F10Aero-cn+hw18

should be expansion waves. The slope of the upper wall should be in the direction of the flow,
w
Since θ is decreasing (expansion process), θw should decrease in the flow direction and reaches 0 at point 1, thus segment 3-1 of the upper wall should be convex (upward-curved). We shall call the zone formed by line 6-7, line 7-3, line 3-2, and line 2-6 as the expansion zone. Near the throat, right running (Family I) waves are generated on the upper wall. These waves should be expansion waves, thus making the upper wall a concave shape (downward-curved). The family I expansion waves will then meet the centerline and be reflected as family II expansion waves, which in turn are reflected on the upper wall, generating a new set of family I expansion waves. The process can go on and on in the flow direction. The end of the process is when one of the family I expansion wave accelerates the flow to the design Mach number on the center line. That wave is wave 3-2. Thus the design process is like this: (1) Select first a downward-curved shape of upper wall 7-3 such that the right-most family I expansion wave meets the center line at point 2 where the flow reaches MD. (2)Then, we can construct the upper wall 3-1 in a way that cancel all the family II waves crossing wave 3-2 and into zone #4. Now, we take a look at how to make the selection in step (1). For the 7 points listed in the table above, we want to know their θ and ω. Points 1 and 2 have the same properties, so do points 6 and 7. Upon further considerations, we have 5 unknowns: θ3, θ5, ω3, ω4, ω5. We need to get as many equations relating these parameters as possible. Points 7 and 5 are on a family II hodograph characteristic curve, θ5 – θ7=ω5 – ω7 θ5 = ω5 Points 5 and 4 are on a family I hodograph characteristic curve, θ4 – θ5 =ω5 – ω4 θ5 =ω4 – ω5 Points 4 and 3 are on a family II hodograph characteristic curve, θ4 – θ3 =ω4 – ω3 θ3 =ω3 – ω4 Points 3 and 2 are on a family I hodograph characteristic curve, θ3 – θ2 =ω2 – ω3 θ3 =ω1 – ω3 Points 2 and 1 have the same flow properties, thus giving us nothing new. Putting them together, θ5 = ω5=ω4 – ω5 θ3 =ω3 – ω4=ω1 – ω3 and we have 4 equations. Recalling that we have 5 unknowns, there would be 1 degree of freedom. We can make θ3 as that unknown to be determined. Once θ3 is known, all the other parameters are determined. ω3 = ω1 – θ3 ω4 = ω1 –2θ3 ≥ 0 θ3 ≤ ω1 /2 ω5 = ω1 /2 – θ3 θ5 = ω1 /2 – θ3 ≤ θ3 θ3 ≥ ω1 /4 thus, ω1 /4 ≤ θ3 ≤ ω1 /2 defines the restriction on the selection of θ3. We can first take a look at the two limiting cases. (a) If we chose θ3 = ω1/4, we get, ω3 = 3ω1/4; ω4 = ω1/2; ω5 = θ5 = ω1/4, which means that θ3 = θ5. It is mentioned that the expansion wall should be downward-curved (θ3 ≥ θ5), we get that the expansion wall should be a straight line with a slope of ω1 /4. (b) If we chose θ3 = ω1 /2, we get, ω3 =ω1/2; ω4 = 0; ω5 = θ5 = 0;
Aerodynamics (Fall 2010) Lecture Note for Class #18; Wednesday, November 3, 2010
Preliminary Aerodynamic Design of the Diverging Portion of a Supersonic Nozzle (Planar 2D)
Task: design the shape (planar 2-D) of the nozzle wall in the diverging portion Considering only a geometrically symmetric nozzle, there must be a geometrical centerline, which is also a streamline. So we can consider the center line as a solid boundary and we need to look at the upper half of the nozzle only.③ ⑦⑤ ①MD⑥④②
Definition of various physical points Point # x y θ ω o ω1=ωD 1 x1 y1 start of t.s. on upper wall 0 o ωD 2 x2 0 start of t.s. on center line 0 on upper wall θ3=? ω3=? 3 x3 y3 on center line 0o ω4=? 4 x4 0 on upper wall θ5=? ω5=? 5 x5 y5 0.0 6 0 0 at throat on center line 0o o on upper wall 0 0.0 7 0 y7 at throat Assuming the sonic line is exactly at 6-7. It is required that the flow in the test section to be uniform and parallel (to the center line). Thus, there is a straight Mach line (1-2) ending at point 1. MD: Design Mach number, MD=2.1339 (ωD=30o) for this example Zone #4: formed by the upper wall of the test section, line 1-2, and the center line segment from 2 to downstream. The flow in this zone is parallel and uniform at the design Mach number. We shall call it the test section zone. Zone #3: enclosed by line 3-2, line 2-1, and upper wall section 1-3. It must be a zone of waves of one family (Family II). Should any family I characteristic wave be present, it would cross over line 1-2 and into the test section and make the test section non-uniform. It is generally true that only a zone of simple waves can be patched to a zone of uniform flow. Since zone #3 is a zone of simple waves, all the family I waves must be straight lines. We shall call zone #3 as the straightening zone. Mach number at point 2 must be MD. When each family I wave meets the upper wall, the slope of the wall must be such that the reflection of the wave is cancelled. Generally, it is undesirable to have compression wave. Should there be any compression wave, there would be adverse pressure gradient, which would make the boundary layer to be much thicker than that when there is favorable pressure gradient. So, all the family I waves in zone #3