Introduction to Fluid Mechanic
Introduction to Fluid Mechanic CE 321
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This 21 page Class Notes was uploaded by Jolie Shields on Saturday September 19, 2015. The Class Notes belongs to CE 321 at Michigan State University taught by Roger Wallace in Fall. Since its upload, it has received 46 views. For similar materials see /class/207369/ce-321-michigan-state-university in Civil Engineering at Michigan State University.
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Date Created: 09/19/15
Lecture 11 Conservation of Mass Solids amp Fluids Conservation Laws Solid mechanics approach Create a free body diagram and analyze the forces on an object Can we use the same approach for fluids Yes but for fluids it is difficult to identify a body of matter readily If we follow this approach as in solid mechanics we deal with a control system or control mass of fluid A control system system can change size and shape but it must always contain the same material Control Mass CM or Control System CS In fluid mechanics the CM or CS approach can be used only in special situations such as fluids at rest For moving fluids such an approach is not very useful since it involves following the fluid particles around A more serious difficulty We are often interested in the forces that a fluid exerts on a fixed surface eg a fan or an airplane so following fluid particles is a wasteful exercise This leads to the concept of Control Volume CV A CV is a designated region in space of fixed shape and volume Its bounding surface called the control surface is fixed The identity and amount of matter contained within the CV may change with time The use of CVs does not permit the direct application of laws of physics from mechanics Why Laws of mechanics are applicable only to a control mass and not to a control volume We need to derive equations for control volumes Two Approaches Lagrangian Follow individual fluid particles and see what they experience 0 Eulerian Observe changes at a fixed point in space Local and Material Rates For a fluid in motion a point fixed in space is occupied by different material particles at different times The rate of change recorded by a probe fixed in space does not give the rate of change of property of any material particle To measure this the probe has to move with the fluid particle The fixed probe measures the local rate of change The rate of change experienced by a fluid particle is called the material or substantive rate ofchange Local and Material Rates The local rate of change ofa X9 t propertyB is denoted by The material rate of change These two are related and this relation is given by the Reynolds Transport Theorem Reynolds Transport Theorem The Reynolds Transport Theorem relates the rate of change of a property associated with a CV to the material rate of change of that property If we consider any property B then Accumurauon Growth mnux EfnuX Er J W IHH W Rate of Increase Rate of Increase Cont butiOn Contribution of B in CV of B of matter due t0 Infl0W due to Out ow Chain Rule of Calculus BEBOCJ DB GB 8x 83 Dr 6 at 6x At Ll E DB a 613 0 u Dt at 8x DB 63 o Bout The rate at which B the property B flows Dt at out m out of the control volume Typical Control Volumes l 5 f a Pipe Fluid flows across part of the control surface but not across all of it b Jet Engine The air that was within the engine at one time is outside at a later time c Balloon A deforming control volume Intensive amp Extensive Properties B any property 9 amount of that property per unit mass m mass Extensive properties depend on the system size eg heat Intensive Properties are normalized to the system size eg density temperature pressure kinetic energy per unit mass Bmb Bmb 2 2 BMW 3191 2 2 Bmgtb1 BmVgtbV DI CV at PzAszbz P1A1V1bl sys Dr 53 2 WW Z poutAoutVoutbout Z pinAinVinbin d a trig 4T5 I 7 1 lt AA quot Fi l39 J I 1 l39 gt P IVF39 I IVl Iquot 39i1 Le I In L I I 39i W rizi i1 m Q 13 mud 39L39E39l IIIUI 5w I mJ anal waiuul mr r Syalum Illllflvllxll lv I Illnu39 r a m m 630V at Bout Bin 1 gtl rem Conservation of Mass The Continuity Equation The continuity equation simply states that the time rate of change of the system mass is equal to zero 4 mum m v 7 p A v 7 pA v 2 2 2 1 1 1 Dr 01 W Net Rate ofFlow ofMass Through Time Rate of Change TIme Rate of Change 0 me Come Surface of ass ofcoincidem Mass ofthe Contents ofthe Coincident Control Volume 5 7 0 dV A V 7 A V 0 Dr OtEip Zpuut out out 2pm 1 71171 Dm Steady State For steady flows the property of interest does not change with time at any given point in space In other words 8 a I Z pouIAoulvoul Z pi Amvm 0 6mW at 2 I pdv 0 For Steady Flows t CV Z pouleulvoul Z pmAmvm 0 prlAl 2 10212142 Only one Inlet and one outlet V1141 V2142 Constant Density Flow How to Select the quotBestquot Control Volume The choice of a control volume is very important Some choices are better than others Practice as many problems as you can Lifu39iljfnfnl an fgemr me Problem 1 The crosssectional area of the test section of a large water tunnel is 100 ftz For a test velocity of 50 fts what vol ume owratc capacity in gal min is needed QAV a oaff Xso ffjfzt kyi S 761quot a u Q 6155M Problem 53 53 Air ows steadily between two cross sections in a long straight section of 025m inside diameter pipe The static tem perature and pressure at each section are indicated in Fig P53 If the average air velocity at section 2 is 320 ms determine the average air velocity at section 1 Section 1 p1 690 kFa abs P2 T1 300 K 63 ll Il NV4 I FIGURE P53 53 Solution Secuon 1 P2 1 p1 690 kPa EDS 27 kPa abs I FIGUR P53 For sfeaa y 0w AeWeen ecfary and2 WI 72 ppRT or A V A V plzleZI ll zz pzzszTz T 15 p 4 ampamp V 32 Z Vquot m pl 101 Tl Again11 ha under 1 Land1 7 0 2 4 m em afr eme a an ideal ya we are e deal ya cgum t on of zze 53 7 76 951 o 1 E z Z IO 3 TL Sectmn 2 53 Solution Com rrhj 557 and Z and very7 14 A A We gev P T V Iz7ieampabaaallt azgm 2 3 2 I l Tz 670 aws 252lt 70 Ln 39 5 lt Problem 55 55 Water ows out through a set of thin closely spaced blades as shown in Fig P55 with a speed of V 10 fts around the entire circumference of the outlet Determine the mass ow rate through the inlet pipe v a I FIGURE P55 an 1 14 cmservav39on of mass ringan M 55 Solution 15 H16 urnn vowne minfwd wi nh 77x1 570 ms Shawn M 176 SAGILCLL MWe quot mlhlef moule7L momHe n FIGURE P55 1 ca 0 FAA13 du iLcf OuHef 094 2 n a 626101 OK0 was 6 H3 366 sujs T a 537739 r LI V w 00 Oldcf vow com Vow