Fluidos

Chapter 3 • Integral Relations for a Control Volume
3.1 Discuss Newton’s second law (the linear momentum relation) in these three forms:

å F = ma

åF =

d (mV ) dt

åF =

ö dæ ç ò Vρ dυ ÷ ÷ dt ç system è ø

Solution: These questions are just to get the students thinking about the basic laws of mechanics. They are valid and equivalent for constant-mass systems, and we can make use ofall of them in certain fluids problems, e.g. the #1 form for small elements, #2 form for rocket propulsion, but the #3 form is control-volume related and thus the most popular in this chapter.

3.2 Consider the angular-momentum relation in the form å MO = ù d é ê ò (r × V) ρ dυ ú dt ê system ú ë û

What does r mean in this relation? Is this relation valid in both solid and fluid mechanics? Isit related to the linear-momentum equation (Prob. 3.1)? In what manner? Solution: These questions are just to get the students thinking about angular momentum versus linear momentum. One might forget that r is the position vector from the moment-center O to the elements ρ dυ where momentum is being summed. Perhaps rO is a better notation.

3.3 For steady laminar flow through a long tube (seeProb. 1.12), the axial velocity distribution is given by u = C(R2 − r2), where R is the tube outer radius and C is a constant. Integrate u(r) to find the total volume flow Q through the tube. Solution: The area element for this axisymmetric flow is dA = 2π r dr. From Eq. (3.7),
Q = ò u dA = ò C ( R 2 − r 2 )2π r dr =
0 R

π CR 4 2

Ans.

Chapter 3 • Integral Relations for a Control Volume149

3.4 Discuss whether the following flows are steady or unsteady: (a) flow near an automobile moving at 55 m/h; (b) flow of the wind past a water tower; (c) flow in a pipe as the downstream valve is opened at a uniform rate; (d) river flow over a spillway of a dam; and (e) flow in the ocean beneath a series of uniform propagating surface waves. Solution: (a) steady (except for vortexshedding) in a frame fixed to the auto. (b) steady (except for vortex shedding) in a frame fixed to the water tower. (c) unsteady by its very nature (accelerating flow). (d) steady except for fluctuating turbulence. (e) Uniform periodic waves are steady when viewed in a frame fixed to the waves.

3.5 A theory proposed by S. I. Pai in 1953 gives the following velocity values u(r) for turbulent(high-Reynolds number) airflow in a 4-cm-diameter tube: r, cm u, m/s 0 6.00 0.25 5.97 0.5 5.88 0.75 5.72 1.0 5.51 1.25 5.23 1.5 4.89 1.75 4.43 2.0 0.00

Comment on these data vis-a-vis laminar flow, Prof. 3.3. Estimate, as best you can, the total volume flow Q through the tube, in m3/s. Solution: The data can be plotted in the figure below.

As seen in the figure, the flat (turbulent) velocities donot resemble the parabolic laminarflow profile of Prob. 3.3. (The discontinuity at r = 1.75 cm is an artifact—we need more

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Solutions Manual • Fluid Mechanics, Fifth Edition

data for 1.75 < r < 2.0 cm.) The volume flow, Q = ò u(2π r)dr, can be estimated by a numerical quadrature formula such as Simpson’s rule. Here there are nine data points: æ ∆r ö Q = 2π (r1u1 + 4r2u2 + 2r3u3 + 4r4u4+ 2r5u5 + 4r6u6 + 2r7u7 + 4r8u8 + r9u9 ) ç ÷ è 3 ø
For the given data, Q ≈ 0.0059 m 3 /s Ans.

3.6 When a gravity-driven liquid jet issues from a slot in a tank, as in Fig. P3.6, an approximation for the exit velocity distribution is u ≈ 2g(h − z), where h is the depth of the jet centerline. Near the slot, the jet is horizontal, two-dimensional, and of thickness 2L, as shown. Find a generalexpression for the total volume flow Q issuing from the slot; then take the limit of your result if L = h.

Fig. P3.6

Solution: Let the slot width be b into the paper. Then the volume flow from Eq. (3.7) is

Q = ò u dA =

+L −L

ò

[2g(h − z)]1/2 b dz =

2b √(2g)[(h + L)3/2 − (h − L)3/2 ] Ans. 3

In the limit of L = h, this formula reduces to

Q ≈ (2Lb) (2gh)

Ans.

3.7 In...