Chapter3_SM - Chapter 3 Integral Relations for a Control...

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Chapter 3 Integral Relations for a Control Volume 3.1 Discuss Newton’s second law (the linear momentum relation) in these three forms: () system dd mm d dt dt ρυ ⎛⎞ ⎜⎟ ∑= ⎝⎠ Fa F V F V 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 of all 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 O system d d dt × Mr V What does r mean in this relation? Is this relation valid in both solid and fluid mechanics? Is it 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 d υ where momentum is being summed. Perhaps r O is a better notation. 3.3 For steady laminar flow through a long tube (see Prob. 1.12), the axial velocity distribution is given by u = C(R 2 r 2 ), 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 = 2 π r dr. From Eq. (3.7), 22 0 2 . R Q u dA C R r r dr Ans == = ∫∫ 4 CR 2
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Chapter 3 Integral Relations for a Control Volume 177 P3.4 A fire hose has a 5-inch inside diameter and is flowing at 600 gal/min. The flow exits through a nozzle contraction at a diameter D n . For steady flow, what should D n be, in inches, to create an exit velocity of 25 m/s? Solution : This is a straightforward one-dimensional steady-flow continuity problem. Some unit conversions are needed: 600 gal/min = 1.337 ft 3 /s; 25 m/s = 82.02 ft/s ; 5 inches = 0.4167 ft The hose diameter (5 in) would establish a hose average velocity of 9.8 ft/s, but we don’t really need this. Go directly to the volume flow: 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 0 0.25 0.5 0.75 1.0 1.25 1.5 1.75 2.0 u, m/s 6.00 5.97 5.88 5.72 5.51 5.23 4.89 4.43 0.00 Comment on these data vis-a-vis laminar flow, Prob. 3.3. Estimate, as best you can, the total volume flow Q through the tube, in m 3 /s. Solution: The data can be plotted in the figure below. . 144 . 0 for Solve ; ) 02 . 82 ( 4 337 . 1 2 2 3 Ans ft D s ft D V A s ft Q n n n n in 1.73 = = = = = π
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Solutions Manual Fluid Mechanics, Fifth Edition 178 As seen in the figure, the flat (turbulent) velocities do not resemble the parabolic laminar- flow profile of Prob. 3.3. (The discontinuity at r = 1.75 cm is an artifact—we need more data for 1.75 < r < 2.0 cm.) The volume flow, Q = u(2 π r)dr, can be estimated by a numerical quadrature formula such as Simpson’s rule. Here there are nine data points: 11 22 33 44 55 66 77 88 99 2( 4 2 4 2 4 2 4 ) 3 ,.
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This note was uploaded on 04/24/2011 for the course MECH 122 taught by Professor M during the Fall '10 term at San Jose State University .

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Chapter3_SM - Chapter 3 Integral Relations for a Control...

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