• Consider two different points on the surface of an airplanewing flying at 80 m/s. The pressure coefficient and flowvelocity at point 1 are 1.5 and 110 m/s, respectively. Thepressure coefficient at point 2 is -0.8. Assumingincompressible flow, calculate the flow velocity at point 2. Table B.2 Properties of the International Standard Atmosphere (ISA) between sea level and 20 kmgeopotential altitude9,0009,500Standard values at altitudeP (Pa)8Alt. (m)T (K)0288.15 1 340.29500 284.90 0.9887 338.371,000 281.65 0.9774 336.431,500 278.40 0.9662 334.492,000 275.15 0.9549 332.532,500 271.90 0.9436 330.563,000 268.65 0.9306 328.583,500 265.40 0.9210326.584,000 262.15 0.9098 324.584,500 258.90 0.8985 322.56255.65 0.8872 320.53252.40 0.8759 318.48249.15 0.8647 316.436,500245.90 0.8534 314.367,000 242.65 0.8421 312.27310.17308.065,0005,5006,0007,500 239.40 0.83088,000 236.15 0.81958,500305.9333,09941,060 0.4052 0.5895 0.481238,251 0.3775 0.5566 0.454435,599 0.3513 0.5252232.90 0.80830.32670.4951229.650.7970 303.7930,742 0.3040 0.4663226.40 0.7857 301.63 28,523 0.2815 0.43890.41270.387710,0000.36390.26550.22680.1851223.150.7744299.46 26,436 0.260910,500 219.90 0.7631 297.27 24,474 0.241511,000 216.65 0.7519 295.07 22,632 0.223412,000 216.65 0.7519 295.07 19,330 0.1908 0.310813,000 216.65 0.7519 295.07 16,510 0.162914,000 216.65 0.,7519 295.07 14,101 0.139215,000 216.65 0.7519 295.07 12,044 0.1189 0.1937 0.158116,000 216.65 0.7519 295.07 10,287 0.1015 0.1654 0.135017,000 216.65 0.7519 295.07 8,786 0.0867 0.1413 0.115318,000 216.65 0.7519 295.07 7,505 0.0741 0.1207 0.098519,000 216.65 0.7519 295.07 6,410 0.0633 0.1031 0.084220,000216.65 0.7519 295.07 5,475 0.0540 0.0880 0.07180a (m/s)p (kg/m³) 0μ/1₂111.22501.1673 0.9529 0.99121.1117 0.9075 0.98231.0581 0.8638 0.97351.0065 0.8216 0.96450.9569 0.7811 0.95560.9091 0.7421 0.94650.8632 0.7055101,325 195,461 0.942189,874 0.887084,556 0.834579,4950.784674,682 0.737170,108 0.691965,764 0.649061,640 0.608357,728 0.569754,020 0.533150,506 0.4985 0.6971 0.5691 0.900647,181 0.4656 0.6597 0.538544,034 0.4346 0.6238 0.50920.90990.8191 0.66860.7768 0.63410.7361 0.60090.93750.92830.91910.25370.2167P₁A₁V₁ = P₂A₂V₂VP₂+PZ=P₁+PZP/= P₁V²C₂T₁+2=C₂T₂+0.8911 V₂ =0.88180.87240.86280.4287 0.85320.4042 0.84360.3807 0.83390.3583 0.82410.81430.33690.31650.80440.29710.7944p=pRTa = √√YRTV₁ =Advanced Aircraft Design: Conceptual Design. Analysis and Optimization of Subsonic Civil Airplanes. First Edition. Egbert Torenbeck.2013 by Tighert Torenbeek. Published 2013 by John Wiley & Sons, Lad.C₂ =h = c₂Te = c₂TP₂P₁PozP₁=0.79440.7944 M0.79442(P₁-P₂)√P[1-(A₂/A₂)²)2(Po-P)P2(Po-P)PsYRY-1-) =)"Y/(-1)17/(-1).(y + 1)²M[4YM-2(y-1)]2 (P₂)(x-1)/y0.79440.7944Y-10.7944 dp = -pvdv0.79440.7944 T₁=To [1 + x2 ¹ Mi ¹0.7944²16) *-₁]Y-2a Po-Pi1-y+2yMY+1PD P² + 1) - / - 1-7/(Y-1)P₁ = Po [(1 + ² = 1¹ M³ ] "2P₁ = Po [(1 + ² = ¹ M²³)]*-1/(-1)(4)Re, =8 =CC₁:8 ="=Ca1 2M² y+15.2xPuVxHoo√Rex0.664a=q=Re1.328REL0.37xRe2ci-ci-c|_ci0.0592Re0.074a=dc₁/daC₂Cp=9004a√M²-1C₂CoatREAR(¹1μ= arcsin(+1)/(-1)1+57.3a0/(ne,AR)1M

Solution: Given Data: u=80 m/sCP,1=-1.5V1=110 m/sCP,2=-0.8 To Find: The flow velocity at point 2.…

Consider two different points on the surface of an airplane wing flying at 80 m/s. The pressure coefficient and flow velocity at point 1 are -1.5 and 110 m/s, respectively. The pressure coefficient at point 2 is -0.8. Assuming incompressible flow, calculate the flow velocity at point 2.

Consider two different points on the surface of an airplane wing flying at 80 m/s. The pressure coefficient and flow velocity at point 1 are -1.5 and 110 m/s, respectively. The pressure coefficient at point 2 is -0.8. Assuming incompressible flow, calculate the flow velocity at point 2.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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