Student Solutions Manual to accompany Engineering Fluid Mechanics, 7th Edition Clayton T. Crowe and Donald F. Elger October 1, 2001 ii prgmea.com Contents 1 Introduction 1 2 Fluid Properties 5 3 Fluid Statics 9 4 Fluids in Motion 21 5 Pressure Variation in Flowing Fluids 35 6 Momentum Principle 45 7 Energy Principle 59 8 Dimensional Analysis and Similitude 69 9 Surface Resistance 79 10 Flow in Conduits 91 11 Drag and Lift 107 12 Compressible Flow 117 13 Flow Measurements 127 14 Turbomachinery 137 15 Varied Flow in Open Channels 143 iii prgmea.com iv CONTENTS prgmea.com Preface This volume presents a variety of example problems for students of fluid me- chanics. It is a companion manual to the text, Engineering Fluid Mechanics, 7th edition, by Clayton T. Crowe, Donald F. Elger and John A. Roberson. Andrew DuBuisson, Steven Ruzich, and Ashley Ater have provided help with editing and with checking the solutions for accuracy. Please transmit any comments or recommendations to Professor Donald F. Elger Mechanical Engineering Department University of Idaho Moscow, ID 83844-0902 [email protected] Phone: (208) 885-7889 FAX: (208) 885-9031 v prgmea.com vi PREFACE Chapter 1 Introduction Problem 1.1 Consider a glass container, half-full of water and half-full of air, at rest on a laboratory table. List some similarities and differences between the liquid (water) and the gas (air). Solution Similarities 1. The gas and the liquid are comprised of molecules. 2. The gas and the liquid are fluids. 3. The molecules in the gas and the liquid are relatively free to move about. 4. The molecules in each fluid are in continual and random motion. 1 2 CHAPTER 1. INTRODUCTION Differences 1. Intheliquidphase,therearestrongattractiveandrepulsiveforcesbetween the molecules; in the gas phase (assuming ideal gas), there are minimal forcesbetweenmoleculesexceptwhentheyareincloseproximity(mutual repulsive forces simulate collisions). 2. A liquid has a definite volume; a gas will expand to fill its container. Since the container is open in this case, the gas will continually exchange molecules with the ambient air. 3. A liquid is much more viscous than a gas. 4. A liquid forms a free surface, whereas a gas does not. 5. Liquids are very difficult to compress (requiring large pressures for small compression), whereas gases are relatively easy to compress. 6. With the exception of evaporation, the liquid molecules stay in the con- tainer. The gas molecules constantly pass in and out of the container. 7. A liquid exhibits an evaporation phenomenon, whereas a gas does not. Comments Most of the differences between gases and liquids can be understood by consid- ering the differences in molecular structure. Gas molecules are far apart, and each molecule moves independently of its neighbor, except when one molecule approaches another. Liquid molecules are close together, and each molecule exerts strong attractive and repulsive forces on its neighbor. Problem 1.2 In an ink-jet printer, the orifice that is used to form ink drops can have a diameter as small as 3 10 6 m. Assuming that ink has the properties of − × water, does the continuum assumption apply? Solution The continuum assumption will apply if the size of a volume, which contains enough molecules so that effects due to random molecular variations average out, is much smaller that the system dimensions. Assume that 104 molecules is sufficient for averaging. If L is the length of one side of a cube that contains 104 molecules and D is the diameter of the orifice, the continuum assumption is satisfied if L 1 D ¿ 3 ThenumberofmoleculesinamoleofmatterisAvogadro’snumber:6.02 1023. × The molecular weight of water is 18, so the number of molecules (N) in a gram of water is 6.02 1023 molecules mole N = × mole 18 g µ ¶µ ¶ molecules =3.34 1012 × g The density of water is 1 g/cm3, so the number of molecules in a cm3 is 3.34 × 1012. The volume of water that contains 104 molecules is 104 molecules Volume = 3.34 1012 molecules × cm3 =3.0 10 19 cm3 − × Since the volume of a cube is L3, where L is the length of a side L= 3 3.0 10 19 cm3 × − =6p.2 10−7 cm × =6.2 10 9 m − × Thus L 6.2 10 9 m = × − D 3.0 10 6 m × − =0.0021 Since L 1, the continuum assumption is quite good. D ¿ 4 CHAPTER 1. INTRODUCTION