AUTO400 Homework Assignment 2 SOLUTION

Department of Mechanical Engineering AMEE401 / AUTO400 Aerodynamics Instructor: Marios M. Fyrillas Email: [email protected]...

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Department of Mechanical Engineering AMEE401 / AUTO400 Aerodynamics

Instructor: Marios M. Fyrillas Email: [email protected] HOMEWORK ASSIGNMENT #2

QUESTION 1

Clearly there are two mechanisms responsible for the drag and the lift, the pressure and the shear stress:

The drag force and the lift force on an object can be obtained by:

 p cos dA    w sin  dA L    p sin  dA    w cos  dA D

LIFT: Most common lift-generating devices (i.e., airfoils, fans, spoilers on cars, etc.) operate in the large Reynolds number range in which the flow has a boundary layer character, with viscous effects confined to the boundary layers and wake regions. For such cases the wall shear stress,  w , contributes little to the lift. Most of the lift comes from the surface pressure distribution, as justified through Bernoulli’s equation. For objects operating in very low Reynolds number regimes (i.e. Re  1 ) viscous effects are important, and the contribution of the shear stress to the lift may be as important as that of the pressure. Such situations include the flight of minute insects and the swimming of microscopic organisms. DRAG: Similarly, when flow separation occurs, i.e. a blunt body or a streamlined body at a large angle of attack, the major component of the drag force is pressure differential due to the low pressure in the flow separation region. On the ultimately streamlined body (a zero thickness flat plate parallel to the flow) the drag is entirely due to the shear stress at the surface (boundary layers) and, is relatively small.

QUESTION 2 Air at standard conditions flows past a flat plate as is indicated in the figure. If the pressure and shear stress distributions on the surface are as indicated (obtained either by experiment or theory), determine the lift and drag on the plate in case: a. The plate is parallel to the upstream flow, and b. It is perpendicular to the upstream flow. c. If the flat plate were oriented at an arbitrary angle relative to the upstream flow as indicated in Figure (c). Assume that the incident angle is small and use the stress profile as defined in (a). To determine an approximate pressure profile, modify the coefficient 35.6 used in (b) appropriately using Bernoulli’s equation. (a) 3.0m 7.6 m/s

1.2m  0.0331/ x N/m 2

 y2  2 p  35.6  1   N/m  0.36   

7.6m/s

The drag force and the lift force on an object can be obtained by:

 p cos dA    w sin  dA L    p sin  dA    w cos  dA D

42.8 N/m2

For the horizontal plate, the angle  is   90o on the top surface and   270o on the bottom surface so that the lift and the drag are given by

 p dA  

L

top

D

bottom

  w dA   top

p dA  0 (because the pressure is p =0)

  w dA (because of symmetry)

 w dA  2

bottom

top

where we have used the fact that because of symmetry the shear stress distribution is the same on the top and the bottom surfaces, as is the pressure also (whether we use gage or absolute pressure). There is no lift generated (the plate does not know up from down). With the given shear stress distribution, we obtain: 1.2

D2

 0

1.2

0.0331 dA  2 x

 0

  35.6  0.6  0.6

D

0.0331 3dx  3*0.0662*2 x



x

1.2 0

 6*0.0662 * 1.2  0.4351 N

  y2  2 2 1   N/m  ( 42.8 N/m ) 3 dy   0.36     0.6

 y3   35.6  42.8 *1.2 *3 - 35.6*3*   240N 3   -0.6

Clearly there are two mechanisms responsible for the drag. On the ultimately streamlined body (a zero thickness flat plate parallel to the flow) the drag is entirely due to the shear stress at the surface (boundary layers) and, in this example, is relatively small. For the ultimately blunted body (a flat plate normal to the upstream flow) the drag is entirely due to the pressure difference between the front and back portions of the object and, in this example, is relatively large. If the flat plate were oriented at an arbitrary angle relative to the upstream flow as indicated in Fig. c, there would be both a lift and a drag, each of which would be dependent on both the shear stress and the pressure. Both the pressure and shear stress distributions would be different for the top and bottom surfaces.

QUESTION 3 Consider inviscid, uniform flow around a rotating circular cylinder. The pressure distribution on the surface can be obtained by Bernoulli’s equation as   2 1 2 sin   2 2 2 . ps  po  U 2  1  4sin 2   2  aU 4 a U   Determine the Drag and the Lift force. Explain your results.



2



2

0

0

cos  d  0,



2

0

sin  d ,

cos  sin  d  0,

Fx   

2

0

Fy  

2

0



2

0



2

0

sin 2  d   ,

cos  sin 2  d  0,





2

0

2

0

cos2  d  

sin 3  d  0

ps cos  d  0

ps sin  d  U 

The Drag force is zero because the flow is assumed to be inviscid. The Lift force is a consequence of the differential pressure between the upper and lower sides of the cylinder.

QUESTION 4

QUESTION 5 a. Describe the character of flow past an object The flow past a blunt object (such as a circular cylinder) varies with Reynolds number. In general, the larger the Reynolds number, the smaller the region (boundary layer) of the flow field in which viscous effects are important. For objects that are not sufficiently streamlined, however, an additional characteristic of the flow is observed. This is termed flow separation and is illustrated in Fig. 9.6. Low Reynolds number flow past a circular cylinder is characterized by the fact that the presence of the cylinder and the accompanying viscous effects are felt throughout a relatively large portion of the flow field. As is indicated in Fig. 9.6a, for Re  0.1 the viscous effects are important several diameters in any direction from the cylinder. A somewhat surprising characteristic of this flow is that the streamlines are essentially symmetric about the center of the cylinder—the streamline pattern is the same in front of the cylinder as it is behind the cylinder. As the Reynolds number is increased, the region ahead of the cylinder in which viscous effects are important becomes smaller, with the viscous region extending only a short distance ahead of the cylinder (boundary layer region). The viscous effects are convected downstream and the flow loses its symmetry. Another characteristic of external flows becomes important - the flow separates from the body at the separation location as indicated in Fig. 9.6b. With the increase in Reynolds number, the fluid inertia becomes more important and at some location on the body, denoted the separation location, the fluid’s inertia is such that it cannot follow the curved path around to the rear of the body. The result is a separation bubble behind the cylinder in which some of the fluid is actually flowing upstream, against the direction of the up-stream flow. At still larger Reynolds numbers, the area affected by the viscous forces is forced farther downstream until it involves only a thin (  D ) boundary layer on the front portion of the cylinder and an irregular, unsteady (perhaps turbulent) wake region that extends far down-stream of the cylinder. The fluid in the region outside of the boundary layer and wake region flows as if it were inviscid. Of course, the fluid viscosity is the same throughout the entire flow field. Whether viscous effects are important or not depends on which region of the flow field we consider. The velocity gradients within the boundary layer and wake regions are much larger than those in the remainder of the flow field. Since the shear stress (i.e., viscous effect) is the product of the fluid viscosity and the velocity gradient, it follows that viscous effects are confined to the boundary layer and wake regions. The characteristics described in Fig 9.6 are typical of flows past blunt bodies. The nature of the flow depends strongly on the Reynolds number. Most familiar flows are similar to the large Reynolds number flows depicted in Fig 9.6c, rather than the low Reynolds number flow situations.

FIGURE 9.6 Character of the steady, viscous flow past a circular cylinder: (a) low Reynolds number flow, (b) moderate Reynolds number flow, (c) large Reynolds number flow.

b. Distinguish the two components of the Drag force and explain their physical origin. For two dimensional objects, the drag is composed of a frictional component, related to viscous shearing in boundary layers, and pressure drag, which is related to the pressure differential between the fore and aft of the body. The character of the flow dictates the relative importance of the two components, i.e. whether the flow separates or remains attached to the object. The main characteristic of a separated flow is a separation bubble/region behind the object in which some of the fluid is actually flowing upstream, against the direction of the upstream flow (see figure). Because of separation, the average pressure on the rear half of the body is considerably less than that on the front half. Thus a large pressure drag is developed and the viscous shear drag may be quite small. c. Explain the meaning of “Streamline” and “Bluff” bodies and give examples. A Streamline body is defined as a body for which the major contribution of the drag force results directly from viscous or skin friction of the fluid on the body. Streamline bodies are slender and do not distort the streamlines of the flow so that the viscous boundary layer is attached over their entire surface. Examples: airfoil, horizontal flat plate, falling droplet. A Bluff (non-streamline) body is defined as the body for which the major contribution to the drag force is due to the low pressure in the separated region (the wake/bubble). The shape of the body produces an adverse pressure gradient which results in flow separation. Examples: Sphere, body of circular or rectangular cross-section, stalled airfoil (airfoil inclined at high angle)

QUESTION 6 At Reynolds numbers Re  105 the drag coefficient of a spherical object is CD  0.4 , while for the object below is CD  0.04 . Give a physical explanation of these experimental results.

The above body is streamlined while the sphere is considered to be a bluff body. See previous question for explanation. Find the ratio of the diameters such that the two objects would experience the same drag force.

10 D

The ratio of the diameters can be obtained as follows: The drag experienced by the two objects must be equal hence 1 1 FD  FD2  U 2 A CD  U 2 A2 CD2 2 2 where subscript 2 corresponds to the droplet-like shape above. 1 1  U 2 A 0.4  U 2 A2 0.04 2 2 The two objects are immerced in a fluid of the same velocity hence A 1   D2  10 D A2 10

QUESTION 7 Do you expect the drag on the object shown in Figure 1 to be less when the wind blows from right to left or when it blows from left to right. Explain.

Figure 1: Object for Question 1

QUESTION 8 Describe the physical mechanism of flow separation. Pressure gradient is one of the factors that influences a flow immensely. It is easy to see that the shear stress caused by viscosity has a retarding effect upon the flow. This effect can however be overcome if there is a negative pressure gradient offered to the flow. A negative pressure gradient is termed a Favourable pressure gradient. Such a gradient enables the flow. A positive pressure gradient has the opposite effect and is termed an Adverse Pressure Gradient. Fluid might find it difficult to negotiate an adverse pressure gradient. Sometimes, we say the fluid has to climb the pressure hill.

Figure 6.4 : Separation of flow over a curved surface

One of the severe effects of an adverse pressure gradient is to separate the flow. Consider flow past a curved surface as shown in Figure 6.4. The geometry of the surface is such that we have a favourable gradient in pressure to start with and up to a point P. The negative pressure gradient will counteract the retarding effect of the shear stress (which is due to viscosity) in the boundary layer. For the geometry considered we have a an adverse pressure gradient downstream of P. Now the adverse pressure gradient begins to retard. This effect is felt more strongly in the regions close to the wall where the momentum is lower than in the regions near the free stream. As indicated in the figure, the velocity near the wall reduces and the boundary layer thickens. A continuous retardation of flow brings the wall shear stress at the point S on the wall to zero. From this point onwards the shear stress becomes negative and the flow reverses and a region of recirculating flow develops. We see that the flow no longer follows the contour of the body. We say that the flow has separated. The point S where the shear stress is zero is called the Point of Separation. Depending on the flow conditions the recirculating flow might terminate and the flow may become reattached to the body. A separation bubble is formed. There are a variety of factors that could influence this reattachment. The pressure gradient may be now favourable due to body geometry and other reasons. The other factor is that the flow initially laminar may undergo transition within the bubble and may become turbulent. A turbulent flow has more energy and momentum than a laminar flow. This can kill separation and the flow may reattach. A short bubble may not be of much consequence.

QUESTION 9 a. The pressure distribution around the surface of a cylinder due to an inviscid uniform flow at the far field is 1 ps  po  U 2 1  4sin 2   . 2 b. Detemine the drag force and comment. Based on the inviscid results there is no drag acting on the cylinder. However, experiments suggest that there is a finite drag due to: (i) viscous effects produced in the boundary layer near the surface giving rise to shear drag and (ii) pressure differences between the front and the back of the body due to flow separation. c. Based on the above result for the pressure gradient, explain the physical mechanisms of flow separation. The pressure distribution indicated in the figure is imposed on the boundary layer flow along the surface of the cylinder. This pressure distribution along the cylinder is such that the stationary fluid at the nose of the cylinder (U fs  0 at  =0) is accelerated to its maximum velocity (U fs  2U at  =90o ) and then decelerated back to zero velocity at the rear of the cylinder (U fs  0 at  =180o ) .

Figure 9.16: Inviscid flow past a circular cylinder: (a) streamlines for the flow if there were no viscous effects, (b) pressure distribution on the cylinder’s surface, (c) free-stream velocity on the cylinder’s surface.

This is accomplished by a balance between pressure and inertia effects; viscous effects are absent for the inviscid flow outside the boundary layer. Physically, in the absence of viscous effects, a fluid particle travelling from the front to the back of the cylinder coasts down the “pressure hill” (from point A to C in the figure) and then back up the hill to (from point C to F) without any loss of energy. There is an exchange between kinetic and pressure energy, but there are no energy losses. The same pressure distribution is imposed on the viscous fluid within the boundary layer. The decrease in pressure in the direction of flow along the front half of the cylinder is termed a favorable pressure gradient. The increase in

pressure in the direction of flow along the rear half of the cylinder is termed an adverse pressure gradient. Consider a fluid particle within the boundary layer. In its attempt to flow from A to F it experiences the same pressure distribution as the particles in the free stream immediately outside the boundary layer—the inviscid flow field pressure. However, because of the viscous effects involved, the particle in the boundary layer experiences a loss of energy as it flows along. This loss means that the particle does not have enough energy to coast all of the way up the pressure hill (from C to F) and to reach point F at the rear of the cylinder. This conclusion can also be obtained from the concept that due to viscous effects the particle at C does not have enough momentum to allow it to coast up the pressure hill to F. The fluid within the boundary layer does not have such an energy supply. Thus, the fluid flows against the increasing pressure as far as it can, at which point the boundary layer separates from (lifts off) the surface. At the separation location, the velocity gradient at the wall and the wall shear stress are zero. Beyond that there is reverse flow in the boundary layer (see Munson pg 570).

QUESTION 10 Describe the dependence of the Reynolds number ( Re ) on the Drag Coefficient (C D ). For very small Reynolds number flows, the drag coefficient varies linearly with the Reynolds number. For most objects, the low Reynolds number flow results are valid up to a Reynolds number of about 1.

QUESTION 11 a. Explain the characteristics of the flow over an airfoil The structure of the flow field past a circular cylinder is completely different for a zero viscosity fluid than it is for a viscous fluid, no matter how small the viscosity is, provided it is not zero. This is due to boundary layer separation. Similar concepts hold for other shaped bodies as well. The flow past an airfoil at zero angle of attack (the angle between the up-stream flow and the axis of the object) is shown in Fig. 9.18a; flow past the same airfoil at a 50 angle of attack is shown in Fig. 9.18b. Over the front portion of the airfoil the pressure decreases in the direction of flow—a favorable pressure gradient. Over the rear portion the pressure increases in the direction of flow—an adverse pressure gradient. If the adverse pressure gradient is not too great (because the body is not too “thick” in some sense), the boundary layer fluid can flow into the slightly increasing pressure region (i.e., from C to the trailing edge in Fig. 9.18a) without separating from the surface. However, if the pressure gradient is too adverse (because the angle of attack is too large, or the body is “thick”), the boundary layer will separate from the surface as indicated in Fig. 9.18b. Such situations can lead to the catastrophic loss of lift called stall, Figure 9.18: Flow visualization photographs of flow past an airfoil (the boundary layer velocity profiles for the points indicated are similar to those indicated in Fig. (b): (a) zero angle of attack, no separation, (b) angle of attack, flow separation. Dye in water.

On aerofoils sometimes the separation occurs near the leading edge and gives rise to a short bubble. What can be dangerous is the separation occurring more towards the trailing edge and the flow not reattaching. In this situation the separated region merges with the wake and may result in stall of the aerofoil (loss of lift).

Figure 6.5 : Separation bubble over an aerofoil

b. What is the effect of flow separation on the drag force and lift force of an airfoil? Similar to flow over a sphere, for a completely attached flow over an airfoil, the pressure acting on the rear surface gives rise to a force in the forward direction which completely counteracts the pressure acting on the front surface producing a force in the rearward direction, resulting in zero pressure drag. However, if the flow is partially separated over the rear surface, the pressure on the rear surface pushing forward will be smaller than the fully attached case, giving rise to a net pressure drag on the airfoil. The aerodynamic lift (vertical force) is derived from the net component of the pressure distribution in the vertical direction. High lift is obtained when the pressure on the bottom surface is large and the pressure on the top surface is small. Separation does not affect the bottom surface pressure distribution but rather produces a relatively higher pressure on the top surface. There is a higher pressure pushing down, hence reducing lift. The loss of lift is called stall. See also: Introduction to Flight, John D. Anderson Jr., 6th Edition, McGraw-Hill, Higher Education, pages 224-228.