流体力学与传热课件1.5 Pipe Flow Systems

1、流体力学与传热课件流体力学与传热课件1.5 Pipe Flow 1.5 Pipe Flow SystemsSystems We will concern two classes of pipe systems: those containing a single pipe and those containing multiple pipes in parallel, series.Pipe system Some of the basic components of a typical pipe system shown in Fig. They include : the pipes th

2、emselves, the various fittings used to connect the individual pipes to form the desired system, metering devices, and the pumps or turbines that add energy to or remove energy from the fluid.1.5.1 Single Pipes For flowing through a pipe three most common types of problems are shown in table in terms

3、 of the parameters involved.Variable type I type II type III Fluid Density given given given Viscosity given given given b. PipeDiameter given given determine Length given given given Roughnessc. FlowFlowrate given determine given d. PressurePressure drop determine given given Example 1Water at 15.6

4、C flows from the basement to the second floor through the 1.91cm diameter copper pipe ( = 0.001524mm ) at a rate of Q=0.756 l/s and exits through a faucet of diameter 1.27cm. As shown in Fig.determine the pressure at point(1) if (a) all losses are neglected, (b) the only losses included are skin fri

5、ction losses, (c) all losses are includedSolution:Since the velocity in the pipe is given by smAQV/64. 20191. 0410756. 02311And the fluid properties are =999.9kg/m3 and =1.1210-3Pas450001012. 10191. 064. 29 .999Re3Thus ,the flow is turbulent flowGoverning equation for either case (a), (b), or (c) is

6、 the mechanical balance equation given byfhgzVpgzVp2222121122Where z1=0, z2=6.1m, p2=0, and outlet velocity is smAQV/97. 50127. 0410756. 02322thusfhVVgzp2212221Where the friction loss is different for each of three cases1(a)If all losses are neglected, Eq.1 givesPap74133264. 297. 59 .99981. 99 .9991

7、 . 6221Note that for this pressure difference, the amount due to elevation change and amount due to the increase in kinetic energy(b)If only losses included are the skin friction losses , the friction loss is 221VDLhfRelative roughness /D=810-5With this/D and Reynolds number (Re=45000), the value of

8、 friction coefficient is obtained from Moody chart as =0.0215. the potential and kinetic energy are the same as part (a). Eq.1 gives PaVDLVVgzp145872264. 20191. 029.180215. 09 .9997413322221212221Of this pressure drop, the amount due to skin friction is 71739 Pa( c) If skin friction losses and minor

9、 losses are included, Eq.1 becomes214587221VKPap222221212221VKVDLVVgzporWhere the 145872Pa contribution is due to elevation change, kinetic energy change, and skin losses, and last term represents the sum of all the minor losses2The loss coefficients of the components (Kf=1.5 for each elbow and 10 f

10、or the wide open globe valve) are given. The loss coefficient of the faucet is 2. thus PaVK62720264. 225 . 14109 .9992223 Note that we did not include an entrance or exit loss because points (1) and (2) are located within the fluid streams, not within an attaching reservoir where the kinetic energy

11、is zero. Thus, by comparing Eqs.2 and 3 we obtain the entire pressure drop as Pap208590627201458721Example 2 Turbine shown in figure extracts 50hp from water flowing through it. The 0.3m diameter, 91.4m-long pipe is assumed to have a friction factor of 0.02. Form losses are negligible. Determine the

12、 flowrate through the pipe and turbine.Solution :The energy equation can be applied between the surface of the lake (point 1) and the outlet of the pipe as sfWHzgVgpzgVgp2222121122Where p1=p2=0,V1=0, z1=27.4m, V2=V1The head loss is given by222311. 081. 923 . 04 .9102. 02VVgVdlHfThe turbine head isVV

13、VdgNWs85.533 . 0414. 381. 910007 .74550422Thus, Eq.1 can be written as VVV85.53311. 081. 924 .27222The velocity of water flowing through pipe can be found as the solution of Eq. 2. Surprisingly, there are two real positive roots: V=2m/s,V=7.49m/s1.5.2 Multiple Pipe Systems In many pipe systems there

14、 is more than one pipe involved. The maze of pipes in a citys water distribution system are typical of such systems. The governing mechanisms for the flow in multiple pipe systems are the same as for the single pipe systems. However, because of numerous unknowns involved, additional complexities may

15、 arise in solving for the flow in multiple systems.Parallel pipe system One of multiple pipe systems is the pipes in parallel, as is shown in figure. 123ABIn this system a fluid particle traveling from A to B may be taken any of the paths available, with the total flowrate equal to the sum of the fl

16、owrates in each pipe.By writing the mechanical energy balance equation it is found that the energy loss experienced by any fluid particle traveling between these locations is the same, independent of the path taken. fABBBBAAAhzVpzVp2222andWhere 22321VKdlhhhhfABfff Thus, the governing equation for pa

17、rallel pipes are321QQQQAThe solution of problem depends on what information is given and what is to be calculatedExample Find the distribution of flow for three parallel pipes arrangement shown in figure. The water kinematic viscosity =10-6m2/s. the total discharge is Q=0.020m3/s.Pipe L/m D/m K 100

18、0.05 0.023 10 150 0.075 0.025 3 200 0.085 0.021 2 123ABSolution For parallel pipes, the continuity equation can be written as321QQQQAnd the governing equation of parallel pipes is 22321VKdlhhhfff12Based on known parameters, Eqs. 2 can be expressed as222321100. 1091. 2262. 3QQQ3Combining Eqs.1 and 3

19、givessmQsmQsmQ/00867. 0;/00629. 0;/00504. 0333231Branching system The flow in a relatively simple looking multiple pipe system may be more complex than it appears initially. The branching system termed the three-reservoir problem shown in figure is such a system. Three reservoir at known elevations

20、are connected together with three pipes of known properties (lengths, diameters, and roughnesses).The problem is to determine the flowrates into or out of the reservoirs. If valve(1) were closed, the fluid would flow from reservoir B to C, and flowrate could be easily calculated. Similar calculation

21、s could be carried out if valve (2) or (3) were closed with the other openWith all valves open, however, it is not necessarily obvious which direction the fluid flowsFor the conditions indicated in figure, it is clear that fluid flows from reservoirs A because the other two reservoirs are lower. Whe

22、ther the fluid flows into or out of the reservoirs B depend on the elevation of reservoirs B and C and properties (length, diameter, roughness) of the three pipes. In general, the flow direction is not obvious, and solution process must include the determination of this direction.exampleThree reserv

23、oirs are connected by three pipes as are shown in figure. For simplicity we assume that the diameter of each pipe is 0.305m, the friction coefficient for each is 0.02, and because of the length-to-diameter ratio, form losses are negligible. Determine the flowrate into or out of each reservoir.Soluti

24、onIt is not obvious which direction the fluid flows in pipe(2). However, we assume that it flows out of the reservoir B, write the governing equations for this case, and check the assumption. The continuity equation requires that Q1+Q2=Q3, which, since the diameters are the same for each pipe, becom

25、es simply321VVV1The energy equation for the fluid flowing from A to C in pipes (1) and (3) can be written as gVDlgVDlzgVgpzgVgpcccAAA2222233332111122By using the fact that pA=pc=0;VA=Vc=0; zc=0, this becomesgVDlgVDlzA222333321111For the given conditions of this problem we obtain23219 .1218 .304305.

26、0181. 9202. 05 .30VV or23214 . 05 .30VV 2Similarly the energy equation for fluid flowing from B and C is gVDlgVDlzgVgpzgVgpcccBBB2222233332222222For the given conditions this can be written as 23224 . 05 . 099. 5VV 3Eqs.1,2 and 3 are the governing equations for this flow, provided the fluid flows fr

27、om reservoir B. it turns out, however, that there is no solution for these equations with positive, real values of the velocities. Although these equations do not appear to be complicated, there is on simple way to solve them directly. Thus, a trial-error solution is suggested.Assuming a value of V1

28、0, calculate V3 from equation 2, and then V2 from Eq.3. it is found that the resulting V1, V2, and V3 trio does not satisfy Eq. 1 for any value of V1 assumed. There is no solution to Eqs. 1,2 and 3 with real, positive values of V1, V2,and V3. thus our original assumption of flow out of reservoir B m

29、ust be incorrect. To obtain the solution, assume the fluid flows into reservoir B and C and out of A. for this case the continuity equation becomesQ1=Q2+Q3Or 321VVV4Application of energy equation between points A and B and A and C givesgVDlgVDlzzBA222222221111andgVDlgVDlzzCA222333321111Which, with t

30、he given data, becomes22215 . 095.23VV and23214 . 094.29VV 56Equations 4,5 and 6 can be solved as follows. By subtracting Eq.5 from 6 we obtain 22325. 198.14VVThus, Eq. 5 can be written as 222222222325 . 025. 198.145 . 095.23VVVVVVor2222275. 297. 825. 198.142VVVWhich, upon squaring both sides, can b

31、e written as040.3164.422242VVBy using the quadratic formula we can solve for to obtain 22V75. 088.412222VorVThus, either V2=6.47m/s or V2=0.867m/s. the value V2=6.47m/s is not a root of the original equations. Thus, V2=0.867m/s and from equation 5, V1=4.85m/s. The corresponding flowrates aresmVDQ/354. 085. 4305. 044321211smVDQ/063. 0867. 0305. 044322222from Ainto BsmQQQ/291. 0063. 0354. 03213Into CIf the friction coe