AE 245 homework #1 solutions

AE 245 homework #1 solutions Tim Smith 24 January 2000 1 Problem1 An aircraft is in steady level flight in a standard atmosphere. 1. Plot the total or stagnation pressure at the nose of the aircraft as a function of the air speed of the aircraft, assuming flight at sea level. Consider airspeeds in the range from 50 mph to 500 mph. 2. Plot the total or stagnation pressure at the nose of the aircraft as a function of the air speed of the aircraft, assuming flight at an altitude of 20,000 ft. Consider airspeeds in the range from 50 mph to 500 mph. 1.1 Sea level flight For subsonic compressible flow, total pressure (Anderson, 3rd ed., Eqn. 4.74) is P o = P 1+ γ, 1 γ γ,1 2 M2 (1) where the Mach number M V =a, sound speed a = p γrt, specific heat ratio γ = 1:4 and the gas constant for air R = 1716 ft-lb/slug- o R. At sea level, standard temperature and pressure are (ibid, p. 77) T = 581:69 o R P = 2116:2lb=ft 2 Though these initial conditions and Eqn. 1 can be readily incorporated into a spreadsheet, the following IDL code lays out the steps leading to Fig. 1 in a more explicit fashion: pro hw1_1a ; Define fundamental constants & conversions. R_air = 1716. ; air gas constant, ft-lb/slug-r gamma = 1.4 ; specific heat ratio c_p/c_v mph2fps = 88./60. ; conversion, 60 mph = 88 ft/s 1

19 total pressure (psi) 18 17 16 15 100 200 300 400 500 air speed (mph) Figure 1: Total pressure as a function of air speed at sea level. ; Define initial conditions. ; Source: Anderson, 3rd ed., p. 77 p = 2116.2 ; static pressure, lb/ftˆ2 T = 518.69 ; static temperature, R a = sqrt(gamma*r_air*t) ; speed of sound, ft/s ; Create speed & Mach number arrays. bins = 101. ; number of array bins u_min = 50.*mph2fps ; minimum speed, ft/s u_max = 500.*mph2fps ; maximum speed, ft/s range = u_max - u_min ; range of speeds, ft/s u = u_min + range*findgen(bins)/(bins-1) ; speed array, ft/s M = u/a ; Mach number array ; Calculate total pressure. alpha = (gamma - 1.)/2. beta = gamma/(gamma - 1.) p_o = p*(1 + alpha*mˆ2)ˆbeta ; total pressure array, lb/ftˆ2 ; Plot total pressure vs. speed. x = u/mph2fps ; air speed, mph y = p_o/144. ; total pressure, psi xtitle = air speed (mph) ; x-axis title 2

ytitle = total pressure (psi) ; y-axis title file = fig1_1a.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, x, y, xtitle=xtitle, $ ytitle=ytitle, font=0, $ xstyle=9, ystyle=9 device, /close set_plot, mac ; set plot device to PostScript ; open file & save EPS ; close the file ; return plot device to Mac end 1.2 Sea level flight 9.0 total pressure (psi) 8.5 8.0 7.5 7.0 100 200 300 400 500 air speed (mph) Figure 2: Total pressure as a function of air speed at 20,000 ft. At an altitude of 20,000 ft, Anderson (Appendix A) gives the new initial conditions T = 447:43 o R P = 973:27 lb=ft 2 3

which, plugged into Eqn. 1, give the plot in Fig. 2. 2 Problem2 At a particular sea level location at a particular time the atmospheric pressure is 14.8 psi and the temperature is 32 degrees F. Using the standard atmospheric model, develop graphs that show how the temperature, pressure, and air density change as a function of altitude, from sea level up to 50,000 ft. Since 50,000 ft. = 15.24 km, this includes the first gradient layer (0 < h < 11 km) and part of the first isothermal layer (11 km < h < 25 km). In the first gradient layer, the temperature is (ibid, Eqn. 3.14) T = T 1 + a 1 h (2) and the pressure is given by (ibid, Eqn. 3.12) P = P 1 T T 1, go a 1 R (3) where the given sea level temperature T 1 =32 o F = 459.97 o R, the given sea level pressure P = 14.8 psi = 2131.2 lb/ft 2, and the temperature gradient a 1 =,6:5 10,3 K/m. Once the temperature and pressure are known, density is given by the perfect gas law ρ = P RT (4) The temperature at h b = 11 km remains constant through the rest of the first isothermal layer, while the pressure for h > 11 km is given by h, h P = b P b exp,g 0 RT b (5) where P b is the pressure at h = 11 km. The following IDL code stacks these layers, giving the plots in Fig. 3. pro hw1_2 ; Define fundamental constants & conversions. ; Source: Van Wylen & Sonntag, _Fundamentals of Classical Thermodynamics_, 3rd ed. R_air = 1716. ; air gas constant, ft-lb/slug-r T_z = 459.67 ; 0 F in R g_o = 32.174 ; std gee, ft/sˆ2 km2ft = 3280.8 ; 1 km in ft m2ft = km2ft/1000. ; 1 m in ft K2R = 1.8 ; 1 K in R ; Define initial conditions. p_sl = 144*14.8 ; sea level pressure, lb/ftˆ2 4

T_sl = 32.0 + T_z ; sea level temperature, R r_sl = p_sl/(r_air*t_sl) ; sea level density, slug/ftˆ3 ; Define break points and lapse rate. h_b = 11.*km2ft ; height of 1st gradient layer, ft h_f = 50000. ; ultimate height, ft a_1 = -6.5e-3*K2R/m2ft ; lapse rate, R/ft ; Fix array sizes. bins = 101. ; number of array bins r_h = h_b/h_f ; height ratio sz_1 = fix(r_h*bins) + 1 ; first array size sz_2 = fix((1-r_h)*bins) + 1 ; second array size ; Create gradient layer altitude, temperature, pressure & density arrays. h_1 = h_b*findgen(sz_1)/(sz_1-1) ; altitude in gradient layer, ft T_1 = T_sl + a_1*h_1 ; temperature in gradient layer, R p_1 = p_sl*(t_1/t_sl)ˆ(-g_o/(a_1*r_air)) ; pressure in gradient layer, lb/ftˆ2 r_1 = p_1/(r_air*t_1) ; density in gradient layer, slug/ftˆ3 T_b = T_1(sz_1-1) ; temperature at break point p_b = p_1(sz_1-1) ; pressure at break point ; Create isothermal layer altitude, temperature, pressure & density arrays. h_2 = h_b + (h_f-h_b)*findgen(sz_2)/(sz_2-1); altitude in isothermal layer, ft T_2 = T_b*(fltarr(sz_1)+1.) ; isothermal temperature, R p_2 = p_b*exp(-g_o*(h_2-h_b)/(r_air*t_b)) ; pressure in isothermal layer, lb/ftˆ2 r_2 = p_2/(r_air*t_b) ; density in isothermal layer, slug/ftˆ3 ; Combine arrays. h = [h_1,h_2] ; combined altitude, ft T = [T_1,t_2] ; combined temperature, R p = [p_1,p_2] ; combined pressure, lb/ftˆ2 rho = [r_1,r_2] ; combined density, slug/ftˆ3 ; Plot temperature vs. altitude. x = T - T_z ; temperature, F y = h/1000. ; altitude, 1000 ft. xtitle = temperature (F) ; x-axis title ytitle = altitude (1000 ft) ; y-axis title file = fig1_2a.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 ; set plot device to PostScript ; open file & save EPS 5

plot, x, y, xtitle=xtitle, $ ytitle=ytitle, font=0, $ xstyle=9, ystyle=9 device, /close ; close the file ; Plot pressure vs. altitude. x = p/144. ; pressure, psi xtitle = pressure (psi) ; x-axis title file = fig1_2b.eps ; file name device,/encapsulated,/preview,filename=file device,/inches, xsize=xsz, ysize=ysz, $ font_size = 7 plot, x, y, xtitle=xtitle, $ ytitle=ytitle, font=0, $ xstyle=9, ystyle=9 device, /close ; open file & save EPS ; close the file ; Plot density vs. altitude. x = rho*1000. ; density, millislug/ftˆ3 xtitle = density (millislug/ftˆ3) ; x-axis title file = fig1_2c.eps ; file name device,/encapsulated,/preview,filename=file device,/inches, xsize=xsz, ysize=ysz, $ font_size = 7 plot, x, y, xtitle=xtitle, $ ytitle=ytitle, font=0, $ xstyle=9, ystyle=9 device, /close set_plot, mac ; open file & save EPS ; close the file ; return plot device to Mac end 6

50 40 altitude (1000 ft) 30 20 10 0-80 -60-40 -20 0 20 temperature ( F) 50 40 altitude (1000 ft) 30 20 10 0 2 4 6 8 10 12 14 pressure (psi) 50 40 altitude (1000 ft) 30 20 10 0 0.5 1.0 1.5 2.0 2.5 density (millislug/ft^3) Figure 3: Atmospheric model for sea-level T = 32 o F, P = 14:8psi. 7

3 Problem3 An F-16 supersonic fighter is in a rapid climb in a standard atmosphere and its time rate of change of altitude (its climb rate) is 5 m/sec. 1. At the instant it passes through an altitude of 10 km, what are the temperature, pressure and density of the surrounding air? 2. At the instant it passes through an altitude of 10 km, what are the time rate of change of temperature, pressure and density of the surrounding air? 3.1 Temperature, pressure and density From Anderson, Appendix A: T = 223:26 K P = 2:6500 10 4 N=m 2 ρ = 0:41351 kg=m 3 3.2 Time rates of change Since temperature is solely a function of altitude, the total temperature differential so that the temperature rise rate dt = T dh (6) h dt dt = T dh h dt (7) Given the lapse rate T = h = a 1 =,6:5 10,3 K/m within the gradient layer and the climb rate dh=dt = 5m/s,the temperature rise rate is dt dt =,6:5 10,3 K m 5 m =,3:250 10,2 K s s (8) Similary, pressure in the isothermal layer is solely a function of temperature, so the pressure rise rate dp dt = P dt T dt (9) First, let s try the analytical method. Taking the derivative of Eqn. 3, 8

P T =,g op 1 T, go ar art T 1 (10) which, substituting Eqn. 3 and Eqn. 4, becomes Referring back to Eqn. 8, the pressure rise rate is P T =,g oρ = 9:807, m=s2 (0:41351 kg=m 3 ) = a 1,6:5 10,3 629:3 K=m dp dt = 629:3 N K m 2,3:250 10,2 K =,20:28 s N K m 2 (11) N s m 2 (12) As a check, let s try a more computational approach. Since temperature is a function of altitude and pressure is a function of temperature, dp dt = P T dh T h dt = P h dh dt (13) We can approximate P= h with a central difference equation, P P(10;100 m), P(9;900 m) = h 10;100 m, 9;900 m 2:6098, 2:6906 10 4 N = 200 m 3,4:040 Substituting into Eqn. 13, the computational approach gives a pressure rise rate of N s m 2 (14) dp dt = P h which is within 0.40% of the earlier answer. dh dt =,4:040 To find the density rise rate, take the derivative of Eqn. 4: N 5 m =,20:20 s m 2 s N s m 2 (15) dρ dt = 1 dp P dt RT dt, RT 2 dt = ρ dp ρ dt P dt, T dt 1 dp = ρ 1 P dt, T dt dt (16) Since we ve already found the pressure and temperature rise rates, dρ dt = 0:41351 kg m 3,20:28,3:250 10,3 1 kg, =,2:563 10,4 2:650 104 223:26 s s m 3 (17) 4 Problem4 An A320 aircraft is cruising at an airspeed of 400 mph. Plot the Mach number of the aircraft as a function of its flight altitude. Use the standard atmosphere model. 9

As before, Mach number M V =a, whereu = 400 mph= 586:67 ft/s, sound speed a = p γrt, specific heat ratio γ = 1:4, the gas constant for air R = 1716 ft-lb/slug- o R and temperature varies as T(0km< h < 11 km) = T 1 + a 1 h T(11 km < h < 25 km) = T b where T 1 = 581:69 o Randa 1 =,6:5 10,3 K/m. The following IDL code (adapted from the code for Problem 2) gives the plot in Fig. 4. 0.60 0.58 Mach number 0.56 0.54 0 10 20 30 40 altitude (1000 ft) Figure 4: Mach number at 400 mph as a function of altitude. pro hw1_4 ; Define fundamental constants & conversions. ; Source: Van Wylen & Sonntag, _Fundamentals of Classical Thermodynamics_, 3rd ed. R_air = 1716. ; air gas constant, ft-lb/slug-r gamma = 1.4 ; specific heat ratio c_p/c_v T_z = 459.67 ; 0 F in R km2ft = 3280.8 ; 1 km in ft m2ft = km2ft/1000. ; 1 m in ft K2R = 1.8 ; 1 K in R mph2fps = 88./60. ; conversion, 60 mph = 88 ft/s ; Define initial conditions. p_sl = 2116.2 ; sea level pressure, lb/ftˆ2 T_sl = 518.69 ; sea level temperature, R u = 400.*mph2fps ; air speed, ft/s 10

; Define break points and lapse rate. h_b = 11.*km2ft ; height of 1st gradient layer, ft h_f = 40000. ; ultimate height, ft a_1 = -6.5e-3*K2R/m2ft ; lapse rate, R/ft ; Fix array sizes. bins = 101. ; number of array bins r_h = h_b/h_f ; height ratio sz_1 = fix(r_h*bins) + 1 ; first array size sz_2 = fix((1-r_h)*bins) + 1 ; second array size ; Create gradient layer altitude & temperature arrays. h_1 = h_b*findgen(sz_1)/(sz_1-1) ; altitude in gradient layer, ft T_1 = T_sl + a_1*h_1 ; temperature in gradient layer, R T_b = T_1(sz_1-1) ; temperature at break point ; Create isothermal layer altitude & temperature arrays. h_2 = h_b + (h_f-h_b)*findgen(sz_2)/(sz_2-1) ; altitude in isothermal layer, ft T_2 = T_b*(fltarr(sz_1)+1.) ; isothermal temperature, R ; Combine into total altitude & temperature arrays. h T = [h_1,h_2] = [T_1,T_2] ; Create Mach number array. a = sqrt(gamma*r_air*t) ; speed of sound array, ft/s M = u/a ; Mach number array ; Plot Mach number vs altitude. x = h/1000. ; altitude, 1000 ft. y = M ; Mach number ytitle = Mach number ; y-axis title xtitle = altitude (1000 ft) ; x-axis title file = fig1_4.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, x, y, xtitle=xtitle, $ ytitle=ytitle, font=0, $ xstyle=9, ystyle=9 device, /close ; set plot device to PostScript ; open file & save EPS ; close the file 11

end 5 Problem5 Consider an airplane flying with a velocity of 100 m/sec at an altitude of 8 km. At a point on the wing where a pitot tube is mounted, the airflow velocity is 140 m/sec. Calculate the total pressure and the static pressure at this point on the wing as measured by the pitot tube. Assume incompressible flow and use the standard atmosphere model. At an altitude of 8 km, Anderson (Appendix A) gives the static parameters T = 236:23 K P = 35;651 N=m 2 ρ = 0:52578 kg=m 3 so the aircraft dynamic pressure q a 1 2 ρv 2 a = 1 2 0:52578 kg m 3 100 m 2 = 2638:9 N s m 2 (18) For incompressible flow, the total pressure P o = P+q a =(35;651+2638:9) N=m 2 = 38;280 N=m 2 (19) while the dynamic pressure at the pitot tube location is q p 1 2 ρv 2 p = 1 2 0:52578 kg m 3 140 m 2 = 5152:6 N s m 2 (20) Since the total pressure remains constant, the static pressure at the pitot tube location is P s = P o + q s =(38;280, 5152:6) N=m 2 = 33;127 N=m 2 (21) 6 Problem6 An Airbus A320 is in steady level flight at an altitude of 10 km in a standard atmosphere. A total pressure of 29,500 N/(sq m) is measured at the nose of the aircraft using a Pitot tube. 1. Assuming incompressible flow, determine the aircraft speed and the aircraft Mach number. 2. Is the assumption of incompressible flow in part (a) justified; explain. 12

6.1 Incompressible flow assumption At an altitude of 10 km, Anderson (Appendix A) gives the static parameters T = 223:26 K P = 26;500 N=m 2 ρ = 0:41351 kg=m 3 For incompressible flow, the air speed is given by s s V 1 = 2 P o, P 29;500, 26;500 = 2 ρ 0:41351 N=m 2 kg=m 3 = 120:46 m s (22) while the sound speed a = p s γrt = 1:4 287: m 2 K m 2 223:26 K = 299:51 m s (23) so the Mach number M 1 u a 120:46 m=s = = 0:4021 (24) 299:51 m=s 6.2 Validity of incompressible flow Incompressible flow is a bad assumption for M > 0:3. Recalculating Mach number with the compressible Pitot tube equation (Anderson, 3rd ed., Eqn. 4.76), 2 M1 2 = 2 γ, 1 4 P0 P γ,1 γ 3, 15 = 5:000 " 29;000 26500 # 0:28571, 1 = 0:15558 (25) so the true Mach number M 1 = 0:39943 is still too large to justify assuming incompressible flow. 13