MECH 314 Problems 1, Spring 2014

MECH 660 Advanced Fluid Mechanics March 17, 2004 American University of Beirut, Spring 2004 Handout # PS2 Solution Problem Set 2 Solution: Inviscid Flow “Advanced Fluid Mechanics Problems” by Shapiro and Sonin Problems 4.4, 4.7, 4.8, 4.9, 4.10, 4.13, 4.18, 4.19, 4.21, 4.23, 4.24, 4.28. Problem 4.4 Refer to Figure 1 for the schematic. The flow is 1D, inviscid and incompressible. Gravi- A1 A(x) x A2 x x+dx A(x) A(x) + dA p n x dS Figure 1: Schematic of Problem 4.4 tational effects are negligible. (a) Applying Bernoulli’s equation along a streamline from location x to location 2, p2 ρ + u2 2 2 = p(x) ρ + u2(x) 2 ⇒ p(x) − p(a) = ρ 2 � u2 2 − u2(x) � ⇒ p(x) − p(a) = ρQ2 2 � 1 A2 2 − 1 A2(x) � (b) We consider a differential volume between x and x+dx. The cross sectional areas are respectively A(x) and A(x) + (dA/dx)dx. The nozzle wall area onto which the pressure acts is dS so that the corresponding force x−component is dFx = p(x)dS ˆn · ˆx = p(x) dA, where ˆn is the unit vector normal to dS and pointing into the wall. The total force component in the x direction is Fx = � 2 1 dFx = � A2 A1 pdA = � A2 A1 ρQ2 2 � 1 A2 2 − 1 A2(x) � dA ⇒ Fx = ρQ2(A1 − A2)2 2A1A2 2 1 (c) The product p dA has always the same sign. • If A ↗, dA > 0 ⇒ u ↘⇒ p ↗ • If A ↘, dA < 0 ⇒ u ↗⇒ p ↘ 2 Problem 4.7 Refer to Figure 2 for the schematic. D d pa, ρa pa ρf ua uf 1 Figure 2: Schematic of Problem 4.7 The fuel-air mass flow rate ratio is α = ˙mf ˙ma = ρf uf (πd2/4) ρa ua (πD2/4) ⇒ d D = � αρa ρf ua uf �1/2 In order to determine the ration ua/uf, we apply Bernoulli’s equation along two stream lines, one for air from the ambient to location 1 in the vicinity of the fuel jet and one for fuel from the reservoir to location 1. for air pa ρa = p1 ρa + u2 a 2 for fuel pa ρf = p1 ρf + u2 f 2 so that ua uf = �ρf ρa �1/2 Then d D = α1/2 �ρa ρf �1/4 3 Problem 4.8 Refer to Figure 3 for the schematic. The flow is inviscid and incompressible. F D1 D2 1 2 u2 u1 Figure 3: Schematic of Problem 4.8 (a) We apply Bernoulli’s equation along a stream line starting from location 1 the top of the bellows and ending at location 2 at the nozzle exit. p1 ρ + u2 1 2 = pa ρ + u2 2 ρ ⇒ p1 − pa = ρ 2 � u2 2 − u2 1 � ⇒ F A1 = ρ 2 � u2 2 − u2 1 � (1) Next we apply the conservation of mass in integral form for a control volume that is moving with the piston d dt (ρV) + ρ u2 πD2 2 4 = 0 ⇒ u2 = − 4 πD2 2 dV dt By also applying conservation of mass in integral form for fixed control volume we find the relation u1 D2 1 = u2 D2 2 ⇒ u1 = − �D2 D1 �2 4 πD2 2 dV dt 4 Substituting expression for u1 and u2 into equation (1) dV dt = −πD2 2 4    2F ρA1 1 1 − � D2 D1 �4    1/2 Integrating from t = 0 to t = τ at which V = 0, τ = V D1 D2 2 � 2ρ πF �1/2 � 1 − �D2 D1 �4�1/2 ≃ V D1 D2 2 � 2ρ πF �1/2 (b) For STP air with V = ∞ liter, D1 = 10 cm, D2 = 1 cm, and F =2kgf: τ = 0.2 s. 5 Problem 4.9 Refer to Figure 4 for the schematic. The flow is inviscid and steady. 1 2 3 h Ta Ta + ∆T H d g Figure 4: Schematic of Problem 4.9 (a) We apply Bernoulli’s equation along the stream line 2 − 3: � 3 2 dp + � 3 2 ρ g z ds + � 3 2 1 2ρ u2 ds = 0 ⇒ (p3 − p2) + ρH g h + 1 2u2 3 ≃ 0 ⇒ g h (ρa − ρH) ≃ 1 2ρH u2 3 ⇒ u3 ≃ � 2 g h � ρa ρH − 1 ��1/2 ⇒ u3 ≃ � 2 g h∆T Ta �1/2 6 where p3 − p2 = −ρa g h, ρ2 = ρ3 = ρH, u2 2 << u2 3, and pa = ρaRTa = ρHR (Ta + ∆T). (b) In the case the cap is closed, � 3 2 dp + � 3 2 ρ g z ds + � 3 2 1 2ρ u2 ds = 0 ⇒ (p3 − p2) + ρH g h ≃ 0 ⇒ p3 ≃ pa − ρHg h ⇒ p3 − p3a ≃ g h (ρa − ρH) ⇒ (∆p)cap ≃ ρa g h ∆T Ta + ∆T where p2 = pa, p3a = pa − ρa g h, ρ2 = ρ3 = ρH, u2 ≃ 0, u3 = 0 and pa = ρaRTa = ρHR (Ta + ∆T). 7 Problem 4.10 Refer to Figure 5 for the schematic. The flow is inviscid and quasi-steady. 1 2 pa V p(t) A u2 Figure 5: Schematic of Problem 4.10 We apply Euler’s equation along a streamline from 1 inside the vessel to 2 at the exit � 2 1 dp ρ + g(z2 − z1) + 1 2 � u2 2 − u2 1 � + � 2 1 ∂u ∂t ds = 0 We neglect gravitational effects. The flow is quasi-steady so that � 2 1 ∂u ∂t ds ≃ 0. At location 1, u1 = 0. Then −2 � 2 1 dp ρ = u2 2 (2) Now we find an expression for u2 by employing the integral form of the conservation of mass for a fixed control volume d dt (ρV) + ρu2A = 0 ⇒ V dρ dt + ρu2A = 0 ⇒ u2 = − V A dt dρ ρ (3) Substituting expression for u2 (equation (3)) into equation (2) � −2 � 2 1 dp ρ �1/2 = − V A dt dρ ρ (4) (a) We employ p = ρ R Ta and noting that dρ/ρ = dp/p then � −2R Ta � 2 1 dp p �1/2 = − V A dt dp p ⇒ � 2R Ta ln p1 p2 �1/2 = − V A dt dp p ⇒ � 2R Ta ln p pa �1/2 = − V A dt dp p ⇒ A V (2R Ta)1/2 dt = − 1 p � ln p pa �1/2dp where p2 = pa, p1 = p(t), p′(t) = p(t) − pa. Integration from t = 0 to t and p from pi to p, we get A V (2R Ta)1/2 t = 2 �� ln pi pa − � ln p pa � 8 (b) We start with equation (4 ) � −2 � 2 1 dp ρ �1/2 = − V A dt dρ ρ Noting that p/pi = (ρ/ρi)γ ⇒ dρ/ρ = (1/γ)dp/p then � 2γ γ − 1 pi ρi �1/2   � p pi � γ−1 γ − �pa pi � γ−1 γ   1/2 = − V γ A dt dp p ⇒ −γ A V � 2γ γ − 1 pi ρi �1/2 dt = p−1   � p pi � γ−1 γ − �pa pi � γ−1 γ   −1/2 dp Performing Taylor series expansion in p′/pa around zero and integrating from t = 0 to t yields −γ A V � 2γ γ − 1 pi ρi �1/2 t = 2 (α pa)1/2 � p′1/2 − p′ i 1/2� ⇒ p′ =  p′ i 1/2 − �pa 2 �1/2 γ A V �pi ρi �1/2 t   2 where α = (γ − 1)/γ. 9 Problem 4.13 Refer to Figure 6 for the schematic. The flow is two-dimensional, inviscid and steady. g air, pa z x ds dx αs α zs - zw h Figure 6: Schematic of Problem 4.13 (a) We consider a stream line along the free surface, Euler’s equation in differential form is 1 ρ dp ds + d ds(gzs) + 1 2 d(u2) ds = 0 where the subscript s denotes the free-surface. Since p ≃ pa on the surface, then gdzs ds + 1 2 d(u2) ds = 0 One the surface ds = dx/ cos αs so that gdzs dx + 1 2 d(u2) dx = 0 Conservation of mass Q = uh ⇒ 1 2 d(u2) dx = 1 2 d dx �Q2 h2 � = −Q2 h3 dh dx = −u2 h dh dx Noting that h ≃ zs − zw then gdzs dx ≃ u2 h �dzs dx − dzw dx � ⇒ dzs dx ≃ u2 gh u2 gh − 1 tan α where tanα = dzw/dx. 10 A Vorticity form of Euler’s equation Euler’s equation in differential form is ∂u ∂t + (u · ∇) u = −1 ρ∇p + g We invoke the identity u × (∇ × u) = 1 2∇ (u · u) − (u · ∇) u Noting that ω = ∇ × u then ∂u ∂t − u × ω = −1 ρ∇p − ∇(gz) − 1 2∇|u|2 Next we take the curl of the above equation ∂ω ∂t − ∇ × u × ω = −∇ × �1 ρ∇p � We invoke another identity ∇ × u × ω = u(∇ · ω) + (ω · ∇)u − ω(∇ · u) − (u · ∇)ω Noting that ∇ · ω = 0 and ∇ · u = − 1 ρ Dρ Dt from the continuity, then ∂ω ∂t + (u · ∇)ω = (ω · ∇)u + ω ρ Dρ Dt − ∇ × �1 ρ∇p � Dω Dt − ω ρ Dρ Dt = (ω · ∇)u − ∇ × �1 ρ∇p � We invoke the identity ∇ × (aA) = a(∇ × A) + (∇a) × A so that ∇ × �1 ρ∇p � = 1 ρ(∇ × ∇p) + ∇ �1 ρ � × ∇p = − 1 ρ2∇ρ × ∇p So that Dω Dt − ω ρ Dρ Dt = (ω · ∇)u + 1 ρ2∇ρ × ∇p Noting that Dω Dt − ω ρ Dρ Dt = ρ D Dt �ω ρ � Then ρ D Dt �ω ρ � = (ω · ∇)u + 1 ρ2∇ρ × ∇p + ∇ × Fb (17) where body force Fb was included in the equation. Analysis of equation (17) 22 • If the body force is conservative (∇ × Fb = 0) and the flow is incompressible, then Dω Dt = (ω · ∇)u (18) • If additionally the flow is steady (ω · ∇)u = 0 (19) • If the body force is conservative (∇×Fb = 0) and the flow is barotropic (∇ρ×∇p = 0), then D Dt �ω ρ � = �ω ρ · ∇ � u (20) 23