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 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

Thrust required in terms of V

SV

kWCSVT DR

2

22

212

10

So, since andCL = L

q Sq = V21

2

Parasite Drag Drag due to Liftvaries with V2 varies with 1/V2

= Wq S

qSCkqSCT LDR2

0Remember

Thrust Required (Parasite)

V (or M)

TR o

r D

TR = ½ V2 S CD,0 +

Let’s Look at Parasite Drag First…

Paras

ite D

rag

Thrust Required (+Induced)

V (or M)

TR o

r D

TR = ½ V2 S CD,0 + 2 kW2 / ( V2 S)

Now add in Induced Drag

Drag due to Lift

Paras

ite D

rag

Thrust Required (Total)

V (or M)

TR o

r D

Drag due to Lift Paras

ite D

rag

Then add them together…Thrust Required = Total Drag

VMin Thrust

TR,MIN

TR = ½ V2 S CD,0 + 2 kW2 / ( V2 S)

Note:Parasite = Induced

at min drag

Minimum Thrust Required

DL

WT

T

W

D

LR

R /

Stated another way:To minimize Thrust Required…

…maximizeLift/Drag

By rearranging we get another useful concept

min,RMax T

W

D

L

L/Dmax is a function of CD0 and k

At Min Drag, parasite drag = induced drag

kC2

1

C2

/kC

C

C

D

L

0,D0,D

D,0

D

L

Max

so: CDmin = CD,0 + CD,i = 2CD,0 = 2kCL

2

solving for CL:

CD,0 = CD,i or CD,0= kCL2

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

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

kCD

L

DMax 0,2

1

CL = (CD,0 /k)1/2 V = (2W/SCL)1/2

TRMIN = CD q SCD = 2CD,0

= 14.8

= 215.3 ft/s

= 405.4 lb

Thrust Available (TA)

Burner

High-Pressure Compressor

High-Pressure Turbine

Afterburner Flameholders Nozzle

Afterburner

Low-Pressure Compressor

Low-Pressure TurbineInlet

Afterburner Fuel Injectors

min

Vin

.mout

Vout

.

Thrust?

Continuity?

Mass flow equation?

min = mout

. .

m = AV.

TA = AV (Vout-Vin)

SLSLA TT

TA = m (Vout-Vin).

Thrust required is a function of the airframeThrust 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:

SLSLA TT

Thrust Available

M) 0.7 + (1

SLSLA TT

Thrust Available vs. Thrust Req’d

V

T

Required

Available-Mil (dry)

Available-Max (wet)

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)

V

T

Required

Available-Mil (dry)

Available-Max (wet)

TA,MIL (at V1)

TR (at V1)

For a given velocity, say V1

Excess Thrust (at V1) = TA,MIL – TR

V

TTR

TA (DRY)

TA (WET)

TXMAXWET

TXMAXDRY

Maximum Excess Thrust

~ low altitude ~ high altitude

Minimum Speed

V

TTR

TA (DRY)

TA (WET)

V

TTR

TA (DRY)

TA (WET)

V

TTR

TA (DRY)

TA (WET)

V

TTR

TA (DRY)

TA (WET)

THRUSTLIMITED

VMIN

STALLLIMITED

VMIN

SL effect lowers the TA

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!