Aero Engineering 315

Lesson 28

Cruise Range and Endurance

T-38 Example

Given:W = 12,000 lbsh = Sea Level

Find: MAX

T-38 Example

Given:W = 12,000 lbsh = Sea Level

Find: ROCMAX

T-38 Ceiling

What happens to TA - TR as we go higher?

Ceilings

Based on maximum climb rates

Absolute Ceiling = 0 ft/min ROC

Service Ceiling = 100 ft/min ROC

Cruise Ceiling = 300 ft/min ROC

Combat Ceiling = 500 ft/min ROC

Cruise performance overview

Thrust Specific Fuel Consumption Average Value Method

Endurance Range

Breguet Equations (conceptual only) Endurance Range

Know points to fly for max range and endurance

Find velocities for max range and endurance from T-38 charts or drag polar

Speaking of Range and Endurance…

0

1

2

3

4

5

6

7

8

9

10

0 1 2 3 4

Flight Mach Number

Thru

st S

peci

fic F

uel C

onsu

mpt

ion,

1/h

r

Rocket

Ramjet

Afterburning Turbofan

AfterburningTurbojet

Turbojet

Low-Bypass-Ratio Turbofan

High-Bypass-Ratio Turbofan

Turboprop

Piston Engine / Propeller

TSFC = Fuel use rate in lb/hr

Thrust output in lb

TSFC = ct = Wf

T

•

Thrust Specific Fuel Consumption

Adjust for altitude

or

SL

ALT

SL tALT t T

T c= c

ct ALT = ct SL (aALT /aSL)

TSFC - Typical Values

Engine type TSFC (1/hr)

Recip Prop 0.25 to 0.60Turboprop 0.35 to 0.60Turbofan 0.35 to 0.60 (high bypass)

0.39 to 0.70 (medium bypass) 0.80 to 1.00 (low bypass)

Turbojet 1.00 to 1.30 1.80 to 2.50 (with afterburner)

Low and High Bypass

FanBurner

High-PressureCompressor

High-Pressure Turbine

Low-Pressure Turbine

Afterburner

NozzleLow-Pressure Compressor

Bypass Duct

Fan

Burner

CompressorHigh-Pressure Turbine

Low-Pressure Turbine

Nozzle

Low Bypass Ratio with Afterburner High Bypass Ratio Bypass Ratio = 0.2 - 1.0 Bypass Ratio = 2.0 - 8.0TSFCDry = 0.8 - 1.3 TSFC = 0.5 - 0.7TSFCWET = 2.2 - 2.7

BPR = Mf / Mc

T-38 Powerplant

Ratings (see Note 1)Power Setting Normal Military Maximum

Power Power PowerAugmentation None None

AfterburnerEngine Speed (Note 2) 96.4 100 100Thrust per engine - lb

No losses 2140 2455 3660Installed 1770 1935 2840

Specific fuel consumption (Note 3)Installed 1.09 1.14 2.64

Notes(1) Sea level static ICAO standard conditions with a fuel specificweight of 6.5 lb/gal.(2) Units are % RPM where 100% = 16,500 RPM.(3) Units are lb/hr per lb thrust.

Supplemental Data

17.3% loss

Endurance—Average Value Method

How long will an airplane fly?

Endurance = Total fuel used in lb

Fuel use rate in lb/hr

E =Wf

• Wf

ct TR

Wf=

ct D

Wf= for SLUF

ct Davg

WfE =

But weight changes cause drag changes, so use the average drag over the segment

Using our average endurance equation:

Maximum Endurance

ct Davg

WfE =

To maximize endurance…

Wf and ct are constant for a

mission segment and altitude

…minimize drag

Example:T-38

Given:W = 11,000 lbhT = 20,000 ftCt = 1.09 (sea level)Fuel burned = 2,000 lb

Find:E for M=.7EMAX

Breguet Equations:Endurance

1

0ln

1

W

W

C

C

cE

D

L

t

where W0=initial weight and W1=final weight

Max E?

For a complete endurance solution, integrate over weight changes

E = Dc

dW

t

1

0

W

W

L/D)max

ct min

(high altitude)

Wfuel max

For our drag polar this means?

Range—Average Value Method

Starting with Endurance

For our average situation

Max Range?

R = EV

ct Davg

Wf VavgR =

Range = Endurance x Velocity

ct (D /V ) avg

WfR =or

Minimize drag/velocity

V

TTR

TA (DRY)

TA (WET)

Range

For a given velocity,say V1

Slope =(D/V1)

= D for (D/V1)

ct (D /V1 )

WfR =

V

TTR

TA (DRY)

TA (WET)

Max Range

Slope= (D/V)MIN, avg

= D for (D/V)MIN, avg

= V for (D/V)MIN, avg

Example:T-38 again

Given:W = 11,000 lbh = 20,000 ftFuel Wt = 2,000 lb

Find: RMAX

Slope tangent at Mavg = 0.63 Davg = 960

Breguet Equations:Range

Dc

dWV

t

R = Vdt = 1

0

W

W

2/11

2/10

D

2/1L

tWW

C

C

c

1

S

22R

For a complete range solution, integrate over weight changes

where W0=initial weight and W1=final weight

Max R?

ct min & min

(High altitude)

W fuel max

)max

CL1/2

CD

Dc

VdW

t

Maximum Range using the Drag Polar

Parasite Drag = Drag due to Lift

so: CD = CD,0 + CD,i = 4 CD,0 /3 = 4 kCL2

solving for CL:

CD,0 = 3CD,i or CD,0= 3k CL2

CL = (CD,0 /3k)1/2

3x

)max

CL1/2

CDoccurs when:

2LDo

MaxD

L

kCC

C

C

Range and Endurance -Aerodynamic Summary

MAX ENDURANCE: MAX RANGE:

2

2/1

3 LDo

MaxD

L

kCC

C

C

Minimum ofThrust Curve

Tangent toThrust Curve

Graphical

Analytical

CL = (CD,0 /3k)1/2CL = (CD,0 /k)1/2

Performance Summary(text p. 173)

*for typical non-afterburning turbojet aircraft

Performance SummaryRelationship Best Case Relation between

Induced and Parasite Drag for best case

Graphical relation for best case

Max Climb Angle CDo = kCL2 - Minimum from TR curve

- Tangent point on a line from origin to PR curve

Max Climb Rate No set relation - Draw a line parallel to PA curve, and move it down till it is tangent to PR curve

Breguet (5.28) or Avg Value (5.27)

Max Jet Powered Range

CDo = 3kCL2 - Tangent point on a line

from origin to TR curve

Breguet (5.26) or Avg Value (5.24)

Max Jet Powered Endurance

CDo = kCL2 - Minimum from TR curve

- Tangent point on a line from origin to PR curve

R = h (L/D)Best Glide Range (angle)

CDo = kCL2 - Minimum from TR curve

- Tangent point on a line from origin to PR curve

Best Glide Endurance (min sink)

3CDo = kCL2 - Minimum from the PR curve

)arcsin(W

DT

W

P

W

DTVCOR x

)(...

W

P

W

VDDOR R...

Climbing

Cruise

Glides

4 5 3 7 6 8 2 1 7 7

Next Lesson (T29)… Prior to class

Read text 5.10 Complete problems #32, 33 and 34 Complete FDP parts a, b, c, d, e, f, i, j, k,

l, r In class

Discuss takeoffs and landings