5-Practice Problem - Chapter 5 Control Volume Approach and...

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Unformatted text preview: Chapter 5 Control Volume Approach and Continuity Principle Problem 5.1 A 10-cm-diameter pipe contains sea water that g32 ows with a mean velocity of 5 m/s. Find the volume g32 ow rate (discharge) and the mass g32 ow rate. Solution The discharge is g84 = g89 g68 where g89 is the mean velocity. Thus g84 = 5 × g28 4 × g61 1 2 = 0 g61 0393 m 3 /s From Table A.4, the density of sea water is 1026 kg/m 3 g61 The mass g32 ow rate is g98 g112 = g29g84 = 1026 × g61 0393 = 40 g61 3 kg/s 29 30 CHAPTER 5. CONTROL VOLUME APPROACH AND CONTINUITY PRINCIPLE Problem 5.2 The velocity pro g31 le of a non-Newtonian g32 uid in a circular conduit is given by g120 g120 max = g23 1 g3 ³ g117 g85 ´ 2 ¸ 1 g64 2 where g120 max is the velocity at the centerline and g85 is the radius of the conduit. Find the discharge (volume g32 ow rate) in terms of g120 max and g85g61 Solution The volume g32 ow rate is g84 = Z g68 g120g103g68 For an axisymmetric duct, this integral can be written as g84 = 2 g28 Z g85 g120g117g103g117 Substituting in the equation for the velocity distribution g84 = 2 g28g120 max Z g85 g23 1 g3 ³ g117 g85 ´ 2 ¸ 1 g64 2 g117g103g117 Recognizing that 2 g117g103g117 = g103g117 2 g62 we can rewrite the integral as g84 = g28g120 max Z g85 g23 1 g3 ³ g117 g85 ´ 2 ¸ 1 g64 2 g103g117 2 = g28g120 max g85 2 Z g85 g23 1 g3 ³ g117 g85 ´ 2 ¸ 1 g64 2 g103 ³ g117 g85 ´ 2 or g84 = g28g120 max g85 2 Z 1 [1 g3 g20 ] 1 g64 2 g103g20 = g3 2 3 g28g120 max g85 2 [1 g3 g20 ] 3 g64 2 | 1 = 2 3 g28g120 max g85 2 31 Problem 5.3 A jet pump injects water at 120 ft/s through a 2-in. pipe into a secondary g32 ow in an 8-in. pipe where the velocity is 10 ft/s. Downstream the g32 ows become fully mixed with a uniform velocity pro g31 le. What is the magnitude of the velocity where the g32 ows are fully mixed? Solution Draw a control volume as shown in the sketch below. Because the g32 ow is steady X g102g118 g29 V · A = Assuming the water is incompressible, the continuity equation becomes X g102g118 V · A = The volume g32 ow rate across station g100 is X g100 V · A = g3 10 × g28 4 μ 8 12 ¶ 2 where the minus sign occurs because the velocity and area vectors have the opposite sense. The volume g32 ow rate across station b is X g101 V · A = g3 120 × g28 4 μ 2 12 ¶ 2 32 CHAPTER 5. CONTROL VOLUME APPROACH AND CONTINUITY PRINCIPLE and the volume g32 ow rate across station c is X g102 V ·...
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This note was uploaded on 12/18/2011 for the course ECON 102 taught by Professor Drfcvvv during the Spring '11 term at Politechnika Wrocławska.

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5-Practice Problem - Chapter 5 Control Volume Approach and...

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