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Introduction to Fluid Mechanic

by: Jolie Shields

Introduction to Fluid Mechanic CE 321

Marketplace > Michigan State University > Civil Engineering > CE 321 > Introduction to Fluid Mechanic
Jolie Shields
GPA 3.85

Roger Wallace

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Roger Wallace
Class Notes
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This 33 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 50 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 16 Energy Equation First Law of Thermodynamics What is energy The ability of an object or the capacity of a physical system to do work Energy can take a variety of forms Units Joules or ergs Joule the amount of energy needed to apply a force of 1 newton over a distance of1 m k k 2 3021ng S S id 1mm 1 What is heat The simple answer is that heat is energy Energy can change from one form to another Different types of energy can be converted into heat energy Light electrical mechanical chemical nuclear and sound energy can cause a substance to heat up by increasing the speed of its molecules Put energy into a system and it heats up Take energy away and it cools Confused about Work and Heat and energy Work and Heat are both forms of energy Work can be converted to heat and heat can be converted to work Energy is never lost First Law of Thermodynamics Example If you pump up a bicycle tire you are performing work to compress the air Both tire and pump warm up If you puncture the tire instead the tireair system has to do work to expand lt loses heat and cools Question How can we heat or cool anything without adding or removing heat Answer Simple By doing work This is the meaning of adiabatic The First Law of Thermodynamics States that energy is never lost but is DE converted from one sys W form to another Dr quot6 quot6 There are many forms E em 2 I 60623 of energy 3 For a system of fluid Sys particles the first law can be written as V2 eEm 7gz e total energy Internal Energy 2 The internal energy of a system is the total of the kinetic energy due to the motion of molecules translational rotational vibrational and the potential energy associated with the vibrational and electric energy of atoms within m0IeCUIes It includes the energy in all the chemical bonds and the energy of the free conduction electrons in metals The First Law of Thermodynamics The time rate OI increase N et time rate OI energy N et time rate OI energy of the total stored energy 2 addition by heat transfer addition by work transfer of the system into the system into the system an ZQMM Zwas e Di 1 gnaw M E I epd v sys Rates Note that we are dealing with heat transfer rate and work transfer rate These are positive going into the system and negative coming out For a control volume that is coincident with the system at an instant of time the sum of the two rates are equal Reynolds Transport Theorem DB 6 Dr E l Z boutpoutAoutVout Z binpinAinVin W W TRANSFER ACROSS BOUNDARIES SYSTEM CONTROL VOLUME the time rate of increase Time rate of increase 2 OI tne property B of the property B of the contents of the control volume the net u of V of the quantity out of the control volume through the control surface Reynolds Transport Theorem for Energy DE 3 Z eoutpoutAoutVout Z einpinAinVin 25 at W TRANSFER ACROSS BOUNDARIES SYSTEM CONTROL VOLUME Time rate of increase the time rate of increase of the oral stored energy of the total stored energy of the system of the contents of the control volume the m u of V of the total stored energy out of the control volume through the control surface First Law for a Control Volume 8 a I epdv ZeaulpoutAautVout ZeinpinAinViiz Qnet in VVIzell n cv Adiabatic Processes Many engineering 0 processes are adiabatic m The work transfer rate is called power Work Transfer Work is transferred across a CV boundary by a moving shaft eg think about turbines fans and propellers WorkForce gtlt Displacement Rate of work PowerForce gtlt Displacement per unit time For a rotating shaft Wsha Tshaft a a IAIlgUIaI39 VGIOCII39y OI U16 S aII Work Transfer In a system it is possible to have more than one shaft cutting across the CS boundary In this case Wsha net in Z Wsha in Z Wsha out Work Transfer Work transfer also occurs at a CS when force associated with fluid pressure acts over a distance p A V VVnormal stressm F normal stressm p VVnormalstTessout Ezormalstress0ut Vout p0utA0utV0ut ermal stress p0utA0utVout Work Transfer Can also occu sue to tangential stress forces But this iS generally zero at quotnets anu outlets as flow is normal to the surfaces First Law for a CV I 60 Z eout 0 out Aout Vout Z em pm Am CV Qnet in Wmajl netm Z pmAm Kn Z pouleul Vout Now let us expand the e terms First Law for a CV 2 2 jepdVZ V gz poonutlOm at 0 2 out 2 V p V Z u g7 g2 Om1me Qnelm 39 Wma nelm For fixed nondeforming control volumes only Simplifications gt m m quotI 7 v v If the time rate of change of the total stored energy 6 of the contents of the CV is zero steady flow loinAinVin pouleulVoul V2 V2 m Z710111 z 7in Mgzouz Zin 390 out 390 in 239 Qnelin Wsha nelin Enthalpy 525143 p 1D Energy Equation for Steady Flows pout10m Vow 2 2 V V V V out m m hour hin 2 gZ0ut Zin Qnet in WSW net in Steady Flow with Zero Shaft Power V V V2 V2 out uin Mgzoul Zin 390 out 390 in 2 Qnel in netin Heat transfer per unit mass I m 2 2 pout V out pin in v v gzout 2 gZin out um gnelin p p x J V loss Steady Incompressible Frictionless Flow out irt qnetin 0 loss Steady Incompressible Flow with rncuon V2 V2 p0 0 gzom 2 pm gzm loss 0 2 This is the energy loss How to compute power loss 553 A water siphon having a constant inside diameter of 3 in is arranged as shown in Fig P553 If the friction loss be tween A and B is 06V22 whens Vis the velocity of flow in he siphon determine the owrale involved 75 aleerrmha hg fawmc amp we use Q AV U1 V 4 I To onh V Wc app7 Hze may eim 39oh 15155 56M P0 14 and I M u AcDA a aHc wl w ff 0U 0 V1 0 0 i 9 1 V 77 le gf gyLyZ gr nefi Vz 2 3 7 5 5 7M 1 p2 32392 8 m a v 08 and MM EZJ a t I794 404543 0 m J 179 72 J Problem 554 554 Water ows through a valve see Fig P554 with a weight owrate r g of 10001bs The pressure just upstream of the valve is 90 psi and the pressure drop across the valve is 5 psi The inside diameters of the valve inlet and exit pipes are 12 and 24 in If the ow through the valve occurs in a horizontal plane determine the loss in available energy across the valve l i l comer VoltMG e SURE 95533 777a con v vaume own in 7 3ch Man i Um We Can Lac 3 5556 7 deermae 6 by I dim74 asapde wi 7 14 Mumarasi c 52447 fm rmly 7 voumc 7514 3 z 055 39Pz f V VI 0 2 55m 7 5 cam crvm 39an of mas pw39nchc v l M 0101 and 397 l quot Z 1 ICE L 77141 2 Ms PIquot L J a 2 0 129 04 Eu lz I VE 1 1 755 Eggp 7 I 7 ii I1 loss 5 JAEMg 7 f M MM 4 539 in 1 2m TEi n l My 2 H m 24quot MM Mawu ZZ quot 1 Aquot 1 M 5 1055 602 1W6 514 Problem 555 555 Water flows steadily from one location to another in the inclined pipe shown in Fig P555 At one section the static pressure is 8 psi At Lhe other secliun the static pressure is 5 psi Which way is the water owing Explain 10 100 R I FIGURE P555 70 defemhe 7716 d n an of Miatr aw ME a ppy it Enerjy egm bn 1555 In 7 m Sfo 39m0lfb and 29w Mm IL72mf5 u 7m ms aAzlaw mm 51555 1 mm mm W fwdMan buf nzyahk FM W haired low dimc 39m m aw 79w Secfem M16203 556 em4 7 00 Z 1 loss 51755 V V9 1quot 73A39 rim or gt 1 neflr39 1 as MK 239 Muz quw quot5 39 153 9 4quot 5M if d All 1 a quot IMI 77 M 514 Fay aw W kc bn J 064 556 kaa 74 loss as s and lay firZ 7L 333 2 syn04 3 WV I 75 M 1quot 93 79quot 5m 3 or IL 22 vhf 1 1 7M waay fpw U rm Sec 39m P 74 fawn4 Problem 557 557 What is the maximum possible power output of he hydroelectric turbine shown in Fig P557 Problem 557 5r OW 740m fe 39on0D IECADrJZ 55 Vivaa P4 7 V 2 lol 1ngl r sz f w hqf olt5 I j Z 7 quot 11d 1h mce 7 quotZ 47 IranI SIM 55quot m R Wm m if in M8 NHL 7 Wm 9 i z Y1 J05 he ow Z 7 rmXmum work a PawCV aw pul I dc l39evwl wile I0550


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