Introduction to Aerospace Engineering

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1 Introduction to Aerospace Engineering Lecture slides Challenge the future

2 3-0-0 Introduction to Aerospace Engineering Aerodynamics 5 & 6 Prof. H. Bijl ir. N. Timmer Delft University of Technology

3 5. Compressibility continued Anderson Ernst Mach

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5 Summary of equations For steady, frictionless, incompressible flow: Continuity equation A V = A V Bernoulli s equation p + ρ V = p + ρ V 4

6 Pitot tube to measure velocity p =p V =V V=0 p + ρv = p + ρv Pt =Pt =Pt p t p = ρ V 5

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9 For steady, isentropic compressible flow (adiabatic and frictionless) : Continuity equation ρ A V = ρ A V Isentropic relations Energy equation γ p ρ γ = = T γ- p ρ T c p T + V = c p T + V Equation of state P = ρrt 8

10 Example : Re-entry Space Shuttle. 9

11 P = 347 N/m T = 953 K P 3 = 40 N/m P 4 = 704 N/m Re-entry Space Shuttle Orbiter Bow shock wave 3 4 Just info: h=30,480 m V= 08 m/s M=4 Boundary layer T=? V=? M=? In 3,4 0

12 Speed of sound p p+dp p p+dp a ρ T ρ +dρ T+dT is equal to ρ T a ρ +dρ T+dT a + da moving sound wave with speed a into a stagnant gas motionless sound wave static observer observer moving with sound wave

13 Speed of sound Apply the continuity equation: ρ A V = ρ A V ρa a = (ρ + dρ) A (a+da) -dimensional flow A = A = A = constant Thus: ρa = (ρ + dρ)(a + da) ρa = ρa + adρ + ρda + dρda da a = - ρ d ρ small, ignore

14 Speed of sound Now apply momentum equation (Euler equation) dp = - ρvdv Substituting this into We obtain a = dp dρ dp = - ρada da = - dp ρa da a = - ρ dρ dp a = + ρ ρ a d ρ Going through the sound wave there is no heat addition, friction is negligible a = dp dρ isentropic 3

15 Speed of sound Now we may apply the isentropic relations. We want to rewrite dp d ρ With isentropic relations we find a = p γ ρ Use the equation of state : p = RT ρ a = γ RT Speed of sound in a perfect gas depends only on T! 4

16 a = γ RT Mach M = V a Ackeret In honor of Ernst Mach the name Mach number was introduced in 99 by Jacob Ackeret 5

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18 Equations for a perfect gas We defined: h=e + pv We found : h=c p T and e= c v T With the eq. of state: pv=rt we find: c p T=c v T+ RT hence: c p =c v + R => c p -c v = R We defined: c γ = c v p R c p = γ γ 7

19 Isentropic flow relations, second form Energy equation: c pt + V = constant c T + V = c T o + V p p o Assume index 0 = stagnation point V o = 0 Substitute: T o V c p T + V = c p T o = + T c T C p p γ R = (from C p Cv = R ) γ T o γ - V = + T γ RT 8

20 Isentropic flow relations, second form with a = γrt To γ - V - =+ M a γ =+ T Bring flow isentropically to rest Isentropic relation can be used γ p ρ γ o = o γ = To - p ρ T γ p o γ - = + M γ- p ρo γ - = + M γ- ρ 9

21 Second form of isentropic relations ρo ρ T γ- = + M T o γ - = + M γ - po γ- = + M p γ γ- 0

22 Compressibility ρ ρ ρ ρ 0 γ - = + M γ Derived from energy equation and isentropic relations M For M < 0.3 the change in density is less than 5 % Thus : for M < 0.3 the flow can be treated as incompressible

23 Supersonic wind tunnel and rocket engine M< M> Rocket motor must be LIGHT Thus the nozzle is short Exhaust of Saturn V that powered the Apollo space missions

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26 Supersonic Wind Tunnel Reservoir Exit p0 T0 V 0 M< Throat M= M > P e T e Subsonic flow Supersonic flow For both the wind tunnel and a rocket motor we can derive an area velocity relation using the continuity and the Euler equation 5

27 Supersonic tunnel and rocket engine ρ VA = const ln ρ + ln A + lnv = ln( const) dρ da dv 0 ρ + A V dp = ρvdv d ρvdv da dv d ρ + + = 0 = = dp dp A V dp a d ρ VdV da dv a A V + + = 0 6

28 Supersonic wind tunnel: area velocity relation Area-Velocity Relation da A = ( -) M dv V Conclusions: Case A. Subsonic flow if dv > 0 then da < 0 and vice versa. Case B. Supersonic flow if dv > 0 then da > 0 and vice versa. Case C. If the flow is sonic (M=) then... 7

29 Supersonic windtunnel: M= in the throat dv da da = =! V - M A 0 A At first glance we see dv = V. But this is not possible on physical basis. Then we must have (for finite dv V ): da = 0 dv = 0 = FINITE! A V 0 If da = 0 A In the THROAT : M =! 8

30 Nozzle flow * * for M= flow conditions (T*,p* ) M T/T 0 T γ- = + M T o X -> P/P 0 po γ- = + M p γ γ- ρ/ρ 0 ρo γ - = + M γ- ρ 9

31 Example 4.9: A supersonic wind tunnel is considered. The air temperature and pressure in the reservoir of the tunnel are: T 0 = 000 K and p 0 = 0 atm. The static temperatures at the throat and exit are: T* = 833 K and T e = 300 K. The mass flow through the tunnel is 0.5 kg/s. For air, c p = 008 J/(kg K). Calculate: a) Velocity at the throat V* b) Velocity at the exit V e c) Area of the throat A* d) Area of the exit A e 30

32 6. Viscous flows Anderson Osborne Reynolds Ludwig Prandtl

33 Subjects lecture 6 Viscous flows Laminar boundary layers - calculation of boundary layer thickness - calculation of skin friction drag 3

34 Viscous flow Up till now we have only dealt with frictionless flow. What is the effect of friction?. D Inviscid flow (No friction) NO DRAG Viscous flow (friction) FINITE DRAG 33

35 Viscous flow In real life the flow at the surface adheres to the surface because of friction between the gas and the solid material: Right at the surface the velocity is zero 34

36 Boundary layer V Friction force Boundary layer (exaggerated) In the vicinity of the surface there is a thin region of retarded flow: the boundary layer The pressure through the boundary layer in a direction perpendicular to the surface is constant 35

37 Viscous flow Inside the boundary layer Bernoulli s law is not valid!!!!!! 36

38 Viscous flows Boundary layer velocity profile Shear stress can be written as : boundary layer thickness, δ τ w du = µ dy y= 0 shear stress, τ w skin friction drag µ = absolute viscosity coefficient or viscosity Air at standard sea level : µ=.789*0-5 kg/ms) 37

39 Viscous flow τ = µ du dy viscosity or dynamic viscosity Fluids for which the shearing stress is linearly related to the rate of shearing strain are called: NEWTONIAN FLUIDS. (Most common fluids (liquids & gases) like are are NEWTONIAN!). Often viscosity appears as µ ν = = KINEMATIC VISCOSITY ρ 38

40 Reynolds number Characteristic velocity Characteristic length Osborne Reynolds 84-9 Re = V L ν (kinematic) viscosity Sail planes Passenger jets Lower leg athlete : Re 0 6, L=wing chord : Re Idem : Re 0 5 L=Diameter leg 39

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42 Viscous flows, some definitions V δ Reynolds number : ρ Re x = dimensionless, and varies linearly with x Laminar flow : streamlines are smooth and regular and a fluid element moves smoothly along a streamline Turbulent flow : streamlines break up and a fluid element moves in a random irregular way V µ x, x 4

43 Laminar boundary layer, boundary layer thickness Consider flat plate flow. What is boundary layer thickness δ and skin friction drag D f at location x? V δ From laminar boundary layer theory : x δ = 5. x Re x Thus δ is proportional to : x (parabolically) 4

44 Laminar boundary layer, skin friction drag x dx τ w L (unit width) Total force = total pressure force + total friction force Total friction force on element dx is: τw(x) dx = τw(x) dx Total skin friction drag is: D f = L o τ w dx 43

45 Laminar boundary layer, skin friction drag For the skin friction coefficient we find from laminar boundary layer theory : c f x = τ ρ w V τ w = q = Re x Thus C f and τ x w decrease as x. The skin friction at the beginning of the plate is larger than near the trailing edge. To calculate the total aerodynamic force we must integrate! 44

46 Laminar boundary layer, skin friction drag L D = f c f.q dx= 0.664q x o L o dx Re x q L = V / ν o dx x Df = q V / ν L =.38q L V L/ ν dx x = x -/ = x Define total skin friction drag coefficient as Cf =.38 ReL L S =.38.L ReL L Cf =.38 ReL Cf = Df q S 45

47 The skin friction coefficient for a laminar boundary layer Cf =.38 ReL Only for low speed incompressible flow (and reasonably accurate for high speed subsonic flow). Re L = Reynolds nr. based on length L. 46

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Introduction to Aerospace Engineering

Introduction to Aerospace Engineering Lecture slides Challenge the future 4-0-0 Introduction to Aerospace Engineering Aerodynamics 3 & 4 Prof. H. Bijl ir. N. Timmer Delft University of Technology Challenge