Development Process - Aerospace Engineering Design - Lecture Slides, Slides of Engineering Dynamics

Download Development Process - Aerospace Engineering Design - Lecture Slides and more Slides Engineering Dynamics in PDF only on Docsity! 1 PRODUCT DEVELOPMENT PROCESS • In last lecture examined – Phase 1: Concept Development – Phase 2: System-Level Design – Contrasted Phase 1 and Phase 2 for ‘product’ vs. ‘research project’ – Check website for many more references (9 examples on line) • Emphasis now on Phase 3: Detailed Design • Simplified calculations to guide Phase 3 Docsity.com 2 WING IN GROUND EFFECT PROJECT Boeing has recently taken interest in the WIG phenomenon and proposed a concept for a massive craft to meet a US Army need for a long-range heavy transport. Called the Pelican, the 500 ft (153 m) span vehicle would carry up to 2,800,000 lb (1,270,060 kg) of cargo while cruising as low as 20 ft (6 m) over water or up to 20,000 ft (6,100 m) over land. Unlike the Soviet concepts, the Pelican would not operate from water, but from conventional runways using a series of 76 wheels as landing gear. Docsity.com 5 EXAMPLES: EKRANOPLAN • Objective/Goal Statements: – “Data gathered from our wind tunnel testing will hopefully allow the team to develop a system of permanent equations that will allow us to mathematically show why ground effect vehicles are more efficient and which designers can use to design WIG vehicles in the future.” – “Use CFD to model ground effect” • Comments: – Great idea, correlate data from wind tunnel and experiment to develop new predictive methods – Already been shown why and how ground effect augments lift – Wind tunnel testing is effective, but can some simply models be developed to understand the phenomena and make predictions? – How does the wing tunnel testing and CFD fit together? Docsity.com 6 EKRANOPLAN: POTENTIAL ANALYSIS • Consider the potential flow generated by a pair of cylinders located in a cross-stream, V∞, each with radius R, and located at x = 0, y = ±A (A>R). • Write the ψ and φ for this flow, and derive the velocity components. Show that there exists a plane of symmetry at y = 0 where vertical velocity is zero. • Show that there exists a streamline that runs along this axis of symmetry. Note: This streamline can be used to represent a ‘ground plane’, and thus this potential flow combination – a source doublet and its ‘image doublet’ represents a cylinder in ground effect. • Derive expressions for the coefficients of lift and drag for the cylinder located about the ground. How do they vary as the cylinder gets closer to the ground (i.e., the cylinder and its image get closer together)? • What are implications of this results for an airplane trying to land on a runway? V∞ A R Docsity.com 7 SPACECRAFT SLOSH • Upper-stages of expendable vehicle fleet undergo orbital maneuvers that may lead to large propellant slosh motions • Slosh motion can affect vehicle performance – Example: reorientation maneuver may cause liquid propellant to exit through pressure relief valves designated for gas venting leading to uncontrollable dynamic instability – Example: cryogenic liquid may splash onto hot sections of tank side walls and dome leading to large boil-off of propellant – Example: NEAR spacecraft interrupted its insertion burn when fuel reaction was larger than anticipated. Prevented NEAR from orbiting Eros and delayed mission – Example: Does Orion need baffles? • Models needed to predict impacts of propellant slosh for various mission scenarios • Limited database to benchmark CFD codes in very low-gravity (low-acceleration) environments Lockheed Martin’s Atlas V 401 Boeing’s Delta IV Heavy Docsity.com 10 EXAMPLE OF CFD CALCULATION • CFD provides additional details • Important to ask whether or not you need these details for the level of fidelity required • Before you do a CFD calculation, ask yourself: – Why am I doing this? – Is there a cheaper, faster way to get what I want? Docsity.com 11 REAL NEED FOR CFD Docsity.com 12 OTHER SPRING-MASS-DAMPER SYSTEMS • Provides: damping ratio, natural frequency, sensitivity to various parameters • Numerous models in existence: car suspension, microphone, control systems, electromechanical systems, etc. Docsity.com 15 SIMPLE PROPELLER: ACTUATOR DISK THEORY 1. Neglect rotation imparted to the flow. 2. Assume the Mach number is low so that the flow behaves as an incompressible fluid. 3. Assume the flow outside the propeller streamtube has constant stagnation pressure (no work is imparted to it). 4. Assume that the flow is steady. Smear out the moving blades so they are one thin steady disk that has approximately the same effect on the flow as the moving blades (the ``actuator disk''). 5. Across the actuator disk, assume that the pressure changes discontinuously, but the velocity varies in a continuous manner. Docsity.com 16 MESSAGES FROM PROPELLER EXAMPLE • When using CFD, be able to say in your report: – We needed a CFD model/solution to get ___________ – The results of the CFD solution were used to __________ – A simpler model was insufficient because it did not capture ________ • When using a simplified model, be able to say in your report: – The assumptions in the simplified model are 1) __________, 2) _________ – Assumption (1) is valid because __________ – Assumption (2) is valid because __________ – The model was developed to predict _________ – The results of the model were used to _________ – This model does not capture the details of __________ Docsity.com 17 COMBUSTION KINETICS EXAMPLE • Many prior senior design projects involved combustion • Consider Methane Combustion • This looks awfully simple – when written as a 1-step process • Equation says nothing about how long the process takes to happen • Approach taken by team from 2005-2006 in project goals: – “Model kinetics of combustion process…” CH4 + 2O2 + 7.52N2 → CO2 + 2H2O + 7.52N2 + heat Fuel Air Products Docsity.com METHANE COMBUSTION (STEPS: 178-262) Table 53 {continued} Table 5.3 (continued) Forward Rate Coefficiomt" Forward Rate Coelficient” No. Reaction A A Nu. Renetion N-Contuinbyy Reactions Containing Reactiaus (eontianed) 1 NNO —R2 +0 FS0E + 2 ac 336 0 CN +O) NCOFO li? N#O;--NO+O TAIE +12 ao 6460 21 CN4H; ~ HON+H ts) N+ OH ~ NO+H TME S13 oo 1120 - 0 KCO+0—NO+CO 235413 au oo 18 N20 +O Na+ 02 JHE +12 oo ale NCO-+ Hl NH#CO SHOE +13 aa oo 182 N;O=0-—NO1 NO 2908 +13 oo 23550 my , 82 NCO+ OH —NO+H+CO 2508 +12 ag iy 1a NiQ+ He Ny +08 4d0E— 14 oo 18.880 r : . ODE + 12 2108 NCOLN~Nr+CO 200E + 13 on 00 Is N;O4+-0H ~ N; + HO; 2WE + 12 oo 21060 " 2 ; 5 x 185 NGM) = No~ (4M) pressure dependent NCO — Oy —~ NO-COy 200E +12 ay 20,000 BOR: eet , NCO+M—N4CO3M 8 ROE +16 -05 48.000 18) HO;—NO-~ NO: OH 21+ 12 oo eo NCO+NO — N04. CO 28SE=17 152 740 we 1 06E = “iat NCO +NO -» N24 €Or 570K +18 -20 00 les 390R 12 g0 240 230 HCN tM H+ CN+M LOGE +29 —33 SS 1 aE 2 1d nO 60 1 HCN4+0—NCO+H UNE 4-08 264 M2 HCN-}O-NH+CO 2778-03 264 SOOR + 00 O68 WA HCN) O-+CN+0H 21SE-- ag LSB S206 +13 oo 330 4 HCN.|OH ~HOCN+H TIDE +06 203 NH con BNO -H 2O0E= 13 on oo 33) HCN + OH - HNCO+H 4 d0B +03 226 NEL OH -- N+: 2008 09 12 80 36 HCN +OH — NH; +CO 1 6OE 4-02 256 x NH +0; - HNO - HOLE +05 20 6.500 B7 He HCN +M 3 H)CNGM J HOE +26 -34 1900 NO OL Ve 100 NH+N— Ny +H 00 on 28 HCN +N Ny + CHy 600R +13 0.0 400 NHS HO — Hy 200E~ oa 13,850 ee Meat 3 bb BY CH Ni CN EN 630E +14 oo © 46,020 MILNO LNOGH dsead “3400 WO CH+ Ny > HON] N 2968 + 0% Wt 20.460 cH N:(+M) (=M) pressure dependent 2) NH -O- OH + ME FONE + 12 0.0 oa POOR 4-13 00 74.000 -~O—H+HNO 4 GUE 4 ES ao oO cn (5) + Nx NH +HCN 1008 f oo 68,000 H-- NH +H 4008415 oo 3.650 U3 NH; +. OH + NH + HO 9 00E = 07 1s 460 C+NO—CN+0 190B +13 oo oo a C+NG—CO+N oo oo MONA Re SEO a i360 CH+NO HCN +0 oo on She a “M— No v= M UE + 30 oe CH+NO = H4NCO 00 00 “ 202 Hrs CH+NO + NaHCO 3008 +13 09 00 2UT) NNH +O ~ OH + No no oo 0 5 + + 28 NNE +O — NH = NO ba op CH: + NO — H+ HNCO 3108 +17 “138 1270 2 NNH SHH oo 00 CH; + NO — OH + HER 2908 +14 069 760 3) NNITLON = #0 7 0 wo CH; + NO = H 4 BEKO 3R0E +13 036 580 MN ON. p 0 NO = HP HNCO 31M +17 “AB 1.270 RL ONMETOHg ~ CHiN, on 8 . . (0 — OH + HCN 290E + 4 “0 69 760 H4NO+M—HNO+M 132 740 =NO — H+ HCNO 3H0E 44 036 380 a5 CH) + NO HCN +130 9608 413 oo 28.80t 13) HNO+O— + OH ne on af th tee hoteaNG n 660 CH) + NO — [CN 4 OH 1008 4 52 00 N75 Ui HNO +0H — NO + 120 1 ey HCNN 40 = CO+H= 220R4 12 00 oe 16 HNO +0) -+ HO) 4 NO po 13.00 HCNN+O—HCN+NO- 200K 4 12 a0 ao 2M cN20-—COEN 710% ao 0a HICNN + 0: ~ 04+ HCO +N; 1 20E + 13 vo 09 + OH -— NCO-FH NDE 4 co 00 HCNN + OF = H+ HCO+D 120E +13 oo 00 N+ HO — HON +0H 8 008 o0 7A HICNN + H+ CHa + Ne 100E + 16 oo oo 262 HNCO4+0= NH+COy 9806-07 lai 8,500 Docsity cong, METHANE COMBUSTION (STEPS: 263-279) Table 5.3 {continued} Forward Rate Coefficient* No. Reaction 4 4 E N-Containing Reactions (continued) 263. HNCO+O-—HNO+CO 1 50E + 08 157 44,000 264 HNCO+O-—NCO+ 0H 2 20E + 06 211 11,400 265 HNCO+H — NH) +CO 3 25E +07 17 3,800 266 HNCO+H— H) +NCO 1 05E +05 25 13,300 267 ~=HNCO+OH — NCO+ H20 465E + 12 06 6,850 268 HNCO+OH — NH2+CO: 155E +12 00 6,850 269 HNCO+M—NH+CO+M L18E + 16 00 84,720 270 HCNO+H — H+HNCO 2 1OE + 15 -0 69 2,850 271. HCNO+H — GH +HCN 270E+11 018 2,120 272 HCNO+H — NH; +CO 1 70E + 14 ~O0 75 2,890 273) HOCN+H —H+HNCO 2 OOF + 07 20 2,000 274 HCCO+NO- HCNO+CO 235E+13 00 00 275) CH, +N—- HiCN+H 6 OE + 14 -O31 290 276 CHy+N— HCN + Hy 3 70E +12 Q15 90 277) NH3+H-— NH) +H 5 ADE +05 24 9915 278 NH; +OH — NH; +H;0 5 Q0E +07 16 955 2799) NH; +O-— NHs + OH 9 40E + 06 1.94 6,460 "The forward rate coefficient k = ATbexp(—E/RT) R is the universal gas constant. T is the temperoture in K The units of A involve gmal/cm’ and s, and those of E, cal/gmol “Clo (5) designates the singlet slate of CH, y Docsity .cony 22 GLOBAL AND QUASI-GLOBAL MECHANISMS • ‘One-step mechanism’ • Parameters: A, Ea/R, m and n • Chosen to provide best fit agreement between • Experimental and predicted flame temperatures, as well as flammability limits • Can be implemented in Excel [ ] [ ] [ ]nmyxayx k yx OHCRT EA dt HCd OHyxCOOyxHC global 2 222 exp 24     −−= +→      ++ Docsity.com 25 MAJOR COMBUSTOR COMPONENTS C om pr es so r Tu rb in e Docsity.com 26 MAJOR COMBUSTOR COMPONENTS • Key Questions: – Why is combustor configured this way? – What sets overall length, volume and geometry of device? C om pr es so r Tu rb in e Fuel Docsity.com 27 APPLICATION TO COMBUSTION SYSTEM MODELING Conceptual model of a gas-turbine combustor using 2 WSRs and 1 PFR Air C om pr es so r Tu rb in e φ ~ 1.0 T~2500 K φ~0.3 Primary Zone Docsity.com 30 OVERALL MESSAGES • Why am I developing a model? • What do I need to get out of it? • How will that result be used for the project? • Can I assume the problem is steady-state? – Can I remove time? – Is the situation that I am looking at replaceable by a static one? • Can I model the problem as 1-D? – What is the dominant direction in which things are happening? • Find similar situations in text books – Don’t expect to find the exact model of what you are doing • Add levels of complexity sequentially – don’t go for the homerun ball right away – Must understand each physical process on their own before looking at interaction – Shows progress Docsity.com 31 HOW DOES IT REALLY WORK? CFD Experiments Simplified Models Docsity.com 32 HOW DOES IT REALLY WORK? CFD Experiments Simplified Models Docsity.com