MAE 101A. Homework 7 - Solutions 3/12/2018

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1 MAE 101A Homework 7 - Solutions 3/12/2018

2 Munson 6.31: The stream function for a two-dimensional, nonviscous, incompressible flow field is given by the expression ψ = 2(x y) where the stream function has the units of ft 2 /s with x and y in feet. (a) Is the continuity equation satisfied? (b) Is the flow field irrotational? If, so, determine the corresponding velocity potential. (c) Determine the pressure gradient in the horizontal x direction at the point x = 2 ft, y = 2 ft. Determine Given Model Verify continuity, ψ = 2(x y) Steady state: / t = 0 Rotation vector ω x = 2 ft Incompressible: Dρ/Dt = 0 Velocity potential, φ y = 2 ft Inviscid: ν 2 V = 0 / x 2D flow: w = 0, / z = 0 Basic Equations ρ t + (ρv ) = 0 ρ DV Dt = p + ρg + μ 2 V u = ψ y and u = φ x and ω = 1 V 2 v = ψ x v = φ y a. Consider conservation of mass ρ t + (ρv ) = 0 1 Dρ ρ Dt + V = 0 Since the flow is incompressible the density of a fluid particle is constant, Dρ/Dt = 0, which implies the flow is divergence free. Since the flow is 2D, w/ z = 0 V = 0 u x + v y + w z = 0 u x + v y = 0 Using the definition of the stream function and the given ψ(x, y). u = ψ ψ = 2 ft/s and v = = 2 ft/s y x x ( ψ y ) + ψ ( y x ) = 0 Eq. 1 The stream function is defined to satisfy conservation of mass for incompressible flows. 1

3 b. The rotation vector is Since the flow is 2D, w = 0 and / z = 0. From Eq. 1: ω = 1 2 V = 1 2 ( v z w y ) i 1 2 ( u z w x ) j ( u y v x ) k ω = 1 2 V = 1 2 ( u y v x ) k ω z = 1 2 ( u y v x ) = 1 (0 0) 2 ω z = 0 Yes, the flow is irrotational since ω = 0. Since the flow is irrotational the velocity potential exists and is defined as, u = φ x and v = φ y u = ψ = 2 so φ = 2x + f(y) y v = ψ = 2 so φ = 2y + g(x) x φ = 2(x + y) + C c. To find the / x consider Navier-Stokes in the x-direction ρ ( u u u u + u + v + w t x y z ) = x + ρg x + μ ( 2 u x u y u z 2) Since flow is steady ( / t = 0), 2D ( / z = 0), inviscid (μ = 0), and g = gk so g x = 0. u u = ρ (u + v x x y ) Since u = 2 ft/s there are no gradients in the velocity field. x = 0 where C is an arbitrary constant Velocity potential (and stream function) simplifies the analysis of a flow by reducing the number of variables needed to describe the flow field. For a 2D planar flow that is incompressible and irrotational, the two components of velocity, u(x, y) and v(x, y), can be replaced with the one function φ(x, y) (or ψ(x, y)) 2

4 Munson 6.45: For a free vortex (see Video V6.4) determine an expression for the pressure gradient (a) along a streamline, and (b) normal to a streamline. Assume that the streamline is in a horizontal plane, and express your answer in terms of the circulation. Determine Model Diagram s and n Free vortex Steady Incompressible Irrotational Basic Equations Inviscid ρ DV = p + ρg + Dt μ 2 V v r = 0 and v θ = 2πr For a free vortex the velocity is defined as v r = 0 and v θ = 2πr a. The streamlines corresponding to a free vortex are concentric circles. As a result 3 = s θ gradient must satisfy conservation of momentum. Consider the θ component of Navier-Stokes. ρ ( v θ t + v v θ r + v θ v θ v rv θ v + v θ r r θ r z ) = 1 z r Plugging in v r = v z = 0, / t = 0, and μ = 0 θ + ρg θ + μ [ 1 r ρ ( v θ v θ ) = 1 r θ r θ + ρg θ The velocity has no theta dependency so v θ / θ = 0, g = gk so g θ = 0 0 = 1 r θ = = 0 s θ (r v θ ) v θ v θ r r r 2 r 2 θ 2. The pressure + 2 r 2 v r θ + 2 v θ z 2 ] As expected there is no streamline pressure gradient. The fluid is traveling in a closed path at steady speed. b. The streamlines corresponding to a free vortex are concentric circles. As a result. The pressure = n r gradient must satisfy conservation of momentum. Consider the r component of Navier-Stokes. ρ ( v r + v v r t r + v θ v r r r v 2 θ θ r + v z v r z ) = Plugging in v r = v z = 0, / t = 0, and μ = 0 ρ ( v θ 2 r + ρg r + μ [ 1 r r ) = r + ρg r (r v r ) v r v r r r r 2 r 2 θ 2 2 r 2 v θ θ + 2 v r z 2 ]

5 Since g = gk, g r = 0 = v 2 θ ρ r r = v r θ 2 ρ = ( r 2πr )2 ρ r = = n r ρ2 4π 2 r 3 A pressure gradient exists in the normal direction to provide the forcing necessary to maintain the circular path. Since / r > 0, pressure increases away from the vortex center, i.e., the vortex center is a pressure minimum. Note: If Euler s Eq. (F = ma) in streamline coordinates is used: The same results are obtained: 1 ρ s 1 ρ n z V g = V s s z + g n = V2 R s = 0 n = ρv2 R = ρ2 4π 2 r 3 4

6 Munson 6.46: (See The Wide World of Fluids article titled Some Hurricane Facts, Section ) Consider a category five hurricane that has a maximum wind speed of 160 mph at the eye wall, 10 mi from the center of the hurricane. If the flow in the hurricane outside of the hurricane s eye is approximated as a free vortex, determine the wind speeds at locations 20 mi, 30 mi, and 40 mi from the center of the storm. Determine Model Diagram Velocity field of Free vortex category 5 hurricane Steady Incompressible Irrotational Inviscid Basic Equations v θ = 2πr For a free vortex the velocity component in the θ direction is v θ = = k where k = 2πr r 2π At the eye wall, r A = 10 mi, the velocity is v θ = 160 mi/. This boundary condition fixes the unknown constant k. k = rv θ k = (10 mi) (160 mi mi2 ) = 1600 As you move away from the center of the hurricane the velocity decreases according to v θ = k/r At r = 20 mi At r = 30 mi At r = 40 mi v θ = (1600 mi2 /) 20 mi v θ = (1600 mi2 /) 30 mi v θ = (1600 mi2 /) 40 mi = 80 mi = 53 mi = 40 mi This simple model shows that the wind speed of a hurricane decays as 1/r. Although there is a 1/r dropoff, a category 5 hurricane is so strong that even at 40 mi the airspeed is still quite large at 40 mph. 5

7 Munson 6.50: Show that the circulation of a free vortex for any closed path that encloses the origin is. Determine Model Diagram Circulation of Free vortex free vortex, Steady Incompressible Irrotational Inviscid Basic Equations = (V n )ds c V = v r e r + v θ e θ = 2πr e θ Consider the circulation around some arbitrary closed curve C = (V n )ds c = (v r n r + v θ n θ )ds c = v θ n θ ds c Only the θ component contributes to the circulation. Locally projecting the curve, ds, onto an arc in the θ direction it can be seen that n θ ds = rdθ. To encircle the origin the curve is integrated from 0 to 2π. = = 2π 0 2π 0 2πr rdθ 2π dθ = 2π [θ] 0 2π = A free vortex has a velocity field in which the fluid is irrotational (except at the center) and has circular streamlines. Since the fluid is irrotational the circulation of any closed curve is a constant value. The magnitude of can be interpreted as the strength of the vortex with larger values resulting in higher velocities and larger normal pressure gradients. dr rdθ ds C 6

8 Munson 6.57: A 15-mph wind flows over a Quonset hut having a radius of 12 ft and a length of 60 ft, as shown in Fig. P6.57. The upstream pressure and temperature are equal to those inside the Quonset hut: psia and 70. Estimate the upward force on the Quonset hut. Find the location θ on the roof of the Quonset hut where the pressure is p. Determine Given Model Upward force, F y V = 15 mph = 22 ft/s Steady flow Locations where R = 12 ft Incompressible p(θ) = p L = 60 ft Inviscid p = psia Irrotational T = 70 Flow over cylinder Basic Equations v r = V cos(θ) (1 R 2 /r 2 ) v θ = V sin(θ) (1 + R 2 /r 2 ) F = pda p = ρrt p + 1 γ 2g V2 + z = constant Diagram Consider the potential flow solution for flow over a circular cylinder, obtained by superposition of a uniform flow and a doublet (source and sink) v r = V cos(θ) (1 R 2 /r 2 ) and v θ = V sin(θ) (1 + R 2 /r 2 ) Along the surface of the cylinder (hut) the coordinate r is the radius of the cylinder (hut): r = R v r = 0 and v θ = 2V sin(θ) As expected v r = 0 there is no flow tough the solid surface. Notice the no slip boundary condition is not satisfied because the fluid is assumed inviscid. Applying Bernoulli s to get the pressure distribution, Neglecting changes in elevation z constant p γ + 1 2g V 2 + z = p γ + 1 2g V2 + z p = p + ρ 2 (V 2 V 2 ) p = p + ρ 2 (V 2 4V 2 sin 2 θ) This pressure distribution is the exterior pressure on the outside surface of the hut p ext = p = p + 1 ρv 2 2 (1 4 sin 2 θ) Consider the net force acting on a differential surface da of the hut roof. df net = (p int p ext )da The vertical component is df y = sin θ df df y = [p p 1 ρv 2 2 (1 4 sin 2 θ)] sin(θ) da 7 df = PdA p ext da p int da df θ df y df x

9 This force is integrated from θ = π to θ = 0, where da = LRdθ Using the trig identity: sin 2 (θ) = 1 cos 2 (θ) Substitute x = cos θ and dx = sin(θ) dθ F y = 1 ρv 2 2 π LR (1 4 sin 2 θ) sin(θ) dθ 0 F y = 1 ρv 2 2 π LR (1 4[1 cos 2 (θ)]) sin(θ) dθ 0 F y = 1 ρv 2 2 π LR [4 cos 2 (θ) 3] sin(θ) dθ 0 F y = 1 ρv LR (4x 2 3) dx 1 F y = 1 2 ρv 2 LR [ 4 3 x3 3x] 1 1 F y = 1 2 ρv 2 LR [ ] = 5 3 ρv 2 LR F y = 5 3 ρv 2 LR The density of the air can be determined using the ideal gas law or simply looked up in Table B3, ρ = P RT = slug ft 3 F y = 5 slug ( ) (22 ft 3 ft 3 s )2 (60 ft)(12 ft) [ 1 lbf s2 ] 1 slug ft F y = lbf The locations where the local net force on the structure vanish can be found by setting df = 0 df = 1 2 ρv 2 (1 4 sin 2 θ)da = 0 The above will be satisfied when the term in parenthesis goes to zero. θ = sin 1 ( 1/4) = 30 and 150 Potential flow theory was applied to derive the equations for the pressure distribution on the surface of a cylinder. One key assumption required to use potential flow is that the fluid is inviscid. In many cases potential flow can be applied since viscous effects are often confined to a thin region called the boundary layer. As the flow decelerates over the back of the cylinder fluid in the boundary layer has insufficient momentum to remain attached to the surface resulting in a wake where viscous effects are relevant. As a result the lift force calculated in this problem should be taken as a rough estimate since the pressure distribution on the rear of the hut is wrong. More will be learned about boundary layers in MAE 101B. 8

10 H7.1: The superposition of a free vortex (with > 0 for CCW flow) and a line sink (withm = Q/b < 0) gives a simple model for a tornado. Determine: a. stream function and velocity potential b. velocity field c. Consider a flow with = ft 2 /s and m = ft 2 /s. If the tornado occurs in sea-level standard air, what is the local velocity (mph) and pressure (psig) at a location 70 ft from the center of the flow? Determine Given Model Stream function ψ = ft 2 /s Steady Velocity potential φ m = ft 2 /s Two-dimensional Velocity field v r and v θ r = 70 ft Incompressible 3 slug v and p at r = 70 ft ρ = ft 3 Inviscid Basic Equations Vortex: φ = 2π θ ψ = 2π ln(r) Sink: φ = m 2π ln(r) ψ = m 2π θ v r = φ r ψ and φ v θ = 1 r θ v θ = ψ r v r = 1 and r θ p + 1 γ 2g V2 + z = constant a. Determine the velocity potential and stream function Sea level STP φ tornado = φ vortex + φ sink φ = 2π θ + m 2π ln(r) ψ tornado = ψ vortex + ψ sink ψ = 2π ln(r) + m 2π θ b. Determine the velocity field v r and v θ v r = φ r = 1 ψ r θ v r = r ( 2π θ + m m ln(r)) = 0 + 2π 2πr v r = m 2πr c. Evaluate the velocity at r = 70 ft 9 v θ = 1 r v θ = 1 φ r θ = ψ r θ ( 2π θ + m 2π ln(r)) = v θ = 2πr 2πr + 0 v r = m 2πr = ft 2 /s 1 mph [ 2π(70 ft) ft/s ] v θ = 2πr = ft 2 /s 1 mph [ 2π(70 ft) ft/s ] v r = mi V = v r 2 + v θ 2 = mi v θ = 26.4 mi

11 Evaluate the pressure p at r = 70 ft by using Bernoulli s Equation. p 1 γ + 1 2g V z 1 = p 2 γ + 1 2g V z 2 At location 2, far away from the tornado, p 2 = p atm and V 2 = 0 p 1 p atm = ρ 2 V 1 2 slug ( p 1,gage = p 1 p atm = 2 ft 3 ) [108.7 mi p 1,gage = psig ]2 [ ft 2 s ] 1 mph [ 1 lbf s2 1 slug ft ] [ 1 ft 12 in ]2 One great advantage of potential flow theory is the principle of superposition, which allows complex flows to be built up from simpler flows with known solutions. It also allows for the use of Bernoulli s Eq since the flow is steady, incompressible, inviscid, and irrotational. A singularity exists in this simple model at the tornado center. Here the model breaks down and an alternative method is needed to find the minimum pressure or maximum velocity. Despite this limitation potential flow theory provided us a model to describe most of the flow field. 10

Offshore Hydromechanics Module 1

Offshore Hydromechanics Module 1 Dr. ir. Pepijn de Jong 4. Potential Flows part 2 Introduction Topics of Module 1 Problems of interest Chapter 1 Hydrostatics Chapter 2 Floating stability Chapter 2 Constant