AE 245 homework #3 solutions

Transcription

1 AE 245 homework #3 olution Tim Smith 4 February Problem1 Conider a low peed airplane with weight of 3100 lb and wing urface area of 300 ft 2. It i powered by a piton engine that deliver a maximum of 260 hp at ea level and a conventional propeller that operate at 75% efficiency. The aircraft aerodynamic propertie are decribed by the following expreion for lift coefficient and drag coefficient C L = C Lα α = 0:08α (1) where angle of attack α i meaured in degree. C D = C Do + KC 2 L = 0:02+0:06 C 2 L (2) 1. Plot the power (in horepower) required to maintain teady level, trimmed flight a a function of the air peed. Your plot hould cover the range for which the required power i le than the maximum available power. 2. Correponding to your reult in part (a), plot the angle of attack required to maintain teady level, trimmed flight a a function of the air peed. 3. What i the maximum air peed which maintain teady level flight at ea level. 4. If the tall angle of attack i α = 20 degree, what i the minimum air peed that the aircraft can maintain without talling? 1.1 Required and available power For teady, level flight, lift equal aircraft weight, L = W; thu, the lift coefficient i given by C L = W qs (3) where S i the wing area and the dynamic preure q 1 2 ρv 2 (4) 1

2 Subtituting Eqn. 3 into Eqn. 2 give the required power for teady, trimmed flight: P r = DV = qsc D V = qsv! W 2 C Do + K qs (5) The available power i given by P a = ηp = 0:75(260) hp = 195 hp (6) where η i the propeller efficiency and P i the maximum deliverable haft horepower power (hp) airpeed (ft/) Figure 1: Required power (olid line) and available power (dotted line) a a function of airpeed. The following IDL code generate Fig. 1: pro hw3_1a ; Define initial condition & converion. hp = ; 1 hp in ft-lb/ rho = e-3 ; ea level denity, lug/ftˆ3 W = ; aircraft weight, lb S = 300. ; wing area, ftˆ2 P_ = 260. ; engine haft power, hp eta = 0.75 ; propeller efficiency C_do = 0.02 ; paraitic drag coefficient K_d = 0.06 ; induced drag factor, 1/pi e AR ; Create air peed array. 2

3 v_min = 14. ; minimum peed, ft/ v_max = 250. ; maximum peed, ft/ bin = fix(v_max - v_min) + 1 ; number of array bin V = v_min + findgen(bin) ; air peed, ft/ ; Create available power array. P_a = eta*p_*(fltarr(bin) + 1.) ; available power, hp ; Calculate required power array. q = rho * Vˆ2/2 ; dynamic preure, pf C_l = W/(q*S) ; lift coefficient C_d = C_do + K_d*C_lˆ2 ; drag coefficient D = q*s*c_d ; drag, lb P_r = D*V/hp ; required power, hp ; Plot required & available power v. airpeed. xtitle = airpeed (ft/) ; x-axi title ytitle = power (hp) ; y-axi title file = fig3_1a.ep ; file name cale = 1.0 xz = 4.0*cale yz = 3.0*cale et_plot, p ; et plot device to PotScript device,/encapulated,/preview,filename=file ; open file & ave EPS device,/inche, xize=xz, yize=yz, $ font_ize = 7 plot, V, P_r, xtitle=xtitle, $ ; olid line i required power ytitle=ytitle, font=0 oplot, V, P_a, linetyle = 1 ; dotted line i available power device, /cloe ; cloe the file end 1.2 Angle of attack Again, lift equal weight for teady, level flight. Equating Eqn. 1 and Eqn. 3 and olving for angle of attack, The following IDL code generate Fig. 2: α = W qsc Lα (7) pro hw3_1b ; Define initial condition & converion. 3

4 angle of attack (deg) airpeed (ft/) Figure 2: Required angle of attack a a function of airpeed, teady level flight at ea level. rho = e-3 ; ea level denity, lug/ftˆ3 W = ; aircraft weight, lb S = 300. ; wing area, ftˆ2 C_La = 0.08 ; lift coefficient lope, 1/degree ; Create air peed array. v_min = 14. ; minimum peed, ft/ v_max = 250. ; maximum peed, ft/ bin = fix(v_max - v_min) + 1 ; number of array bin V = v_min + findgen(bin) ; air peed, ft/ ; Calculate required angle of attack array. q = rho * Vˆ2/2 ; dynamic preure, pf alpha = W/(q*S*C_La) ; angle of attack, deg ; Plot required & available power v. airpeed. xtitle = airpeed (ft/) ; x-axi title ytitle = angle of attack (deg) ; y-axi title file = fig3_1b.ep ; file name cale = 1.0 xz = 4.0*cale yz = 3.0*cale et_plot, p device,/encapulated,/preview,filename=file ; et plot device to PotScript ; open file & ave EPS 4

5 device,/inche, xize=xz, yize=yz, $ font_ize = 7 plot, V, alpha, xtitle=xtitle, $ ; olid line i alpha ytitle=ytitle, font=0 device, /cloe ; cloe the file end Note that angle of attack α blow up a airpeed V! 0. Thi i the ort of illine that reult from extending linear aumption (a in Eqn. 1) into nonlinear regime (i.e., tall). 1.3 Maximum air peed Maximum air peed occur at the higher-peed interection of the required power curve P r and the maximum available power curve P a. A reaonable firt etimate can be caled off Fig. 1: V max 241 ft= (8) while a imple biection routine 1 give V max = 241:633 ft= (9) 1.4 Minimum air peed Given a tall angle α = 20, we can equate Eqn. 4 and Eqn. 7 to give Solving for minimum air peed, 2W V min = = ρsc Lα α q min 1 2 ρv 2 min = W SC Lα α (10) 2(3100) 2:376910,3 ft= = 73:717 ft= (11) (300)0:08(20) which i coniderably higher than the peed at the lower-peed interection of P r and P a. 1 Not included; contact me if you want a copy. Thi in t a numerical method cla! 5

6 2 Problem2 Conider a jet airplane with weight of 15,850 lb and wing urface area of 360 ft 2. It i powered by a jet engine that provide a maximum thrut of 6000 lb at ea level. The aircraft aerodynamic propertie are decribed by the following expreion for lift coefficient C L = 0:184α (12) and drag coefficient C D = 0:02+0:04C 2 L (13) where angle of attack α i meaured in degree. 1. Plot the thrut (in lb) required to maintain teady level, trimmed flight a a function of the air peed. Your plot hould cover the range for which the required thrut i le than the maximum available thrut. 2. Correponding to your reult in part (a), plot the angle of attack required to maintain teady level, trimmed flight a a function of the air peed. 3. What i the maximum air peed which maintain teady level flight at ea level? 4. If the tall angle of attack i α = 20 degree, what i the minimum air peed that the aircraft can maintain without talling? 2.1 Required and available thrut A before, lift equal aircraft weight, L = W; ubtituting Eqn. 3 into Eqn. 2 give the required thrut (i.e., the drag) for teady, trimmed flight: T r = D = qsc D = qs! W 2 C Do + K qs (14) The available thrut i defined a T a = 6;000 lb. The following IDL code generate Fig. 3: pro hw3_2a ; Define initial condition & converion. rho = e-3 ; ea level denity, lug/ftˆ3 W = ; aircraft weight, lb S = 360. ; wing area, ftˆ2 T_ = ; engine haft power, hp C_do = 0.02 ; paraitic drag coefficient K_d = 0.04 ; induced drag factor, 1/pi e AR ; Create air peed array. 6

7 force (lb) airpeed (ft/) Figure 3: Thrut (olid line) and drag (dotted line) a a function of airpeed. v_min = 60. ; minimum peed, ft/ v_max = 900. ; maximum peed, ft/ bin = fix(v_max - v_min) + 1 ; number of array bin V = v_min + findgen(bin) ; air peed, ft/ ; Create available thrut array. T_a = T_*(fltarr(bin) + 1.) ; available thrut, lb ; Calculate required thrut (or drag) array. q = rho * Vˆ2/2 ; dynamic preure, pf C_l = W/(q*S) ; lift coefficient C_d = C_do + K_d*C_lˆ2 ; drag coefficient D = q*s*c_d ; drag, lb ; Plot required & available power v. airpeed. xtitle = airpeed (ft/) ; x-axi title ytitle = force (lb) ; y-axi title file = fig3_2a.ep ; file name cale = 1.0 xz = 4.0*cale yz = 3.0*cale et_plot, p device,/encapulated,/preview,filename=file device,/inche, xize=xz, yize=yz, $ ; et plot device to PotScript ; open file & ave EPS 7

8 font_ize = 7 plot, V, D, xtitle=xtitle, $ ; olid line i drag ytitle=ytitle, font=0 oplot, V, T_a, linetyle = 1 ; dotted line i thrut device, /cloe ; cloe the file end 2.2 Angle of attack A in Problem 1, Eqn. 7 give angle of attack a a function of airpeed angle of attack (deg) airpeed (ft/) Figure 4: Required angle of attack a a function of airpeed, teady level flight at ea level. The following IDL code generate Fig. 4: pro hw3_2b ; Define initial condition & converion. rho = e-3 ; ea level denity, lug/ftˆ3 W = ; aircraft weight, lb S = 360. ; wing area, ftˆ2 C_La = ; lift coefficient lope, 1/degree ; Create air peed array. v_min = 60. ; minimum peed, ft/ v_max = 900. ; maximum peed, ft/ 8

9 bin = fix(v_max - v_min) + 1 ; number of array bin V = v_min + findgen(bin) ; air peed, ft/ ; Calculate required angle of attack array. q = rho * Vˆ2/2 ; dynamic preure, pf alpha = W/(q*S*C_La) ; angle of attack, deg ; Plot required & available power v. airpeed. xtitle = airpeed (ft/) ; x-axi title ytitle = angle of attack (deg) ; y-axi title file = fig3_2b.ep ; file name cale = 1.0 xz = 4.0*cale yz = 3.0*cale et_plot, p device,/encapulated,/preview,filename=file device,/inche, xize=xz, yize=yz, $ font_ize = 7 plot, V, alpha, xtitle=xtitle, $ ; olid line i alpha end ytitle=ytitle, font=0 device, /cloe ; et plot device to PotScript ; open file & ave EPS ; cloe the file 2.3 Maximum air peed Maximum air peed occur when drag equal thrut; ubtituting T a for T a in Eqn. 14, Multiply through by (qs) 2 and rearrange into the quadratic equation T a qs = W 2 C Do + K (15) qs which can be olved by the quadratic formula: C Do (qs) 2, T a (qs)+kw 2 = 0 (16) qs = T a p Ta 2, 4C Do KW 2 = 6000p (6000) 2, 4(0:02)0:04(15850) 2 lb =(150;000148;320) lb (17) 2C Do 2(0:02) Subtituting the upper olution, (qs) max = 298;320 lb, into Eqn. 4 give the maximum air peed 2(qS) V max = max = ρs 2(298;320) 2:376910,3 (360) ft = 835:02 ft (18) 9

10 A quick glance at Fig. 3 confirm that thi i a reaonable air peed for the high-peed interection of the lift and drag curve. 2.4 Minimum air peed Subtituting new value into Eqn. 11 give the minimum air peed 2W V min = = ρsc Lα α 2(15;850) 2:376910,3 ft= = 100:33 ft= (19) (360)0:184(20) A quick glance at Fig. 4 confirm that thi i a reaonable air peed for a tall angle of α =