1. Lesson 24 Performance— Thrust Required and Thrust Available
Aero Engineering 315
Lesson 24 Performance—
Thrust Required and
Thrust Available

2. Aircraft performance in the news

3. Thrust req’d & available objectives
Given a T-38 thrust required (TR) chart
Find stall Mach, thrust req’d, min drag Mach
Find L/Dmax from thrust req’d or drag polar
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

4. Thrust required in terms of V
Remember
CL = L
q S = W
q S
q = rV2 1 2
So, since and
Parasite Drag
Drag due to Lift
varies with V2
varies with 1/V2

5. Thrust Required (Parasite)
Let’s Look at Parasite Drag First…
TR = ½ r V2 S CD,0 +
Parasite Drag
TR or D
V (or M)

6. Thrust Required (+Induced)
Now add in Induced Drag
TR = ½ r V2 S CD,0 + 2 kW2 / (r V2 S)
Parasite Drag
Drag due to Lift
TR or D
V (or M)

7. Thrust Required (Total)
Then add them together…Thrust Required = Total Drag
TR = ½ r V2 S CD,0 + 2 kW2 / (r V2 S)
Parasite Drag
TR or D
Drag due to Lift
TR,MIN
Note: Parasite = Induced at min drag
VMin Thrust
V (or M)

8. Minimum Thrust Required
Stated another way:
To minimize Thrust Required…
…maximize
Lift/Drag
By rearranging we
get another useful
concept

9. L/Dmax is a function of CD0 and k
At Min Drag, parasite drag = induced drag
CD,0 = CD,i or CD,0= kCL2
CL = (CD,0 /k)1/2
so: CDmin = CD,0 + CD,i = 2CD,0 = 2kCL2
solving for CL:

10. Example: T-37 CL = (CD,0 /k)1/2 CD = 2CD,0 = 14.8 = 215.3 ft/s
Using CD = CL2 (from whole aircraft lesson), S = 184 ft2 and W = 6,000 SL (SA)
Find L/Dmax,, TRMIN, and TRMIN
= 14.8
V = ft/s; q = 206 lb/ft2; CD =
CD,0 = ½ CD =
L/Dmax = ½ {CD,0 / (p eo AR)}-0.5
CL = (CD,0 /k)1/2
V = (2W/rSCL)1/2
= ft/s
CD = 2CD,0
TRMIN = CD q S
= lb

11. . . . . . Thrust Available (TA) min Vin mout Vout TA = m (Vout-Vin)
Continuity?
Mass flow equation?
min = mout .
TA = rAV (Vout-Vin)
m = rAV .

12. 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:
Maximum Thrust: full thrust with afterburner
Depends on altitude and Mach number:

13. Thrust Available vs. Thrust Req’d
Available-Max (wet) T
VMAX(WET)
VMAX(DRY)
Available-Mil (dry)
Required V

14. 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!

15. Excess Thrust (TX) T V TA,MIL (at V1)
Available-Max (wet) T
TA,MIL (at V1)
Excess Thrust (at V1) = TA,MIL – TR
Available-Mil (dry)
For a given velocity, say V1
TR (at V1)
Required V

16. Maximum Excess Thrust TR T
TA (WET)
TA (DRY)
TXMAXDRY
TXMAXWET V

17. Minimum Speed r/rSL effect lowers the TA VMIN VMIN ~ low altitude
A (WET)
A (WET) T T
A (WET)
A (WET) T T
A (DRY)
A (DRY)
STALL
LIMITED
VMIN T T
THRUST
LIMITED
VMIN
A (DRY)
A (DRY) V V V V
~ low altitude
~ high altitude

18. Use our T-38 Find: MMIN MMIN = 0.28 limited by stall
Given: W = 10,000 lbs.
h = 10,000 ft
Find: MMIN
MMIN = 0.28
limited by stall

19. How About Higher? Find: MMIN MMIN = 0.45 limited by thrust
Given: W = 10,000 lbs.
h = 30,000 ft
Find: MMIN
MMIN = 0.45
limited by thrust

20. Even Higher? Find: MMIN for Mil Thrust, Max Thrust
Given: W = 10,000 lbs.
h = 40,000 ft
Find: MMIN for Mil Thrust, Max Thrust

21. 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 m2, aspect ratio is 7.5, Oswald efficiency factor is 0.9, and zero-lift drag coefficient CD0 is 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.

22. 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)

23. Next Lesson (T25)… Prior to class In 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!