Fundamentals of Fluid Mechanics

16.121 ANALYTICAL SUBSONIC AERODYNAMICS, MASSACHUSETTS INSTITUTE OF TECHNOLOGY Fundamentals of Fluid Mechanics 1 F UNDAMENTALS OF F LUID M ECHANICS 1.1 A SSUMPTIONS 1. Fluid is a continuum 2. Fluid is inviscid 3. Fluid is adiabatic 4. Fluid is a perfect gas 5. Fluid is a constant-density fluid 6. Discontinuities (shocks, waves, vortex sheets) are treated as separate and serve as boundaries for continuous portions of the flow 1.2 N OTATION 0 p = pressure (static) ρ = density T = temperature (absolute) Q = velocity vector of fluid particles Q = U i + V j + Wk V = control volume 0 S = surface surrounding V σ = impermeable body n = normal directed into the fluid R = gas constant F = body force per unit mass F = ∇Ω Ω = potential of the force field Gravity field: F = −g k; Ω = −g z p h = enthalpy per unit mass; h = e + ρ c p = specific heat at constant pressure c v = specific heat at constant volume γ = c p /c v e = internal energy per unit mass s = entropy per unit mass 0 1.3 C ONTINUITY E QUATION ∂ρ + ∇(ρQ) = 0 ∂t Dρ + ρ∇Q = 0 Dt Ñ Ó ∂ρ ρ(Qn)d s 0 = 0 dV 0 + V 0 ∂t S 0 +V Ñ h i ∂ρ + ∇(ρQ) dV 0 = 0 V 0 ∂t 1 1.4 C ONSERVATION OF M OMENTUM X Ñ Fi = i V0 DQ ∇p =F − Dt ρ Ó ∂ (ρQ)dV 0 + ρQ(Qn)d s 0 ∂t S 0 +V 1.5 C ONSERVATION OF T HERMODYNAMIC E NERGY D h Q2 i ∇ · (pQ) e+ =− + F ·Q Dt 2 ρ D h Q 2 i ∂p ρ h+ = + ρF · Q Dt 2 ∂t 1.6 E QUATION OF S TATE p = RρT (thermally perfect gas) c p , c v = constants (calorically perfect gas) 2 P RESSURE DISTRIBUTION AND COMPRESSIBILITY 2.1 A SSUMPTIONS 1. Steady flow 2. Inviscid fluid 3. No discontinuities (shocks) 4. Perfect gas 5. One-dimensional motion 6. Adiabatic flow 7. F ≡ 0 8. Isentropic 2.2 N OTATION ( )0 = stagnation conditions, Q = 0 ( )∞ = free stream conditions, Q = u c = u ∞ c ( ) = conditions on body surface (airfoil) Q = u 0 i + u 0 j + ωk u 0 = u ∞ + γu 2 2.3 E NERGY E QUATIONS h=e+ p ρ i h 1 d h + Q2 = 0 2 (Heat content plus kinetic energy is constant) 2.4 P ERFECT G AS R ELATIONS p = ρRT pV = RT V≡ 1 ρ Can show, without effort: ρV γ = constant ³ 1 ´γ p = constant ρ p a 2 = γ , a = speed of sound ρ r ³ ´ Q= 2c p T0 − T h ³ p ´ γ−1 i T i γ = T0 1 − T0 − T = T0 1 − T0 p0 n h ³ p ´ γ−1 io 1 γ 2 Q = 2c p T0 1 − p0 h 2.5 M ACH N UMBER M2 = M2 = 2 2c p (T0 − T ) 2c p (T0 − T ) Q = = p 2 a γRT γρ 2c p ³T 0 ´ −1 = ´ 2 ³ T0 −1 (γ − 1) T γ(c p − c v ) T T0 h γ − 1 2i = 1+ M = β(γ, M ) T 2 γ γ p 0 ³ T0 ´ γ−1 = = β γ−1 p T ³ 1 ρ0 T0 ´ γ−1 1 = = β γ−1 ρ T 3 2.6 OTHER USEFUL FORMS , EXPRESSIONS Q 2 = 2c p (T0 − T ) p0 = γRT0 ρ0 Q 2 2c p ³ T ´ 2 ³ T ´ = 1 − = 1 − γR T0 γ−1 T0 a 02 ³ ´ T γ−1 Q 2 = 1− 2 a0 T0 γ h p γ − 1 ³ Q ´2 i γ−1 = 1− 2 a0 p0 1 ρ h γ − 1 ³ Q ´2 i γ−1 = 1− ρ0 2 a0 a 02 = γ a 2 = a 02 − γ−1 2 Q 2 2.7 P RESSURE , VELOCITY RELATIONS IN ISENTROPIC FLOW With some effort, one may show: γ h p γ−1 2 ³ Q 2 ´i γ−1 = 1+ M∞ 1 − 2 2 u∞ p∞ Expanding the right-hand side: p γ³ Q2 ´ 2 γ ³ Q 2 ´2 4 γ(2 − γ) ³ Q 2 ´3 6 γ(2 − γ)(3 − 2γ) ³ Q 2 ´4 8 = 1+ 1− 2 M ∞ + 1− 2 M ∞ + 1− 2 M ∞ + 1− 2 M ∞ +... 2 u∞ 8 u∞ 48 u∞ 384 u∞ p∞ Obtain an expression for cp = p − p∞ 1 2 2 ρ ∞ u∞ Let Q = u ∞ + γV, γV ¿1 U∞ Find c p and discuss its limitations. 3 S IMILARITY OF FLOWS 3.1 R EQUIREMENTS FOR SIMILARITY OF FLOWS 1. Similarity in boundary geometry Boundary of one flow can be made to coincide with that of another if its linear dimensions are multiplied by a constant 2. Dynamic constraint Dependent variables of one flow are proportional to those of another at the corresponding points. 4 Example Problem - Illustration Consider the dynamics of an incompressible fluid flow with constant. Equation of incompressibility: D p ∂ρ ∂ρ =0 = + ui Dt ∂t ∂x i Equation of continuity: ∂u i =0 ∂x i Introduce dimensionless variables: 0 ui = ui , U 0 ρ = ρ , ρ0 0 p = p , ρ 0U 2 0 xi = xi , L 0 t = tU L U , ρ 0 , L − reference quantities 3.2 L INEAR MOMENTUM 0 0 ³∂ 0 0 0 ∂ ´ 0 ∂p ρL γ ∂2 ρ + uα 0 ui = − 0 + 2 Fi + 0 0 ui U U L ∂x α ∂x a ∂t ∂x ∂x α i ∂ρ ∂t Froude no: F = pU −→ Reynolds no: gL Re = UγL 0 0 0 + uα ∂ρ 0 0 0 ∂x α =0 ∂u α 0 ∂x a =0 inertia forces gravity force −→ inertia force viscous force F and Re must be the same for both flows. This is sufficient for dynamic similarity along with similar boundary geometry. U , ρ 0 , L may be different for both flows. 4 E QUATIONS GOVERNING IRROTATIONAL FLOWS OF A HOMENTROPIC GAS For this class of flows the simplification is through the introduction of the velocity potential, φ, where Q = ∇φ or ui = ∂φ ∂x i and the vorticity is zero: ω = ∇ ×Q = ∇ × ∇φ = 0 where ω is the vorticity vector. The unsteady Bernoulli equation may be written, for this class of flows: ³1 ´ ∂Q 1 + ∇ Q 2 − Qxω = − ∇p ∂t 2 ρ 5 since, p = p(ρ), ω=0 ³1 ´ 1 ∂Q + ∇ Q 2 + ∇p = 0 ∂t 2 ρ or Z ³ ∂φ 1 ∂p ´ =0 ∇ + Q2 + ∂t 2 ρ therefore ∂φ 1 2 + Q + ∂t 2 Absorb f (t ) into φ and obtain ∂φ 1 2 + Q + ∂t 2 Z Z dp = f (t ) ρ dp = constant ρ Differentiate above equation with respect to time, t : ∂Q 1 ∂ρ ∂2 φ +Q · + a2 =0 2 ∂t ρ ∂t ∂t Expressing the continuity equation in terms of φ: 1 ∂p 1 + ∇2 φ + Q · ∇ρ = 0 ρ ∂t ρ Linear momentum equation rewritten yields Q· o 1 1 1 1 n ∂Q ∇ρ = 2 Q · ∇p = 2 Q − − (Q · ∇)Q ρ a ∂t ρ a Combining the above three equations yields: h i 1 ∂2 Φ 2 ∂Q 1 2 + Q · φ − = ∇ Q · (Q · ∇)Q a 2 ∂t 2 a 2 ∂t a2 since u i = ∂Φ ∂x i , the above equation may be written: ³ u 2 ´ ∂2 Φ ³ v 2 ´ ∂2 Φ ³ w 2 ´ ∂2 Φ uv ∂2 Φ v w ∂2 Φ uw ∂2 Φ ∗ 1− 2 + 1 − + 1 − − 2 − 2 − 2 = a ∂x 2 a 2 ∂y 2 a 2 ∂z 2 a 2 ∂x∂y a 2 ∂y∂z a 2 ∂x∂z u ∂2 φ v ∂2 φ w ∂2 φ 1 ∂2 φ + 2 + 2 + 2 a 2 ∂x∂t a 2 ∂y∂t a 2 ∂z∂t a 2 ∂t 2 where u= ∂Φ ∂x v= ∂Φ ∂y w= ∂Φ ∂z For steady flow of a calorically perfect gas: h 0 = constant Q2 = c p T0 2 γ−1 2 a 2 = a 02 − (Φx + Φ2y + Φ2z ) 2 Equation * is the potential-flow equation. cp T + 6 5 S MALL P ERTURBATION T HEORY We will consider a slender body immersed in a uniform flow, viz., in the uniform flow: Q = U∞ i in the perturbed flow: Q = ui + v j + w k 0 0 0 j k Q = (U∞ + u )i + v + w Q = ∇φ Now define a perturbation velocity potential, φ(x, y, z), where ∂φ ∂x 0 u = ∂φ ∂y 0 v = 0 w = ∂φ ∂z ∴ Φ(x, y, z) = U∞ x + φ(x, y, z) Using the notation in eqn(*): 0 u = U∞ + w = 0 v =v = 0 w =w = ∂Φ ∂φ = U∞ + ∂x ∂x ∂Φ ∂φ = ∂y ∂y ∂Φ ∂φ = ∂z ∂z Φxx = ∂2 φ = φxx ∂x 2 Φy y = ∂2 φ = φy y ∂y 2 Φzz = ∂2 φ = φzz ∂z 2 Φx y = ∂2 φ = φx y ∂x∂y Φy z = ∂2 φ = φy z ∂y∂z 7 ∂2 φ = φxz ∂x∂z Substituting Φ = U∞ x + φ and multiplying eqn(*) by a 2 we obtain the perturbation equation or perturbation velocity potential equation, for steady flow: Φxz = ∗∗[a 2 −(U∞ +φx )2 ]φxx +[a 2 −(φ y )2 ]φ y y +[a 2 −(φz )2 ]φzz −2(U∞ +φx )φ y φx y −2(U∞ +φx )φz φxz −2φ y φz φ y z = 0 Note a 2 may be expressed as: 0 0 0 0 γ−1 (2u U∞ + u 2 + v 2 + w 2 ) 2 γ − 1 2 a 2 = a∞ − (2φx U∞ + (φx )2 + (φ y )2 + (φz )2 ) 2 Also, note that eqn(**) is exact! It is also non-linear. 2 a 2 = a∞ − 5.1 P ERTURBATIONS Assume the perturbations are small, viz., 0 u ¿ 1; U∞ 0 v ¿ 1; U∞ 0 w ¿1 U∞ In the limit of small perturbations, we may neglect the terms containing squares of the perturbation velocities in comparison to those containing first powers. Eqn(**) with a 2 substituted becomes 2 2 (1−M ∞ )φxx +φ y y +φzz = M ∞ (γ+1) φy φx φx 2 2 2 φz φxx +M ∞ (γ−1) (φ y y +φzz )+2M ∞ φx y +2M ∞ φxz U∞ U∞ U∞ U∞ Note that each term on the right-hand side is non-linear. Each term on the right-hand side contains a perturbation velocity (φx , φ y , or φz ). Hence, we may neglect the right-hand side in comparison to the left-hand side. We obtain 2 (1 − M ∞ )φxx + φ y y + φzz = 0 5.2 QUESTIONS 1. What is the equation where M ∞ → 1? 2. What is the equation where M ∞ À 1? 6 B OUNDARY C ONDITIONS 1. The body surface is a stream line. (inviscid, irrotational flows) 2. Flow velocity must be tangent to body surface 3. Velocity vector has to be orthogonal to the unit normal of the body surface 8 The body surface is described by f (x, y, z) f (x, y, z) = 0 Boundary condition is expressed as Q · ∇ f (x, y, z) = 0 or ui ∂f =0 ∂x i Introducing the perturbation velocities u = U∞ + u v =v 0 w =w Substituting, 0 (U∞ + u ) 0 0 0 ∂f ∂f ∂f +v +w =0 ∂x ∂y ∂z 0 Since u ¿ U∞ , we may write: U∞ 0 ∂f 0 ∂f ∂f +v +w =0 ∂x ∂y ∂z This equation must be satisfied on the surface of the body. Consider the two-dimensional case: 0 w =0 ∂f =0 ∂z We obtain: Therefore u −1 U∞ 0 v ∂ f /∂x d y =− = ∂ f /∂y d x U∞ is the slope of the body (approximately) the slope of the streamline. Recall that 0 u = ∂φ dy ´ = U∞ ∂y d x BODY 0 Now for thin bodies, a small angle of attack, y BODY ≈ 0: this suggests an expansion of v (x, y) in a powers of y: ³ ∂v 0 ´ 0 0 v (x, y) = v (x, 0) + y + ... ∂y y=0 ³d y ´ 0 ∴ v(x, y) = v (x, 0) ∼ = U∞ d x BODY For three-dimensional planar flows ∂f ∼ =0 ∂z and the boundary condition becomes ³ ∂y ´ 0 v (x, 0, z) = U∞ ∂x BODY at infinity: 0 u →0 0 v →0 0 w →0 0 0 0 or w , v , and w are finite. 9 7 L INEARIZED P RESSURE C OEFFICIENT Let’s revisit the pressure coefficient, c p : cp ≡ p − p∞ 1 2 2 ρ ∞U∞ where p is the pressure (local, static) at the location or point of interest in the flow field. Note that c p is dimensionless. Since, U2 γ γ 1 1 γp ∞ 2 2 2 ρ ∞U∞ = ρ ∞U∞ = p ∞ 2∞ = p ∞ M ∞ 2 2 γp ∞ 2 a∞ 2 then cp = i 2 h p − 1 2 p γM ∞ ∞ for an inviscid, adiabatic, isentropic, steady flow and 0 0 0 Q = (U∞ + u )i + v j + w k We show that 1 1 2 h + Q 2 = h ∞ + U∞ 2 2 which for a calorically perfect gas leads directly to 2 −Q2 T γ − 1 U∞ = 2 2 a∞ T∞ 0 0 0 0 γ−1 = 1 − 2 [2u U∞ + u 2 + v 2 + w 2 ] 2a ∞ Isentropic flow conditions lead to: h T i γ p γ−1 = T∞ p∞ γ h p γ − 1 2 ³ 2u 0 u 02 + v 02 + w 02 ´i γ−1 = 1− M∞ + 2 2 U∞ U∞ p∞ In the case of small velocity perturbations, u0 ¿1 U∞ ³ u 0 ´2 U∞ n1 ³ v 0 ´2 n1 U∞ ³ w 0 ´2 n1 U∞ Using the binomial expansion, we show that p γ 2 ³ u0 u 02 + v 02 + w 02 ´ 2 + + ... = 1 − M∞ 2 2 U∞ U∞ p∞ therefore: 2u 0 U∞ Discuss the limitations implied in the above expression for c p . cp = − 10 8 C ROCCO ’ S T HEOREM Consider the motion of a fluid element. The fluid element may both translate and rotate. Let: v = translational velocity w = rotational velocity ω = angular velocity where 1 w = ∇×v 2 ∇ × v ≡ vorticity Combine Euler’s equation, first and second laws of thermodynamics: ρ ∂v + ρ(v · ∇)∇ = −∇p ∂t T ∇s = ∇h − v∇p = ∇h − h = h0 − We obtain: T ∇s = ∇h 0 − v × (∇ × v) + ∂v ∂t ∇p ρ v2 2 n Crocco’s Theorem For steady flow, we obtain T ∇s = ∇h 0 − v × (∇ × v) or v × (∇ × v) = ∇h 0 − T ∇s For two-dimensional, steady flows: 2w = 1 ³ ∂s ∂h 0 ´ T − v ∂n ∂n Vorticity =⇒ rates of change of entropy and stagnation enthalpy normal to the streamlines Flow over a supersonic blunt body: 11 For this flow, h 0 = constant ∂h 0 =0 h ∂s 6= 0 ∂n (why?) 12