Aero Engineering 315
Lesson 28
Cruise Range and Endurance
T-38
Example
Given:
W = 12,000 lbs
h = Sea Level
Find: gMAX
T-38
Example
Given:
W = 12,000 lbs
h = 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…
10
Thrust Specific Fuel
Consumption
9
R ocket
T h ru s t S p e c if ic F u e l C o n s u m p t io n , 1 /h r
8
7
Fuel use rate in lb/hr
TSFC =
Thrust output in lb
•
Wf
TSFC = ct =
T
A fterburning T urbofan
6
R am jet
5
4
A fterburning
T urbojet
3
2
Adjust for altitude
ct
T urbojet
T urboprop
L ow -B ypass-R atio T urbofan
1
Piston E ngine / Propeller
0
1
2
Flig h t M ach Nu m b e r
3
= ct
SL
T SL
or
H igh-B ypass-R atio T urbofan
0
ALT
T ALT
4
ct ALT = ct SL (aALT /aSL)
TSFC - Typical Values
Engine type
TSFC (1/hr)
Recip Prop
Turboprop
Turbofan
0.25 to 0.60
0.35 to 0.60
0.35 to 0.60 (high bypass)
0.39 to 0.70 (medium bypass)
0.80 to 1.00 (low bypass)
1.00 to 1.30
1.80 to 2.50 (with afterburner)
Turbojet
Low and High Bypass
Fan
L ow -P ressure C om pressor
F an
B ypass D uct
L ow -P ressure T urbine
B urner
H igh-P ressure
C om pressor
H igh-P ressure
T urbine
N ozzle
B urner
L ow -Pressure T urbine
A fterburner
N ozzle
C om pressor
H igh-Pressure T urbine
Low Bypass Ratio with Afterburner
Bypass Ratio = 0.2 - 1.0
TSFCDry = 0.8 - 1.3
TSFCWET = 2.2 - 2.7
High Bypass Ratio
Bypass Ratio = 2.0 - 8.0
TSFC = 0.5 - 0.7
BPR = Mf / Mc
T-38 Powerplant
Supplemental Data
Ratings (see Note 1)
Power Setting
Normal
Power
None
96.4
Military
Power
None
100
Maximum
Power
Afterburner
100
Augmentation
Engine Speed (Note 2)
17.3% loss
Thrust per engine - lb
No losses
2140
2455
3660
Installed
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 specific
weight of 6.5 lb/gal.
(2) Units are % RPM where 100% = 16,500 RPM.
(3) Units are lb/hr per lb thrust.
Endurance—
Average Value Method
How long will an airplane fly?
Total fuel used in lb
Fuel use rate in lb/hr
Endurance =
E=
D Wf
•
Wf
=
D Wf
ct TR
=
D Wf
ct D
for SLUF
But weight changes cause drag changes,
so use the average drag over the segment
E=
D Wf
ct Davg
Maximum Endurance
Using our average endurance equation:
To maximize
endurance…
E=
DWf
ct Davg
DWf
c
…minimize drag
and t are constant for a
mission segment and altitude
Example:
T-38
Given:
W = 11,000 lb
hT = 20,000 ft
Ct = 1.09 (sea level)
Fuel burned =
2,000 lb
Find:
E for M=.7
EMAX
Breguet Equations:
Endurance
For a complete endurance solution, integrate over weight changes
W1
E =
 dW

W0
ct D
Max E?
ct min
(high altitude)
E 
1 CL
ct C D
W 0
ln 

W1
where W0=initial weight
and W1=final weight
Wfuel max
L/D)max
For our drag polar this means?
Range—
Average Value Method

Starting with Endurance
Range = Endurance x Velocity
R = EV

For our average situation
R=

DWf Vavg
ct Davg
or
R=
D Wf
ct (D /V ) avg
Max Range? Minimize drag/velocity
T
Range
TR
TA (WET)
R=
= D for
(D/V1)
D Wf
ct (D /V1 )
TA (DRY)
Slope =
(D/V1)
For a given velocity,
say V1
V
T
Max Range
TR
TA (WET)
Slope
= (D/V)MIN, avg
TA (DRY)
= D for
(D/V)MIN, avg
= V for (D/V)MIN, avg
V
Example:
T-38 again
Given:
W = 11,000 lb
h = 20,000 ft
Fuel Wt = 2,000 lb
Find: RMAX
Slope tangent at
Mavg = 0.63
Davg = 960
Breguet Equations:
Range
For a complete range solution, integrate over weight changes
 VdW
W1
R =
 Vdt
=

Max R?
ct min & min
ct D
W0
(High altitude)
 V dW
ct D
R  2
2
1 C
1/ 2
L
 S ct C D
W
1/ 2
0
W
1/ 2
1

W fuel max
CL1/2
where W0=initial weight
and W1=final weight
CD
)
max
Maximum Range using the Drag Polar
CL1/2
CD
)
max
occurs when:
Parasite Drag = 3x Drag due to Lift
CD,0 = 3CD,i
or
CD,0= 3k CL2
so: CD = CD,0 + CD,i = 4 CD,0 /3 = 4 kCL2
solving for CL:
CL = (CD,0 /3k)1/2
Range and Endurance Aerodynamic Summary
MAX ENDURANCE:
MAX RANGE:
Graphical
C L  Minimum of
 Thrust Curve
C D  Max
C
1/ 2
L
CD
 Tangent to
 Thrust Curve

 Max
Analytical
C Do  kC
2
L
CL = (CD,0 /k)1/2
C Do  3 kC
CL = (CD,0 /3k)1/2
2
L
Performance Summary
(text p. 173)
*for typical non-afterburning turbojet aircraft
Performance Summary
Best Case
Relation between
Induced and
Parasite Drag for
best case
Graphical relation for
best case
Max Climb Angle
CDo = kCL2
-
W
V (T  D ) Px
R .O .C . 

W
W
Max Climb Rate
No set relation
-
Breguet (5.28) or
Avg Value (5.27)
Max Jet
Powered Range
CDo = 3kCL2
-
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
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
Relationship
g  arcsin(
Climbing
Cruise
T D
)
R = h (L/D)
Glides
R .O . D . 
VD
W

PR
W
Minimum from TR curve
- Tangent point on a line
from origin to PR curve
Draw a line parallel to PA
curve, and move it down till
it is tangent to PR curve
Tangent point on a line
from origin to TR curve
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