MECH 314 Final, Fall 2006

MECH 411 Introduction to Fluid Mechanics February 1, 2006 American University of Beirut, Fall 2005-2006 Handout # Final Final Exam • This exam is an open book exam. • All problems have equal weight. • You have 3 hours. Problem 1 Air is blown downwards at a volume flow rate Q through the pipe of diameter d as shown in Fig 1. After leaving the pipe, air flows radially outwards (in an axisymmetric fashion) between two parallel disks of diameter D and separation h. Given: D = 10 cm, h = 5 mm, d = 1 cm, Q = 0.05 m3/s, ρ = 1.225 kg/m3, µ = 1.789 × 10−5 Pa.s. (a) Find the average velocity V (r) as a function of radial position r. What is Reynolds number (based on channel height h) as a function of radial position r? For the given parameters, is the flow laminar or turbulent? Would the flow be modeled as a viscous or inviscid flow ? (b) Find the pressure distribution p(r) as a function of radial position r. (c) What is the force (due to pressure) acting on the lower disk ? [Hint: account only for pressure variation from r = d/2 to r = D/2.] Q d D atmosphere pa Q h disk of mass M supported by the flow fixed r z Figure 1: Schematic for Problem 1. 1 Problem 2 A piston in the cylinder of an internal combustion engine slides back and forth on a thin film of oil that fills that gap between the piston surface and the cylinder wall, as depicted in Fig. 2. The gap thickness h is uniform along the length L and circumference of the piston and is much smaller than both of these dimensions. We wish to determine the velocity of the oil in the gap. Assume that the inertia of the oil is negligible compared to the pressure and viscous forces, and choose a reference system fixed in the moving piston as shown in the figure. In this coordinate system, the cylinder wall moves with the speed uc: uc = um sin ωt where um is the maximum piston speed and ω is the angular frequency of the crankshaft. The pressure below the piston is atmosphere pressure pa while that above the piston in the combustion chamber, pc, is pc = pa + pm 2 (1 − cos ωt) where pm is the maximum pressure increase in the cylinder, which occurs at ωt = π. Assuming that the oil flow in the gap is a combination of plane Couette and Poiseuille flows, (a) derive an expression for the velocity distribution u(y, t) in terms of the parameters pm, um, ω, h, L and the viscosity µ. (b) Derive an expression for the time-averaged volumetric flow rate ¯Q of oil past the piston. pc pa L h pc pa h L y x uc Figure 2: Schematic for Problem 2. 2 Problem 3 The drag (F) on a submarine moving well below the free surface is to be determined by a test on a model scaled to 1/20 th of the prototype. The test is to be carried out in a water tunnel. Determine the ratio of model to prototype drag (Fm/Fp) needed for finding the prototype drag when the speed of the prototype is 2.57 m/s. The kinematic viscosity of sea water is 1.3×10−6 m2/s, and the density is 1010 kg/m3 at the depth of the prototype. The water in the tunnel has a temperature of 50 ◦C. If the model submarine were to be tested in a wind tunnel where, at free stream, p = 2000 kPa, and T = 50◦C, what will the ratio Fm/Fp have to be ? Note: Only the Reynolds number and the Euler number (∆p/ρV 2 ≡ (F/L2)/ρV 2) are im- portant for dynamic similarity, where L =submarine length. Explain briefly why the Mach number and the Froude number may be neglected. 3 Problem 4 Water at 10◦C is to flow from reservoir A to reservoir B through a cast-iron pipe of length 20 m at a rate of Q = 0.0020 m3/s as shown in Fig. 3. The system contains a sharp-edged entrance and six regular threaded 90 ◦ elbows. Determine the pipe diameter needed, taking Kelbow = 1.5 [Hint: the friction factor lies between 0.030 and 0.035.] B A total length = L = 20 m Elevation z2 = 0 m Elevation z1 = 2 m D (2) (1) Figure 3: Schematic for Problem 4. 4 Problem 1 Solution (a) Taking control volume as shown, conservation of mass Q = 2πrhV (r) ⇒ V (r) = Q 2πrh Reynolds number is then given by Reh(r) = ρV (r)h µ = ρQ 2πµr At r = R, Reh(r = R) = ρQ 2πµR = 10898 also Reh(r = d/2) = 108980. (b) Bernoulli along a streamline from point (1) at radius r to point (2) at radius D/2 p(r) + 1 2ρV (r)2 = pa + 1 2ρV 2 2 From part (a), V2 = V (r = D/2) = Q πDh, then p(r) − pa = ρQ2 2π2h2 � 1 D2 − 1 4r2 � (c) The upward force on the disk due to pressure is F = − � D/2 r0 (p(r) − pa)2πr dr = − � D/2 d/2 ρQ2 2π2h2 � 1 D2 − 1 4r2 � 2πr dr = − ρQ2 4πh2 �1 2 � 1 − d2 D2 � + ln d D � ≃ ρQ2 4πh2 � ln D d − 1 2 � leading to F = 17.57 N. 5 Problem 2 Solution The flow is fully developed with v = 0, so that p = p(x). Momentum in x−direction, after neglecting the interia term, and ∂2u/∂x2: dp dx = µ∂2u ∂y2 ⇒ u = 1 2µ dp dxy2 + C1y + C2 The boundary conditions are: u = 0 at y = 0 and u = uc at y = h, we C2 = 0 uc = 1 2µ dp dxh2 + C1h ⇒ C1 = − 1 2µ dp dxh + uc h The velocity field is then given by u(y, t) = 1 2µ dp dx(y2 − hy) + uc y h The volume flow rate is Q(t) = πDp � h 0 u(y, t) dy = πDp � − h3 12µ dp dx + uc h 2 � with −dp/dx = pm 2L (1 − cos ωt) and uc = um sin ωt, Q(t) = πDp � h3 12µ pm 2L(1 − cos ωt) + umh 2 sin ωt � The time averaged volume flow rate is obtained by averaging over one period ¯Q = ω 2π � 2π/ω 0 Q(t) dt = πDp h3 12µ pm 2L 6