AE 245 homework #4 solutions

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1 AE 245 homework #4 solutions Tim Smith 14 February Problem1 Consider a general aviation aircraft with weight of 3300 lbs and wing surface area of 310 ft 2. It is powered by a piston engine that provides a maximum of 250 hp at sea level and a conventional propeller that operates at 80% efficiency. The aerodynamic properties of the aircraft are described by a parabolic drag polar with C D = 0:02 + 0:06C 2 L (1) 1. What is the power level required to maintain a steady flight path angle of 3 degrees and an air speed of 220 ft/s at sea level? 2. Assume the aircraft flies with a flight path angle of 3 degrees using maximum available power. Determine all possible air speeds which maintain steady flight at sea level. 3. What is the minimum power level required to maintain steady climb rate of 10 ft/s at sea level? What is the corresponding air speed? 4. Plot the maximum steady climb rate of the aircraft as it depends on altitude from sea level to 20,000 ft. Plot the corresponding air speed as it depends on altitude. 5. Plot the minimum glide angle which maintains steady gliding flight as it depends on altitude from sea level to 20,000 ft. Plot the corresponding air speed as it depends on altitude. 1.1 Required power at γ = 3, 220 ft/s At V = 220 ft/s and sea level, the dynamic pressure and the lift coefficient q 1 2 ρv 2 1 = 2 (2: ,3 )(220) 2 lb lb = 57:521 ft2 ft 2 (2) so the required power for steady level flight C W 3300 L = qs = = 0:18507 (3) 57:521(310) 1

2 P r = qsv, C Do + KCL 2 57:521(310)220, = 0: :06(0:18507) hp = 157:31 hp (4) 550 Since lift equals weight for steady climbing flight, excess power P x is related to rate of climb RC = V sinγ by P x P a, P r = WV sin γ = 3300(220)sin3 550 hp = 69:083 hp (5) Thus, the engine shaft horsepower required for climbing flight at the given flight path angle γ and air speed V is P s = P r + P x η = 157: :083 0:8 which is 13.2% higher than the maximum available shaft horsepower. hp = 282:99 hp (6) 1.2 Air speed at γ = 3, P max For a maximum available power P a;max = ηp s;max = 0:8(250) hp = 200 hp (7) the possible air speeds for climbing flight are given by finding the roots of f (V )=P a;max, qsv, C Do + KC 2 L,WV sinγ (8) The two roots of this, as shown in Fig 1, are at V, = 16:588 ft=s V+ = 207:30 ft=s The following IDL code plots this function : pro hw4_1b ; Define initial conditions & conversions. hp = 550. ; 1 hp in ft-lb/s rho = e-3 ; sea level density, slug/ftˆ3 W = ; aircraft weight, lb S = 310. ; wing area, ftˆ2 P_s = 250. ; engine shaft power, hp eta = 0.80 ; propeller efficiency C_do = 0.02 ; parasitic drag coefficient K_d = 0.06 ; induced drag factor, 1/pi e AR 2

3 excess power (hp) airspeed V (ft/s) Figure 1: Function f (V ) (Eqn. 8) for sea level flight. gamma = 3*!dtor ; flight path angle, radians ; Create air speed array. v_min = 10. ; minimum speed, ft/s v_max = 250. ; maximum speed, ft/s bins = fix(v_max - v_min) + 1 ; number of array bins V = v_min + findgen(bins) ; air speed, ft/s ; Create available power array. P_a = eta*p_s*(fltarr(bins) + 1.) ; available power, hp ; Calculate required power (for level flight) array. q = rho * Vˆ2/2 ; dynamic pressure, psf 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 ; Calculate climb power array. h_dot = V*sin(gamma) ; rate of climb, ft/s P_c = W*h_dot/hp ; climbing power, hp ; Calculate and plot excess power for root-finding. P_x = P_a - P_c - P_r 3

4 xtitle = airspeed V (ft/s) ; x-axis title ytitle = excess power (hp) ; y-axis title file = fig4_1b.eps ; file name scale = 1.0 xsz = 4.0*scale ysz = 3.0*scale set_plot, ps device,/encapsulated,/preview,filename=file device,/inches, xsize=xsz, ysize=ysz, $ font_size = 7 plot, V, P_x, xtitle=xtitle, $ ytitle=ytitle, font=0 device, /close ; set plot device to PostScript ; open file & save EPS ; close the file end 1.3 Minimum power for 10 ft/s climb rate For a given climb rate RC, the required shaft horsepower P s = 1 η, WRC+ qsv CDo + KC 2 L (9) is minimized when P s = V = 0; since RC is a constant, this becomes From the definition of dynamic pressure, KqSV C2, L V + CDo + KCL 2 S q +V q V = 0 (10) q V = ρv = 2q V (11) while the lift coefficient for steady flight (Eqn. 3) gives CL 2 V = 2C L V W =,2 WC L q qs q 2 S V =,4C2 L V (12) Substituting into Eqn. 10 and rearranging,, 3CDo, KCL 2 qs = 0 (13) The only nontrivial solution is 3C Do = KC 2 L (14) 4

5 which, of course, is the condition for minimum required power (Anderson, 3rd ed,, Eqn. 6.29). This leads to the dynamic pressure q W r s K :06 = = S 3C Do 310 3(0:02) lb lb = 10:645 ft2 ft 2 (15) while the air speed V = s r 2q 2(10:645) ft ft ρ = 2: ,3 = 94:642 s s (16) Substituting this into Eqn. 9 gives the minimum shaft horsepower for a 10 ft/s climb rate. P s = WRC, qsv[4c Do] η = 3300(10)+10:645(310)94:642 [4(0:02)] hp = 131:78 hp (17) 0:8(550) 1.4 Maximum climb rate The air speed for maximum climb rate (Eqn. 16) varies with density as p ρ o =ρ where ρ o is the density at sea level. However, the dynamic pressure (Eqn. 15) is not a function of altitude, nor is lift coefficient at maximum climb C L = W q S = r 3CDo K (18) The required power for maximum climb rate P r;max = V q S(C Do + KC L 2 ) (19) thus also varies with density as p ρ o =ρ, while the available power varies as (Anderson, 3rd ed., Example 6.4) The maximum rate of climb P a;max = ηp s ρ ρ o (20) RC max = P a;max, P r;max W (21) and air speed for maximum climb rate V given in Fig. 2 are plotted by the following IDL code: pro hw4_1d ; Define fundamental constants & conversions. 5

6 airspeed V* (ft/s) altitude (1000 ft) rate of climb (ft/s) altitude (1000 ft) Figure 2: Maximum climb rate and airspeed at max climb as a function of altitude. R = ; air gas constant, ft-lb/slug-r T_z = ; 0 F in R g_o = ; std gee, ft/sˆ2 km2ft = ; 1 km in ft m2ft = km2ft/1000. ; 1 m in ft K2R = 1.8 ; 1 K in R hp = 550. ; 1 hp in ft-lb/s ; Define initial conditions. p_sl = ; sea level pressure, lb/ftˆ2 T_sl = ; sea level temperature, R rho_sl = e-3 ; sea level density, slug/ftˆ3 a_1 = -6.5e-3*K2R/m2ft ; lapse rate, R/ft 6

7 W = ; aircraft weight, lb S = 310. ; wing area, ftˆ2 P_s = 250. ; engine shaft power, hp eta = 0.80 ; propeller efficiency C_do = 0.02 ; parasitic drag coefficient K_d = 0.06 ; induced drag factor, 1/pi e AR ; Create gradient layer altitude, temperature, pressure & density arrays. bins = 201. ; number of array bins h = 100.*findgen(bins) ; altitude in gradient layer, ft T = T_sl + a_1*h ; temperature in gradient layer, R p = p_sl*(t/t_sl)ˆ(-g_o/(a_1*r)) ; pressure in gradient layer, lb/ftˆ2 rho = p/(r*t) ; density in gradient layer, slug/ftˆ3 ; Calculate and plot air speed for max steady climb as function of altitude. r1 = sqrt(k_d/(3.*c_do)) r2 = (2.*W)/(rho*S) V_star = sqrt(r1*r2) q_star = W*r1/S ; max climb air speed, ft/s ; max climb dynamic pressure, lb/ftˆ2 xtitle = altitude (1000 ft) ; x-axis title ytitle = airspeed V* (ft/s) ; y-axis title file = fig4_1dv.eps ; file name scale = 1.0 xsz = 4.0*scale ysz = 3.0*scale set_plot, ps device,/encapsulated,/preview,filename=file device,/inches, xsize=xsz, ysize=ysz, $ font_size = 7 plot, h/1000., V_star, xtitle=xtitle, $ ytitle=ytitle, font=0, $ xstyle = 1, ystyle = 1 device, /close ; set plot device to PostScript ; open file & save EPS ; close the file ; Calculate and plot required power, available power and maximum rate of climb. C_L = W/(q_star*S) ; max climb lift coefficient C_D = C_do + K_d*C_Lˆ2 ; max climb drag coefficient P_r = q_star*s*c_d*v_star ; required power, ft-lb/s P_a = (rho/rho_sl)*eta*p_s*hp ; available power, ft-lb/s RC = (P_a - P_r)/W ; max rate of climb, ft/s ytitle = rate of climb (ft/s) ; y-axis title file = fig4_1dr.eps ; file name scale = 1.0 xsz = 4.0*scale ysz = 3.0*scale set_plot, ps device,/encapsulated,/preview,filename=file device,/inches, xsize=xsz, ysize=ysz, $ ; set plot device to PostScript ; open file & save EPS 7

8 font_size = 7 plot, h/1000., RC, xtitle=xtitle, $ ytitle=ytitle, font=0, $ xstyle = 1, ystyle = 1 device, /close ; close the file end 1.5 Minimum glide angle The glide angle γ g =,γ is given (for small γ)by Setting the derivative with respect to air speed to zero, γ g = D W = qs W (C Do + KC 2 L ) (22) γ g S V = W C q qs D V + W KS we can substitute derivatives from Eqn. 11 and Eqn. 12, yielding 2C L C L V = 0 (23) The only nontrivial solution is at γ g 2qS, V = CD, 2KC 2 WV L = 0 (24) C D = C Do + KC 2 L = 2KC2 L (25) Thus, the lift coefficient for minimum glide angle is C L = r CDo K (26) Since Eqn. 3 holds for all steady linear flight (level, climbing or descending), the dynamic pressure so the minimum glide angle q = W S r K C Do (27) γ g;min = W S r K C Do S W (2C Do) =2 p KC Do = 6:928 10,2 rad = 3:970 (28) Note that neither the dynamic pressure nor the minimum glide angle are functions of density (and thus altitude). However, the velocity at minimum glide angle 8

9 V = s s r 2q 2W K ρ = (29) ρs C Do does vary with density, as shown in Fig airspeed V (ft/s) altitude (1000 ft) rate of climb (ft/s) altitude (1000 ft) Figure 3: Minimum glide angle and airspeed at minimum glide angle as a function of altitude. pro hw4_1d ; Define fundamental constants & conversions. R = ; air gas constant, ft-lb/slug-r T_z = ; 0 F in R g_o = ; std gee, ft/sˆ2 km2ft = ; 1 km in ft 9

10 m2ft = km2ft/1000. ; 1 m in ft K2R = 1.8 ; 1 K in R hp = 550. ; 1 hp in ft-lb/s ; Define initial conditions. p_sl = ; sea level pressure, lb/ftˆ2 T_sl = ; sea level temperature, R rho_sl = e-3 ; sea level density, slug/ftˆ3 a_1 = -6.5e-3*K2R/m2ft ; lapse rate, R/ft W = ; aircraft weight, lb S = 310. ; wing area, ftˆ2 P_s = 250. ; engine shaft power, hp eta = 0.80 ; propeller efficiency C_do = 0.02 ; parasitic drag coefficient K_d = 0.06 ; induced drag factor, 1/pi e AR ; Create gradient layer altitude, temperature, pressure & density arrays. bins = 201. ; number of array bins h = 100.*findgen(bins) ; altitude in gradient layer, ft T = T_sl + a_1*h ; temperature in gradient layer, R p = p_sl*(t/t_sl)ˆ(-g_o/(a_1*r)) ; pressure in gradient layer, lb/ftˆ2 rho = p/(r*t) ; density in gradient layer, slug/ftˆ3 ; Calculate and plot air speed for max steady climb as function of altitude. r1 = sqrt(k_d/(3.*c_do)) r2 = (2.*W)/(rho*S) V_star = sqrt(r1*r2) q_star = W*r1/S ; max climb air speed, ft/s ; max climb dynamic pressure, lb/ftˆ2 xtitle = altitude (1000 ft) ; x-axis title ytitle = airspeed V* (ft/s) ; y-axis title file = fig4_1dv.eps ; file name scale = 1.0 xsz = 4.0*scale ysz = 3.0*scale set_plot, ps device,/encapsulated,/preview,filename=file device,/inches, xsize=xsz, ysize=ysz, $ font_size = 7 plot, h/1000., V_star, xtitle=xtitle, $ ytitle=ytitle, font=0, $ xstyle = 1, ystyle = 1 device, /close ; set plot device to PostScript ; open file & save EPS ; close the file ; Calculate and plot required power, available power and maximum rate of climb. C_L = W/(q_star*S) ; max climb lift coefficient C_D = C_do + K_d*C_Lˆ2 ; max climb drag coefficient P_r = q_star*s*c_d*v_star ; required power, ft-lb/s P_a = (rho/rho_sl)*eta*p_s*hp ; available power, ft-lb/s RC = (P_a - P_r)/W ; max rate of climb, ft/s 10

11 ytitle = rate of climb (ft/s) ; y-axis title file = fig4_1dr.eps ; file name scale = 1.0 xsz = 4.0*scale ysz = 3.0*scale set_plot, ps device,/encapsulated,/preview,filename=file device,/inches, xsize=xsz, ysize=ysz, $ font_size = 7 plot, h/1000., RC, xtitle=xtitle, $ ytitle=ytitle, font=0, $ xstyle = 1, ystyle = 1 device, /close ; set plot device to PostScript ; open file & save EPS ; close the file end 2 Problem2 Consider a jet aircraft with weight of 16,850 lbs and wing surface area of 350. sq ft. It is powered by a turbojet engine that provides a maximum thrust of 6300 lbs at sea level. The aerodynamic properties of the aircraft are described by a parabolic drag polar with C D = 0:02 + 0:04C 2 L (30) 1. What is the thrust level required to maintain a steady flight path angle of 3 degrees and an air speed of 350 ft/s at sea level? 2. Assume the aircraft flies with a flight path angle of 3 degrees using maximum available thrust. Determine all possible air speeds which maintain steady flight at sea level. 3. What is the minimum thrust level required to maintain steady climb rate of 10 ft/s at sea level? What is the corresponding air speed? 4. Plot the maximum flight path angle of the aircraft as it depends on altitude from sea level to 20,000 ft. Plot the corresponding air speed as it depends on altitude. 5. Plot the minimum glide angle which maintains steady gliding flight as it depends on altitude from sea level to 20,000 ft. Plot the corresponding air speed as it depends on altitude. 2.1 Required thrust at γ = 3, 350 ft/s At sea level, the air speed V = 350 ft/s has a dynamic pressure so Eqn. 3 gives the lift coefficient q 1 = 2 ρv 2 1 = 2 (2: ,3 )(350) 2 lb lb = 145:59 ft2 ft 2 (31) 11

12 C W 16;850 L = qs = = 0:33067 (32) 145:59(350) Continuing the chain of calculations, the drag coefficient C D = C Do + KC 2 L = 0:02 + 0:04(0:33067)2 = 2: ,2 (33) so the thrust required for steady climb at γ = 3 and 350 ft/s is T = W sinγ + qsc D = 16;850sin :59(350)2: ,2 lb = 2123:9 lb (34) 2.2 Air speed at γ = 3, T max For steady climbing flight at a flight path angle of 3 and maximum thrust, the thrust balance is W sinγ = T, qsc D (35) Susbtituting the drag polar and the lift coefficient for steady flight and rearranging, W 2 qsc do + qsk = T,W sinγ (36) qs Multiplying by qs and rearranging, we have the quadratic equation C Do (qs) 2 +(W sinγ, T )(qs)+kw 2 = 0 (37) which can be solved by the quadratic formula, Inserting numerical values gives two roots, qs = T,W sinγ p (W sinγ, T ) 2, 4C Do KW 2 2C Do (38) (qs), = 2112:5lb (qs)+ = 268;790 lb Substituting this into Eqn. 2 yields the air speeds for 6,300 lb thrust and γ = 3 V, = 71:265 ft=s V+ = 803:870 ft=s 12

13 2.3 Minimum thrust for 10 ft/s climb rate At a constant rate of climb RC = V sinγ = 10 ft/s, the thrust balance sinγ RC T = V =, D W (39) Rearranging and expanding the drag term, T = RC V W + qsc Do + KC 2 L (40) which has minima at Substituting in the derivatives of Eqn. 11 and Eqn. 12, T V =,RC W V V + SC q D V + 2qSKC C L L V = 0 (41) T V =,RC W 2qS V V + V C 4qS D, V KC2 L = 0 (42) If there s an analytical solution to this, it s beyond my patience. Instead, the following IDL code returns Fig. 4, a plot of T = V as a function of air speed, while a secant-method root-finding routine finds that the minimum thrust for a 10 ft/s climb rate is at V = 279:63 ft=s T = 1602:3lb pro hw4_2c ; Define initial conditions & conversions. rho = e-3 ; sea level density, slug/ftˆ3 W = ; aircraft weight, lb S = 350. ; wing area, ftˆ2 T_a = ; engine shaft power, hp C_do = 0.02 ; parasitic drag coefficient K = 0.04 ; induced drag factor, 1/pi e AR RC = 10. ; rate of climb, ft/s ; Create air speed array. v_min = 200. ; minimum speed, ft/s v_max = 300. ; maximum speed, ft/s bins = fix(v_max - v_min) + 1 ; number of array bins V = v_min + findgen(bins) ; air speed, ft/s ; Calculate slope of thrust curve, dt/dv. 13

14 2 0 dt/dv (lb-s/ft) airspeed V (ft/s) Figure 4: Thrust slope T = V for steady 10 ft/s climb rate. q = rho * Vˆ2/2 ; dynamic pressure, psf C_l = W/(q*S) ; lift coefficient C_d = C_do + K*C_lˆ2 ; drag coefficient dtdv = - RC*W/Vˆ2 + (2*q*S*C_d)/V - (4*q*S*K*C_lˆ2)/V ; Plot dt/dv as a function of airspeed for minimum-finding. xtitle = airspeed V (ft/s) ; x-axis title ytitle = dt/dv (lb-s/ft) ; y-axis title file = fig4_2c.eps ; file name scale = 1.0 xsz = 4.0*scale ysz = 3.0*scale set_plot, ps device,/encapsulated,/preview,filename=file device,/inches, xsize=xsz, ysize=ysz, $ font_size = 7 plot, V, dtdv, xtitle=xtitle, $ ytitle=ytitle, font=0 device, /close ; set plot device to PostScript ; open file & save EPS ; close the file end 14

15 2.4 Maximum flight path angle The thrust at altitude is a function of density (Anderson, 3rd ed., Example 6.4), The flight path angle at a given thrust T is (for small γ) T = T o ρ ρ o (43) γ = T, qsc D W (44) At a constant altitude (so T is constant), this has a maximum at Substituting in the derivatives of Eqn. 11 and Eqn. 12, γ S q V =, C D W V + qk C2 L = 0 (45) V The only nontrivial solution is at γ 2qS V =, WV (C D, 2KCL 2 ) (46) C D = C Do + KC 2 L = 2KC2 L (47) Solving for lift coefficient at maximum climb, C L = W q S = r CDo K (48) so the dynamic pressure q = W S r K C Do (49) and the airspeed at maximum flight path angle V = s r 2W K (50) ρs C Do This airspeed and the resulting thrust are plotted in Fig. 5 as a function of altitude by the following IDL code: pro hw4_2d ; Define fundamental constants & conversions. 15

16 airspeed V* (ft/s) altitude (1000 ft) flight path angle (deg) altitude (1000 ft) Figure 5: Maximum flight path angle and airspeed at max climb angle as a function of altitude. R = ; air gas constant, ft-lb/slug-r T_z = ; 0 F in R g_o = ; std gee, ft/sˆ2 km2ft = ; 1 km in ft m2ft = km2ft/1000. ; 1 m in ft K2R = 1.8 ; 1 K in R hp = 550. ; 1 hp in ft-lb/s ; Define initial conditions. p_sl = ; sea level pressure, lb/ftˆ2 T_sl = ; sea level temperature, R rho_sl = e-3 ; sea level density, slug/ftˆ3 a_1 = -6.5e-3*K2R/m2ft ; lapse rate, R/ft W = ; aircraft weight, lb 16

17 S = 350. ; wing area, ftˆ2 T_o = ; engine thrust, lb C_do = 0.02 ; parasitic drag coefficient K = 0.04 ; induced drag factor, 1/pi e AR ; Create gradient layer altitude, temperature, pressure & density arrays. bins = 201. ; number of array bins h = 100.*findgen(bins) ; altitude in gradient layer, ft T = T_sl + a_1*h ; temperature in gradient layer, R p = p_sl*(t/t_sl)ˆ(-g_o/(a_1*r)) ; pressure in gradient layer, lb/ftˆ2 rho = p/(r*t) ; density in gradient layer, slug/ftˆ3 ; Calculate and plot air speed for max steady climb as function of altitude. r1 = sqrt(k/c_do) r2 = (2.*W)/(rho*S) V_star = sqrt(r1*r2) q_star = W*r1/S ; max climb air speed, ft/s ; max climb dynamic pressure, lb/ftˆ2 xtitle = altitude (1000 ft) ; x-axis title ytitle = airspeed V* (ft/s) ; y-axis title file = fig4_2dv.eps ; file name scale = 1.0 xsz = 4.0*scale ysz = 3.0*scale set_plot, ps device,/encapsulated,/preview,filename=file device,/inches, xsize=xsz, ysize=ysz, $ font_size = 7 plot, h/1000., V_star, xtitle=xtitle, $ ytitle=ytitle, font=0, $ xstyle = 1, ystyle = 1 device, /close ; set plot device to PostScript ; open file & save EPS ; close the file ; Calculate and plot required power, available power and maximum rate of climb. C_L = W/(q_star*S) ; max climb lift coefficient C_D = C_do + K*C_Lˆ2 ; max climb drag coefficient D = q_star*s*c_d ; drag, lb T_a = (rho/rho_sl)*t_o ; available thrust, lb gamma = (T_a - D)*!radeg/W ; max flight path angle, deg ytitle = flight path angle (deg) ; y-axis title file = fig4_2dr.eps ; file name scale = 1.0 xsz = 4.0*scale ysz = 3.0*scale set_plot, ps device,/encapsulated,/preview,filename=file device,/inches, xsize=xsz, ysize=ysz, $ font_size = 7 plot, h/1000., gamma, xtitle=xtitle, $ ; set plot device to PostScript ; open file & save EPS 17

18 ytitle=ytitle, font=0, $ xstyle = 1, ystyle = 1 device, /close ; close the file end 2.5 Minimum glide angle As shown in section 1.5, the minimum glide angle γ g;min = 2 p KC Do = 2 p 0:04(0:02)=5:657 10,2 rad = 3:241 (51) is not a function of altitude, but the velocity at minimum glide angle V g;min = s r 2W K (52) ρs C Do is, as shown in Fig. 6. pro hw4_2e ; Define fundamental constants & conversions. R = ; air gas constant, ft-lb/slug-r T_z = ; 0 F in R g_o = ; std gee, ft/sˆ2 km2ft = ; 1 km in ft m2ft = km2ft/1000. ; 1 m in ft K2R = 1.8 ; 1 K in R ; Define initial conditions. p_sl = ; sea level pressure, lb/ftˆ2 T_sl = ; sea level temperature, R rho_sl = e-3 ; sea level density, slug/ftˆ3 a_1 = -6.5e-3*K2R/m2ft ; lapse rate, R/ft W = ; aircraft weight, lb S = 350. ; wing area, ftˆ2 C_do = 0.02 ; parasitic drag coefficient K_D = 0.04 ; induced drag factor, 1/pi e AR ; Create gradient layer altitude, temperature, pressure & density arrays. bins = 201. ; number of array bins h = 100.*findgen(bins) ; altitude in gradient layer, ft T = T_sl + a_1*h ; temperature in gradient layer, R p = p_sl*(t/t_sl)ˆ(-g_o/(a_1*r)) ; pressure in gradient layer, lb/ftˆ2 rho = p/(r*t) ; density in gradient layer, slug/ftˆ3 18

19 airspeed V (ft/s) altitude (1000 ft) rate of climb (ft/s) altitude (1000 ft) Figure 6: Minimum glide angle and airspeed at minimum glide angle as a function of altitude. ; Calculate and plot air speed for min steady glide as function of altitude. q = (W/S)*sqrt(K_D/C_do) ; min glide dynamic pressure, lb/ftˆ2 V = sqrt(2*q/rho) ; min glide air speed, ft/s xtitle = altitude (1000 ft) ; x-axis title ytitle = airspeed V (ft/s) ; y-axis title file = fig4_2ev.eps ; file name scale = 1.0 xsz = 4.0*scale ysz = 3.0*scale set_plot, ps device,/encapsulated,/preview,filename=file device,/inches, xsize=xsz, ysize=ysz, $ ; set plot device to PostScript ; open file & save EPS 19

20 font_size = 7 plot, h/1000., V, xtitle=xtitle, $ ytitle=ytitle, font=0, $ xstyle = 1, ystyle = 1 device, /close ; close the file ; Calculate and plot minimum glide angle as a function of altitude. gamma = 2*sqrt(K_D*C_do)*(fltarr(bins) + 1) ; glide angle, radians gamma = gamma*!radeg ; convert to degrees ytitle = rate of climb (ft/s) ; y-axis title file = fig4_2eg.eps ; file name scale = 1.0 xsz = 4.0*scale ysz = 3.0*scale set_plot, ps device,/encapsulated,/preview,filename=file device,/inches, xsize=xsz, ysize=ysz, $ font_size = 7 plot, h/1000., gamma, xtitle=xtitle, $ ytitle=ytitle, font=0, $ yrange=[3.,5.] device, /close ; set plot device to PostScript ; open file & save EPS ; force plot y-range ; close the file end 3 Problem3 Assume that an aircraft has weight of 5600 lbs and wing surface area of 330 ft 2. The maximum power available from the engine propulsion system at sea level is P a = 122:2 + 0:556V (53) and the power required for steady level flight at sea level is P r = 357:8, 2:4375V + 0:005625V 2 (54) where power is in hp and air speed is in ft/s. The lift coefficient is C L = 0:16α, 0:00533α 2 (55) where angle of attack is in degrees, this expression is valid up to stall. 1. Develop graphical plots for the maximum power and thrust available from the engine propulsion system and the power and thrust required for steady level flight as functions of the air speed. Assume sea level flight. 20

21 2. If the aircraft engine delivers 300 hp, what is the air speed (or air speeds) for which the aircraft maintains steady level flight at sea level? What is the required throttle as a percentage of the maximum power available? What is the required angle of attack? 3. If the aircraft engine delivers a constant 500 lbs of thrust, what is the air speed (or speeds) of the aircraft for which steady level flight is maintained at sea level? What is the required throttle as a percentage of the maximum thrust available? What is the required angle of attack? 4. What are the power level and the thrust level required to maintain steady level flight at an air speed of 350 ft/s at sea level? What is the required throttle as a percentage of the maximum power available? What is the required value of the angle of attack? 5. What is the maximum air speed that can be maintained in steady level flight at sea level? What are the required power level and the required thrust level? What is the required throttle as a percentage of the maximum power available? What is the required value of the angle of attack? 6. What is the minimum air speed that can be maintained in steady level flight at sea level? What are the required power level and the required thrust level? What is the required throttle as a percentage of the maximum power available? What is the required value of the angle of attack? 7. What is the maximum rate of climb that can be maintained at sea level? What are the required power level and the required thrust level? What are the required values of the air speed and of the angle of attack? 8. What is the minimal glide slope angle at sea level, assuming the engine is turned off? What are the required values of the air speed and of the angle of attack? 3.1 Available and required thrust and power This simply entails plotting Eqns. 53 and 54 for power, and using the relation T = P=V (56) for thrust. Figure 7 shows these plots. 3.2 Air speed at P d = 300 hp Given a delivered power of P d = 300 hp, possible air speeds lie at the roots of 300 hp =(357:8, 2:4375u + 0:005625u 2 ) hp (57) where u V =V o is a nondimensionalization (V o = 1 ft/s) that lets us avoid confusion about what our dimensions are. Rearranging and solving by the quadratic formula gives V, = 25:179 ft=s V+ = 408:16 ft=s which is confirmed by a glance at Fig. 7. The available power at these airspeeds is 21

22 force (lb) airspeed (ft/s) power (hp) airspeed (ft/s) Figure 7: Required (solid line) and available (dotted line) thrust and power as a function of air speed. 22

23 P, = (122:2 + 0:556[25:179]) hp = 136:20 hp P+ = (122:2 + 0:556[408:16]) hp = 349:14 hp so the percent power throttle θ p; = P d =P is θ p;, = θ p;+ = :20 = 220:27% :14 = 85:925% To find the possible angle of attack, start by calculating dynamic pressures and lift coefficients q 1, = 2 (2: ,3 )(25:179) 2 lb lb = 0:75346 ft2 ft 2 q+ 1 = 2 (2: ,3 )(408:16) 2 lb lb = 197:99 ft2 ft 2 C L, = C L+ = :75346(330) = 22: = 8: ,2 197:99(330) corresponding to the possible airspeeds. Substituting into Eqn. 55, 0 = 0:00533α 2,, 0:16α, + 22:522 0 = 0:00533α 2 +, 0:16α + + 8: ,2 The low-speed solution has no real roots, α, =(15:009 63:248 p,1) (58) while the high-speed solution has two real roots, α+;1 = 0:5450 α+;2 = 29:473 The lower angle of attack at high speed is more physically reasonable, as is well beyond stall for most airfoils, but both are allowed by the given mathematical model. 23

24 3.3 Air speed at T d = 500 lb Given a delivered thrust of T d = 500 hp, possible air speeds lie at the roots of 500V lb =(357:8, 2:4375u + 0:005625u 2 ) hp (59) Dividing through by V o = 1 ft/s and using the identity 550 ft-lb/s = 1 hp, this becomes 500u = 196;790, 1340:6u + 3:0938u 2 (60) Solving this by the quadratic formula gives the air speeds V, = 139:75 ft=s V+ = 455:21 ft=s which is confirmed by a glance at Fig. 7. The available thrust at these airspeeds is T, = T+ = 550 (122:2 + 0:556[139:75]) lb = 786:73 hp 139: (122:2 + 0:556[455:21]) hp = 453:45 hp 455:21 so the percent thrust throttle θ t ; = T d =T is θ t ;, = θ t ;+ = :73 = 63:554% :45 = 110:27% To find the possible angle of attack, start by calculating dynamic pressures and lift coefficients q 1, = 2 (2: ,3 )(139:75) 2 lb lb = 23:211 ft2 ft 2 q+ 1 = 2 (2: ,3 )(455:21) 2 lb lb = 246:27 ft2 ft 2 C L, = C L+ = :211(330) = 0: = 6: ,2 246:27(330) 24

25 corresponding to the possible airspeeds. Substituting into Eqn. 55, 0 = 0:00533α 2,, 0:16α, + 0: = 0:00533α 2 +, 0:16α + + 6: ,2 The low-speed solution has roots at α,;1 = 5:6220 α,;2 = 24:396 while the high-speed solution has two real roots, α+;1 = 0:43700 α+;2 = 29:581 Again, all 4 angles are allowed by the given mathematical model. 3.4 Power and thrust for 350 ft/s air speed At V = 350 ft/s, the required power P r =(357:8, 2:4375[350]+0:005625[350] 2 ) hp = 193:75 hp (61) so the required thrust The available power at V = 350 ft/s is T r = P r V = 193:75 hp 350 ft=s 550 ft lb = 304:44 lb (62) 1hps P a =(122:2 + 0:556[350]) hp = 316:80 hp (63) so the percent power throttle is A glance at Fig. 7 confirms these values. The dynamic pressure is θ 193:75 p = = 61:155% (64) 316:80 q 1 = 2 (2: ,3 )(350) 2 lb lb = 145:59 ft2 ft 2 (65) 25

26 while substituting the lift coefficient C L = 5600 = 0:11656 (66) 145:59(330) into Eqn. 55, 0 = 0:00533α 2,, 0:16α, + 0:11656 (67) gives the roots α 1 = 0:74666 α 2 = 29: Maximum air speed In steady level flight, maximum air speed is at the upper root of P a = P r : Combining terms and cancelling units, (122:2 + 0:556u) hp =(357:8, 2:4375u + 0:005625u 2 ) hp (68) 0 = 235:6, 2:9935u + 5:625 10,3 u 2 (69) which has roots at V, = 96:04 ft=s V+ = 436:14 ft=s Thus, the maximum air speed V max = 436:14 ft/s; the required throttle is, of course, 100%. The dynamic pressure at V max is so the lift coefficient q 1 = 2 (2: ,3 )(436:14) 2 lb lb = 226:06 ft2 ft 2 (70) Substituting this into Eqn. 55, C L = :06(330) = 7: ,2 (71) 0 = 0:00533α 2,, 0:16α, + 7: ,2 (72) 26

27 gives the roots α 1 = 0:47630 α 2 = 29: Minimum air speed As shown in the last section, V, = 96:04 ft/s is the lower intersection of the required and available power curves. The dynamic pressure at V min is so the lift coefficient q 1 = 2 (2: ,3 )(96:04) 2 lb lb = 10:962 ft2 ft 2 (73) C L = 5600 = 1:5481 (74) 10:962(330) Substituting this into Eqn. 55, 0 = 0:00533α 2,, 0:16α, + 1: ,2 (75) gives the complex roots α =(15:009 8:0727 p,1) (76) In physical terms, this means that the airfoil is incapable of reaching a high enough C L to support steady level flight at V,.TheC L curve has a maximum at C L α so α max = 15:009, giving a maximum lift coefficient = 0:16, 0:01066α = 0 (77) C Lmax = 0:16(15:009), 0:00533(15:009) 2 = 1:2007 (78) The minimum dynamic pressure q max = W SC Lmax = :2007(330) lb lb = 14:133 ft2 ft 2 (79) yielding a minimum air speed V min = s r 2q 2(14:133) ft ft ρ = 2: ,3 = 109:05 s s (80) 27

28 The required power at minimum air speed P r =(357:8, 2:4375[109:05]+0:005625[109:05] 2 ) hp = 158:88 hp (81) while the available power P a =(122:2 + 0:556[109:05]) hp = 182:83 hp (82) so the percent power throttle is θ 158:88 p = = 86:900% (83) 182: Maximum rate of climb The rate of climb has a maximum where Rearranging and taking the two derivatives, the nontrivial solution is at RC 1 Pa P V = W V, r = 0 (84) V 0:566 =,2: :125 10,2 V (85) Solving, the air speed at maximum rate of climb V RCmax = 0: :4375 ft ft 1:125 10,2 = 266:98 s s (86) The available power P a =(122:2 + 0:556[266:98]) hp = 270:64 hp (87) while the required power P r =(357:8, 2:4375[266:98]+0:005625[266:98] 2 ) hp = 107:98 hp (88) so that the maximum rate of climb RC max = (270:15, 107:98) hp 5600 lb 550 ft lb = 15:928 ft 1hps s (89) To find angle of attack, start with dynamic pressure 28

29 q 1 = 2 (2: ,3 )(266:98) 2 lb lb = 84:711 ft2 ft 2 (90) so the lift coefficient C L = 5600 = 0:20033 (91) 84:711(330) Substituting this into Eqn. 55, 0 = 0:00533α 2,, 0:16α, + 0:20167 (92) gives the possible angles of attack at maximum climb α 1 = 1:3087 α 2 = 28: Minimum glide angle The glide slope (for small γ g ) γ g = D W = P r WV (93) has a minimum when γ g V = 1 W V Pr 1 1 P r P = V W V V, r V 2 = 0 (94) The only nontrivial solution is at 1 P r P V V = r V 2 (95) Multiplying by V 2 and expanding P r, V (,2: : ,2 V )=(357:8, 2:4375V + 5:625 10,3 V 2 ) (96) Regrouping, we have 0 = 357:8, 5:625 10,3 V 2 (97) so the air speed at minimum glide angle 29

30 V min = r 357:8 ft ft 5:625 10,3 = 252:21 s s (98) The required power at minimum glide angle is P r =(357:8, 2:4375[252:21]+0:005625[252:21] 2 ) hp = 100:84 hp (99) so that minimum drag is D min = P r V min = 100:84 hp s 252:21 ft 550 ft lb = 219:90 lb (100) 1hps and minimum glide slope γ gmin = As ever, to find angle of attack we start with dynamic pressure 219:90 lb 5600 lb = 3: ,2 rad = 2:2499 (101) so the lift coefficient q 1 = 2 (2: ,3 )(252:21) 2 lb lb = 75:597 ft2 ft 2 (102) C L = 5600 = 0:22448 (103) 75:597(330) Substituting this into Eqn. 55, 0 = 0:00533α 2,, 0:16α, + 0:22448 (104) gives the possible angles of attack at maximum climb α 1 = 1:4751 α 2 = 28:534 and we re done. (Whew!) 30