REPORT Design Helicopter 2 seat

REPORT Design Helicopter 2 seat tài liệu, giáo án, bài giảng , luận văn, luận án, đồ án, bài tập lớn về tất cả các lĩnh...

REPORT: HELICOPTER DESIGN TEAM: HELICOPTER Guider : Professor Mr.Vũ Ngọc Ánh Member: Hoàng Tiến Nam_G1002009 Nguyễn Triệu Nhật Thanh_G1002942 Nguyễn Quang Vinh_G1004005 Đặng Trí Tâm_G1002835  Content: I/ Aerodynamic of helicopter II/ Performance of helicopter a b c d e Power losses Result form helicopter design Vertical drag in hover Tail rotor-fin interference in hover Parasite drag in forward flight I/ Aerodynamic of helicopter 1.Main rotor in level flight Input parameter: 𝜎 = 0.0475 , 𝜃1 = −8.7 , 𝐴𝑏 = 125𝑓𝑡 , 𝑓 = 19.3𝑓𝑡 , 𝑀𝑑𝑟3𝑐𝑙=0 = 0.8 (𝑁𝐴𝐶𝐴 0012) 𝑜𝑓 𝑡𝑖𝑝 𝑎𝑖𝑟𝑓𝑜𝑖𝑙, 𝛾 = 8.1, 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑜𝑢𝑛𝑑 = Find: 𝐶𝑄 𝐶𝐻 𝜎 , 𝜎 1116𝑓𝑡 𝑓𝑡 𝑉 𝐶𝑇 , Ω𝑅 = 672 , 𝜇 = = 0.35 , = 0.025 𝑠 𝑠 Ω𝑅 𝜎 , 𝜃0 , 𝛼 𝑇𝑃𝑃 , 𝐴1 − 𝑏1𝑠 , 𝐵1 + 𝑎1𝑠 Produre: a Calculate 𝑓 𝐴𝑏 + 𝜎 𝜇4 (𝐶𝑇 /𝜎)2 = 19.3 125 + 0.0475 0.354 (0.025)2 = 0.156 Note: 1/𝜇 is already included in the chart ordinate b Find 𝜃0 from top chart of first page 𝜃1 =5 𝜃0 𝜃1 =50 = 20 c Find 𝐶𝑄 /𝜎 from bottom chart 𝐶𝑄 = 0.0059 𝜎 d e f g h Is rotor operating beyond limit line of Δ𝐶𝑄 /𝜎 = 0.004? Yes Calculate effective shift of limit lines due to swist diferent than chart value Δ𝐶𝑇 = −0.003(𝜃1 + 50 ) = −0.003(−8.7 + 5) = 0.0111 𝜎 Estimate new value of Δ𝐶𝑄 /𝜎 with limit lines and point of sudden change of curvature shifted to the right by Δ𝐶𝑇 /𝜎 𝐶𝑄 = 0.0056 𝜎 Calculate advancing tip Mach number, 𝑀1,90 and drag rise Mach number 𝑀𝑑𝑟3 (1 + 𝜇)Ω𝑅 (1 + 0.35)672 𝑀1,90 = = = 0.813 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑜𝑢𝑛𝑑 1116 𝑀𝑑𝑟3 = 𝑀𝑑𝑟3𝑐𝑙=0 [1 − 6(𝐶𝑇 /𝜎)2 ] = 0.8[1 − 6(0.025)2 ] = 0.79 If 𝑀1,90 > 𝑀𝑑𝑟3 , use figure 3.43 to find compressibility loss 𝑀1,90 = 1.03 𝑀𝑑𝑟3  Δ𝐶𝑄 /𝜎𝑐𝑜𝑚𝑝 𝑀Ω𝑅 = 0.0005 i Find 𝐶𝐻 𝜎 , 𝜆′ , 𝐴1 − 𝑏1𝑠 , 𝐵1 + 𝑎1𝑠 from charts 𝐶𝐻 = 0.002, 𝜆′ = −0.025, 𝜎 𝐴1 − 𝑏1𝑠 = (0) = 0, 8.1 𝐵1 + 𝑎1𝑠 = 7𝑑𝑒𝑔 j Find Tip path plane 𝛼 𝑇𝑃𝑃 from 𝜆′ 𝜎𝐶𝑇 /𝜎 −0.025 0.0475 × 0.025 −1 −1 𝛼𝑇𝑃𝑃 = 𝑡𝑎𝑛 [ + ] = 𝑡𝑎𝑛 [ + ] = −4 𝑑𝑒𝑔 𝜇 2𝜇 0.35 × 0.352 k Find 𝜃0 from 𝜃0 = 𝜃0 𝜃1 =50 − 0.75(𝜃1 + 50 ) = 20 − 0.75(−8.7 + 5) = 22.8 𝑑𝑒𝑔 Main rotor in level flight with compressibility losses Given: 𝜎 = 0.0475 , 𝜃1 , 𝐴𝑏 , 𝑓, 𝑀𝑑𝑟3𝑐𝑙=0 𝑜𝑓 𝑡𝑖𝑝 𝑎𝑖𝑟𝑓𝑜𝑖𝑙, 𝛾, 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑜𝑢𝑛𝑑, Ω𝑅, 𝜇, 𝐶𝑇 /𝜎 Ω𝑅 = 772𝑓𝑡/𝑠 Find: 𝐶𝑄 𝐶𝐻 𝜎 , 𝜎 , 𝜃0 , 𝐴1 − 𝑏1𝑠 , 𝐵1 + 𝑎1𝑠 Procedure: Same as example up to step g g Calculate advancing tip Mach number,, 𝑀1,90 and drag rise Mach number 𝑀𝑑𝑟3 (1 + 0.35)772 𝑀1,90 = = 0.934 1116 𝑀𝑑𝑟3 = 𝑀𝑑𝑟3𝑐𝑙=0 [1 − 6(𝐶𝑇 /𝜎)2 ] = 0.7995 h calculate 𝑀1,90 / 𝑀𝑑𝑟3 𝑀1,90 /𝑀𝑑𝑟3 = 1.168 i find Δ𝐶𝑄 /𝜎𝑐𝑜𝑚𝑝 𝑀Ω𝑅 from figure 3.43 Δ𝐶𝑄 /𝜎𝑐𝑜𝑚𝑝 = 0.0056 𝑀Ω𝑅 Δ𝐶𝑄 j find 𝜎 𝑐𝑜𝑚𝑝 Δ𝐶𝑄 772 = 0.0056 ( ) = 0.0019 𝜎𝑐𝑜𝑚𝑝 1116 k find new value of 𝐶𝑄 /𝜎 𝐶𝑄 = 0.0047 + 0.0019 = 0.0066 𝜎 remainder same as example Tail rotor Parameter 𝜎 = 0.0475 , 𝜃1 = −8 , 𝛼𝑠 = 0, 𝑀𝑑𝑟3 = 0.764, 𝑜𝑓 𝑡𝑖𝑝 𝑎𝑖𝑟𝑓𝑜𝑖𝑙, 𝛾 = 8.1, 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑜𝑢𝑛𝑑 = Ω𝑅 = 1116𝑓𝑡 , 𝑠 672𝑓𝑡 , 𝜇 = 0.35, 𝑠 𝐿𝑇 = 1.23, 𝑅𝑀 Find 𝐶𝑇 𝐶𝐻 𝜎 , 𝜎 , 𝜃0 , 𝐶𝑄 𝜌 = 0.0049, = 1, 𝛿3 = 𝜎𝑀 𝜌0 [𝜌0 𝐴𝑏 (Ω𝑅) ] 𝑇 = 10895, 𝜌0 𝐴𝑏 (Ω𝑅)3 [ ] = 163940 550 𝑀 𝐶𝑄 𝜎 Procedure: a Find 𝑇𝑇 from: 550 𝜌0 𝐴𝑏 (Ω𝑅)3 𝐶𝑄 ( )[ ] Ω𝑅𝑀 𝐿𝑇 550 𝜎 𝑀 𝑀 𝑅𝑀 550 × 163940 × 0.0049 𝑇𝑇 = = 535 672 × 1.23 𝑇𝑇 = 𝐶 𝑇𝑇 𝑏 (Ω𝑅) b Find 𝜎𝑇 = 𝜌𝐴 𝑇 𝐶𝑇 𝜎𝑇 = 𝜌𝐴 𝑇𝑇 535 𝑏 (Ω𝑅) = 10895 = 0.0491 c For tail rotor, 𝛼 𝑇𝑃𝑃 = 𝑎1 𝑎𝑛𝑑 𝐵1 = Plot 𝛼 𝑇𝑃𝑃 and 𝐵1 + 𝑎1𝑠 vs 𝜃0 , find intersection 𝜆′ 𝜎𝐶𝑇 /𝜎 𝛼 𝑇𝑃𝑃 = 𝑡𝑎𝑛−1 [ + ] = 2.9𝑑𝑒𝑔 𝜇 2𝜇 d Find 𝐶𝑄 𝜎 from chart e Calculate advancing tip Mach no and drag rise Mach no (1 + 𝜇)Ω𝑅 (1 + 0.35)672 𝑀1,90 = = = 0.813 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑜𝑢𝑛𝑑 1116 𝑀𝑑𝑟3 = 𝑀𝑑𝑟3𝑐𝑙=0 [1 − 6(𝐶𝑇 /𝜎)2 ] = 0.8[1 − 6(0.0047)2 ] = 0.789 f If 𝑀1,90 > 𝑀𝑑𝑟3 , use figure 3.43 to find compressibility correction 𝑀1,90 = 1.0304 𝑀𝑑𝑟3 𝑀1,90 𝑀𝑑𝑟3 = 1.0304 => Very small => Δ𝐶𝑄 /𝜎𝑐𝑜𝑚𝑝 𝑀Ω𝑅 ~0 =>no compressibility loss g Find 𝐶𝐻 𝜎 from 𝐶𝐻 𝐶𝐻 𝐶𝑇 𝐶𝑇 = ( − 𝑡𝑎𝑛𝑎1𝑠 ) + 𝑡𝑎𝑛𝑎1𝑠 𝜎 𝜎 𝜎 𝜎 𝐶𝐻 = 0.0008 + 0.047𝑡𝑎𝑛2.9 = 0.0029 𝜎 h Find 𝜃0 from 𝜃0 = 𝜃0 𝜃1 =50 − 0.75(𝜃1 + 50 ) 𝜃0 = 6.7 − 0.75(−8 + 5) = 8.95𝑑𝑒𝑔 Helicopter in Autorotation Given: Main rotor;  ,1 , M dr3 of tip airfoil, R, Ab Tail rotor; power to run at flat pitch 15h.p Transmission and accessory losses 10 h.p Airframe aerodynamics; Lift and drag aversus F Flight Conditions; G.W.;  ,  / o , speed of sound Find: Rate of descent and trim conditions Procedure: a Convert tail rotor and transmission and accessory to CQ /  for main rotor 15  10  0.00015 164146 b Find compressibility losses as in example CQ /    c Estimate  F , find LF/q and calculate LF  F  0, LF / q  4.5; LF  295.97 d CT /   0.023 e Find o 5 using CT /  and CQ /  o 5  '  f Find    tan g  '  0.025 from chart    tan 1 1 0.025  4o 0.35 h Find new values of LF / q, LF , CT /  LF / q  3.1, LF  204, CT /   0.014 i Find new values of o 5 o 5  6.5 j   '  CT /   Find TPP  tan 1    2    0.025 0.0475  0.014  TPP  tan 1     4  0.352   0.35 k Find f from Figure 2, Appendix A l Calculate CDF /    f / Ab CDF /   0.008 m Find CH /  (assume CH /   0.0004 als  ) f  18 e Parasite drag in forward flight Parasite drag = streamline drag + bluff body drag The primary component of streamline drag is skin friction 𝑅𝑁 = 𝜌𝑉𝐿 = 6400 𝑉𝐿 𝑎𝑡 𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙 𝜇 We will determine drag coefficient of plat plate This is aproximility the coefficient of our design Length of helicopter: 18 ft Velociy: 140 knot Code matlab for thickness of boundary layer clc clear V=140; u=0.000198; h=input ('Nhap vao gia tri h') %height of imperfection for i=0:1.8:18 r=(V*i)/u;%Reynold% tb=(0.154*i)/(r^(1/7));%thickness of the boundary% CDa/CDf=0.75*(h/tb)^(1/3),%Drag ratio actual/freestream% end Result: tb=[0 0.0372 0.0674 0.0954 0.1221 0.1478 0.1729 0.1973 0.2212 0.2447 0.2678] c(0.0042)=[inf 0.3615 0.2965 0.2641 0.2433 0.2282 0.2167 0.2073 0.1996 0.1930 0.1872 ] c(0.021)=[inf 0.6098 0.5002 0.4455 0.4104 0.3850 0.3655 0.3497 0.3366 0.3255 0.3158] clc clear tb=[0 0.0372 0.0674 0.0954 0.1221 0.1478 0.1729 0.1973 0.2212 0.2447 0.2678] i=0:1.8:18 plot(i,tb,'+' ') clc clear c1=[1 0.3615 0.2965 0.2641 0.2433 0.2282 0.2167 0.2073 0.1996 0.1930 0.1872 ] c2=[1 0.6098 0.5002 0.4455 0.4104 0.3850 0.3655 0.3497 0.3366 0.3255 0.3158] i=0:1.8:18 plot(i,c1,'+',i,c2,'*') h=0.021 ft h=0.0042 ft The plot show boundary layer thickness over 18ft with 10 distance It show the ratio of actual drag coefficient to free stream drag coefficient for surface imperfections with heights of 0.0042 and 0.021ft The plot show that even when the boundary layer is 0.2 ft thick, a rivet head with a height of 0.021 ft has a drag coefficient that is 35% off the what it would be in the free stream  Estimate parameter for design *Estimate Drag Input parameter: Three-view drawing,drag data,reference speed Estimate :equivalent flat plate area at zero angle of attack a/Basic fuselage: -Determine frontal area ,AF from three view, AF=75.58ft2 -Determine fineness ratio l/d=6.5 -Find CDF from Figure 4.17 => CDF  0.076 =>fF=AFCDF=5.7ft2 b/Main rotor hub and shaft: -Determine frontal area of hub AMH  3.8 ft -Determine frontal area of shaft AMS  1.108 ft -Determine diameter of shaft DMS  0.6 ft -From Table 4.2, estimate CD MH RPM=0 CDMH  1.1 -From figure 4.22 find drag ratio  s  0, RPM  100%, D.R  1.00 / 0.95  1.05 -Calculate CD MH corr => CDMH  1.1 corr fMH=AMHCD MH corr=3.8.1.1=4.18 ft2 -Calculate R.N of shaft (140knot) R.N  6400 140 1.6  0.6  0.8 106 -Determine CD MS from figure 4.23 CDMS  0.3 => f MS  AMS CDMS => f MS  0.31 -Find Ki from figure 4.24 -Determine Z / Wp  3.8 / 7.8  0.487 => Ki  0.18( F  3 ) => fM=(1+Ki)fM => f M  1.18  4.49  5.3 ft Total f M  f MH  f MS =4.18 +0.31=4.49 c/Tail rotor hub and shaft -Determine frontal area AT  0.7 ft -From Table 4.2, estimate CD TH CDTH  1.1 -From figure 4.22, find drag ratio  S  0, RPM  100%, D.R  1/ 0.95  1.05 -Calculate CD TH corr CDTH  1.11.05  1.16 corr fT  AT CDTHcorr fT  0.7 1.16  0.812 ft d/Main landing gear: -Determine frontal area AMLG  4.58 ft -From figure 4.26, estimate CD MLG => CDMLG  0.4 =>fMLG = AMLG CD MLG=1.83 ft2 e/ Horizontal stabilizer -Area ,AH(Appendix A) A  20 ft -Span, bH bH  10 ft Aspect ratio ,A.R =b2H/AH A.R  -Thickness ratio,t/c t / c  0.12 -Mean aerodynamic chord, MAC MAc  ft -Reynolds number at 140 knots R.N  106 -Estimate CD0 from figure 4.15 CDo  0.009 -Estimate CL H from trim conditions(see chapter 8) CLH  3 -Estimate span efficiency factor ,   0.2 -Calculate induced drag coeff CDi  CLH (1   )   A.R  0.007 -Calculate root thickness t  0.12   0.24 ft -Estimate junction drag coeff, CDj from Figure 4.21 => CD j  0.072 -Compute equiv junction drag coeff CD j equiv => CD  CD j (t )   2   AH  jequiv  0.072  0.0242   2   0.004 20   -Total drag coefficient=CD0+CDi+CDj equiv => CDH  0.0164 -Estimate qH qH q q  0.75 -Calculate f H  qH (CDH AH ) q f H  0.75  0.0164  20  0.246 ft f/Vertical stabilizer -Area,AV AV  24.22 ft -Thickness ratio,t/c t / c  0.12 -Mean aerodynamic chord,MAC MAc  ft -Reynolds number at 140 knots R.N  4.5 106 -Estimate CD0 from Figure 4.15 CDo  0.009 -Estimate qV q qV q  0.75 -Calculate fV=qV/q(CDvAv) fV  0.75  0.01 24.22  0.18 ft g/Rotor fuselage interference drag -From figure 4.25 estimate deta => CD  0.018 fint  CD AF F  h/Exhaust drag -Ask engine manufacture to estimate net exhaust thrust Tnet  2(11)  22 -For engine installation on example helicopter fex  Tnet / q => f ex   22  0.3 ft 0.5  0.00238  2362 i/miscellaneous drag items -Estimate total drag of antennas, door handles, lights, steps, skin gaps, cooling leakeag, ventilation, f misc  0.5 ft Total equivalent flat plate area ftot  f F  f N  f M  fT  f MLG  f NLG  f H  fV  fint  f ex  f misc ftot  5.74  1.1  5.3  0.812  1.83  0.25  0.18  1.36  0.3  0.5  16.272 ft ... losses 0.0059 0.0066 0.0 02 0.0 02 -0. 025 22 .8 -4 -0. 025 17.8 -4.6 Entire helicopter in level flight helicoptr in helicopter autorotation in climb 0.0059 0.0 029 8.95 2. 9 9 .28 21 .7 1.5 hp_main(hp) hp_tail(hp)... and f LF / q  22 .5, f  22 k Calculate new value of CT /   0. 023 l Find new value of