17-1 Design of UAV Systems Standard atmospherec 2002 LM Corporation Lesson objective - to discuss...

17-1

Design of UAV Systems

Standard atmospherec 2002 LM Corporation

Lesson objective - to discuss another

UAV Operating Environment

•The atmosphere

Expectations - You will understand how this environment drives UAV design and operations

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Why is it important

A UAV operates in two important environments

1. Physical

- The atmosphere in which it flies

- Varies with altitude (primary) and weather (secondary)

2. Functional

- The rules for how it operates

- Originally established for manned aircraft

- Now imposed on unmanned aircraft

If you don’t understand both environments, you can’t design for them

Lesson 6

This lesson

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• Indicated air speed (IAS) - the speed sensed by the aircraft (proportional to dynamic pressure)

• Equivalent air speed (EAS) - indicated air speed corrected for instrument and installation errors and compressibility effects (important when M > 0.5)

• True air speed (TAS) - the actual speed through the air• MSL Altitude - Altitude above mean sea level (the

altitude used for pilotage and air traffic control)• AGL - Altitude above ground level• Indicated altitude - The altitude sensed by the aircraft• Pressure altitude - Indicated altitude corrected to sea

level standard conditions• Geometric altitude - Actual altitude (measured - RF, etc)

Definitions

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Defined by internationally agreed models on how pressure, temperature and density vary with altitude above mean sea level (MSL). Used widely for:

- Altimeter calibration- Air traffic control- Aircraft design- Performance computation- Etc.

Models describe a “Standard Day” and provide variations for other conditions

- “Hot” day- “Tropical” day- “Cold” day- “Polar” day

The atmosphere

For simplicity we will use the Standard Day only

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Sea level standard (SLS) conditions (in British units)

Static pressure (P0) - 2116.22 lb/ft^2 (psf) or…...

- 29.92 in. Hg

Temperature (T0) - 518.67 degR or…...- 59.0 degF

Assumptions -

- No moisture- Obeys perfect gas law

Standard atmosphere

For simplicity (mine) we will use English units

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Key definitions

Ta = Ambient temperature (at altitude)Pa = Ambient pressure (at altitude)M = Mach numberc = Speed of sound (kts)q = Dynamic pressure (lb/ft^2) or (psf)h = Pressure altitude (in thousands of feet)

Static pressure ratio () = Pa/P0

Static temperature ratio () = Ta/T0

Density ratio () = /

Dynamic pressure (q) = 0.7*Pa*M^2

Speed of sound in kts (c)= 661.5*sqrt ()

True air speed in kts (KTAS) = M*cEquivalent air speed in kts (KEAS) = KTAS*sqrt()

(17.1)(17.2) (17.3)(17.4)(17.5)(17.6)(17.7)

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q - the alternate form

• The standard form

q = [/2]V^2 • But by definition (basic fluid mechanics)

V = Mc where c = speed of sound

c = [gRT]^0.5 where ≈ 1.4, g = gravity, R = universal gas

constant P = gRT

Thereforeq = [/2][Mc]^2 = P[/2]M^2 = 2116.220.7 *M^2

or q = 1481.35M^2

where = Pa/P0

(17.4a)We will use this form extensively

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Atmosphere characteristics

• Sea level - 36089 ft(troposphere)

- Exponential ambient pressure decrease

- Linear ambient temperature decrease

• 36089 - 65617 ft (stratosphere)

- Logarithmic ambient pressure decrease

- Constant ambient temperature

• 65617 - 104987 ft

- Exponential ambient pressure decrease

- Linear ambient temperature increase

1962 U.S. Atmosphere

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0 50 100

Altitude (Kft)

Am

bie

nt/

Se

a L

ev

el

Ra

tio

Pressure (delta)

Temperature (theta)

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Spread sheet equations

Altitude (Kft) h Ambient pressure ratio Ambient temperature ratio

0-36.089 delta = (1-6.875586*h/1000)^5.255912 theta = 1-1.9812*h/288.16

30 0.2970 0.7937

36.089-65.617 delta = 0.2234*exp((h-36.089)*(-4.80637*10^-2)) theta = .7519

40 0.1851 0.7519

theta = 0.6826+h*1.0566*10^-365.617-104.987 delta = exp(0.15116-h*0.04678)

80 0.0276 0.7671

Input Outputs

Approximate 1962 U.S. Standard Atmosphere

8 0 0 .0 2 7 6 0 .7 6 7 1 5 7 9 .4 2 9 .5

(17.8)

(17.10)

(17.9)

(17.13)(17.12)

(17.11)

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Typical calculations

• Throughout this course we will use the standard atmosphere for performance, aerodynamic and propulsion calculations. Some examples:

1. A UAV is flying at M = 0.75 and h = 65 Kft. What is the dynamic pressure?- From (17.4a) q = 1481.35**M^2 - From (17.10) = .05567- q = 1481.35*0.05567*0.75^2 = 46.4 psf

2. A UAV is flying at KTAS = 350 Kts. What Mach will it fly at h = 30, 50 and 70Kft? How about KEAS?

- From (17.4-7) KTAS = M*c = M*661.5*sqrt () (17.14)

and KEAS = KTAS*sqrt () = M*661.5*sqrt () (17.15)

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Calculations - cont’d

2. cont’d• From (17.15) M = KTAS/661.5*sqrt () M@h=30Kft = 350/(661.5*sqrt (.7938)) = 0.59 M@h=50Kft = 350/(661.5*sqrt (.7519)) = 0.61 M@h=70Kft = 350/(661.5*sqrt (.7565)) = 0.61

• From (17.16) KEAS = M*661.5*sqrt () M@h=30Kft = 0.59*661.5*sqrt (.2970)) = 213 KEAS

M@h=50Kft = 0.61*661.5*sqrt (.1145)) = 136.5 KEAS M@h=70Kft = 0.61*661.5*sqrt (.0438)) = 84.4 KEAS

3. A UAV with a wing loading of 60 psf wants to loiter at a Cl of 0.8 for L/Dmax. What M will it fly at 30/50/70 Kft?

• At loiter L ≈ W = Cl*q*S (See RayAD Chapter 5.3)W/S wing loading = Cl*q or q = (W/S)/Cl (17.16)

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3. cont’d• Combining (17.14) and (17.16) we find

M = sqrt((W/S)/(Cl*1481.35*)) (17.17)

M@h=30Kft = sqrt(60/(0.8*1481.35*0.2970)) = 0.41 M@h=50Kft = sqrt(60/(0.8*1481.35*0.1145)) = 0.66 M@h=70Kft = sqrt(60/(0.8*1481.35*0.0438)) = 1.07

- Is there something wrong at 70 Kft?No - it can’t fly at subsonic speeds at 70Kft at this wing loading. It either needs a bigger wing or has to lose weight (burn fuel - and a lot)

4. Assume the UAV in problem 3 has a loiter Mach number of 0.6. What will its loiter altitude be?

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4. cont’d• From (17.16)

= (W/S)/(Cl*1481.35*M^2) (17.18)

@ M=0.6 = 60/(0.8*1481.35*0.36) = 0.141

- At what altitude will = 0.141?Answer - From chart 17.7, we can see that the

altitude is somewhere between 40 and 50 Kft. If we solve (17.10) we will find that hlo = 45.7 Kft

Note – We will program equations (17.8) through (17.13). We will use them a lot.

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Implications for design

• At a given air speed, dynamic pressure drops about an order of magnitude by 50Kft. It drops about another order of magnitude by 100Kft.

- Generating lift above 65K ft is a major challenge- Simple aerodynamic analysis show why

• Altitude is also a problem for the engine even though drag goes down with q

- Engine power required to run generators does not go down (environmental control requirements can go from cooling to heating)

• Above 65K ft. lift and power are major design issues

- This lesson will address the lift issue

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Global Hawk Example

http://www.fas.org/irp/program/collect/global_hawk.htm

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Global Hawk Assessment

•Takeoff weight = 25,600 lbs, wing area (S) = 540 ft^2, fuel = 14,500 lbs•Assuming 15% fuel consumption for takeoff and climb to initial cruise altitude (50Kft), the initial cruise wing loading (W/S) would be

- W/S = (25,600-0.15*14,500)/540 = 43.4 psf•At a nominal cruise Mach of 0.6, cruise q = 61.0 psf . If L = W, what would the cruise lift coefficient (Cl-cruise) be?

Answer : Cl-cruise = (W/S)/q = 0.71•At mid-mission (loiter with 50% fuel) the wing loading would be

- W/S = (25,600-0.5*14,500)/540 = 34.0 psf at a required loiter altitude of 65Kft. Assuming the loiter speed is also Mach 0.6 (q = 29.7 psf), what would Cl-loiter have to be?

Answer : Cl-loiter = 34/29.68 = 1.15 (near or at stall speed?)

In order to operate efficiently at high altitude, Global Hawk has to fly at high cruise and loiter lift coefficients. What if it had to loiter even higher - say 70Kft?

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70Kft Assessment

• Option 1 - Loiter at a higher speed• For Cl-loiter = 0.84 and W/S = 34.0 psf as before

- Required q-loiter = 34/0.84 = 40.5 or M = 0.8 (too high for a thick unswept wing!)

• Option 2 - Loiter at a higher Cl• For M = 0.6 (q = 23.35) and W/S = 34.0 psf as before

- Required Cl-loiter = 34/23.35 = 1.45 (way too high!)

• Option 3 - Loiter at a lighter weight• Assume M = 0.6 (q = 23.35) and Cl-loiter = 0.84 as before

- Allowable W/S = 23.35*0.84 = 19.6 (less than empty weight )

• Option 4 - Combination of Options 1-3• Assume M = 0.65 (q = 27.4) and Cl-loiter = 1.1

- Allowable W/S = 27.4*1.1 = 30.1 (35% fuel remaining)

A 70Kft requirement would be a significant challenge!

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Summary - Atmosphere

• Simple atmospheric models in combination with

basic aerodynamic analysis are invaluable for first-

order evaluation of air vehicle requirements and

concepts• High altitude requirements drive air vehicle design• Very high altitude (>65Kft) is a difficult (but not

impossible) design space• Simple analysis of existing aircraft can provide

very useful insight for pre-concept and conceptual

design• Secondary power at high altitude is a major issue

that we will address in subsequent sessions

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Expectations

You should now understand how to model a standard atmosphere

- Equations for pressure and temperature are all you need to fully characterize the standard day physical environment

- You will use these models extensively throughout this course (and your career if you plan to work in air vehicle design)

- See spreadsheet SSC.StdAtmosphere.xls

You should also understand the issues associated with UAV operating environments

- Very high altitude will be a major design challenge- Simple atmospheric and aerodynamic analysis can

identify the problems



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Homework (extra credit)Design of UAV Systems


1. (Individual grade) Use spreadsheet ASE261.StdAtmosphere.xls to calculate true airspeed at the WAS altitude required for your project UAV assuming a wing loading of 35 psf - Assess one lift coefficient per team member

- Cl = 0.3, 0.9,1.2 or 1.52. (team grade) Which of the four (4) lift coefficients

would be most appropriate for your WAS UAV at loiter? Explain your answer.

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Intermission

17-1 Design of UAV Systems Standard atmospherec 2002 LM Corporation Lesson objective - to discuss...

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