ht c - 1L H L p t t t gt Igt-1Met H0.3L H L h t 1-t t t 1êet t P t9 P 0 p P 9 P r p d p c p b p t p n H0.5LH7.8244LH0.8788LH10LH0.92LH0.1900LH0.96L
5 MAE113_solutions_HW4.nb 5 M 9 T 9 T 0 2 g t -1 BJP t9 P 9 N Hg t-1lêg t - 1F t l t t HP t9 êp 9 L Igt-1Mëgt c pc c pt F m 0 a 0 BH1+fL V 9 a 0 -M 0 +H1+fL R tt 9 êt 0 R c V 9 êa 0 V 9 a 0 M 9 g t R t T 9 R c T H1-P 0 êp 9 L F 295AH L H L h T a 0 2 AH1+fLHV 9 êa 0 L 2 -M 0 2 E 2 fh PR h P S f Fêm 0 2 V 0 HFêm 0L a 0 2 AH1+fLHV 9 êa 0 L 2 -M 0 2 E kgës H0.2859L H0.2869L mgês H295mêsL2 AH LH3.0945L E 1000 J` 2H1LH LJ kjêkg N 2H1L 590H L kj AH LH3.0945L E h 0 h P h T % H1-0.5L E kgês % % In Example 7.1, F m mgês, f , and S Here, we had a 46.8% decrease in specific 0 kgês kg thrust, a 37.0% decrease in fuel to air ratio, and thus, a 18.5% increase in thrust specific fuel consumption.
6 6 MAE113_solutions_HW4.nb You are required to design an internal compression inlet with a fixed capture area A c 14 ft 2 1.3m 2 that will operate for a cruise Mach number M with an inlet throat Mach number M t 1.2. a) Determine and A c ê for cruise Note that for ideal flight, A c, as shown on page 702 A c A * A * and since we know the Mach numbers at 0 and t, we can find the AêA * ratios by looking up in an isentropic flow table or using equation We get A c H LI M and now we can find A c ft ft m 2 b) If the throat area is held constant, can this inlet be started for M 0 4 or less? When M 0 4, using equation 10.7, A * , and P AI A A * M P 02 E M0 H LH0.1388L So A c would have to be or less in order for the engine to start. However, based on part a, that ratio is , so the inlet will not start since the throat area is too small. c) If it is desired to start the inlet at M 0 2.0, find the required values of A c and. At M 0 2, such that and A * , and P AI A A * M P 02 E M0 H1.6875LH0.7209L A c A c ft ft m 2
7 MAE113_solutions_HW4.nb An aircraft with a turbojet engine uses a fixed contraction ratio, internal compression inlet designed for M The aircraft is flying at M with the inlet started, and the shock is optimally positioned when the inlet suddenly unstarts, popping the shock. a) Find the maximum pressure recovery for the started and unstarted conditions. When M 0 3.0, When M and started, we already know the A c ratio, so we can use A * , and P AI A A * M P 02 E M0 H4.2346LH0.3283L A c A * A * A * H1.6875L A* A * For this AêA * ratio, from the isentropic table, M t A shock with this initial Mach number will have a pressure ratio P For the unstarted case, the shock has the initial Mach number of M2, and thus P b) Find the ratio of unstarted inlet mass flow rate to the started inlet mass flow rate. Fraction Spilled A cê - ê A c ê %
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