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192 Chapter 5 ■ Finite Control Volume Analysis integral involves mass flowrates at sections 112 and 122 so that from Eq. 5.4 we get or cs rV nˆ dA m # 2 m # 1 0 m # 1 m # 2 and from Eqs. 1, 5.6, and 5.7 we obtain r 1 A 1 V 1 r 2 A 2 V 2 or since A 1 A 2 V 1 r 2 r 1 V 2 Air at the pressures and temperatures involved in this example problem behaves like an ideal gas. The ideal gas equation of state 1Eq. 1.82 is r p (4) RT (1) (2) (3) Thus, combining Eqs. 3 and 4 we obtain V 1 p 2T 1 V 2 p 1 T 2 118.4 psia21540 °R211000 ft s2 1100 psia21453 °R2 219 fts (Ans) COMMENT We learn from this example that the continuity equation 1Eq. 5.52 is valid for compressible as well as incompressible flows. Also, nonuniform velocity distributions can be handled with the average velocity concept. Significant average velocity changes can occur in pipe flow if the fluid is compressible. E XAMPLE 5.3 Conservation of Mass—Two Fluids GIVEN The inner workings of a dehumidifier are shown in Fig. E5.3a. Moist air 1a mixture of dry air and water vapor2 enters the dehumidifier at the rate of 600 lbmhr. Liquid water drains out of the dehumidifier at a rate of 3.0 lbmhr. A simplified sketch of the process is provided in Fig. E5.3b. FIND Determine the mass flowrate of the dry air and the water vapor leaving the dehumidifier. Fan Section (1) m • 4 m • 5 Control volume Cooling coil Motor Cooling coil F I G U R E E5.3a m • 1 = 600 lbm/hr Condensate (water) F I G U R E E5.3b m • 2 = ? Fan Section (2) Section (3) m • 3 = 3.0 lbm/hr SOLUTION The unknown mass flowrate at section 122 is linked with the known flowrates at sections 112 and 132 with the control volume designated with a dashed line in Fig. E5.3b. The contents of the control volume are the air and water vapor mixture and the condensate 1liquid water2 in the dehumidifier at any instant. Not included in the control volume are the fan and its motor, and the condenser coils and refrigerant. Even though the flow in the vicinity of the fan blade is unsteady, it is unsteady in a cyclical way. Thus, the flowrates at sections 112, 122, and 132 appear steady and the time rate of change of the mass of the contents of
5.1 Conservation of Mass—The Continuity Equation 193 the control volume may be considered equal to zero on a timeaverage basis. The application of Eqs. 5.4 and 5.5 to the control volume contents results in rV nˆ dA m # 1 m # 2 m # 3 0 cs or m # 2 m # 1 m # 3 600 lbmhr 3.0 lbmhr 597 lbmhr (Ans) COMMENT Note that the continuity equation 1Eq. 5.52 can be used when there is more than one stream of fluid flowing through the control volume. The answer is the same with a control volume which includes the cooling coils to be within the control volume. The continuity equation becomes m # 2 m # 1 m # 3 m # 4 m # 5 where m # 4 is the mass flowrate of the cooling fluid flowing into the control volume, and m # 5 is the flowrate out of the control volume through the cooling coil. Since the flow through the coils is steady, it follows that m # 4 m # 5. Hence, Eq. 1 gives the same answer as obtained with the original control volume. (1) E XAMPLE 5.4 Conservation of Mass—Nonuniform Velocity Profile GIVEN Incompressible, laminar water flow develops in a straight pipe having radius R as indicated in Fig. E5.4a. At section (1), the velocity profile is uniform; the velocity is equal to a constant value U and is parallel to the pipe axis everywhere. At section (2), the velocity profile is axisymmetric and parabolic, with zero velocity at the pipe wall and a maximum value of u max at the centerline. FIND (a) How are U and u max related? (b) How are the average velocity at section (2), V 2 , and u max related? Section (1) u 1 = U Control volume Pipe dA 2 = 2 πr dr R r F I G U R E E5.4a Section (2) [ ( ) ] u 2 = u max 1 - _ r 2 R SOLUTION these infinitesimal areas the fluid velocity is denoted as u . (a) An appropriate control volume is sketched (dashed lines) in 2u max R c 1 a r 2 (5) R b d r dr A 1 U 0 0 dA 2 Fig. E5.4a. The application of Eq. 5.5 to the contents of this control volume yields 0 (flow is steady) 2 Thus, in the limit of infinitesimal area elements, the summation is replaced by an integration and the outflow through section (2) is given by 0 r dV (1) 0t cv rV nˆ dA 0 rV ˆn dA r 2 R u 2 2pr dr cs 122 0 (3) At the inlet, section (1), the velocity is uniform with V 1 U so that By combining Eqs. 1, 2, and 3 we get 112rV ˆn dA r 1 A 1 U (2) 2 R u 2 2r dr 1 A 1 U 0 0 (4) At the outlet, section (2), the velocity is not uniform. However, the net flowrate through this section is the sum of flows through numerous small washer-shaped areas of size dA 2 2r dr Since the flow is considered incompressible, 1 2 . The parabolic velocity relationship for flow through section (2) is used in Eq. 4 to yield as shown by the shaded area element in Fig. E5.4b. On each of dr Integrating, we get from Eq. 5 r F I G U R E E5.4b 2u max a r 2 2 r4 R 4R 2b R 2 U 0 0
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Sixth Edition F undamentals of Flui
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About the Authors vii Bruce R. Muns
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Preface xi Life Long Learning Probl
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C ontents 1 INTRODUCTION 1 Learning
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Contents xix 6.4.3 Irrotational Flo
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12.6 Axial-Flow and Mixed-Flow Pump
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30 Chapter 1 ■ Introduction fluid
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32 Chapter 1 ■ Introduction Secti
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2 Fluid Statics CHAPTER OPENING PHO
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I ndex Absolute pressure, 13, 48 Ab
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■ TABLE 1.4 Conversion Factors fr
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■ TABLE 1.7 Approximate Physical
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