Solutions Manual for Fluid Mechanics Seventh Edition in SI Units Potential Flow and Computational Fluid Dynamics PROPRIETARY AND CONFIDENTIAL PDF

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Solutions Manual for Fluid Mechanics Seventh Edition in SI Units Frank M. White Chapter 8 Potential Flow and Computational Fluid Dynamics PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state...

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Solutions Manual for

Fluid Mechanics Seventh Edition in SI Units

Frank M. White Chapter 8 Potential Flow and Computational Fluid Dynamics

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of the McGraw-Hill. © 2011 by The McGraw-Hill Companies, Inc. Limited distribution only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

8.1 Prove that the streamlines ψ (r, θ) in polar coordinates, from Eq. (8.10), are orthogonal to the potential lines φ (r, θ ). Solution: The streamline slope is represented by

Since the ψ − slope = −1/(φ − slope), the two sets of lines are orthogonal. Ans.

8.2 The steady plane flow in the figure has the polar velocity components vθ = Ωr and vr = 0. Determine the circulation Γ around the path shown. Solution: Start at the inside right corner, point A, and go around the complete path:

Fig. P8.2

Γ = ∫ V ⋅ ds = 0(R2 − R1 ) + ΩR2 (π R2 ) + 0(R1 − R2 ) + ΩR1 (−π R1 )

(

or: Γ = π Ω R22 − R12

)

Ans.

8.3 Why is a problem of potential flow more difficult to solve when it involves a boundary on which the pressure is prescribed? Solution: If the pressure is prescribed on a boundary, the problem will involve the Bernoulli’s equation, which is nonlinear in the velocity. The problem then becomes nonlinear, and is more difficult to solve than a problem purely in terms of the velocity potential.

8.4 Using cartesian coordinates, show that each velocity component (u, v, w) of a potential flow satisfies Laplace’s equation separately if ∇2φ = 0. Solution: This is true because the order of integration may be changed in each case:

2

8.5 Is the function 1/r a legitimate velocity potential in plane polar coordinates? If so, what is the associated stream function ψ ( r, θ ) ? Solution: Evaluation of the laplacian of (1/r) shows that it is not legitimate:

P8.6

A proposed harmonic function F(x, y, z) is given by

(a) If possible, find a function f(y) for which the Laplacian of F is zero. If you do indeed solve part (a), can your final function F serve as (b) a velocity potential, or (c) a stream function? Solution: Evaluate ∇2F and see if we can find a suitable f(y) to make it zero:

Solving for f(y) eliminated y3, which is not a harmonic function. (b) Since ∇2F = 0, it can indeed be a velocity potential, although the writer does not think it is a realistic flow pattern. (c) Since F is three-dimensional, it cannot be a stream function.

8.7 Given the plane polar coordinate velocity potential φ = Br2cos(2θ), where B is a constant. (a) Show that a stream function also exists. (b) Find the algebraic form of ψ(r, θ). (c) Find any stagnation points in this flow field. Solution: (a) First find the velocities from φ and then check continuity:

Yes, continuity is satisfied, so ψ exists. Ans.(a) (b) Find ψ from its definition:

3

(c) By checking to see where vr and vθ = 0 from part (a), we find that the only stagnation point is at the origin of coordinates, r = 0. These functions define plane stagnation flow, Fig. 8.19b.

8.8 Consider a flow with constant density and viscosity. If the flow possesses a velocity potential as defined by Eq. (8.1), show that it exactly satisfies the full Navier-Stokes equation (4.38). If this is so, why do we back away from the full Navier-Stokes equation in solving potential flows? Solution: If V = ∇φ, the full Navier-Stokes equation is satisfied identically:

The viscous (final) term drops out identically for potential flow, and what remains is

The Bernoulli relation is an exact solution of Navier-Stokes for potential flow. We don’t exactly “back away,” we need also to solve ∇2φ = 0 in order to find the velocity potential.

8.9 Which of the following are necessary conditions for the applicability of Bernoulli’s equation? (a) steady flow, (b) inviscid flow, (c) incompressible flow, (d) irrotational flow, and (e) along a streamline. Solution: Necessary conditions for applicability of Bernoulli’s equation: (1) inviscid flow (zero viscosity) (2) incompressible flow (constant density) (3) along a streamline (Bernoulli’s constant is streamline specific unless the flow is irrotational)

P8.10

For the velocity distribution u = - B y, v = + B x, w = 0, evaluate the circulation Γ about the rectangular closed curve defined by (x, y) = (1,1), (3,1), (3,2), and (1,2). Interpret your result, especially vis-à-vis the velocity potential. Solution: Given that Γ = ∫V·ds around the curve, divide the rectangle into (a, b, c, d) pieces as shown.

4

Γ =

∫ u ds + ∫ v ds + ∫ u ds + ∫ v ds = (−B)(2) +(3B)(1)+(−2B)(2)+(−B)(1) = + 4B a

b

c

Ans.

d

The flow is rotational. Check |curlV| = 2B = constant, so Γ = (2B)Aregion = (2B)(2) = 4B.

8.11 Consider the two-dimensional flow u = –Ax, v = +Ay, where A is a constant. Evaluate the circulation Γ around the rectangular closed curve defined by (x, y) = (1, 1), (4, 1), (4, 3), and (1, 3). Interpret your result especially vis-a-vis the velocity potential.

Fig. P8.11

Solution: Given Γ = ∫ V · ds around the curve, divide the rectangle into (a, b, c, d) pieces as shown:

The circulation is zero because the flow is irrotational: curl V ≡ 0, Γ = ∫ dφ ≡ 0.

8.12 A two-dimensional Rankine half-body, 8 cm thick, is placed in a water tunnel at 20°C. The water pressure far up-stream along the body centerline is 105 kPa. (a) What is the nose radius of the half-body? (b) At what tunnel flow velocity will cavitation bubbles begin to form on the surface of the body? Solution: (a) The nose radius is the distance a in Fig. 8.6, the Rankine half-body:

5

(b) At 20°C the vapor pressure of water is 2337 Pa. Maximum velocity occurs, as shown, on the upper surface at θ ≈ 63°, where z ≈ 2.04a ≈ 2.6 cm and V ≈ 1.26U∞. Write the Bernoulli’s equation between upstream and V, assuming the surface pressure is just vaporizing at the minimum-pressure point:

8.13 A power-plant discharges cooling water through the manifold in Fig. P8.13, which is 55 cm in diameter and 8 m high and is perforated with 25,000 holes 1 cm in diameter. Does this manifold simulate a line source? If so, what is the equivalent source strength m? Solution: With that many small holes, equally distributed and presumably with equal flow rates, the manifold does indeed simulate a line source of strength

Fig. P8.13

8.14 Consider the flow due to a vortex of strength K at the origin. Evaluate the circulation from Eq. (8.23) about the clockwise path from (a, 0) to (2a, 0) to (2a, 3π /2) to (a, 3π /2) and back to (a, 0). Interpret your result. Solution: Break the path up into (1, 2, 3, 4) as shown. Then

6

Fig. P8.14

There is zero circulation about all closed paths which do not enclose the origin.

P8.15 Starting at the stagnation point in Fig. 8.6, the fluid acceleration along the half-body surface rises to a maximum and eventually drops off to zero far downstream. (a) Does this maximum occur at the point in Fig. 8.6 where Umax = 1.26U? (b) If not, does the maximum acceleration occur before or after that point? Explain. Solution: Since the flow is steady, the fluid acceleration along the half-body surface is convective, dU/dt = U(dU/ds), where s is along the surface. (a) At the point of maximum velocity in Fig. 8.6, dU/ds = 0, hence dU/dt = 0, so answer (a) is No. (b) A plot of U(x) along the surface is shown in Fig. 8.9b, and we see that the slope dU/ds is rather small (and negative) downstream of the maximum-velocity point. Therefore maximum acceleration occurs before the point of maximum velocity, at about s/a ≈ 0.9. The actual value of maximum acceleration (not asked) is approximately (V/U)[d(V/U/)d(s/a)] ≈ 0.51. A graph (not requested) is as follows:

Fig. P8.15

7

8.16 A tornado may be modeled as the circulating flow shown in Fig. P8.16, with υr = υz = 0 and υθ ( r) such that

Fig. P8.16

Determine whether this flow pattern is irrotational in either the inner or outer region. Using the r-momentum equation (D.5) of App. D, determine the pressure distribution p(r) in the tornado, assuming p = p∞ as r → ∞. Find the location and magnitude of the lowest pressure. Solution: The inner region is solid-body rotation, the outer region is irrotational:

The pressure is found by integrating the r-momentum equation (D-5) in the Appendix:

In the inner region, we integrate the radial pressure gradient and match at r = R:

The minimum pressure occurs at the origin, r = 0:

8

8.17 A category-3 hurricane on the Saffir-Simpson scale (www.encyclopedia.com) has a maximum velocity of 130 mi/h. Let the match-point radius be R = 18 km (see Fig. P8.16). Assuming sea-level standard conditions at large r, (a) find the minimum pressure; (b) find the pressure at the match-point; and (c) show that both minimum and match-point pressures are independent of R. Solution: Convert 130 mi/h = 58.1 m/s = ω R. Let ρ = 1.22 kg/m3. (a) From Prob. 8.16,

(b) Again from Prob. 8.16, the match pressure only drops half as low as the minimum pressure:

(c) We see from above that both pmin and pmatch have R in their formulas, but only in conjunction with ω. That is, these pressures depend only upon Vmax , wherever it occurs.

8.18 Air flows at 1.2 m/s along a flat wall when it meets a jet of air issuing from a slot at A. The jet volume flow is 0.4 m3/s per m of width into the paper. If the jet is approximated as a line source, (a) locate the stagnation point S. (b) How far vertically will the jet flow extend?

Fig. P8.18

Solution: This is only half a source. The equivalent source strength is m = 2(0.4m3/s) / [2π(1m)] = 0.127 m2/s. Then, as in Fig. 8.9a, the stagnation point S is at a = m/U = (0.127 m2/s) / (1.2 m/s) = 0.106 m to the left of A. Ans. (a) The effective half-body, shown as a dashed line in Fig. P8.18, extends above the wall to a distance equal to πa = π(0.106) = 0.333 m. Ans.(b) 8.19 Find the position (x, y) on the upper surface of the half-body in Fig. 8.9a for which the local velocity equals the uniform stream velocity. What should the pressure be at this point? Solution: The surface velocity and surface contour are given by Eq. (8.18):

9

If V = U∞, then a2/r2 = –2a cosθ /r, or cosθ = –a/(2r). Combine with the surface profile above, and we obtain an equation for θ alone: tan θ = −2(π − θ ). The final solution is:

Since the velocity equals U∞ at this point, the surface pressure p = p∞ . Ans.

8.20 Find the pressure as a function of the radial coordinate for (a) a source at the origin, (b) an irrotational vortex at the origin, and (c) a source plus an irrotational vortex at the origin, where the pressure at infinity is zero. Is the pressure in (c) equal to the sum of those in (a) and (b)? What can you comment on this result? Is it always true that the pressure of a flow is equal to the sum of the pressures of its basic components? Solution: m , Vθ = 0 2π r

Source

Vr =

r-momentum:

m 1 ∂p ∂V = −ρVr r = ρ   3  2π  r ∂r ∂r

2

Irrotational vortex r-momentum:

Source + vortex

τ 2π r 2  τ 2 1 ∂p V = ρ θ = ρ  3  2π  r r ∂r

Vr = 0, Vθ =

Vr =

m , 2π r

Vθ =

τ 2π r

 ∂p ∂V V 2  = ρ −Vr r + θ  ∂r r  ∂r   m 2  τ 2  1 = ρ   +    3  2 π   2 π   r r-momentum:

which is the sum of the pressures due to source and vortex alone. This is a special case in which the linear superposition of pressure works. The two velocity fields are orthogonal to each other, and the inertia terms due to these basic flow fields do not interact with each other. The linear superposition of pressures does not work in general, even though it works for the velocity.

10

8.21 Plot the streamlines and potential lines of the flow due to a line source of strength m at (a, 0) plus a line source 3m at (–a, 0). What is the flow pattern viewed from afar?

Fig. P8.21

Solution: The pattern viewed close-up is shown above. The pattern viewed from afar is at right and represents a single source of strength 4m. Ans.

8.22 A two-dimensional flow field is generated by two sources or sinks of unequal strengths located at (1, 0) and (–1, 0). Find the position of any stagnation point when the strengths at these two points are, respectively, (a) 1 and 4, (b) –1 and –4, and (c) 1 and –4. Solution: The stagnation point, if any, is to be located on the x-axis, say, at (x, 0) Velocity due to m1 and m2

V=

m1 m2 + at (x, 0) 2 π (x −1) 2 π (x +1)

(a) m1 = 1, m2 = 4, both sources, stagnation point lies between them −1 < x < 1 4 1 + = 0 ⇒ x = 0.6 2 π (x −1) 2 π (x +1)

(b) m1 = −1, (c) m1 = 1,

m2 = −4, both sinks, stagnation is located, same as (a), at x = 0.6. m2 = −4, source and stronger sink, stagnation point lies to the right side of m1,

4 1 − =0 2 π (x −1) 2 π (x +1)

⇒

x = 1.67

11




8.23 Plot the streamlines and potential lines of the flow due to a line source of strength 3m at (a, 0) plus a line sink of strength –m at (–a, 0). What is the pattern viewed from afar?

Fig. P8.23

Solution: The pattern viewed close-up is shown at upper right—there is a stagnation point to the left of the sink, at (x, y) = (–2a, 0). The pattern viewed from afar is at right and represents a single source of strength +2m. Ans.

8.24 Plot the streamlines of the flow due to a line vortex of strength +K at (0, +a) plus a line vortex of strength –K at (0, –a). What is the pattern viewed from afar? Solution: The pattern viewed close-up is shown at right (see Fig. 8.17b of the text). The pattern viewed from afar represents little or nothing, since the two vortices cancel strengths and cause no flow at ∞. Ans.

Fig. P8.24

12

P8.25 At point A in Fig. P8.25 is a clockwise line vortex of strength K = 12 m2/s. At point B is a line source of strength m = 25 m2/s. Determine the resultant velocity induced by these two at point C . Solution: The vortex induces a velocity K/rAC = (12m2/s)/4m = 3 m/s vertically at point C. The source induces a velocity m/rBC = (25m2/s)/5m = 5 m/s down to the left at a 37° angle. Sum vertical and horizontal velocities from the sketch at right: Fig. P8.25

The resultant induced velocity is 4 m/s to the left of point C.

8.26 Consider inviscid stagnation flow, ψ = Kxy (see Fig. 8.19b), Superimposed with a source at the origin of strength m. Plot the resulting streamlines in the upper half plane, using the length scale a = (m/K)1/2. Give a physical explanation of the flow pattern. Solution: The sum of stagnation flow plus a line source at the origin is

The plot is below, using MATLAB, and represents stagnation flow toward a bump of height a.

Fig. P8.26

13

8.27 Find the resultant velocity vector induced at point A in Fig. P8.27 due to the combination of uniform stream, vortex, and line source. Solution: The velocities caused by each term—stream, vortex, and sink—are shown below. They have to be added together vectorially to give the final result:

Fig. P8.27

8.28 Line sources of equal strength m = Ua, where U is a reference velocity, are placed at (x, y) = (0, a) and (0, –a). Sketch the stream and potential lines in the upper half plane. Is y = 0 a “wall”? If so, sketch the pressure coefficient

along the wall, where po is the pressure at (0, 0). Find the minimum pressure point and indicate where flow separation might occur in the boundary layer. Solution: This problem is an “image” flow and is sketched in Fig. 8.17a of the text. Clearly y = 0 is a “wall” where

From Bernoulli, p + ρ u 2/2 = po,

14

The minimum pressure coefficient is Cp,min = –1.0 at x = a, as shown in the figure. Beyond this point, pressure increases (adverse gradient) and separation is possible.

Fig. P8.28

8.29 Let the vortex/sink flow of Eq. (8.16) simulate a tornado as in Fig. P8.29. Suppose that the circulation about the tornado is Γ = 8500 m2/s and that the pressure at r = 40 m is 2200 Pa less than the far-field pressure. Assuming inviscid flow at sea-level density, estimate (a) the appropriate sink strength –m, (b) the pressure at r = 15 m, and (c) the angle β at which the streamlines cross the circle at r = 40 m (see Fig. P8.29).

Fig. P8.29

Solution: The given circulation yields the circumferential velocity at r = 40 m:

Assuming sea-level density ρ = 1.225 kg/m3, we use Bernoulli to find the radial velocity: 1.225  ρ ρ p∞ + (0)2 = (p∞ − Δp) + vθ2 + v2r = p∞ − 2200 + (33.8)2 + v2r  2  2 2

(

)

With circumferential and radial (inward) velocity known, the streamline angle β is

(b) At r = 15 m, compute v r = m/r = 1980/15 ≈ 132 m/s (unrealistically high) and v θ = Γ/2 π r = 8500/[2 π ( 15)] ≈ 90 m/s (high again, there is probably a viscous core here). Then we use Bernoulli again to compute the pressure at r = 15 m:

If we assume sea-level pressure of 101 kPa at ∞, then pabsolute = 101 – 16 ≈ 85 kPa.

15

8.30 A coastal power plant takes in cooling water through a vertical perforated manifold, as in Fig. P8.30. The total volume flow intake is 110 m3/s. Currents of 25 cm/s flow past the manifold, as shown. Estimate (a) how far downstream and (b) how far normal to the paper the effects of the intake are felt in the ambient 8-m-deep waters.

Fig. P8.30

Solution: The sink strength m leads to the desired dimensions. The distance downstream from the sink is a and the distance normal to the paper is pa (see Fig. 8.6):

8.31 Water at 20°C flows past a halfbody as shown in Fig. P8.31. Measured pressures at points A and B are 160 kPa and 90 kPa, respectively, with uncertainties of 3 kPa each. Estimate the stream velocity and its uncertainty. Solution: Since Eq. (8.18) is for the upper surface, use it by noting that VC = VB in the figure:

Fig. P8.31

  2 2  2 rC π − π /2 π 2 2 2 = = , VC = VB = U∞ 1+   + cos(π /2) = 1.405U∞2 a sin(π /2) 2   π  (π /2) 

ρ ρ 998 Bernoulli: pA + VA2 = 160000 + 0 = pB + VB2 = 90000 + 1.405U∞2 2 2 2 Solve for U∞ ≈ 10.0 m / s Ans.

(

)

The uncertainty in (pA – pB) is as high as 6000 Pa, hence the uncertainty in U∞ is ±0.4 m/s. Ans.

16

8.32 Sources of equal strength m are placed at the four symmetric positions (a, a), (–a, a), (a, –a), and (–a...