Chapt07 - The turbulent wake behind a bluff body immersed...

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Unformatted text preview: The turbulent wake behind a bluff body immersed in a stream flow is a subject of the present chapter. This is a digitized video image showing the distribution of tracer-dye concentration in the wake of the body. Compare with Fig. 5.2a of the text, which is a laminar wake. (Courtesy of R. Balachandar, by permission of the American Society of Mechanical Engineers.) | v v 426 | e-Text Main Menu | Textbook Table of Contents | Study Guide Chapter 7 Flow Past Immersed Bodies Motivation. This chapter is devoted to “external” flows around bodies immersed in a fluid stream. Such a flow will have viscous (shear and no-slip) effects near the body surfaces and in its wake, but will typically be nearly inviscid far from the body. These are unconfined boundary-layer flows. Chapter 6 considered “internal” flows confined by the walls of a duct. In that case the viscous boundary layers grow from the sidewalls, meet downstream, and fill the entire duct. Viscous shear is the dominant effect. For example, the Moody chart of Fig. 6.13 is essentially a correlation of wall shear stress for long ducts of constant cross section. External flows are unconfined, free to expand no matter how thick the viscous layers grow. Although boundary-layer theory (Sec. 7.3) is helpful in understanding external flows, complex body geometries usually require experimental data on the forces and moments caused by the flow. Such immersed-body flows are commonly encountered in engineering studies: aerodynamics (airplanes, rockets, projectiles), hydrodynamics (ships, submarines, torpedos), transportation (automobiles, trucks, cycles), wind engineering (buildings, bridges, water towers, wind turbines), and ocean engineering (buoys, breakwaters, pilings, cables, moored instruments). This chapter provides data and analysis to assist in such studies. 7.1 Reynolds-Number and Geometry Effects The technique of boundary-layer (BL) analysis can be used to compute viscous effects near solid walls and to “patch” these onto the outer inviscid motion. This patching is more successful as the body Reynolds number becomes larger, as shown in Fig. 7.1. In Fig. 7.1 a uniform stream U moves parallel to a sharp flat plate of length L. If the Reynolds number UL/ is low (Fig. 7.1a), the viscous region is very broad and extends far ahead and to the sides of the plate. The plate retards the oncoming stream greatly, and small changes in flow parameters cause large changes in the pressure distribution along the plate. Thus, although in principle it should be possible to patch the | v v 427 | e-Text Main Menu | Textbook Table of Contents | Study Guide 428 Chapter 7 Flow Past Immersed Bodies U δ≈L ;; ;; u = 0.99 U Large viscous displacement effect u<U L U ReL = 10 x Viscous region Inviscid region U (a) Small displacement effect δ L U U u<U ReL = 10 7 x Laminar BL Turbulent BL Fig. 7.1 Comparison of flow past a sharp flat plate at low and high Reynolds numbers: (a) laminar, low-Re flow; (b) high-Re flow. Viscous Inviscid region U (b) viscous and inviscid layers in a mathematical analysis, their interaction is strong and nonlinear [1 to 3]. There is no existing simple theory for external-flow analysis at Reynolds numbers from 1 to about 1000. Such thick-shear-layer flows are typically studied by experiment or by numerical modeling of the flow field on a digital computer [4]. A high-Reynolds-number flow (Fig. 7.1b) is much more amenable to boundary-layer patching, as first pointed out by Prandtl in 1904. The viscous layers, either laminar or turbulent, are very thin, thinner even than the drawing shows. We define the boundarylayer thickness as the locus of points where the velocity u parallel to the plate reaches 99 percent of the external velocity U. As we shall see in Sec. 7.4, the accepted formulas for flat-plate flow are | v v x | e-Text Main Menu | 5.0 Re1/2 x 0.16 Re1/7 x laminar (7.1a) turbulent (7.1b) Textbook Table of Contents | Study Guide 7.1 Reynolds-Number and Geometry Effects 429 where Rex Ux/ is called the local Reynolds number of the flow along the plate surface. The turbulent-flow formula applies for Rex greater than approximately 106. Some computed values from Eq. (7.1) are 104 ( /x)lam 105 106 0.050 Rex 0.016 0.005 ( /x)turb 107 0.016 0.022 108 0.011 The blanks indicate that the formula is not applicable. In all cases these boundary layers are so thin that their displacement effect on the outer inviscid layer is negligible. Thus the pressure distribution along the plate can be computed from inviscid theory as if the boundary layer were not even there. This external pressure field then “drives” the boundary-layer flow, acting as a forcing function in the momentum equation along the surface. We shall explain this boundary-layer theory in Secs. 7.4 and 7.5. For slender bodies, such as plates and airfoils parallel to the oncoming stream, we conclude that this assumption of negligible interaction between the boundary layer and the outer pressure distribution is an excellent approximation. For a blunt-body flow, however, even at very high Reynolds numbers, there is a discrepancy in the viscous-inviscid patching concept. Figure 7.2 shows two sketches of flow past a two- or three-dimensional blunt body. In the idealized sketch (7.2a), there is a thin film of boundary layer about the body and a narrow sheet of viscous wake in the rear. The patching theory would be glorious for this picture, but it is false. In the Beautifully behaved but mythically thin boundary layer and wake Red = 10 5 Thin front boundary layer (a) Outer stream grossly perturbed by broad flow separation and wake Red = 10 5 | v v Fig. 7.2 Illustration of the strong interaction between viscous and inviscid regions in the rear of bluntbody flow: (a) idealized and definitely false picture of blunt-body flow; (b) actual picture of bluntbody flow. | e-Text Main Menu (b) | Textbook Table of Contents | Study Guide Chapter 7 Flow Past Immersed Bodies actual flow (Fig. 7.2b), the boundary layer is thin on the front, or windward, side of the body, where the pressure decreases along the surface (favorable pressure gradient). But in the rear the boundary layer encounters increasing pressure (adverse pressure gradient) and breaks off, or separates, into a broad, pulsating wake. (See Fig. 5.2a for a photograph of a specific example.) The mainstream is deflected by this wake, so that the external flow is quite different from the prediction from inviscid theory with the addition of a thin boundary layer. The theory of strong interaction between blunt-body viscous and inviscid layers is not well developed. Flows like that of Fig. 7.2b are normally studied experimentally. Reference 5 is an example of efforts to improve the theory of separated-boundary-layer flows. Reference 6 is a textbook devoted to separated flow. EXAMPLE 7.1 A long, thin flat plate is placed parallel to a 20-ft/s stream of water at 20°C. At what distance x from the leading edge will the boundary-layer thickness be 1 in? Solution Since we do not know the Reynolds number, we must guess which of Eqs. (7.1) applies. From Table 1.4 for water, 1.09 10 5 ft2/s; hence U 1.09 With 1 in 1 12 1.84 106 ft 1 5 (Ux/ )1/2 x 2 or 20 ft/s 10 5 ft2/s ft, try Eq. (7.1a): Laminar flow: (U/v) 52 x ( 112 ft)2(1.84 106 ft 1) 511 ft 25 Now we can test the Reynolds number to see whether the formula applied: Rex Ux (20 ft/s)(511 ft) 1.09 10 5 ft2/s 108 9.4 This is impossible since the maximum Rex for laminar flow past a flat plate is 3 try again with Eq. (7.1b): Turbulent flow: or x (U/v)1/ 7 0.16 7/6 ( 112 ft)(1.84 106 ft 1)1/7 7/6 (4.09)7/6 0.16 Rex 106. So we 0.16 (Ux/ )1/7 x Test (20 ft/s)(5.17 ft) 1.09 10 5 ft2/s 9.5 5.17 ft Ans. 106 This is a perfectly proper turbulent-flow condition; hence we have found the correct position x on our second try. | v v 430 | e-Text Main Menu | Textbook Table of Contents | Study Guide 7.2 Momentum-Integral Estimates 431 7.2 Momentum-Integral Estimates When we derived the momentum-integral relation, Eq. (3.37), and applied it to a flatplate boundary layer in Example 3.11, we promised to consider it further in Chap. 7. Well, here we are! Let us review the problem, using Fig. 7.3. A shear layer of unknown thickness grows along the sharp flat plate in Fig. 7.3. The no-slip wall condition retards the flow, making it into a rounded profile u(y), which merges into the external velocity U constant at a “thickness” y (x). By utilizing the control volume of Fig. 3.11, we found (without making any assumptions about laminar versus turbulent flow) in Example 3.11 that the drag force on the plate is given by the following momentum integral across the exit plane (x) D(x) b u(U u) dy (7.2) 0 where b is the plate width into the paper and the integration is carried out along a vertical plane x constant. You should review the momentum-integral relation (3.37) and its use in Example 3.11. Kármán’s Analysis of the Flat Plate Equation (7.2) was derived in 1921 by Kármán [7], who wrote it in the convenient form of the momentum thickness D(x) u 1 U bU2 0 u dy U (7.3) Momentum thickness is thus a measure of total plate drag. Kármán then noted that the drag also equals the integrated wall shear stress along the plate x D(x) w(x) b 0 dD dx or b (7.4) w Meanwhile, the derivative of Eq. (7.3), with U dD dx dx bU2 constant, is d dx By comparing this with Eq. (7.4) Kármán arrived at what is now called the momentumintegral relation for flat-plate boundary-layer flow y U U p = pa δ (x) τ w (x) u (x, y) x | v v Fig. 7.3 Growth of a boundary layer on a flat plate. | x=0 e-Text Main Menu x=L | Textbook Table of Contents | Study Guide Chapter 7 Flow Past Immersed Bodies d dx U2 w (7.5) It is valid for either laminar or turbulent flat-plate flow. To get a numerical result for laminar flow, Kármán assumed that the velocity profiles had an approximately parabolic shape u(x, y) U y2 2y 0 2 y (x) (7.6) which makes it possible to estimate both momentum thickness and wall shear y2 2y 2 0 y2 2y 1 u y w 2 2 15 dy (7.7) 2U y0 By substituting (7.7) into (7.5) and rearranging we obtain d where ing edge 15 U dx (7.8) / . We can integrate from 0 to x, assuming that 1 2 x 5.5 0 at x 0, the lead- 15 x U 2 1/2 or 5.5 Re1/2 x Ux (7.9) This is the desired thickness estimate. It is all approximate, of course, part of Kármán’s momentum-integral theory [7], but it is startlingly accurate, being only 10 percent higher than the known exact solution for laminar flat-plate flow, which we gave as Eq. (7.1a). By combining Eqs. (7.9) and (7.7) we also obtain a shear-stress estimate along the plate cf 2w U2 8 15 Rex 1/2 0.73 Re1/2 x (7.10) Again this estimate, in spite of the crudeness of the profile assumption (7.6) is only 10 percent higher than the known exact laminar-plate-flow solution cf 0.664/Re1/2, x treated in Sec. 7.4. The dimensionless quantity cf, called the skin-friction coefficient, is analogous to the friction factor f in ducts. A boundary layer can be judged as “thin” if, say, the ratio /x is less than about 0.1. This occurs at /x 0.1 5.0/Re1/2 or at Rex 2500. For Rex less than 2500 we can x estimate that boundary-layer theory fails because the thick layer has a significant effect on the outer inviscid flow. The upper limit on Rex for laminar flow is about 3 106, where measurements on a smooth flat plate [8] show that the flow undergoes transition to a turbulent boundary layer. From 3 106 upward the turbulent Reynolds number may be arbitrarily large, and a practical limit at present is 5 109 for oil supertankers. | v v 432 | e-Text Main Menu | Textbook Table of Contents | Study Guide 7.2 Momentum-Integral Estimates 433 y = h +δ * y U U U y=h h Outer streamline h u Fig. 7.4 Displacement effect of a boundary layer. Displacement Thickness δ* 0 Simulated effect x Another interesting effect of a boundary layer is its small but finite displacement of the outer streamlines. As shown in Fig. 7.4, outer streamlines must deflect outward a distance *(x) to satisfy conservation of mass between the inlet and outlet h Ub dy ub dy h * (7.11) 0 0 The quantity * is called the displacement thickness of the boundary layer. To relate it to u(y), cancel and b from Eq. (7.11), evaluate the left integral, and slyly add and subtract U from the right integrand: Uh (U u U) dy U(h *) 0 (u U) dy 0 or * u dy U 1 0 (7.12) Thus the ratio of */ varies only with the dimensionless velocity-profile shape u/U. Introducing our profile approximation (7.6) into (7.12), we obtain by integration the approximate result * 1 3 1.83 Re1/2 x * x (7.13) These estimates are only 6 percent away from the exact solutions for laminar flat-plate 1.721x/Re1/2. Since * is much smaller than x flow given in Sec. 7.4: * 0.344 x for large Rex and the outer streamline slope V/U is proportional to *, we conclude that the velocity normal to the wall is much smaller than the velocity parallel to the wall. This is a key assumption in boundary-layer theory (Sec. 7.3). We also conclude from the success of these simple parabolic estimates that Kármán’s momentum-integral theory is effective and useful. Many details of this theory are given in Refs. 1 to 3. EXAMPLE 7.2 | v v Are low-speed, small-scale air and water boundary layers really thin? Consider flow at U 1 ft/s past a flat plate 1 ft long. Compute the boundary-layer thickness at the trailing edge for (a) air and (b) water at 20°C. | e-Text Main Menu | Textbook Table of Contents | Study Guide 434 Chapter 7 Flow Past Immersed Bodies Solution Part (a) From Table A.3, air 1.61 E-4 ft2/s. The trailing-edge Reynolds number thus is UL ReL (1 ft/s)(1 ft) 1.61 E-4 ft2/s 6200 Since this is less than 106, the flow is presumed laminar, and since it is greater than 2500, the boundary layer is reasonably thin. From Eq. (7.1a), the predicted laminar thickness is 5.0 x or, at x Part (b) 1 ft, From Table A.2 6200 0.0634 ft water 0.0634 0.76 in Ans. (a) 1.08 E-5 ft2/s. The trailing-edge Reynolds number is (1 ft/s)(1 ft) 1.08 E-5 ft2/s ReL 92,600 This again satisfies the laminar and thinness conditions. The boundary-layer thickness is 5.0 x or, at x 1 ft, 92,600 0.0164 ft 0.0164 0.20 in Ans. (b) Thus, even at such low velocities and short lengths, both airflows and water flows satisfy the boundary-layer approximations. | In Chaps. 4 and 6 we learned that there are several dozen known analytical laminarflow solutions [1 to 3]. None are for external flow around immersed bodies, although this is one of the primary applications of fluid mechanics. No exact solutions are known for turbulent flow, whose analysis typically uses empirical modeling laws to relate timemean variables. There are presently three techniques used to study external flows: (1) numerical (digital-computer) solutions, (2) experimentation, and (3) boundary-layer theory. Computational fluid dynamics (CFD) is now well developed and described in advanced texts such as that by Anderson et al. [4]. Thousands of computer solutions and models have been published; execution times, mesh sizes, and graphical presentations are improving each year. Both laminar- and turbulent-flow solutions have been published, and turbulence modeling is a current research topic [9]. Except for a brief discussion of computer analysis in Chap. 8, the topic of CFD is beyond our scope here. Experimentation is the most common method of studying external flows. Chapter 5 outlined the technique of dimensional analysis, and we shall give many nondimensional experimental data for external flows in Sec. 7.6. The third tool is boundary-layer theory, first formulated by Ludwig Prandtl in 1904. We shall follow Prandtl’s ideas here and make certain order-of-magnitude assumptions to greatly simplify the Navier-Stokes equations (4.38) into boundary-layer equations which are solved relatively easily and patched onto the outer inviscid-flow field. v v 7.3 The Boundary-Layer Equations | e-Text Main Menu | Textbook Table of Contents | Study Guide 7.3 The Boundary-Layer Equations 435 One of the great achievements of boundary-layer theory is its ability to predict the flow separation illustrated in Fig. 7.2b. Before 1904 no one realized that such thin shear layers could cause such a gross effect as flow separation. Unfortunately, even today theory cannot accurately predict the behavior of the separated-flow region and its interaction with the outer layer. This is the weakness of boundary-layer theory, which we hope will be overcome by intensive research into the dynamics of separated flows [6]. Derivation for Two-Dimensional Flow We consider only steady two-dimensional incompressible viscous flow with the x direction along the wall and y normal to the wall, as in Fig. 7.3.1 We neglect gravity, which is important only in boundary layers where fluid buoyancy is dominant [2, sec. 4.13]. From Chap. 4, the complete equations of motion consist of continuity and the x- and y-momentum relations u x u u 0 y (7.14a) u y x p x 2 u x2 2 p y 2 2 y u x 2 x u y2 y2 (7.14b) (7.14c) These should be solved for u, , and p subject to typical no-slip, inlet, and exit boundary conditions, but in fact they are too difficult to handle for most external flows. In 1904 Prandtl correctly deduced that a shear layer must be very thin if the Reynolds number is large, so that the following approximations apply: Velocities: u u x Rates of change: u y (7.15a) x y (7.15b) Our discussion of displacement thickness in the previous section was intended to justify these assumptions. Applying these approximations to Eq. (7.14c) results in a powerful simplification p y 0 or p p(x) only (7.16) In other words, the y-momentum equation can be neglected entirely, and the pressure varies only along the boundary layer, not through it. The pressure-gradient term in Eq. (7.14b) is assumed to be known in advance from Bernoulli’s equation applied to the outer inviscid flow p x dp dx U dU dx 1 (7.17) | v v For a curved wall, x can represent the arc length along the wall and y can be everywhere normal to x with negligible change in the boundary-layer equations as long as the radius of curvature of the wall is large compared with the boundary-layer thickness [1 to 3]. | e-Text Main Menu | Textbook Table of Contents | Study Guide 436 Chapter 7 Flow Past Immersed Bodies Presumably we have already made the inviscid analysis and know the distribution of U(x) along the wall (Chap. 8). Meanwhile, one term in Eq. (7.14b) is negligible due to Eqs. (7.15) 2 2 u x2 u y2 (7.18) However, neither term in the continuity relation (7.14a) can be neglected—another warning that continuity is always a vital part of any fluid-flow analysis. The net result is that the three full equations of motion (7.14) are reduced to Prandtl’s two boundary-layer equations u x Continuity: Momentum along wall: u where u x 0 y u y dU dx U u y u y (7.19a) 1 (7.19b) y laminar flow u turbulent flow These are to be solved for u(x, y) and (x, y), with U(x) assumed to be a known function from the outer inviscid-flow analysis. There are two boundary conditions on u and one on : At y At y 0 (wall): u (x) (outer stream): u 0 U(x) (no slip) (7.20a) (patching) (7.20b) Unlike the Navier-Stokes equations (7.14), which are mathematically elliptic and must be solved simultaneously over the entire flow field, the boundary-layer equations (7.19) are mathematically parabolic and are solved by beginning at the leading edge and marching downstream as far as you like, stopping at the separation point or earlier if you prefer.2 The boundary-layer equations have been solved for scores of interesting cases of internal and external flow for both laminar and turbulent flow, utilizing the inviscid distribution U(x) appropriate to each flow. Full details of boundary-layer theory and results and comparison with experiment are given in Refs. 1 to 3. Here we shall confine ourselves primarily to flat-plate solutions (Sec. 7.4). 7.4 The Flat-Plate Boundary Layer The classic and most often used solution of boundary-layer theory is for flat-plate flow, as in Fig. 7.3, which can represent either laminar or turbulent flow. 2 | v v For further mathematical details, see Ref. 2, sec. 2.8. | e-Text Main Menu | Textbook Table of Contents | Study Guide 7.4 The Flat-Plate Boundary Layer Laminar Flow 437 For laminar flow past the plate, the boundary-layer equations (7.19) can be solved exactly for u and , assuming that the free-stream velocity U is constant (dU/dx 0). The solution was given by Prandtl’s student Blasius, in his 1908 dissertation from Göttingen. With an ingenious coordinate transformation, Blasius showed that the dimensionless velocity profile u/U is a function only of the single composite dimensionless variable (y)[U/( x)]1/2: u U f( ) U vx y 1/2 (7.21) where the prime denotes differentiation with respect to . Substitution of (7.21) into the boundary-layer equations (7.19) reduces the problem, after much algebra, to a single third-order nonlinear ordinary differential equation for f 1 2 f ff 0 (7.22) The boundary conditions (7.20) become At y 0: f (0) As y → : f (0) 0 (7.23a) f ( ) → 1.0 (7.23b) This is the Blasius equation, for which accurate solutions have been obtained only by numerical integration. Some tabulated values of the velocity-profile shape f ( ) u/U are given in Table 7.1. Since u/U approaches 1.0 only as y → , it is customary to select the boundarylayer thickness as that point where u/U 0.99. From the table, this occurs at 5.0: U 1/2 5.0 99% x or Table 7.1 The Blasius Velocity Profile [1 to 3] x 5.0 Re1/2 x Blasius (1908) v v | u/U y[U/( x)]1/2 u/U 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 | y[U/( x)]1/2 0.0 0.06641 0.13277 0.19894 0.26471 0.32979 0.39378 0.45627 0.51676 0.57477 0.62977 0.68132 0.72899 0.77246 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 4.6 4.8 5.0 0.81152 0.84605 0.87609 0.90177 0.92333 0.94112 0.95552 0.96696 0.97587 0.98269 0.98779 0.99155 1.00000 e-Text Main Menu | Textbook Table of Contents | Study Guide (7.24) Chapter 7 Flow Past Immersed Bodies With the profile known, Blasius, of course, could also compute the wall shear and displacement thickness 0.664 Re1/2 x cf 1.721 Re1/2 x * x (7.25) Notice how close these are to our integral estimates, Eqs. (7.9), (7.10), and (7.13). When cf is converted to dimensional form, we have 0.332 w(x) 1/2 1/2 U1.5 1/2 x 1/2 The wall shear drops off with x because of boundary-layer growth and varies as velocity to the 1.5 power. This is in contrast to laminar pipe flow, where w U and is independent of x. If w(x) is substituted into Eq. (7.4), we compute the total drag force x D(x) b w(x) 0 dx 0.664b 1/2 1/2 U1.5x1/2 (7.26) The drag increases only as the square root of the plate length. The nondimensional drag coefficient is defined as 2D(L) U2bL CD 1.328 1/2 ReL 2cf (L) (7.27) Thus, for laminar plate flow, CD equals twice the value of the skin-friction coefficient at the trailing edge. This is the drag on one side of the plate. Kármán pointed out that the drag could also be computed from the momentum relation (7.2). In dimensionless form, Eq. (7.2) becomes CD 2 L 0 u 1 U u dy U (7.28) This can be rewritten in terms of the momentum thickness at the trailing edge CD Computation of 2 (L) L (7.29) from the profile u/U or from CD gives x 0.664 Re1/2 x laminar flat plate (7.30) Since is so ill defined, the momentum thickness, being definite, is often used to correlate data taken for a variety of boundary layers under differing conditions. The ratio of displacement to momentum thickness, called the dimensionless-profile shape factor, is also useful in integral theories. For laminar flat-plate flow H * 1.721 0.664 2.59 (7.31) A large shape factor then implies that boundary-layer separation is about to occur. | v v 438 | e-Text Main Menu | Textbook Table of Contents | Study Guide 7.4 The Flat-Plate Boundary Layer 439 1.0 Turbulent 0.8 Seventh root profile, Eq. (7.39) 10 5 = Rex 10 6 10 7 0.6 u U 0.4 Exact Blasius profile for all laminar Rex ( Table 7.1) 0.2 Parabolic approximation, Eq. ( 7.6) 0 0.2 Fig. 7.5 Comparison of dimensionless laminar and turbulent flat-plate velocity profiles. 0.4 0.6 0.8 1.0 y δ If we plot the Blasius velocity profile from Table 7.1 in the form of u/U versus y/ , we can see why the simple integral-theory guess, Eq. (7.6), was such a great success. This is done in Fig. 7.5. The simple parabolic approximation is not far from the true Blasius profile; hence its momentum thickness is within 10 percent of the true value. Also shown in Fig. 7.5 are three typical turbulent flat-plate velocity profiles. Notice how strikingly different in shape they are from the laminar profiles. Instead of decreasing monotonically to zero, the turbulent profiles are very flat and then drop off sharply at the wall. As you might guess, they follow the logarithmic-law shape and thus can be analyzed by momentum-integral theory if this shape is properly represented. Transition to Turbulence The laminar flat-plate boundary layer eventually becomes turbulent, but there is no unique value for this change to occur. With care in polishing the wall and keeping the free stream quiet, one can delay the transition Reynolds number to Rex,tr 3 E6 [8]. However, for typical commercial surfaces and gusty free streams, a more realistic value is Rex,tr 5 E5. EXAMPLE 7.3 | v v A sharp flat plate with L 1 m and b 3 m is immersed parallel to a stream of velocity 2 m/s. Find the drag on one side of the plate, and at the trailing edge find the thicknesses , *, and for (a) air, 1.23 kg/m3 and 1.46 10 5 m2/s, and (b) water, 1000 kg/m3 and 6 2 1.02 10 m /s. | e-Text Main Menu | Textbook Table of Contents | Study Guide Chapter 7 Flow Past Immersed Bodies Solution Part (a) The airflow Reynolds number is VL (2.0 m/s)(1.0 m) 1.46 10 5 m2/s 137,000 106, we assume that the boundary layer is laminar. From Eq. (7.27), Since this is less than 3 the drag coefficient is 1.328 (137,000)1/2 CD 0.00359 Thus the drag on one side in the airflow is 1 CD 2 U2bL D 0.00359( 1 )(1.23)(2.0)2(3.0)(1.0) 2 0.0265 N Ans. (a) The boundary-layer thickness at the end of the plate is 5.0 Re1/2 L L 5.0 (137,000)1/2 0.0135(1.0) or 0.0135 0.0135 m 13.5 mm Ans. (a) We find the other two thicknesses simply by ratios: 1.721 5.0 * * 2.59 4.65 mm 1.79 mm Ans. (a) Notice that no conversion factors are needed with SI units. Part (b) The water Reynolds number is ReL 2.0(1.0) 1.02 10 6 1.96 106 This is rather close to the critical value of 3 106, so that a rough surface or noisy free stream might trigger transition to turbulence; but let us assume that the flow is laminar. The water drag coefficient is CD and D 1.328 (1.96 106)1/2 0.000949 0.000949( 1 )(1000)(2.0)2(3.0)(1.0) 2 5.70 N Ans. (b) The drag is 215 times more for water in spite of the higher Reynolds number and lower drag coefficient because water is 57 times more viscous and 813 times denser than air. From Eq. (7.26), in laminar flow, it should have (57)1/2(813)1/2 7.53(28.5) 215 times more drag. The boundary-layer thickness is given by L (1.96 5.0 106)1/2 0.00357(1000 mm) or 0.00357 3.57 mm Ans. (b) By scaling down we have | v v 440 | e-Text Main Menu | Textbook Table of Contents | Study Guide 7.4 The Flat-Plate Boundary Layer * 1.721 5.0 * 2.59 1.23 mm 0.48 mm 441 Ans. (b) The water layer is 3.8 times thinner than the air layer, which reflects the square root of the 14.3 ratio of air to water kinematic viscosity. Turbulent Flow There is no exact theory for turbulent flat-plate flow, although there are many elegant computer solutions of the boundary-layer equations using various empirical models for the turbulent eddy viscosity [9]. The most widely accepted result is simply an integal analysis similar to our study of the laminar-profile approximation (7.6). We begin with Eq. (7.5), which is valid for laminar or turbulent flow. We write it here for convenient reference: U2 w(x) d dx (7.32) From the definition of cf, Eq. (7.10), this can be rewritten as cf 2 d dx (7.33) Now recall from Fig. 7.5 that the turbulent profiles are nowhere near parabolic. Going back to Fig. 6.9, we see that flat-plate flow is very nearly logarithmic, with a slight outer wake and a thin viscous sublayer. Therefore, just as in turbulent pipe flow, we assume that the logarithmic law (6.21) holds all the way across the boundary layer u u* 1 ln yu* B w u* 1/2 (7.34) with, as usual, 0.41 and B 5.0. At the outer edge of the boundary layer, y and u U, and Eq. (7.34) becomes U u* 1 u* ln B (7.35) But the definition of the skin-friction coefficient, Eq. (7.10), is such that the following identities hold: U u* 2 cf 1/2 u* Re cf 2 1/2 (7.36) Therefore Eq. (7.35) is a skin-friction law for turbulent flat-plate flow 2 cf cf 2 1/2 2.44 ln Re 1/2 5.0 (7.37) It is a complicated law, but we can at least solve for a few values and list them: v | | 104 105 106 107 cf v Re 0.00493 0.00315 0.00217 0.00158 e-Text Main Menu | Textbook Table of Contents | Study Guide Chapter 7 Flow Past Immersed Bodies Following a suggestion of Prandtl, we can forget the complex log friction law (7.37) and simply fit the numbers in the table to a power-law approximation cf 0.02 Re 1/6 (7.38) This we shall use as the left-hand side of Eq. (7.33). For the right-hand side, we need an estimate for (x) in terms of (x). If we use the logarithmic-law profile (7.34), we shall be up to our hips in logarithmic integrations for the momentum thickness. Instead we follow another suggestion of Prandtl, who pointed out that the turbulent profiles in Fig. 7.5 can be approximated by a one-seventh-power law u U 1/7 y (7.39) turb This is shown as a dashed line in Fig. 7.5. It is an excellent fit to the low-Reynoldsnumber turbulent data, which were all that were available to Prandtl at the time. With this simple approximation, the momentum thickness (7.28) can easily be evaluated: y 1/7 1/7 y 1 dy 0 7 72 (7.40) We accept this result and substitute Eqs. (7.38) and (7.40) into Kármán’s momentum law (7.33) cf or Re 0.02 Re 1/6 9.72 1/6 2 d dx d dx Re 0.16 Re6/7 x 7 72 d(Re ) d(Rex) 9.72 Separate the variables and integrate, assuming 0 at x or x (7.41) 0: 0.16 Re1/ 7 x (7.42) Thus the thickness of a turbulent boundary layer increases as x6/ 7, far more rapidly than the laminar increase x1/2. Equation (7.42) is the solution to the problem, because all other parameters are now available. For example, combining Eqs. (7.42) and (7.38), we obtain the friction variation cf 0.027 Re1/ 7 x (7.43) Writing this out in dimensional form, we have 0.0135 w,turb 1/ 7 6/ 7 U13/ 7 (7.44) 1/ 7 x Turbulent plate friction drops slowly with x, increases nearly as insensitive to viscosity. We can evaluate the drag coefficient from Eq. (7.29) CD | v v 442 | e-Text Main Menu | 0.031 Re1/ 7 L and U2, and is rather 7 cf (L) 6 Textbook Table of Contents (7.45) | Study Guide 7.4 The Flat-Plate Boundary Layer 443 Then CD is only 16 percent greater than the trailing-edge skin friction [compare with Eq. (7.27) for laminar flow]. The displacement thickness can be estimated from the one-seventh-power law and Eq. (7.12): * 1/ 7 y 1 dy 0 1 8 (7.46) The turbulent flat-plate shape factor is approximately H 1 8 7 72 * 1.3 (7.47) These are the basic results of turbulent flat-plate theory. Figure 7.6 shows flat-plate drag coefficients for both laminar-and turbulent-flow conditions. The smooth-wall relations (7.27) and (7.45) are shown, along with the effect of wall roughness, which is quite strong. The proper roughness parameter here is x/ or L/ , by analogy with the pipe parameter /d. In the fully rough regime, CD is independent of the Reynolds number, so that the drag varies exactly as U2 and is inde- 0.014 200 Fully rough Eq. ( 7.48b) L ε = 300 0.012 500 0.010 1000 0.008 2000 CD 0.006 5000 10 4 2 × 10 4 0.004 v v | | 2 × 10 5 10 6 Transition 0.002 Fig. 7.6 Drag coefficient of laminar and turbulent boundary layers on smooth and rough flat plates. This chart is the flat-plate analog of the Moody diagram of Fig. 6.13. 5 × 10 4 Turbulent smooth Eq. ( 7.45 ) Laminar: Eq. ( 7.27 ) 0 10 5 e-Text Main Menu Eq. ( 7.49 ) 10 6 10 7 10 8 ReL | Textbook Table of Contents | Study Guide 10 9 Chapter 7 Flow Past Immersed Bodies pendent of . Reference 2 presents a theory of rough flat-plate flow, and Ref. 1 gives a curve fit for skin friction and drag in the fully rough regime: 2.5 x cf 2.87 1.58 log CD 1.89 1.62 log (7.48a) L 2.5 (7.48b) Equation (7.48b) is plotted to the right of the dashed line in Fig. 7.6. The figure also shows the behavior of the drag coefficient in the transition region 5 105 ReL 8 107, where the laminar drag at the leading edge is an appreciable fraction of the total drag. Schlichting [1] suggests the following curve fits for these transition drag curves depending upon the Reynolds number Retrans where transition begins: CD CD 0.031 Re1/ 7 L 1440 ReL Retrans 5 105 (7.49a) 0.031 Re1/ 7 L 8700 ReL Retrans 3 106 (7.49b) EXAMPLE 7.4 A hydrofoil 1.2 ft long and 6 ft wide is placed in a water flow of 40 ft/s, with 1.99 slugs/ft3 and 0.000011 ft2/s. (a) Estimate the boundary-layer thickness at the end of the plate. Estimate the friction drag for (b) turbulent smooth-wall flow from the leading edge, (c) laminar turbulent flow with Retrans 5 105, and (d) turbulent rough-wall flow with 0.0004 ft. Solution Part (a) The Reynolds number is ReL UL (40 ft/s)(1.2 ft) 0.000011 ft2/s 106 4.36 Thus the trailing-edge flow is certainly turbulent. The maximum boundary-layer thickness would occur for turbulent flow starting at the leading edge. From Eq. (7.42), (L) L or (4.36 0.16 106)1/7 0.018(1.2 ft) 0.018 0.0216 ft Ans. (a) This is 7.5 times thicker than a fully laminar boundary layer at the same Reynolds number. Part (b) For fully turbulent smooth-wall flow, the drag coefficient on one side of the plate is, from Eq. (7.45), CD | v v 444 | e-Text Main Menu | 0.031 (4.36 106)1/7 0.00349 Textbook Table of Contents | Study Guide 7.5 Boundary Layers with Pressure Gradient 445 Then the drag on both sides of the foil is approximately D Part (c) 2CD( 1 U2)bL 2 2(0.00349)( 1 )(1.99)(40)2(6.0)(1.2) 2 CD 1440 4.36 106 0.00349 Ans. (b) 105, Eq. (7.49a) applies: 5 With a laminar leading edge and Retrans 80 lb 0.00316 The drag can be recomputed for this lower drag coefficient: 2CD( 1 U2)bL 2 D Part (d) 72 lbf Ans. (c) Finally, for the rough wall, we calculate L From Fig. 7.6 at ReL tion (7.48b) applies: 3000 106, this condition is just inside the fully rough regime. Equa- 4.36 CD 1.2 ft 0.0004 ft (1.89 1.62 log 3000) 2.5 0.00644 and the drag estimate is D 1 2CD( 2 U2)bL 148 lbf Ans. (d) This small roughness nearly doubles the drag. It is probable that the total hydrofoil drag is still another factor of 2 larger because of trailing-edge flow-separation effects. 7.5 Boundary Layers with Pressure Gradient3 The flat-plate analysis of the previous section should give us a good feeling for the behavior of both laminar and turbulent boundary layers, except for one important effect: flow separation. Prandtl showed that separation like that in Fig. 7.2b is caused by excessive momentum loss near the wall in a boundary layer trying to move downstream against increasing pressure, dp/dx 0, which is called an adverse pressure gradient. The opposite case of decreasing pressure, dp/dx 0, is called a favorable gradient, where flow separation can never occur. In a typical immersed-body flow, e.g., Fig. 7.2b, the favorable gradient is on the front of the body and the adverse gradient is in the rear, as discussed in detail in Chap. 8. We can explain flow separation with a geometric argument about the second derivative of velocity u at the wall. From the momentum equation (7.19b) at the wall, where 0, we obtain u y wall u y 2 u y2 2 or 2 U wall wall dU dx dp dx 1 dp dx 3 | v v This section may be omitted without loss of continuity. | e-Text Main Menu | Textbook Table of Contents | Study Guide (7.50) 446 Chapter 7 Flow Past Immersed Bodies for either laminar or turbulent flow. Thus in an adverse gradient the second derivative of velocity is positive at the wall; yet it must be negative at the outer layer (y ) to merge smoothly with the mainstream flow U(x). It follows that the second derivative must pass through zero somewhere in between, at a point of inflection, and any boundarylayer profile in an adverse gradient must exhibit a characteristic S shape. Figure 7.7 illustrates the general case. In a favorable gradient (Fig. 7.7a) the profile U U u u PI (a) Favorable gradient: dU >0 dx dp <0 dx (b) Zero gradient: dU =0 dx dp =0 dx No separation, PI inside wall No separation, PI at wall dp >0 dx U U U u u u PI PI Backflow PI (c) Weak adverse gradient: τw = 0 (d ) Critical adverse gradient: (e) Excessive adverse gradient: dU <0 dx v v | Backflow at the wall: dp >0 dx Fig. 7.7 Effect of pressure gradient on boundary-layer profiles; PI point of inflection. Zero slope at the wall: Separation Separated flow region No separation, PI in the flow | e-Text Main Menu | Textbook Table of Contents | Study Guide 7.5 Boundary Layers with Pressure Gradient 447 is very rounded, there is no point of inflection, there can be no separation, and laminar profiles of this type are very resistant to a transition to turbulence [1 to 3]. In a zero pressure gradient (Fig. 7.7b), e.g., flat-plate flow, the point of inflection is at the wall itself. There can be no separation, and the flow will undergo transition at Rex no greater than about 3 106, as discussed earlier. In an adverse gradient (Fig. 7.7c to e), a point of inflection (PI) occurs in the boundary layer, its distance from the wall increasing with the strength of the adverse gradient. For a weak gradient (Fig. 7.7c) the flow does not actually separate, but it is vulnerable to transition to turbulence at Rex as low as 105 [1, 2]. At a moderate gradient, a critical condition (Fig. 7.7d) is reached where the wall shear is exactly zero ( u/ y 0). This is defined as the separation point ( w 0), because any stronger gradient will actually cause backflow at the wall (Fig. 7.7e): the boundary layer thickens greatly, and the main flow breaks away, or separates, from the wall (Fig. 7.2b). The flow profiles of Fig. 7.7 usually occur in sequence as the boundary layer progresses along the wall of a body. For example, in Fig. 7.2a, a favorable gradient occurs on the front of the body, zero pressure gradient occurs just upstream of the shoulder, and an adverse gradient occurs successively as we move around the rear of the body. A second practical example is the flow in a duct consisting of a nozzle, throat, and diffuser, as in Fig. 7.8. The nozzle flow is a favorable gradient and never separates, nor Separation point τw = 0 Boundary layers Profile point of inflection Backflow δ ( x) Nearly inviscid core flow x U(x) δ ( x) Separation Nozzle: Decreasing pressure and area v v | e-Text Main Menu | Diffuser: Increasing pressure and area Velocity constant Decreasing velocity Favorable gradient Fig. 7.8 Boundary-layer growth and separation in a nozzle-diffuser configuration. Throat: Constant pressure and area Increasing velocity | U(x) Zero gradient Adverse gradient (boundary layer thickens) Textbook Table of Contents | Study Guide Dividing streamline 448 Chapter 7 Flow Past Immersed Bodies does the throat flow where the pressure gradient is approximately zero. But the expanding-area diffuser produces low velocity and increasing pressure, an adverse gradient. If the diffuser angle is too large, the adverse gradient is excessive, and the boundary layer will separate at one or both walls, with backflow, increased losses, and poor pressure recovery. In the diffuser literature [10] this condition is called diffuser stall, a term used also in airfoil aerodynamics (Sec. 7.6) to denote airfoil boundary-layer separation. Thus the boundary-layer behavior explains why a large-angle diffuser has heavy flow losses (Fig. 6.23) and poor performance (Fig. 6.28). Presently boundary-layer theory can compute only up to the separation point, after which it is invalid. New techniques are now developed for analyzing the strong interaction effects caused by separated flows [5, 6]. Laminar Integral Theory Both laminar and turbulent theories can be developed from Kármán’s general twodimensional boundary-layer integral relation [7], which extends Eq. (7.33) to variable U(x) 1 cf 2 w U2 d dx (2 H) dU U dx (7.51) *(x)/ (x) is the shape factor. From where (x) is the momentum thickness and H(x) Eq. (7.17) negative dU/dx is equivalent to positive dp/dx, that is, an adverse gradient. We can integrate Eq. (7.51) to determine (x) for a given U(x) if we correlate cf and H with the momentum thickness. This has been done by examining typical velocity profiles of laminar and turbulent boundary-layer flows for various pressure gradients. Some examples are given in Fig. 7.9, showing that the shape factor H is a good indicator of the pressure gradient. The higher the H, the stronger the adverse gradient, and separation occurs approximately at 3.5 2.4 H laminar flow turbulent flow (7.52) The laminar profiles (Fig. 7.9a) clearly exhibit the S shape and a point of inflection with an adverse gradient. But in the turbulent profiles (Fig. 7.9b) the points of inflection are typically buried deep within the thin viscous sublayer, which can hardly be seen on the scale of the figure. There are scores of turbulent theories in the literature, but they are all complicated algebraically and will be omitted here. The reader is referred to advanced texts [1, 2, 9]. For laminar flow, a simple and effective method was developed by Thwaites [11], who found that Eq. (7.51) can be correlated by a single dimensionless momentumthickness variable , defined as 2 v dU dx (7.53) Using a straight-line fit to his correlation, Thwaites was able to integrate Eq. (7.51) in closed form, with the result | v v 2 | e-Text Main Menu | 2 0 U0 U 6 0.45 U6 x U5 dx (7.54) 0 Textbook Table of Contents | Study Guide 7.5 Boundary Layers with Pressure Gradient 1.0 449 1.0 Flat plate Favorable gradients: 0.9 0.8 2.2 = H = 0.8 δ* θ 2.4 2.6 (Flat plate) 0.6 0.6 u 0.5 U 2.9 3.2 0.4 δ * = 1.3 θ 4 1. .5 1 .6 1 .7 8 1 1. 9 1. 1 4 . .0 2 .3 2. 2 .2 2 2 0.7 2.7 u U H= Separation 0.4 3.5 (Separation) 0.3 Points of inflection (adverse gradients) 0.2 0 0.2 0.4 0.6 0.2 0.1 0.8 1.0 y δ 0 0.1 0.5 y δ (a) (b) 0.2 0.3 0.4 0.6 0.7 0.8 0.9 1.0 Fig. 7.9 Velocity profiles with pressure gradient: (a) laminar flow; (b) turbulent flow with adverse gradients. where 0 is the momentum thickness at x 0 (usually taken to be zero). Separation (cf 0) was found to occur at a particular value of Separation: 0.09 (7.55) Finally, Thwaites correlated values of the dimensionless shear stress S , and his graphed result can be curve-fitted as follows: S( ) w 0.09)0.62 ( U w /( U) with (7.56) This parameter is related to the skin friction by the identity S 1 2 cf Re (7.57) Equations (7.54) to (7.56) constitute a complete theory for the laminar boundary layer with variable U(x), with an accuracy of 10 percent compared with exact digital-computer solutions of the laminar-boundary-layer equations (7.19). Complete details of Thwaites’ and other laminar theories are given in Refs. 2 and 3. As a demonstration of Thwaites’ method, take a flat plate, where U constant, 0, and 0 0. Equation (7.54) integrates to | v v 2 | e-Text Main Menu | 0.45 x U Textbook Table of Contents | Study Guide Chapter 7 Flow Past Immersed Bodies or 0.671 Re1/2 x x (7.58) This is within 1 percent of Blasius’ exact solution, Eq. (7.30). With 0, Eq. (7.56) predicts the flat-plate shear to be w (0.09)0.62 U or 0.671 Re1/2 x 2w U2 cf 0.225 (7.59) This is also within 1 percent of the Blasius result, Eq. (7.25). However, the general accuracy of this method is poorer than 1 percent because Thwaites actually “tuned” his correlation constants to make them agree with exact flat-plate theory. We shall not compute any more boundary-layer details here, but as we go along, investigating various immersed-body flows, especially in Chap. 8, we shall use Thwaites’ method to make qualitative assessments of the boundary-layer behavior. EXAMPLE 7.5 In 1938 Howarth proposed a linearly decelerating external-velocity distribution U(x) x L U0 1 (1) as a theoretical model for laminar-boundary-layer study. (a) Use Thwaites’ method to compute the separation point xsep for 0 0, and compare with the exact digital-computer solution xsep/L 0.119863 given by H. Wipperman in 1966. (b) Also compute the value of cf 2 w/( U2) at x/L 0.1. Solution Part (a) First note that dU/dx U0/L constant: Velocity decreases, pressure increases, and the pressure gradient is adverse throughout. Now integrate Eq. (7.54) 2 0.45v x/L)6 U 6(1 0 x 0 Then the dimensionless factor v | | e-Text Main Menu L U0 0.075 1 U0 L x L 1 x L 6 1 (2) 6 1 (3) 0.09 for separation 0.09 xsep L | 0.075 dx 2 From Eq. (7.55) we set this equal to or 5 is given by dU v dx sep x L U5 1 0 2 v 450 0.075 1 1 (2.2) 1/6 xsep L 6 1 0.123 Textbook Table of Contents Ans. (a) | Study Guide 7.6 Experimental External Flows 451 This is less than 3 percent higher than Wipperman’s exact solution, and the computational effort is very modest. Part (b) To compute cf at x/L 0.1 (just before separation), we first compute (x 0.1L) 0.075[(1 0.1) 6 1] at this point, using Eq. (3) 0.0661 Then from Eq. (7.56) the shear parameter is S(x 0.1L) ( 0.0661 0.09)0.62 0.099 1 2f c Re (4) We can compute Re in terms of ReL from Eq. (2) or (3) 2 0.0661 UL/ L2 or 0.0661 ReL 0.257 Re1/2 L Re at x L 0.1 Substitute into Eq. (4): 0.099 or cf 1 2 cf (0.257 Re1/2) L 0.77 Re1/2 L ReL UL Ans. (b) We cannot actually compute cf without the value of, say, U0 L/ . 7.6 Experimental External Flows Boundary-layer theory is very interesting and illuminating and gives us a great qualitative grasp of viscous-flow behavior, but, because of flow separation, the theory does not generally allow a quantitative computation of the complete flow field. In particular, there is at present no satisfactory theory for the forces on an arbitrary body immersed in a stream flowing at an arbitrary Reynolds number. Therefore experimentation is the key to treating external flows. Literally thousands of papers in the literature report experimental data on specific external viscous flows. This section gives a brief description of the following externalflow problems: 1. Drag of two-and three-dimensional bodies a. Blunt bodies b. Streamlined shapes 2. Performance of lifting bodies a. Airfoils and aircraft b. Projectiles and finned bodies c. Birds and insects | v v For further reading see the goldmine of data compiled in Hoerner [12]. In later chapters we shall study data on supersonic airfoils (Chap. 9), open-channel friction (Chap. 10), and turbomachinery performance (Chap. 11). | e-Text Main Menu | Textbook Table of Contents | Study Guide 452 Chapter 7 Flow Past Immersed Bodies Drag of Immersed Bodies Any body of any shape when immersed in a fluid stream will experience forces and moments from the flow. If the body has arbitrary shape and orientation, the flow will exert forces and moments about all three coordinate axes, as shown in Fig. 7.10. It is customary to choose one axis parallel to the free stream and positive downstream. The force on the body along this axis is called drag, and the moment about that axis the rolling moment. The drag is essentially a flow loss and must be overcome if the body is to move against the stream. A second and very important force is perpendicular to the drag and usually performs a useful job, such as bearing the weight of the body. It is called the lift. The moment about the lift axis is called yaw. The third component, neither a loss nor a gain, is the side force, and about this axis is the pitching moment. To deal with this three-dimensional force-moment situation is more properly the role of a textbook on aerodynamics [for example, 13]. We shall limit the discussion here to lift and drag. When the body has symmetry about the lift-drag axis, e.g., airplanes, ships, and cars moving directly into a stream, the side force, yaw, and roll vanish, and the problem reduces to a two-dimensional case: two forces, lift and drag, and one moment, pitch. A final simplification often occurs when the body has two planes of symmetry, as in Fig. 7.11. A wide variety of shapes such as cylinders, wings, and all bodies of revolution satisfy this requirement. If the free stream is parallel to the intersection of these two planes, called the principal chord line of the body, the body experiences drag only, with no lift, side force, or moments.4 This type of degenerate one-force drag data is what is most commonly reported in the literature, but if the free stream is not parallel to the chord line, the body will have an unsymmetric orientation and all three forces and three moments can arise in principle. In low-speed flow past geometrically similar bodies with identical orientation and relative roughness, the drag coefficient should be a function of the body Reynolds number CD f (Re) (7.60) Lift force Yawing moment Arbitrary body Drag force Rolling moment Pitching moment Fig. 7.10 Definition of forces and moments on a body immersed in a uniform flow. V Freestream velocity Side force | v v 4 In bodies with shed vortices, such as the cylinder in Fig. 5.2, there may be oscillating lift, side force, and moments, but their mean value is zero. | e-Text Main Menu | Textbook Table of Contents | Study Guide 7.6 Experimental External Flows 453 Vertical plane of symmetry Horizontal plane of symmetry Drag only if V is parallel to chord line Principal chord line V Fig. 7.11 Only the drag force occurs if the flow is parallel to both planes of symmetry. Doubly symmetric body The Reynolds number is based upon the free-stream velocity V and a characteristic length L of the body, usually the chord or body length parallel to the stream VL Re (7.61) For cylinders, spheres, and disks, the characteristic length is the diameter D. Characteristic Area Drag coefficients are defined by using a characteristic area A which may differ depending upon the body shape: CD 1 2 drag V2A (7.62) The factor 1 is our traditional tribute to Euler and Bernoulli. The area A is usually one 2 of three types: 1. Frontal area, the body as seen from the stream; suitable for thick, stubby bodies, such as spheres, cylinders, cars, missiles, projectiles, and torpedoes. 2. Planform area, the body area as seen from above; suitable for wide, flat bodies such as wings and hydrofoils. 3. Wetted area, customary for surface ships and barges. In using drag or other fluid-force data, it is important to note what length and area are being used to scale the measured coefficients. Friction Drag and Pressure Drag As we have mentioned, the theory of drag is weak and inadequate, except for the flat plate. This is because of flow separation. Boundary-layer theory can predict the separation point but cannot accurately estimate the (usually low) pressure distribution in the separated region. The difference between the high pressure in the front stagnation region and the low pressure in the rear separated region causes a large drag contribution called pressure drag. This is added to the integrated shear stress or friction drag of the body, which it often exceeds: | v v CD | e-Text Main Menu | CD,press Textbook Table of Contents CD,fric | Study Guide (7.63) 454 Chapter 7 Flow Past Immersed Bodies The relative contribution of friction and pressure drag depends upon the body’s shape, especially its thickness. Figure 7.12 shows drag data for a streamlined cylinder of very large depth into the paper. At zero thickness the body is a flat plate and exhibits 100 percent friction drag. At thickness equal to the chord length, simulating a circular cylinder, the friction drag is only about 3 percent. Friction and pressure drag are about equal at thickness t/c 0.25. Note that CD in Fig. 7.12b looks quite different when based upon frontal area instead of planform area, planform being the usual choice for this body shape. The two curves in Fig. 7.12b represent exactly the same drag data. Figure 7.13 illustrates the dramatic effect of separated flow and the subsequent failure of boundary-layer theory. The theoretical inviscid pressure distribution on a circular cylinder (Chap. 8) is shown as the dashed line in Fig. 7.13c: Cp p 1 2 p V2 1 4 sin2 (7.64) where p and V are the pressure and velocity, respectively, in the free stream. The actual laminar and turbulent boundary-layer pressure distributions in Fig. 7.13c are startlingly different from those predicted by theory. Laminar flow is very vulnerable to the adverse gradient on the rear of the cylinder, and separation occurs at 82°, which Friction drag percent 100 Data scatter 50 0 (a) Percentage of pressure drag 3 0 0.2 0.4 t c 0.6 0.8 1.0 Circular cylinder 0.3 CD based upon frontal area (t b) 0.2 CD based upon planform area (c b) CD Width b 0.1 | v v Fig. 7.12 Drag of a streamlined two-dimensional cylinder at Rec 106: (a) effect of thickness ratio on percentage of friction drag; (b) total drag versus thickness when based upon two different areas. t Flat plate (b) | 0 0 e-Text Main Menu V 0.2 | 0.4 0.6 t Thickness ratio c Textbook Table of Contents c 0.8 | 1.0 Study Guide 7.6 Experimental External Flows 455 Separation Separation θ θ 82° V p∞ Broad wake 120° V p∞ CD = 1.2 CD = 0.3 (a) Narrow wake (b) 1.0 0.0 p – p∞ Laminar – 1.0 Cp = ρ V 2/ 2 Turbulent Inviscid theory – 2.0 Cp = 1 – 4 sin2 θ Fig. 7.13 Flow past a circular cylinder: (a) laminar separation; (b) turbulent separation; (c) theoretical and actual surface-pressure distributions. – 3.0 0° 45° 90° 135° 180° θ (c) | v v certainly could not have been predicted from inviscid theory. The broad wake and very low pressure in the separated laminar region cause the large drag CD 1.2. The turbulent boundary layer in Fig. 7.13b is more resistant, and separation is delayed until 120°, with a resulting smaller wake, higher pressure on the rear, and 75 percent less drag, CD 0.3. This explains the sharp drop in drag at transition in Fig. 5.3. The same sharp difference between vulnerable laminar separation and resistant turbulent separation can be seen for a sphere in Fig. 7.14. The laminar flow (Fig. 7.14a) separates at about 80°, CD 0.5, while the turbulent flow (Fig. 7.14b) separates at 120°, CD 0.2. Here the Reynolds numbers are exactly the same, and the turbulent boundary layer is induced by a patch of sand roughness at the nose of the ball. Golf balls fly in this range of Reynolds numbers, which is why they are deliberately dimpled to induce a turbulent boundary layer and lower drag. Again we would find the actual pressure distribution on the sphere to be quite different from that predicted by inviscid theory. | e-Text Main Menu | Textbook Table of Contents | Study Guide 456 Chapter 7 Flow Past Immersed Bodies Fig. 7.14 Strong differences in laminar and turbulent separation on an 8.5-in bowling ball entering water at 25 ft/s: (a) smooth ball, laminar boundary layer; (b) same entry, turbulent flow induced by patch of nose-sand roughness. (U.S. Navy photograph, Ordnance Test Station, Pasadena Annex.) (a ) (b ) In general, we cannot overstress the importance of body streamlining to reduce drag at Reynolds numbers above about 100. This is illustrated in Fig. 7.15. The rectangular cylinder (Fig. 7.15a) has rampant separation at all sharp corners and very high drag. Rounding its nose (Fig. 7.15b) reduces drag by about 45 percent, but CD is still high. Streamlining its rear to a sharp trailing edge (Fig. 7.15c) reduces its drag another 85 percent to a practical minimum for the given thickness. As a dramatic contrast, the circular cylinder (Fig. 7.15d) has one-eighth the thickness and one-threehundredth the cross section (c) (Fig. 7.15c), yet it has the same drag. For high-performance vehicles and other moving bodies, the name of the game is drag reduction, for which intense research continues for both aerodynamic and hydrodynamic applications [20, 39]. The drag of some representative wide-span (nearly two-dimensional) bodies is shown versus the Reynolds number in Fig. 7.16a. All bodies have high CD at very low (creeping flow) Re 1.0, while they spread apart at high Reynolds numbers according to CD = 2.0 V | v v Fig. 7.15 The importance of streamlining in reducing drag of a body (CD based on frontal area): (a) rectangular cylinder; (b) rounded nose; (c) rounded nose and streamlined sharp trailing edge; (d) circular cylinder with the same drag as case (c). CD = 1.1 V (b) (a) V CD = 0.15 V (c) | e-Text Main Menu (d ) | Textbook Table of Contents | Study Guide 7.6 Experimental External Flows 457 their degree of streamlining. All values of CD are based on the planform area except the plate normal to the flow. The birds and the sailplane are, of course, not very twodimensional, having only modest span length. Note that birds are not nearly as efficient as modern sailplanes or airfoils [14, 15]. Table 7.2 gives a few data on drag, based on frontal area, of two-dimensional bodies of various cross section, at Re 104. The sharp-edged bodies, which tend to cause flow separation regardless of the character of the boundary layer, are insensitive to the Reynolds number. The elliptic cylinders, being smoothly rounded, have the laminarto-turbulent transition effect of Figs. 7.13 and 7.14 and are therefore quite sensitive to whether the boundary layer is laminar or turbulent. 100 Smooth circular cylinder L =∞ D =5 10 Plate normal to stream Square cylinder 1 CD 0.1 0.01 Pigeon Seagull Smooth flat plate parallel to stream Vulture Sailplane Airfoil Transition 0.001 0.1 1 10 10 3 Re (a) 100 10 4 10 5 10 6 10 7 100 10 CD 1 Stokes' law: 24/Re Disk Sphere 2:1 ellipsoid 0.1 | v v Fig. 7.16 Drag coefficients of smooth bodies at low Mach numbers: (a) two-dimensional bodies; (b) three-dimensional bodies. Note the Reynolds-number independence of blunt bodies at high Re. | Airship hull 0.01 0.1 e-Text Main Menu 1 | 10 100 10 3 Re (b) Textbook Table of Contents 10 4 | 10 5 10 6 Study Guide 10 7 458 Chapter 7 Flow Past Immersed Bodies Table 7.2 Drag of TwoDimensional Bodies at Re CD based on frontal area 104 Shape Square cylinder: CD based on frontal area Shape CD based on frontal area Shape Plate: Plate: Half-cylinder: 2.1 1.6 2.0 1.2 1.7 Half tube: Thin plate normal to a wall: 1.4 Equilateral triangle: 1.6 1.2 Hexagon: 2.0 2.3 Shape 1.0 0.7 CD based on frontal area Rounded nose section: H L/H:: L/H CD: 0.5 1.16 1.0 0.90 0.4 2.3 0.7 2.7 2.0 0.70 1.2 2.1 4.0 0.68 6.0 0.64 L Flat nose section Rounded nose section: L/H: L/H: CD: D H 0.1 1.9 2.0 1.8 2.5 1.4 L Laminar Turbulent 1:1 1.2 0.3 2:1 0.6 0.2 4:1 0.35 0.15 8:1 0.25 0.1 | v v Elliptical cylinder: | e-Text Main Menu | Textbook Table of Contents | Study Guide 3.0 1.3 6.0 0.9 7.6 Experimental External Flows 459 EXAMPLE 7.6 A square 6-in piling is acted on by a water flow of 5 ft/s that is 20 ft deep, as shown in Fig. E7.6. Estimate the maximum bending exerted by the flow on the bottom of the piling. h = 6 in 5 ft/s L = 20 ft E7.6 Solution Assume seawater with 1.99 slugs/ft3 and kinematic viscosity ing width of 0.5 ft, we have Reh (5 ft/s)(0.5 ft) 0.000011 ft2/s 2.3 0.000011 ft2/s. With a pil- 105 This is the range where Table 7.2 applies. The worst case occurs when the flow strikes the flat side of the piling, CD 2.1. The frontal area is A Lh (20 ft)(0.5 ft) 10 ft2. The drag is estimated by F CD( 1 V2A) 2 2.1( 1 )(1.99 slugs/ft3)(5 ft/s)2(10 ft2) 2 522 lbf If the flow is uniform, the center of this force should be at approximately middepth. Therefore the bottom bending moment is M0 FL 2 522(10) 5220 ft lbf Ans. According to the flexure formula from strength of materials, the bending stress at the bottom would be S M0y I (5220 ft lb)(0.25 ft) 1 4 12 (0.5 ft) 251,000 lbf/ft2 1740 lbf/in2 to be multiplied, of course, by the stress-concentration factor due to the built-in end conditions. | v v Some drag coefficients of three-dimensional bodies are listed in Table 7.3 and Fig. 7.16b. Again we can conclude that sharp edges always cause flow separation and high drag which is insensitive to the Reynolds number. Rounded bodies like the ellipsoid have drag which depends upon the point of separation, so that both the Reynolds num- | e-Text Main Menu | Textbook Table of Contents | Study Guide 460 Chapter 7 Flow Past Immersed Bodies Table 7.3 Drag of Three-Dimensional Bodies at Re CD based on frontal area Body 104 Body CD based on frontal area Cone: Cube: 1.07 θ: CD: θ 0.81 10˚ 0.30 20˚ 0.40 30˚ 0.55 40˚ 0.65 60˚ 0.80 75˚ 1.05 90˚ 1.15 L /D: CD: 1 0.64 2 0.68 3 0.72 5 0.74 10 0.82 20 0.91 40 0.98 Short cylinder, laminar flow: L D Cup: 1.4 Porous parabolic dish [23]: Porosity: CD: CD: 0.4 0 1.42 0.95 0.1 1.33 0.92 0.2 1.20 0.90 0.3 1.05 0.86 0.4 0.95 0.83 0.5 0.82 0.80 Average person: Disk: CD A ≈ 9 ft 2 1.17 Parachute (Low porosity): Pine and spruce trees [24]: 1.2 Body U, m/s: CD: CD based on frontal area Ratio C D A ≈ 1.2 ft 2 10 1.2 ± 0.2 30 0.7 ± 0.2 40 0.5 ± 0.2 Ratio Body Rectangular plate: 20 1.0 ± 0.2 CD based on frontal area L /d 0.5 1 2 4 8 1.15 0.90 0.85 0.87 0.99 Flat-faced cylinder: b/h 1 5 10 20 ∞ h b h 1.18 1.2 1.3 1.5 2.0 Laminar Ellipsoid: L /d 0.75 1 2 4 8 d | v v L | 0.5 0.47 0.27 0.25 0.2 d L Turbulent 0.2 0.2 0.13 0.1 0.08 e-Text Main Menu | Textbook Table of Contents | Study Guide ∞ 1.20 7.6 Experimental External Flows 461 ber and the character of the boundary layer are important. Body length will generally decrease pressure drag by making the body relatively more slender, but sooner or later the friction drag will catch up. For the flat-faced cylinder in Table 7.3, pressure drag decreases with L/d but friction increases, so that minimum drag occurs at about L/d 2. Aerodynamic Forces on Road Vehicles Automobiles and trucks are now the subject of much research on aerodynamic forces, both lift and drag [21]. At least one textbook is devoted to the subject [22]. Consumer, manufacturer, and government interest has cycled between high speed/high horsepower and lower speed/lower drag. Better streamlining of car shapes has resulted over the years in a large decrease in the automobile drag coefficient, as shown in Fig. 7.17a. Modern cars have an average drag coefficient of about 0.35, based upon the frontal | v v Fig. 7.17 Aerodynamics of automobiles: (a) the historical trend for drag coefficients [From Ref. 21]; (b) effect of bottom rear upsweep angle on drag and downward lift force [From Ref. 25]. | e-Text Main Menu | Textbook Table of Contents | Study Guide 462 Chapter 7 Flow Past Immersed Bodies Horsepower required area. Since the frontal area has also decreased sharply, the actual raw drag force on cars has dropped even more than indicated in Fig. 7.17a. The practical minimum, shown tentatively for the year 2000, is CD 0.15 for a tear-shaped vehicle, which can be achieved any time the public is willing to purchase such a shape. Note that basing CD on the frontal area is awkward, since one would need an accurate drawing of the automobile to estimate its frontal area. For this reason, some technical articles simply report the raw drag in newtons or pound-force, or the product CDA. Many companies and laboratories have automotive wind tunnels, some full-scale and/or with moving floors to approximate actual kinematic similarity. The blunt shapes of most automobiles, together with their proximity to the ground, cause a wide variety of flow and geometric effects. Simple changes in part of the shape can have a large influence on aerodynamic forces. Figure 7.17b shows force data by Bearman et al. [25] for an idealized smooth automobile shape with upsweep in the rear of the bottom section. We see that by simply adding an upsweep angle of 25°, we can quadruple the downward force, gaining tire traction at the expense of doubling the drag. For this study, the effect of a moving floor was small — about a 10 percent increase in both drag and lift compared to a fixed floor. It is difficult to quantify the exact effect of geometric changes on automotive forces, since, e.g., changes in a windshield shape might interact with downstream flow over the roof and trunk. Nevertheless, based on correlation of many model and full-scale tests, Ref. 26 proposes a formula for automobile drag which adds separate effects such as front ends, cowls, fenders, windshield, roofs, and rear ends. Figure 7.18 shows the horsepower required to drive a typical tractor-trailer truck at speeds up to 80 mi/h (117 ft/s or 36 m/s). The rolling resistance increases linearly and the air drag quadratically with speed (CD 1.0). The two are about equally important at 55 mi/h, which is the nominal speed limit in the United States. As shown in Fig. 7.18b, air drag can be reduced by attaching a deflector to the top of the tractor. If the angle of the deflector is adjusted to carry the flow smoothly over the top and around the sides of the trailer, the reduction in CD is about 20 percent. Thus, at 55 mi/h the total resistance is reduced 10 percent, with a corresponding reduction in | v v Fig. 7.18 Drag reduction of a tractor-trailer truck: (a) horsepower required to overcome resistance; (b) deflector added to cab reduces air drag by 20 percent. (Uniroyal Inc.) 550 500 450 400 350 300 250 200 150 100 50 0 Gross engine horsepower required Air resistance Rolling resistance, hp 0 10 20 30 40 50 60 70 Vehicle speed, mi/h 80 (a) | e-Text Main Menu | Textbook Table of Contents (b) | Study Guide 7.6 Experimental External Flows 463 fuel costs and/or trip time for the trucker. This type of applied fluids engineering can be a large factor in many of the conservation-oriented transportation problems of the future. EXAMPLE 7.7 A high-speed car with m 2000 kg, CD 0.3, and A 1 m2 deploys a 2-m parachute to slow down from an initial velocity of 100 m/s (Fig. E7.7). Assuming constant CD, brakes free, and no rolling resistance, calculate the distance and velocity of the car after 1, 10, 100, and 1000 s. For air assume 1.2 kg/m3, and neglect interference between the wake of the car and the parachute. dp = 2 m V0 = 100 m/s x E7.7 Solution Newton’s law applied in the direction of motion gives m Fx dV dt Fc 1 V 2(CDcAc 2 Fp CDp Ap) where subscript c denotes the car and subscript p the parachute. This is of the form K2 V m dV dt K CD A 2 Separate the variables and integrate V dV V2 V0 or V0 1 K m V t dt 0 K t m 1 Rearrange and solve for the velocity V: V V0 (K/m)V0t 1 (CDc Ac K CDp Ap) 2 (1) We can integrate this to find the distance traveled: S V0 ln (1 K V0 m t) 1.2; hence Now work out some numbers. From Table 7.3, CDp | v v CDc Ac | e-Text Main Menu | CDp Ap 0.3(1 m2) Textbook Table of Contents (2) 1.2 | 4 (2 m)2 4.07 m2 Study Guide 464 Chapter 7 Flow Past Immersed Bodies 1 2 K V0 m Then (4.07 m2)(1.2 kg/m3)(100 m/s) 2000 kg 0.122 s 1 Now make a table of the results for V and S from Eqs. (1) and (2): t, s 1 10 100 1000 V, m/s 89 45 7.6 0.8 S, m 94 654 2110 3940 Air resistance alone will not stop a body completely. If you don’t apply the brakes, you’ll be halfway to the Yukon Territory and still going. Other Methods of Drag Reduction Sometimes drag is good, for example, when using a parachute. Do not jump out of an airplane holding a flat plate parallel to your motion (see Prob. 7.81). Mostly, though, drag is bad and should be reduced. The classical method of drag reduction is streamlining (Figs. 7.15 and 7.18). For example, nose fairings and body panels have produced motorcycles which can travel over 200 mi/h. More recent research has uncovered other methods which hold great promise, especially for turbulent flows. 1. Oil pipelines introduce an annular core of water to reduce the pumping power [36]. The low-viscosity water rides the wall and reduces friction up to 60 percent. 2. Turbulent friction in liquid flows is reduced up to 60 percent by dissolving small amounts of a high-molecular-weight polymer additive [37]. Without changing pumps, the Trans-Alaska Pipeline System (TAPS) increased oil flow 50 percent by injecting small amounts of polymer dissolved in kerosene. 3. Stream-oriented surface vee-groove microriblets reduce turbulent friction up to 8 percent [38]. Riblet heights are of order 1 mm and were used on the Stars and Stripes yacht hull in the Americas Cup races. Riblets are also effective on aircraft skins. 4. Small, near-wall large-eddy breakup devices (LEBUs) reduce local turbulent friction up to 10 percent [39]. However, one must add these small structures to the surface. 5. Air microbubbles injected at the wall of a water flow create a low-shear bubble blanket [40]. At high void fractions, drag reduction can be 80 percent. 6. Spanwise (transverse) wall oscillation may reduce turbulent friction up to 30 percent [41]. Drag reduction is presently an area of intense and fruitful research and applies to many types of airflows and water flows for both vehicles and conduits. The drag data above, such as Tables 7.2 and 7.3, are for bodies “fully immersed” in a free stream, i.e., with no free surface. If, however, the body moves at or near a free liquid surface, wave-making drag becomes important and is dependent upon both the Reynolds number and the Froude number. To move through a water surface, a ship | v v Drag of Surface Ships | e-Text Main Menu | Textbook Table of Contents | Study Guide 7.6 Experimental External Flows 465 must create waves on both sides. This implies putting energy into the water surface and requires a finite drag force to keep the ship moving, even in a frictionless fluid. The total drag of a ship can then be approximated as the sum of friction drag and wavemaking drag: F Ffric Fwave or CD CD,fric CD,wave The friction drag can be estimated by the (turbulent) flat-plate formula, Eq. (7.45), based on the below-water or wetted area of the ship. Reference 27 is an interesting review of both theory and experiment for wakemaking surface ship drag. Generally speaking, the bow of the ship creates a wave system whose wavelength is related to the ship speed but not necessarily to the ship length. If the stern of the ship is a wave trough, the ship is essentially climbing uphill and has high wave drag. If the stern is a wave crest, the ship is nearly level and has lower drag. The criterion for these two conditions results in certain approximate Froude numbers [27]: Fr V gL 0.53 N high drag if N low drag if N 1, 3, 5, 7, . . . ; (7.65) 2, 4, 6, 8, . . . where V is the ship’s speed, L is the ship’s length along the centerline, and N is the number of half-lengths, from bow to stern, of the drag-making wave system. The wave drag will increase with the Froude number and oscillate between lower drag (Fr 0.38, 0.27, 0.22, . . .) and higher drag (Fr 0.53, 0.31, 0.24, . . .) with negligible variation for Fr 0.2. Thus it is best to design a ship to cruise at N 2, 4, 6, 8. Shaping the bow and stern can further reduce wave-making drag. Figure 7.19 shows the data of Inui [27] for a model ship. The main hull, curve A, shows peaks and valleys in wave drag at the appropriate Froude numbers 0.2. Introduction of a bulb protrusion on the bow, curve B, greatly reduces the drag. Adding a second bulb to the stern, curve C, is still better, and Inui recommends that the design speed of this two-bulb ship be at N 4, Fr 0.27, which is a nearly “waveless” condition. In this figure CD,wave is defined as 2Fwave/( V 2L2) instead of using the wetted area. The solid curves in Fig. 7.19 are based on potential-flow theory for the belowwater hull shape. Chapter 8 is an introduction to potential-flow theory. Modern digital computers can be programmed for numerical CFD solutions of potential flow over the hulls of ships, submarines, yachts, and sailboats, including boundary-layer effects driven by the potential flow [28]. Thus theoretical prediction of flow past surface ships is now at a fairly high level. See also Ref. 15. | v v Body Drag at High Mach Numbers | All the data presented above are for nearly incompressible flows, with Mach numbers assumed less than about 0.5. Beyond this value compressibility can be very important, with CD fcn(Re, Ma). As the stream Mach number increases, at some subsonic value Mcrit 1 which depends upon the body’s bluntness and thickness, the local velocity at some point near the body surface will become sonic. If Ma increases beyond Macrit, shock waves form, intensify, and spread, raising surface pressures near the front of the body and therefore increasing the pressure drag. The effect can be dramatic with CD e-Text Main Menu | Textbook Table of Contents | Study Guide 466 Chapter 7 Flow Past Immersed Bodies A Main hull (without bulb) B With bow-bulb C With bow and stern-bulbs 0.002 Potential-flow theory CD, wave 0.001 A B C 0 0.10 Fig. 7.19 Wave-making drag on a ship model. (After Inui [27].) Note: The drag coefficient is defined as CDW 2F/( V2L2). 0.20 0.30 0.40 Design speed 0.50 0.60 V Fr = √Lg increasing tenfold, and 70 years ago this sharp increase was called the sonic barrier, implying that it could not be surmounted. Of course, it can be — the rise in CD is finite, as supersonic bullets have proved for centuries. Figure 7.20 shows the effect of the Mach number on the drag coefficient of various body shapes tested in air.5 We see that compressibility affects blunt bodies earlier, with Macrit equal to 0.4 for cylinders, 0.6 for spheres, and 0.7 for airfoils and pointed projectiles. Also the Reynolds number (laminar versus turbulent boundary-layer flow) has a large effect below Macrit for spheres and cylinders but becomes unimportant above Ma 1. In contrast, the effect of the Reynolds number is small for airfoils and projectiles and is not shown in Fig. 7.20. A general statement might divide Reynolds- and Mach-number effects as follows: Ma 0.4: Reynolds number important, Mach number unimportant 0.4 Ma 1: both Reynolds and Mach numbers important Ma 1.0: Reynolds number unimportant, Mach number important At supersonic speeds, a broad bow shock wave forms in front of the body (see Figs. 9.10b and 9.19), and the drag is mainly due to high shock-induced pressures on the front. Making the bow a sharp point can sharply reduce the drag (Fig. 9.28) but does not eliminate the bow shock. Chapter 9 gives a brief treatment of compressibility. References 30 and 31 are more advanced textbooks devoted entirely to compressible flow. 5 | v v There is a slight effect of the specific-heat ratio k which would appear if other gases were tested. | e-Text Main Menu | Textbook Table of Contents | Study Guide 7.6 Experimental External Flows 467 Biological Drag Reduction A great deal of engineering effort goes into designing immersed bodies to reduce their drag. Most such effort concentrates on rigid-body shapes. A different process occurs in nature, as organisms adapt to survive high winds or currents, as reported in a series of papers by S. Vogel [33, 34]. A good example is a tree, whose flexible structure allows it to reconfigure in high winds and thus reduce drag and damage. Tree root systems have evolved in several ways to resist wind-induced bending moments, and trunk cross sections have become resistant to bending but relatively easy to twist and reconfigure. We saw this in Table 7.3, where tree drag coefficients [24] reduced by 60 percent as wind velocity increased. The shape of the tree changes to offer less resistance. The individual branches and leaves of a tree also curl and cluster to reduce drag. Figure 7.21 shows the results of wind tunnel experiments by Vogel [33]. A tulip tree leaf, Fig. 7.21(a), broad and open in low wind, curls into a conical low-drag shape as wind increases. A compound black walnut leaf group, Fig. 7.21(b), clusters into a lowdrag shape at high wind speed. Although drag coefficients were reduced up to 50 percent by flexibility, Vogel points out that rigid structures are sometimes just as effective. An interesting recent symposium [35] was devoted entirely to the solid mechanics and fluid mechanics of biological organisms. Forces on Lifting Bodies Lifting bodies (airfoils, hydrofoils, or vanes) are intended to provide a large force normal to the free stream and as little drag as possible. Conventional design practice has evolved a shape not unlike a bird’s wing, i.e., relatively thin (t/c 0/18) with a rounded leading edge and a sharp trailing edge. A typical shape is sketched in Fig. 7.22. For our purposes we consider the body to be symmetric, as in Fig. 7.11, with the 2.0 1.8 Cylinder in cross flow: 1.6 Laminar, Re ≈ 1 E5 Turbulent, Re ≈ 1 E6 1.4 1.2 CD 1.0 Sphere Laminar, Re ≈ 1 E5 Turbulent, Re ≈ 1 E6 0.8 0.6 Pointed body of revolution 0.4 | v v Fig. 7.20 Effect of the Mach number on the drag of various body shapes. (Data from Refs. 23 and 29.) | Airfoil 0.2 0.0 0.0 e-Text Main Menu 1.0 | 2.0 Mach number 3.0 Textbook Table of Contents 4.0 | Study Guide 468 Chapter 7 Flow Past Immersed Bodies 5 m/s 5 m/s 10 m/s 10 m/s 20 m/s Fig. 7.21 Biological adaptation to wind forces: (a) a tulip tree leaf curls into a conical shape at high velocity; (b) black walnut leaves cluster into a low-drag shape as wind increases. (From Vogel, Ref. 33.) 20 m/s (a) (b) free-stream velocity in the vertical plane. If the chord line between the leading and trailing edge is not a line of symmetry, the airfoil is said to be cambered. The camber line is the line midway between the upper and lower surfaces of the vane. The angle between the free stream and the chord line is called the angle of attack . The lift L and the drag D vary with this angle. The dimensionless forces are defined with respect to the planform area Ap bc: Lift coefficient: CL Drag coefficient: CD L 1 2 (7.66a) V2Ap D 1 2 (7.66b) V2Ap Planform area = bc Lift Drag Angle of attack t = thickness α b = span V c = cho | v v Fig. 7.22 Definition sketch for a lifting vane. rd | e-Text Main Menu | Textbook Table of Contents | Study Guide 7.6 Experimental External Flows 469 If the chord length is not constant, as in the tapered wings of modern aircraft, Ap c db. For low-speed flow with a given roughness ratio, CL and CD should vary with and the chord Reynolds number CL f ( , Rec) or CD f ( , Rec) where Rec Vc/ . The Reynolds numbers are commonly in the turbulent-boundarylayer range and have a modest effect. The rounded leading edge prevents flow separation there, but the sharp trailing edge causes a separation which generates the lift. Figure 7.23 shows what happens when a flow starts up past a lifting vane or an airfoil. Just after start-up in Fig. 7.23a the streamline motion is irrotational and inviscid. The rear stagnation point, assuming a positive angle of attack, is on the upper surface, and there is no lift; but the flow cannot long negotiate the sharp turn at the trailing edge: it separates, and a starting vortex forms in Fig. 7.23b. This starting vortex is shed downstream in Fig. 7.23c and d, and a smooth streamline flow develops over the wing, leaving the foil in a direction approximately parallel to the chord line. Lift at this time is fully developed, and the starting vortex is gone. Should the flow now cease, a stopping vortex of opposite (clockwise) sense will form and be shed. During flight, increases or decreases in lift will cause incremental starting or stopping vortices, always with the effect of maintaining a smooth parallel flow at the trailing edge. We pursue this idea mathematically in Chap. 8. At a low angle of attack, the rear surfaces have an adverse pressure gradient but not enough to cause significant boundary-layer separation. The flow pattern is smooth, as in Fig. 7.23d, and drag is small and lift excellent. As the angle of attack is increased, the upper-surface adverse gradient becomes stronger, and generally a separation bub- (a) | v v Fig. 7.23 Transient stages in the development of lift: (a) start-up: rear stagnation point on the upper surface: no lift; (b) sharp trailing edge induces separation, and a starting vortex forms: slight lift; (c) starting vortex is shed, and streamlines flow smoothly from trailing edge: lift is now 80 percent developed; (d) starting vortex now shed far behind, trailing edge now very smooth: lift fully developed. | e-Text Main Menu (b) (c) (d ) | Textbook Table of Contents | Study Guide 470 Chapter 7 Flow Past Immersed Bodies Fig. 7.24 At high angle of attack, smoke-flow visualization shows stalled flow on the upper surface of a lifting vane. [From Ref. 19, Illustrated Experiments in Fluid Mechanics (The NCFMF Book of Film Notes), National Committee for Fluid Mechanics Films, Education Development Center, Inc., copyright 1972.] ble begins to creep forward on the upper surface.6 At a certain angle 15 to 20°, the flow is separated completely from the upper surface, as in Fig. 7.24. The airfoil is said to be stalled: Lift drops off markedly, drag increases markedly, and the foil is no longer flyable. Early airfoils were thin, modeled after birds’ wings. The German engineer Otto Lilienthal (1848 – 1896) experimented with flat and cambered plates on a rotating arm. He and his brother Gustav flew the world’s first glider in 1891. Horatio Frederick Phillips (1845 – 1912) built the first wind tunnel in 1884 and measured the lift and drag of cambered vanes. The first theory of lift was proposed by Frederick W. Lanchester shortly afterward. Modern airfoil theory dates from 1905, when the Russian hydrodynamicist N. E. Joukowsky (1847 – 1921) developed a circulation theorem (Chap. 8) for computing airfoil lift for arbitrary camber and thickness. With this basic theory, as extended and developed by Prandtl and Kármán and their students, it is now possible to design a low-speed airfoil to satisfy particular surface-pressure distributions and boundary-layer characteristics. There are whole families of airfoil designs, notably those developed in the United States under the sponsorship of the NACA (now NASA). Extensive theory and data on these airfoils are contained in Ref. 16. We shall discuss this further in Chap. 8. Figure 7.25 shows the lift and drag on a symmetric airfoil denoted as the NACA 0009 foil, the last digit indicating the thickness of 9 percent. With no flap extended, 6 | v v For some airfoils the bubble leaps, not creeps, forward, and stall occurs rapidly and dangerously. | e-Text Main Menu | Textbook Table of Contents | Study Guide 7.6 Experimental External Flows 471 this airfoil, as expected, has zero lift at zero angle of attack. Up to about 12° the lift coefficient increases linearly with a slope of 0.1 per degree, or 6.0 per radian. This is in agreement with the theory outlined in Chap. 8: CL,theory 2h c 2 sin (7.67) where h/c is the maximum camber expressed as a fraction of the chord. The NACA 0009 has zero camber; hence CL 2 sin 0.11 , where is in degrees. This is excellent agreement. The drag coefficient of the smooth-model airfoils in Fig. 7.25 is as low as 0.005, which is actually lower than both sides of a flat plate in turbulent flow. This is misleading inasmuch as a commercial foil will have roughness effects; e.g., a paint job will double the drag coefficient. The effect of increasing Reynolds number in Fig. 7.25 is to increase the maximum lift and stall angle (without changing the slope appreciably) and to reduce the drag coefficient. This is a salutary effect, since the prototype will probably be at a higher Reynolds number than the model (107 or more). For takeoff and landing, the lift is greatly increased by deflecting a split flap, as shown in Fig. 7.25. This makes the airfoil unsymmetric (or effectively cambered) and changes the zero-lift point to 12°. The drag is also greatly increased by the flap, but the reduction in takeoff and landing distance is worth the extra power needed. A lifting craft cruises at low angle of attack, where the lift is much larger than the drag. Maximum lift-to-drag ratios for the common airfoils lie between 20 and 50. Some airfoils, such as the NACA 6 series, are shaped to provide favorable gradients over much of the upper surface at low angles. Thus separation is small, and transition to turbulence is delayed; the airfoil retains a good length of laminar flow even CL Rec = 6 × 10 6 1.6 0.04 α CD Split flap 1.2 With flap at 60° 0.03 Rec = 9 × 10 6 With flap at 60° 0.02 6 × 10 6 3 × 10 6 0.8 Rec = 3 × 10 6 6 × 10 6 9 × 10 6 No flap 0.4 | v v Fig. 7.25 Lift and drag of a symmetric NACA 0009 airfoil of infinite span, including effect of a split-flap deflection. Note that roughness can increase CD from 100 to 300 percent. | 0.01 No flap –12 –8 e-Text Main Menu –4 0 4 8 12 16 –8 α , deg | Textbook Table of Contents –4 0 4 α , deg | Study Guide 8 12 16 472 Chapter 7 Flow Past Immersed Bodies Stall 1.2 NACA 0009 Stall 0009 with split flap NACA 63 – 009 0.8 CL Low-drag “bucket” 0.4 0 Fig. 7.26 Lift-drag polar plot for standard (0009) and a laminar-flow (63-009) NACA airfoil. 0 0.008 0.016 0.024 CD at high Reynolds numbers. The lift-drag polar plot in Fig. 7.26 shows the NACA 0009 data from Fig. 7.25 and a laminar-flow airfoil, NACA 63 – 009, of the same thickness. The laminar-flow airfoil has a low-drag bucket at small angles but also suffers lower stall angle and lower maximum lift coefficient. The drag is 30 percent less in the bucket, but the bucket disappears if there is significant surface roughness. All the data in Figs. 7.25 and 7.26 are for infinite span, i.e., a two-dimensional flow pattern about wings without tips. The effect of finite span can be correlated with the dimensionless slenderness, or aspect ratio, denoted (AR), b2 Ap AR b c (7.68) where c is the average chord length. Finite-span effects are shown in Fig. 7.27. The lift slope decreases, but the zero-lift angle is the same; and the drag increases, but the zero-lift drag is the same. The theory of finite-span airfoils [16] predicts that the effective angle of attack increases, as in Fig. 7.27, by the amount CL AR (7.69) When applied to Eq. (7.67), the finite-span lift becomes CL The associated drag increase is CD | v v CD | e-Text Main Menu | 2 sin ( 2h/c) 1 2/AR CL sin CD (7.70) CL , or 2 CL (7.71) AR Textbook Table of Contents | Study Guide 7.6 Experimental External Flows 473 where CD is the drag of the infinite-span airfoil, as sketched in Fig. 7.25. These correlations are in good agreement with experiments on finite-span wings [16]. The existence of a maximum lift coefficient implies the existence of a minimum speed, or stall speed, for a craft whose lift supports its weight L or W 2 CL,max( 1 V s Ap) 2 1/2 2W Vs (7.72) CL,max Ap The stall speed of typical aircraft varies between 60 and 200 ft/s, depending upon the weight and value of CL,max. The pilot must hold the speed greater than about 1.2Vs to avoid the instability associated with complete stall. The split flap in Fig. 7.25 is only one of many devices used to secure high lift at low speeds. Figure 7.28a shows six such devices whose lift performance is given in 7.28b along with a standard (A) and laminar-flow (B) airfoil. The double-slotted flap achieves CL,max 3.4, and a combination of this plus a leading-edge slat can achieve CL,max 4.0. These are not scientific curiosities; e.g., the Boeing 727 commercial jet aircraft uses a triple-slotted flap plus a leading-edge slat during landing. Also shown as C in Fig. 7.28b is the Kline-Fogleman airfoil [17], not yet a reality. The designers are amateur model-plane enthusiasts who did not know that conventional aerodynamic wisdom forbids a sharp leading edge and a step cutout from the trailing edge. The Kline-Fogleman airfoil has relatively high drag but shows an amazing continual increase in lift out to 45°. In fact, we may fairly say that this airfoil does not stall and provides smooth performance over a tremendous range of flight conditions. No explanation for this behavior has yet been published by any aerodynamicist. This airfoil is under study and may or may not have any commercial value. Another violation of conventional aerodynamic wisdom is that military aircraft are beginning to fly, briefly, above the stall point. Fighter pilots are learning to make quick maneuvers in the stalled region as detailed in Ref. 32. Some planes can even fly continuously while stalled — the Grumman X-29 experimental aircraft recently set a record 67°. by flying at AR = ∞ CL AR = b2 Ap CD AR = ∆ CD ≈ 2 CL π AR b2 Ap AR = ∞ CL ∆α ≈ π AR | v v Fig. 7.27 Effect of finite aspect ratio on lift and drag of an airfoil: (a) effective angle increase; (b) induced drag increase. | e-Text Main Menu CD ∞ α –β (a) | Textbook Table of Contents (b) | Study Guide α 474 Chapter 7 Flow Past Immersed Bodies 4 D Optimum but cumbersome combination CL Plain flap or aileron H 3 E G Split flap F KlineFogleman airfoil E, D F External airfoil flap 2 I G Slotted flap A 1 H Fig. 7.28 Performance of airfoils with and without high-lift devices: A NACA 0009; B NACA 63009; C Kline-Fogleman airfoil (from Ref. 17); D to I shown in (a): (a) types of high-lift devices; (b) lift coefficients for various devices. C B Double-slotted flap α, deg I Leading edge slat – 10 0 10 New Aircraft Designs 20 30 40 50° (b) (a) The Kline-Fogleman airfoil in Fig. 7.28 is a departure from conventional aerodynamics, but there have been other striking departures, as detailed in a recent article [42]. These new aircraft, conceived presently as small models, have a variety of configurations, as shown in Fig. 7.29: ring-wing, cruciform, flying saucer, and flap-wing. A saucer configuration (Fig. 7.29c), with a diameter of 40 in, has been successfully flown by radio control, and its inventor, Jack M. Jones, plans for a 20-ft two-passenger version. Another 18in-span microplane called the Bat (not shown), made by MLB Co., flies for 20 min at 40 mi/h and contains a video camera for surveillance. New engines have been reduced to a 10- by 3-mm size, producing 20 W of power. At the other end of the size spectrum, Boeing and NASA engineers have proposed a jumbo flying-wing jetliner, similar in shape to the stealth bomber, which would carry 800 passengers for a range of 7000 mi. Further information on the performance of lifting craft can be found in Refs. 12, 13, and 16. We discuss this matter again briefly in Chap. 8. (a) Ring-wing | v v Fig. 7.29 New aircraft designs do not necessarily look like your typical jetliner. (From Ref. 42.) (b) Cruciform delta (c) Flying saucer | e-Text Main Menu (d) Flap-wing dragonfly | Textbook Table of Contents | Study Guide 7.6 Experimental External Flows 475 EXAMPLE 7.8 An aircraft weighs 75,000 lb, has a planform area of 2500 ft2, and can deliver a constant thrust of 12,000 lb. It has an aspect ratio of 7, and CD 0.02. Neglecting rolling resistance, estimate the takeoff distance at sea level if takeoff speed equals 1.2 times stall speed. Take CL,max 2.0. Solution 0.00237 slug/ft3, is The stall speed from Eq. (7.72), with sea-level density 2W Vs 1/2 2(75,000) 2.0(0.00237)(2500) CL,max Ap 1.2Vs Hence takeoff speed V0 1/2 112.5 ft/s 135 ft/s. The drag is estimated from Eq. (7.71) for AR CD 2 CL 7 0.02 0.02 7 as 2 0.0455C L A force balance in the direction of takeoff gives Fs m dV dt thrust drag kV2 T k Since we are looking for distance, not time, we introduce dV/dt rate variables, and integrate S0 0 or S0 V0 m 2 dS 0 d(V2) T kV2 T m ln 2 T kV 0 2k k 1 2 CD Ap (1) V dV/ds into Eq. (1), sepa- const T m ln T D0 2k (2) where D0 kV 2 is the takeoff drag. Equation (2) is the desired theoretical relation for takeoff 0 distance. For the particular numerical values, take 75,000 32.2 m CL0 CD0 k 1 2 1 2 W V 2 Ap 0 75,000 (0.00237)(135)2(2500) 0.0455(CL0)2 0.02 CD0 Ap 1 2 2329 slugs 0.108 1 2 ( )(0.108)(0.00237)(2500) D0 kV 2 0 1.39 0.319 slug/ft 5820 lb Then Eq. (2) predicts that S0 2329 slugs 12,000 ln 2(0.319 slug/ft) 12,000 5820 3650 ln 1.94 2420 ft Ans. | v v A more exact analysis accounting for variable k [13] gives the same result to within 1 percent. | e-Text Main Menu | Textbook Table of Contents | Study Guide 476 Chapter 7 Flow Past Immersed Bodies Summary This chapter has dealt with viscous effects in external flow past bodies immersed in a stream. When the Reynolds number is large, viscous forces are confined to a thin boundary layer and wake in the vicinity of the body. Flow outside these “shear layers” is essentially inviscid and can be predicted by potential theory and Bernoulli’s equation. The chapter begins with a discussion of the flat-plate boundary layer and the use of momentum-integral estimates to predict the wall shear, friction drag, and thickness of such layers. These approximations suggest how to eliminate certain small terms in the Navier-Stokes equations, resulting in Prandtl’s boundary-layer equations for laminar and turbulent flow. Section 7.4 then solves the boundary-layer equations to give very accurate formulas for flat-plate flow at high Reynolds numbers. Rough-wall effects are included, and Sec. 7.5 gives a brief introduction to pressure-gradient effects. An adverse (decelerating) gradient is seen to cause flow separation, where the boundary layer breaks away from the surface and forms a broad, low-pressure wake. Boundary-layer theory fails in separated flows, which are commonly studied by experiment. Section 7.6 gives data on drag coefficients of various two- and threedimensional body shapes. The chapter ends with a brief discussion of lift forces generated by lifting bodies such as airfoils and hydrofoils. Airfoils also suffer flow separation or stall at high angles of incidence. Problems cm, a chord length of 1.5 m, and a wingspan of 12 m. What is the appropriate value of the Reynolds number for correlating the lift and drag of this wing? Explain your selection. Equation (7.1b) assumes that the boundary layer on the plate is turbulent from the leading edge onward. Devise a scheme for determining the boundary-layer thickness more accurately when the flow is laminar up to a point Rex,crit and turbulent thereafter. Apply this scheme to computation of the boundary-layer thickness at x 1.5 m in 40 m/s flow of air at 20°C and 1 atm past a flat plate. Compare your result with Eq. (7.1b). Assume Rex,crit 1.2 E6. Air at 20°C and 1 atm flows at 15 m/s past a flat plate with Rex,crit 1 E6. At what point x will the boundarylayer thickness be 8 mm? Why do Eqs. (7.1) seem to fail? Make a sketch illustrating the discrepancy; then use the ideas in Prob. 7.3 to complete this problem correctly. SAE 30 oil at 20°C flows at 1.8 ft3/s from a reservoir into a 6-in-diameter pipe. Use flat-plate theory to estimate the position x where the pipe-wall boundary layers meet in the center. Compare with Eq. (6.5), and give some explanations for the discrepancy. For the laminar parabolic boundary-layer profile of Eq. (7.6), compute the shape factor H and compare with the exact Blasius result, Eq. (7.31). Air at 20°C and 1 atm enters a 40-cm-square duct as in Fig. P7.7. Using the “displacement thickness” concept of Fig. 7.4, estimate (a) the mean velocity and (b) the mean pressure in the core of the flow at the position x Most of the problems herein are fairly straightforward. More difficult or open-ended assignments are labeled with an asterisk. Problems labeled with an EES icon will benefit from the use of the Engineering Equation Solver (EES), while problems labeled with a computer disk may require the use of a computer. The standard end-of-chapter problems 7.1 to 7.124 (categorized in the problem list below) are followed by word problems W7.1 to W7.12, fundamentals of engineering exam problems FE7.1 to FE7.10, comprehensive problems C7.1 to C7.4, and design project D7.1. P7.3 Problem Distribution Section 7.1 7.2 7.3 7.4 7.4 7.5 7.6 7.6 7.6 Topic Problems P7.4 Reynolds-number and geometry Momentum-integral estimates The boundary-layer equations Laminar flat-plate flow Turbulent flat-plate flow Boundary layers with pressure gradient Drag of two-dimensional bodies Drag of three-dimensional bodies Lifting bodies — airfoils 7.1 – 7.5 7.6 – 7.12 7.13 – 7.15 7.16 – 7.29 7.30 – 7.46 7.47 – 7.52 7.53 – 7.63 7.64 – 7.114 7.115 – 7.124 P7.5 P7.6 P7.2 For flow at 20 m/s past a thin flat plate, estimate the distances x from the leading edge at which the boundary-layer thickness will be either 1 mm or 10 cm for (a) air and (b) water at 20°C and 1 atm. Air, equivalent to that at a standard altitude of 4000 m, flows at 450 mi/h past a wing which has a thickness of 18 | v v P7.1 | e-Text Main Menu | P7.7 Textbook Table of Contents | Study Guide Problems 477 3 m. (c) What is the average gradient, in Pa/m, in this section? 40 × 40 cm square duct Boundary layers 2 m /s Ucore 3m for the parabolic flat-plate profile of Eq. (7.3). Yet when this new profile is used in the integral analysis of Sec. 7.3, we get the lousy result /x 9.2/Re1/2, which is 80 perx cent high. What is the reason for the inaccuracy? [Hint: The answer lies in evaluating the laminar boundary-layer momentum equation (7.19b) at the wall, y 0.] P7.13 Derive modified forms of the laminar boundary-layer equations (7.19) for the case of axisymmetric flow along the outside of a circular cylinder of constant radius R, as in Fig. P7.13. Consider the two special cases (a) R and (b) R. What are the proper boundary conditions? P7.7 y P7.8 1.2 kg/m3 and 1.8 E-5 kg/(m s), flows at Air, 10 m/s past a flat plate. At the trailing edge of the plate, the following velocity profile data are measured: y, mm 0 0.5 1.0 2.0 3.0 4.0 5.0 0 1.75 3.47 6.58 8.70 9.68 10.0 δ (x) u 6.0 u, m/s U r p ≈ constant 10.0 R P7.9 If the upper surface has an area of 0.6 m2, estimate, using momentum concepts, the friction drag, in N, on the upper surface. Repeat the flat-plate momentum analysis of Sec. 7.2 by replacing the parabolic profile, Eq. (7.6), with a more accurate sinusoidal profile: u U P7.13 P7.14 Show that the two-dimensional laminar-flow pattern with dp/dx 0 y sin x 2 u Compute momentum-integral estimates of cf, /x, */x, and H. P7.10 Repeat Prob. 7.9, using the polynomial profile suggested by K. Pohlhausen in 1921: u U 2 y 2 y3 y4 3 4 Does this profile satisfy the boundary conditions of laminar flat-plate flow? P7.11 Find the correct form for a cubic velocity-profile polynomial u A By Cy2 Dy3 | v v to replace Eq. (7.6) in a flat-plate momentum analysis. Find the value of / for this profile, but do not pursue the analysis further. P7.12 The velocity profile shape u/U 1 exp ( 4.605y/ ) is a smooth curve with u 0 at y 0 and u 0.99U at y and thus would seem to be a reasonable substitute | e-Text Main Menu | U0(1 e Cy) 0 0 is an exact solution to the boundary-layer equations (7.19). Find the value of the constant C in terms of the flow parameters. Are the boundary conditions satisfied? What might this flow represent? P7.15 Discuss whether fully developed laminar incompressible flow between parallel plates, Eq. (4.143) and Fig. 4.16b, represents an exact solution to the boundary-layer equations (7.19) and the boundary conditions (7.20). In what sense, if any, are duct flows also boundary-layer flows? P7.16 A thin flat plate 55 by 110 cm is immersed in a 6-m/s stream of SAE 10 oil at 20°C. Compute the total friction drag if the stream is parallel to (a) the long side and (b) the short side. P7.17 Helium at 20°C and low pressure flows past a thin flat plate 1 m long and 2 m wide. It is desired that the total friction drag of the plate be 0.5 N. What is the appropriate absolute pressure of the helium if U 35 m/s? Textbook Table of Contents | Study Guide 478 Chapter 7 Flow Past Immersed Bodies P7.18 In Prob. 7.7, when the duct is perfectly square, the core velocity speeds up. Suppose we wish to hold the core velocity constant by slanting the upper and lower walls while keeping the front and rear walls parallel. What angle of slant will be the best approximation? For this condition, what will be the total friction drag on the duct walls? P7.19 Program a method of numerical solution of the Blasius flat-plate relation, Eq. (7.22), subject to the conditions in (7.23). You will find that you cannot get started without knowing the initial second derivative f (0), which lies between 0.2 and 0.5. Devise an iteration scheme which starts at f (0) 0.2 and converges to the correct value. Print out u/U f ( ) and compare with Table 7.1. P7.20 Air at 20°C and 1 atm flows at 20 m/s past the flat plate in Fig. P7.20. A pitot stagnation tube, placed 2 mm from the wall, develops a manometer head h 16 mm of Meriam red oil, SG 0.827. Use this information to estimate the downstream position x of the pitot tube. Assume laminar flow. Boundary layer 20 m/s 2 mm so, determine the correct dimensionless form for , assuming that 0 at the wall, y 0. P7.23 Suppose you buy a 4- by 8-ft sheet of plywood and put it on your roof rack. (See Fig. P7.23.) You drive home at 35 mi/h. (a) Assuming the board is perfectly aligned with the airflow, how thick is the boundary layer at the end of the board? (b) Estimate the drag on the sheet of plywood if the boundary layer remains laminar. (c) Estimate the drag on the sheet of plywood if the boundary-layer is turbulent (assume the wood is smooth), and compare the result to that of the laminar boundary-layer case. P7.23 *P7.24 Air at 20°C and 1 atm flows past the flat plate in Fig. P7.24 under laminar conditions. There are two equally spaced pitot stagnation tubes, each placed 2 mm from the wall. The manometer fluid is water at 20°C. If U 15 m/s and L 50 cm, determine the values of the manometer readings h1 and h2, in mm. Boundary layer x h U P7.20 L 2 mm L P7.21 For the experimental setup of Fig. P7.20, suppose the stream velocity is unknown and the pitot stagnation tube is traversed across the boundary layer of air at 1 atm and 20°C. The manometer fluid is Meriam red oil, and the following readings are made: 2 mm h1 h2 P7.24 y, mm 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 h, mm 1.2 4.6 9.8 15.8 21.2 25.3 27.8 29.0 29.7 29.7 | v v Using these data only (not the Blasius theory) estimate (a) the stream velocity, (b) the boundary-layer thickness, (c) the wall shear stress, and (d) the total friction drag between the leading edge and the position of the pitot tube. P7.22 For the Blasius flat-plate problem, Eqs. (7.21) to (7.23), does a two-dimensional stream function (x, y) exist? If | e-Text Main Menu | P7.25 Modify Prob. 7.24 to the following somewhat more difficult scenario. Let the known data be U 15 m/s and h1 8 mm of water. Use this information to determine (a) L, in cm, (b) h2, in mm. P7.26 Consider laminar boundary-layer flow past the squareplate arrangements in Fig. P7.26. Compared to the friction drag of a single plate 1, how much larger is the drag of four plates together as in configurations (a) and (b)? Explain your results. Textbook Table of Contents | Study Guide Problems 479 1 3 2 4 1 P7.26 1 2 3 (a) 4 (b) *P7.27 A thin smooth disk of diameter D is immersed parallel to a uniform stream of velocity U. Assuming laminar flow and using flat-plate theory as a guide, develop an approximate formula for the drag of the disk. P7.28 Flow straighteners are arrays of narrow ducts placed in wind tunnels to remove swirl and other in-plane secondary velocities. They can be idealized as square boxes constructed by vertical and horizontal plates, as in Fig. P7.28. The cross section is a by a, and the box length is L. Assuming laminar flat-plate flow and an array of N N boxes, derive a formula for (a) the total drag on the bundle of boxes and (b) the effective pressure drop across the bundle. Using strip theory, derive a formula for the drag coefficient of this plate. Compare this result with the drag of the same plate immersed in a uniform stream U0. y U L y =δ δ u( y) x a P7.32 U0 a P7.33 An alternate analysis of turbulent flat-plate flow was given by Prandtl in 1927, using a wall shear-stress formula from pipe flow L P7.28 w 0.0225 U2 1/4 U Let the flow straighteners in Fig. P7.28 form an array of 20 20 boxes of size a 4 cm and L 25 cm. If Show that this formula can be combined with Eqs. (7.33) the approach velocity is U0 12 m/s and the fluid is and (7.40) to derive the following relations for turbulent sea-level standard air, estimate (a) the total array drag flat-plate flow: and (b) the pressure drop across the array. Compare with 0.072 0.0577 Sec. 6.6. 0.37 cf CD P7.30 Repeat Prob. 7.16 if the fluid is water at 20°C and the plate Re1/5 Re1/5 Re1/5 x x L x surface is smooth. These formulas are limited to Rex between 5 105 and P7.31 Repeat Prob. 7.16 if the fluid is water at 20°C and the plate 107. has an average roughness height of 1.5 mm. P7.32 A flat plate of length L and height is placed at a wall *P7.34 A thin equilateral-triangle plate is immersed parallel to a 12 m/s stream of water at 20°C, as in Fig. P7.34. Assumand is parallel to an approaching boundary layer, as in Fig. ing Retr 5 105, estimate the drag of this plate. P7.32. Assume that the flow over the plate is fully turbuP7.35 The solutions to Prob. 7.26 are (a) F 2.83F1-plate and (b) lent and that the approaching flow is a one-seventh-power F 2.0F1-plate. Do not reveal these results to your friends. law Repeat Prob. 7.26 assuming the boundary-layer flow is tury 1/7 bulent, and comment on the striking increase in numerical u(y) U0 values. | v v P7.29 | e-Text Main Menu | Textbook Table of Contents | Study Guide P7.36 EES P7.37 P7.38 P7.39 P7.40 P7.41 Chapter 7 Flow Past Immersed Bodies wall turbulent-flow theory to estimate the position x of the probe, in m. 2m *P7.42 A four-bladed helicopter rotor rotates at n r/min in air with properties ( , ). Each blade has chord length C and extends from the center of rotation out to radius R (the hub 2m size is neglected). Assuming turbulent flow from the leading edge, develop an analytical estimate for the power P required to drive this rotor. 2m P7.43 In the flow of air at 20°C and 1 atm past a flat plate in Fig. P7.43, the wall shear is to be determined at position x by a 12 m/s floating element (a small area connected to a strain-gage P7.34 force measurement). At x 2 m, the element indicates a shear stress of 2.1 Pa. Assuming turbulent flow from the leading edge, estimate (a) the stream velocity U, (b) the boundary-layer thickness at the element, and (c) the boundA ship is 125 m long and has a wetted area of 3500 m2. ary-layer velocity u, in m/s, at 5 mm above the element. Its propellers can deliver a maximum power of 1.1 MW to seawater at 20°C. If all drag is due to friction, estimate U the maximum ship speed, in kn. Floating element with A wind tunnel has a test section 1 m square and 6 m long x negligible gap with air at 20°C moving at an average velocity of 30 m/s. It is planned to slant the walls outward slightly to account for the growing boundary-layer displacement thickness on the four walls, thus keeping the test-section velocity constant. At what angle should they be slanted to keep V constant between x 2 m and x 4 m? Atmospheric boundary layers are very thick but follow forP7.43 mulas very similar to those of flat-plate theory. Consider wind blowing at 10 m/s at a height of 80 m above a smooth beach. Estimate the wind shear stress, in Pa, on the beach P7.44 Extensive measurements of wall shear stress and local velocity for turbulent airflow on the flat surface of the Uniif the air is standard sea-level conditions. What will the versity of Rhode Island wind tunnel have led to the folwind velocity striking your nose be if (a) you are standlowing proposed correlation: ing up and your nose is 170 cm off the ground and (b) you are lying on the beach and your nose is 17 cm off the y2 w uy 1.77 0.0207 ground? 2 A hydrofoil 50 cm long and 4 m wide moves at 28 kn in seawater at 20°C. Using flat-plate theory with Retr 5 E5, Thus, if y and u(y) are known at a point in a flat-plate boundestimate its drag, in N, for (a) a smooth wall and (b) a ary layer, the wall shear may be computed directly. If the rough wall, 0.3 mm. answer to part (c) of Prob. 7.43 is u 27 m/s, determine Hoerner [12, p. 3.25] states that the drag coefficient of a whether the correlation is accurate for this case. flag in winds, based on total wetted area 2bL, is approxi- P7.45 A thin sheet of fiberboard weighs 90 N and lies on a mated by CD 0.01 0.05L/b, where L is the flag length rooftop, as shown in Fig. P7.45. Assume ambient air at in the flow direction. Test Reynolds numbers ReL were 1 20°C and 1 atm. If the coefficient of solid friction between E6 or greater. (a) Explain why, for L/b 1, these drag valboard and roof is 0.12, what wind velocity will genues are much higher than for a flat plate. Assuming seaerate enough fluid friction to dislodge the board? level standard air at 50 mi/h, with area bL 4 m2, find (b) P7.46 A ship is 150 m long and has a wetted area of 5000 m2. the proper flag dimensions for which the total drag is apIf it is encrusted with barnacles, the ship requires 7000 hp proximately 400 N. to overcome friction drag when moving in seawater at 15 Repeat Prob. 7.20 with the sole change that the pitot probe kn and 20°C. What is the average roughness of the barnais now 10 mm from the wall (5 times higher). Show that cles? How fast would the ship move with the same power the flow there cannot possibly be laminar, and use smoothif the surface were smooth? Neglect wave drag. | v v 480 | e-Text Main Menu | Textbook Table of Contents | Study Guide Problems 481 2m 3m xsep, θ sep 1m x 1.5 m θ U0 U Fiberboard R 2m P7.50 thin, show that the core velocity U(x) in the diffuser is given approximately by Roof P7.45 U P7.47 As a case similar to Example 7.5, Howarth also proposed the adverse-gradient velocity distribution U U0(1 x2/L2) and computed separation at xsep/L 0.271 by a series-expansion method. Compute separation by Thwaites’ method and compare. P7.48 In 1957 H. Görtler proposed the adverse-gradient test cases U (1 Constant width b θ U? U? x U0 U(x) θ W L P7.51 *P7.52 In Fig. P7.52 a slanted upper wall creates a favorable pressure gradient on the upper surface of the flat plate. Use Thwaites’ theory to estimate P7.49 CD *P7.50 For flow past a cylinder of radius R as in Fig. P7.50, the theoretical inviscid velocity distribution along the surface is U 2U0 sin (x/R), where U0 is the oncoming stream velocity and x is the arc length measured from the nose (Chap. 8). Compute the laminar separation point xsep and sep by Thwaites’ method, and compare with the digitalcomputer solution xsep/R 1.823 ( sep 104.5°) given by R. M. Terrill in 1960. P7.51 Consider the flat-walled diffuser in Fig. P7.51, which is similar to that of Fig. 6.26a with constant width b. If x is measured from the inlet and the wall boundary layers are v | | e-Text Main Menu U0 (2x tan )/W where W is the inlet height. Use this velocity distribution with Thwaites’ method to compute the wall angle for which laminar separation will occur in the exit plane when diffuser length L 2W. Note that the result is independent of the Reynolds number. U0 x/L)n and computed separation for laminar flow at n 1 to be xsep/L 0.159. Compare with Thwaites’ method, assuming 0 0. P7.49 Based strictly upon your understanding of flat-plate theory plus adverse and favorable pressure gradients, explain the direction (left or right) for which airflow past the slender airfoil shape in Fig. P7.49 will have lower total (friction pressure) drag. v 1 | 1 2 F U 2 bL 0 Slanted wall U0 h0 U(x) x L, b Flat plate P7.52 Textbook Table of Contents | Study Guide 1 2 h0 482 Chapter 7 Flow Past Immersed Bodies P7.53 P7.54 P7.55 P7.56 on the upper plate surface if U0 L/ 105. Compare with Eq. (7.27). From Table 7.2, the drag coefficient of a wide plate normal to a stream is approximately 2.0. Let the stream conditions be U and p . If the average pressure on the front of the plate is approximately equal to the free-stream stagnation pressure, what is the average pressure on the rear? A chimney at sea level is 2 m in diameter and 40 m high. When it is subjected to 50 mi/h storm winds, what is the estimated wind-induced bending moment about the bottom of the chimney? A ship tows a submerged cylinder, which is 1.5 m in diameter and 22 m long, at 5 m/s in fresh water at 20°C. Estimate the towing power, in kW, required if the cylinder is (a) parallel and (b) normal to the tow direction. A delivery vehicle carries a long sign on top, as in Fig. P7.56. If the sign is very thin and the vehicle moves at 55 mi/h, estimate the force on the sign (a) with no crosswind and (b) with a 10 mi/h crosswind. how fast can Joe ride into the head wind? (c) Why is the result not simply 10 5.0 5.0 m/s, as one might first suspect? P7.60 A fishnet consists of 1-mm-diameter strings overlapped and knotted to form 1- by 1-cm squares. Estimate the drag of 1 m2 of such a net when towed normal to its plane at 3 m/s in 20°C seawater. What horsepower is required to tow 400 ft2 of this net? P7.61 A filter may be idealized as an array of cylindrical fibers normal to the flow, as in Fig. P7.61. Assuming that the fibers are uniformly distributed and have drag coefficients given by Fig. 7.16a, derive an approximate expression for the pressure drop p through a filter of thickness L. Filter section U p+∆p 8m Phil's Pizza: 555-5748 U p 60 cm Array of cylinders (fibers) P7.61 P7.56 | v v P7.57 The main cross-cable between towers of a coastal suspension bridge is 60 cm in diameter and 90 m long. Estimate the total drag force on this cable in crosswinds of 50 mi/h. Are these laminar-flow conditions? P7.58 A long cylinder of rectangular cross section, 5 cm high and 30 cm long, is immersed in water at 20°C flowing at 12 m/s parallel to the long side of the rectangle. Estimate the drag force on the cylinder, per unit length, if the rectangle (a) has a flat face or (b) has a rounded nose. P7.59 Joe can pedal his bike at 10 m/s on a straight level road with no wind. The rolling resistance of his bike is 0.80 N s/m, i.e., 0.80 N of force per m/s of speed. The drag area (CDA) of Joe and his bike is 0.422 m2. Joe’s mass is 80 kg and that of the bike is 15 kg. He now encounters a head wind of 5.0 m/s. (a) Develop an equation for the speed at which Joe can pedal into the wind. [Hint: A cubic equation for V will result.] (b) Solve for V, i.e., | e-Text Main Menu | P7.62 A sea-level smokestack is 52 m high and has a square cross section. Its supports can withstand a maximum side force of 90 kN. If the stack is to survive 90-mi/h hurricane winds, what is its maximum possible width? P7.63 The cross section of a cylinder is shown in Fig. P7.63. Assume that on the front surface the velocity is given by potential theory (Sec. 8.4), V 2U sin , from which the surface pressure is computed by Bernoulli’s equation. In V = 2 U∞ sin θ R θ p∞ , U∞ Separated-flow wake P7.63 Textbook Table of Contents | Study Guide Problems 483 the separated flow on the rear, the pressure is assumed equal to its value at 90°. Compute the theoretical drag coefficient and compare with Table 7.2. P7.64 A parachutist jumps from a plane, using an 8.5-mdiameter chute in the standard atmosphere. The total mass of the chutist and the chute is 90 kg. Assuming an open chute and quasi-steady motion, estimate the time to fall from 2000- to 1000-m altitude. P7.65 The drag of a sphere at very low Reynolds numbers ReD 1 was given analytically by G. G. Stokes in 1851: F 3 VD [2, pp. 175 – 178]. This formula is an example of Stokes’ law of creeping motion, where inertia is negligible. Show that the drag coefficient in this region is CD 24/ReD. A 1-mm-diameter sphere falls in 20°C glycerin at 2.5 mm/s. Is this a creeping motion? Compute (a) the Reynolds number, (b) the drag, and (c) the specific gravity of the sphere. P7.66 A sphere of density s and diameter D is dropped from rest in a fluid of density and viscosity . Assuming a constant drag coefficient Cd 0, derive a differential equation for the fall velocity V(t) and show that the solution is V C P7.68 P7.69 P7.70 3gCd 0(S 1) 4S2D 1/2 tanh Ct 1/2 where S s / is the specific gravity of the sphere material. Apply the theory of Prob. 7.66 to a steel sphere of diameter 2 cm, dropped from rest in water at 20°C. Estimate the time required for the sphere to reach 99 percent of its terminal (zero-acceleration) velocity. A baseball weighs 145 g and is 7.35 cm in diameter. It is dropped from rest from a 35-m-high tower at approximately sea level. Assuming a laminar-flow drag coefficient, estimate (a) its terminal velocity and (b) whether it will reach 99 percent of its terminal velocity before it hits the ground. Two baseballs from Prob. 7.68 are connected to a rod 7 mm in diameter and 56 cm long, as in Fig. P7.69. What power, in W, is required to keep the system spinning at 400 r/min? Include the drag of the rod, and assume sealevel standard air. A baseball from Prob. 7.68 is batted upward during a game at an angle of 45° and an initial velocity of 98 mi/h. Neglect spin and lift. Estimate the horizontal distance traveled, (a) neglecting drag and (b) accounting for drag in a numerical (computer) solution with a transition Reynolds number ReD,crit 2.5 E5. | v v P7.67 4gD(S 1) 3Cd 0 | e-Text Main Menu | 28 cm 28 cm Baseball Ω Baseball P7.69 P7.71 A football weighs 0.91 lbf and approximates an ellipsoid 6 in in diameter and 12 in long (Table 7.3). It is thrown upward at a 45° angle with an initial velocity of 80 ft/s. Neglect spin and lift. Assuming turbulent flow, estimate the horizontal distance traveled, (a) neglecting drag and (b) accounting for drag with a numerical (computer) model. P7.72 A settling tank for a municipal water supply is 2.5 m deep, and 20°C water flows through continuously at 35 cm/s. Estimate the minimum length of the tank which will ensure that all sediment (SG 2.55) will fall to the bottom for particle diameters greater than (a) 1 mm and (b) 100 m. P7.73 A balloon is 4 m in diameter and contains helium at 125 kPa and 15°C. Balloon material and payload weigh 200 N, not including the helium. Estimate (a) the terminal ascent velocity in sea-level standard air, (b) the final standard altitude (neglecting winds) at which the balloon will come to rest, and (c) the minimum diameter ( 4 m) for which the balloon will just barely begin to rise in sea-level standard air. P7.74 If D 4 m in Prob. 7.73 and the helium remains at 15°C and 125 kPa, estimate the time required for the balloon to rise through the standard atmosphere from sea level to its equilibrium altitude, which is approximately 4000 m. P7.75 The helium-filled balloon in Fig. P7.75 is tethered at 20°C and 1 atm with a string of negligible weight and drag. The EES diameter is 50 cm, and the balloon material weighs 0.2 N, D = 50 cm U θ P7.75 Textbook Table of Contents | Study Guide 484 P7.76 P7.77 P7.78 P7.79 P7.80 Chapter 7 Flow Past Immersed Bodies not including the helium. The helium pressure is 120 kPa. Estimate the tilt angle if the airstream velocity U is (a) 5 m/s or (b) 20 m/s. P7.83 Extend Prob. 7.75 to make a smooth plot of tilt angle versus stream velocity U in the range 1 U 12 mi/h. (A spreadsheet is recommended for this task.) Comment on the effectiveness of this system as an air-velocity instrument. Modify Prob. 7.75 as follows: Let the fluid be water at P7.84 20°C, and let the 50-cm-diameter sphere be solid cork (SG 0.16). Estimate the tilt angle if the water velocity U is 3 m/s. Apply Prob. 7.61 to a filter consisting of 300- mdiameter fibers packed 250 per square centimeter in the plane of Fig. P7.61. For air at 20°C and 1 atm flowing at 1.5 m/s, estimate the pressure drop if the filter is 5 cm thick. Assume that a radioactive dust particle approximates a sphere of density 2400 kg/m3. How long, in days, will it take such a particle to settle to sea level from an altitude of 12 km if the particle diameter is (a) 1 m or (b) 20 m? A heavy sphere attached to a string should hang at an angle when immersed in a stream of velocity U, as in Fig. P7.80. Derive an expression for as a function of the sphere and flow properties. What is if the sphere is steel (SG 7.86) of diameter 3 cm and the flow is sea-level standard air at U 40 m/s? Neglect the string *P7.85 drag. θ 1.2 ft2 falling feet first [12, p. 313]. What are the minimum and maximum terminal speeds that can be achieved by a skydiver at 5000-ft standard altitude? A high-speed car has a drag coefficient of 0.3 and a frontal area of 1 m2. A parachute is to be used to slow this car from 80 to 40 m/s in 8 s. What should the chute diameter be? What distance will be traveled during this deceleration? Take m 2000 kg. A Ping-Pong ball weighs 2.6 g and has a diameter of 3.8 cm. It can be supported by an air jet from a vacuum cleaner outlet, as in Fig. P7.84. For sea-level standard air, what jet velocity is required? P7.84 An aluminum cylinder (SG 2.7) slides concentrically down a taut 1-mm-diameter wire as shown in Fig. P7.85. Its length L 8 cm, and its radius R 1 cm. A 2-mm-diameter hole down the cylinder center is lubricated by SAE 30 oil at 20°C. Estimate the terminal fall velocity V of the cylinder if ambient air drag is (a) neglected and (b) included. Assume air at 1 atm and 20°C. Oil film U D, ρs R P7.80 A typical U.S. Army parachute has a projected diameter of 28 ft. For a payload mass of 80 kg, (a) what terminal velocity will result at 1000-m standard altitude? For the same velocity and net payload, what size dragproducing “chute” is required if one uses a square flat plate held (b) vertically and (c) horizontally? (Neglect the fact that flat shapes are not dynamically stable in free fall.) P7.82 The average skydiver with parachute unopened weighs 175 lbf and has a drag area CD A 9 ft2 spread-eagled and | v v P7.81 | e-Text Main Menu | L V P7.85 Textbook Table of Contents | Study Guide Problems 485 P7.86 Hoerner [Ref. 12, pp. 3 – 25] states that the drag coefficient of a flag of 2 1 aspect ratio is 0.11 based on planform area. The University of Rhode Island has an aluminum flagpole 25 m high and 14 cm in diameter. It flies equalsized national and state flags together. If the fracture stress of aluminum is 210 MPa, what is the maximum flag size that can be used yet avoids breaking the flagpole in hurricane (75 mi/h) winds? P7.87 A tractor-trailer truck has a drag-area CD A 8 m2 bare and 6.7 m2 with an aerodynamic deflector (Fig. 7.18b). Its rolling resistance is 50 N for each mile per hour of speed. Calculate the total horsepower required at sea level with and without the deflector if the truck moves at (a) 55 mi/h and (b) 75 mi/h. P7.88 A pickup truck has a clean drag-area CD A of 35 ft2. Estimate the horsepower required to drive the truck at 55 mi/h (a) clean and (b) with the 3- by 6-ft sign in Fig. P7.88 installed if the rolling resistance is 150 lbf at sea level. torque of 0.004 N m. Making simplifying assumptions to average out the time-varying geometry, estimate and plot the variation of anemometer rotation rate with wind velocity U in the range 0 U 25 m/s for sea-level standard air. D = 5 cm Ω 15 cm D = 5 cm 15 cm U 6 ft P7.91 Eat at Joe's 3 ft P7.92 A 1500-kg automobile uses its drag area CD A 0.4 m2, plus brakes and a parachute, to slow down from 50 m/s. Its brakes apply 5000 N of resistance. Assume sea-level standard air. If the automobile must stop in 8 s, what diameter parachute is appropriate? P7.93 A hot-film probe is mounted on a cone-and-rod system in a sea-level airstream of 45 m/s, as in Fig. P7.93. Estimate the maximum cone vertex angle allowable if the flow-induced bending moment at the root of the rod is not to exceed 30 N cm. Ho t film 3 cm 45 m/s P7.88 | v v P7.89 A water tower is approximated by a 15-m-diameter sphere mounted on a 1-m-diameter rod 20 m long. Estimate the bending moment at the root of the rod due to aerodynamic forces during hurricane winds of 40 m/s. P7.90 In the great hurricane of 1938, winds of 85 mi/h blew over a boxcar in Providence, Rhode Island. The boxcar was 10 EES ft high, 40 ft long, and 6 ft wide, with a 3-ft clearance above tracks 4.8 ft apart. What wind speed would topple a boxcar weighing 40,000 lbf? *P7.91 A cup anemometer uses two 5-cm-diameter hollow hemispheres connected to 15-cm rods, as in Fig. P7.91. Rod drag is negligible, and the central bearing has a retarding | e-Text Main Menu | 20 cm 5-mm dia. P7.93 P7.94 A rotary mixer consists of two 1-m-long half-tubes rotating around a central arm, as in Fig. P7.94. Using the drag from Table 7.2, derive an expression for the torque T required to drive the mixer at angular velocity in a fluid Textbook Table of Contents | Study Guide 486 Chapter 7 Flow Past Immersed Bodies mobile of mass 1500 kg and frontal area 2 m2, the following velocity-versus-time data are obtained during a coastdown: of density . Suppose that the fluid is water at 20°C and the maximum driving power available is 20 kW. What is the maximum rotation speed r/min? t, s Ω V, m/s *P7.98 R=1m D = 5 cm P7.94 P7.95 An airplane weighing 12 kN, with a drag-area CD A 5 m2, lands at sea level at 55 m/s and deploys a drag parachute 3 m in diameter. No other brakes are applied. (a) How long will it take the plane to slow down to 30 m/s? (b) How far will it have traveled in that time? *P7.96 A Savonius rotor (Fig. 6.29b) can be approximated by the two open half-tubes in Fig. P7.96 mounted on a central axis. If the drag of each tube is similar to that in Table 7.2, derive an approximate formula for the rotation rate as a function of U, D, L, and the fluid properties ( , ). 0 10 20 30 40 27.0 24.2 21.8 19.7 17.9 Estimate (a) the rolling resistance and (b) the drag coefficient. This problem is well suited for digital-computer analysis but can be done by hand also. A buoyant ball of specific gravity SG 1 dropped into water at inlet velocity V0 will penetrate a distance h and then pop out again, as in Fig. P7.98. Make a dynamic analysis of this problem, assuming a constant drag coefficient, and derive an expression for h as a function of the system properties. How far will a 5-cm-diameter ball with SG 0.5 and CD 0.47 penetrate if it enters at 10 m/s? Diameter D (SG < 1) V0 h Ω P7.98 Axis D P7.99 Two steel balls (SG 7.86) are connected by a thin hinged rod of negligible weight and drag, as in Fig. P7.99. A stop prevents the rod from rotating counterclockwise. Estimate the sea-level air velocity U for which the rod will first begin to rotate clockwise. D L U D = 2 cm L U 10 cm Hinge Stop 45˚ P7.96 10 cm A simple measurement of automobile drag can be found by an unpowered coastdown on a level road with no wind. Assume constant rolling resistance. For an auto- | v v P7.97 | e-Text Main Menu | D = 1 cm P7.99 Textbook Table of Contents | Study Guide Problems 487 P7.100 In creeping motion or Stokes’ flow at velocity U past a sphere of diameter D, density (fluid inertia) is unimportant and the drag force is F 3 UD. This formula is valid if ReD 1.0 (see Fig. 7.16b). (a) Verify that Stokes’ formula is equivalent to CD 24/ReD. (b) Determine the largest diameter raindrop whose terminal fall velocity follows Stokes’ formula in sea-level standard air. P7.101 Icebergs can be driven at substantial speeds by the wind. Let the iceberg be idealized as a large, flat cylinder, D L, with one-eighth of its bulk exposed, as in Fig. P7.101. Let the seawater be at rest. If the upper and lower drag forces depend upon relative velocities between the iceberg and the fluid, derive an approximate expression for the steady iceberg speed V when driven by wind velocity U. D> L > U L/8 V 7L / 8 Iceberg P7.101 | v v P7.102 Sand particles (SG 2.7), approximately spherical with diameters from 100 to 250 m, are introduced into an upward-flowing stream of water at 20°C. What is the minimum water velocity which will carry all the sand particles upward? P7.103 When immersed in a uniform stream V, a heavy rod hinged at A will hang at Pode’s angle , after an analysis by L. Pode in 1951 (Fig. P7.103). Assume that the cylinder has normal drag coefficient CDN and tangential coefficient CDT which relate the drag forces to VN and VT, respectively. Derive an expression for Pode’s angle as a function of the flow and rod parameters. Find for a steel rod, L 40 cm, D 1 cm, hanging in sea-level air at V 35 m/s. P7.104 Suppose that the body in Fig. P7.103 is a thin plate weighing 0.6 N, of length L 20 cm and width b 4 cm into the paper. For sea-level standard air, plot Pode’s angle versus velocity V in the range 0 V 40 m/s. Would this device make a good air-velocity meter (anemometer)? | e-Text Main Menu | A θ CD N , VN CD T , VT V L, D, ρs P7.103 P7.105 A ship 50 m long, with a wetted area of 800 m2, has the hull shape tested in Fig. 7.19. There are no bow or stern bulbs. The total propulsive power available is 1 MW. For seawater at 20°C, plot the ship’s velocity V kn versus power P for 0 P 1 MW. What is the most efficient setting? P7.106 A smooth steel 1-cm-diameter sphere (W 0.04 N) is fired vertically at sea level at the initial supersonic velocity V0 1000 m/s. Its drag coefficient is given by Fig. 7.20. Assuming that the speed of sound is constant at a 343 m/s, compute the maximum altitude of the projectile (a) by a simple analytical estimate and (b) by a digitalcomputer program. P7.107 Repeat Prob. 7.106 if the body is a 9-mm steel bullet (W 0.07 N) which approximates the “pointed body of revolution” in Fig. 7.20. P7.108 The data in Fig. P7.108 are for the lift and drag of a spinning sphere from Ref. 12, pp. 7 – 20. Suppose that a tennis ball (W 0.56 N, D 6.35 cm) is struck at sea level with initial velocity V0 30 m/s and “topspin” (front of the ball rotating downward) of 120 r/s. If the initial height of the ball is 1.5 m, estimate the horizontal distance traveled before it strikes the ground. P7.109 Repeat Prob. 7.108 if the ball is struck with “underspin” (front of the ball rotating upward). P7.110 A baseball pitcher throws a curveball with an initial velocity of 65 mi/h and a spin of 6500 r/min about a vertical axis. A baseball weighs 0.32 lbf and has a diameter of 2.9 in. Using the data of Fig. P7.108 for turbulent flow, estimate how far such a curveball will have deviated from its straight-line path when it reaches home plate 60.5 ft away. *P7.111 A table tennis ball has a mass of 2.6 g and a diameter of 3.81 cm. It is struck horizontally at an initial velocity of 20 m/s while it is 50 cm above the table, as in Fig. P7.111. For sea-level air, what spin, in r/min, will cause the ball to strike the opposite edge of the table, 4 m away? Make an analytical estimate, using Fig. P7.108, and account for the fact that the ball decelerates during flight. Textbook Table of Contents | Study Guide 488 Chapter 7 Flow Past Immersed Bodies Figu re U nava ilab le P7.108 4m 20 m/s ω? 50 cm ? P7.111 | v v P7.112 Repeat Prob. 7.111 by making a detailed digital-computer solution for the flight path of the ball. Use Fig. P7.108 for lift and drag. P7.113 An automobile has a mass of 1000 kg and a drag-area CD A 0.7 m2. The rolling resistance of 70 N is approximately constant. The car is coasting without brakes at 90 km/h as it begins to climb a hill of 10 percent grade (slope tan 1 0.1 5.71°). How far up the hill will the car come to a stop? *P7.114 Suppose that the car in Prob. 7.113 is placed at the top of the 10 percent grade hill and released from rest to coast EES down without brakes. What will be its speed, in km/h, after dropping a vertical distance of 20 m? | e-Text Main Menu | P7.115 An airplane weighs 180 kN and has a wing area of 160 m2 and a mean chord of 4 m. The airfoil properties are given by Fig. 7.25. If the airplane cruises at 250 mi/h at 3000-m standard altitude, what propulsive power is required to overcome wing drag? P7.116 The airplane of Prob. 7.115 is designed to land at V0 1.2Vstall, using a split flap set at 60°. What is the proper landing speed in mi/h? What power is required for takeoff at the same speed? P7.117 Suppose that the airplane of Prob. 7.115 takes off at sea level without benefit of flaps, with CL constant so that the takeoff speed is 100 mi/h. Estimate the takeoff distance if the thrust is 10 kN. How much thrust is needed to make the takeoff distance 1250 m? *P7.118 Suppose that the airplane of Prob. 7.115 is fitted with all the best high-lift devices of Fig. 7.28. What is its minimum stall speed in mi/h? Estimate the stopping distance if the plane lands at V0 1.25Vstall with constant CL 3.0 and CD 0.2 and the braking force is 20 percent of the weight on the wheels. P7.119 An airplane has a mass of 5000 kg, a maximum thrust of 7000 N, and a rectangular wing with aspect ratio 6.0. It EES takes off at sea level with a 60° split flap whose twodimensional properties are shown in Fig. 7.25. Assume all lift and all drag are due to the wing. What is the proper wing size if the takeoff distance is to be 1 km? P7.120 Show that if Eqs. (7.70) and (7.71) are valid, the maximum lift-to-drag ratio occurs when CD 2CD . What are Textbook Table of Contents | Study Guide Fundamentals of Engineering Exam Problems (L/D)max and for a symmetric wing when AR 5 and CD 0.009? P7.121 In gliding (unpowered) flight, the lift and drag are in equilibrium with the weight. Show that if there is no wind, the aircraft sinks at an angle tan drag lift For a sailplane of mass 200 kg, wing area 12 m2, and aspect ratio 11, with an NACA 0009 airfoil, estimate (a) the stall speed, (b) the minimum gliding angle, and (c) the maximum distance it can glide in still air when it is 1200 m above level ground. P7.122 A boat of mass 2500 kg has two hydrofoils, each of chord 30 cm and span 1.5 m, with CL,max 1.2 and CD 0.08. Its engine can deliver 130 kW to the water. For seawater at 20°C, estimate (a) the minimum speed for which the 489 foils support the boat and (b) the maximum speed attainable. P7.123 In prewar days there was a controversy, perhaps apocryphal, about whether the bumblebee has a legitimate aerodynamic right to fly. The average bumblebee (Bombus terrestris) weighs 0.88 g, with a wing span of 1.73 cm and a wing area of 1.26 cm2. It can indeed fly at 10 m/s. Using fixed-wing theory, what is the lift coefficient of the bee at this speed? Is this reasonable for typical airfoils? *P7.124 The bumblebee can hover at zero speed by flapping its wings. Using the data of Prob. 7.123, devise a theory for flapping wings where the downstroke approximates a short flat plate normal to the flow (Table 7.3) and the upstroke is feathered at nearly zero drag. How many flaps per second of such a model wing are needed to support the bee’s weight? (Actual measurements of bees show a flapping rate of 194 Hz.) Word Problems W7.1 W7.2 W7.3 W7.4 W7.5 W7.6 W7.7 How do you recognize a boundary layer? Cite some physical properties and some measurements which reveal appropriate characteristics. In Chap. 6 the Reynolds number for transition to turbulence in pipe flow was about Retr 2300, whereas in flatplate flow Retr 1 E6, nearly three orders of magnitude higher. What accounts for the difference? Without writing any equations, give a verbal description of boundary-layer displacement thickness. Describe, in words only, the basic ideas behind the “boundary-layer approximations.” What is an adverse pressure gradient? Give three examples of flow regimes where such gradients occur. What is a favorable pressure gradient? Give three examples of flow regimes where such gradients occur. The drag of an airfoil (Fig. 7.12) increases considerably if you turn the sharp edge around 180° to face the stream. Can you explain this? W7.8 In Table 7.3, the drag coefficient of a spruce tree decreases sharply with wind velocity. Can you explain this? W7.9 Thrust is required to propel an airplane at a finite forward velocity. Does this imply an energy loss to the system? Explain the concepts of thrust and drag in terms of the first law of thermodynamics. W7.10 How does the concept of drafting, in automobile and bicycle racing, apply to the material studied in this chapter? W7.11 The circular cylinder of Fig. 7.13 is doubly symmetric and therefore should have no lift. Yet a lift sensor would definitely reveal a finite root-mean-square value of lift. Can you explain this behavior? W7.12 Explain in words why a thrown spinning ball moves in a curved trajectory. Give some physical reasons why a side force is developed in addition to the drag. Fundamentals of Engineering Exam Problems | v v FE7.1 A smooth 12-cm-diameter sphere is immersed in a stream of 20°C water moving at 6 m/s. The appropriate Reynolds number of this sphere is approximately (a) 2.3 E5, (b) 7.2 E5, (c) 2.3 E6, (d) 7.2 E6, (e) 7.2 E7 FE7.2 If, in Prob. FE7.1, the drag coefficient based upon frontal area is 0.5, what is the drag force on the sphere? (a) 17 N, (b) 51 N, (c) 102 N, (d) 130 N, (e) 203 N FE7.3 If, in Prob. FE7.1, the drag coefficient based upon frontal area is 0.5, at what terminal velocity will an aluminum sphere (SG 2.7) fall in still water? (a) 2.3 m/s, (b) 2.9 m/s, (c) 4.6 m/s, (d) 6.5 m/s, (e) 8.2 m/s | e-Text Main Menu | FE7.4 For flow of sea-level standard air at 4 m/s parallel to a thin flat plate, estimate the boundary-layer thickness at x 60 cm from the leading edge: (a) 1.0 mm, (b) 2.6 mm, (c) 5.3 mm, (d) 7.5 mm, (e) 20.2 mm FE7.5 In Prob. FE7.4, for the same flow conditions, what is the wall shear stress at x 60 cm from the leading edge? (a) 0.053 Pa, (b) 0.11 Pa, (c) 0.16 Pa, (d) 0.32 Pa, (e) 0.64 Pa FE7.6 Wind at 20°C and 1 atm blows at 75 km/h past a flagpole 18 m high and 20 cm in diameter. The drag coefficient, Textbook Table of Contents | Study Guide 490 Chapter 7 Flow Past Immersed Bodies FE7.9 An airplane has a mass of 19,550 kg, a wing span of 20 m, and an average wing chord of 3 m. When flying in air of density 0.5 kg/m3, its engines provide a thrust of 12 kN against an overall drag coefficient of 0.025. What is its approximate velocity? (a) 250 mi/h, (b) 300 mi/h, (c) 350 mi/h, (d) 400 mi/h, (e) 450 mi/h FE7.10 For the flight conditions of the airplane in Prob. FE7.9 above, what is its approximate lift coefficient? (a) 0.1, (b) 0.2, (c) 0.3, (d) 0.4, (e) 0.5 based upon frontal area, is 1.15. Estimate the windinduced bending moment at the base of the pole. (a) 9.7 kN m, (b) 15.2 kN m, (c) 19.4 kN m, (d) 30.5 kN m, (e) 61.0 kN m FE7.7 Consider wind at 20°C and 1 atm blowing past a chimney 30 m high and 80 cm in diameter. If the chimney may fracture at a base bending moment of 486 kN m, and its drag coefficient based upon frontal area is 0.5, what is the approximate maximum allowable wind velocity to avoid fracture? (a) 50 mi/h, (b) 75 mi/h, (c) 100 mi/h, (d) 125 mi/h, (e) 150 mi/h FE7.8 A dust particle of density 2600 kg/m3, small enough to satisfy Stokes’ drag law, settles at 1.5 mm/s in air at 20°C and 1 atm. What is its approximate diameter? (a) 1.8 m, (b) 2.9 m, (c) 4.4 m, (d) 16.8 m, (e) 234 m Comprehensive Problems C7.1 Jane wants to estimate the drag coefficient of herself on her bicycle. She measures the projected frontal area to be 0.40 m2 and the rolling resistance to be 0.80 N s/m. The mass of the bike is 15 kg, while the mass of Jane is 80 kg. Jane coasts down a long hill which has a constant 4° slope. (See Fig. C7.1.) She reaches a terminal (steady state) speed of 14 m/s down the hill. Estimate the aerodynamic drag coefficient CD of the rider and bicycle combination. C7.2 Air at 20°C and 1 atm flows at Vavg 5 m/s between long, smooth parallel heat-exchanger plates 10 cm apart, as in Fig. C7.2. It is proposed to add a number of widely spaced 1-cmlong interrupter plates to increase the heat transfer, as shown. Although the flow in the channel is turbulent, the boundary layers over the interrupter plates are essentially laminar. Assume all plates are 1 m wide into the paper. Find (a) the pressure drop in Pa/m without the small plates present. Then find (b) the number of small plates per meter of channel length which will cause the pressure drop to rise to 10.0 Pa/m. C7.3 A new pizza store is planning to open. They will, of course, offer free delivery, and therefore need a small delivery car with a large sign attached. The sign (a flat plate) is 1.5 ft U Interrupter plates L 1 cm 5 m/s C7.2 high and 5 ft long. The boss (having no feel for fluid mechanics) mounts the sign bluntly facing the wind. One of his drivers is taking fluid mechanics and tells his boss he can save lots of money by mounting the sign parallel to the wind. (See Fig. C7.3.) (a) Calculate the drag (in lbf) on the sign alone at 40 mi/h (58.7 ft/s) in both orientations. (b) Suppose the car without any sign has a drag coefficient of 0.4 and a frontal area of 40 ft2. For V 40 mi/h, calculate the total drag of the car-sign combination for both orientations. (c) If the car has a rolling resistance of 40 lbf at 40 mi/h, calculate the horsepower required by the engine to drive the car at 40 mi/h in both orientations. (d) Finally, if the engine can deliver 10 hp for 1 h on a gallon of gasoline, calculate the fuel efficiency in mi/gal for both orientations at 40 mi/h. V | v v C7.1 | e-Text Main Menu | Textbook Table of Contents | Study Guide References 491 C7.3 C7.4 Consider a pendulum with an unusual bob shape: a hemispherical cup of diameter D whose axis is in the plane of oscillation, as in Fig. C7.4. Neglect the mass and drag of the rod L. (a) Set up the differential equation for the oscillation (t), including different cup drag (air density ) in each direction, and (b) nondimensionalize this equation. (c) Determine the natural frequency of oscillation for small 1 rad. (d) For the special case L 1 m, D 10 cm, m 50 g, and air at 20°C and 1 atm, with (0) 30°, find (numerically) the time required for the oscillation amplitude to drop to 1°. L Air m Cup shape C7.4 Design Project als—are to be selected through your analysis. Make suitable assumptions about the instantaneous drag of the cups and rods at any given angle (t) of the system. Compute the instantaneous torque T(t), and find and integrate the instantaneous angular acceleration of the device. Develop a complete theory for rotation rate versus wind speed in the range 0 U 50 mi/h. Try to include actual commercial bearing-friction properties. D7.1 It is desired to design a cup anemometer for wind speed, similar to Fig. P7.91, with a more sophisticated approach than the “average-torque” method of Prob. 7.91. The design should achieve an approximately linear relation between wind velocity and rotation rate in the range 20 U 40 mi/h, and the anemometer should rotate at about 6 r/s at U 30 mi/h. All specifications—cup diameter D, rod length L, rod diameter d, the bearing type, and all materi- References H. Schlichting, Boundary Layer Theory, 7th ed., McGrawHill, New York, 1979. 2. F. M. White, Viscous Fluid Flow, 2d ed., McGraw-Hill, New York, 1991. 3. L. Rosenhead (ed.), Laminar Boundary Layers, Oxford University Press, London, 1963. 4. D. A. Anderson, J. C. Tannehill, and R. H. Pletcher, Computational Fluid Mechanics and Heat Transfer, 2d ed., Hemisphere, New York, 1997. 5. R. H. Pletcher and C. L. Dancey, “A Direct Method of Calculating through Separated Regions in Boundary Layer Flow,” J. Fluids Eng., September 1976, pp. 568 – 572. | v v 1. | e-Text Main Menu | 6. P. K. Chang, Control of Flow Separation, McGraw-Hill, New York, 1976. See also P. K. Chang, Recent Development in Flow Separation, Pang Han Pub. Co., Seoul, Korea, 1983. 7. T. von Kármán, “On Laminar and Turbulent Friction,” Z. Angew. Math. Mech., vol. 1, 1921, pp. 235 – 236. 8. G. B. Schubauer and H. K. Skramstad, “Laminar Boundary Layer Oscillations and Stability of Laminar Flow,” Natl. Bur. Stand. Res. Pap. 1772, April 1943 (see also J. Aero. Sci., vol. 14, 1947, pp. 69 – 78, and NACA Rep. 909, 1947). 9. W. Rodi, Turbulence Models and Their Application in Hydraulics, Brookfield Pub., Brookfield, VT, 1984. Textbook Table of Contents | Study Guide Chapter 7 Flow Past Immersed Bodies 10. P. W. Runstadler, Jr., et al., “Diffuser Data Book,” Creare Inc., Tech. Note 186, Hanover, NH, May 1975. B. Thwaites, “Approximate Calculation of the Laminar Boundary Layer,” Aeronaut. Q., vol. 1, 1949, pp. 245 – 280. S. F. Hoerner, Fluid Dynamic Drag, published by the author, Midland Park, NJ, 1965. J. D. Anderson, Fundamentals of Aerodynamics, 2d ed., McGraw-Hill, New York, 1991. V. Tucker and G. C. Parrott, “Aerodynamics of Gliding Flight of Falcons and Other Birds,” J. Exp. Biol., vol. 52, 1970, pp. 345 – 368. Eric Tupper, Introduction to Naval Architecture, Butterworth, Boston, 1996. I. H. Abbott and A. E. von Doenhoff, Theory of Wing Sections, Dover, New York, 1959. R. L. Kline and F. F. Fogleman, Airfoil for Aircraft, U.S. Patent 3,706,430 Dec. 19, 1972. T. Y. Yu et al. (eds.), Swimming and Flying in Nature: Proceedings of a Symposium, Plenum, New York, 1975. National Committee for Fluid Mechanics Films, Illustrated Experiments in Fluid Mechanics, M.I.T. Press, Cambridge, MA, 1972. E. M. Uram and H. E. Weber (eds.), “Laminar and Turbulent Boundary Layers,” ASME Symp. Proc., vol. I00167, February 1984. G. Sovran, T. Morel, and W. T. Mason, Jr. (eds.), Aerodynamic Drag Mechanisms of Bluff Bodies and Road Vehicles, Plenum, New York, 1978. W. H. Hucho, Aerodynamics of Road Vehicles, Butterworth, Boston, 1986. R. D. Blevins, Applied Fluid Dynamics Handbook, van Nostrand Reinhold, New York, 1984. R. C. Johnson, Jr., G. E. Ramey, and D. S. O’Hagen, “Wind Induced Forces on Trees,” J. Fluids Eng., vol. 104, March 1983, pp. 25 – 30. P. W. Bearman et al., “The Effect of a Moving Floor on WindTunnel Simulation of Road Vehicles,” Paper No. 880245, SAE Transactions, J. Passenger Cars, vol. 97, sec. 4, 1988, pp. 4.200 – 4.214. CRC Handbook of Tables for Applied Engineering Science, 2d ed., CRC Press, Boca Raton, FL, 1973. T. Inui, “Wavemaking Resistance of Ships,” Trans. Soc. Nav. Arch. Marine Engrs., vol. 70, 1962, pp. 283 – 326. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. | v v 492 | e-Text Main Menu | 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. L. Larsson et al., “A Method for Resistance and Flow Prediction in Ship Design,” Trans. Soc. Nav. Arch. Marine Engrs., vol. 98, 1990, pp. 495 – 535. R. L. Street, G. Z. Watters, and J. K. Vennard, Elementary Fluid Mechanics, 7th ed., Wiley, New York, 1995. J. D. Anderson, Jr., Modern Compressible Flow, with Historical Perspective, 2d ed., McGraw-Hill, New York, 1990. J. D. Anderson, Jr., Hypersonic and High Temperature Gas Dynamics, McGraw-Hill, New York, 1989. J. Rom, High Angle of Attack Aerodynamics: Subsonic, Transonic, and Supersonic Flows, Springer-Verlag, New York, 1992. S. Vogel, “Drag and Reconfiguration of Broad Leaves in High Winds,” J. Exp. Bot., vol. 40, no. 217, August 1989, pp. 941 – 948. S. Vogel, Life in Moving Fluids, Princeton University Press, Princeton, NJ, 1981. J. A. C. Humphrey (ed.), Proceedings 1st International Symposium on Mechanics of Plants, Animals, and Their Environment, Engineering Foundation, New York, January 1998. D. D. Joseph, R. Bai, K. P. Chen, and Y. Y. Renardy, “CoreAnnular Flows,” Annu. Rev. Fluid Mech., vol. 29, 1997, pp. 65 – 90. J. W. Hoyt and R. H. J. Sellin, “Scale Effects in Polymer Solution Pipe Flow,” Experiments in Fluids, vol. 15, no. 1, June 1993, pp. 70 – 74. S. Nakao, “Application of V-shape Riblets to Pipe Flows,” J. Fluids Eng., vol. 113, December 1991, pp. 587 – 590. P. R. Bandyopadhyay, “Review: Mean Flow in Turbulent Boundary Layers Disturbed to Alter Skin Friction,” J. Fluids Eng., vol. 108, 1986, pp. 127 – 140. N. K. Madavan, S. Deutsch, and C. L. Merkle, “Measurements of Local Skin Friction in a Microbubble Modified Turbulent Boundary Layer,” J. Fluid Mech., vol. 156, 1985, pp. 237 – 256. S. M. Trujillo, D. G. Bogard, and K. S. Ball, “Drag Reduction in a Turbulent Boundary Layer by Wall Oscillation,” Paper AIAA 97-1870, AIAA Shear Flow Control Conference, Austin, TX, December 1997. J. G. Chandler, “Microplanes,” Popular Science, January 1998, pp. 54 – 59. Textbook Table of Contents | Study Guide ...
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