Wednesday, December 12, 2018
'Flow Past Aerofoil\r'
'ME2135E research lab business relationship Flow Past an Aerofoil by LIN SHAODUN Lab Group Date A0066078X 2B 10th Feb 2011 TABLE OF limit EXPERIMENTAL DATA â⬠TABLE 1, 2, 3 2 represent â⬠? 4 GRAPH â⬠? 5 GRAPH â⬠6 SAMPLE enumeration 7 tidings 8 1 EXPERIMENTAL DATA remit 1: Coordinate of squelch Tapping Tapping No. 1 2 3 4 5 6 7 8 9 10 11 Note: panel 2: atmospheric repulse per unit area Readings Manometer propensity: Pressure Readings Pitot Pressure Static Pressure atmospherical Pressure Atmospheric Temperature Stall angle: At the end of the experiment 474 mm 497 mm euchre mm 29ðC (mm) 0. 0 2. 5 5. 0 10 20 30 40 50 60 70 80 (mm) 0. 000 3. 268 4. 443 5. 853 7. 172 7. 502 7. 254 6. 617 5. 04 4. 580 3. 279 0 0. 025 0. 049 0. 098 0. 197 0. 295 0. 394 0. 492 0. 591 0. 689 0. 787 0 0. 032 0. 044 0. 058 0. 071 0. 074 0. 071 0. 065 0. 056 0. 045 0. 032 At the beginning of the experiment 474 mm 497 mm vitamin D mm 29ðC 2 Manometer Readings at versatile Tappi ng 1 2 3 4 5 6 7 8 9 10 11 478 489 494 501 505 506 506 505 502 501 500 496 478 484 492 498 500 502 502 500 499 499 495 475 478 486 494 497 499 500 498 498 498 493 476 475 480 488 493 495 498 496 496 498 486 540 532 528 522 518 516 514 507 503 502 509 562 550 546 526 522 518 514 508 504 502 495 523 520 520 518 517 516 516 515 515 515 498 516 514 515 516 515 514 514 512 513 514Table 3: Pressure Coefficients ( ) Free Stream fastness v v v ( ) Reynolds Number 3 Coefficients at various Tapping 1 2 3 4 5 6 7 8 9 10 11 -0. 956 -0. 478 -0. 261 0. 043 0. 217 0. 261 0. 261 0. 217 0. 087 0. 043 0. 000 -0. 174 -0. 956 -0. 696 -0. 348 -0. 087 0. 000 0. 087 0. 087 0. 000 -0. 043 -0. 043 0. 783 -1. 087 -0. 956 -0. 609 -0. 261 -0. one hundred thirty -0. 043 0. 000 -0. 087 -0. 087 -0. 087 2. 174 -1. 043 -1. 087 -0. 869 -0. 522 -0. 304 -0. 217 -0. 087 -0. 174 -0. 174 -0. 087 -0. 609 1. 739 1. 391 1. 217 0. 956 0. 783 0. 696 0. 609 0. 304 0. 130 0. 087 0. 391 2. 695 2. 74 2. 000 1. 130 0. 956 0. 783 0. 609 0. 348 0. 174 0. 087 -0. 217 1. 000 0. 869 0. 869 0. 783 0. 739 0. 696 0. 696 0. 652 0. 652 0. 652 -0. 087 0. 696 0. 609 0. 652 0. 696 0. 652 0. 609 0. 609 0. 522 0. 565 0. 609 GRAPH ? 3. 0 2. 5 2. 0 1. 5 CPL , processor against X/C @ 4ð Cpl 3. 0 2. 5 CPL , CPU against X/C @ 8ð Cpl Cpu 2. 0 1. 5 Cpu CPL ,CPU CPL ,CPU 1. 0 0. 5 0. 0 -0. 5 -1. 0 -1. 5 0. 0 0. 1 0. 2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 0. 9 1. 0 1. 0 0. 5 0. 0 -0. 5 X/C -1. 0 -1. 5 0. 0 0. 1 0. 2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 X/C 0. 9 1. 0 orbital cavity = 0. 437 playing area = 0. 813 4 3. 2. 5 2. 0 1. 5 CPL , CPU against X/C @ 12ð 3. 0 2. 5 CPL , CPU against X/C @ 16ð Cpl Cpu 2. 0 1. 5 Cpl Cpu CPL ,CPU CPL ,CPU X/C 0. 0 0. 1 0. 2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 0. 9 1. 0 1. 0 0. 5 0. 0 -0. 5 -1. 0 -1. 5 1. 0 0. 5 0. 0 -0. 5 -1. 0 -1. 5 0. 0 0. 1 0. 2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 X/C 0. 9 1. 0 Area = 0. 858 GRAPH ? Area = 0. 729 3. 0 2. 5 2. 0 1. 5 CPF , kiss of life against Y/C @ 4ð Cpf Cpr 3. 0 2. 5 2. 0 1. 5 CPF , CPR against Y/C @ 8ð Cpf Cpr CPF ,CPR 1. 0 0. 5 0. 0 -0. 5 -1. 0 CPU ,CPR Y/C 1. 0 0. 5 0. 0 -0. 5 -1. 0 Y/C -1. 5 -0. 10 -0. 08 -0. 06 -0. 4 -0. 02 0. 00 0. 02 0. 04 0. 06 0. 08 0. 10 -1. 5 -0. 10 -0. 08 -0. 06 -0. 04 -0. 02 0. 00 0. 02 0. 04 0. 06 0. 08 0. 10 Area = 0. 032 Area = 0. 079 5 3. 0 2. 5 2. 0 CPF , CPR against Y/C @ 12ð Cpf Cpr 3. 0 2. 5 2. 0 1. 5 CPF , CPR against Y/C @ 16ð Cpf Cpr 1. 5 CPL ,CPU CPL ,CPU Y/C 1. 0 0. 5 0. 0 1. 0 0. 5 0. 0 -0. 5 -1. 0 -0. 5 -1. 0 Y/C -1. 5 -0. 10 -0. 08 -0. 06 -0. 04 -0. 02 0. 00 0. 02 0. 04 0. 06 0. 08 0. 10 -1. 5 -0. 10 -0. 08 -0. 06 -0. 04 -0. 02 0. 00 0. 02 0. 04 0. 06 0. 08 0. 10 Area = -0. 038 GRAPH Area = -0. 053 0. 437 0. 813 0. 858 0. 729 0. 32 0. 079 -0. 038 -0. 053 0. 434 0. 794 0. 847 0. 715 0. 062 0. 191 0. 141 0. 150 0. 439 0. 877 1. 316 1. 755 1. 8 1. 6 1. 4 1. 2 Cl Cd CL , CD against ? CL,CD, 2?? 1. 0 0. 8 0. 6 0. 4 0. 2 0. 0 2*Pi*a -0. 2 0. 0 2. 0 4. 0 6. 0 8. 0 ? 10. 0 12. 0 14. 0 16. 0 6 SAMP LE CALCULATION The sample calculation is based on Tapping 2 & Table 1: Coordinate of Pressure Tapping Table 3: Pressure Coefficients 1. Air Density at 29ðC ( ) ( ) 2. Free Stream Velocity v v v ( ) 3. Reynolds Number 4. Pressure Coefficient ( ) ( ) ( ) ( ) 5. facelift and pouf Coefficient 7 DISCUSSION 1.Plot CL and CD against Please refer to Page 6. on the same graph. 2. Compare the experimentally measured CL with the Thin Aerofoil Theory prediction of . establish the similarity and discrepancy observed. The graph shows at fine attack angle (4ð and 8ð), the measured Lift coefficient is quite close to theoretical predicted value , this is because at small attack angle, air stream flows along the control surface surface smoothly without flow insularity, which fulfills the prefatorial assumption of Thin Aerofoil Theory, thence the experimental result matches with theoretical value well.When further growing attack angle, the streamline become highly curved, until at certain angle the streamline is no long-lasting attached to the surface surface and flow separation is occurred, massive turbulence wake appears on surface upper surface, which greatly reduce the lift. At this spot aerofoil is actually ââ¬Å"blockingââ¬Â the air flow, hence the Lift coefficient is significantly reduced after kick in Stall angle, and can no long acquire the theoretical predicted value . 3.What would you expect the lift and swing out puff to be when At , since the 0015 aerofoil is symmetrical, the jam on upper and g note surface of aerofoil is the same, hence it will not produce some(prenominal) lift quarter. The of 0015 aerofoil is 0. 0147 at (when Re=80000), so there is small dishevel force even out at 8 4. Does the why. which you have obtained gives the total drag on the aerofoil? Explain Total Drag of aerofoil is contributed by Parasite Drag and bring on Drag, the Parasite Drag is related to , while experience drag is a byproduct of lift.Ind uce drag is a drag force occurs when aerofoil redirects the airflow overture at it. Refer to below diagram, the lift force is normal to chord of aerofoil, when decompose the lift force to horizontal and vertical component, the horizontal component , which is in the same direction of drag. 5. Explain from the pressure dissemination why there is a lift force. development as example, the pressure distribution diagram shows the lower surface of aerofoil has lesser pressure subside ? igher pressure, while upper surface of aerofoil has oftentimes higher pressure terminate, result in lower pressure. The integration of pressure drop along the aerofoil is the area under the curve, which represents force in a unit length of aerofoil, compare the area enfold for upper and lower surface, we can see the there is a resultant lift force produced. 3. 0 2. 5 2. 0 1. 5 CPL , CPU against X/C @ 8ð Cpl Cpu CPL ,CPU 1. 0 0. 5 0. 0 -0. 5 -1. 0 -1. 5 0. 0 0. 1 0. 2 0. 3 0. 0. 5 0. 6 0. 7 0. 8 0. 9 1. 0 X/C 9 6. Comment on the pressure distribution on the aerofoil when stall is reached. using as example, when stall angle is reached, the pressure drop of upper surface become insignificant repayable to massive turbulent wake, hence the lift force is greatly reduce and stall happens. 3. 0 2. 5 2. 0 1. 5 CPL , CPU against X/C @ 16ð Cpl Cpu CPL ,CPU 1. 0 0. 5 0. 0 -0. 5 -1. 0 -1. 5 0. 0 0. 1 0. 2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 X/C 0. 9 1. 0 10\r\n'
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