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Method of characteristic for bell nozzle design
- 1. Characteristic Method Inviscid &Compressible Flow Method of obtaining nozzle contours: 1-Characteristic 2-Finite difference 3-Time dependence Velocity potential equation for irrotational flow in (3-D) and (2-D): 1 − 1 𝑎2 𝜕𝜙 𝜕𝑥 2 𝜕2𝜙 𝜕𝑥2 + 1 − 1 𝑎2 𝜕𝜙 𝜕𝑦 2 𝜕2𝜙 𝜕𝑦2 − 2 𝑎2 𝜕𝜙 𝜕𝑥 𝜕𝜙 𝜕𝑦 𝜕2𝜙 𝜕𝑥𝜕𝑦 = 0 10/11/2021 1 1 − ∅𝑥 2 𝑎2 ∅𝑥𝑥 + 1 − ∅𝑦 2 𝑎2 ∅𝑦𝑦 + 1 − ∅𝑧 2 𝑎2 ∅𝑧𝑧 − 2∅𝑥∅𝑦 𝑎2 ∅𝑥𝑦 − 2∅𝑥∅𝑧 𝑎2 ∅𝑥𝑧 − 2∅𝑧∅𝑦 𝑎2 ∅𝑧𝑦 = 0 𝑢 = 𝜕𝜙 𝜕𝑥 & v= 𝜕𝜙 𝜕𝑦 & z= 𝜕𝜙 𝜕𝑧
- 2. 𝜕𝜙 𝜕𝑥 = 𝑢, 𝜕𝜙 𝜕𝑦 =v 10/11/2021 2 1− 𝑢2 𝑎2 𝜕𝑢 𝜕𝑥 + 1 − 𝑣2 𝑎2 𝜕𝑣 𝜕𝑦 − 2𝑢𝑣 𝑎2 𝜕𝑢 𝜕𝑦 = 0 𝜕𝜙 𝜕𝑥 = 𝑓(𝑥, 𝑦) 𝑑𝑓 = 𝜕𝑓 𝜕𝑥 ⅆx + 𝜕𝑓 𝜕𝑦 ⅆ𝑦 𝑑 𝜕𝜙 𝜕𝑥 = 𝑑𝑢 = 𝜕2 𝜙 𝜕𝑥2 ⅆx + 𝜕2 𝜙 𝜕𝑥𝜕𝑦 ⅆ𝑦 𝑑 𝜕𝜙 𝜕𝑦 = 𝑑𝑢 = 𝜕2 𝜙 𝜕𝑥𝜕𝑦 ⅆx + 𝜕2 𝜙 𝜕𝑦2 ⅆy 𝜕𝜙 𝜕𝑦 = 𝑓(𝑥, 𝑦)
- 3. • Therefore for2-D: 𝜕𝑢 𝜕𝑥 𝑖,𝑗 = 2𝑢𝑣 𝑎2 𝜕𝑢 𝜕𝑦 − (1 − 𝑣2 𝑎2) 𝜕𝑣 𝜕𝑦 1 − 𝑢2 𝑎2 Velocity of the fluid in the flow field: 10/11/2021 3
- 4. • Writing Taylorseries for D: 𝑢𝑖+1,𝑗 = 𝑢𝑖,𝑗 + 𝜕𝑢 𝜕𝑥 𝑖,𝑗 Δ𝑥 + 1 2 𝜕2 𝑢 𝜕𝑥2 (Δ𝑥)2+ ⋯ By ignoring higher orders, 𝜕𝑢 𝜕𝑥 𝑖,𝑗 will be obtain. The determinator should not be equal to zero to the fraction is to be determinable. In Mach line definition axial velocity is sonic and then: 1 − 𝑢2 𝑎2=0 so u=a=sonic 10/11/2021 4
- 5. • u=a v 𝑉 𝜇 𝜇 Therefore sin 𝜇= 𝑎 𝑣 = 1 𝑀 so this is a Mach line meaning So Mach line is characteristic line 10/11/2021 5
- 6. 1 − 𝑢2 𝑎2 𝜙𝑥𝑥 − 2𝑢𝑣 𝑎2 𝜙𝑥𝑦+ 1 − 𝑣2 𝑎2 𝜙𝑦𝑦 = 0 • 𝑑𝑥𝜙𝑥𝑥 + 𝑑𝑦 𝜙𝑥𝑦 = 𝑑𝑢 solve for 𝜙𝑥𝑦 • 𝑑𝑥 𝜙𝑥𝑦 + 𝑑𝑦𝜙𝑦𝑦 = 𝑑𝑣 𝜕2𝜙 𝜕𝑥𝜕𝑦 = 1 − 𝑢2 𝑎2 0 1 − 𝑣2 𝑎2 𝑑𝑥 𝑑𝑢 0 0 𝑑𝑣 𝑑𝑦 1 − 𝑢2 𝑎2 − 2𝑢𝑣 𝑎2 1 − 𝑣2 𝑎2 𝑑𝑥 𝑑𝑦 0 0 𝑑𝑥 𝑑𝑦 = 𝑁 𝐷 10/11/2021 6
- 7. • This fractionwill become indeterminate when D=0. • To have an Char-Line, 𝜙𝑥𝑦 had to be indeterminate. • For given incremental changes, that the denominator goes to zero then 𝜙𝑥𝑦= discontinuous. • If D=0,so: 𝑑𝑦 𝑑𝑥 = − 𝑢𝑣 𝑎2±( 𝑢2+𝑣2 𝑎2 −1) 1− 𝑢2 𝑎2 : Char-Line eq have to be solved * 𝑢2 + 𝑣2=𝑉2so 𝑢2+𝑣2 𝑎2 = 𝑉2 𝑎2 = 𝑀2 then 𝑑𝑦 𝑑𝑥 = − 𝑢𝑣 𝑎2±( 𝑀2−1) 1− 𝑢2 𝑎2 10/11/2021 7
- 8. 𝐶+ 𝐶− 𝑉 𝜇 𝜇 𝜃 Stream line A • 𝑢= 𝑉 cos 𝜃 & 𝑣 = 𝑉 sin 𝜃 By substituting and simplifying, 𝑑𝑦 𝑑𝑥 can be rewritten as: • 𝑑𝑦 𝑑𝑥 = tan(𝜃 ∓ 𝜇) (char-line Eq) D tells us how char-lines are located, what is their orientation and local velocity, but N does essentially tell us the relationship of the properties and how they change over the char-line. 10/11/2021 8
- 9. • We have3 conditions: 1. M>1: two real roots, two char-line which means super sonic flow, we got hyperbolic (PDE). 2. M=1: one real root, one char-line/ Sonic case/ Parabolic PDE. 3. M<1: Imaginary roots/ elliptic PDE. If N=0: To 𝜙𝑥𝑦 =𝑑u/𝑑y = 𝑑𝑣/𝑑x be finite. Since 0<𝑑u/𝑑x<C 𝑑𝑣 𝑑𝑢 = ∓ 𝑀2−1 𝑑𝑣 𝑣 Which is exactly means P-M expansion angle, i.e., 𝑑𝑣 𝑑𝑢 = ∓ 𝑀2−1 𝑑𝑣 𝑣 = 𝑑𝜃: this is compatibility EQ 10/11/2021 9
- 10. • According tocompatibility Eq and based on char-line Eq: 𝑑𝜃 = − 𝑀2−1 𝑑𝑣 𝑣 : 𝐶− : 𝑅𝑅𝑤 𝑑𝜃 = + 𝑀2−1 𝑑𝑣 𝑣 : 𝐶+: 𝐿𝑅𝑤 By integrating: 𝐶− ≡ 𝜃 + 𝛾 𝑀 = Constant=𝐾− 𝐶+ ≡ 𝜃 − 𝛾 𝑀 = Constant=𝐾+ Where K is a constant parameter along each char-line. 10/11/2021 10
- 11. • P-M expansionwave angle: 𝜃 = 𝛾 𝑀2 − 𝛾 𝑀1 Where 𝛾 is a P-M function: 𝛾(M)= 𝛾+1 𝛾−1 tan−1 𝛾−1 𝛾+1 (𝑀2 − 1) − tan−1 𝑀2 − 1 𝜃 = 1 2 (𝐾−+ 𝐾+) & 𝛾 = 1 2 (𝐾− − 𝐾+) 10/11/2021 11
- 12. • Nozzle definitiontypes according to char-line: Gradually expansion nozzle Minimum length nozzle 10/11/2021 12
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- 14. P A B • Consider theintersection of two characteristic lines A and B at point P, then we have: 𝑚1 = tan( 𝜃 − 𝑎 𝐴 + 𝜃 − 𝑎 𝑃 2 ) 𝑚11 = tan( 𝜃 − 𝑎 𝐵 + 𝜃 − 𝑎 𝑃 2 ) And 𝑦𝑃 = 𝑦𝐴 + 𝑚1(𝑥𝑃 − 𝑥𝐴) 𝑦𝑃 = 𝑦𝐵 + 𝑚11(𝑥𝑃 − 𝑥𝐵) 𝑥𝑃 = 𝑦1 − 𝑦𝐵 + 𝑚11𝑥𝐵 − 𝑚1𝑥𝐴 𝑚11 − 𝑚1 10/11/2021 14
- 15. • Inviscid Machcontour • Diverging length at Mach 3 • RS-25 contour is on MATLAB (press Ctrl + Clink on it) Also Solidworks part is here 10/11/2021 15
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- 18. • 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑟𝑎𝑡𝑖𝑜: 𝑜𝑢𝑡𝑠𝑖𝑑𝑒𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑐ℎ𝑎𝑚𝑏𝑒𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑇𝑒𝑚𝑝 𝑟𝑎𝑡𝑖𝑜: (𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑟𝑎𝑡𝑖𝑜) 𝛾−1 𝛾 • 𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑡ℎ𝑟𝑜𝑎𝑡 𝑇𝑒𝑚𝑝 = 2𝛾𝑅.𝑐ℎ𝑎𝑚𝑏𝑒𝑟 𝑇𝑒𝑚𝑝 𝛾−1 𝐶ℎ𝑒𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑡ℎ𝑟𝑜𝑎𝑡 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 2 𝛾+1 𝛾 𝛾−1 ∗ 2.088 • 𝐶ℎ𝑒𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑡ℎ𝑟𝑜𝑎𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 2𝛾𝑅.𝐶ℎ𝑎𝑚𝑏𝑒𝑟 𝑇𝑒𝑚𝑝 𝛾+1 • 𝐸𝑥𝑖𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑇ℎ𝑟𝑜𝑎𝑡 𝑇𝑒𝑚𝑝. (1 − 𝑇𝑒𝑚𝑝 𝑟𝑎𝑡𝑖𝑜) • 𝐸𝑥𝑖𝑡 𝑇𝑒𝑚𝑝 = 𝐶ℎ𝑎𝑚𝑏𝑒𝑟 𝑇𝑒𝑚𝑝 . ( 𝑃𝑒𝑥𝑖𝑡 𝑃𝑐ℎ𝑎𝑚𝑏𝑒𝑟 ) 𝛾−1 𝛾 • 𝐸𝑥𝑖𝑡 𝑠𝑜𝑢𝑛𝑑 𝑠𝑝𝑒𝑒𝑑 = 𝛾𝑅𝑇𝑒𝑥𝑖𝑡 • 𝐸𝑥𝑖𝑡 𝑀𝑎𝑐ℎ = 𝑒𝑥𝑖𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑒𝑥𝑖𝑡 𝑠𝑜𝑢𝑛𝑑 𝑠𝑝𝑒𝑒𝑑 • 𝑃𝑀 = 𝛾+1 𝛾−1 × tan−1 𝛾−1 𝛾+1 𝑀2 − 1 − tan−1 𝑀2 − 1 • 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑤𝑎𝑙𝑙 𝑎𝑛𝑔𝑙𝑒 = 1 2 {𝑃𝑀(𝑒𝑥𝑖𝑡 𝑀𝑎𝑐ℎ)} • 𝐴𝑛𝑔𝑙𝑒 𝑖𝑛𝑐𝑟𝑒𝑚𝑒𝑛𝑡 = 2. ( 90 − 𝑀𝑎𝑥 𝑎𝑛𝑔𝑙𝑒 − 𝑓𝑖𝑥 90 − 𝑀𝑎𝑥 𝑎𝑛𝑔𝑙𝑒 ) 10/11/2021 18
- 19. • MATLAB CodingMethod: 10/11/2021 19 Exit pressure Pressure ratio Throat T,P,V Exit V,T,a,M Prandtl-Mayer function to finding expansion angle-its recursive Maximum Wall angle Number of division Calculate position of the center line Calculate the wall positions- its recursive Export X,Y into CAD
- 20. For more informationread Rocket Science book which is written by Mahdi Hossein gholi nejad and Professor Mofid Gorji. With special thanks from 10/11/2021 20