Question

There is a subsonic compressible flow given in cartesian coordinates in meters. It has the velocity potential:

$\phi (x,y)=V_{\propto }x+\frac{70}{\sqrt{1-(M_{\propto })^{2}}}e^{-2\pi \sqrt{1-(M_{\propto })^{2}y}}sin2\pi x$

Given an altitude of 10 km and a velocity of 240 m/s, find M,p, and T with the location (0.10 m, 0.15 m)

Answer 1

PD 2TTU Po= I atem do ax -21 4016811 e Cos (27T XO:10) 0,8090 $ (414) - Vox + 70 -217. (11-M6%.4). 11-1² Vor 240ms Atom Too = 300k xiy = 0.100.15m Mo-Vo 240 90 Step 2 90 JORT 1.480.287X103 x 300 347.188 ms Mo=240 0.6912 -M02 01 72.266 347.188 Istep 2 → Ve fu?+12 u=dd fre Әр ay 1-10². Y u-3, (+ ) = 2 + 2 (To) No + 217 (79). COS 27TX JT-MO? 21171-016912.(0,15) ..U= 240 + 217(70) ocos (2180110), 11-01619122 -217 41-0169122 ) 0.15 0,5061 fos t36) Qeve gees or protecten U: 240 + 608161 (0.8990) (0.5061) peepee, Place M/S 14489.91771 - 70 (J1-M3) 4. 217 21- Mg2). sing. 277 J1-mo? ay • 21 (0.72266). sìn ( 2178001), é 217 (172266X015 +690.8638).(206689)(015061) = –130.8277 m/s = 1 Vai Juļ+02 - 1169-177)2+ (-130-827)2 7|5061369 m/s =V · step3 - Find stagnation temp. at (814)-(0.100.15) Co-ordinate in Subsonic compressible frow, T(814), To = temp. o free stream To = stag. temp (214) at Mes = 0. 6912 0.6912 - 0562 - 1.077 + (1.082-6-077) 1.0948 0.64 -0.62 = 1,0948x300 k 1 328144 k = To sound 9 speed corresponds to stagi temp at (x!4) JWRTo3 1.48287 x 328.44 => 363+273 ms=20 Mach number at (214) MeV a=ao- Wol. ve 363.273 - 7:44x 5061 369 V= 2 (p) e Iva - to 1-006912 q 1 Το Too To= progus Too P2 40- 2 to a 2 2 a = 32 mis M = 506.369 1.577 SM 821 steps -staffe pressure at (214) Po Po Doby 0.62 & Po HOBETE at m 91.577 2.48325 Р 10:3707 2 p=2 Po X Poo Х Po Po 1.37078 1 atm = 10055197 atm =p 2.48825 p=static pressure at ary Po = pressure in free stream Po= stagnation pressure step 6: stafie temp at ay To at M 1:577 1.3611 T T- I To To X1.09 48X300K = 241.305k 113611 .: static femp. at (814) (0.110:15) Coordinate in culbenelle compressible flow. i's 2411305k Scanned with CamScanner

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