fluid_mechanics
488 Chapter 9 ■ Flow over Immersed Bodies E XAMPLE 9.7 Drag on a Flat Plate GIVEN The water ski shown in Fig. E9.7a moves through 70 °F water with a velocity U. FIND Estimate the drag caused by the shear stress on the bottom of the ski for 0 6 U 6 30 fts. SOLUTION Clearly the ski is not a flat plate, and it is not aligned exactly parallel to the upstream flow. However, we can obtain a reasonable approximation to the shear force by using the flat plate results. That is, the friction drag, d f , caused by the shear stress on the bottom of the ski 1the wall shear stress2 can be determined as With A /b 4 ft 0.5 ft 2 ft 2 , r 1.94 slugsft 3 , and m 2.04 10 5 lb # sft 2 1see Table B.12 we obtain where d f and U are in pounds and fts, respectively. The friction coefficient, C Df , can be obtained from Fig. 9.15 or from the appropriate equations given in Table 9.3. As we will see, for this problem, much of the flow lies within the transition regime where both the laminar and turbulent portions of the boundary layer flow occupy comparable lengths of the plate. We choose to use the values of C Df from the table. For the given conditions we obtain Re / rU/ m d f 1 2 rU 2 /bC Df d f 1 211.94 slugsft 3 212.0 ft 2 2U 2 C Df 1.94 U 2 C Df 11.94 slugs ft 3 214 ft2U 2.04 10 5 lb # sft 2 3.80 105 U where U is in fts. With U 10 fts, or Re / 3.80 10 6 , we obtain from Table 9.3 C Df 0.4551log Re / 2 2.58 1700Re / (1) 0.00308. From Eq. 1 the corresponding drag is d f 1.941102 2 10.003082 0.598 lb By covering the range of upstream velocities of interest we obtain the results shown in Fig. E9.7b. (Ans) COMMENTS If Re f 1000, the results of boundary layer theory are not valid—inertia effects are not dominant enough and the boundary layer is not thin compared with the length of the plate. For our problem this corresponds to U 2.63 10 3 fts. For all practical purposes U is greater than this value, and the flow past the ski is of the boundary layer type. The approximate location of the transition from laminar to turbulent boundary layer flow as defined by Re cr rUx crm 5 10 5 is indicated in Fig. E9.7b. Up to U 1.31 fts the entire boundary layer is laminar. The fraction of the boundary layer that is laminar decreases as U increases until only the front 0.18 ft is laminar when U 30 fts. For anyone who has water skied, it is clear that it can require considerably more force to be pulled along at 30 fts than the 2 4.88 lb 9.76 lb 1two skis2 indicated in Fig. E9.7b. As is discussed in Section 9.3, the total drag on an object such as a water ski consists of more than just the friction drag. Other components, including pressure drag and wave-making drag, add considerably to the total resistance. 5 4 Entire boundary layer laminar 5 4 U x = 0 x b = width = 0.5 ft x = 4 ft = f , lb 3 2 x cr f 3 2 x cr , ft (a) 1 1 F I G U R E E9.7 0 0 5 10 15 20 25 30 U, ft/s (b) 9.2.6 Effects of Pressure Gradient The boundary layer discussions in the previous parts of Section 9.2 have dealt with flow along a flat plate in which the pressure is constant throughout the fluid. In general, when a fluid flows past an object other than a flat plate, the pressure field is not uniform. As shown in Fig. 9.6, if the Reynolds number is large, relatively thin boundary layers will develop along the surfaces. Within
9.2 Boundary Layer Characteristics 489 U fs1 U fs2 p 1 p 2 The free-stream velocity on a curved surface is not constant. U (1) U fs = U U fs = 0 (2) U fs = 2U these layers the component of the pressure gradient in the streamwise direction 1i.e., along the body surface2 is not zero, although the pressure gradient normal to the surface is negligibly small. That is, if we were to measure the pressure while moving across the boundary layer from the body to the boundary layer edge, we would find that the pressure is essentially constant. However, the pressure does vary in the direction along the body surface if the body is curved, as shown by the figure in the margin. The variation in the free-stream velocity, U fs , the fluid velocity at the edge of the boundary layer, is the cause of the pressure gradient in this direction. The characteristics of the entire flow 1both within and outside of the boundary layer2 are often highly dependent on the pressure gradient effects on the fluid within the boundary layer. For a flat plate parallel to the upstream flow, the upstream velocity 1that far ahead of the plate2 and the free-stream velocity 1that at the edge of the boundary layer2 are equal— U U fs . This is a consequence of the negligible thickness of the plate. For bodies of nonzero thickness, these two velocities are different. This can be seen in the flow past a circular cylinder of diameter D. The upstream velocity and pressure are U and p 0 , respectively. If the fluid were completely inviscid 1m 02, the Reynolds number would be infinite 1Re rUDm 2 and the streamlines would be symmetrical, as are shown in Fig. 9.16a. The fluid velocity along the surface would vary from U fs 0 at the very front and rear of the cylinder 1points A and F are stagnation points2 to a maximum of U fs 2U at the top and bottom of the cylinder 1point C2. This is also indicated in the figure in the margin. The pressure on the surface of the cylinder would be symmetrical about the vertical midplane of the cylinder, reaching a maximum value of p 0 rU 2 2 1the stagnation pressure2 at both the front and back of the cylinder, and a minimum of p 0 3rU 2 2 at the top and bottom of the cylinder. The pressure and free-stream velocity distributions are shown in Figs. 9.16b and 9.16c. These characteristics can be obtained from potential flow analysis of Section 6.6.3. Because of the absence of viscosity 1therefore, t w 02 and the symmetry of the pressure distribution for inviscid flow past a circular cylinder, it is clear that the drag on the cylinder is zero. Although it is not obvious, it can be shown that the drag is zero for any object that does not produce a lift 1symmetrical or not2 in an inviscid fluid 1Ref. 42. Based on experimental evidence, however, we know that there must be a net drag. Clearly, since there is no purely inviscid fluid, the reason for the observed drag must lie on the shoulders of the viscous effects. To test this hypothesis, we could conduct an experiment by measuring the drag on an object 1such as a circular cylinder2 in a series of fluids with decreasing values of viscosity. To our initial surprise we would find that no matter how small we make the viscosity 1provided it is not precisely zero2 we would measure a finite drag, essentially independent of the value of m. As was noted in Section 6.6.3, this leads to what has been termed d’Alembert’s paradox—the drag on an U fs C A θ F (a) U, p 0 A 1 p 0 + ρU 2 2 p 0 F 2U C p 1 p 0 – ρU 2 2 p 0 – ρU 2 U fs U 3 p 0 – ρU 2 2 C 0 90 180 θ , degrees (b) A F 0 0 90 180 θ , degrees (c) F I G U R E 9.16 Inviscid flow past a circular cylinder: (a) streamlines for the flow if there were no viscous effects, (b) pressure distribution on the cylinder’s surface, (c) free-stream velocity on the cylinder’s surface.
- Page 462 and 463: 438 Chapter 8 ■ Viscous Flow in P
- Page 464 and 465: 440 Chapter 8 ■ Viscous Flow in P
- Page 466 and 467: 442 Chapter 8 ■ Viscous Flow in P
- Page 468 and 469: 444 Chapter 8 ■ Viscous Flow in P
- Page 470 and 471: 446 Chapter 8 ■ Viscous Flow in P
- Page 472 and 473: 448 Chapter 8 ■ Viscous Flow in P
- Page 474 and 475: 450 Chapter 8 ■ Viscous Flow in P
- Page 476 and 477: 452 Chapter 8 ■ Viscous Flow in P
- Page 478 and 479: WARNING Stand clear of Hazard areas
- Page 480 and 481: 456 Chapter 8 ■ Viscous Flow in P
- Page 482 and 483: 458 Chapter 8 ■ Viscous Flow in P
- Page 484 and 485: 460 Chapter 8 ■ Viscous Flow in P
- Page 486 and 487: 462 Chapter 9 ■ Flow over Immerse
- Page 488 and 489: 464 Chapter 9 ■ Flow over Immerse
- Page 490 and 491: 466 Chapter 9 ■ Flow over Immerse
- Page 492 and 493: 468 Chapter 9 ■ Flow over Immerse
- Page 494 and 495: 470 Chapter 9 ■ Flow over Immerse
- Page 496 and 497: 472 Chapter 9 ■ Flow over Immerse
- Page 498 and 499: 474 Chapter 9 ■ Flow over Immerse
- Page 500 and 501: 476 Chapter 9 ■ Flow over Immerse
- Page 502 and 503: 478 Chapter 9 ■ Flow over Immerse
- Page 504 and 505: 480 Chapter 9 ■ Flow over Immerse
- Page 506 and 507: 482 Chapter 9 ■ Flow over Immerse
- Page 508 and 509: 484 Chapter 9 ■ Flow over Immerse
- Page 510 and 511: 486 Chapter 9 ■ Flow over Immerse
- Page 514 and 515: 490 Chapter 9 ■ Flow over Immerse
- Page 516 and 517: 492 Chapter 9 ■ Flow over Immerse
- Page 518 and 519: 494 Chapter 9 ■ Flow over Immerse
- Page 520 and 521: 496 Chapter 9 ■ Flow over Immerse
- Page 522 and 523: 498 Chapter 9 ■ Flow over Immerse
- Page 524 and 525: 500 Chapter 9 ■ Flow over Immerse
- Page 526 and 527: 502 Chapter 9 ■ Flow over Immerse
- Page 528 and 529: 504 Chapter 9 ■ Flow over Immerse
- Page 530 and 531: 506 Chapter 9 ■ Flow over Immerse
- Page 532 and 533: 508 Chapter 9 ■ Flow over Immerse
- Page 534 and 535: 510 Chapter 9 ■ Flow over Immerse
- Page 536 and 537: 512 Chapter 9 ■ Flow over Immerse
- Page 538 and 539: 514 Chapter 9 ■ Flow over Immerse
- Page 540 and 541: 516 Chapter 9 ■ Flow over Immerse
- Page 542 and 543: 518 Chapter 9 ■ Flow over Immerse
- Page 544 and 545: 520 Chapter 9 ■ Flow over Immerse
- Page 546 and 547: 522 Chapter 9 ■ Flow over Immerse
- Page 548 and 549: 524 Chapter 9 ■ Flow over Immerse
- Page 550 and 551: 526 Chapter 9 ■ Flow over Immerse
- Page 552 and 553: 528 Chapter 9 ■ Flow over Immerse
- Page 554 and 555: ∋ 530 Chapter 9 ■ Flow over Imm
- Page 556 and 557: 532 Chapter 9 ■ Flow over Immerse
- Page 558 and 559: 10 Open-Channel Flow CHAPTER OPENIN
- Page 560 and 561: 536 Chapter 10 ■ Open-Channel Flo
488 Chapter 9 ■ Flow over Immersed Bodies<br />
E XAMPLE 9.7<br />
Drag on a Flat Plate<br />
GIVEN The water ski shown in Fig. E9.7a moves through<br />
70 °F water with a velocity U.<br />
FIND Estimate the drag caused by the shear stress on the bottom<br />
of the ski for 0 6 U 6 30 fts.<br />
SOLUTION<br />
Clearly the ski is not a flat plate, and it is not aligned exactly<br />
parallel to the upstream flow. However, we can obtain a reasonable<br />
approximation to the shear force by using the flat plate results.<br />
That is, the friction drag, d f , caused by the shear stress on<br />
the bottom of the ski 1the wall shear stress2 can be determined as<br />
With A /b 4 ft 0.5 ft 2 ft 2 , r 1.94 slugsft 3 , and m <br />
2.04 10 5 lb # sft 2 1see Table B.12 we obtain<br />
where d f and U are in pounds and fts, respectively.<br />
The friction coefficient, C Df , can be obtained from Fig. 9.15 or<br />
from the appropriate equations given in Table 9.3. As we will see,<br />
for this problem, much of the flow lies within the transition<br />
regime where both the laminar and turbulent portions of the<br />
boundary layer flow occupy comparable lengths of the plate. We<br />
choose to use the values of C Df from the table.<br />
For the given conditions we obtain<br />
Re / rU/<br />
m<br />
d f 1 2 rU 2 /bC Df<br />
d f 1 211.94 slugsft 3 212.0 ft 2 2U 2 C Df<br />
1.94 U 2 C Df<br />
<br />
11.94 slugs ft 3 214 ft2U<br />
2.04 10 5 lb # sft 2 3.80 105 U<br />
where U is in fts. With U 10 fts, or Re / 3.80 10 6 , we<br />
obtain from Table 9.3 C Df 0.4551log Re / 2 2.58 1700Re / <br />
(1)<br />
0.00308. From Eq. 1 the corresponding drag is<br />
d f 1.941102 2 10.003082 0.598 lb<br />
By covering the range of upstream velocities of interest we obtain<br />
the results shown in Fig. E9.7b.<br />
(Ans)<br />
COMMENTS If Re f 1000, the results of boundary layer<br />
theory are not valid—inertia effects are not dominant enough and<br />
the boundary layer is not thin compared with the length of the<br />
plate. For our problem this corresponds to U 2.63 10 3 fts.<br />
For all practical purposes U is greater than this value, and the flow<br />
past the ski is of the boundary layer type.<br />
The approximate location of the transition from laminar to turbulent<br />
boundary layer flow as defined by Re cr rUx crm <br />
5 10 5 is indicated in Fig. E9.7b. Up to U 1.31 fts the entire<br />
boundary layer is laminar. The fraction of the boundary layer that<br />
is laminar decreases as U increases until only the front 0.18 ft is<br />
laminar when U 30 fts.<br />
For anyone who has water skied, it is clear that it can require<br />
considerably more force to be pulled along at 30 fts than the<br />
2 4.88 lb 9.76 lb 1two skis2 indicated in Fig. E9.7b. As is<br />
discussed in Section 9.3, the total drag on an object such as a water<br />
ski consists of more than just the friction drag. Other components,<br />
including pressure drag and wave-making drag, add considerably<br />
to the total resistance.<br />
5<br />
4<br />
Entire boundary<br />
layer laminar<br />
5<br />
4<br />
U<br />
x = 0<br />
x<br />
b = width = 0.5 ft<br />
x = 4 ft = <br />
f , lb<br />
3<br />
2<br />
x cr<br />
f<br />
3<br />
2<br />
x cr , ft<br />
(a)<br />
1<br />
1<br />
F I G U R E E9.7<br />
0<br />
0<br />
5 10 15 20 25 30<br />
U, ft/s<br />
(b)<br />
9.2.6 Effects of Pressure Gradient<br />
The boundary layer discussions in the previous parts of Section 9.2 have dealt with flow along a<br />
flat plate in which the pressure is constant throughout the <strong>fluid</strong>. In general, when a <strong>fluid</strong> flows past<br />
an object other than a flat plate, the pressure field is not uniform. As shown in Fig. 9.6, if the<br />
Reynolds number is large, relatively thin boundary layers will develop along the surfaces. Within