流体力学英文版课后习题问题详解.doc

《流体力学英文版课后习题问题详解.doc》由会员分享，可在线阅读，更多相关《流体力学英文版课后习题问题详解.doc（20页珍藏版）》请在课桌文档上搜索。

1、wordWhat will be the (a) the gauge pressure and (b) the absolute pressure of water at depth 12m below the surface? water = 1000 kg/m3, and Patmosphere = 101kN/m2.Solution:RSet the pressure of atmosphere to be zero, then the gauge pressure at depth 12m below the surface is Absolute pressure of water

2、at depth 12m 1.3 A differential manometer as shown in Fig. is sometimes used to measure small pressure difference. When the reading is zero, the levels in two reservoirs are equal. Assume that fluid B is methane甲烷, that liquid C in the reservoirs is kerosene (specific gravity = 0.815), and that liqu

3、id A in the U tube is water. The inside diameters of the reservoirs and U tube are 51mm and , respectively. If the reading of the manometer is145mm., what is the pressure difference over the instrumentIn meters of water, (a) when the change in the level in the reservoirs is neglected, (b) when the c

4、hange in the levels in the reservoirs is taken into account? What is the percent error in the answer to the part (a)?Solution：pa=1000kg/m3 pc=815kg/m3pb=/m3 D/d=8 R=When the pressure difference between two reservoirs is increased, the volumetric changes in the reservoirs and U tubes (1)so (2)and hyd

5、rostatic equilibrium gives following relationship (3)so (4)substituting the equation (2) for x into equation (4) gives (5)awhen the change in the level in the reservoirs is neglected, bwhen the change in the levels in the reservoirs is taken into accounterror=1.4 There are two U-tube manometers fixe

6、d on the fluid bed reactor, as shown in the figure. The readings of two U-tube manometersareR1=400mm，R2=50mm, respectively. The indicating liquid is mercury. The top of the manometer is filled with the water to prevent from the mercury vapor diffusing into the air, and the height R3=50mm. Try to cal

7、culate the pressure at point A and B.Solution: There is a gaseous mixture in the U-tube manometer meter. The densities of fluids are denoted by , respectively.The pressure at point A is given by hydrostatic equilibrium is small and negligible in parison withand H2O, equation above can be simplified=

8、7161N/m=7161+136009.810.4=60527N/mDdpapaHhA1.5 Water discharges from the reservoir through the drainpipe, which thethroat diameter is d. The ratio of D to d equals 1.25. The verticaldistance h between the tank A and axis of the drainpipe is 2m. What height H from the centerline of the drainpipe to t

9、he water level in reservoir is required for drawing the water from the tank A to the throat of the pipe? Assume that fluid flow is a potential flow.The reservoir, tank A and the exit of drainpipe are all open to air.Solution:Bernoulli equation is written between stations 1-1 and 2-2, with station 2-

10、2 being reference plane:Where p1=0, p2=0, and u1=0, simplification of the equation 1The relationship between the velocity at outlet and velocity uo at throat can be derived by the continuity equation: 2Bernoulli equation is written between the throat and the station 2-2 3bining equation 1,2,and 3 gi

11、vesSolving for H1.6 A liquid with a constant density kg/m3 is flowing at an unknown velocity V1 m/s through a horizontal pipe of cross-sectional area A1 m2 at a pressure p1 N/m2, and then it passes to a section of the pipe in which the area is reduced gradually to A2 m2 and the pressure is p2. Assum

12、ing no friction losses, calculate the velocities V1 and V2 if the pressure difference (p1 - p2) is measured.Solution: In Fig, the flow diagram is shown with pressure taps to measure p1 and p2. From the mass-balance continuity equation , for constant where 1 = 2 = ,For the items in the Bernoulli equa

13、tion , for a horizontal pipe,z1=z2=0Then Bernoulli equation bees, after substituting for V2,Rearranging,Performing the same derivation but in terms of V2,1.7 A liquid whose coefficient of viscosity is flows below the critical velocity for laminar flow in a circular pipe of diameter d and with mean v

14、elocity V. Show that the pressure loss in a length of pipe is .Oil of viscosity 0.05 Pas flows through a pipe of diameter 0.1m with a average Solution:The average velocity V for a cross section is found by summing up all the velocities over the cross section and dividing by the cross-sectional area

15、1From velocity profile equation for laminar flow 2substituting equation 2 for u into equation 1 and integrating 3rearranging equation 3 gives1.8. In a vertical pipe carrying water, pressure gauges are inserted at points A and B where the pipe diameters are 0.15m and 0.075m respectively. The point B

16、is 2.5m below A and when the flow rate down the pipe is 0.02 m3/s, the pressure at B is 14715 N/m2 greater than that at A. Assuming the losses in the pipe between A and B can be expressed as where V is the velocity at A, find the value of k.If the gauges at A and B are replaced by tubes filled with

17、water and connected to a U-tube containing mercury of relative density 13.6, give a sketch showing how the levels in the two limbs of the U-tube differ and calculate the value of this difference in metres. Solution:dA=0.15m; dBzA-zB=lQm3/s,pB-pA=14715 N/m2When the fluid flows down, writing mechanica

18、l balance equationmaking the static equilibrium1.9The liquid vertically flows down through the tube from the station a to the station b, then horizontally through the tube from the station c to the station d, as shown in figure. Two segments of the tube, both abandcd，have the same length, the diamet

19、er and roughness.Find:1the expressions of , hfab, and hfcd, respectively.2the relationship between readings R1and R2 in the U tube.Solution:(1) From Fanning equationandsoFluid flows from station a to station b, mechanicalenergy conservation giveshence 2from station c to station dhence 3From static e

20、quationpa-pb=R1-g -lg 4pc-pd=R2-g 5Substituting equation 4 in equation 2 ，thentherefore 6Substituting equation 5 in equation 3 ，then 7ThusR1=R21.10 Waterpasses through a pipe of diameter d4 m with the average velocity m/s, as shown in Figure. 1) What is the pressure drop DP when water flows through

21、the pipe length L=2 m, in m H2O column?Lr2) Find the maximum velocity and point r at which it occurs. 3) Find the point r at whichthe average velocity equals the local velocity.4if kerosene flows through this pipe，how do the variables above change？the viscosity and density of Water are 0.001 Pas and

22、1000 kg/m3，respectively；and the viscosity and density of keroseneare 3 Pas and800 kg/m3，respectivelysolution:1from Hagen-Poiseuille equation2maximum velocity occurs at the center of pipe, from equation 9so umax374) kerosene:1.12 As shown in the figure, the water level in the reservoir keeps constant

23、. A steel drainpipe(with the inside diameter of 100mm) is connected to the bottom of the reservoir.One arm of theU-tubemanometeris connectedto the drainpipe at the position 15m away from the bottom of the reservoir, and the other is opened to the air, the U tube is filled with mercury and the left-s

24、ide arm of the U tube above the mercury is filled with water. The distance between the upstream tap and the outlet of the pipeline is 20m.a) When the gate valve is closed, R=600mm, h=1500mm; when the gate valve is opened partly, R=400mm, h=1400mm.The friction coefficient is 0.025, and the loss coeff

25、icient of the entranceis 0.5. Calculate the flow rate of water when the gate valve is opened partly. (in m/h)b) When the gate valve is widely open, calculate the static pressure at the tap (in gauge pressure, N/m). le/d15 when the gate valve is widely open, and the friction coefficient is still 0.02

26、5.Solution：(1) When the gate valve is opened partially, the water discharge isSet up Bernoulli equation between the surface of reservoir 11and the section of pressure point 22，and take the center of section 22as the referring plane, then aIn the equation (the gauge pressure) When the gate valve is f

27、ully closed, the height of water level in the reservoir can berelated to h (the distance between the center of pipe and the meniscus of left arm of U tube).b where h=R=Substitute the known variables into equation bSubstitute the known variables equation a9.816.66=the velocity is V the flow rate of w

28、ater is 2 the pressure of the point where pressure is measured when the gate valve is wide-open. Write mechanical energy balance equation between the stations 11and 3-3，then csince input the above data into equation c，the velocity is: V=3.51 m/sWrite mechanical energy balance equation between thesta

29、tions 11and 22, for the same situation of water level dsince input the above data into equation d，9.816.66=the pressure is: 1.14 Water at 20 passes through a steel pipe with an inside diameter of 300mm and 2m long. There is a attached-pipe(603.5mm) which is parallel with the main pipe. The total len

30、gth including the equivalent length of all form lossesof the attached-pipe is 10m. A rotameter is installed in the branch pipe. When the reading of 3/h, try to calculate the flow rate in the main pipe and the total flow rate, respectively. The frictional coefficient of the main pipe and the attached

31、-pipe is 0.018 and 0.03, respectively.Solution：The variables of main pipe are denoted by a subscript1, and branch pipe bysubscript 2. The friction loss for parallel pipelines isThe energy loss in the branch pipe isIn the equation input the data into equation cThe energy loss in the main pipe isSo Th

32、e water discharge of main pipe isTotal water discharge is1.16 A Venturimeter is used for measuring flow of water along a pipe. The diameter of the Venturi throat is two fifths the diameter of the pipe. The inlet and throat are connected by water filled tubes to a mercury U-tube manometer. The veloci

33、ty of flow along the pipe is found to be m/s, where R is the manometer reading in metres of mercury. Determine the loss of head between inlet and throat of the Venturi when R is 0.49m. (Relative density of mercury is 13.6).Solution:Writing mechanical energy balance equation between the inlet 1 and t

34、hroat o for Venturi meter1rearranging the equation above, and set (z2-z1)=x2from continuity equation 3substituting equation 3 for Vo into equation 2 gives 4from the hydrostatic equilibrium for manometer 5substituting equation 5 for pressure difference into equation 4 obtains 6rearranging equation 6

35、1.17.Sulphuric acid of specific gravity 1.3 is flowing througha pipe of 50 mm internal diameter. A thin-lipped orifice, 10mm, is fitted in the pipe and the differential pressure shown by a mercury manometer is 10cm. Assuming that the leads to the manometer are filled with the acid, calculate (a)the

36、weight of acid flowing per second, and (b) the approximate friction loss in pressure caused by the orifice.The coefficient of the orifice may be taken as 0.61, the specific gravity of mercury as 13.6, and the densityof water as 1000 kg/m3Solution:a)b) approximate pressure dropPapressure difference d

37、ue to increase of velocity in passing through the orifice pressure drop caused by friction loss Water is used to test for the performances of pump. The gauge pressure at the discharge connection is 152kPaand the reading of vacuum gauge at the suction connection of the pump is 24.7 kPa as the flow ra

38、te is 26m3/h.The while the centrifugal pump operates at the speed of 2900r/min. If the vertical distance between the suction connection and discharge connection is 0.4m, the diameters of both the suction and discharge line are the same. Calculate the mechanical efficiency of pump and list the perfor

39、mance of the pump under this operating condition.Solution:Write the mechanical energy balance equation between the suction connection and discharge connectionwheretotal heads of pump is efficiency of pump is since Then mechanical efficiencyThe performance of pump is Flow rate ，m/h262.2 Water is tran

40、sported by a pump from reactor, which has 200 mm Hg vacuum, to the tank, in which the gauge pressure is 0.5 kgf/cm2, as shown in Fig. The total equivalent length of pipe is 200 m including all local frictional loss. The pipeline is f573.5 mm , the orifice coefficient of Co and orifice diameter do ar

41、e 0.62 and 25 mm, respectively. Frictional coefficient l.Calculate: Developed head H of pump, in m (the reading R of U pressure gauge in orifice meter is 168 mm Hg)Solution:Equation(1.6-9)Mass flow rate2) Fluid flow through the pipe from the reactor to tank, the Bernoulli equation is as followsfor V

42、1=V2Dz=10m Dp/rThe relation between the hole velocity and velocityof pipeFriction lossso +10+Hg2.3 . A centrifugal pump is to be used to extract water from a condenser in which the vacuum is 640 mm of mercury, as shown in figure. At the rated discharge, the net positive suction head must be at least

43、 3m above the cavitation vapor pressure of 710mm mercury vacuum. If losses in the suction pipe accounted for a head of 1.5m. What must be the least height of the liquid level in the condenser above the pump inlet?Solution:From an energy balance,Where Po=760-640=120mmHgPv=760-710=50mmHgUse of the equ

44、ation will give the minimum height Hg as2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half, what will the new flowrate be ? Density of acid 1840kg/m3 Viscosity of acid 2510-3 PasSolution:Velocity of

45、 acid in the pipe:Reynolds number:from Fig.1.22 for a smooth pipe when Re=6109if the pressure drop falls to /2=kPaso2u22.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half on assumption that the change of friction factor is negligible, what will the new flowrate be ? Density of acid 1840kg/m3 Viscosity of acid