流体力学英文版课后习题问题详解.docx

1、流体力学英文版课后习题问题详解1.1 What 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:Rearranging the equation 1.1-4Set the pressure of atmosphere to be zero, then the gauge pressure at depth 12m belo

2、w the surface is Absolute pressure of water 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 kero

3、sene (specific gravity = 0.815), and that liquid A in the U tube is water. The inside diameters of the reservoirs and U tube are 51mm and 6.5mm , respectively. If the reading of the manometer is145mm., what is the pressure difference over the instrument In meters of water, (a) when the change in the

4、 level in the reservoirs is neglected, (b) when the change 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/m3 pb=0.77kg/m3 D/d=8 R=0.145mWhen the pressure difference between two reservoirs is increased, th

5、e volumetric changes in the reservoirs and U tubes (1)so (2)and hydrostatic equilibrium gives following relationship (3)so (4)substituting the equation (2) for x into equation (4) gives (5)（a）when the change in the level in the reservoirs is neglected, （b）when the change in the levels in the reservo

6、irs is taken into accounterror=1.4 There are two U-tube manometers fixed on the fluid bed reactor, as shown in the figure. The readings of two U-tube manometers are R1=400mm，R2=50mm, respectively. The indicating liquid is mercury. The top of the manometer is filled with the water to prevent from the

7、 mercury vapor diffusing into the air, and the height R3=50mm. Try to calculate the pressure at point A and B. Figure for problem 1.4Solution: 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 hydrosta

8、tic equilibrium is small and negligible in comparison withand H2O , equation above can be simplified= =10009.810.05+136009.810.05=7161N/m=7161+136009.810.4=60527N/m1.5 Water discharges from the reservoir through the drainpipe, which the throat diameter is d. The ratio of D to d equals 1.25. The vert

9、ical distance h between the tank A and axis of the drainpipe is 2m. What height H from the centerline of the drainpipe to the 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 th

10、e exit of drainpipe are all open to air.Solution:Bernoulli equation is written between stations 1-1 and 2-2, with station 2-2 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

11、the continuity equation: 2Bernoulli equation is written between the throat and the station 2-2 3Combining equation 1,2,and 3 givesSolving for HH=1.39m1.6 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 press

12、ure 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. Assuming no friction losses, calculate the velocities V1 and V2 if the pressure difference (p1 - p2) is measured.Solution: In Fig1.6, the flow diagram is shown with pressu

13、re taps to measure p1 and p2. From the mass-balance continuity equation , for constant where 1 = 2 = ,For the items in the Bernoulli equation , for a horizontal pipe,z1=z2=0Then Bernoulli equation becomes, after substituting for V2,Rearranging,Performing the same derivation but in terms of V2,1.7 A

14、liquid whose coefficient of viscosity is flows below the critical velocity for laminar flow in a circular pipe of diameter d and with mean velocity 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 velocity of 0.6m/s

15、. Calculate the loss of pressure in a length of 120m.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 1From velocity profile equation for laminar flow 2substituting equation 2 for u into equa

16、tion 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 is 2.5m below A and when the flow rate down the pipe is 0.02 m3/s, the pressure at B is 1

17、4715 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 water and connected to a U-tube containing mercury of relative density 13.6, give a sketc

18、h 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; dB=0.075mzA-zB=l=2.5mQ=0.02 m3/s,pB-pA=14715 N/m2When the fluid flows down, writing mechanical balance equation0.295making the static equilibrium1.9The liquid vert

19、ically 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 ab and cd，have the same length, the diameter and roughness.Find:（1）the expressions of , hfab, and hfcd, r

20、espectively.（2）the relationship between readings R1and R2 in the U tube.Solution:(1) From Fanning equationandsoFluid flows from station a to station b, mechanical energy conservation gives hence 2from station c to station dhence 3From static equationpa-pb=R1（-）g -lg 4pc-pd=R2（-）g 5Substituting equat

21、ion 4 in equation 2 ，thentherefore 6Substituting equation 5 in equation 3 ，then 7ThusR1=R21.10 Water passes through a pipe of diameter di=0.004 m with the average velocity 0.4 m/s, as shown in Figure. 1) What is the pressure drop P when water flows through the pipe length L=2 m, in m H2O column?2) F

22、ind the maximum velocity and point r at which it occurs. 3) Find the point r at which the average velocity equals the local velocity.4）if kerosene flows through this pipe，how do the variables above change？（the viscosity and density of Water are 0.001 Pas and 1000 kg/m3，respectively；and the viscosity

23、 and density of kerosene are 0.003 Pas and 800 kg/m3，respectively）solution:1） from Hagen-Poiseuille equation2）maximum velocity occurs at the center of pipe, from equation 1.4-19so umax=0.42=0.8m3）when u=V=0.4m/s Eq. 1.4-174) kerosene:1.12 As shown in the figure, the water level in the reservoir keep

24、s constant. A steel drainpipe (with the inside diameter of 100mm) is connected to the bottom of the reservoir. One arm of the U-tube manometer is connected to 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 mercur

25、y and the left-side 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, a

26、nd the loss coefficient of the entrance is 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 coeffici

27、ent is still 0.025.Figure for problem 1.12Solution：(1) When the gate valve is opened partially, the water discharge is Set up Bernoulli equation between the surface of reservoir 11 and the section of pressure point 22，and take the center of section 22 as the referring plane, then （a）In the equation

28、(the gauge pressure) When the gate valve is fully closed, the height of water level in the reservoir can be related to h (the distance between the center of pipe and the meniscus of left arm of U tube). （b） where h=1.5mR=0.6mSubstitute the known variables into equation b Substitute the known variabl

29、es equation a 9.816.66=the velocity is V =3.13m/s the flow rate of water 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 11 and 3-3，then （c）since input the above data into equation c， 9.81the v

30、elocity is: V=3.51 m/sWrite mechanical energy balance equation between thestations 11 and 22, for the same situation of water level （d）since 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. Ther

31、e is a attached-pipe (603.5mm) which is parallel with the main pipe. The total length including the equivalent length of all form losses of the attached-pipe is 10m. A rotameter is installed in the branch pipe. When the reading of the rotameter is 2.72m3/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-pipe is 0.018 and 0.03, respect