We are using blunt shaped nose in blended wing,instead of blunt,if we use needle shaped nose what would happen ,it reduce shockwave or it produce bow shock?
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asked Jul 24, 2015 at 17:10
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1$\begingroup$ This question might need more information, such as a context. Are we talking about the X-48 BWB testbed, or some other test aircraft, or an experimental design of yours, or a model aircraft? $\endgroup$– KeithSJul 24, 2015 at 17:39
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$\begingroup$ The short answer is that a needle-shaped nose would reduce the leading-edge stagnation point of the aircraft, potentially reducing drag at low angles of attack. However, if it's just the nose, there will be additional stagnation points along the leading edge of the fatter wing. You could build a BWB aircraft similar in theory to the SR-71 (in fact the SR-71 was a BWB design in many ways), but there is a tradeoff; the thin leading edge will cause the airfoil to stall at a lower critical AOA, reducing the performance envelope of the aircraft. $\endgroup$– KeithSJul 24, 2015 at 17:46
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$\begingroup$ This is my experimental design. $\endgroup$– Thangaraj SundaramoorthyJul 24, 2015 at 17:53
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4$\begingroup$ Sounds like you should run your experiment, then report back to us! $\endgroup$– FreeManJul 24, 2015 at 17:58
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1 Answer
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I assume you talk about the airfoil's nose, not the fuselage nose.
A pointed nose will cause flow separation in subsonic flow if the angle of attack is only slightly different from the angle of the leading edge. This flow separation will mightily increase drag because now no suction peak at the nose will develop. This suction peak is causing what is sometimes called "leading edge thrust" (PDF!). Without it, the aerodynamic forces will act in a direction orthogonal to the wing's surface and not (almost) orthogonal to the flow direction.
In pure supersonic flow the pointed leading edge is superior, because it causes less wave drag. However, it is preferable to sweep the wing such that the flow around the leading edge will stay subsonic even in supersonic flight, precisely because it still allows leading edge thrust to manifest itself.
answered Oct 4, 2015 at 11:25