气相中的稳态分子扩散 (5).pdf

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1、Fundamental equation of fluid flowPrinciples of Chemical Engineering-CDUT1、Material balanceApplication of law of conservation of mass in chemical engineering.During the stable process：incoming material=discharged material2、Heat balanceApplication of law of conservation of energy in chemical engineer

2、ing.During the stable process：amount of incoming material=amount of discharged material3、BalanceState the direction and limits of the processOn the condition of T and P，the concentration of two phases ownsa definite relationship at phase equilibrium.4、Process rateIt reflects the speed and relates to

3、 the size of equipment in production process.Process impetus Process resistance Process impetusProcess rateProcess resistanceFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTResearch area：Taking the production process or production unit equipment as the research area,the mate

4、rial at its inlet and outlet is quantitatively calculated.Unsteady processinputoutputaccumulationsystemOutputInputaccumulationSteady-state processaccumulation0inputoutputsystemOutputInputFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTContinuity equation is the specific expr

5、ession of law of conservation of mass in fluid flow system.Law of conservation of mass：210ssdmwwdWhen it is steady-state flow,dm/d=0，and：12ssswwwswuA222111AuAuExtended to any section of the piping system：111222=constantswu Au AuAContinuity equation Steady-state flowContinuous fluidApplication condit

6、ionFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTContinuity equationFor incompressible fluid，=constant，so：1122=constantsVu Au AuAFor the incompressible fluid flowing with steady state incircular pipeline:222214=41dudu21221)(=dduuFundamental equation of fluid flowPrinciples

7、 of Chemical Engineering-CDUTTotal energy balance equationCalculation rangeCross section:1-1 and 2-2Calculation basis1 kg The input and output energies of 1 kg fluid in and out of the system are as follows：Internal energy：the sum of the internal energies of a substanceU1：internal energy that 1 kg fl

8、uid inputU2：internal energy that 1 kg fluid outputJ/kgChange of internal energy in system：U=U2-U1Fluid:Fundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTTotal energy balance equationPotential energy：the energy produced by the action of gravity on fluidgZ1：potential energy that

9、 1 kg fluid inputgZ2：potential energy that 1 kg fluid outputJ/kgChange of potential energy in system：gZ=gZ2-gZ1 Kinetic energy：the energy produced by a fluid moving at a given speed1/2u12：kinetic energy that 1 kg fluid input1/2u22：kinetic energy that 1 kg fluid outputJ/kg22221Change of internal ener

10、gy2 in syste2m2uuuFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTTotal energy balance equationPressure energy：the energy from a certain pressure of fluidis called the pressure energy.Physical meaning:it is a ability that compressed fluid particleexpand outward and work.p1v1

11、：static pressure energy that 1 kg fluid inputp2v2：static pressure energy that 1 kg fluid outputJ/kgChange of static pressure energy in system：pv=p2v2-p1v1Static pressure energy of 1 kg fluid=J/kgpVpvmFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTTotal energy balance equati

12、onThermal Qe：the energy that heat exchanger provides for 1 kgfluid,J/kg。0eQ heat exchanger provides energyheater0eQ heat exchanger removes energycoolerExternal work We：the energy gained by 1 kg fluid throughpump(or other conveying machinery)，J/kg。double-pipe heat exchangerFundamental equation of flu

13、id flowPrinciples of Chemical Engineering-CDUTTotal energy balance equationAccording to the law of conservation of energy：Energy inputEnergy outputEnergy input：21111 12eeuUgzp vQW2222222uUgzp vEnergy output:Total energy balance equation of flow system:2()2eeuUg zpvQW Fundamental equation of fluid fl

14、owPrinciples of Chemical Engineering-CDUTMechanical-Energy Balance Equation for flow systemThe conversion and loss of mechanical energyTotal energyMechanical energyInternal energyThermal energyKinetic energyPotential energyStatic pressure energy External workMechanical energy can converted with each

15、 other in the transported process of fluidBecause of the viscosity of fluid,partial mechanical energy loses and converts to internal energyFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTMechanical-Energy Balance Equation for flow systemThe conversion and loss of mechanical

16、energyThe conversion and lossof mechanical energyABupKinetic energy converts to static pressure energy BCu keeps constantpBecause of viscous friction,static pressure energy converts to inertial energy Fundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTAccording to the first law

17、 of thermodynamics：21 J/kgvevUQpdveefQQhThe fluid obtain thermal energyin flow process.The volume expansion work is caused by the heat of the fluidduring its flow.It is the thermal energyexchanged between fluid and environment.To overcome flow resistance,theenergy loses and converts into thermal ene

18、rgy,raise the temperatureof the fluid slightly.Mechanical-Energy Balance Equation for flow systemMechanical-Energy Balance Equation for flow systemFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUT21vevUQpdveefQQh21vefvUQhpdv2()2eeuUg zpvQW 212()-2vefvug zpvpdvWh 221121()=()vp

19、vppvd pvpdvvdp212-2pefpug zvdpWh Mechanical-Energy Balance Equation for flow systemFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTFor incompressible fluid:=constantv=constant2121()pppvdpv pp2-2efupg zWh 221211221122efppugzWugzhSuitable for incompressible fluidsEquation of m

20、echanical energy balance for flow systemMechanical-Energy Balance Equation for flow systemFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUT+21=+212222211121fehpgzuWpgzuMechanicalenergyand effective workof 1kg fluid inputpipelineMechanical energy andmechanical energy lossof1kg

21、fluidoutputpipeline Calculating on the base of per unit mass fluidunit：J/kgunit：mFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUT+21=+212222211121fehpgzuWpgzuunit：J/kgghgpzgugWgpzgufe+2=+22222211121gWHee=Set:ghHff=feHgpzguHgpzgu+2=+22222211121dynamic headpotential headstatic

22、 headhead of delivery(external head)loss of headunit：m Calculating on the base of per unit mass fluid Calculating on the base of per unit weight fluidFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUT+21=+212222211121fehpgzuWpgzu Calculating on the base of per unit mass fluidu

23、nit：J/kgghgpzgugWgpzgufe+2=+22222211121gWHee=令令ghHff=feHgpzguHgpzgu+2=+22222211121unit：m+21=+2122221121fehpgzuWpgzu Calculating on the base of per unit volume fluidunit：Pa Calculating on the base of per unit weight fluidFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTDiscuss

24、ion about Bernoulli Equation Compressible fluidFor the flow of compressible fluid,if the change of absolute pressure between the two sections is less than 20%of the original absolute pressure,for,it also can be calculated with Bernoulli equation.%20-121pppThe density of fluid should be expressed as

25、the average densitymbetween two fluid sections.2+=21mBernoulli equation still suitable for any moment in an unsteady flow system.Fundamental equation of fluid flowPrinciples of Chemical Engineering-CDUT Steady-state fluidIdeal fluid flows steady-state in pipeline without external work,as hf=0,We=022

26、22211121+21=+21pgzupgzu Calculating on the base of per unit mass fluidgpzgugpzgu2222211121+2=+2 Calculating on the base of per unit weight fluid Effective power The effective work that conveying equipment do on fluid per unit time is called the effective power.eesNW weseNVWJ/s or WImportant basis fo

27、r selecting fluidtransport equipment.Discussion about Bernoulli EquationFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUT Static fluidIf the fluid is static in system，so u=0,hf=0，We=0，222111+=+pgzpgzFluid statics basic equationto ensure the flowrate of fluid(u)to sure the rel

28、ative position between equipment(Z)to sure the effective power of conveying equipment(We)to sure the pressure of fluid in pipeline(p)Discussion about Bernoulli EquationApplication of Bernoulli EquationFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUT1.Points of solving proble

29、m with Mechanical-Energy Balance Equation To ensure the drawing and calculating rangeDraw Schematic diagram of flow system and indicate the flow direction；Determine the upper and lower cross sections and ensure calculating range.upper sectionlower sectionwater pumpelevated tanklower tankFundamental

30、equation of fluid flowPrinciples of Chemical Engineering-CDUT Selection for sectionsThe section should be perpendicular to the flow direction,and the fluid must be continuous between the two sections；The unknown quantity should be on or between two sections；The physical quantity on sections,except f

31、or those that need to be solved,should be known or can be calculated.1.Points of solving problem with Mechanical-Energy Balance Equationwater pumpelevated tanklower tankFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUT To sure datum planeThe datum plane can be chosen arbitrar

32、ily,but must be parallel to the ground；The value of Z refers to the vertical distance between the center of the section and the datum plane；Take the datum plane to make it through the lower relative position of the two cross sectionselevated tanklower tankwater pump1.Points of solving problem with M

33、echanical-Energy Balance Equationdatum planeFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUT Consistency of unit Each physical quantity in equation should adopt a consistent unit.Expression for pressure The expression for pressure on two sections should be consistent.p=p2-p1

34、absolute pressuregauge pressure absolute pressure=atmospheric pressure+gauge pressure 1.Points of solving problem with Mechanical-Energy Balance EquationFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTExample 1-12,P31 Calculate the effective power of the conveying machineThe

35、 water(at normal temperature)in liquid storage tank is conveyed to the top of absorption tower,but keeping the volume of liquid of storage tank constant at same time.The relative position of each part is shown as figure.The diameter of water pipe is ,the pressure on the nozzle joint of the outlet of

36、 drain pipe is (gauge pressure),the water volume per hour of conveying is 34.5m3/h.When water flow through all pipes(except nozzle),if the loss of energy is 160 J/kg,please work out the efficient energy of pump.76mm 3mm46.15 10 PaAnswer:Set the liquid surface of storage tank as the topper section 1-

37、1 which is also chose for datum plane.Set the nozzle joint of the outlet of drain pipe as 2-2.Apply the Mechanical-Energy Balance Equation at two sections:22112212gg22efupupzWzh22121212-g-2efu up pWz zhor(1)Fundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTExample 1-12,P31 Cal

38、culate the effective power of the conveying machineThere:z1=0,z2=26,p1=0(gauge pressure),p1=(gauge pressure),.Because the section area of storage tank is much larger than the area of pipe section,thus .46.15 10 Pa160/fhJKg10u,2234.5m/s=2.49m/s3600/40.07V squA Use the values mentioned above in eq(1),

39、and the density of water=1000 Kg/m3,thus:242.496.15 1026 9.81160/479.7/21000eWJ kgJ kgEfficient energy of pump Ne:34.510009.583/3600ssvkg s 479.7 9.5834596.97eesNWW Fundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTExample1-13,P32 Calculate the pressure of pipe sectionAs shown

40、 in figure,water flows with steady state in siphon whose diameter is uniform.On the condition of ignoring the energy lose in pipeline,please try to solve two problems:(1)the flow velocity of water in pipe;(2)the pressure at section 2-2,3-3,4-4,5-5.The known condition is that the atmospheric pressure

41、 is ,and the units in figure are mm.51.0133 10 PaAnswer:(1)Apply the Mechanical-Energy Balance Equation between the section 1-1 and the internal section of pipe outlet 6-6,and set section 6-6 as datum plane.Because ,thus:0fh22661112gg22upupzzthere,z1=2 m,z6=0,p1=0(gauge pressure),u1=0.(1)So269.8 11=

42、2uu6=4.43m/sBecause the diameter of pipe is constant,so the flow velocity of water at any section is 4.43m/s.Fundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTExample1-13,P32 Calculate the pressure of pipe section(2)Because there is no pump and the energy lose also can be igno

43、red,hence,the total mechanical energy at any section is equal.Take the section 1-1 as an example(setting the section 2-2 as datum plane).2111101330=g=(9.81 3+)J/kg=130.8J/kg21000upEz Fundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTExample1-15 P33 To ensure the relative posit

44、ion between equipmentIt is a elevated tank system for conveying water in figure and the volume of water in water tank keeps constant.The diameter of water pipe is ,the volume of conveying water per hour is 15.4m3/h.When water flows through all of pipes(except pipe outlet),the loss of energy ,where u

45、 represents the average velocity of water in pipe.Please try to solve two problems:(1)the height(h)of water surface in water tank above the outlet of drain pipe;(2)if the volume of conveying water per hour increases 5%,the diameter of pipe and other layout keep invariantly,and the energy loss of pip

46、e can be calculated as the above equation,how much should the water in the tank rise?60mm 3mm215fhuAnswer:(1)Set the water surface in water tank as upper section 1-1.Set the internal section of drain pipe outlet as lower section 2-2 whose center line is datum plane.Apply the Mechanical-Energy Balanc

47、e Equation between two sections:22112212gg22efupupzWzh(1)Fundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTExample1-15 P33 To ensure the relative position between equipmentthere,z1=h,z2=0,p1=p2=0(gauge pressure).Because the section area of storage tank is much larger than the

48、area of pipe section,thus .10u,2215.4m/s=1.87m/s3600/40.054V squA 221515 1.87/52.45/fhuJ kgJ kgUse the values mentioned above in eq(1),21.87/252.45/9.81 m5.52mh(2)If the volume of conveying water per hour increases 5%,the diameter of keeps invariantly,thus the velocity of water in pipe also increase

49、s 5%.Hence,the velocity of water u2 after increasing the volume of conveying water per hour can be calculated with following equations.221.051.05 1.87 m/s=1.96m/suuFundamental equation of fluid flowPrinciples of Chemical Engineering-CDUTExample1-15 P33 To ensure the relative position between equipment2221515 1.96/57.62/fhuJ kgJ kg2=1.96/257.62/9.81 m=6.07mhSo,when the volume of conveying water per hour increases 5%,the rise of water in the tank is calculated:6.07m-5.52m=0.55mFundamental equation of fluid flow