Lesson 24 Performance— Thrust Required and Thrust Available 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 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

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)

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)

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)

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

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:

Example: T-37 CL = (CD,0 /k)1/2 CD = 2CD,0 = 14.8 = 215.3 ft/s 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 = 14.8 V = 484.4 ft/s; q = 206 lb/ft2; CD = 0.0243 CD,0 = ½ CD = 0.0121 L/Dmax = ½ {CD,0 / (p eo AR)}-0.5 CL = (CD,0 /k)1/2 V = (2W/rSCL)1/2 = 215.3 ft/s CD = 2CD,0 TRMIN = CD q S = 405.4 lb

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

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:

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

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) 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

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

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

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

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

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

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 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!