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AE 245 homework #1 solutions
Tim Smith
24 January 2000
1 Problem 1
An aircraft is in steady level flight in a standard atmosphere.
- 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.
- 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
Po = P
γ 1 2
M^2
where the Mach number M V = a , sound speed a =
p γ RT , 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 :2 lb=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
100 200 300 400 500 air speed (mph)
15
16
17
18
19
total pressure (psi)
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/ftˆ T = 518.69 ; static temperature, R a = sqrt(gammaR_airT) ; 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 + alphaMˆ2)ˆbeta ; total pressure array, lb/ftˆ
; 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
which, plugged into Eqn. 1, give the plot in Fig. 2.
2 Problem 2
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
a^ go
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
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
P = Pb exp
g 0
h h (^) b RTb
where Pb is the pressure at h = 11 km. The following IDL code stacks these layers, giving the plots in Fig. 3.
pro hw1_
; 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/sˆ 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/ftˆ
T_sl = 32.0 + T_z ; sea level temperature, R r_sl = p_sl/(R_air*T_sl) ; sea level density, slug/ftˆ
; 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-3K2R/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_hbins) + 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_bfindgen(sz_1)/(sz_1-1) ; altitude in gradient layer, ft T_1 = T_sl + a_1h_1 ; temperature in gradient layer, R p_1 = p_sl(T_1/T_sl)ˆ(-g_o/(a_1R_air)) ; pressure in gradient layer, lb/ftˆ r_1 = p_1/(R_air*T_1) ; density in gradient layer, slug/ftˆ
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_bexp(-g_o(h_2-h_b)/(R_airT_b)) ; pressure in isothermal layer, lb/ftˆ r_2 = p_2/(R_airT_b) ; density in isothermal layer, slug/ftˆ
; 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/ftˆ rho = [r_1,r_2] ; combined density, slug/ftˆ
; 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. xsz = 4.0scale ysz = 3.0scale
set_plot, ’ps’ ; set plot device to PostScript device,/encapsulated,/preview,filename=file ; open file & save EPS device,/inches, xsize=xsz, ysize=ysz, $ font_size = 7
-80 -60 -40 -20 0 20 temperature (¡F)
0
10
20
30
40
50
altitude (1000 ft)
2 4 6 8 10 12 14 pressure (psi)
0
10
20
30
40
50
altitude (1000 ft)
0.5 1.0 1.5 2.0 2. density (millislug/ft^3)
0
10
20
30
40
50
altitude (1000 ft)
Figure 3: Atmospheric model for sea-level T = 32 o^ F, P = 14 :8 psi.
3 Problem 3
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.
- At the instant it passes through an altitude of 10 km, what are the temperature, pressure and density of the surrounding air?
- 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
dT =
∂ T
∂ h
dh (6)
so that the temperature rise rate
dT dt
∂ T
∂ h
dh dt
Given the lapse rate ∂ T =∂ h = a 1 = 6 : 5 10 ^3 K/m within the gradient layer and the climb rate dh = dt = 5 m/s, the temperature rise rate is
dT dt = 6 : 5 10 ^3
K
m
m s
= 3 : 250 10 ^2
K
s
Similary, pressure in the isothermal layer is solely a function of temperature, so the pressure rise rate
dP dt
∂ P
∂ T
dT dt
First, let’s try the analytical method. Taking the derivative of Eqn. 3,
As before, Mach number M V = a , where u = 400 mph= 586 :67 ft/s, sound speed a =
p γ RT , specific heat ratio γ = 1 :4, the gas constant for air R = 1716 ft-lb/slug- o R and temperature varies as
T (0 km < h < 11 km) = T 1 + a 1 h T (11 km < h < 25 km) = Tb
where T 1 = 581 : 69 o R and a 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 10 20 30 40 altitude (1000 ft)
Mach number
Figure 4: Mach number at 400 mph as a function of altitude.
pro hw1_
; 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/ftˆ T_sl = 518.69 ; sea level temperature, R u = 400.*mph2fps ; air speed, ft/s
; 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-3K2R/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_hbins) + 1 ; first array size sz_2 = fix((1-r_h)bins) + 1 ; second array size
; Create gradient layer altitude & temperature arrays.
h_1 = h_bfindgen(sz_1)/(sz_1-1) ; altitude in gradient layer, ft T_1 = T_sl + a_1h_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 = [h_1,h_2] T = [T_1,T_2]
; Create Mach number array.
a = sqrt(gammaR_airT) ; 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. xsz = 4.0scale ysz = 3.0scale
set_plot, ’ps’ ; set plot device to PostScript device,/encapsulated,/preview,filename=file ; open file & save EPS device,/inches, xsize=xsz, ysize=ysz, $ font_size = 7 plot, x, y, xtitle=xtitle, $ ytitle=ytitle, font=0, $ xstyle=9, ystyle= device, /close ; close the file
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
V 1 =
s
Po P ρ
s
29 ; 500 26 ; 500 0 : 41351
N=m^2 kg=m^3
m s
while the sound speed
a =
p
γ RT =
s
m 2 K m 2
223 :26 K = 299 : 51
m s
so the Mach number
M 1
u a
120 :46 m=s 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),
M^21 =
γ 1
P 0
P
γ γ^1
1
1
so the true Mach number M 1 = 0 :39943 is still too large to justify assuming incompressible flow.
