Aero Engineering 315
Lesson 24
Performance—
Thrust Required and
Thrust Available
Aircraft performance in the news
Thrust req’d & available objectives

Given a T-38 thrust required (TR) chart


Find L/Dmax from thrust req’d or drag polar








Find stall Mach, thrust req’d, min drag Mach
Understand importance of L/Dmax
Know relationship of induced & parasite drag at L/Dmax
Find velocity for L/Dmax from thrust req’d or drag polar
State at what point max excess thrust occurs
Sketch thrust available (TA) versus velocity for mil and
AB
Calculate changes in TA for changes in altitude
From T-38 charts find: TA, excess thrust (TX), max
excess thrust, max Mach, and min Mach
State whether T-38 min Mach is thrust or stall limited
Thrust required in terms of V
Remember
So, since
TR  CD0 qS  k CL qS
2
L
W
=
CL =
qS
qS
1
2
TR   V S C D 0
2
Parasite Drag
varies with V2
and

1
q =
V2
2
kW
2
1
2
V S
2
Drag due to Lift
varies with 1/V2
Thrust Required (Parasite)
Let’s Look at Parasite Drag First…
TR or D
TR = ½  V2 S CD,0 +
V (or M)
Thrust Required (+Induced)
Now add in Induced Drag
TR or D
TR = ½  V2 S CD,0 + 2 kW2 / ( V2 S)
V (or M)
Thrust Required (Total)
Then add them together…Thrust Required = Total Drag
TR or D
TR = ½  V2 S CD,0 + 2 kW2 / ( V2 S)
TR,MIN
Note:
Parasite = Induced
at min drag
VMin Thrust
V (or M)
Minimum Thrust Required
Stated another way:
To minimize Thrust Required…
L W
W

 TR 
D TR
L/ D
By rearranging we
get another useful
concept
L
W


D  Max TR , min
…maximize
Lift/Drag
L/Dmax is a function of CD0 and k
At Min Drag, parasite drag = induced drag
CD,0 = CD,i
or
CD,0= kCL2
so: CDmin = CD,0 + CD,i = 2CD,0 = 2kCL2
solving for CL:
CL = (CD,0 /k)1/2
L
CL
CD,0/k
1




D  Max CD
2CD ,0
2 C D ,0 k
Example: T-37
Using CD = 0.02 + 0.057CL2 (from whole aircraft
lesson), S = 184 ft2 and W = 6,000 lb. @ SL (SA)
Find L/Dmax,, TRMIN, and V @ TRMIN
L
1


D  Max 2 CD,0k
= 14.8
CL = (CD,0 /k)1/2
V = (2W/SCL)1/2 = 215.3 ft/s
CD = 2CD,0
TRMIN = CD q S
= 405.4 lb
Thrust Available (TA)
Inlet
.m
Low-Pressure
Compressor
Burner
Low-Pressure
Turbine
Afterburner Flameholders
Nozzle
.m
in
Vin
Vout
High-Pressure
Compressor
.
T = m (Vout-Vin)
Thrust?
.
.
Continuity? min = m
. = AV
Mass flow m
A
High-Pressure
Turbine
Afterburner
Afterburner Fuel Injectors
TA = AV (Vout-Vin)
out
equation?
out
  

T A  T SL 
  SL 
Thrust Available
 Thrust
required is a function of the airframe
 Thrust available is a function of the engine(s)
 i.e. the amount of thrust the engine(s)
produce
 Military Thrust: full thrust without afterburner
 Depends on altitude:
  

T A  T SL 
  SL 

Maximum Thrust: full thrust with afterburner
 Depends on altitude and Mach number:
  
 (1 + 0.7 M)
T A  T SL 
  SL 
Thrust Available vs. Thrust Req’d
Available-Max (wet)
T
Available-Mil (dry)
Required
V
VMAX(DRY)
VMAX(WET)
TA vs. TR
This viper is flying at Edwards right now. GE132
motor in it, and it would not go above 0.98M at
~20,000 ft, level. It was configured with external
tanks, conformal tanks, and 2 X 2000 lb bombs,
~45,000 GW viper. Lots o’ drag = lots of thrust
required!
Excess Thrust (TX)
Available-Max (wet)
T
TA,MIL (at V1)
Excess Thrust (at V1) =
TA,MIL – TR
Required
Available-Mil (dry)
TR (at V1)
V
For a given velocity, say V1
Maximum Excess Thrust
TR
T
TXMAX
TA (WET)
TA (DRY)
WET
TXMAX
DRY
V
Minimum Speed
TR
T
TA (WET)
T
TR
/SL effect
lowers the TA
TA (WET)
TA (DRY)
TA (DRY)
STALL
LIMITED
VMIN
V
~ low altitude
THRUST
LIMITED
VMIN
V
~ high altitude
Use our T-38
Given: W = 10,000 lbs.
h = 10,000 ft
Find: MMIN
MMIN = 0.28
limited by stall
How About
Higher?
Given: W = 10,000 lbs.
h = 30,000 ft
Find: MMIN
MMIN = 0.45
limited by
thrust
Even Higher?
Given: W = 10,000 lbs.
h = 40,000 ft
Find: MMIN for Mil
Thrust, Max Thrust
Homework #26
Consider an airplane patterned after
the twin-engine Beechcraft Queen Air
executive transport. The airplane
weight is 38,220 N, wing area is
27.3 m2, aspect ratio is 7.5, Oswald
efficiency factor is 0.9, and zero-lift
drag coefficient CD0 is 0.03. Calculate
the thrust required to fly at a velocity
of 350 km/hr at (a) standard sea level
and (b) an altitude of 4.5 km.
Homework #27
Given an 8,000 lb T-38 flying at 10,000 ft, determine:

Thrust required (TR) at Mach 0.5

Thrust available in military power (TADRY) at Mach 0.5

Thrust available in maximum power (TAWET) at Mach
0.5

Excess thrust (TX) at Mach 0.5 (assume military
power setting)

Mach number for minimum drag

Minimum drag

Minimum Mach number and what causes this limit
(thrust or stall)

Maximum Mach number (assume maximum power
setting)
Next Lesson (T25)…

Prior to class



Read 5.5 – 5.6
Complete problems 26, 27 and 28
In class


Discuss power required and power
available
This is different from thrust required!