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ENGG 2230 (29)
Lecture

# Ch 3 Solutions.pdf

136 Pages
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School
Department
Engineering
Course
ENGG 2230
Professor
Bahram Gharabaghi
Semester
Winter

Description
Chapter 3 Integral Relations for a Control Volume P3.1 Discuss Newtons second law (the linear momentum relation) in these three forms: dd mmd ()FaFVFV dtdt system Solution: These questions are just to get the students thinking about the basic laws of mechanics. They are valid and equivalent for constantmass 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 controlvolume related and thus the most popular in this chapter. P3.2 Consider the angularmomentum relation in the form d d ()Mr V O dt system What does r mean in this relation? Is this relation valid in both solid and fluid mechanics? Is it related to the linearmomentum 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 momentcenter O to the elements d where momentum is being summed. Perhaps r is a better notation. O P3.3 For steady laminar flow through a long tube (see Prob. 1.12), the axial velocity 22 distribution is given by u C(R r), 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 r dr. From Eq. (3.7), R 224 () 2QudACRrrdrAnsCR . 2 0 Chapter 3 Integral Relations for a Control Volume 177 P3.4 A fire hose has a 5inch inside diameter and is flowing at 600 galmin. The flow . For steady flow, what should D be, exits through a nozzle contraction at a diameter D nn in inches, to create an exit velocity of 25 ms? Solution: This is a straightforward onedimensional steadyflow continuity problem. Some unit conversions are needed: 3 s; 25 ms = 82.02 fts ; 5 inches = 0.4167 ft 600 galmin = 1.337 ft The hose diameter (5 in) would establish a hose average velocity of 9.8 fts, but we dont really need this. Go directly to the volume flow: 3 ftft 22 .144 .0for Solve ; ) 02 . 82 (337 .1AnsftDDVAQin 1.73 nnnn 4ss P3.5 Water at 20C flows through a 5inchdiameter smooth pipe at a high 18 Reynolds number, for which the velocity profile is given by u U(yR), where U is oo the centerline velocity, R is the pipe radius, and y is the distance measured from the wall toward the centerline. If the centerline velocity is 25 fts, estimate the volume flow rate in gallons per minute. Solution: The formula for average velocity in this powerlaw case was given in Example 3.4: 22ft 0.8370.837(25)20.92VUUU avooo (1)(2)(1)(2)mm s 1818 3 2.5ftftgal 2 [20.92]()2.85.ThusQVAftAns 1280 avpipe 12ssmin Solutions Manual Fluid Mechanics, Fifth Edition 178 flow Q issuing from the slot; then take the limit of your result if P3.6 When a gravitydriven liquid jet .Lh issues from a slot in a tank, as in Fig. P3.6, an approximation for the exit velocity distribution is 2(),ughz where h is the depth of the jet centerline. Near the slot, the jet is horizontal, twodimensional, and of thickness 2L, as shown. Find a general expression for the total volume Fig. P3.6 Solution: Let the slot width be b into the paper. Then the volume flow from Eq. (3.7) is +L 2b 12 3232 QudA[2g(hz)]bdz.Ans (2g)[(hL)(hL)] 3 L L<
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