流體力學(xué)chapter-5-control-volume-analysis

1、Chapter 5 Flow Analysis Using Control Volume,,2,MAIN TOPICS,5.1 Conservation of Mass5.2 Newton’s Second Law – Linear Momentum Equations 5.3 First Law of Thermodynamics – Energy Equation5.4 Second Law

2、of Thermodynamics – Irreversible Flow,3,5.1 Conservation of Mass–Continuity Equation_1,Basic Law for Conservation of MassFor the system and a fixed, nondeforming control volume that are coincident at an instant of ti

3、me, the Reynolds Transport Theorem leads to,B=M and b =1,Time rate of change of the mass of the coincident system,Time rate of change of the mass of the content of the coincident control volume,Net rate of flow of mass t

4、hrough the control surface,=,+,Intensive and Extensive,Intensive Property: Fluid property does not depend on the amount of material present in the system, e.g., V, a, P, T, etc.Extensive Property: Fluid property depend

5、s on the amount of material present in the system, e.g., mass, momentum, energy, etc.,5,Conservation of Mass–Continuity Equation_2,System and control volume at three different instances of time. (a) System and control v

6、olume at time t – δt. (b) System and control volume at time t, coincident condition. (c) System and control volume at time t + δt.,6,Conservation of Mass–Continuity Equation_3,For a fixed, nondeforming control volume, t

7、he control volume formulation of the conservation of mass: The continuity equation,,,Rate of increaseOf mass in CV,Net influx of mass,,7,Conservation of Mass–Continuity Equation_4,Incompressible FluidsFor Steady fl

8、ow,The mass flow rate into a control volume must be equal to the mass flow rate out of the control volume.,,8,Other Definition,Mass flowrate through a section of control surfaceThe average velocity,9,Fixed, Nondeformi

9、ng Control Volume_1,When the flow is steadyWhen the flow is steady and incompressible When the flow is not steady,,“+” : the mass of the contents of the control volume is increasing“-” : the mass of the content

10、s of the control volume is decreasing.,,10,Fixed, Nondeforming Control Volume_2,When the flow is uniformly distributed over the opening in the control surface (one dimensional flow)When the flow is nonuniformly dis

11、tributed over the opening in the control surface,11,Ex 5.2 Steady, Compressible Flow,Air flows steadily between two sections in a long, straight portion of 10cm. inside diameter as indicated in Figure E5.2. The uniforml

12、y distributed temperature and pressure at each section are given. If the average air velocity (Nonuniform velocity distribution) at section (2) is 305m/s, calculate the average air velocity at section (1).,12,Example 5.

13、2 Solution,,Flow is Steady,The continuity equation,Since A1=A2,,,The ideal gas equation,,13,Example 5.4,Incompressible, laminar water flow develops in a straight pipe having radius R as indicated in Figure E5.4. At secti

14、on (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

15、the pipe wall and a maximum value of umax at the centerline. How are U and umax related? How are the average velocity at section (2), , and umax related?,14,Example 5.4 Solution,Flow is Steady,,With incompressible co

16、ndition,The continuity equation,,,15,Example 5.5,A bathtub is being filled with water from a faucet. The rate of flow from the faucet is steady at 34L/min. The tub volume is approximated by a rectangular space as indicat

17、e Figure E5.5(a). Estimate the time rate of change of the depth of water in the tub, , in in./min at any instant.,16,Example 5.5 Solution1/2,The continuity equation,17,Example 5.5 Solution2/2,,18,Moving, Nond

18、eforming Control Volume,When a moving control volume is used, the fluid velocity relative to the moving control is an important variable.W is the relative fluid velocity seen by an observer moving with the control volum

19、e. Vcv is the control volume velocity as seen from a fixed coordinate system. V is the absolute fluid velocity seen by a stationary observer in a fixed coordinate system.,,,,19,5.2 The Linear Momentum Equations_1,Newt

20、on’s second law for a system moving relative to an inertial coordinate system.,Time rate of change of the linear momentum of the system,=,Sum of external forces acting on the system,20,The Linear Momentum Equations_2,Fo

21、r the system and a fixed, nondeforming control volume that are coincident at an instant of time, the Reynolds Transport Theorem leads to,,Time rate of change of the linear momentum of the coincident system,Time rate of c

22、hange of the linear momentum of the content of the coincident control volume,Net rate of flow of linear momentum through the control surface,=,+,,,,,21,The Linear Momentum Equations_4,For a fixed and nondeforming contro

23、l volume, the control volume formulation of Newton’s second law,Contents of the coincidentcontrol volume,,Linear momentum equation,22,Example 5.10,As shown in Figure E5.10 (a), a horizontal jet of water exits a nozzle

24、with a uniform speed of V1=3m/s, strike a vane, and is turned through an angleθ. Determine the anchoring force needed to hold the vane stationary. Neglect gravity and viscous effects.,Determine the anchoring force needed

25、 to hold the vane stationary.,23,Example 5.10 Solution,The x and z direction components of linear momentum equation,24,Example 5.11,Determine the anchoring force required to hold in place a conical nozzle attached to the

26、 end of a laboratory sin faucet when the water flowrate is 0.6 liter/s. The nozzle mass is 0.1kg. The nozzle inlet and exit diameters are 16mm and 5mm, respectively. The nozzle axis is vertical and the axial distance bet

27、ween section (1) and (2) is 30mm. The pressure at section (1) is 464 kPa. to hold the vane stationary. Neglect gravity and viscous effects.,25,Example 5.11 Solution1/3,26,Example 5.11 Solution2/3,27,Example 5.11 Solution

28、3/3,,28,Example 5.12,Water flows through a horizontal, 180° pipe bend. The flow cross-section area is constant at a value of 0.01m2 through the bend. The magnitude of the flow velocity everywhere in the bend is axia

29、l and 15m/s. The absolute pressure at the entrance and exit of the bend are 207kPa and 165kPa, respectively. Calculate the horizontal (x and y) components of the anchoring force required to hold the bend in place.,29,Exa

30、mple 5.12 Solution1/2,The x direction component of linear moment equation,,,At section (1) and (2), the flow is in the y direction and therefore u=0 at both sections.,The y direction component of linear moment equation,3

31、0,Example 5.12 Solution2/2,For one-dimensional flow,,,31,Example 5.14,If the flow of Example 5.4 is vertically upward, develop an expression for the fluid pressure drop that occurs between sections (1) and (2).,32,Exampl

32、e 5.14 Solution,The axial component of linear moment equation,,,33,Example 5.15 - Trust,A static thrust as sketched in Figure E5.15 is to be designed for testing a jet engine. The following conditions are known for a typ

33、ical test: Intake air velocity = 200 m/s; exhaust gas velocity= = 500 m/s; intake cross-section area = 1m2; intake static pressure = -22.5 kPa=78.5 kPa (abs); intake static temperature = 268K; exhaust static pressure =0

34、kPa=101 kPa (abs). Estimate the normal trust for which to design.,34,Example 5.15 Solution,The x direction component of linear moment equation,,,35,Example 5.16,A sluice gate across a channel of width b is shown in the c

35、losed and open position in Figure E5.16(a) and (b). Is the anchoring force required to hold the gate in place larger when the gate is closed or when it is open?,36,Example 5.16 Solution,When the gate is open, the horizon

36、tal forces acting on the contents of the control volume are identified in Figure E5.16 (d).,When the gate is closed, the horizontal forces acting on the contents of the control volume are identified in Figure E5.16 (c).,

37、,37,Moving CV,Contents of the coincidentcontrol volume,,,Contents of the coincidentcontrol volume,,Chapter 4: Reynolds transport equation for a control volume moving with constant velocity is,,,,,38,..,Example 5.17,,A

38、vane on wheels move with a constant velocity V0 when a stream of water having a nozzle exit velocity of V1 is turned 45° by the vane as indicated in Figure E5.17(a). Note that this is the same moving vane considered

39、 in Section 4.4.6 earlier . Determine the magnitude and direction of the force, F, exerted by the stream of water on the vane surface. The speed of the water jet leaving the nozzle is 30m/s, and the vane is moving to the

40、 right with a constant speed of 6m/s.,39,..,Example 5.17,,40,..,Example 5.17 Solution1/2,The x direction component of linear moment equation,,The z direction component of linear moment equation,,41,..,Example 5.17 Solut

41、ion2/2,,42,5.3 First Law of Thermodynamics – The Energy Equation_1,The first law of thermodynamics for a system is,Time rate of increase of the total stored energy of the system,Net time rate of energy addition by heat

42、transfer into the system,Net time rate of energy addition by work transfer into the system,=,+,Total stored energy per unit mass for each particle in the system,,“+” going into system“-” coming out,,The net rate of hea

43、t transfer into the system,,The net rate of work transfer into the system,43,First Law of Thermodynamics – The Energy Equation_2,For the system and the contents of the coincident control volume that is fixed and nondefo

44、rming -- Reynolds Transport Theorem leads to,Time rate of increase of the total stored energy of the system,Net time rate of increase of the total stored energy of the contents of the control volume,The net rate of flow

45、of the total stored energy out of the control volume through the control surface,=,+,44,First Law of Thermodynamics – The Energy Equation_3,For the control volume that is coincident with the system at an instant of time

46、.The control volume formula for the first law of thermodynamics:,45,Rate of Work done by CV,Shaft work : the rate of work transferred into through the CS by the shaft work ( negative for work transferred

47、out, positive for work input required) Work done by normal stresses at the CS:Work done by shear stresses at the CS:Other work ?,,,46,First Law of Thermodynamics – The Energy Equation_4,Energy equation,,47,Applicat

48、ion ofEnergy Equation_1,When the flow is steady,The integral of,???,,Uniformly distribution,Only one stream entering and leaving,,48,Application ofEnergy Equation_2,If shaft work is involved….,One-dimensional energy e

49、quation for steady-in-the-mean flow,Enthalpy,,The energy equation is written in terms of enthalpy.,49,Energy Equation vs. Bernoulli Equation_1,For steady, incompressible flow…One-dimensional energy equation,,where,For s

50、teady, incompressible, frictionless flow…,Bernoulli equation,,,Frictionless flow…,50,Energy Equation vs. Bernoulli Equation_2,For steady, incompressible, frictional flow…,,Defining “useful or available energy”…,Defining

51、 “l(fā)oss of useful or available energy”…,Frictional flow…,51,Energy Equation vs. Bernoulli Equation_3,For steady, incompressible flow with friction and shaft work…,,,,Head loss,52,Application of Energy Equation to Nonuni

52、form Flows_1,If the velocity profile at any section where flow crosses the control surface is not uniform…,For one stream of fluid entering and leaving the control volume….,Where ? is the kinetic energy coefficient and V